mm-4859 === Subject: Re: #343 *Archimedean Postulate* as per division; concede to LWalk-- AP-Reals NonArchimedean; new book 2nd edition: New True Mathematics > The major difference between Old Reals and AP-Reals is that AP-Reals > are consecutive, having holes between them and the symmetrical opposite > of AP-adics. The Old Reals were designed around absolute continuity > for which there is no evidence of that in Physics. In fact Planck's > Constant contradicts absolute-continuity. Physics also contradicts infinity. The observable universe is not infinite. And yet your AP-reals and AP-adics have infinite numbers of digits. So how can your AP-reals and AP-adics be based on physics? === Subject: Re: #343 *Archimedean Postulate* as per division; concede to LWalk-- AP-Reals NonArchimedean; new book 2nd edition: New True Mathematics posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > The major difference between Old Reals and AP-Reals is that AP-Reals > are consecutive, having holes between them and the symmetrical opposite > of AP-adics. The Old Reals were designed around absolute continuity > for which there is no evidence of that in Physics. In fact Planck's > Constant contradicts absolute-continuity. > Physics also contradicts infinity. The observable universe > is not infinite. And yet your AP-reals and AP-adics have > infinite numbers of digits. So how can your AP-reals and > AP-adics be based on physics? And so I've just caught Tribble in a trap! It happens ever so often that an argument that a standard analyst uses against one so-called crank is also an argument that a different crank can use against the standard analyst! In this case, Tribble's argument that the AP-reals and AP-reals don't describe physics because the observable universe is finite, actually supports WM-style ultrafinitism, and that ultrafinitism describes physics better than standard analysis for the exact same reason -- because the observable universe is finite! So we can convert Tribble's argument, repeated below: > Physics also contradicts infinity. The observable universe > is not infinite. And yet your AP-reals and AP-adics have > infinite numbers of digits. So how can your AP-reals and > AP-adics be based on physics? into: Physics also contradicts infinity. The observable universe is not infinite. And yet the standard reals have infinite numbers of digits, and also there exist infinitely many standard naturals. So how can the standard reals and standard naturals be based on physics? Game, set, match! If one can't use AP-reals and AP-adics to describe physics, then one can't use standard reals and standard naturals to describe physics either! Of course, the standard analysts want to have it both ways, due to this two-front attack on standard analysis. Infinite objects, such as AP-adics, which don't exist in standard analysis are struck down because the observable universe is finite. But they are quite happy to allow infinite objects, like standard N, which do exist in standard analysis. The argument that the universe is finite is only used against objects that don't exist in standard analysis, not objects that do standardly exist -- even though the exact same argument can be used to show that neither AP-adics nor standard N exists! Suppose AP became a true ultrafinitist, and his upper limit of 10^500 became not only the largest possible number, but the largest number, period. Would that make the AP-adics more physically acceptable to Tribble? === Subject: Re: #357 Infinity in Physics-- electromagnetic potential; new book \ 2nd edition: New True Mathematics posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > For example, neither you [Tim Little] or Tribble ever realized that > Euclidean > Geometry was finite because the Reals are finite leftwards. He's got us there Tim, because, you see, we fail to see that because all the reals are finite, there can only be a finite number of reals, obviously. Pointless to argue with logic like that. > In fact, if I look through Wikipedia for Tim Little I find nothing. > You are not even worth a second rate basketball player as far as being > notable. So you are doing alot [sic] of wrong things and > your intelligence is thus of a [sic] below average intelligence. > [...] > Lost again. You may know a little bit about Algebra, but you know next > to nothing about Geometry, same for Tribble. > [...] > Laughable that Tim nor Tribble knows much if anything about Physics > and yet > here those two are lecturing about it. Both of you are far more > arrogant and > ignorant than you are smart. Neither one of you probably ever had a > University > course in Physics. Probably both of you would not have faired [sic] well, > and lucky [sic] > to have passed a physics course. > You need to improve on your manners, and you consistently display your > shoe size rather than your IQ. I looked up Archimedes Plutonium on Wikipedia. === Subject: Re: #357 Infinity in Physics-- electromagnetic potential; new book \ 2nd edition: New True Mathematics sha1:wdVkEZ+gIssFsDR0tmmMvgvM51g= > For example, neither you [Tim Little] or Tribble ever realized that > Euclidean > Geometry was finite because the Reals are finite leftwards. He's got us there Tim, because, you see, we fail to see that because > all the reals are finite, there can only be a finite number of > reals, obviously. Pointless to argue with logic like that. Probably the third greatest syllogism of all time. It's not as good as: Everything is made up of atoms. The universe is a thing. So, the universe is one big atom. But, it could be third. -- Jesse F. Hughes We will run this with the same kind of openness that we've run Windows. Steve Ballmer, speaking about MS's new .Net project. === Subject: Re: #343 *Archimedean Postulate* as per division; concede to LWalk-- AP-Reals NonArchimedean; new book 2nd edition: New True Mathematics posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > Suppose AP became a true ultrafinitist, and his upper limit > of 10^500 became not only the largest possible number, but > the largest number, period. Would that make the AP-adics > more physically acceptable to Tribble? If AP could demonstrate that his AP numbers actually follow the rules of a consistent algebra (which means fixing the multitude of inconsistencies, imprecise definitions, and contradictions that have been pointed out to him), *that* would make them more physically acceptable to me, or anyone else for that matter. (Substitute logically for physically.) Oh, and to correct your characterization of AP: he's not trying to design an alternative theory to the standard reals, he's trying to create an extension to the reals and thereby show how the standard reals are flawed and always have been. === Subject: Re: #343 *Archimedean Postulate* as per division; concede to LWalk-- AP-Reals NonArchimedean; new book 2nd edition: New True Mathematics posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > The major difference between Old Reals and AP-Reals is that AP-Reals > are consecutive, having holes between them and the symmetrical opposite > of AP-adics. The Old Reals were designed around absolute continuity > for which there is no evidence of that in Physics. In fact Planck's > Constant contradicts absolute-continuity. > Physics also contradicts infinity. The observable universe > is not infinite. And yet your AP-reals and AP-adics have > infinite numbers of digits. So how can your AP-reals and > AP-adics be based on physics? And so I've just caught Tribble in a trap! > So we can convert Tribble's argument, > into: > Physics also contradicts infinity. The observable universe > is not infinite. And yet the standard reals have infinite > numbers of digits, and also there exist infinitely many > standard naturals. So how can the standard reals and > standard naturals be based on physics? Game, set, match! If one can't use AP-reals and AP-adics > to describe physics, then one can't use standard reals > and standard naturals to describe physics either! You have destroyed, rather brilliantly I might add, a simple straw man argument I never made. Bravo. Brilliant. If you read closely, you'll see that I'm asking AP, who believes that math is a subset of physics and therefore must be inherently based on physics, how his AP-reals, which utilize infinite digit sequences, can be derived from that same finite physics. I never said I believe that math is derived from physics, because I don't. I did say that you can't do physics without math, but you can do math without physics. And, naturally, the meaning of this was completely lost to AP. I'm curious how he proposes to do any physics calculations without using math. But to be fair, there is a small but valid point we can extract from your brilliance. Since universe is not continuous at its quantum level, then yes, using infinite-precision reals to model such physics is overkill; a small subset of the rationals is really all that's necessary for an accurate physical model. And remember that the rationals are built up from the naturals. (And in fact, that's exactly what physicists do.) But this has nothing to do with what I was asking AP. === Subject: Re: #357 Infinity in Physics-- electromagnetic potential; new book 2nd edition: New True Mathematics <87bprhvzht.fsf@phiwumbda.org> <49D4028E.8060208@hotmail.com> <49D45B5A.6070601@hotmail.com> nothing. That is not surprising. If I appeared on such a page I would have to dispute the notability criterion for any such appearance. > I find though a athlete who runs one race and comes in second or > third place is entered in Wikipedia as notable. If the race were an international Olympic event, I think that would be notable, yes. Certainly I have never come close to second or third place in any sort of international physics olympics. - Tim === Subject: NPR Story on History of Science NPR hints at importance of Monte Carlo Method http://www.npr.org/templates/story/story.php?storyId=102748024 === Subject: Re: NPR Story on History of Science > NPR hints at importance of Monte Carlo Method http://www.npr.org/templates/story/story.php?storyId=102748024 Don't pay it any attention. NPR has an obnoxious knack of dummy downing history, science, math, news, music and anything else it can get its corporate tainted hands on. === Subject: Re: NPR Story on History of Science > NPR hints at importance of Monte Carlo Method > http://www.npr.org/templates/story/story.php?storyId=102748024 Don't pay it any attention. NPR has an obnoxious knack of dummy downing > history, science, math, news, music and anything else it can get its > corporate tainted hands on. This below is far more detailed on the Monte Carlo Method and the Manhattan \ Project: < www.fas.org/sgp/othergov/doe/lanl/pubs/00326867.pdf > David Bernier === Subject: Re: NPR Story on History of Science > NPR hints at importance of Monte Carlo Method > http://www.npr.org/templates/story/story.php?storyId=102748024 Don't pay it any attention. NPR has an obnoxious knack of dummy downing > history, science, math, news, music and anything else it can get its > corporate tainted hands on. to explain what Stanislaw Ulam did before and after he was recruited to work on the H-bomb project (I don't know about the A-bomb project). I imagine they might have used the Monte Carlo method to simulate nuclear reactions and processes with no known closed-form solution. But not even nuclear reactions were mentioned. David Bernier === Subject: Re: NPR Story on History of Science Content-ID: <20090404225549.W88516@agora.rdrop.com> --------------------------------------------------------------------- Content-ID: <20090404225549.S88516@agora.rdrop.com On Apr 4, \ 9:06[NonBreakingSpace]pm, William Elliot > NPR hints at importance of Monte Carlo Method > Don't pay it any attention. [NonBreakingSpace]NPR has an obnoxious knack of dummy > downing history, science, math, news, music and anything else it can > get its corporate tainted hands on. They also allow the cold hard truth to sneak into their interviews and > subsequent NPR productions. Are you saying those interviews are bogus? Are you saying NPR knowingly lies via obfuscation? > I'm saying that they're worse than ever. They used to regularly have cold hard truth speakers. Not anymore. More and more they interview people without substance yet with Bush bashed, they may have attempted a sligh redress. For cold hard truth, I have to go elsewhere such as KBOO. As of 2008, they have deteriorated beyond any hope. I've removed my bequeath to NPR and am giving it to KBOO, thus doubling my bequeath to KBOO. How about that, a bunch of broke amateurs can put out better news and music than NPR's host of affluent professionals. Riddle of the day. Why do US give the mega-rich banksters who caused this current Mother of all Depressions, trillions of our dollars to support them in the lavish manner they consider their divine right, instead of simply imprisoning them and giving the money to ourselves? === Subject: Re: NPR Story on History of Science Content-ID: <20090405065842.G39462@agora.rdrop.com On Apr 4, 10:11, William \ Elliot > NPR and PBS may not be entirely cartel/cabal free, but they is far > better than 99.9% of all the rest. Perhaps your KBOO is in the 0.1% > good guys category. KBOO is listener sponsored with no corporate sponsor/advertisers, government funding or politically appointed heads of staff. NPR on the other hand is blessed with all of the above including now a lousy corporate model of properly censored public radio. ---- === Subject: Re: NPR Story on History of Science Content-ID: <20090405115927.U76708@agora.rdrop.com> --------------------------------------------------------------------- Content-ID: <20090405115927.T76708@agora.rdrop.com On Apr 4, 10:11, William \ Elliot > NPR and PBS may not be entirely cartel/cabal free, but they is far > better than 99.9% of all the rest. [NonBreakingSpace]Perhaps your KBOO is in the 0.1% > good guys category. > KBOO is listener sponsored with no corporate sponsor/advertisers, > government funding or politically appointed heads of staff. > NPR on the other hand is blessed with all of the above including > now a lousy corporate model of properly censored public radio. > ---- Federal and state funding into NPR is severely limited, as almost > nonexistent. They are sort of on the NO FLY list with the feds. NPR isn't perfect, but it's still far better than most other > alternatives. Your KBOO is fine for those within range of its signal, and don't > bother giving us that internet access crap, because the internet is > simply too bogged down with live video commercials, smut and games to > allow any reasonable audio downloads of KBOO, much less live coverage. Obviously you can afford nothing but the very best in internet access > and its data throughput. Lucky for you. > I'm lucky that I'm in an area where I can receive KBOO. I can also receive KLCC which has a bunch of NPR and some other stuff. I can also receive a radical liberal and a radical conservative station that are comercial bastards to which I seldom listen. We the people have met the enemy and it is republican. Riddle of the day. When you gonna get ya free million dollar stimulus? === Subject: Re: NPR Story on History of Science <20090405112425.M70278@agora.rdrop.com> Content-ID: <20090405204305.E40776@agora.rdrop.com> --------------------------------------------------------------------- Content-ID: <20090405204305.D40776@agora.rdrop.com On Apr 5, 12:04, William \ Elliot > I'm lucky that I'm in an area where I can receive KBOO. [NonBreakingSpace]I can > also receive KLCC which has a bunch of NPR and some other stuff. > I can also receive a radical liberal and a radical conservative > station that are comercial bastards to which I seldom listen. > We the people have met the enemy and it is republican. > Riddle of the day. [NonBreakingSpace]When you gonna get ya free million dollar stimulus? The mostly republican Mafia cabal and their Big Energy puppet masters > are still in charge of way too much. Guess which religion they mostly > suck up to and/or hide within? > Zionism? mega mania merged monsters In our deregulated, out of control, ultra-modern, hyper-commercial, socially engineered society of the dehumanizing capitalists' cultural revolution wherein the measure of all things is money, are you an anybody corporate dog wage slave, within that compulsively competitive, spiritually and socially bankrupt, free trade, media-prattle reality? Riddle of the day. When all about you is bull, how do you extract the bull from the ? === Subject: Re: NPR Story on History of Science > On Apr 4, 9:06 pm, William Elliot > NPR hints at importance of Monte Carlo Method Don't pay it any attention. NPR has an obnoxious knack of dummy > downing history, science, math, news, music and anything else it can > get its corporate tainted hands on. > They also allow the cold hard truth to sneak into their interviews and > subsequent NPR productions. > Are you saying those interviews are bogus? > Are you saying NPR knowingly lies via obfuscation? > I'm saying that they're worse than ever. They used to > regularly have cold hard truth speakers. Not anymore. More and more they interview people without substance yet > with Bush bashed, they may have attempted a sligh redress. For cold hard truth, I have to go elsewhere such as KBOO. As of 2008, they have deteriorated beyond any hope. > I've removed my bequeath to NPR and am giving it to KBOO, > thus doubling my bequeath to KBOO. How about that, a bunch of broke amateurs can put out better > news and music than NPR's host of affluent professionals. Riddle of the day. Why do US give the mega-rich banksters who caused this > current Mother of all Depressions, trillions of our dollars to support > them in the lavish manner they consider their divine right, instead of > simply imprisoning them and giving the money to ourselves? I agree with you. NPR and PBS is no more public than AIG. The public gets invited to watch their TV programs but never to speak and the interviewees \ are almost always from the corporate side of the isle. And we know what those think of the word public. cg === Subject: Re: NPR Story on History of Science > On Apr 4, 9:06 pm, William Elliot > NPR hints at importance of Monte Carlo Method Don't pay it any attention. NPR has an obnoxious knack of dummy > downing history, science, math, news, music and anything else it can > get its corporate tainted hands on. > They also allow the cold hard truth to sneak into their interviews and > subsequent NPR productions. > Are you saying those interviews are bogus? > Are you saying NPR knowingly lies via obfuscation? > I'm saying that they're worse than ever. They used to > regularly have cold hard truth speakers. Not anymore. More and more they interview people without substance yet > with Bush bashed, they may have attempted a sligh redress. For cold hard truth, I have to go elsewhere such as KBOO. As of 2008, they have deteriorated beyond any hope. > I've removed my bequeath to NPR and am giving it to KBOO, > thus doubling my bequeath to KBOO. How about that, a bunch of broke amateurs can put out better > news and music than NPR's host of affluent professionals. Riddle of the day. Why do US give the mega-rich banksters who caused this > current Mother of all Depressions, trillions of our dollars to support > them in the lavish manner they consider their divine right, instead of > simply imprisoning them and giving the money to ourselves? > I agree with you. NPR and PBS is no more public than AIG. The public gets \ > invited to watch their TV programs but never to speak and the interviewees > are almost always from the corporate side of the isle. And we know what > those think of the word public. What is hilarious is how this comes as a shock to people. Everything is hunky-dory until they begin to contradict someone's cherished beliefs, and then Omigod! They went over to the darkside. Newsflash, Loretta: they always were the darkside. -- Michael Press === Subject: Re: NPR Story on History of Science <20090404224724.O88516@agora.rdrop.com> <7oqdnf7L4eWwM0XUnZ2dnUVZ_vjinZ2d@golden.net> Content-ID: <20090405113817.L70278@agora.rdrop.com> --------------------------------------------------------------------- Content-ID: <20090405113817.F70278@agora.rdrop.com> On Apr 4, 9:06 pm, \ William Elliot > Are you saying NPR knowingly lies via obfuscation? > I'm saying that they're worse than ever. [NonBreakingSpace]They used to > regularly have cold hard truth speakers. Not anymore. > More and more they interview people without substance yet > with Bush bashed, they may have attempted a sligh redress. > For cold hard truth, I have to go elsewhere such as KBOO. > As of 2008, they have deteriorated beyond any hope. > I've removed my bequeath to NPR and am giving it to KBOO, > thus doubling my bequeath to KBOO. > How about that, a bunch of broke amateurs can put out better > news and music than NPR's host of affluent professionals. > I agree with you. NPR and PBS is no more public than AIG. The public > gets invited to watch their TV programs but never to speak and [NonBreakingSpace]the > interviewees are almost always from the corporate side of the isle. And > we know what those think of the word public. What national media network or publications (if any) suit your mindset? > Haven't you noticed, ever since RayGun, NPR is another national media network promoting government and corporate propaganda. All that NPR has left is that it's merely the best of the worst. === Subject: Re: NPR Story on History of Science <20090404224724.O88516@agora.rdrop.com> <7oqdnf7L4eWwM0XUnZ2dnUVZ_vjinZ2d@golden.net> Content-ID: <20090405204839.T40776@agora.rdrop.com> --------------------------------------------------------------------- Content-ID: <20090405204839.Q40776@agora.rdrop.com> On Apr 4, 9:06 pm, \ William Elliot > Are you saying NPR knowingly lies via obfuscation? > I'm saying that they're worse than ever. [NonBreakingSpace]They used to > regularly have cold hard truth speakers. Not anymore. > More and more they interview people without substance yet > with Bush bashed, they may have attempted a sligh redress. > For cold hard truth, I have to go elsewhere such as KBOO. > As of 2008, they have deteriorated beyond any hope. > I've removed my bequeath to NPR and am giving it to KBOO, > thus doubling my bequeath to KBOO. > How about that, a bunch of broke amateurs can put out better > news and music than NPR's host of affluent professionals. > I agree with you. NPR and PBS is no more public than AIG. The public > gets invited to watch their TV programs but never to speak and [NonBreakingSpace]the > interviewees are almost always from the corporate side of the isle. And > we know what those think of the word public. > What national media network or publications (if any) suit your mindset? > Haven't you noticed, ever since RayGun, NPR is another national media > network promoting government and corporate propaganda. [NonBreakingSpace]All that NPR has > left is that it's merely the best of the worst. NPR and PBS are no failsafe house of corrupt republicans and their > faith-based Big Energy puppet masters, though this doesn.92t mean they > don.92t have a politically correct policy of obfuscation. If you can.92t > manage to read between the lines, perhaps you should stay in bed, and > not listen to anyone (especially yourself). > If you believe everything you read, then you better not read. Japanese Proverb Half the lies our enemies tell about us aren't true! Sir Boyle Roche === Subject: Re: NPR Story on History of Science NPR hints at importance of Monte Carlo Method http://www.npr.org/templates/story/story.php?storyId=102748024 > and the mathematicians of this newsgroup need to hear it from NPR before they convince themselves of the importance of Monte Carlo Method? === Subject: Re: NPR Story on History of Science posting-account=tNh-2AoAAACqdWo2IikcW-FF_IGUAvWf AppleWebKit/525.27.1 (KHTML, like Gecko) Version/3.2.1 Safari/525.27.1,gzip(gfe),gzip(gfe) > NPR hints at importance of Monte Carlo Method http://www.npr.org/templates/story/story.php?storyId=102748024 and the mathematicians of this newsgroup need to hear it from NPR before > they convince themselves of the importance of Monte Carlo Method? Curiously George, you so took the sentiment right out of my mind! NPR. Also on NPR, Steven Weinberg claimed some of John Wheeler's ideas were theological. One could, almost had to, conclude that Physics deals with the Whole Universe in a restricted sense. Sort of like the over exaggerated claim of an NPR show titled All Things Considered where the unstated ... Well Almost is ever present but always underplayed. http://twitter.com/scifri gives the guest list of last Friday. http://www.interactiveinsightsgroup.com/blog1/how-to-search-the-social-web-u\ ltimate-toolkit/ is an interesting link related to search of ... Enjo(y)... -- Mahipal ñme always changesî (A Poetic Force of Nature) http://infiniteperspectivesmachines.blogspot.com/ === Subject: Re: Regular polygon posting-account=em7BlwoAAADOhRXv8AilqN49TjuokE7m Trident/4.0; InfoPath.2; FDM; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) I was wondering how to get it, I had got that formula using top down approach so wanted to know about a proof. === Subject: Re: Cantor's argument is erroneous posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) >Cantor's argument is erroneous and its adoption leads to unsound >mathematics. >The basic idea in the argument is that there is no bijection between >the set of counting numbers and the set of infinite binary strings. >But such a bijection exists, it can be expressed in terms of limit >points, or by transfinite induction; informally, it can be defined as >the correspondence between the paths and the leaf (i.e. limit) nodes >in the infinite binary tree. This invalidates all results relating to >Cantor's transfinite. > This is nonsense. Being charitable about the notion of limit node: > Yes, there's a bijection between the limit nodes and the paths. > Do tell -- in the charitable interpretation, what _is_ a limit node? You guys can be sloppy: http://en.wikipedia.org/wiki/Limit_point Please note that I say limit points and only later, qualified as >informal, I say limit nodes. So what? First, _you_ haven't bothered to explain > what those limit points _are_ - pointing to that > wikipedia page doesn't answer that question > until you specify what topological space > you're talking about. So what? You are sloppy, that's what. If I have to qualify a topology, that's a thing: that you guys keep wander about crankyness and limit nodes for tens of posts is your sloppyness, not mine. -LV === Subject: Re: Cantor's argument is erroneous posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/20080201 Firefox/2.0.0.12 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > So what? You are sloppy, that's what. If I have to qualify a topology, > that's a thing: that you guys keep wander about crankyness and limit > nodes for tens of posts is your sloppyness, not mine. You DON'T have to qualify a topology. You DO have TO KNOW what the you are talking about! AND YOU DON'T! You don't know what soundness is, you don't know what a limit point is, and you don't know what a proof is. If you DID know what a proof was, you would know that the problem with Cantor's theorem is NOT how you construct the bijection in the first place, BUT WHAT HAPPENS AFTER you construct it. What happens AFTER you construct is that ANYbody can THEN construct, FROM IT, a string that is NOT mapped to ANY node, a subset of N that is NOT mapped to any natural n in N. === Subject: Re: Cantor's argument is erroneous posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) So what? You are sloppy, that's what. If I have to qualify a topology, > that's a thing: that you guys keep wander about crankyness and limit > nodes for tens of posts is your sloppyness, not mine. You DON'T have to qualify a topology. > You DO have TO KNOW what the you are talking about! > AND YOU DON'T! > You don't know what soundness is, you don't know what a limit point > is, > and you don't know what a proof is. > If you DID know what a proof was, you would know that the problem > with Cantor's theorem is NOT how you construct the bijection in the > first place, BUT WHAT HAPPENS AFTER you construct it. I know that very well, rather some other people around here don't, as they go from the dogmatic if you give a list it can be diagonalised to the as well dogmatic and completely unwarranted such list cannot exist at all: tell them, not me, and stop this gratuitus ad-hominem. The sloppiness, at least in the approach, is surely not mine: rather, self-indulgence is the other side of these standard inquisitors. > What happens AFTER you construct is that ANYbody can THEN > construct, FROM IT, a string that is NOT mapped to ANY node, Go on then, give me an explicit construction of the string that is supposed not to be in the following list: (Constructively valid, not the usual BS of a string that differs from any other etc. by dogmatic fiat. I specify this because: 1) that's all I would accept and care for, 2) because Virgil keeps claiming Cantor's arguments are constructive, which is of course plain bull given that they are *all* not even predicative.) (empty) (0) 1(0) 01(0) 11(0) 001(0) 011(0) 101(0) 111(0) 0001(0) 0011(0) 0101(0) 0111(0) 1001(0) 1011(0) 1101(0) 1111(0) 00001(0) ... (1) The algo roughly goes as: after the first few entries, just enumerate all binary combinations for increasing string lengths, terminated by 1 (0). The limit point is (1). This being simply a way as any other (i.e. modulo arbitrary remapping) to enumerate all paths in the binary tree without repetitions. > a subset of N that is NOT mapped to any natural n in N. n in N*, this is the TRANSFINITE! I might very well be wrong on many things, but didn't you notice my many complaints already about this invalid mixing of the finite and the transfinite? (And don't just tell me that's in the original argument because, indeed, that that argument is invalid is my very point.) -LV === Subject: Re: Cantor's argument is erroneous >You guys can be sloppy: http://en.wikipedia.org/wiki/Limit_point Please note that I say limit points and only later, qualified as >informal, I say limit nodes. So what? First, _you_ haven't bothered to explain > what those limit points _are_ - pointing to that > wikipedia page doesn't answer that question > until you specify what topological space > you're talking about. So what? You are sloppy, that's what. No! It is LV's combination of sloppiness and arrogance which leads him to speak of limit points without knowing enough about them. Thus LV's arrogance reveals his ignorance. === Subject: What are the chances to win in a lottery game? posting-account=fPUT4goAAABJ1knpUmWe7fFdWT7SyGS2 Gecko/2009032609 Firefox/2.0.0.12;MEGAUPLOAD 1.0,gzip(gfe),gzip(gfe) What are the chances to win 5 numbers in a lottery game 6 from 45, having x combinations of 6 numbers? For ex., for x=1 chance is 1/34808. === Subject: Re: growth of | zeta(1/2 +it) | *ou* croissance de | zeta(1/2 + it) | > Large values of the modulus of zeta on the critical line is an intriguing > subject. If the Riemann Hypothesis is true, then the Lindelof conjecture \ is > true. The Lindelof conjecture implies some O-bounds for zeta on the > critical line Re(s) = 1/2 : > - Lindelof conjecture ==> | zeta(1/2 + it) | = O(t^c) for any c>0, where t>0 . > principle implies the result zeta(x - iy) = Conjugate( zeta( x + \ iy)), x, y real. > So it is common to consider only zeta in the upper half-plane, for zeros and > other results. > Zeta has trivial zeros in -2,-4...What is important is the critical > strip x /in (0,1), and y /in (-oo,oo), z = x + iy, all the other zeros > must be there. doesn't compute Z(t) accurately enough to check RH up to some height. My interest in large values of | zeta(1/2 + it) | dates from the time I first heard audio files based on Z(t) by Ken Takusagawa: < http://web.mit.edu/kenta/www/six/parallel/2-Final-Report.html > --> riemann_low.mp3 and riemann_hi.mp3 . riemann_hi.mp3 is quite bumpy. This is more apparent if played at 1/2 normal speed or less. David Bernier === === === Subject: Re: Bounded Variation <15787512.26548.1238867426622.JavaMail.jakarta@nitrogen.mathforum.org> , > Hey, How to show that f(x)=x^a sin (1/x^b) (x is not 0) and f(0)=0 is bounded > variation on [-1,1] (a>b). (b>0, a is real number) You have problems with defining x^a, x^b for x < 0. Best to stick to [0, 1]. > I need a hint to start, I start with a partition for the interval by having > x_i = -1 + 2i/n. (n is fixed) and then I tried to bounded the total variation, but I don't know how to > continue. thanx Hint: If g is continuously differentiable on [c, d], then the total variation of g on [c, d] is int_[c,d] |g'(x)| dx. === Subject: Re: Bounded Variation > Hey, How to show that f(x)=x^a sin (1/x^b) (x is not 0) and f(0)=0 is bounded variation on [-1,1] (a>b). (b>0, a is real number) I need a hint to start, I start with a partition for the interval by having x_i = -1 + 2i/n. (n is fixed) and then I tried to bounded the total variation, but I don't know how to continue. Since x can be negative, I would take the domain of f to be [0, oo[ . [ For example, what is (-1/2)^pi ? ] You can change variables: u:= 1/x^b . So x = u^(-1/b), x^a = 1/[ u^(a/b) ]. Is there a relation between the maximum [i.e.supremum of] variation of f in \ (0, 1] and the maximum variation of g(u):= sin(u)/u^(a/b) in [1, oo) ? David Bernier === Subject: Re: Uniform convergence I'm having trouble on where to begin with this problem. Suppose I have > a function f: R -> R continuous on [0,1], with f(1) = 0. I need to > show that the sequence of functions f(x)x^n converges uniformly on > [0,1]. Assume r > 0, let b = sup{ |f(x)| : x in [0,1] }, > restrict x to [0,1] and show the following: some d < 1 with for all x > d, |f(x)| < r some n in N with for all j > n, x <= d, |f(x)| x^j <= bd^j < r. For a guy always complainging about how others write, you certainly post a lot of gibberish. > My first impression is to split [0, 1] into two subintervals. On the > left half, it would appear that the sequence of functions converge > uniformly to 0 (take the maximum of f(x) on [0, c], and a big enough n, > then f(x)x^n can be as small as you want). On the right half (ie. on > [c, 1]), I have to somehow leverage the fact that f is continuous and > f(1) = 0 to get f(x)x^n as small as possible. Am I on the right track? I think you're more or less on the right tract. === Subject: Re: Uniform convergence a function f: R -> R continuous on [0,1], with f(1) = 0. I need to > show that the sequence of functions f(x)x^n converges uniformly on > [0,1]. > Assume r > 0, let b = sup{ |f(x)| : x in [0,1] }, > restrict x to [0,1] and show the following: > some d < 1 with for all x > d, |f(x)| < r > some n in N with for all j > n, x <= d, |f(x)| x^j <= bd^j < r. For a guy always complainging about how others write, you certainly > post a lot of gibberish. > I don't pass it off as English, only pseudo-symbolic logic. There is however some ambiguity in the second statement. Clearer is: some n in N with for all j > n, for all x <= d, |f(x)| x^j <= bd^j < r. Is there anything you don't understand about the two statements or the two lines setting up the proof suggestion? On the other hand, I did leave out details left to the reader. BTW, W^3 scores a spelling typo. === Subject: Re: Uniform convergence > I'm having trouble on where to begin with this problem. Suppose I have > a function f: R -> R continuous on [0,1], with f(1) = 0. I need to > show that the sequence of functions f(x)x^n converges uniformly on > [0,1]. > Assume r > 0, let b = sup{ |f(x)| : x in [0,1] }, > restrict x to [0,1] and show the following: > some d < 1 with for all x > d, |f(x)| < r > some n in N with for all j > n, x <= d, |f(x)| x^j <= bd^j < r. For a guy always complainging about how others write, you certainly > post a lot of gibberish. I don't pass it off as English, only pseudo-symbolic logic. > There is however some ambiguity in the second statement. Clearer is: some n in N with for all j > n, for all x <= d, > |f(x)| x^j <= bd^j < r. Is there anything you don't understand about the two statements > or the two lines setting up the proof suggestion? I don't try to understand it. Why waste my time? The point is: You've complained to the high heavens about writing deficiencies in others over the years, all the while merrily posting garbage like the above. When exactly does your hypocrisy-meter kick in? > On the other hand, I did leave out details left to the reader. BTW, W^3 scores a spelling typo. Brilliant. === Subject: Re: Closed subset of a product >Hi all, Let E be a topological space. For each _x_ in E and each _r_ in (0,+oo), >let N(x,r) be a neighborhood of _x_. Consider on the set of all >functions from E to R the topology of simple convergence, that is, the >product topology in prod_{x in E}R. I would like to prove that the set F >of all functions from E to R such that when _x_ in E, when r > 0, when _y_ in N(x,r), |f(y) - f(x)| <= r is a closed set. Actually, I am able to do it easily using nets or >filters, but what I would like to have is a proof of that fact which >consisted in proving that the complement of F is an open set. It looks >quite simple, so I suppose that, by being unable to do it, I am missing >something that should be obvious. Any ideas? Let's say a quadruple (I, J, x, y) is 'good' if I and J are open intervals, x,y in E, and there exists a positive real r such that y is in N(x,r), and |a-b| > r for all a in I and b in J. Then the complement of F is just the union over good quadruples of the sets U(I, J, x, y) = { g : g(x) in I and g(y) in J }. RS === Subject: Re: Closed subset of a product > Let E be a topological space. For each _x_ in E and each _r_ in (0,+oo), > let N(x,r) be a neighborhood of _x_. Consider on the set of all > functions from E to R the topology of simple convergence, that is, the > product topology in prod_{x in E}R. I would like to prove that the set F > of all functions from E to R such that > when _x_ in E, when r > 0, when _y_ in N(x,r), |f(y) - f(x)| <= r > is a closed set. Actually, I am able to do it easily using nets or > filters, but what I would like to have is a proof of that fact which > consisted in proving that the complement of F is an open set. It looks > quite simple, so I suppose that, by being unable to do it, I am missing > something that should be obvious. > Any ideas? Let's say a quadruple (I, J, x, y) is 'good' if I and J are open > intervals, x,y in E, and there exists a positive real r such that > > y is in N(x,r), and |a-b| > r for all a in I and b in J. > > Then the complement of F is just the union over good quadruples > of the sets U(I, J, x, y) = { g : g(x) in I and g(y) in J }. Jose Carlos Santos === Subject: Re: Magnify form posting-account=15T5jAkAAACpO9KgK0Ue212ywoAPQtp7 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; MDDC),gzip(gfe),gzip(gfe) > snip Ben you should read up on the history of the Calculus. It originated > with the concept of infinitely small or infinitesimals. Mitch Raemsch- Hide quoted text - - Show quoted text - Mathematics has made advances since those days. Infinitesimals are as > obsolete as phlogiston. We do everything now with deltas and epsilons. > Look up analysis as a mathematical discipline. Uncle Ben > You are all bull and bluster, Uncle Bonehead. Calculus today > is the same calculus it was in Newton's day. Look up head > as an academic crank.- Hide quoted text - - Show quoted text - Again you reveal your ignorance, dear Androcles. Since Newton and > Leibnitz invented calculus, mathematicians fretted about the lack of > rigour in discussions of limits, infinite series, and your precious > infinitesimals. They, unlike you, knew back then that on an open > interval, there is no number closest to the boundary. Vague handwaving about mathematicians is a typical ignorant > Bonehead argument. Quote/ > 1. Assume that there exists a smallest number greater than zero: Call it > h. > 2. But h/2 is smaller than h > /unquote -- Uncle Bonehead Having assumed the existence of h as defined by Bonehead, > k = h/2 cannot exist. Assume a spherical cow... Lemma 1: Therefore Uncle Bonehead is illogical. > Lemma 2: Therefore Uncle Bonehead is not a mathematician or > even a physicist, he hallucinates he divide photons. Again you ignorantly contradict yourself, Bonehead. Look up > contradiction as a wit. sqrt(2) = 1 + 1/(2+1/(1/(2+1/(1/(2+1/(1/(2+1/(1/(2+1/(1+...))))))))) > (Extended fraction, Xeno style)- Hide quoted text - - Show quoted text - This one I will leave as a classic fumble! All the best to you Dr. Parker! (I think that should be continued fraction.) Uncle Ben > You don't know how to think, and no, it is an extended fraction > (which obviously you've never heard of). > http://tinyurl.com/d6ps8y- Hide quoted text - - Show quoted text - extended fraction. According to Google, the name continued fraction is 500 times more common. UB === Subject: Re: Magnify form > snip Ben you should read up on the history of the Calculus. It originated > with the concept of infinitely small or infinitesimals. Mitch Raemsch- Hide quoted text - - Show quoted text - Mathematics has made advances since those days. Infinitesimals are as > obsolete as phlogiston. We do everything now with deltas and epsilons. > Look up analysis as a mathematical discipline. Uncle Ben > You are all bull and bluster, Uncle Bonehead. Calculus today > is the same calculus it was in Newton's day. Look up head > as an academic crank.- Hide quoted text - - Show quoted text - Again you reveal your ignorance, dear Androcles. Since Newton and > Leibnitz invented calculus, mathematicians fretted about the lack of > rigour in discussions of limits, infinite series, and your precious > infinitesimals. They, unlike you, knew back then that on an open > interval, there is no number closest to the boundary. Vague handwaving about mathematicians is a typical ignorant > Bonehead argument. Quote/ > 1. Assume that there exists a smallest number greater than zero: Call it > h. > 2. But h/2 is smaller than h > /unquote -- Uncle Bonehead Having assumed the existence of h as defined by Bonehead, > k = h/2 cannot exist. Assume a spherical cow... Lemma 1: Therefore Uncle Bonehead is illogical. > Lemma 2: Therefore Uncle Bonehead is not a mathematician or > even a physicist, he hallucinates he divide photons. Again you ignorantly contradict yourself, Bonehead. Look up > contradiction as a wit. sqrt(2) = 1 + 1/(2+1/(1/(2+1/(1/(2+1/(1/(2+1/(1/(2+1/(1+...))))))))) > (Extended fraction, Xeno style)- Hide quoted text - - Show quoted text - This one I will leave as a classic fumble! All the best to you Dr. Parker! (I think that should be continued fraction.) Uncle Ben > You don't know how to think, and no, it is an extended fraction > (which obviously you've never heard of). > http://tinyurl.com/d6ps8y- Hide quoted text - - Show quoted text - extended fraction. According to Google, the name continued fraction is 500 times more common. UB I'd wager you teach kids to call the bonnet of a car the hood and the boot a trunk in the little rot schulehaus in the back forty of Skin-neck-tady, too. So which of two definitions is correct, h is the smallest number or k = h/2 is smaller than h? Oh wait, k is always a constant and i, j are always integers. It must be i = 1, since that is the smallest integer > 0 that exists. C'mon, Dr. Bonehead, you like to argue trivial semantics such as reductio-ad-absurdum conclusions being called postulates. What was it again? Oh yes... the speed of light is c in all inertial frames... -- bwhahahaha! Amazing what the crazy colonists could come up with. Here's a real challenge to your intellect, Uncle Bonehead: http://www.androcles01.pwp.blueyonder.co.uk/Lightcurve.xls I call it a real challenge because that light curve is empirical data. Change the value in cell Q2 to 3 to see it modelled. Change it back to 0 and you can play with the values in cells K2-P2 and see the effect they have. And please, no stupid lectures about closing speed, this is called research. If we knew what we were doing it would not be called research, would it? -- Einstein Only two things are infinite; the universe and human stupidity - and I'm not sure about green Uncle Dr. Bonehead -- Einstein. === Subject: Re: Magnify form posting-account=15T5jAkAAACpO9KgK0Ue212ywoAPQtp7 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; MDDC),gzip(gfe),gzip(gfe) > snip Ben you should read up on the history of the Calculus. It originated > with the concept of infinitely small or infinitesimals. Mitch Raemsch- Hide quoted text - - Show quoted text - Mathematics has made advances since those days. Infinitesimals are as > obsolete as phlogiston. We do everything now with deltas and epsilons. > Look up analysis as a mathematical discipline. Uncle Ben > You are all bull and bluster, Uncle Bonehead. Calculus today > is the same calculus it was in Newton's day. Look up head > as an academic crank.- Hide quoted text - - Show quoted text - Again you reveal your ignorance, dear Androcles. Since Newton and > Leibnitz invented calculus, mathematicians fretted about the lack of > rigour in discussions of limits, infinite series, and your precious > infinitesimals. They, unlike you, knew back then that on an open > interval, there is no number closest to the boundary. Vague handwaving about mathematicians is a typical ignorant > Bonehead argument. Quote/ > 1. Assume that there exists a smallest number greater than zero: Call it > h. > 2. But h/2 is smaller than h > /unquote -- Uncle Bonehead Having assumed the existence of h as defined by Bonehead, > k = h/2 cannot exist. Assume a spherical cow... Lemma 1: Therefore Uncle Bonehead is illogical. > Lemma 2: Therefore Uncle Bonehead is not a mathematician or > even a physicist, he hallucinates he divide photons. Again you ignorantly contradict yourself, Bonehead. Look up > contradiction as a wit. sqrt(2) = 1 + 1/(2+1/(1/(2+1/(1/(2+1/(1/(2+1/(1/(2+1/(1+...))))))))) > (Extended fraction, Xeno style)- Hide quoted text - - Show quoted text - This one I will leave as a classic fumble! All the best to you Dr. Parker! (I think that should be continued fraction.) Uncle Ben > You don't know how to think, and no, it is an extended fraction > (which obviously you've never heard of). >http://tinyurl.com/d6ps8y-Hide quoted text - - Show quoted text - extended fraction. According to Google, the name continued > fraction is 500 times more common. UB I'd wager you teach kids to call the bonnet of a car the hood > and the boot a trunk in the little rot schulehaus in the back forty > of Skin-neck-tady, too. > So which of two definitions is correct, h is the smallest number > or k = h/2 is smaller than h? > Oh wait, k is always a constant and i, j are always integers. It must > be i = 1, since that is the smallest integer > 0 that exists. C'mon, Dr. Bonehead, you like to argue trivial semantics such as > reductio-ad-absurdum conclusions being called postulates. What was > it again? Oh yes... the speed of light is c in all inertial frames... > -- bwhahahaha! > Amazing what the crazy colonists could come up with. Here's a real challenge to your intellect, Uncle Bonehead: > http://www.androcles01.pwp.blueyonder.co.uk/Lightcurve.xls > I call it a real challenge because that light curve is empirical data. > Change the value in cell Q2 to 3 to see it modelled. > Change it back to 0 and you can play with the values in cells K2-P2 > and see the effect they have. > And please, no stupid lectures about closing speed, this is called > research. > If we knew what we were doing it would not be called research, > would it? -- Einstein > Only two things are infinite; the universe and human stupidity - and I'm > not sure about green Uncle Dr. Bonehead -- Einstein.- Hide quoted text - - Show quoted text - If proof by contradiction befuddles you, John, I'm sorry to have taxed your brain. The way it goes, you see, is as follows: One proves something false by assuming it to be true and then showing that that assumption leads to a contradiction. As in the following: Know ye that in mathematics, 0 < x < 1 is an open interval, and 0 <= x <= 1 is a closed interval. 1. Assume that John Parker has a Ph. D. in mathematics. 2. All Ph. D. mathematicians know that in an open interval, there is neither a least nor a greatest real number. 3. But John Parker does not know that, which contradicts line 2. 4. Therefore the assumption in line 1 is false. Uncle Ben === Subject: Re: Magnify form > snip Ben you should read up on the history of the Calculus. It > originated > with the concept of infinitely small or infinitesimals. Mitch Raemsch- Hide quoted text - - Show quoted text - Mathematics has made advances since those days. Infinitesimals are > as > obsolete as phlogiston. We do everything now with deltas and > epsilons. > Look up analysis as a mathematical discipline. Uncle Ben > You are all bull and bluster, Uncle Bonehead. Calculus today > is the same calculus it was in Newton's day. Look up head > as an academic crank.- Hide quoted text - - Show quoted text - Again you reveal your ignorance, dear Androcles. Since Newton and > Leibnitz invented calculus, mathematicians fretted about the lack of > rigour in discussions of limits, infinite series, and your precious > infinitesimals. They, unlike you, knew back then that on an open > interval, there is no number closest to the boundary. Vague handwaving about mathematicians is a typical ignorant > Bonehead argument. Quote/ > 1. Assume that there exists a smallest number greater than zero: Call > it > h. > 2. But h/2 is smaller than h > /unquote -- Uncle Bonehead Having assumed the existence of h as defined by Bonehead, > k = h/2 cannot exist. Assume a spherical cow... Lemma 1: Therefore Uncle Bonehead is illogical. > Lemma 2: Therefore Uncle Bonehead is not a mathematician or > even a physicist, he hallucinates he divide photons. Again you ignorantly contradict yourself, Bonehead. Look up > contradiction as a wit. sqrt(2) = 1 + 1/(2+1/(1/(2+1/(1/(2+1/(1/(2+1/(1/(2+1/(1+...))))))))) > (Extended fraction, Xeno style)- Hide quoted text - - Show quoted text - This one I will leave as a classic fumble! All the best to you Dr. Parker! (I think that should be continued fraction.) Uncle Ben > You don't know how to think, and no, it is an extended fraction > (which obviously you've never heard of). >http://tinyurl.com/d6ps8y-Hide quoted text - - Show quoted text - extended fraction. According to Google, the name continued > fraction is 500 times more common. UB I'd wager you teach kids to call the bonnet of a car the hood > and the boot a trunk in the little rot schulehaus in the back forty > of Skin-neck-tady, too. > So which of two definitions is correct, h is the smallest number > or k = h/2 is smaller than h? > Oh wait, k is always a constant and i, j are always integers. It must > be i = 1, since that is the smallest integer > 0 that exists. C'mon, Dr. Bonehead, you like to argue trivial semantics such as > reductio-ad-absurdum conclusions being called postulates. What was > it again? Oh yes... the speed of light is c in all inertial frames... > -- bwhahahaha! > Amazing what the crazy colonists could come up with. Here's a real challenge to your intellect, Uncle Bonehead: > http://www.androcles01.pwp.blueyonder.co.uk/Lightcurve.xls > I call it a real challenge because that light curve is empirical data. > Change the value in cell Q2 to 3 to see it modelled. > Change it back to 0 and you can play with the values in cells K2-P2 > and see the effect they have. > And please, no stupid lectures about closing speed, this is called > research. > If we knew what we were doing it would not be called research, > would it? -- Einstein > Only two things are infinite; the universe and human stupidity - and I'm > not sure about green Uncle Dr. Bonehead -- Einstein.- Hide quoted > text - - Show quoted text - If proof by contradiction befuddles you, John, It doesn't, so If proof by Einstein-said-so befuddles you, Dr. Bonehead, I'm not sorry to have taxed your brain, you obviously don't have one and would argue trivialities. The way it goes, you see, is as follows: One proves something false by assuming it to be true and then showing that that assumption leads to a contradiction, such as light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body leading to the speed of light is c in all inertial frames -- about which Uncle Bonehead says Harry is right. And Androcles did indeed read Einstein's paper more carefully than I did. --bwhahahaha! Of course, Uncle Bonehead is so smart he knew all along that Einstein was disproving his own postulate by reductio-ad-absurdum, but Uncle Bonehead is too stupid to realise Androcles has outsmarted him with something as trivial as Quote/ 1. Assume that there exists a smallest number greater than zero: Call it h. 2. But h/2 is smaller than h /unquote -- Uncle Bonehead Therefore line 2 is false, a proof by contradiction. A. Assume that the green Dr. Bonehead has a Ph. D. in mathematics. Therefore the assumption in line A is false, he didn't claim to and he's ignorant of the subject. What he DID claim was a Ph. D. in physics. Here's a real challenge to your intellect, green Uncle Dr. Bonehead: http://www.androcles01.pwp.blueyonder.co.uk/Lightcurve.xls I call it a real challenge because that light curve is empirical data. Change the value in cell Q2 to 3 to see it modelled. Change it back to 0 and you can play with the values in cells K2-P2 and see the effect they have. And please, no stupid lectures about closing speed, this is called research and defending a thesis. If we knew what we were doing it would not be called research, would it? -- Einstein Only two things are infinite; the universe and human stupidity - and I'm not sure about green Uncle Dr. Bonehead the MIT fraudulent lecturer -- Einstein === Subject: Re: Magnify form posting-account=15T5jAkAAACpO9KgK0Ue212ywoAPQtp7 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; MDDC),gzip(gfe),gzip(gfe) snip Ben you should read up on the history of the Calculus. It originated > with the concept of infinitely small or infinitesimals. Mitch Raemsch- Hide quoted text - - Show quoted text - Mathematics has made advances since those days. Infinitesimals are as > obsolete as phlogiston. We do everything now with deltas and epsilons. > Look up analysis as a mathematical discipline. Uncle Ben > You are all bull and bluster, Uncle Bonehead. Calculus today > is the same calculus it was in Newton's day. Look up head > as an academic crank.- Hide quoted text - - Show quoted text - Again you reveal your ignorance, dear Androcles. Since Newton and > Leibnitz invented calculus, mathematicians fretted about the lack of > rigour in discussions of limits, infinite series, and your precious > infinitesimals. They, unlike you, knew back then that on an open > interval, there is no number closest to the boundary. Vague handwaving about mathematicians is a typical ignorant > Bonehead argument. Quote/ > 1. Assume that there exists a smallest number greater than zero: Call it > h. > 2. But h/2 is smaller than h > /unquote -- Uncle Bonehead Having assumed the existence of h as defined by Bonehead, > k = h/2 cannot exist. Assume a spherical cow... Lemma 1: Therefore Uncle Bonehead is illogical. > Lemma 2: Therefore Uncle Bonehead is not a mathematician or > even a physicist, he hallucinates he divide photons. Again you ignorantly contradict yourself, Bonehead. Look up > contradiction as a wit. sqrt(2) = 1 + 1/(2+1/(1/(2+1/(1/(2+1/(1/(2+1/(1/(2+1/(1+...))))))))) > (Extended fraction, Xeno style)- Hide quoted text - - Show quoted text - This one I will leave as a classic fumble! All the best to you Dr. Parker! (I think that should be continued fraction.) Uncle Ben > You don't know how to think, and no, it is an extended fraction > (which obviously you've never heard of). > http://tinyurl.com/d6ps8y-Hide quoted text - - Show quoted text - extended fraction. According to Google, the name continued > fraction is 500 times more common. UB- Hide quoted text - - Show quoted text - At 0.2% frequency, extended fraction is probably a mistranslation from another language by a translator who didn't know about continued fractions. === Subject: Re: Magnify form snip Ben you should read up on the history of the Calculus. It > originated > with the concept of infinitely small or infinitesimals. Mitch Raemsch- Hide quoted text - - Show quoted text - Mathematics has made advances since those days. Infinitesimals are > as > obsolete as phlogiston. We do everything now with deltas and > epsilons. > Look up analysis as a mathematical discipline. Uncle Ben > You are all bull and bluster, Uncle Bonehead. Calculus today > is the same calculus it was in Newton's day. Look up head > as an academic crank.- Hide quoted text - - Show quoted text - Again you reveal your ignorance, dear Androcles. Since Newton and > Leibnitz invented calculus, mathematicians fretted about the lack of > rigour in discussions of limits, infinite series, and your precious > infinitesimals. They, unlike you, knew back then that on an open > interval, there is no number closest to the boundary. Vague handwaving about mathematicians is a typical ignorant > Bonehead argument. Quote/ > 1. Assume that there exists a smallest number greater than zero: Call > it > h. > 2. But h/2 is smaller than h > /unquote -- Uncle Bonehead Having assumed the existence of h as defined by Bonehead, > k = h/2 cannot exist. Assume a spherical cow... Lemma 1: Therefore Uncle Bonehead is illogical. > Lemma 2: Therefore Uncle Bonehead is not a mathematician or > even a physicist, he hallucinates he divide photons. Again you ignorantly contradict yourself, Bonehead. Look up > contradiction as a wit. sqrt(2) = 1 + 1/(2+1/(1/(2+1/(1/(2+1/(1/(2+1/(1/(2+1/(1+...))))))))) > (Extended fraction, Xeno style)- Hide quoted text - - Show quoted text - This one I will leave as a classic fumble! All the best to you Dr. Parker! (I think that should be continued fraction.) Uncle Ben > You don't know how to think, and no, it is an extended fraction > (which obviously you've never heard of). > http://tinyurl.com/d6ps8y-Hide quoted text - - Show quoted text - extended fraction. According to Google, the name continued > fraction is 500 times more common. UB- Hide quoted text - - Show quoted text - At 0.2% frequency, extended fraction is probably a mistranslation from another language by a translator who didn't know about continued fractions. You never heard of adding the velocity of light to the velocity of its source, either. You were thoroughly indoctrinated, Dr. , and incapable of original thought. (Or even thought of any kind.) Some of us call that brainwashed. === Subject: Re: Orders and Quotients >[...] > Perhaps someone there can find out, where this criterion > in Blyth's book appeared for the first time. To what criterion are you referring? The condition that, given the order in X together with an equivalence relation in X, every closed chain is contained in an equivalence class. Of course, in the case of ker(f) this is automatically satisfied. Did you consider the case of preorders, i.e. reflexive and transitive > but not necessarily antisymmetric? > For these, you always have quotients, so it might give a nicer theory > anyway. Yes I did notice that preorders have no problem, not even a problem > with final preorders that I can see. Preorders however are messy. Yet with a final or quotient preorder, my first write up would > have no problem. So here's the score, topology has final quotient topologies > and preorder theory has final quotient preorders while > algebra and order theory do not have final factor groups > nor final quotient orders. Can one weaken the definition of a group to have final factor groups? Weaken in what way? In general, in algebraic settings, if A is some algebra and X is a set, together with a surjective map (of sets) f: A --> X, then the existence of an algebra structure is equivalent to the condition that ker(f) is a congruence relation, which means that ker(f) is a subalgebra of A x A. Hence, whenever you have equivalence relations which are not congruences, such lifted structures will fail to exist. What is final? In topology, it's the strongest topology. > In preorder theory, it's the weakest preorder. In both cases it is a best lifting of a set map u: |A| --> X to a morphism f: A --> B with |f| = u where | | is the forgetful functor PrOrd --> Set or Top --> Set. The two cases just appear opposite, because the conditions on maps to be morphism go in opposite directions. For example, if you have two topologies t and t' on the same set X, then the identity map (X,t) --> (X,t') is continuous iff t' is contained in t, whereas when you have two orders a and b on X, the identity map (X,a) --> (X,b) is order preserving if a is contained in b. -- Marc === Subject: Re: The complete infinite binary tree has only countably many infinite paths. posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) If you can't understand, look at my proof C. C) Consider the edges of the complete binary tree (an edge connects > two subsequent nodes of a path). Take all edges and put them on one > and the same level of the tree, side by side, such that the tree > now > is an array of parallel edges: |||||||... This array limits the > number > of possible paths of the tree. Yes, Each possible path correponds to a subset of the array, so there > can't be more paths than subsets of the array, There are > uncountably many subsets of the array (not all correspond to paths, > however removing these does not change the upper > limit, as removing an uncountable number > of subsets may still leave an uncounable number of subsets), But here it does not. You are completely right that *all* infinite subsets of the set of edges form an uncountable set. We know this from the set N and its infinite subsets. That is because these subsets can have gaps like 1,2,3,5,6,7,... 1,2,8,9,10,... 2,4,6,... and so on. The linear subsets 1,2,3,... 2,3,4,... 3,4,5... and so on form a countable set. I told you already: If it was poss.92ble that two paths could use a set of edges together, then split and after a while come together again, then you were right. But that is not the case in the tree. Therefore the upper limit of paths is countable. === Subject: Re: The complete infinite binary tree has only countably many infinite paths. > If you can't understand, look at my proof C. C) Consider the edges of the complete binary tree (an edge connects > two subsequent nodes of a path). Take all edges and put them on one > and the same level of the tree, side by side, such that the tree > now > is an array of parallel edges: |||||||... This array limits the > number > of possible paths of the tree. Yes, Each possible path correponds to a subset of the array, so there > can't be more paths than subsets of the array, There are > uncountably many subsets of the array (not all correspond to paths, > however removing these does not change the upper > limit, as removing an uncountable number > of subsets may still leave an uncounable number of subsets), But here it does not. > You are completely right that *all* infinite subsets of the set of > edges form an uncountable set. Correct. So until you discuss which subsets actually do correspond to paths you are left with an upper limit of uncountable. The form of words you want at this point is In my putative proof C I do not discuss which subsets of the set of edges correspond to paths. Thus, as it stands my putative proof C only provides an upper limit of uncountable. I now wish to add something to this putative proof. - William Hughes === Subject: Re: The complete infinite binary tree has only countably many infinite paths. posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Yes, Each possible path correponds to a subset of the array, so there > can't be more paths than subsets of the array, There are > uncountably many subsets of the array (not all correspond to paths, > however removing these does not change the upper > limit, as removing an uncountable number > of subsets may still leave an uncounable number of subsets), But here it does not. > You are completely right that *all* infinite subsets of the set of > edges form an uncountable set. Correct. So until you discuss which subsets actually do > correspond to paths you are left with an upper limit > of uncountable. As an easy introductory example, consider the set of edges {a, b, c} in 0.111... 0. / a 0 1 / /b 0 1 0 1 / / / / c 01010101 Its linear subsets {a} , {a, b}, and {a, b, c} appear in all paths starting with at least one or two or three nodes of value 1, respectively. But there is no path in the binary tree that contains only the following subsets of nodes: {b}, {c}, {a, c}, {b, c} . Together with the empty set the four latter subsets complete the power set of {a, b, c}. But they are not in the tree. And those linear subsets which in fact occur in the tree, form only a countable set. Now you might argue that nearly half of the power set is present and that nearly half of uncountable is uncountable nonetheless. But that would be wrong. If you look at longer sets of edges {1, 2, 3, ..., n} with n instead of only 3 elements, then you will find that only their linear subsets [*] {1} {1, 2} ... {1, 2, 3, ..., n} conribute to paths. Those subsets which would be required to yield the power set, like, for example, {1, 2, 3, ..., k, k+j, ..., n} do not contribute to paths. It should be not too difficult to see that all subsets of edges in the tree that contribute to paths, form a countable set, because A) all subsets of the set of edges of the form {1, 2, 3, ..., n} form a countable set, and B) all linear subsets [*] of these subsets contribute to a cartesian product of countable sets. Therefore all subsets of the set of edges in the binary tree form a countable set. === Subject: Re: The complete infinite binary tree has only countably many infinite paths. Yes, Each possible path correponds to a subset of the array, so there > can't be more paths than subsets of the array, There are > uncountably many subsets of the array (not all correspond to paths, > however removing these does not change the upper > limit, as removing an uncountable number > of subsets may still leave an uncounable number of subsets), But here it does not. > You are completely right that *all* infinite subsets of the set of > edges form an uncountable set. Correct. So until you discuss which subsets actually do > correspond to paths you are left with an upper limit > of uncountable. As an easy introductory example, consider the set of edges > {a, b, c} in 0.111... 0. > / a > 0 1 > / /b > 0 1 0 1 > / / / / c > 01010101 Its linear subsets {a} , {a, b}, and {a, b, c} appear in all paths > starting with at least one or two or three nodes of value 1, > respectively. But there is no path in the binary tree that contains > only the following subsets of nodes: > {b}, {c}, {a, c}, {b, c} . > Together with the empty set the four latter subsets complete the power > set of {a, b, c}. But they are not in the tree. And those linear > subsets which in fact occur in the tree, form only a countable set. > Now you might argue that nearly half of the power set is present and > that nearly half of uncountable is uncountable nonetheless. But that > would be wrong. If you look at longer sets of edges {1, 2, 3, ..., n} > with n instead of only 3 elements, then you will find that only their > linear subsets [*] > {1} > {1, 2} > ... > {1, 2, 3, ..., n} > conribute to paths. Those subsets which would be required to yield the > power set, like, for example, > {1, 2, 3, ..., k, k+j, ..., n} > do not contribute to paths. It should be not too difficult to see that all subsets of edges in the > tree that contribute to paths, form a countable set, because > A) all subsets of the set of edges of the form > {1, 2, 3, ..., n} > form a countable set, and > B) all linear subsets [*] of these subsets contribute to a cartesian > product of countable sets. Therefore all subsets of the set of edges > in the binary tree form a countable set. That, as most of WM's claims, only holds for the set of all finite subsets, which excludes all paths in the complete tree. === Subject: Re: The complete infinite binary tree has only countably many infinite paths. posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I Yes, Each possible path correponds to a subset of the array, so there > can't be more paths than subsets of the array, There are > uncountably many subsets of the array (not all correspond to paths, > however removing these does not change the upper > limit, as removing an uncountable number > of subsets may still leave an uncounable number of subsets), But here it does not. > You are completely right that *all* infinite subsets of the set of > edges form an uncountable set. Correct. So until you discuss which subsets actually do > correspond to paths you are left with an upper limit > of uncountable. No, before we can talk about which sets actually do correspond to paths you have to note that your original proof ignored this vital aspect. - William Hughes === Subject: Re: The complete infinite binary tree has only countably many infinite paths. posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Yes, Each possible path correponds to a subset of the array, so there > can't be more paths than subsets of the array, There are > uncountably many subsets of the array (not all correspond to paths, > however removing these does not change the upper > limit, as removing an uncountable number > of subsets may still leave an uncounable number of subsets), But here it does not. > You are completely right that *all* infinite subsets of the set of > edges form an uncountable set. Correct. So until you discuss which subsets actually do > correspond to paths you are left with an upper limit > of uncountable. No, before we can talk about which sets actually > do correspond to paths you have to note that > your original proof ignored this vital > aspect. 1) It is an obvious fact. I did not expect that it was necessary to explain. 2) With no regard to this obvious fact my proof (C) uses the fact that every path of the complete binary tree must have an edge which distinguishes this path from every other path. Otherwise you have only potential infinity saying: For every path p there is a path q such that on level n q has another edge than p. That is not sufficient to obtain actual infinity, i.e., the irrational limit of a sequence of rationals as a path in the tree. If pi exists and if pi can be the diagonal of Cantor's list, then every digit must exist. Then, in the binary tree, every edge must exist. This means there is an edge of pi that differs from every other path. This is different from the Cantor's list. There is no digit of pi that differs from every other entry of the list. There we have only: For every number q =/= pi there is a digit of pi, that is different from the corresponding digit of pi. In the tree we have (if pi exists as a complete path): There is an edge of pi that is not in any other path. The same is true for every real number that is represented by a path in the tree. Therefore my proof (C) holds. === Subject: Re: The complete infinite binary tree has only countably many infinite paths. > Yes, Each possible path correponds to a subset of the array, so > there > can't be more paths than subsets of the array, There are > uncountably many subsets of the array (not all correspond to paths, > however removing these does not change the upper > limit, as removing an uncountable number > of subsets may still leave an uncounable number of subsets), But here it does not. > You are completely right that *all* infinite subsets of the set of > edges form an uncountable set. Correct. So until you discuss which subsets actually do > correspond to paths you are left with an upper limit > of uncountable. No, before we can talk about which sets actually > do correspond to paths you have to note that > your original proof ignored this vital > aspect. 1) It is an obvious fact. I did not expect that it was necessary to > explain. Since you do not accept without explanation what to all of us are obvious facts, or even accept them when given exhaustive explanation, for that matter, we are not required to accept without explanation what you claim to be obvious facts. So, no explanation from you means unacceptable to us. > 2) With no regard to this obvious fact my proof (C) uses the fact that > every path of the complete binary tree must have an edge which > distinguishes this path from every other path. False! Only a terminal edge can distinguish a path from all other paths, and in a complete infinite binary tree there are no terminal edges. In a complete infinite binary tree it takes infinitely many edges to distinguish any one path from all others. WM imagination is crippled by still being limited to what happens in finite sets and trees. > Otherwise you have only > potential infinity saying: > For every path p there is a path q such that on level n q has another > edge than p. WM may be so limited, but we are not. > That is not sufficient to obtain actual infinity, i.e., the irrational > limit of a sequence of rationals as a path in the tree. If pi exists and if pi can be the diagonal of Cantor's list, then > every digit must exist. Since pi is not between 0 and 1, and such listings only list numbers between 0 and 1, trying to list pi is a fools game, and WM is just the fool to try it. > There we have > only: For every number q =/= pi there is a digit of pi, that is > different from the corresponding digit of pi. WM may have this in his weird world of MathUnrealism, but we don't. In the tree we have (if pi exists as a complete path): There is an > edge of pi that is not in any other path. Lets work with 1/pi as a complete path. Then every edge in the path of 1/pi is an edge in uncountably many other paths. In a complete infinite binary tree, the sub-tree rooted at any node is tree-isomorphic to the whole tree rooted at the root node. That is just one of the many things that WM does not understand about such trees. > The same is true for every > real number that is represented by a path in the tree. Therefore my > proof (C) holds. That proof holds no better than a sieve holds water. They are both full of holes. === Subject: Re: The complete infinite binary tree has only countably many infinite paths. posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I Yes, Each possible path correponds to a subset of the array, so there > can't be more paths than subsets of the array, There are > uncountably many subsets of the array (not all correspond to paths, > however removing these does not change the upper > limit, as removing an uncountable number > of subsets may still leave an uncounable number of subsets), But here it does not. > You are completely right that *all* infinite subsets of the set of > edges form an uncountable set. Correct. So until you discuss which subsets actually do > correspond to paths you are left with an upper limit > of uncountable. No, before we can talk about which sets actually > do correspond to paths you have to note that > your original proof ignored this vital > aspect. 1) It is an obvious fact. Nope. Subsets which do not correspond to paths {1,4,6,..} {1,2,3,...} {2,3,4,...} {7,8,9,...} {2,5,10,...} Subsets which do correspond to paths {1,3,7,...} {2,6,14,...} {2,5,11,...} {1,4,10,...} It is not obvious which subsets go where. Before we can talk about this non-obvious question you have to note that your original proof ignored this vital aspect. - William Hughes === Subject: Re: The complete infinite binary tree has only countably many infinite paths. > Yes, Each possible path correponds to a subset of the array, so there > can't be more paths than subsets of the array, There are > uncountably many subsets of the array (not all correspond to paths, > however removing these does not change the upper > limit, as removing an uncountable number > of subsets may still leave an uncounable number of subsets), But here it does not. > You are completely right that *all* infinite subsets of the set of > edges form an uncountable set. Correct. So until you discuss which subsets actually do > correspond to paths you are left with an upper limit > of uncountable. No, before we can talk about which sets actually > do correspond to paths you have to note that > your original proof ignored this vital > aspect. 1) It is an obvious fact. Nope. Subsets which do not correspond to paths {1,4,6,..} > {1,2,3,...} > {2,3,4,...} > {7,8,9,...} > {2,5,10,...} Subsets which do correspond to paths {1,3,7,...} > {2,6,14,...} > {2,5,11,...} > {1,4,10,...} This is a misunderstanding. I did not use the enumeration of edges as you seem to do. Compare the sketch containing the edges a, b, c, of my last posting. The set {a, b, c} corresponds to a path, namely to all paths that start with 0.111.... corresponds to a path too. If you wish, you can also denote it by {E1, E2, E3, ..., En}. It corresponds to every path that shares the first n edges witha given path. As the given path you can select any path of the tree. > It is not obvious which subsets go where. > Before we can talk about this non-obvious > question you have to note that > your original proof ignored this vital > aspect. It is really so simple that mentioning seemed unnecessary to me. Why do you think that I chose the tree instad of general sets or lists? In the tree there are only linear subsets. That is the point. I will explain it agian: Consider any finite set of edges that is within one path of the tree. Example: Consider n edges turning right for the path 0.111... Denote these edges by {E1, E2, E3, ..., En} Which subsets of this set are in the tree? There are all linear subsets {E1}, {E1, E2} ,{E1, E2, E3}, ..., {E1, E2, E3, ..., En} There is no nonlinear subset like {E2, E3} or {E3} or {E2, E3, E4, ..., En}. This holds for the whole tree: Every finite set of edges belongs to a countable set (because all nodes are countable). And every finite subset of such a set belongs to a countable set of subsets. So we have a countable set of countable subsets. And there are not more subsets in the tree. === Subject: Re: The complete infinite binary tree has only countably many infinite paths. posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I Yes, Each possible path correponds to a subset of the array, so there > can't be more paths than subsets of the array, There are > uncountably many subsets of the array (not all correspond to paths, > however removing these does not change the upper > limit, as removing an uncountable number > of subsets may still leave an uncounable number of subsets), But here it does not. > You are completely right that *all* infinite subsets of the set of > edges form an uncountable set. Correct. So until you discuss which subsets actually do > correspond to paths you are left with an upper limit > of uncountable. No, before we can talk about which sets actually > do correspond to paths you have to note that > your original proof ignored this vital > aspect. 1) It is an obvious fact. Nope. Subsets which do not correspond to paths {1,4,6,..} > {1,2,3,...} > {2,3,4,...} > {7,8,9,...} > {2,5,10,...} Subsets which do correspond to paths {1,3,7,...} > {2,6,14,...} > {2,5,11,...} > {1,4,10,...} This is a misunderstanding. I did not use the enumeration of edges as > you seem to do. You did not do anything. You completely ignored the question of which subsets belong to paths. Your excuse for not doing anything was that it was obvious. However, no matter what how you label the edges, there is no obvious way to divide the subsets of edges that correpond to paths from the subsets that do not correspond to paths. - William Hughes === Subject: Re: The complete infinite binary tree has only countably many infinite paths. posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > You completely > ignored the question of which subsets belong > to paths. Your excuse for not > doing anything was that it was obvious. > However, no matter what how you label the edges, there > is no obvious way to divide the subsets of edges that correpond > to paths from the subsets that do not correspond to paths. By this easy introductory example of a set {a, b, c} of edges of the path 0.111... 0. / a 0 1 / /b 0 1 0 1 / / / / c 01010101 ... I showed that all subsets that belong to this set of edges and are components of at least one path in the tree are linear, namely {a} , {a, b}, and {a, b, c}. But there is no path in the binary tree that contains only one of the following subsets of nodes: {b}, {c}, {a, c}, {b, c}. This holds for every set of edges that belong to one path of he tree. You argued that paths are subsets of edges. I showed that all subsets of edes that can belong to paths form linear subsets and hence are countable. === Subject: Re: The complete infinite binary tree has only countably many infinite paths. You completely > ignored the question of which subsets belong > to paths. Your excuse for not > doing anything was that it was obvious. > However, no matter what how you label the edges, there > is no obvious way to divide the subsets of edges that correpond > to paths from the subsets that do not correspond to paths. By this easy introductory example of a set {a, b, c} of edges of the > path 0.111... 0. > / a > 0 1 > / /b > 0 1 0 1 > / / / / c > 01010101 > ... > I showed that all subsets that belong to this set of edges and are > components of at least one path in the tree are linear, namely > {a} , {a, b}, and {a, b, c}. But there is no path in the binary tree > that contains only one of the following subsets of nodes: > {b}, {c}, {a, c}, {b, c}. This holds for every set of edges that belong to one path of he tree. > You argued that paths are subsets of edges. > I showed that all subsets of edes that can belong to paths form linear > subsets and hence are countable. Given that any path is a set of infinitely many edges, the set of its subsets is uncountable. So that WM is WRONG! AGAIN! AS USUAL! === Subject: Re: The complete infinite binary tree has only countably many infinite paths. posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I You completely > ignored the question of which subsets belong > to paths. Your excuse for not > doing anything was that it was obvious. > However, no matter what how you label the edges, there > is no obvious way to divide the subsets of edges that correpond > to paths from the subsets that do not correspond to paths. By this easy introductory example of a set {a, b, c} of edges of the > path 0.111... Nope, you only label one subset of edges. In order to separate all subsets that correspond to paths from all subsets that do not correspond to paths you need consider all subsets, i.e. you need to label all the edges. (Note, assuming you already know all subsets which correspond to paths is begging the question.) Try again. - William Hughes === Subject: Re: The complete infinite binary tree has only countably many infinite paths. posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > You completely > ignored the question of which subsets belong > to paths. Your excuse for not > doing anything was that it was obvious. > However, no matter what how you label the edges, there > is no obvious way to divide the subsets of edges that correpond > to paths from the subsets that do not correspond to paths. By this easy introductory example of a set {a, b, c} of edges of the > path 0.111... Nope, you only label one subset of edges. I show by a general argument that holds for *every path* that the set of all its subsets is countable. You wanted to argue that you could form uncountably many paths by such subsets. I showed that the subsets required for an uncountable set are not subsets that are in the binary tree. > In order to separate all subsets that correspond to paths > from all subsets that do not correspond to paths you need > consider all subsets, i.e. you need to label all the edges. I have labelled all edges. They form a countable set. Then I have shown that every subset of edges that lie between the root node and one of the labelled edges is countable. So we have a countable set of countable sets. > (Note, assuming you already know all subsets which correspond > to paths is begging the question.) Nevertheless it is simply true and unavoidable. There is no chance to have more subsets. That is why the binary tree is such a wonderful simple example to prove that there are not uncountably many real numbers in the unit interval. You can't see the forest because there are so many trees hindering you? === Subject: Re: The complete infinite binary tree has only countably many infinite paths. > You completely > ignored the question of which subsets belong > to paths. Your excuse for not > doing anything was that it was obvious. > However, no matter what how you label the edges, there > is no obvious way to divide the subsets of edges that correpond > to paths from the subsets that do not correspond to paths. By this easy introductory example of a set {a, b, c} of edges of the > path 0.111... Nope, you only label one subset of edges. I show by a general argument that holds for *every path* that the set > of all its subsets is countable. If *any path* contains a set of infinitely many edges, then that set of edges has uncountably many subsets. If *no path* contains a set of infinitely many edges, then the tree is finite. > You wanted to argue that you could > form uncountably many paths by such subsets. No! Just that there are uncountably many paths if a path is defined to be ANY maximal totally ordered sets of nodes (or edges) ordered by ancestors preceding posterity. The only way to limit the number of paths to less than uncountably many in an infinite binary tree is to deny pathhood to some of what qualify as paths by the above definition. Which makes the tree incomplete. > I showed that the subsets > required for an uncountable set are not subsets that are in the binary > tree. No you didn't. One can easily see that every function from N to {left-child, right-child} determines a unique path (chain of nodes) different from that of any different such function. And Cantor showed that there are uncountably many such functions. > In order to separate all subsets that correspond to paths > from all subsets that do not correspond to paths you need > consider all subsets, i.e. you need to label all the edges. I have labelled all edges. They form a countable set. Then I have > shown that every subset of edges that lie between the root node and > one of the labelled edges is countable. But none of these is a path, so you are not counting paths, and your count is irrelevant. > So we have a countable set of > countable sets. Irrelevant as you are not counting paths. > You can't see the forest because there are so many trees hindering > you? We see a good deal further into that forest than WM can see. WM's mathematical myopia includes being unable to see past finite sets. And is compounded by his quantifier dyslexia, among other things. === Subject: Re: The complete infinite binary tree has only countably many infinite paths. posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I You completely > ignored the question of which subsets belong > to paths. Your excuse for not > doing anything was that it was obvious. > However, no matter what how you label the edges, there > is no obvious way to divide the subsets of edges that correpond > to paths from the subsets that do not correspond to paths. By this easy introductory example of a set {a, b, c} of edges of the > path 0.111... Nope, you only label one subset of edges. I show Nope, you did nothing at all, you claim that the argument is obvious. However, you have not shown any way, let alone an obvious way to divide all the subsets of edges that correspond to paths from all the subsets that do not correspond to paths. > by a general argument that holds for *every path* that the set > of all its subsets is countable The total number of subsets is the number of paths times the number of subsets in each path which correpond to a path. So if there are a countable number of paths then there are a countable number of subsets and if there are an uncountable number of paths there are an uncountable number of subsets. Yes, this is obvious, it is also useless. - William Hughes === Subject: Re: The complete infinite binary tree has only countably many infinite paths. posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > You completely > ignored the question of which subsets belong > to paths. Your excuse for not > doing anything was that it was obvious. > However, no matter what how you label the edges, there > is no obvious way to divide the subsets of edges that correpond > to paths from the subsets that do not correspond to paths. By this easy introductory example of a set {a, b, c} of edges of the > path 0.111... Nope, you only label one subset of edges. I show Nope, you did nothing at all, you claim that > the argument is obvious. However, you have not shown > any way, let alone an obvious way to divide all the subsets of > edges that correspond to paths from all the subsets > that do not correspond to paths. Attempted change of topic. You demanded: > Yes, Each possible path correponds to a subset of the array, so there > can't be more paths than subsets of the array, There are > uncountably many subsets of the array (not all correspond to paths, > however removing these does not change the upper > limit, as removing an uncountable number > of subsets may still leave an uncounable number of subsets), All subsets of the array that are in the tree as proper subsets of paths are 1) finite 2) linear. There are only a countable number of finite subsets. There are only a countable number of linear subsets of every finite subset. by a general argument that holds for *every path* that the set > of all its subsets is countable The total number of subsets is the number of paths According to your original argument paths are unions of subsets of edges. In fact they are unions of finite subsets of edges. Therefore it is sufficient to count all finite subsets that can be in the tree and can contribute to paths. This is what I have done. Your demanding would be senseless if we had to start with a given number of paths. > times the number of subsets in each path > which correpond to a path. That is nonsense. If we start from paths then we do not need subsets at all. > So if there are a countable number of paths then there > are a countable number of subsets and if there > are an uncountable number of paths there are an > uncountable number of subsets. Yes, this is obvious, > it is also useless. Exactly. Then considering subsets would be usesless. You seem to have caught yourself in your own trap. We must determine all possible subsets that can contribute. These subsets are countable and have only linear, hence countably many, sub- subsets. === Subject: Re: The complete infinite binary tree has only countably many infinite paths. posting-account=-eQqtQoAAACZVM-kNEsOn3k7GSvoJoS4 GTB5; .NET CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; \ .NET CLR 3.5.21022; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > By this easy introductory example of a set {a, b, c} of edges of the > path 0.111... 0. > / a > 0 1 > / /b > 0 1 0 1 > / / / / c > 01010101 > ... > I showed that all subsets that belong to this set of edges and are > components of at least one path in the tree are linear, namely > {a} , {a, b}, and {a, b, c}. But there is no path in the binary tree > that contains only one of the following subsets of nodes: > {b}, {c}, {a, c}, {b, c}. This holds for every set of edges that belong to one path of he tree. > You argued that paths are subsets of edges. > I showed that all subsets of edes that can belong to paths form linear > subsets and hence are countable. This is hare-brained, even by WM's standards. Of *course* the nodes and/or edges that make up *one path* can be thought of as a linear set and are countable. Nobody has any problems with that. Now *if the number of all paths were countable*, you could even use this to prove that the number of paths must be countable. Congratulations on finding yet another circular argument, Herr Professor. === Subject: Re: The complete infinite binary tree has only countably many infinite paths. Yes, Each possible path correponds to a subset of the array, so > there > can't be more paths than subsets of the array, There are > uncountably many subsets of the array (not all correspond to > paths, > however removing these does not change the upper > limit, as removing an uncountable number > of subsets may still leave an uncounable number of subsets), But here it does not. > You are completely right that *all* infinite subsets of the set of > edges form an uncountable set. Correct. So until you discuss which subsets actually do > correspond to paths you are left with an upper limit > of uncountable. No, before we can talk about which sets actually > do correspond to paths you have to note that > your original proof ignored this vital > aspect. 1) It is an obvious fact. Nope. Subsets which do not correspond to paths {1,4,6,..} > {1,2,3,...} > {2,3,4,...} > {7,8,9,...} > {2,5,10,...} > Subsets which do correspond to paths {1,3,7,...} > {2,6,14,...} > {2,5,11,...} > {1,4,10,...} It is not obvious which subsets go where. > Before we can talk about this non-obvious > question you have to note that > your original proof ignored this vital > aspect. > - William Hughes There is a simple, but different, notation in which EVERY subset of N corresponds uniquely to a path: Each edge occurs at a level indexed by a natural, so one has two edges at level 1, 4 edges at level 2, 8 edges at level 3, and so on, and each path can be unambiguously identified by the set of levels at which it branches left rather than right. In this notation, there is a bijection between the set of subsets of N and the set of paths in a complete infinite binary tree, each such set of naturals being the set of level indices of the set of left branchings of the path. === Subject: Re: The complete infinite binary tree has only countably many infinite paths. posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) CORRECTION If pi exists and if pi can be the diagonal of Cantor's list, then > every digit must exist. Then, in the binary tree, every edge must > exist. This means there is an edge of pi that differs from every other > path. This is different from the Cantor's list. There is no digit of > pi that differs from every other entry of the list. There we have > only: For every number q =/= pi there is a digit of pi, that is > different from the corresponding digit of q. === Subject: Re: The complete infinite binary tree has only countably many infinite paths. CORRECTION If pi exists and if pi can be the diagonal of Cantor's list Cantor's list is a list of binary sequences provided by anyone else for Cantor to test. Originally (by Cantor himself), the terms for such sequences were taken from {m,w}, not {0,1}, and so were not numbers at all. When the argument was modified to have sequences of 0's and 1's, the values of such sequences had to be between 0 and 1, which pi is not. Therefore, pi cannot be in any list or a diagonal for any list. Do you want to try 1/pi, WM? === Subject: Re: The complete infinite binary tree has only countably many infinite paths. > If you can't understand, look at my proof C. C) Consider the edges of the complete binary tree (an edge connects > two subsequent nodes of a path). Take all edges and put them on one > and the same level of the tree, side by side, such that the tree > now > is an array of parallel edges: |||||||... This array limits the > number > of possible paths of the tree. Yes, Each possible path correponds to a subset of the array, so there > can't be more paths than subsets of the array, There are > uncountably many subsets of the array (not all correspond to paths, > however removing these does not change the upper > limit, as removing an uncountable number > of subsets may still leave an uncounable number of subsets), But here it does not. > You are completely right that *all* infinite subsets of the set of > edges form an uncountable set. Correct. So until you discuss which subsets actually do > correspond to paths you are left with an upper limit > of uncountable. The form of words you want at this point is In my putative proof C I do not discuss which subsets of the set > of edges correspond to paths. Thus, as it stands my putative proof > C only provides an upper limit of uncountable. I now wish > to add something to this putative proof. - William Hughes Actually, admitting uncountably many subsets to N, or to any countable set, allows direct proof of the existence of uncountably many paths, at least in a COMPLETE binary tree. As I show in another post. === Subject: Re: The complete infinite binary tree has only countably many infinite paths. posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Actually, admitting uncountably many subsets to N, or to any countable > set, allows direct proof of the existence of uncountably many paths, at > least in a COMPLETE binary tree. As I show in another post.- But there are only countably many countable sets of edges in the tree. As I show in another post. === Subject: Re: The complete infinite binary tree has only countably many infinite paths. Actually, admitting uncountably many subsets to N, or to any countable > set, allows direct proof of the existence of uncountably many paths, at > least in a COMPLETE binary tree. As I show in another post.- But there are only countably many countable sets of edges in the > tree. Only if one is restricted to finite sets of edges, if there are countably many edges, then there are uncountably many sets of edges. As I show in another post. WM's showings are always in another post. Jam every other day, means jam yesterday and jam tomorrow but never jam today. Every infinite sequence of 0's and 1's denotes a separate set of edges forming a path in the complete infinite binary tree with the position of the 0 or 1 in the sequence denoting the position of the edge in the sequence of edges and the value, 0 or 1, denoting a left or right branching, respectively. Every set of naturals then can denote that sequence which branches left at those levels ennumerated in the set. So every set of naturals denotes a path different from the path denoted by any other set of naturals. Uncountably many sets of naturals implies uncountably many sets of edges and uncountably many paths. === Subject: Re: The complete infinite binary tree has only countably many infinite paths. posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > Actually, admitting uncountably many subsets to N, or to any countable > set, allows direct proof of the existence of uncountably many paths, > Indeed. Possibly more importantly, we have an uncountable number of things distinguished by a countable number of things. This is something WM claims is impossible, and is thus a direct rather than an indirect contradiction. - William Hughes === Subject: Re: The complete infinite binary tree has only countably many infinite paths. posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Actually, admitting uncountably many subsets to N, or to any countable > set, allows direct proof of the existence of uncountably many paths, Indeed. > Possibly more importantly, we have an uncountable number of things > distinguished by a countable number of things. This is something > WM claims is impossible, I do not claim that in general (although more generally there is no uncountability at all) but in the binary tree. The power set of N may consist of uncountably many elements that all are distinguished by presence or absence of countably many naturals. As I have tried to explain to you, this is not possible in the binary tree. That's why I applied the binary tree (to show that there is no uncountability at all and also no actual infinity and hence no complete binary tree). === Subject: Re: The complete infinite binary tree has only countably many infinite paths. Actually, admitting uncountably many subsets to N, or to any countable > set, allows direct proof of the existence of uncountably many paths, Indeed. > Possibly more importantly, we have an uncountable number of things > distinguished by a countable number of things. This is something > WM claims is impossible, I do not claim that in general (although more generally there is no > uncountability at all) but in the binary tree. Where, as usual, WM is wrong. > The power set of N may consist of uncountably many elements that all > are distinguished by presence or absence of countably many naturals. > As I have tried to explain to you, this is not possible in the binary > tree. On the contrary. Every path has a first edge and a second and so on for all natural number levels. And every path is uniquely identified by the set of levels at which the branching is a left, rather than right, branching. Thus for every subset of N a unique path distinguished from the path of any other subset of N. I.e., a bijection between subsets of N, of which there are uncountably many, and paths in the complete infinite binary tree, of which this proves there are also uncountably many. >That's why I applied the binary tree (to show that there is no > uncountability at all and also no actual infinity and hence no > complete binary tree). At which you failed miserably. > === Subject: Re: Distinguish nested brackets in Mathematica > A low-tech solution: > Artificially write the expression so that each nesting level is on a > separate line, with parentheses indented in increments of 3 spaces, > say, to show the level of nesting. > quasi >solution using colors? (Sorry, I didn't mean to run down your answer.) One can try to use a modification of the method Whitehead and Russell used in _Principia_. They used more dots for the stronger separator, but used the same number for what can be called both right and left parentheses. Accordingly, one can set up a notation so that each bracket gets a label, with the labels of the matching brackets the same. Your idea of colors would be one way to do this, but sometimes we need black and white. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Question abut the QFT Applicability of Maxwell / Faraday's Integral Equations posting-account=c6T4LggAAACAqdk8Zx_zXI_oErOujh3M Gecko/20050915 Firefox/1.0.7,gzip(gfe),gzip(gfe) Jay, a bit to add. > Hi Jay, > I'll quote you, > Where I am headed with this, is a discussion of QCD confinement, > which > also centers around what does and does not flow across a baryon > surface. > But classical and quantum electrodynamics seems a good place from > which > to germinate this discussion. A study of Baryon inter-action in the interior of a > Neutron Star follows, The possible reaction is, > neutron + neutron + Energy => neutron + antineutron. > The Energy = 2*neutron, and this > Energy = Pressure*Area*distance. So the Area is equal > to the relation of two neutrons and the distance is the > crushing of the two neutrons that forces the Pauli Exclusion > Principle into effect and convert one of the neutrons into > an anti-neutron, and then decay to neutrino's etc. These are very close using a BOE (Back Of Envelope) > computation such as in the QBasic program, > ++++++++++++++++++++++ > CLS > pi = 3.1416 > c = 3 * 10 ^ 10 '(cm/sec) 'Neutron Star (center) Pressure (N/m^2) > P = 9.91 ^ 34 '(Newtons/meter^2) 'Neutron radius > R = .4 * 10 ^ -15 '(meters) 'Neutron Area > A = pi * R ^ 2 '(m^2) 'Neutron mass > M = 1.67 * 10 ^ -24 '(gm) 'Neutron Energy > E = M * c ^ 2 '(ergs) 'Force on a Neutron > F = P * A 'Newtons > F = F * 10 ^ 5 'dynes PRINT Displacement S by Force F to obtain Neutron Energy E > S = E / F 'cm > S = S / 100 'meters > PRINT S =; S; Meters > PRINT PRINT The value for S is theorized to be the Neutron Radius > PRINT S/R =; S / R ; This quantity should be =1 > ------------------------------------------------------------ In the above BOE calculation, a Pressure P was numerically the program. Please see this thread, +++ === Subject: neutron star interior pressure? --- Given the huge exponents and the coarseness of the calculations, the interior N-star Pressure P and the Baryon decay pressure were well within a magnitude. That (IMHO) is beyond mere coincidence. It goes without saying I can slightly adjust parameters such as N-star radius, neutron radius and area etc. If for example one does create a Neutron model with a finite crushing pressure, that theory can be tested by the maximum size of N-stars. You might of course end up predicting Black Holes evaporate prior to the event horizon crush, but a possible decision is whether Baryon consvervation *always* holds forever and ever, and BH's are the result of that theory, or there are circumstances in which Baryon conservation does not hold. Ken S. Tucker PS: Our friend Mr. Hawking should enjoy this post. === Subject: Blending three functions at the intersection of three regions? posting-account=30TueAoAAAAeRakeyTKqut8-jTCzw9_e SV1),gzip(gfe),gzip(gfe) Hi all, I am getting stuck with this problem. I have a function y=f(x,y) if (x,y) belong to the 2-D region Z1 or y =h(x,y) if (x,y) belong to Z2 or y=g(x,y) if (x,y) belong to Z3. Z1,Z2 and Z3 are three partitions or R^2. The problem is that y is not continuous at the boundaries of these regions (Z1,Z2, and Z3). I want to apply a smoothing technique in the intersection of three regions: Z1, Z2 and Z3. However it is quite tough for me. Do you guys know what simple technique to handle this? P/S for the boundary between each pairs of the neigboring regions (for example Z1 and Z2, or Z2 and Z3,..) I try to set a weighted technique and it is OK, however for the region joined of three regions, it is getting tough). Please help, === Subject: Re: Blending three functions at the intersection of three regions? >Hi all, I am getting stuck with this problem. >I have a function y=f(x,y) if (x,y) belong to the 2-D region Z1 or y =h(x,y) if (x,y) belong to Z2 or y=g(x,y) if (x,y) belong to Z3. > I suppose you mean z = on the left side of each of those equations. >Z1,Z2 and Z3 are three partitions or R^2. What are the boundaries between these like? Do the regions overlap or just touch in some curve or straight line dividing them? >The problem is that y is not continuous at the boundaries of these >regions (Z1,Z2, and Z3). I want to apply a smoothing technique in the >intersection of three regions: Z1, Z2 and Z3. Do the regions overlap? Do they intersect at a single point? Do you have a particular Z1, Z2, Z3 in mind? However it is quite tough for me. Do you guys know what simple >technique to handle this? P/S for the boundary between each pairs of the neigboring regions (for >example Z1 and Z2, or Z2 and Z3,..) I try to set a weighted technique >and it is OK, however for the region joined of three regions, it is >getting tough). Please help, > How about giving more details about where the problem comes from and what you are trying to accomplish? --Lynn http://math.asu.edu/~kurtz === Subject: Re: Blending three functions at the intersection of three regions? posting-account=30TueAoAAAAeRakeyTKqut8-jTCzw9_e SV1),gzip(gfe),gzip(gfe) Hi Kurtz, > I suppose you mean z = on the left side of each of those equations. Sorry, I do not understand this question. But my question posted earlier is correct. y is defined in R^2 but in each partion, it equals different function. Specifically: pseudo-code of y is as follows: For given three different h(x,y), g (x,y), f(x,y) if x^2+y^2 <=A i.e. Region Z1) y= f(x,y); elseif y>=0 i.e. Region Z2) y=g(x,y); elseif y<0 i.e. Region Z3) y=h(x,y); end Where A is a constant. I do not have a problem to smooth out the boundary between Z2 and Z3 (i.e. the x axis). But I get a problem at the boundary between Z1,Z2, and Z3. (i.e. the point with coordinate (x=sqrt(A), y=0). Z1,Z2 and Z3 are three partitions or R^2. What are the boundaries between these like? Do the regions overlap or > just touch in some curve or straight line dividing them? Z1 The problem is that y is not continuous at the boundaries of these >regions (Z1,Z2, and Z3). I want to apply a smoothing technique in the >intersection of three regions: Z1, Z2 and Z3. Do the regions overlap? Do they intersect at a single point? Do you > have a particular Z1, Z2, Z3 in mind? However it is quite tough for me. Do you guys know what simple >technique to handle this? P/S for the boundary between each pairs of the neigboring regions (for >example Z1 and Z2, or Z2 and Z3,..) I try to set a weighted technique >and it is OK, however for the region joined of three regions, it is >getting tough). Please help, How about giving more details about where the problem comes from and > what you are trying to accomplish? --Lynn http://math.asu.edu/~kurtz === Subject: Re: Blending three functions at the intersection of three regions? Top posting corrected. Please don't top post. My response is at the bottom. It is easier to follow if you post immediately after what you are responding to... >Z1,Z2 and Z3 are three partitions or R^2. > What are the boundaries between these like? Do the regions overlap or > just touch in some curve or straight line dividing them? Z1 >The problem is that y is not continuous at the boundaries of these >regions (Z1,Z2, and Z3). I want to apply a smoothing technique in the >intersection of three regions: Z1, Z2 and Z3. > Do the regions overlap? Do they intersect at a single point? Do you > have a particular Z1, Z2, Z3 in mind? >However it is quite tough for me. Do you guys know what simple >technique to handle this? >P/S for the boundary between each pairs of the neigboring regions (for >example Z1 and Z2, or Z2 and Z3,..) I try to set a weighted technique >and it is OK, however for the region joined of three regions, it is >getting tough). >Please help, > How about giving more details about where the problem comes from and > what you are trying to accomplish? > --Lynn > http://math.asu.edu/~kurtz >Hi Kurtz, > I suppose you mean z = on the left side of each of those equations. Sorry, I do not understand this question. But my question posted >earlier is correct. y is defined in R^2 but in each partion, it equals >different function. When you write y = f(x,y) you are using y as both the dependent and independent variable, which I think isn't what you mean. Such an equation might represent a curve in the xy plane. Suppose you had a temperature at each region in the xy plane. T = f(x,y) on one region, T = g(x,y) on the second, and T = h(x,y) on the third. The temperature is discontinuous between the regions, and you want to smooth it out. Isn't that the type of question you are asking? If so, you don't want to use y on both sides. If that isn't the type of question you are asking, it doesn't make any sense to me. > Specifically: pseudo-code of y is as follows: For given three different h(x,y), g >(x,y), f(x,y) >if x^2+y^2 <=A i.e. Region Z1) y= f(x,y); elseif y>=0 i.e. Region Z2) y=g(x,y); elseif y<0 i.e. Region Z3) y=h(x,y); end Where A is a constant. I do not have a problem to smooth out the boundary between Z2 and Z3 >(i.e. the x axis). You mean the x axis outside the disk Z1, I presume. >But I get a problem at the boundary between Z1,Z2, and Z3. (i.e. the >point with coordinate (x=sqrt(A), y=0). Is Z2 the region in the upper half plane but exterior to the disk? If so, have you smoothed it at the upper half circle boundary between Z1 and Z2? And similarly for Z3 and the lower half of the circle boundary? --Lynn http://math.asu.edu/~kurtz === Subject: Re: Blending three functions at the intersection of three regions? posting-account=30TueAoAAAAeRakeyTKqut8-jTCzw9_e SV1),gzip(gfe),gzip(gfe) Hi Lynn: Sorry, yes I indeed did not understand the top posting. Do you want me to answer right below your question? If so, here it is: > Is Z2 the region in the upper half plane but exterior to the disk? Yes, it is. If> so, have you smoothed it at the upper half circle boundary between Z1 > and Z2? And similarly for Z3 and the lower half of the circle boundary? Yes, I am thinking of it. I think it is similar to the case of boundary between Z2 and Z3. A weighted technique may help. === Subject: Re: Blending three functions at the intersection of three regions? >Hi Lynn: Sorry, yes I indeed did not understand the top posting. Do you want >me to answer right below your question? If so, here it is: > Is Z2 the region in the upper half plane but exterior to the disk? >Yes, it is. If> so, have you smoothed it at the upper half circle boundary between >Z1 > and Z2? And similarly for Z3 and the lower half of the circle boundary? Yes, I am thinking of it. I think it is similar to the case of >boundary between Z2 and Z3. A weighted technique may help. > OK. I'm going to assume you have it smoothed on the x axis part between Z2 and Z3 and between Z2 and Z3 and Z1 on the circular boundary part, at least away from your two problem points at x = +- sqrt(A). Now imagine a small circles around each of your two problem points. So everywhere exterior to these two circles you have a smooth approximation to your functions. I don't know how complicated you want to make it or how smooth, but here's something you might try. Let's call the small circle centered at A by the name C. You have your approximating function which I will call S(x,y) which is smooth exterior to C. Take the average of the values of S(x,y) on the boundary of C and use that for S(A,0), the value at the center. Now extend the definition of S from the center to the boundary linearly along each radius. This will give you a continuous function, although it may not be differentiable at the center of the circle C. Will that work for you? --Lynn http://math.asu.edu/~kurtz === Subject: Re: Blending three functions at the intersection of three regions? posting-account=30TueAoAAAAeRakeyTKqut8-jTCzw9_e SV1),gzip(gfe),gzip(gfe) Hi Kurtz: Oh yes, I saw my mistake. Yes, the notation y should be changed to something like T as you pointed out. Do you have any idea how to smooth it out? Actually, I intend to define a boundary region and then apply a weighted approach on this boundary. While it is easy to do it between Z2 and Z3. It is not easy at all for the joint between three Z1,Z2, and Z3. I am looking for some surface blending approaches in the geometrical modeling but it seems too complex to me. > Top posting corrected. Please don't top post. My response is at the > bottom. It is easier to follow if you post immediately after what you > are responding to... >Z1,Z2 and Z3 are three partitions or R^2. > What are the boundaries between these like? Do the regions overlap or > just touch in some curve or straight line dividing them? Z1 >The problem is that y is not continuous at the boundaries of these >regions (Z1,Z2, and Z3). I want to apply a smoothing technique in the >intersection of three regions: Z1, Z2 and Z3. > Do the regions overlap? Do they intersect at a single point? Do you > have a particular Z1, Z2, Z3 in mind? >However it is quite tough for me. Do you guys know what simple >technique to handle this? >P/S for the boundary between each pairs of the neigboring regions (for >example Z1 and Z2, or Z2 and Z3,..) I try to set a weighted technique >and it is OK, however for the region joined of three regions, it is >getting tough). >Please help, > How about giving more details about where the problem comes from and > what you are trying to accomplish? > --Lynn >http://math.asu.edu/~kurtz >Hi Kurtz, > I suppose you mean z = on the left side of each of those equations. Sorry, I do not understand this question. But my question posted >earlier is correct. y is defined in R^2 but in each partion, it equals >different function. When you write y = f(x,y) you are using y as both the dependent and > independent variable, which I think isn't what you mean. Such an > equation might represent a curve in the xy plane. Suppose you had a > temperature at each region in the xy plane. T = f(x,y) on one region, > T = g(x,y) on the second, and T = h(x,y) on the third. The temperature > is discontinuous between the regions, and you want to smooth it out. > Isn't that the type of question you are asking? If so, you don't want > to use y on both sides. If that isn't the type of question you are > asking, it doesn't make any sense to me. Specifically: pseudo-code of y is as follows: For given three different h(x,y), g >(x,y), f(x,y) >if x^2+y^2 <=A i.e. Region Z1) y= f(x,y); elseif y>=0 i.e. Region Z2) y=g(x,y); elseif y<0 i.e. Region Z3) y=h(x,y); end Where A is a constant. I do not have a problem to smooth out the boundary between Z2 and Z3 >(i.e. the x axis). You mean the x axis outside the disk Z1, I presume. But I get a problem at the boundary between Z1,Z2, and Z3. (i.e. the >point with coordinate (x=sqrt(A), y=0). > Is Z2 the region in the upper half plane but exterior to the disk? If > so, have you smoothed it at the upper half circle boundary between Z1 > and Z2? And similarly for Z3 and the lower half of the circle > boundary? --Lynn http://math.asu.edu/~kurtz- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - === Subject: Re: Blending three functions at the intersection of three regions? >Hi Kurtz: >Oh yes, I saw my mistake. Yes, the notation y should be changed to >something like T as you pointed out. > Did you understand the part about not top posting (obviously not). I asked you several specific questions about your regions and boundaries, and I had a reason for asking them. If the problem is not clear to me, how can I hope to help you? Instead of top posting generalities, include the questions in your response and answer them where they are asked. Otherwise this is going to be a short conversation. --Lynn http://math.asu.edu/~kurtz === Subject: Re: sequence question <49d5beb3$0$24820$426a74cc@news.free.fr> posting-account=ShKz4goAAAB-5fxA1JyErZXtNgWVpdct Gecko/2009033100 Ubuntu/9.04 (jaunty) Firefox/3.0.8,gzip(gfe),gzip(gfe) === Subject: The Big Bang is the Exploding Lump Theory posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) God is Creator. I want to know how God created this universe. Albert Einstein === Subject: Re: The Big Bang is the Exploding Lump Theory If Einstein said this, he was most arrogant and hence most stupid to even think that he could know how God created the universe, he did not understand \ that his own understanding was mathematically not in the same plane as God. The purpose here is not to discredit Papa Einstein , but to score home the \ mathematical point to all who search, is that N^N is always going to be greater than our maximum proportionate equation , X^2 =3 by proportion ( our \ Derivative of the Brahama Guptas equation, (NX^2 +1= Y^2) are limited by that equation that our proportion is fixed and limited by that fixed value \ of y in the equation. Gods value may be -4 or more and our maximum \ understanding of proportion is -1 i.e X^2=-1= inverse zero. So in effect we cannot go past \ the inverse zero. Gods understanding of his creation i.e God measure of least proportion and thus his accuracy is beyond our comprehension, and Einstein was one of us. Your sincerely, Inverse 19. P.S Big bang will be easy to disprove, because of inverse proportion, unless the proportion was controlled , in which case we are all exploding currently. Also I am trying to form a group within this discussion, centered \ on ideas like that of NICKS and our inverse mathematics 19 development.I would prefer this group for its creativity rather than more of very smart Einsteins ideas which are passe. === Subject: Re: The Big Bang is the Exploding Lump Theory posting-account=A_fxOQkAAAC3cvPrAtLES_UPpvsB8VF2 Trident/4.0; GTB5; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; yie8),gzip(gfe),gzip(gfe) > God is Creator. I want to know how God created this universe. Albert Einstein A paper entitled The Origin of the Universe as Interpreted by Model Mechanics is available in my website: http://www.geocities.com/kn seto/index.htm Ken Seto === Subject: Re: The Big Bang is the Exploding Lump Theory Morontheist BURT: > God is Creator. And his name is Odin. Prove me wrong. I dare you. -- To his friend a man a friend shall prove, and gifts with gifts requite; But men shall mocking with mockery answer, and fraud with falsehood meet. (The Poetic Edda) Must have been written with fundies in mind... My personal judgment of monotheism: http://www.carcosa.de/nojebus === Subject: Re: The Big Bang is the Exploding Lump Theory posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) God is Creator. [snip crap] 1)http://www.godchecker.com/ > 2) idiot > 3)http://www.mazepath.com/uncleal/god.jpg > 4) idiot > 5)http://www.mazepath.com/uncleal/evolve.gif > 6) idiot > 7) Can God make a collection plate so vast that even He cannot fill > it? Sure! ALL OF THEM. > 8) idiot > 9) Orthodoxy means not thinking - not needing to think. Orthodoxy > is unconsciousness. George Orwell, 1984 > 10) idiot > -- > Uncle Alhttp://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals)http://www.mazepath.com/uncleal/lajos.htm#a2 Were you a lump at the beginning of time Al? === Subject: Re: The Big Bang is the Exploding Lump Theory > God is Creator. I want to know how God created this universe. Albert Einstein Funny how some people can see a quote and miss the point completly! The word god is for me nothing more than the expression and product of human weaknesses, the Bible a collection of honourable, but still primitive \ legends which are nevertheless pretty childish. No interpretation no matter \ how subtle can (for me) change this. A. Einstein Also I do not believe in a personal God and I have never denied this but have expressed it clearly. If something is in me which can be called religious then it is the unbounded admiration for the structure of the world so far as \ our science can reveal it. And It was, of course, a lie what you read about my religious convictions, a lie which is being systematically repeated. I do not believe in a personal God and I have never denied this but have expressed it clearly. If something \ is in me which can be called religious then it is the unbounded admiration for the structure of the world so far as our science can reveal it. .................................... Slatts === Subject: Re: The Big Bang is the Exploding Lump Theory > God is Creator. > I want to know how God created this universe. Albert Einstein Funny how some people can see a quote and miss the point completly! The word god is for me nothing more than the expression and product of >human weaknesses, 2.2 billion people disagree with you. The Dukester, American-American ***** The Mass is the most perfect form of Prayer. Pope Paul VI ***** === Subject: Re: The Big Bang is the Exploding Lump Theory God is Creator. I want to know how God created this universe. Albert Einstein > Funny how some people can see a quote and miss the point completly! > The word god is for me nothing more than the expression and product > of human weaknesses, 2.2 billion people disagree with you. Yes they do - just like a few hundred years ago they would have said the earth was flat or the sun goes round the earth! Slatts === Subject: Re: The Big Bang is the Exploding Lump Theory > God is Creator. > I want to know how God created this universe. Albert Einstein Funny how some people can see a quote and miss the point completly! The word god is for me nothing more than the expression and product > of human weaknesses, > 2.2 billion people disagree with you. Yes they do - just like a few hundred years ago they would have said the >earth was flat or the sun goes round the earth! Then the telescope was invented. The Dukester, American-American ***** The Mass is the most perfect form of Prayer. Pope Paul VI ***** === Subject: Re: The Big Bang is the Exploding Lump Theory > God is Creator. I want to know how God created this universe. Albert Einstein > Funny how some people can see a quote and miss the point completly! > The word god is for me nothing more than the expression and > product of human weaknesses, 2.2 billion people disagree with you. > Yes they do - just like a few hundred years ago they would have said > the earth was flat or the sun goes round the earth! Then the telescope was invented. > You don't need a telescope to know the earth is not flat or the earth orbits \ the sun! Learn a little science! Slatts === Subject: Re: The Big Bang is the Exploding Lump Theory > Then the telescope was invented. >You don't need a telescope to know the earth is not flat or the earth orbits >the sun! >Learn a little science! Then why did the revelation come with the invention of the telescope? The Dukester, American-American ***** The Mass is the most perfect form of Prayer. Pope Paul VI ***** === Subject: Re: The Big Bang is the Exploding Lump Theory posting-account=ez3aLwkAAACMTwN_mr_iuSKc7vvn-MPK SV1),gzip(gfe),gzip(gfe) > God is Creator. > I want to know how God created this universe. Albert Einstein Funny how some people can see a quote and miss the point completly! The word god is for me nothing more than the expression and product of >human weaknesses, 2.2 billion people disagree with you. The Dukester, American-American portion reacted too > ***** > The Mass is the most perfect form of Prayer. > Pope Paul VI > ***** end of portion reacted too at least 5 billion people disagree with that (and you know it) Love, Peter van Velzen April 2009 Amstelveen The Netherland === Subject: Re: The Big Bang is the Exploding Lump Theory > God is Creator. > I want to know how God created this universe. Albert Einstein >Funny how some people can see a quote and miss the point completly! >The word god is for me nothing more than the expression and product of >human weaknesses, > 2.2 billion people disagree with you. >at least 5 billion people disagree with that >(and you know it) What I know, for sure, is that 4% of the world is atheist, and that means you're a minority to Christians 8:1, 13.5:1 for people that believe in the one almighty God, and 26:1 who are not atheistic. Smile, it only hurts until you accept it. The Dukester, American-American ***** The Mass is the most perfect form of Prayer. Pope Paul VI ***** === Subject: Re: The Big Bang is the Exploding Lump Theory <6oeht4hb0u3mn6q8k6l3flq0jbete4g7ji@4ax.com> posting-account=ez3aLwkAAACMTwN_mr_iuSKc7vvn-MPK SV1),gzip(gfe),gzip(gfe) God is Creator. I want to know how God created this universe. Albert Einstein >Funny how some people can see a quote and miss the point completly! >The word god is for me nothing more than the expression and product of >human weaknesses, > 2.2 billion people disagree with you. >at least 5 billion people disagree with that >(and you know it) What I know, for sure, is that 4% of the world is atheist, and that means you're > a minority to Christians 8:1, 13.5:1 for people that believe in the one almighty > God, and 26:1 who are not atheistic. Smile, it only hurts until you accept it. The Dukester, American-American > ***** > The Mass is the most perfect form of Prayer. > Pope Paul VI > *****- Tekst uit oorspronkelijk bericht niet weergeven - - Tekst uit oorspronkelijk bericht weergeven - I have no problem with being more intelligent than 98 % of the people, I do have problems when trying to teach that 98% something. Some of them apperantly lack the ability to understand. But with you I suspect it is not lack of intelligence, but rather unwillingness, Love, === Subject: Re: The Big Bang is the Exploding Lump Theory > What I know, for sure, is that 4% of the world is atheist, and that means \ you're > a minority to Christians 8:1, 13.5:1 for people that believe in the one almighty > God, and 26:1 who are not atheistic. > Smile, it only hurts until you accept it. >I have no problem with being more intelligent than 98 % of the people, >I do have problems when trying to teach that 98% something. Yet you 4%'ers don't have any support to be found for your beliefs. The Dukester, American-American ***** The Mass is the most perfect form of Prayer. Pope Paul VI ***** === Subject: Re: The Big Bang is the Exploding Lump Theory <6oeht4hb0u3mn6q8k6l3flq0jbete4g7ji@4ax.com > God is Creator. > I want to know how God created this universe. Albert Einstein Funny how some people can see a quote and miss the point completly! The word god is for me nothing more than the expression and product of >human weaknesses, 2.2 billion people disagree with you. > What I know, for sure, is that 4% of the world is atheist, and that means > you're a minority to Christians 8:1, 13.5:1 for people that believe in the > one almighty God, and 26:1 who are not atheistic. Smile, it only hurts until you accept it. You know, theist addicts like you have a talent for kicking themselves to the curb in short order. Your turn is approaching. -- Patrick L. The Chief Instigator Humphrey (patrick@io.com) Houston, Texas www.io.com/~patrick/aeros.php (TCI's 2008-09 Houston Aeros) AA#2273 === Subject: Re: The Big Bang is the Exploding Lump Theory > God is Creator. > I want to know how God created this universe. Albert Einstein >Funny how some people can see a quote and miss the point completly! >The word god is for me nothing more than the expression and product of >human weaknesses, > 2.2 billion people disagree with you. > What I know, for sure, is that 4% of the world is atheist, and that means > you're a minority to Christians 8:1, 13.5:1 for people that believe in the > one almighty God, and 26:1 who are not atheistic. > Smile, it only hurts until you accept it. You know, theist addicts like you have a talent for kicking themselves to >the curb in short order. Your turn is approaching. Your's too, west loop. And then you and I will both stand before almighty God to answer for that which we did in the flesh. Are you ready? The Dukester, American-American ***** The Mass is the most perfect form of Prayer. Pope Paul VI ***** === Subject: Re: The Big Bang is the Exploding Lump Theory posting-account=XKjFtwkAAABSbI0JT7M5JDLtG3v3d1uB Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) On Apr 5, 8:19 am, pba...@worldonline.nl God is Creator. > I want to know how God created this universe. Albert Einstein Funny how some people can see a quote and miss the point completly! The word god is for me nothing more than the expression and product of >human weaknesses, 2.2 billion people disagree with you. The Dukester, American-American portion reacted too> ***** > The Mass is the most perfect form of Prayer. > Pope Paul VI > ***** end of portion reacted too at least 5 billion people disagree with that > (and you know it) Peter, take a moment or two to enjoy the irony of Chicken Duke responding to a charge of being weak by showing how weak he is in once again failing to support a single thing he says! Enjoy, if you will the weakness of his argument, too, in that he claims to understand science, yet simultaneously insists that scientific truths are decided not by the evidence but by a large number of non-scientists agreeing to blindly believe in something! LoL! Ultimately, Chicken Duke's weakness is in his head as he repeatedly demonstrates. Budikka === Subject: Re: The Big Bang is the Exploding Lump Theory >Peter, take a moment or two to enjoy the irony of Chicken Duke >responding to a charge of being weak by showing how weak he is in once >again failing to support a single thing he says! The only thing you ever tried to support, miss leather teddie of 2009, is the non-existence of the biblical flood, and you had to plagiarize that in your desperate attempt to avoid embarrassment. The Dukester, American-American ***** The Mass is the most perfect form of Prayer. Pope Paul VI ***** === Subject: Re: The Big Bang is the Exploding Lump Theory posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) > God is Creator. I want to know how God created this universe. Albert Einstein Funny how some people can see a quote and miss the point completly! The word god is for me nothing more than the expression and product of > human weaknesses, the Bible a collection of honourable, but still primitive > legends which are nevertheless pretty childish. No interpretation no matter > how subtle can (for me) change this. A. Einstein Also > I do not believe in a personal God and I have never denied this but have > expressed it clearly. If something is in me which can be called religious > then it is the unbounded admiration for the structure of the world so far \ as > our science can reveal it. > And > It was, of course, a lie what you read about my religious convictions, a > lie which is being systematically repeated. I do not believe in a personal > God and I have never denied this but have expressed it clearly. If something > is in me which can be called religious then it is the unbounded admiration > for the structure of the world so far as our science can reveal it. .................................... > Slatts Einstein wasn't an atheist. He believed in an impersonal Creator. Mitch Raemsch === Subject: Re: The Big Bang is the Exploding Lump Theory I didn't know that - got a reference? I thought he said The word God is for me nothing more than the expression and product of human weaknesses, .... Slatts === Subject: Re: The Big Bang is the Exploding Lump Theory posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) Einstein wasn't an atheist. He believed in an impersonal Creator. I didn't know that - got a reference? I thought he said The word God is for me nothing more than the expression > and product of human weaknesses, .... I believe he was refering to the bible god. > Slatts I have place to reference but I believe you can find this quote of Einsteins: I want to know how God created this universe. I want to know His thoughts. All the rest are just details Mitch Raemsch === Subject: Re: The Big Bang is the Exploding Lump Theory > I didn't know that - got a reference? > I thought he said The word God is for me nothing more than the > expression and product of human weaknesses, .... I believe he was refering to the bible god. > Slatts I have place to reference but I believe you can find this quote of > Einsteins: > I want to know how God created this universe. I want to know His > thoughts. All the rest are just details We will have to chose to differ then, because he said he believed in the god \ of Spinoza which, as I understand it is not a creator but the aspect of nature which we describe as laws. Slatts === Subject: Re: The Big Bang is the Exploding Lump Theory posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) > Einstein wasn't an atheist. He believed in an impersonal Creator. > I didn't know that - got a reference? > I thought he said The word God is for me nothing more than the > expression and product of human weaknesses, .... I believe he was refering to the bible god. > Slatts I have place to reference but I believe you can find this quote of > Einsteins: > I want to know how God created this universe. I want to know His > thoughts. All the rest are just details We will have to chose to differ then, because he said he believed in the god > of Spinoza which, as I understand it is not a creator but the aspect of > nature which we describe as laws. Slatts- Hide quoted text - - Show quoted text - Spinoza saw that God reveals himself only through the universe Einstein siad Creator Mitch Raemsch === Subject: Re: adding a boundary to a Riemann domain I found out the answer to my question below. A Riemann domain (M,P) is a Hausdorff space with a local homeomorphism P: M -> C^n. So for each point Z in M there's an open neighborhood U of Z on which P is a homeomorphism. Given M, what is the usual way of adding a boundary dM to M? If P is injective on M, so M is homeomorphic to a domain D in C^n, does P extend to a homeomorphism from M+dM to D + boundary(D) in C^n? Laura === Subject: Re: f(n) = (L+1+2*sqrt(L+1/2))^[n] ,simplify William Elliot a .8ecrit : > Good Morning, > what does mean this iterated expression. > Since f is a real function, n any non negative integer, > we consider nth iterate of L+1+2*sqrt(L+1/2) , > then after iterations L takes value -1/2 > F.i f(0) = L = -1/2 > f(1) = -1/2+1+2*sqrt(-1/2+1/2) = 1/2 > .... Let g(x) = x + 1 + 2.sqr(x + 1/2). g^n is defined by induction as, > g^0 = identity function > g^(n+1) = g o g^n h(x) = lim(n->oo) g^n(x) = -1/2 > Dont be afraid. Alain always tries to make complex very simple things. His question is : how to compute f(n) when f(0)=-1/2 and f(n+1)=f(n)+1+2sqr(f(n)+1/2) And a basic induction gives f(n)=n^2 - 1/2 === Subject: Re: f(n) = (L+1+2*sqrt(L+1/2))^[n] ,simplify <20090404113053.Y13131@agora.rdrop.com> <49d7e00c$0$2763$ba4acef3@news.orange.fr> posting-account=06BQLAoAAADoC7Y4z9FWcUwGvMa7xMG9 7.4),gzip(gfe),gzip(gfe) > William Elliot a .8ecrit : > Good Morning, > what does mean this iterated expression. > Since f is a real function, n any non negative integer, > we consider nth iterate of L+1+2*sqrt(L+1/2) , > then after iterations L takes value -1/2 > F.i f(0) = L = -1/2 > f(1) = -1/2+1+2*sqrt(-1/2+1/2) = 1/2 > .... Let g(x) = x + 1 + 2.sqr(x + 1/2). g^n is defined by induction as, > g^0 = identity function > g^(n+1) = g o g^n h(x) = lim(n->oo) g^n(x) = -1/2 Dont be afraid. Alain always tries to make complex very simple things. His question is : how to compute f(n) when > f(0)=-1/2 and f(n+1)=f(n)+1+2sqr(f(n)+1/2) And a basic induction gives f(n)=n^2 - 1/2- Masquer le texte des messages pr.8ec.8edents - - Afficher le texte des messages pr.8ec.8edents - Happy Day, Speaking about 'a basic induction' might at least be a proof but never an explanation. What I am interested in is not' making complex very simple things'. I do like building mecchanical objects from my own. So the process is available to lots of other functions. Alain === Subject: Re: f(n) = (L+1+2*sqrt(L+1/2))^[n] ,simplify alainverghote@gmail.com a .8ecrit : > What I am interested in is not' making complex very simple things'. > I do like building mecchanical objects from my own. So the process is available to lots of other functions. > :) === Subject: Re: f(n) = (L+1+2*sqrt(L+1/2))^[n] ,simplify <20090404113053.Y13131@agora.rdrop.com> <49d7e00c$0$2763$ba4acef3@news.orange.fr> <49d88c0d$0$17763$ba4acef3@news.orange.fr> posting-account=06BQLAoAAADoC7Y4z9FWcUwGvMa7xMG9 7.4),gzip(gfe),gzip(gfe) > alainvergh...@gmail.com a .8ecrit :> What I am interested in is not' making complex very simple things'. > I do like building mecchanical objects from my own. So the process is available to lots of other functions. :) Well, An other one : f(n) = sqrt(L^2+2*sqrt(L^2-1)+1)^[n] ,L=1 Alain === Subject: Re: f(n) = (L+1+2*sqrt(L+1/2))^[n] ,simplify alainverghote@gmail.com a .8ecrit : > alainvergh...@gmail.com a .8ecrit :> What I am interested in is not' making complex very simple things'. > I do like building mecchanical objects from my own. > So the process is available to lots of other functions. > :) Well, An other one : > f(n) = sqrt(L^2+2*sqrt(L^2-1)+1)^[n] ,L=1 Alain Once again, writing f(n) = sqrt(L^2+2*sqrt(L^2-1)+1)^[n] ,L=1 is a complex way to write a classical problem. The normal method (the one any reader may understand) is to write : Find the general term of the sequence : u_0=1, u_(n+1)=sqrt((u_n)^2+2*sqrt((u_n)^2-1)+1) then, writing v_n=(u_n)^2 and w_n=sqrt(v_n - 1), it's easy to find w_n = n and so u_n = sqrt(n^2 + 1) So f(n) = sqrt (n^2 +1) But WHY do you use this crazy notation f(n) = sqrt(L^2+2*sqrt(L^2-1)+1)^[n] ,L=1 instead of the normal, classical, one : Find the general term of the sequence : u_0=1, u_(n+1)=sqrt((u_n)^2+2*sqrt((u_n)^2-1)+1) === Subject: Re: Mathematica Home Edition posting-account=NmmPRwoAAAAgZNQJ-bKLq6F8YgOlfxLE 3.2.0; GTB5; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) think this is a sales pitch targetting thehobbyist market sector - possibly an ardent base ofhard-core enthusiasts hwo will be willing to fork outadditional funds in the future to keep up to date.Perhaps folk who have recently forked out the Home Editionmay care to comment?Note there is a very active Mathematica support forumwhich simply does not exisy with the other freewareproducts. === Subject: Re: Mathematica Home Edition > think this is a sales pitch targetting thehobbyist market sector - > possibly an ardent base ofhard-core enthusiasts hwo will be willing to > fork outadditional funds in the future to keep up to date.Perhaps folk > who have recently forked out the Home Editionmay care to comment?Note > there is a very active Mathematica support forumwhich simply does not > exisy with the other freewareproducts. There is an active, but very slooooow Mathematica support forum. comp.soft-sys.math.mathematica Ask a question on there, and it can take a day to appear. Then if someone replies, it takes time for that to appear. Not even employees of Wolfram Research can post to comp.soft-sys.math.mathematica without the message being approved. The result is that solving a complex problem can take considerable time. Another problem is lots of people answer the same question, which is a waste of peoples time. The result is that people post Mathematica problems to sci.math.symbolic, which irritates a few (I can think of one prof in particular). But asking for Mathematica support on sci.math.symbolic is one of the better ways to get quick answers. There are more people reading comp.soft-sys.math.mathematica with a better knowledge of Mathematica than those on sci.math.symbolic. Some WRI employees post on sci.math.symbolic, but less than on comp.soft-sys.math.mathematica. Unfortunately, sci.math.symbolic has quite a low signal to noise ratio, which comp.soft-sys.math.mathematica does not. Moderation does have some benefits. I, along with a member of WRI staff, thought that a 'whitelist' of people who could post without delay would be useful. Anyone who regularly posts sensible things would then not incur delays. But that idea was not liked by the moderator. There are many messages each day on the Sage developers list, so to say the free ones don't have this is not true. Sage has been in existence for a lot shorter time than Mathematica, but with funding from companies like Microsoft, Google and others, Sage seems to be going forward at quite a rate. I would also add the S/N ratio is just as high as on comp.soft-sys.math.mathematica - the idiots who tend to dominate sci.math.symbolic have thankfully not appeared on the Sage groups. -- I respectfully request that this message is not archived by companies as unscrupulous as 'Experts Exchange' . In case you are unaware, 'Experts Exchange' take questions posted on the web and try to find idiots stupid enough to pay for the answers, which were posted freely by others. They are leeches. === Subject: Re: Mathematica Home Edition > For a Windows application, 32-bit means it runs on computers with a 32-bit \ > processor (Intel X68-based processors). 64-bit (native) means it runs on \ > the newer Intel 64-bit processors. If you don't know what you have, you > very likely have an x86 32-bit machine. The software is in no way > crippled, it is just compiled differently for the different bit-size > architectures. The 32- and 64-bit versions are the same; they just run on \ > different CPUs. Indeed 64-bit machines have larger registers and other > greater hardware specifications, but this is all a function of the > electronics and software which can utilize it. On Windows, execute: Start -> Run -> msinfo32 Virtually all the processors sold now are 64-bit, but few computers come with a 64-bit operating system. Despite all the 64-bit hype, for most people (probably 99%), there is little benefit in 64-bits - in fact, it has its disadvantages. My laptop has a 64-bit processor, but came with a 32-bit version of Vista. I've since removed Vista and stuck Solaris on it, which is 64-bit OS. (The only benefit is that Solaris is a better OS than Vista - it's nothing to do with the 64-bits) > Look at the System Summary page and the System Type enrty. It will probably > say X86-based PC. This is 32-bit. I don't have a Windows PC handy, but I know you are right, so I'm not going to bother firing one up. > The story is essentially the same but the processors have different names \ > for the Mac and Sun machines. Linux is very often found on Intel > processors. Look up 32-bit application and 64-bit application on > wikipedia for some more detail. Hope this helps. Again all new Suns come with 64-bit processors. This box must be close to 10 years old, but is 64-bit bash-3.00$ isainfo -v 64-bit sparcv9 applications vis2 vis 32-bit sparc applications vis2 vis v8plus div32 mul32 My point is, the professional version of Mathematica will make full use of a 64-bit version of Vista or XP - the home edition will not. So I think WRI are not exactly honest when they say its the same as the pro version. As I said above, I don't believe there is any benefit for 99% of users in having a 64-bit processor, 64-bit OS or 64-bit applications. However, I would imagine that people playing with Mathematica are the sort of people who could well fall into the 1% that would. Mathematica users are some computations. I'm sure if you go back to some advertising literate on the first 64-bit version of Mathematica, you will find all the advantages listed. Now the home edition is 32-bit, and WRI skip over the issue somewhat. They do say its 32-bit, but they also claim its the same as the pro version, which it is not. -- I respectfully request that this message is not archived by companies as unscrupulous as 'Experts Exchange' . In case you are unaware, 'Experts Exchange' take questions posted on the web and try to find idiots stupid enough to pay for the answers, which were posted freely by others. They are leeches. === Subject: Re: Mathematica Home Edition ... > My point is, the professional version of Mathematica will make full use > of a 64-bit version of Vista or XP - the home edition will not. So I > think WRI are not exactly honest when they say its the same as the pro > version. > Contrast this with my professional and home editions of Mathematica 7 for Linux. They are bit-for-bit identical. (I still have to download the home version in order to get a serial number.) -- Ron Bruck === Subject: Re: Mathematica Home Edition <49d697e3@212.67.96.135> <49d7c32e@212.67.96.135> posting-account=fuODWQoAAACwKFkiR1AZOPqJMnEnSkk2 rv:1.8.1.20) Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) > For a Windows application, 32-bit means it runs on computers with a 32-bit > processor (Intel X68-based processors). 64-bit (native) means it runs on > the newer Intel 64-bit processors. If you don't know what you have, you > very likely have an x86 32-bit machine. The software is in no way > crippled, it is just compiled differently for the different bit-size > architectures. The 32- and 64-bit versions are the same; they just run on > different CPUs. Indeed 64-bit machines have larger registers and other > greater hardware specifications, but this is all a function of the > electronics and software which can utilize it. On Windows, execute: Start -> Run -> msinfo32 Virtually all the processors sold now are 64-bit, but few computers come > with a 64-bit operating system. Despite all the 64-bit hype, for most > people (probably 99%), there is little benefit in 64-bits - in fact, it > has its disadvantages. My laptop has a 64-bit processor, but came with a 32-bit version of > Vista. I've since removed Vista and stuck Solaris on it, which is > 64-bit OS. (The only benefit is that Solaris is a better OS than Vista - > it's nothing to do with the 64-bits) Look at the System Summary page and the System Type enrty. It will probably > say X86-based PC. This is 32-bit. I don't have a Windows PC handy, but I know you are right, so I'm not > going to bother firing one up. The story is essentially the same but the processors have different names > for the Mac and Sun machines. Linux is very often found on Intel > processors. Look up 32-bit application and 64-bit application on > wikipedia for some more detail. Hope this helps. Again all new Suns come with 64-bit processors. This box must be close > to 10 years old, but is 64-bit bash-3.00$ isainfo -v > 64-bit sparcv9 applications > vis2 vis > 32-bit sparc applications > vis2 vis v8plus div32 mul32 My point is, the professional version of Mathematica will make full use > of a 64-bit version of Vista or XP - the home edition will not. So I > think WRI are not exactly honest when they say its the same as the pro > version. As I said above, I don't believe there is any benefit for 99% of users > in having a 64-bit processor, 64-bit OS or 64-bit applications. However, > I would imagine that people playing with Mathematica are the sort of > people who could well fall into the 1% that would. Mathematica users are > some computations. I'm sure if you go back to some advertising literate on the first 64-bit > version of Mathematica, you will find all the advantages listed. Now the > home edition is 32-bit, and WRI skip over the issue somewhat. They do > say its 32-bit, but they also claim its the same as the pro version, > which it is not. -- > I respectfully request that this message is not archived by companies as > unscrupulous as 'Experts Exchange' . In case you are unaware, > 'Experts Exchange' take questions posted on the web and try to find > idiots stupid enough to pay for the answers, which were posted freely > by others. They are leeches. For programs such as Mathematica (or Maple) 64bit can mean more than just the limits on addressing space or sizes of data structures. Does Mathematica use gmp (GnuMP) for big integer and/or high precision floating-point calculations? Does not gmp run significantly faster, on the same machine, when compiled as 64bit than as 32bit? Does Mathematica use Intel's MKL, or ATLAS, for double precision BLAS operations? Again, does not ATLAS run significantly faster compiled as 64bit than as 32bit on the same machine? For Maple, experiment indicates that the answer is yes for these questions. It's plausible that the situation for Mathematica is the same. It's not too difficult to set up a Linux machine with a 64bit OS along with 32bit runtime libraries as optional packages. I believe that SuSE and ubuntu provide for that. Or a machine can be set up as dual-boot, for Windows 32&64, or for Linux 32&64. Thus, the very same machine could be used to compare various operations -- selected to hit gmp, or mpfr, or BLAS, etc. There might also be a difference in the memory allocation. There could even be a different cost for moving things around in memory. And symbolic objects might be the sorts of things for which the difference in width has such effects. In that case there could be a trade off between speed and memory use. The situation could also be quite complicated, and depend on the particular type of computation. Here's a small example I found, running in both 32bit and 64bit commandline Maple on the very same machine, under 64bit Linux with 32bit runtime optionally installed. $ maple12 -q -s -binary IBM_INTEL_LINUX -c print(time(12345678!)); - c quit 111.709 $ maple12 -q -s -binary X86_64_LINUX -c print(time(12345678!)); -c quit 71.890 === Subject: Re: Mathematica Home Edition > I'm sure if you go back to some advertising literate on the first 64-bit > version of Mathematica, you will find all the advantages listed. Now the > home edition is 32-bit, and WRI skip over the issue somewhat. They do > say its 32-bit, but they also claim its the same as the pro version, > which it is not. > For programs such as Mathematica (or Maple) 64bit can mean more than > just the limits on addressing space or sizes of data structures. I mentioned that issue earlier. The term 64-bit is not without some ambiguity. > Does Mathematica use gmp (GnuMP) for big integer and/or high precision > floating-point calculations? Does not gmp run significantly faster, on > the same machine, when compiled as 64bit than as 32bit? Does > Mathematica use Intel's MKL, or ATLAS, for double precision BLAS > operations? Again, does not ATLAS run significantly faster compiled as > 64bit than as 32bit on the same machine? I covered that issue in a previous post. I would find it amazing if use was not made of 64-bit registers for big integer calculations. I assumed WRI would do that whenever the processor supports that, since as you say, it would make a massive difference in speed. I suspect the difference between the 32-bit and 64-bit versions of Mathematica would be just the address space. > For Maple, experiment indicates that the answer is yes for these > questions. It's plausible that the situation for Mathematica is the > same. It's not too difficult to set up a Linux machine with a 64bit OS along > with 32bit runtime libraries as optional packages. I believe that SuSE > and ubuntu provide for that. Or a machine can be set up as dual-boot, > for Windows 32&64, or for Linux 32&64. Thus, the very same machine > could be used to compare various operations -- selected to hit gmp, or > mpfr, or BLAS, etc. I can't be bothered to search for it now, as I have something much more important to do (prepare for a chess match), but there is someone who compiles a list of the benchmark results from Mathematica. I suspect he has examples of people submitting 64-bit versions of Windows and 32-bit versions on similar similar hardware. If anyone has more inclination than me, they could probably use some sort of hex viewer to verify if the 32-bit versions make use of 64-bit registers for integer calculations. It's a long time since I did any assembly language programming, but I know the 16-bit registers were names things like AX, while the 32-bit versions had names like EAX for 'extended'. I assume the 64-bit ones have some sort of name like that. One would only need to find the opcode (I think that is the right expression) to see if such registers were used. It would not surprise me if someone has written a tool to give you a yes/no answer for that question. But I can't be bothered myself to do it. I have seen some negative comments from the GMP developers about speed on SPARC processors. I don't believe there is an instruction which can multiple two 64-bit numbers and give a 128-bit result on a 64-bit SPARC processor. Other processor families could do that, by returning the upper and lower 64-bits of the 128-bit number in two registers. I don't think Mathematica's performance on SPARC processors is very good at all compared to x86 hardware, which could partially be explained by this. > There might also be a difference in the memory allocation. There could > even be a different cost for moving things around in memory. And > symbolic objects might be the sorts of things for which the difference > in width has such effects. In that case there could be a trade off > between speed and memory use. The situation could also be quite > complicated, and depend on the particular type of computation. Yes. As I said above, there are disadvantages to 64-bit too. Although I did not state it, I was thinking of the extra bytes used in address references would result in a lower cache hit rate for any give cache size. executables, not 64-bit ones, simply because Sun realise there is no advantage in making them 64-bit. Whenever I write software, I rarely add the compiler flag to produce 64-bit binaries. > Here's a small example I found, running in both 32bit and 64bit > commandline Maple on the very same machine, under 64bit Linux with > 32bit runtime optionally installed. $ maple12 -q -s -binary IBM_INTEL_LINUX -c print(time(12345678!)); - > c quit > 111.709 $ maple12 -q -s -binary X86_64_LINUX -c print(time(12345678!)); -c > quit > 71.890 I don't know Maple. but I assume you are returning the time to compute 12345678 factorial. That rather suggests that 64-bit registers are not used in the big number libraries, as I doubt the address space would make any difference. It would be good if someone from Wolfram Research would say *exactly* what is the difference between the 32 and 64-bit versions of Mathematica on Windows. As I say, I think WRI are being slightly devious when they say the home and pro versions are the same. Then one reads that the Windows version is 32-bit. I'm sure George Orwell would have something to say on the matter, as it appears that the home and pro versions are equal, but one is more equal than the other. -- I respectfully request that this message is not archived by companies as unscrupulous as 'Experts Exchange' . In case you are unaware, 'Experts Exchange' take questions posted on the web and try to find idiots stupid enough to pay for the answers, which were posted freely by others. They are leeches. === Subject: Re: Mathematica Home Edition > If anyone has more inclination than me, they could probably use some > sort of hex viewer to verify if the 32-bit versions make use of 64-bit > registers for integer calculations. That is impossible. In 32 bit mode you can't access any of the 64 bit registers. The opcodes will be rejected as illegal instructions. The machine must run in 64 bit mode to be able to use the 64 bit instructions and registers. > It's a long time since I did any > assembly language programming, but I know the 16-bit registers were > names things like AX, while the 32-bit versions had names like EAX for > 'extended'. I assume the 64-bit ones have some sort of name like that. Yes, the name is Rax > One would only need to find the opcode (I think that is the right > expression) to see if such registers were used. You can't use any 64 bit opcode in 32 bit mode. -- jacob navia jacob at jacob point remcomp point fr logiciels/informatique http://www.cs.virginia.edu/~lcc-win32 === Subject: was: Re: Mathematica Home Edition now 32 bit vs 64 bit <49d697e3@212.67.96.135> <49d85afc@212.67.96.135> posting-account=wy-WGAoAAABUfYlyvtPTFsKujAiLOBd1 Gecko/2009032609 Firefox/3.0.8 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) I doubt that the speed of bignum arithmetic makes much difference to the vast majority of users of Mathematica. I also question whether, in the current line of workstation computers, 64-bit addresses help for computer algebra systems. Here is an (admittedly very crude) analysis for why I say this. Assume that you have 4 gigabytes of memory, about all that you are likely to be able to stuff into your system, filled with computer algebra system data that is mostly indirection, pointers, small integers, etc. That is, you have 2^32 bytes. or 256 megawords (actually, 268,435,456....) of 32 bits each. Now switch to a 64-bit address situation. You now have effectively halved the memory down to 128 megawords, since each datum is twice as big. (yeah, yeah, not everything is data, some is program. But I told you this was crude.) So in order to have equivalent memory of objects, you need to stuff about 8 gigabytes of memory into a 64-bit system. As has been previously noted, using this system will be slower because caches will be filled while holding only half the amount of data. Is there a case where 64-bit systems will definitely win? Yes, if you run out of memory address space on a 32-bit system and would crash. Numerical linear algebra using ginormous arrays of floats is subject to different analysis. === Subject: Re: Mathematica Home Edition > I doubt that the speed of bignum arithmetic makes much difference to > the vast majority of users of Mathematica. I can't work out why you changed the subject - it only confuses people, so I put it back. The discussion was very much specific to Mathematica. I would have thought a 32 vs 64-bit argument was not appropriate for sci.math.symbolic anyway! I would have thought that a lot of home users of Mathematica might be interested in things related to large integers. There was quite a bit of interest in projects to find large primes and similar. > I also question whether, in the current line of workstation computers, > 64-bit addresses help for computer algebra systems. Here is an (admittedly very crude) analysis for why I say this. Assume that you have 4 gigabytes of memory, about all that you are > likely to be able to stuff into your system, That's not true now. This 7-year old machine I'm using (Sun Blade 2000) 16 GB in it. One of my mates stuffed 8 GB in his Intel based PC in order to improve the performance of Vista - apparently, it does work better with 8 GB than 4 GB. > filled with computer > algebra system data that is mostly indirection, pointers, small > integers, etc. > That is, you have 2^32 bytes. or 256 megawords (actually, > 268,435,456....) of 32 bits each. > Now switch to a 64-bit address situation. You now have effectively > halved the memory down to 128 megawords, since each datum is twice as > big. (yeah, yeah, not everything is data, some is program. But I > told you this was crude.) I don't follow your logic there. The fact you have a 64-bit system does not mean you have to put all your small integers into spaces capable of holding large (64 bit) numbers. You can just as easily put them in integers (4 bytes) or shorts (2 bytes). About the only exception I've met is a very old Cray, where I found that the size of a 'short' was 8 bytes. > So in order to have equivalent memory of objects, you need to stuff > about 8 gigabytes of memory into a 64-bit system. You might be right, but it does not make sence to me. > As has been previously noted, using this system will be slower because > caches will be filled while holding only half the amount of data. years back in Personal Computer World (PC magazine aimed at the masses in the UK). There with this idiot singing its praises, as if it was a *must-have* and the next best thing since sliced bread. (Not sure if you use that expression in the USA). machine (Ultra 1) more than a decade earlier. I doubt the Ultra 1 was the first 64-bit machine (it just happened to be the first I knew of). I also mentioned this issue of the cache, the fact that for 99% of uses it would be no benefit. Funny, Personal Computer World chose not to publish my letter! Clearly there is no advantage in 64-bits for the masses, but some users (big graphics people using Photoshop) it is an advantage. I would have thought some Mathematica users would fall into that category too, and they would include home users. I've known the program to use tons of memory doing some symbolic things too. > Is there a case where 64-bit systems will definitely win? Yes, if you run out of memory address space on a 32-bit system and > would crash. > Numerical linear algebra using ginormous arrays of floats is subject > to different analysis. That sort of thing I would suspect home users are less likely to want. I somethings think Wolfram Research don't realise that even their 'home' users are not typical people and likely to have untypical computer systems. I recall a long time ago asking WRI for a new password to move Mathematic from an old Sun to a newer one at work (a university). Somehow it come up in the conversation with the helpful Wolfram Research fellow, that the site license we had allowed any member of staff to use the program at home. The stupid guy [See note #1 below] in the procurement department had never made us aware of that. So I asked Wolfram for a SPARC license for home use, but was initially told A Sun is not a home computer so the license was declined. I pointed out to the Wolfram employee that plenty of people used Suns at home. (If you don't believe me, take a look on comp.unix.solaris). Wolfram Research accepted that and gave me a SPARC license for home use. I note the home edition is not available for Solaris - either x86 or SPARC. There are quite a few people running Solaris on home machines now. The fact Solaris is now free, open-source, and runs on x86 hardware has dramatically increased its usage. But there are plenty of home users using SPARCs too. -------- Note #1 To give you some idea of the idiot we had in procurement department, this is the conversation I had when I wanted to buy a copy of MATLAB for home use (allowed by the license) * I went to his department to pick up some work related things and offered him the cash (£100 I think) for MATLAB. Sorry he says, we \ can't take cash. * OK I said, here is a cheque instead. We can accept that, but you will have to wait 2 weeks for it to clear. (I was a member of staff, so not much chance of me hiding if the cheque bounced. And it was hardly a fortune.) * Do you have a cheque book on you I asked him? Yes he replied. So I said I'll give you the £100 cash, you write the cheque, and I take the software. Sorry we can't do that either. I then said to his colleague Is he always this awkward? To which his colleague replied Tom was born awkward and done a degree in Mathematica site license, but is no doubt just as helpful on other matters. -- I respectfully request that this message is not archived by companies as unscrupulous as 'Experts Exchange' . In case you are unaware, 'Experts Exchange' take questions posted on the web and try to find idiots stupid enough to pay for the answers, which were posted freely by others. They are leeches. === Subject: Re: Et Tu, Bott? : Unimodularity of Intersection Form in 4-D Sorry, forgot to say that the book is : Differential Forms in Algebraic Topology (strange to combine differential with algebraic topology, which does not use tangent planes.) === Subject: Re: what is Math ? Just to comment that you may say that there is more than one mathematics, mathematical system. Every single time you run into an undecidable statement, and choose how to deal with it, you may say that mathematics branches out. If you, e.g., refuse to accept A. of Choice, the mathematics you do will be sighificantly different than if you did accept it. === Subject: Re: what is Math ? <260715.27311.1238887433658.JavaMail.jakarta@nitrogen.mathforum.org If you, \ e.g., refuse to accept A. of Choice, the mathematics you do > will be sighificantly different than if you did accept it. Yes, and even more than that: it will be different again if you accept the negation, accept a weaker version, accept a stronger form of the negation, or some combination of the last two. - Tim === Subject: Re: #345 either the AP-Reals or Old Reals but not both; new book 2nd edition: New True Mathematics <87hc17qoum.fsf@phiwumbda.org> <87r60bp8th.fsf@phiwumbda.org> posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Both sides would have to concede something for this to work. The > question I ask is, what would need to be conceded so that the > theory counts as a good compromise. > One thing I find irksome about your posts is that you spend too much > time responding to nonmath posters, so you do not filter out enough > of the noise. What's ironic about this statement of AP's is that it's one of the few statements on which AP and the standard analysts agree. Both AP and the standard analysts believe that I spend too much time responding to nonmath posters. Of course, AP considers the standard analysts to be the nonmath posters, but to the standard analysts, AP and the other so-called cranks are the posters who have no math to contribute. > The people who do not belong in sci.math > and it is your too friendly nature that you talk to them. Once again, the standard analysts believe that the cranks do not belong on sci.math, and it is my too friendly nature that I talk to the cranks. > Trouble is that it takes up too much time. Once again, both AP and the standard analysts believe that it takes up too much time for me to talk to their opponents. > perhaps a compromise can be made so that > both AP and the standard analysts can agree that the theory is > a valid one. > No compromise will do. I reluctantly agree -- if both AP and the standard analysts are going to accuse their opponents of being nonmath posters, then surely no compromise will do. First, it would be necessary for both AP and the standard analysts to acknowledge that their opponents do have math to contribute. > So when you talk to the cabal, you end up with > friends, but you probably end up empty on new ideas, new discoveries. And I don't want to end up empty on new ideas and discoveries. Thus, I will listen to AP to learn about his new theory, but I will also listen to the standard analysts to see what their arguments against the theory are, to gain a better idea of whether AP's theory would ever be accepted as well as what changes AP would need to make to his theory in order for it to be accepted. So now, back to the theory: > So, here is a question for you, that is on my mind. Is there anything > in ZFC or any axioms of Reals > that disallows a Coordinate System where the negative Reals are > embedded alongside the positive > Reals as a checkerboard pattern? So that 0d9999...99999 then > 1d0000....0000 there is a (-)1d000...000 > sandwiched in between those two positive Reals. > So the entire Cartesian Coordinate System is replaced by this > checkerboard coordinate system. > So is there any objections by the old math to this new Coordinate > System? Interesting question. Since I know that AP will prefer a geometric argument, let me therefore argue geometrically. I've mentioned the Birkhoff Ruler Postulate in many posts before. It was the first postulate in my high school geometry text book. We can see a statement of the postulate at this link: http://www.rio.k12.wi.us/MATH/geo1.html Ruler Postulate: The points on a line can be matched, one-to-one, with the set of real numbers. The real number that corresponds to a point is the coordinate of the point. The distance, AB, between two points, A and B, on a line is equal to the absolute value of the difference between the coordinates of A and B. Of course, for AP's theory we won't use the standard reals, but the AP-reals instead. But the part I want to focus on is the second part of the postulate, which I repeat: The distance, AB, between two points, A and B, on a line is equal to the absolute value of the difference between the coordinates of A and B. So going back to AP's example, let A, B, C be three points whose coordinates are given as follows: A = 0d999...999 B = (-)1d000...000 C = 1d000...000 Then we calculate the distances: AB = abs(0d999...999-(-)1d000...000) = abs(1d999...999) = 1d999...999 BC = abs((-)1d000...000-1d000...000) = abs((-)2d000...000) = 2d000...000 AC = abs(0d999...999-1d000...000) = abs((-)0d000...0001) = 0d000...0001 So far, so good. Nothing in the distance part of the Ruler Postulate forbids what I've written so far. Now the second postulate from my old high school geometry book is the Segment Addition Postulate, which goes as follows: Segment Addition Postulate If B is between A and C, then AB + BC = AC. And that postulate is the killer. Using only the Ruler and Segment Postulates, we can come up with an objection to AP's proposed coordinate system. Theorem: (-1)d000...000 is not between 0d999...999 and 1d000...000. Proof: Assume towards a contradiction that B = (-1)d000...000 really is between A = 0d999...999 and C = 1d000...000. Then by the Segment Addition Postulate, we have: AB + BC = AC. BC, AC as follows: AB + BC = AC 1d999...999 + 2d000...000 = 0d000...0001 3d999...999 = 0d000...0001. which is an apparent contradiction. It's obvious that 3d999...999 isn't the same as 0d000...0001. Thus, the assumption that B is between A and C leads to a contradiction. So (-)1d000...000 isn't between 0d999...999 and 1d000...000. QED Notice that by combining the Segment Addition with Hilbert's II.3: II.3: Of any three points situated on a straight line, there is always one and only one which lies between the other two. we have that for any points, A, B, and C, exactly one of: AB + BC = AC AC + BC = AB AB + AC = BC must be true. Checking these for out three points, we have: AC + BC = AB 0d000...0001 + 2d000...000 = 1d999...999 2d000...0001 = 1d999...999 (ruled out) AB + AC = BC 1d999...999 + 0d000...0001 = 2d000...000 2d000...000 = 2d000...000 (checks out) So 0d999...999 is between (-)1d000...000 and 1d000...000. Indeed, we can prove the following theorem: Theorem: If A has a positive coordinate and B has a negative coordinate, then 0 lies between A and B (and therefore, all the positive coordinate lie on one side of 0 and all the negative coordinates lie on the other side). Proof: Let A have coordinate a and B have coordinate -b, for unsigned (positive) AP-reals a and b. C has coordinate 0. So we have: AB = abs(a-(-b)) = abs(a+b) = a+b AC = abs(a-0) = abs(a) = a BC = abs(-b-0) = abs(-b) = b AB + BC = AC (a+b) + b = a a + 2b = a (fails since b is not 0) AB + AC = BC (a+b) + a = b 2a + b = b (fails since a is not 0) AC + BC = AB a + b = (a+b) (checks out) Therefore 0 is between every positive point and negative point. QED Notice that with the standard reals in standard geometry, we ultimately prove that the order of the coordinates must be exactly the same as the standard order of the reals. === Subject: Re: #359 according to standard analysis trigonometry is not ordered where 270 < 0; new book 2nd edition: New True Mathematics <87hc17qoum.fsf@phiwumbda.org> <87r60bp8th.fsf@phiwumbda.org> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > Okay, earlier today Tim Little argues that Field theory requires > postulates of Order: Refer back to that page again: one of the defining properties of a > total order is that *every* pair of elements a, b satisfies exactly > one of a = b, a < b, or a > b. If there are any elements for which > this is not true, for example by being undefined, then the field fails > to be an ordered field. I am saying, okay, you can have ordering postulates but with no hidden > assumptions. > So that your Ordering is a independent concept, not a dependent one. Ordinarily, '<' would be primitive for the first order theory of ORDERED fields. > That Ordering first must be given an absolute-value and then ranked as > to order. > So that 0 subtract 3 is negative 3 which is not a Independent Concept > for > Ordering. The absolute value of -3 is 3 and where 3 is > 0. So the Field Theory at present has hidden assumptions and its Ordering > is a > dependent concept. No, ordinarily, '<' is primitive for the first order theory of ORDERED fields. What you said about absolute value would be disentangled for you if you merely looked at the primitives and axioms of the first order theory of ordered fields and the definititions then made. So why don't you do that by looking in a basic book on this subject? MoeBlee === Subject: Re: #359 according to standard analysis trigonometry is not ordered where 270 < 0; new book 2nd edition: New True Mathematics posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > Okay, earlier today Tim Little argues that Field theory requires > postulates of Order: Refer back to that page again: one of the defining properties of a > total order is that *every* pair of elements a, b satisfies exactly > one of a = b, a < b, or a > b. If there are any elements for which > this is not true, for example by being undefined, then the field fails > to be an ordered field. I am saying, okay, you can have ordering postulates but with no hidden > assumptions. > So that your Ordering is a independent concept, not a dependent one. Ordinarily, '<' would be primitive for the first order theory of > ORDERED fields. That Ordering first must be given an absolute-value and then ranked as > to order. So that 0 subtract 3 is negative 3 which is not a Independent Concept > for > Ordering. The absolute value of -3 is 3 and where 3 is > 0. So the Field Theory at present has hidden assumptions and its Ordering > is a > dependent concept. No, ordinarily, '<' is primitive for the first order theory of ORDERED > fields. What you said about absolute value would be disentangled for > you if you merely looked at the primitives and axioms of the first > order theory of ordered fields and the definititions then made. So why > don't you do that by looking in a basic book on this subject? > Don't you have one crumb of logic inside that worthless math brain of > yours? Get back to me when you have an insult to hurl with an actual basis. > If the definitions of ZFC and Field Theory allow only the Cartesian > Coordinate > System to the exclusion of any other Coordinate System for the Numbers > of > mathematics, I don't know what you mean by the Numbers of mathematics. There are many different number systems defined in ZFC; and I made no claim that ZFC excludes the existence of whatever particular number system you have in mind. And there are many models of the first order theory of ordered fields. And, again, I made no claim that whatever particular number system you have in mind is or is not an ordered field. > then to someone with a tiny stitch of math brains > realizes that > Ordering is flawed for ZFC and Field theory. Only you can say that what you mean by ordering is flawed. In any case, your response here seems based on your fabrication of the import of my own remarks, particularly as to what number systems ZFC provides for and what number systems are ordered fields. > Now run home and start reading some books on logic, you imbecile nitwit Your insults have far greater verbal force than your actual reasoning on the subject being discussed, which is what makes your insults so relatively feeble. MoeBlee === Subject: Re: #359 according to standard analysis trigonometry is not ordered where 270 < 0; new book 2nd edition: New True Mathematics posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > You are so daft that you even believe Cantor extracted a new Real with > his > diagonal technique. Please do not put words in my mouth. I've never said anything about a new real. MoeBlee === Subject: Re: Virgil, Moeblee, come defend your Cantor Re: #359 ; new book 2nd edition: New True Mathematics posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > So, I give MoeBlee and Virgil two digits 0 and 1 and to two place > value 00 01 01 11 So, as per Cantor, hand me a new third number different from those > four > and which have only two digits. There are only 4 binary sequences of length 2. I don't know what you think that has to do with any refuting any comment I've ever posted. MoeBlee === Subject: Re: Virgil, Moeblee, come defend your Cantor Re: #359 ; new book 2nd edition: New True Mathematics posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > Do you accept the Cantor diagonal method? What does accept mean? In such theories as Z set theory, et. al., the famous diagonal arguments are formalized and are indeed first order proofs from the axioms. And at a pre-formal level, the famous diagonal arguments are correct in the sense of using only principles and reasoning that are common to ordinary mathematics. These are not matters of opinion or beliefs that are merely personal. Rather, in the first sense (formal proof), for each diagonal argument, it is a finitistic fact that there is a certain finite sequence of formulas that has the property of being a first order proof from certain given axioms. And, in the second sense (informal proof), it is open to inspection that every principle and inference method used is ordinary (of course, relative to a reasonable sense of 'ordinary'), despite that the method of diagonalization itsself is an unexpected use of said principles and methods. Those remarks I made do not entail any particular overall view of Cantor's work nor do they bear on any notion of acceptance other than of the facts mentioned. MoeBlee === Subject: Re: Virgil, Moeblee, come defend your Cantor Re: #359 ; new book 2nd edition: New True Mathematics posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) Do you accept the Cantor diagonal method? What does accept mean? In such theories as Z set theory, et. al., the famous diagonal > arguments are formalized and are indeed first order proofs from the > axioms. And at a pre-formal level, the famous diagonal arguments are > correct in the sense of using only principles and reasoning that are > common to ordinary mathematics. These are not matters of opinion or beliefs that are merely personal. > Rather, in the first sense (formal proof), for each diagonal argument, > it is a finitistic fact that there is a certain finite sequence of > formulas that has the property of being a first order proof from > certain given axioms. And, in the second sense (informal proof), it is > open to inspection that every principle and inference method used is > ordinary (of course, relative to a reasonable sense of 'ordinary'), > despite that the method of diagonalization itsself is an unexpected > use of said principles and methods. Those remarks I made do not entail any particular overall view of > Cantor's work nor do they bear on any notion of acceptance other > than of the facts mentioned. P.S. You still have not said what the fact that there are only 4 binary sequences of length 2 has to do with refuting anything I've said. MoeBlee === Subject: Re: Virgil, Moeblee, come defend your Cantor Re: #359 ; new book 2nd edition: New True Mathematics posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > What I despise about you and Virgil and most every math poster is that > your > minds are not good enough for independent discrimination of truth. > That your > minds are so little, that they depend on a outside chorus of a > thousand textbooks > that say that Cantor was right when in fact Cantor was full of > baloney. I refer to textbooks but I do not defer to them in place of my independent verification of the steps in a purported proof. Also, I strongly suspect that you only think you know my thoughts regarding Cantor. If you have some difference with anything I've said about Cantor himself or about his work, which is all pre-formal, then please say what exact remarks of mine you have in mind. In any case, whatever criticisms you imagine regarding the extent of my knowledge, understanding, creativity, or originality, those criticisms aren't refutations of any remarks I've made to you. In particular, you will find in my posts above from today, a few corrections and explanations in response to your posts. And no matter your irrelevent estimations of me, you will find that those remarks of mine are correct. MoeBlee === Subject: Re: #358 Lwalk-- questions on whether I need the Archimedean Postulate for a continuity interval; new book 2nd edition: New True \ Mathematics <87hc17qoum.fsf@phiwumbda.org> <87r60bp8th.fsf@phiwumbda.org> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > Is there a postulate of ZFC or Field theory that says there must be an > infinitude > of numbers to make a Field? No. And, if I'm not mistaken, the first order theory of fields is not even able to discuss finitude/infinitude. However, in ZFC we prove various theorems about finitude and infinitude of different kinds of fields. > Is there a postulate that says at least > one interval must have > an infinity of numbers between its endpoints? If you mean intervals (between two different points) of real numbers, then it is not a postulate of ZFC, but rather a theorem of ZFC that any such interval has infinitely many members. > Is the Archimedean > Postulate that very > postulate that puts an infinity of numbers in at least one interval of > a set? The Archimedian principle for the real numbers is proven as a theorem in ZFC. However I don't think we need to go through that result just to prove that every interval (between two different points) of reals is infinite. MoeBlee === Subject: Re: #352 ; new book 2nd edition: New True Mathematics <87hc17qoum.fsf@phiwumbda.org> <87r60bp8th.fsf@phiwumbda.org> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) >absolute > value is a concept that must be reckoned with that is missing > from the ZFC and the Field axioms of mathematics. In ZFC and/or in the first order theory of fields, one may find a definition of 'absolute value' > What I was going to post before I saw your post, LWalk, was > the idea that a concept in ZFC or Field-theory maybe what is > called a secondary concept or a dependent concept or a contingent > concept or a > composite-concept as compared to what can be thought of > as a primitive concept or a prime concept. There are primitive symbols and defined symbols. Corresponding to the symbols, in an informal way, are concepts. > The concept of greater than or less than, < or >. Is that concept > a barebones concept or is it a secondary concept, dependent > on other concepts? It depends on the particular theory. In ordinary set theory, the only primitives are 'e' (epsilon symbol) and '=' (or even just 'e' if we like). In the first order theory of fields, as I would think ordinarily axiomatized, '<' is primitive. > Membership in set theory is a primitive independent concept. Addition > is a primitive concept. Addition is not primitive in set theory. MoeBlee === Subject: Re: #352 ; new book 2nd edition: New True Mathematics <87hc17qoum.fsf@phiwumbda.org> <87r60bp8th.fsf@phiwumbda.org> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > In ZFC and/or in the first order theory of fields, one may find a > definition of 'absolute value' I meant to say 'first order theory of ordered fields'. > In the first order theory of fields, as I would think > ordinarily axiomatized, '<' is primitive. Again, I meant to say 'the first order theory of ordered fields'. MoeBlee === Subject: Re: #345 either the AP-Reals or Old Reals but not both; new book 2nd edition: New True Mathematics <87hc17qoum.fsf@phiwumbda.org> <87r60bp8th.fsf@phiwumbda.org> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > I will listen to AP to learn about his new theory You keep referring to this theory. What are its logic, primitives, axioms, and definitions? Oh, that's right. Neither you nor AP have a declared logic, primitives, axioms, or system of definitions for these various notions and notations of AP. So, I suppose we must take your use of 'theory' in a broad sense of that word. MoeBlee === Subject: #363 why the Number Systems must be a Well-Ordered Field and why superior over the Old Reals?; new book 2nd edition: New True Mathematics I suggested this idea and indicated that a number system like the Old Reals which are Complete and Archimedean and Total Ordered Fields, are still flawed because they fail to be a Well Ordered Field. So why am I calling the highest form of Algebra of a Field as that of a Well Ordered Field and why does that make the Old Reals inferior and decayed? The answer I have is that any Field less than a Well Ordered Field is flawed because it is not maximally full as a container of water for example when full is maximum. The Old Reals, because they are Dedekindean, do not allow for a BackView which means that the Old Reals have numbers such as 0.333..... but not numbers such as 0d333....33333 and 0d333....33334. In other words, Old Reals have less numbers than does AP-Reals. And this is paradoxical, because Old Reals have a Real in between any two given Reals, yet AP-Reals usually have a new third Real between any two given AP-Reals except for consecutive AP-Reals such as 0d3333...3333 and 0d333...33334 where there is no new third AP-Real between them. So it strikes me as paradoxical that when you have absolute-continuity in Old Reals, and a gap ridden continuity in AP-Reals, yet the AP-Reals have more members than the Old Reals. How to solve the paradox? The answer is simple. The Old Reals are a fake set and the AP-Reals are the True Reals. And ditto for the AP-adics which replace the Peano Naturals as a fake set of numbers. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: #363 why the Number Systems must be a Well-Ordered Field and why superior over the Old Reals?; new book 2nd edition: New True Mathematics <49D98D16.5020405@hotmail.com> posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG 1.1.4322; .NET CLR 2.0.50727; MS-RTC LM 8; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > The Old Reals, because they are Dedekindean, do not allow for a BackView > which means that the Old Reals have numbers such as 0.333..... but not > numbers such as 0d333....33333 and 0d333....33334. In other words, Old > Reals have less numbers than does AP-Reals. And this is paradoxical, > because Old Reals have a Real in between any two given Reals, yet > AP-Reals usually have a new third Real between any two given AP-Reals > except for consecutive AP-Reals such as 0d3333...3333 and 0d333...33334 > where there is no new third AP-Real between them. If there are more AP-reals than reals, then there should be an AP-real for every real, and then some left over. So which AP-real corresponds to the real 0.333... (1/3)? Which AP-real corresponds to the real sqrt(2) = 1.414... ? === Subject: Re: #363 why the Number Systems must be a Well-Ordered Field and why superior over the Old Reals?; new book 2nd edition: New True \ Mathematics sha1:bOxYJ/JxzjYwXs7tkf8XkTW0iT4= > How to solve the paradox? The answer is simple. The Old Reals are a fake > set and the AP-Reals are the True Reals. And ditto for the AP-adics > which replace the Peano Naturals as a fake set of numbers. Sure. But what's the BackView of sqrt(2)? I know the FrontView (I think), but I'm having trouble with the BackView (tm). -- Jesse F. Hughes Until the patch is released, Microsoft said computer users should be careful not to visit unfamiliar Web sites. === Subject: #362 Ring, multiplicative inverses and AP-adics; new book 2nd edition: New True Mathematics Something I been itching and dying to do is to see if every AP-adics has a multiplicative inverse so that the AP-adics is not just a Ring but a Field. Usually it is the lack of a multiplicative inverse that disqualifies most Rings from becoming a Field. So let me just try out one of these to see. So the multiplicative identity is the South Pole or 999...999 +1 which has one more place value than all the other AP-adics as that of _1_0000...00000 and that is a good way of representing it. It would be like if all the numbers were from 99 to 0 and where 100 is the last number. And this South Pole and North Pole numbers are imaginary within AP-adics. So let me just work out one example of a multiplicative inverse. Try 757575...7575 So I have N x 7575...7575 = _1_000....0000 thus I have N = _1_000...0000 / 7575....7575 So I do the finite approximations: 100/75 = 1.33 10000/7575 = 1.3201 1000000/ 757575 = 1.320001 So it appears that the multiplicative inverse of the AP-adic of 7575...7575 is the AP-adic of 1H320000....00001r Now the H indicates the inverse is in the second hemisphere beyond the South Pole and is 32% of the distance away from the South Pole going to the North Pole. It may or may not have a radix value, depending on how composite the number 7575....7575 is. All I wanted to do is satisfy myself as to how easy it is to have a multiplicative inverse. I was bothered by the fact that the radix is a finite string and whether all AP-adics had a multiplicative inverse, but this example calms me somewhat. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: #345 either the AP-Reals or Old Reals but not both; new book 2nd edition: New True Mathematics sha1:4EoWn73GuCJqnWCvrOqoD38Wu4s= > Once again, the standard analysts believe that the cranks do not > belong on sci.math, and it is my too friendly nature that I talk to > the cranks. Simply not so. I read sci.math in large part because I find the crank wars entertaining (and occasionally thought provoking). Although I am sometimes exasperated, I also find some enjoyment in discussions involving Han, David Petry (though he's a distasteful fellow in many ways) and Newberry. Also, JSH is obviously endlessly entertaining, and sci.math is much duller when he's on a self-imposed exile. Given the popularity of the crank threads, I don't think I'm alone in these somewhat embarrassing opinions. There are, of course, folks who don't like anyone to respond to trolls (and many folk think that our cranks are mostly trolls as well). But who has accused you of having a too friendly nature? You like making things up, don't you? -- Jesse F. Hughes This notion that the United States is getting ready to attack Iran is simply ridiculous. Having said that, all options are on the table, -- George W. Bush === Subject: Re: JSH: Weird blind spots with Pell's Equation >there are weird blind spots in what is commonly reported related to >results > what I call the alternates to Pell's Equation: > The alternates are easier to solve in general >there is NO good reason to work on Pell's Equation directly in general. >But things are weirder than that!!! >So there's no point in records with Pell's Equation > My guess on what may have occurred is that with the continuing > fractions solution in hand, math people don't just play with the full > equation, and are taught to just solve Pell's Equation using > continuing fractions or something. Dumbass. ....continuing fractions or something fking Dumbass. Euler and Fermat ***may*** not have cared to talk as much about the > alternates for ***their own weird** reasons. Guess JSH knows better than Euler, what Euler thinks. > Remember they were involved in challenges and stuff. It was an > intellectual activity for them which was somewhat about passing the > time which could be a big deal for members of their class. Cause they didn't have cable or nintendo ?? > And Fermat was a lawyer. For him math was just that thing he did out > of boredom (or he was driven for reasons he didn't fully understand). Liar, You don't know what he knew. Like can you imagine? no. > He was shown the rational parameterization of > Pell's Equation but couldn't be bothered with it. Did you show him? >Yet conics are a > HUGE area, and rational parameterizations have practical application. The oddities of some ego games by some great men long dead may have > skewed history in a fascinating way. Go learn some math, skewed egotistical NPDer JSH. Your spin are a set of lies to justify your ego. > James Harris === Subject: Re: JSH: Weird blind spots with Pell's Equation > You think you can insult me in the face of the stupidity of your > society Admit it: you want to be in that society. You have to know the secret password, and I will not tell you that it is Please. DRAT! -- Michael Press === Subject: Re: JSH: Weird blind spots with Pell's Equation posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) there are weird blind spots in what is commonly reported related to >results > what I call the alternates to Pell's Equation: > The alternates are easier to solve in general >there is NO good reason to work on Pell's Equation directly in general. >But things are weirder than that!!! So there's no point in records with Pell's Equation My guess on what may have occurred is that with the continuing > fractions solution in hand, math people don't just play with the full > equation, and are taught to just solve Pell's Equation using > continuing fractions or something. Dumbass. ....continuing fractions or something fking Dumbass. Euler and Fermat ***may*** not have cared to talk as much about the > alternates for ***their own weird** reasons. Guess JSH knows better than Euler, what Euler thinks. Remember they were involved in challenges and stuff. It was an > intellectual activity for them which was somewhat about passing the > time which could be a big deal for members of their class. Cause they didn't have cable or nintendo ?? And Fermat was a lawyer. For him math was just that thing he did out > of boredom (or he was driven for reasons he didn't fully understand). Liar, You don't know what he knew. Like can you imagine? no. He was shown the rational parameterization of > Pell's Equation but couldn't be bothered with it. Did you show him? Yet conics are a > HUGE area, and rational parameterizations have practical application. The oddities of some ego games by some great men long dead may have > skewed history in a fascinating way. Go learn some math, skewed egotistical NPDer JSH. Your spin are a set of lies to justify your ego. LOL. You people are stupid. Your insults are more than funny now. They are HILARIOUS. You think you can insult me in the face of the stupidity of your > society not re-discovering things that Fermat and Euler certainly > thought were trivial so you could work harder on the scale of a > freaking SQUARE??!!! Your group is completely in its own little world. A world where rank stupidity about basic math gives you the right to > still insult me after I point it out. You know what the rest of the world would call you though? LOSERS James Harris You are a little bitch! Did you pee yourself when you were in school and were afraid of all the little girls? You are so delusional and make ridiculous claims to account for peeing yourself. You are also a narcissist! Crank, troll and kook! See crank.net for your own page. It has gotten to the point where an entire website can be attributed to your lunacy and ramblings, oh wait, we already have that with yours. Cheat, liar and charlatan - go sell your snake oil to someone who will buy it - but not sci.math! Have a nice day! === Subject: Re: JSH: Weird blind spots with Pell's Equation posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) there are weird blind spots in what is commonly reported related to >results > what I call the alternates to Pell's Equation: > The alternates are easier to solve in general >there is NO good reason to work on Pell's Equation directly in general. >But things are weirder than that!!! So there's no point in records with Pell's Equation My guess on what may have occurred is that with the continuing > fractions solution in hand, math people don't just play with the full > equation, and are taught to just solve Pell's Equation using > continuing fractions or something. Dumbass. ....continuing fractions or something fking Dumbass. Euler and Fermat ***may*** not have cared to talk as much about the > alternates for ***their own weird** reasons. Guess JSH knows better than Euler, what Euler thinks. Remember they were involved in challenges and stuff. It was an > intellectual activity for them which was somewhat about passing the > time which could be a big deal for members of their class. Cause they didn't have cable or nintendo ?? And Fermat was a lawyer. For him math was just that thing he did out > of boredom (or he was driven for reasons he didn't fully understand). Liar, You don't know what he knew. Like can you imagine? no. He was shown the rational parameterization of > Pell's Equation but couldn't be bothered with it. Did you show him? Yet conics are a > HUGE area, and rational parameterizations have practical application. The oddities of some ego games by some great men long dead may have > skewed history in a fascinating way. Go learn some math, skewed egotistical NPDer JSH. Your spin are a set of lies to justify your ego. LOL. You people are stupid. Your insults are more than funny now. They are HILARIOUS. You think you can insult me in the face of the stupidity of your > society not re-discovering things that Fermat and Euler certainly > thought were trivial so you could work harder on the scale of a > freaking SQUARE??!!! Your group is completely in its own little world. A world where rank stupidity about basic math gives you the right to > still insult me after I point it out. You know what the rest of the world would call you though? LOSERS James Harris You are a little bitch! I'm the person who made the connection between Pell's Equation and a > rational parameterization of 3 of 4 conic sections in rationals. Given x^2 - Dy^2 = 1, in rationals: y = 2t/(D - t^2) and x = (D + t^2)/(D - t^2) showing it more traditionally versus the way that results from my own > re-derivation. And you get a parameterization for hyperbolas with integer D>0, for > circles with D=-1, and for ellipses with D<-1. > You get an ellipse for any D < 0 (with the circle being a special case of ellipse). But this is all trivial and well-known - so obvious, trivial and well-known that most people wouldn't even bother to mention it. You on the other hand are so desperate to salvage SOMETHING out of your Pell's Equation Fiasco that you try to hyperinflate even the tiniest observation to the level of something that Gauss or Newton might have done. And you don't need to express anything in terms of an auxiliary variable. Marcus. > Your histrionics in your reply make you so much more like what you > describe than me. After all, I have results. You have cursing at the researcher. Maybe your society calls that something to be proud of, but again, to > the rest of the world--for the people living in the real world--it > spells out: L - O - S - E - R James Harris === Subject: Re: JSH: Weird blind spots with Pell's Equation posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) there are weird blind spots in what is commonly reported related to >results > what I call the alternates to Pell's Equation: > The alternates are easier to solve in general >there is NO good reason to work on Pell's Equation directly in general. >But things are weirder than that!!! So there's no point in records with Pell's Equation My guess on what may have occurred is that with the continuing > fractions solution in hand, math people don't just play with the full > equation, and are taught to just solve Pell's Equation using > continuing fractions or something. Dumbass. ....continuing fractions or something fking Dumbass. Euler and Fermat ***may*** not have cared to talk as much about the > alternates for ***their own weird** reasons. Guess JSH knows better than Euler, what Euler thinks. Remember they were involved in challenges and stuff. It was an > intellectual activity for them which was somewhat about passing the > time which could be a big deal for members of their class. Cause they didn't have cable or nintendo ?? And Fermat was a lawyer. For him math was just that thing he did out > of boredom (or he was driven for reasons he didn't fully understand). Liar, You don't know what he knew. Like can you imagine? no. He was shown the rational parameterization of > Pell's Equation but couldn't be bothered with it. Did you show him? Yet conics are a > HUGE area, and rational parameterizations have practical application. The oddities of some ego games by some great men long dead may have > skewed history in a fascinating way. Go learn some math, skewed egotistical NPDer JSH. Your spin are a set of lies to justify your ego. LOL. You people are stupid. Your insults are more than funny now. They are HILARIOUS. You think you can insult me in the face of the stupidity of your > society not re-discovering things that Fermat and Euler certainly > thought were trivial so you could work harder on the scale of a > freaking SQUARE??!!! Your group is completely in its own little world. A world where rank stupidity about basic math gives you the right to > still insult me after I point it out. You know what the rest of the world would call you though? LOSERS James Harris You are a little bitch! I'm the person who made the connection between Pell's Equation and a > rational parameterization of 3 of 4 conic sections in rationals. Given x^2 - Dy^2 = 1, in rationals: y = 2t/(D - t^2) and x = (D + t^2)/(D - t^2) showing it more traditionally versus the way that results from my own > re-derivation. And you get a parameterization for hyperbolas with integer D>0, for > circles with D=-1, and for ellipses with D<-1. You get an ellipse for any D < 0 (with the circle being a > special case of ellipse). Yup. That is a meaningful correction. However, I think to a lot of > people the circle is kind of separate from ellipses, so I went with > that usage. But this is all trivial and well-known - so obvious, trivial > and well-known that most people wouldn't even bother to mention > it. You on the other hand are so desperate to salvage Well yes, most MATH PEOPLE wouldn't. But there's something wrong with > a lot of you. Notice how you jumped from something correct--the circle is an > ellipse--to something completely different. That's typical behavior from you. It betrays a logic problem in how > your mind works. SOMETHING out of your Pell's Equation Fiasco that you try > to hyperinflate even the tiniest observation to the level of > something that Gauss or Newton might have done. And you > don't need to express anything in terms of an auxiliary variable. Marcus. Hey, you can fight to win here, but that means that textbooks and > references continue to not notice that there is a rational > parameterization for circles, ellipses and hyperbolas from the well- > known Pell's Equation. Given x^2 - Dy^2 = 1, in rationals: y = 2t/(D - t^2) and x = (D + t^2)/(D - t^2) > I cannot see why you go through this stuff with the two equations for x and y in terms of D and t. I mean, just look at: x^2 - Dy^2 = 1. It's trivial to note that: 1. When D = -1, this becomes x^2 + y^2 = 1, a circle 2. When D > 0, it is the equation of a hyperbola 3. When D < 0, it is the equation of an ellipse. And, of course, your parametrization does not provide a COMPLETE characterization of conics, not even a complete characterization of ellipses and hyperbolas. You might at some point want to look up 'quadratic forms' and terms like 'positive definite' and 'negative definite', etc. > showing it more traditionally versus the way that results from my own > re-derivation. And you get a parameterization for hyperbolas with integer D>0, for > circles with D=-1, and for ellipses with D<-1. > Why do you keep leaving out -1 < D < 0 ? It's ellipses there also. > There is NO WAY that people who truly loved mathematics, or knowledge > for that matter, would fight against proper acknowledgment of such > information!!! > But it's TRIVIAL!!! Nobody cares, or should care! > You lie. Where did I lie? > As I've said. You lie and then you ask people to love you anyway, and trust you on > mathematical topics, when I say, if you lie, they're fools to do so. > I don't ask anybody to do anything, let alone 'love me' or trust me, and unlike you I don't try to tell them what to think. People judge for themselves strictly on the merit of what I say. They can disbelieve it or not as they see fit. It should stand on its own without all the propaganda and the predictions of apocalypse, etc. That's my secret method. You should try it. > Nothing your society says about mathematics can be taken at face value > without thorough checking as you demonstrate a complete contempt for > the truth!!! > I say your observation on this topic is trivial. That is when you get down to it a matter of opinion: 'trivial' is subjective. It's not a matter of 'truth'. People are perfectly free to disagree with me on this and agree with you. It is just a question of who is the more persuasive. > REMEMBER!!! Your fight here is to STOP the mathematical literature > from reflecting a beautiful mathematical result. > Self-described 'beautiful'. Self-described 'freaking' 'cool', 'nifty', whatever kind of self-promoting trendy hype you want to apply to it. You desperately want massive credit for something that is the mathematical equivalent of a teaspoonful of spit. > You are as I've said like arsonists who become firefighters. You are anti-mathematicians, pretending to be something else. > More social whining and pompous pontificating from the Old Master. You must actually think this works. Has it, ever? I would say, stick to the math, but obviously that is not going to help you here at all. Marcus. > James Harris === Subject: Re: JSH: Weird blind spots with Pell's Equation > There is NO WAY that people who truly loved mathematics, or knowledge > for that matter, would fight against proper acknowledgment of such > information!!! If that is the case, how did it take you so long to realise that your factorisation algorithm was so badly broken? Why can you not acknowledge for once and for all that your FLT proof is broken? It fails even /your/ half-arsed definition of mathematical proof. Why can you not bring yourself (as you admit) to actually check your work like a real mathematician would? You come here asking for responses to your brainstorming, then go mental when you get it. If you don't like the responses, stop prompting for them. > You lie. As I've said. Make up something else. This one is getting boring. Unless you LIKE sounding like an 8-year-old with the sulks. M === Subject: Re: JSH: Weird blind spots with Pell's Equation Maybe your society calls that something to be proud > of, but again, to > the rest of the world--for the people living in the > real world--it > spells out: L - O - S - E - R James Harris > I tried to talk to you in a civil way and you did not take up the chance. So here it goes: How does 14 years, without friends, without any accomplishement, without a degree, without a job spell?. How about the prospect of dying alone in a straight jacket, with no one visiting you? S-E-L-F D-E-L-U-S-I-O-N L-O-S-E-R Nobody likes you. They know you well. === Subject: Re: JSH: Weird blind spots with Pell's Equation posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.1),gzip(gfe),gzip(gfe) Nothing your society says about mathematics can be taken at face value > without thorough checking as you demonstrate a complete contempt for > the truth!!! James Harris- You wouldn't know truth if it jumped up and bit your ass up. You create your own truth to compensate for you ludicrous mathurbating! You are a delusional narcissist and have NOTHING. 14 years of nothing but lies and deciet. Get your head out of your ass! === Subject: Re: JSH: Weird blind spots with Pell's Equation posting-account=sKfmEQkAAAC8kI3Pv6_U_nt9sVsxZ_ou 1.1.4322),gzip(gfe),gzip(gfe) > Looking over the literature over Pell's Equation available on the web, > it appears that there are weird blind spots in what is commonly blah, blah, blah,.... ad inf. Alte Kack' im alten Frack === Subject: Re: JSH: Weird blind spots with Pell's Equation > Looking over the literature over Pell's Equation available on the web, > it appears that there are weird blind spots in what is commonly blah, blah, blah,.... ad inf. Alte Kack' im alten Frack > you can get a good blind spot mirror at any auto part store. The Pell's equation blind spot mirrors appear weird to an observer cruising well below \ the speed of light. So to eliminate the weirdness, you may want to floor it. \ === Subject: Cause of the Big Bang posting-account=FtughAoAAABXWy1b_Ucoiu1tjtfMtYS1 AppleWebKit/525.19 (KHTML, like Gecko) Chrome/1.0.154.53 Safari/525.19,gzip(gfe),gzip(gfe) The Big Bang was caused by a trough of the 22.2 billion year EUWS cycle -- one of a long sequence of physical cycles that can be found throughout the geological and climatic records. These cycles originate from outside our universe. Go to http://www.uct-news.com/page9.html to see charts of EUWS cycles between 3.38 million years and 821 million years. === Subject: Re: more math by tommy1729 >more math by tommy1729 10^999 + 13 = semiprime How do you know? quasi === === posting-account=d-ESTAkAAAAG0l03yI1WJgsTVXx4ebeJ Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) >http://tinyurl.com/crp8as You need to be more observant if you are going to do this kind of work! 1. NOT the same shape! Google is equilateral triangle. BBC image is longer on the sides than the tail, plus the tail seems to not be cut off straight but has a slight protrusion out the back. 2. Same color (black) but all that means is it's projected against a evidence of same color. 3. And VERY clearly the Lights are NOT the same. Google has bushes or lights on vertexes of triangle, a light somewhat in from each vertex and one in the center. All lights are Yellow. The BBC image has lights under triangle arranged in square patterns two white, two red and one green. No lights at vertexes. 4.And lastly the BBC object clearly appears to be flying. The Google image appears to be on the ground. Hence only the BBC image fits the F in UFO. If you are going to try to prove anything here, you had better start gathering details like some CSI rather than some half-baked major media reporter telling a good story to please his editor. === <49D7712B.9060305@somewhere.no> posting-account=d-ESTAkAAAAG0l03yI1WJgsTVXx4ebeJ Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) On Apr 4, 9:39 am, Paul B. Andersen But you seem to have forgotten that USA isn't the whole world. > Do the PRIVATE BANKING COMPANIES who control the pentagon / military industrial complex > control the rest of the world as well? And you seem to have forgotten that the private banking companies (You think the Federal Reserve is not private? Go spend a bit 'o time on Never heard of the new world order I take it or who is behind it... Hey Paul, Time to wake up... === posting-account=d-ESTAkAAAAG0l03yI1WJgsTVXx4ebeJ Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > [1] There is no evidence the trails are cow paths. True. But they are the size to be trails or perhaps erosion washes. Of particular interest, however is the way the trail at the lower right vertex bends around it as if there is a fence there and cows or people are perhaps avoiding a fence surrounding the triangular area. > [2] Compare the trail widths with other objects like trees, cabins . How about just compare with the scale on the map? Duh. I read about 2-5 feet. given the low resolution of the image this certainly seems to fit trails or perhaps erosion washes. > [3] You DO see trails from outside the UFO go under the UFO. I don't. > [4] The UFO is translucent. My research indicates this translucent > effect arises from the local deformations of space and time. Space and > time exist and distort but differently to what GR erroneously predicts. What I do see (space-time speculation for which there is even LESS evidence than for the cow paths) are ground features outside the triangle that smoothly join similar features inside the area. That seems a better explanation than space-time translucent to me. If this really IS a UFO then if we look at the same area on one of the other online satellites it shouldn't be there, unless it's permanently parked out in the weeds. I have no problem with UFOs. I've seen them myself. My problem here is wild speculation that this object is even flying at all in addition to being unknown. Since it looks suspiciously like a triangular patch of dirt with a fence around it, I find having it fly a bit of a stretch without more evidence. === posting-account=ZROLyAoAAAA_aSfFuK3O5IO8JswlpHSk Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1) ; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 1.1.4322; .NET CLR 3.5.30729; .NET CLR 3.0.30618; WWTClient2),gzip(gfe),gzip(gfe) [1] There is no evidence the trails are cow paths. True. But they are the size to be trails or perhaps erosion washes. > Of particular interest, however is the way the trail at the lower > right vertex bends around it as if there is a fence there and cows > or people are perhaps avoiding a fence surrounding the triangular > area. [2] Compare the trail widths with other objects like trees, cabins . How about just compare with the scale on the map? Duh. I read about > 2-5 feet. given the low resolution of the image this certainly seems > to fit trails or perhaps erosion washes. > [3] You DO see trails from outside the UFO go under the UFO. I don't. I've marked them below: http://img11.imageshack.us/img11/1189/ufowithtrailmarkings.png Here is the unmarked image: http://img11.imageshack.us/img11/7382/ufowithouttrailmarkings.png > [4] The UFO is translucent. My research indicates this translucent > effect arises from the local deformations of space and time. Space and > time exist and distort but differently to what GR erroneously predicts. What I do see (space-time speculation for which there is even LESS > evidence than for the cow paths) are ground features outside the > triangle that smoothly join similar features inside the area. That > seems a better explanation than space-time translucent to me. If > this really IS a UFO then if we look at the same area on one of the > other online satellites it shouldn't be there, unless it's permanently > parked out in the weeds. So if ground trails from outside the area join inside the area the object cannot be a building of some sort (tower, cabin). The evidence suggests it is either [1] A ground-etched glyph - like a crop circle or man-made figure [2] A translucent UFO. Don't forget Occams razor - the simplest explanation that accounts for ALL the evidence. What about all the other triangular UFOs that have been seen, photographed of same shape, symmetry, color and light configuration? What about the NASA STS-80 footage which shows a TRANSLUCENT UFO flying right by it? http://www.youtube.com/watch?v=naVjGEcZbYo&feature=related (skip to minute 1:12). Witness testimony and heresay evidence given by government personel (like Bob Lazar) does suggest these things are real. > I have no problem with UFOs. I've seen them myself. My problem here is > wild speculation that this object is even flying at all in addition > to being unknown. Since it looks suspiciously like a triangular > patch of dirt with a fence around it, I find having it fly a bit of > a stretch without more evidence. of a UFO. It gives credence to the hundreds of thousands of UFO reports, witness testimony, photographic and video evidence suggesting extra- terrestrials are constantly buzzing this place. By the way, the military no longer refer to UFOs as UFOs, they call them ETVs - extra-terrstrial vehicles. Their passengers are reffered to as EBEs - extra-terrestrial biological entities. This was the terminology used in Official Army UFO manual leaked used in the 50s (leaked in the 80s): http://www.scribd.com/doc/12873313/Army-Official-UFO-Manual === posting-account=d-ESTAkAAAAG0l03yI1WJgsTVXx4ebeJ Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > I've marked them below: http://img11.imageshack.us/img11/1189/ufowithtrailmarkings.png Here is the unmarked image: http://img11.imageshack.us/img11/7382/ufowithouttrailmarkings.png These marked paths are far to wide to match the paths like say those going around the lower right vertex. They look more like ground features to me. > So if ground trails from outside the area join inside the area the > object cannot be a building of some sort (tower, cabin). The evidence > suggests it is either [1] A ground-etched glyph - like a crop circle or man-made figure Yep, I'd agree with that. > [2] A translucent UFO. Remote possibility, but like I said, before, I'd want people seeing it FLYING before I'd start to think seriously along those lines. Every triangular ground feature isn't a triangular UFO even though there are clearly LOTS of sightings of such alleged craft that seem remarkably uniform in characteristics (being a strong indicator of some reality behind the stories). > Don't forget Occams razor - the simplest explanation that accounts for > ALL the evidence. Don't forget Einstein: Explanations should be as simple as possible, but not simpler! > What about all the other triangular UFOs that have been seen, > photographed of same shape, symmetry, color and light configuration? What about them? I mentioned that above. I presume one has to ask that each case stands on it's own and then perhaps one can take the best cases and start to speculate from that. > Witness testimony and hearsay evidence given by government personnel > (like Bob Lazar) does suggest these things are real. Yes there is much to suggest these things are real, but the topic here trust ANYTHING Bob Lazar says. I suspect shill. He's just one more Adamski shill in my humble opinion. Ever talk to any of the once rabid Adamski followers? They are all quite embarrassed now by his wild venusian claims. Sorry bunky, it's all just more chaff in the wind! You don't investigate UFOs like scientific research, there are PEOPLE and INTELLIGENCE behind these issues with agendas. You have to approach it like a murder investigation. You have to assume the guilty will be doing their best to mislead, cover-up and lie about everything. And in this case the resources of the guilty are immense! > of a UFO. Nope. It's an image SUGGESTIVE of a UFO. The F in UFO stands for flying. I have seen no evidence that this is any more than a ground feature. > It gives credence to the hundreds of thousands of UFO reports, witness > testimony, photographic and video evidence suggesting extra- > terrestrials are constantly buzzing this place. No it doesn't. It gives credence to nothing unless somehow this image can be tied to the other observations by more than it being a triangle with spots! The credence of all other observations etc. have to stand just as this one does on the FACTS (and that doesn't mean plausible explanations) of each one. > By the way, the military no longer refer to UFOs as UFOs, they call > them ETVs - extra-terrstrial vehicles. Their passengers are reffered > to as EBEs - extra-terrestrial biological entities. This was the > terminology used in Official Army UFO manual leaked used in the 50s > (leaked in the 80s): http://www.scribd.com/doc/12873313/Army-Official-UFO-Manual Just how do you know WHAT the military now refers to UFOs as? You use some manual from the 50's? Please! If you want to find out what terms they now use, perhaps you'd better talk to someone presently in the service and engaged in such work. You think they'll tell you? You might as well be some major media reporter shoving a mic in the murder defendant's face asking him if he really did it! What's he going to say? Yeah, sure, I did it! They all deserved to die? Gimme some kind of break here! === > I've marked them below: > http://img11.imageshack.us/img11/1189/ufowithtrailmarkings.png > Here is the unmarked image: > http://img11.imageshack.us/img11/7382/ufowithouttrailmarkings.png These marked paths are far to wide to match the paths like say >those going around the lower right vertex. They look more like ground >features to me. > So if ground trails from outside the area join inside the area the > object cannot be a building of some sort (tower, cabin). The evidence > suggests it is either > [1] A ground-etched glyph - like a crop circle or man-made figure Yep, I'd agree with that. > [2] A translucent UFO. Remote possibility, but like I said, before, I'd want people seeing it >FLYING before I'd start to think seriously along those lines. Every >triangular ground feature isn't a triangular UFO even though there are >clearly LOTS of sightings of such alleged craft that seem remarkably >uniform in characteristics (being a strong indicator of some reality >behind the stories). > Don't forget Occams razor - the simplest explanation that accounts for > ALL the evidence. Don't forget Einstein: Explanations should be as simple as possible, >but not simpler! > What about all the other triangular UFOs that have been seen, > photographed of same shape, symmetry, color and light configuration? What about them? I mentioned that above. I presume one has to ask that >each case stands on it's own and then perhaps one can take the best >cases and start to speculate from that. > Witness testimony and hearsay evidence given by government personnel > (like Bob Lazar) does suggest these things are real. Yes there is much to suggest these things are real, but the topic here >trust ANYTHING Bob Lazar says. I suspect shill. He's just one more >Adamski shill in my humble opinion. Ever talk to any of the once >rabid Adamski followers? They are all quite embarrassed now by his >wild venusian claims. Sorry bunky, it's all just more chaff in the >wind! You don't investigate UFOs like scientific research, there are PEOPLE >and INTELLIGENCE behind these issues with agendas. You have to >approach it like a murder investigation. You have to assume the guilty >will be doing their best to mislead, cover-up and lie about >everything. And in this case the resources of the guilty are >immense! > of a UFO. Nope. It's an image SUGGESTIVE of a UFO. The F in UFO stands for >flying. I have seen no evidence that this is any more than a ground >feature. > It gives credence to the hundreds of thousands of UFO reports, witness > testimony, photographic and video evidence suggesting extra- > terrestrials are constantly buzzing this place. No it doesn't. It gives credence to nothing unless somehow this image >can be tied to the other observations by more than it being a triangle >with spots! The credence of all other observations etc. have to stand >just as this one does on the FACTS (and that doesn't mean plausible >explanations) of each one. > By the way, the military no longer refer to UFOs as UFOs, they call > them ETVs - extra-terrstrial vehicles. Their passengers are reffered > to as EBEs - extra-terrestrial biological entities. This was the > terminology used in Official Army UFO manual leaked used in the 50s > (leaked in the 80s): > http://www.scribd.com/doc/12873313/Army-Official-UFO-Manual Just how do you know WHAT the military now refers to UFOs as? You >use some manual from the 50's? Please! If you want to find out what >terms they now use, perhaps you'd better talk to someone presently in >the service and engaged in such work. You think they'll tell you? You >might as well be some major media reporter shoving a mic in the murder >defendant's face asking him if he really did it! What's he going >to say? Yeah, sure, I did it! They all deserved to die? Gimme some >kind of break here! Very nice post there. Pleased to meet you. Have a Great Day! === sha1:M1kTuNliJUUsu78bP/CDGwRU44s= > [4] The UFO is translucent. My research indicates this translucent > effect arises from the local deformations of space and time. Space and > time exist and distort but differently to what GR erroneously predicts. What I do see (space-time speculation for which there is even LESS > evidence than for the cow paths) are ground features outside the > triangle that smoothly join similar features inside the area. That > seems a better explanation than space-time translucent to me. If > this really IS a UFO then if we look at the same area on one of the > other online satellites it shouldn't be there, unless it's permanently > parked out in the weeds. But he has *research*. *RESEARCH*. All you have are eminently plausible explanations that are consistent with our daily observations. How does that compare with *research*? By John Schoenfeld, no less. -- Out of the rubbles of Trent Lott's house -- he's lost his entire house -- there's going to be a fantastic house. And I'm looking forward to sitting on the porch. -- George W. Bush consoles Katrina victims, Sep 2, 2005. === Subject: -- Induced homomorphisms (of groups) If G and H are groups, of which N and K are normal subgroups respectively, then we know that any homomorphism f : G ---> H with f(N) < K induces a homomorphism F : G/N ---> H/K given by gN |-> f(g)K. My question is: does *every* homomorphism G/N --> H/K arise in this way? Also, can distinct homomorphisms f, g : G ---> H with f(N) < K and g(N) < K induce the *same* homomorphism G/N ---> H/K as above? In other words, is there some sort of 1-1 correspondence between homomorphisms G/N --> H/K and homomorphisms f : G ---> H with f(N) < K ? I don't think there is (or, at least, I haven't been able to show that there is one), but I still think it's instructive to ask such a question :) === Subject: Re: -- Induced homomorphisms (of groups) posting-account=_l4K0QkAAAC09JhOoK_ZfoJKXOmr_jZf Gecko/2009032609 Firefox/3.0.8 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > If G and H are groups, of which N and K are normal subgroups respectively, then we know that any homomorphism f : G ---> H with f(N) < K induces a homomorphism F : G/N ---> H/K given by gN |-> f(g)K. My question is: does *every* homomorphism G/N --> H/K arise in this way? No. Consider a simple case where N is trivial; an affirmative answer would imply that every map from G to H/K can be lifted to a map from G to H. Now take G=Z/2Z, the additive cyclic group of order 2, H=Z, the infinite additive cyclic group, N={0}, and K=2Z. Then the map sending 1+2Z to 1+2Z is a map from G/N to H/K, but the only map from G to H is the zero map, which does not induce the identity from G to H/K. > Also, can distinct homomorphisms f, g : G ---> H with f(N) < K and g(N) < K induce the *same* homomorphism G/N ---> H/K as above? Yes. Take G = H = Z, N=K=2Z, and let f be the identity map and g be the map that sends 1 to -1. Then the induced maps G/N-->H/K are both equal, but f and g are not equal. -- Arturo Magidin === Subject: Re: Polynomial approximation of Randomness ? <21228413.26759.1238872223834.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=LChCFQoAAACR0FoxHzVn6GGERsr9zp8c Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) > simple ; a random polynomial aahhh...hhmm....you mean, No one had ever tried this in history days or at present. some interesting lines of the day: There is a famous anecdote inspired by Euler's arguments with secular philosophers over religion, which is set during Euler's second stint at the St. Petersburg academy. The French philosopher Denis Diderot was visiting Russia on Catherine the Great's invitation. However, the Empress was alarmed that the philosopher's arguments for atheism were influencing members of her court, and so Euler was asked to confront the Frenchman. Diderot was later informed that a learned mathematician had produced a proof of the existence of God: he agreed to view the proof as it was presented in court. Euler appeared, advanced toward Diderot, and in a tone of perfect conviction announced, => Sir, a+b^n / z=x, hence God existsreply!. Diderot, to whom (says the story) all mathematics was gibberish, stood dumbstruck as peals of laughter erupted from the court. Embarrassed, he asked to leave Russia, a request that was graciously granted by the Empress. However amusing the anecdote may be, it is apocryphal, given that Diderot was a capable mathematician who had published mathematical treatises. by Leonhard Euler === Subject: How can I show that L_infinity[a,b] is not separable? posting-account=FCi6GQoAAABQxBq_scLdgl_V5oEmWhw3 .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) How can I show that L_infinity[a,b] is not separable? I assure you that this is not homework (classes are done, preparing for exam). === Subject: Re: How can I show that L_infinity[a,b] is not separable? How can I show that L_infinity[a,b] is not separable? I assure you that this is not homework (classes are done, preparing >for exam). This has been asked and answered previously. See, for example, http://mathforum.org/kb/message.jspa?messageID=5670643&tstart=0 --Lynn http://math.asu.edu/~kurtz === Subject: Re: a riddle How about this. It is the highest volume to aluminum ratio for a cylinder given a set volume and minimum wall stiffness. If you make the can shorter \ and fatter, you take the bend out of the wall and make the can more flimsy, \ requiring a thicker can to maintain stiffness. The answer is likely just good size for the hand, but thought I'd throw another angle out. === Subject: Re: a riddle 'thicker can' should read thicker can wall === Subject: Re: How to discretize this ? posting-account=LChCFQoAAACR0FoxHzVn6GGERsr9zp8c Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) > hi..., => u(.) = f ; where u is unknown, f is known function => u(x) = f {we will take unknown function in 1 variable} => u(xi) = f i {doubt:- how to discretize and sample the function} thnx a lot :) > for the help > > Why should I refuse a good dinner > simply because I don't understand > the digestive processes involved? Oliver Heaviside => please any clue...? I got this hint: using Collocation or weighted Residuals[ orthogonality principals ] => I'm waiting for your help..? thnx a lot for the help :) I had no need of that hypothesis ( Je n'avais pas besoin de cette hypoth.8fse-l.88, as a reply to Napoleon, who had asked why he hadn't mentioned God in his book on astronomy. ) by Pierre-Simon Laplace === Subject: Re: How to discretize this ? posting-account=LChCFQoAAACR0FoxHzVn6GGERsr9zp8c Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) > hi..., > => u(.) = f ; where u is unknown, f is known function > => u(x) = f {we will take unknown function in 1 variable} > => u(xi) = f i {doubt:- how to discretize and sample the function} > thnx a lot :) > for the help > > Why should I refuse a good dinner > simply because I don't understand > the digestive processes involved? > Oliver Heaviside => please any clue...? That's not possible - the question you ask is so vague > and general it approaches meaninglessness. > Awesome; I intentionally write the question like that and exactly the same happened what I think, coming to the point you said that question is Vague; that's absolutely correct but, what my point is Numerical solution of PDE's in Science & Engineering by Lapidus and Pinder [1982] A portion extracted from that:- => u(.) = f (1) => U(.)-f = R(.) { u=~ U = sig j=1 N aj Bj(.) } =>[ sig j=1 N aj Bj(.)] -f = R(.) (2) => and then invoking method weighted residuals, Our objective is to find the set of expansion coefficients {aj} that minimizes the residual R in the region of interest consisting of the union of the non-overlapping sub-domains for a given set of basis functions {Bj(.)} this is what I'm talking; If he(Galerkin) can discretize the mesh like that; For my simple function[ it just one variable only] Can't I discretize it like that ? OR do you need more...? please write it I'll provide all the details. > I'll give you a clue about something else, though: > Heaviside was an excellent mathematician - he > wasn't using mathematics that he didn't understand > at all, he was simply talking about using an > intuitive/informal notion of distributions > (or generalized functions) without knowing > the formal mathematical justification for them. > The things he did with distributions were correct. I got this hint: using Collocation or >weighted Residuals[ orthogonality principals ] => I'm waiting for your help..? thnx a lot for the help :) I had no need of that hypothesis ( Je n'avais pas besoin de cette hypoth.8fse-l.88, as a reply to >Napoleon, >who had asked why he hadn't mentioned God in his book on astronomy. ) by Pierre-Simon Laplace David C. Ullrich Understanding Godel isn't about following his formal proof. > That would make a mockery of everything Godel was up to. > (John Jones, My talk about Godel to the post-grads. > in sci.logic.) There is a famous anecdote inspired by Euler's arguments with secular philosophers over religion, which is set during Euler's second stint at the St. Petersburg academy. The French philosopher Denis Diderot was visiting Russia on Catherine the Great's invitation. However, the Empress was alarmed that the philosopher's arguments for atheism were influencing members of her court, and so Euler was asked to confront the Frenchman. Diderot was later informed that a learned mathematician had produced a proof of the existence of God: he agreed to view the proof as it was presented in court. Euler appeared, advanced toward Diderot, and in a tone of perfect conviction announced, => Sir, a+b^n / z=x, hence God existsreply!. Diderot, to whom (says the story) all mathematics was gibberish, stood dumbstruck as peals of laughter erupted from the court. Embarrassed, he asked to leave Russia, a request that was graciously granted by the Empress. However amusing the anecdote may be, it is apocryphal, given that Diderot was a capable mathematician who had published mathematical treatises. by Leonhard Euler === Subject: Starting a new paper on Baryons and Confinement I guess I have been getting warmed up the last few days with some preliminary threads here and elsewhere just to get a few things straight in my mind. Today, I began work on a new paper dealing with the Yang-Mills foundations of baryons and QCD confinement. The first draft is linked below, and I will provide updates as they develop. http://jayryablon.files.wordpress.com/2009/03/confinement-10.pdf I may get diverted a bit by my US tax filing the next few days, and I am quite busy at work right now so this will mostly be a weekend and after-midnight project, but I do hope to get this paper, which I hope will synthesize many individual insights I have had and subjects I have studied over the past several years, into a something of value for others. Constructive comments are always appreciated. and insights posted on the various newsgroups. Jay. ____________________________ Jay R. Yablon Email: jyablon@nycap.rr.com co-moderator: sci.physics.foundations Weblog: http://jayryablon.wordpress.com/ Web Site: http://home.roadrunner.com/~jry/FermionMass.htm === Subject: Re: What prevents protons and electrons from coming together? posting-account=5GUrzQkAAADun29oaK3p_W_saUVxxHUF Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) They are attractive by the electric force. Mitch Raemsch Classically, there's a centrifugal barrier that prevents orbiting > bodies from crashing into each other. Quantum mechanically, this > translates into the notion that the probability of locating the > electron in the immediate vicinity of the nucleus is very close to > zero. ------------- mumbler parrot the electron starts from the nuc and extends outwards only to certain specific directions and there is a difference between the behaviour of electrons that are located at the poles of the nuc and those that are located vertically along the an axis parallel to its longitudinal axis 2 those at the poles and connected to protons those that are perpendicular to thje longitudinal axis and connected to protons there are ee;ctron cupturs so my suggested (unprecedented) explanation to that is that *there is some electric* circle*stream betwen the poles ---and the main body of the Atom and nuc * it gets out from the poles ans *comes back* at the main body of the Atom are beta emitters dont forget that even the Atom is a sort of a magnet!! see in and look for the Iron 3d description (it has the 3D and 2D scheme of the same atom that i used 9the 2D) for simpler(more economic' ) descriptions Betta emitters *and their location* are marked there in brackets and that sytem and *trend* goes on all along the periodic table!! Iron was just one example among the whole lot ATB Y.Porat ---------------------------------------- === Subject: Re: What prevents protons and electrons from coming together? posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) [snip all] I'm now starting to wonder if my energies would be better directed at > making your employer understand that you are wasting hours a day at > work by harassing other people on the internet for the past seven > months. Do you think that'd be a good thing for me to do, David > Strich? I think so. Do you think so, ? Let's ponder the workplace presense that is you, David Strich. We all know your name is David Strich from the past postings where you > arrogantly signed your name with your 200 IQ. We can also see that > you have been posting from the same IP address for the past seven > months. Given that you have been posting from 9am to 5pm Eastern, with the > obvious hour for lunch, it is quite clear now that despite the fact > you are clearly ed in the head, that your place of work is at a VA > hospital. Given the information found out about a certain david strich who > works at KPMG in NY, it sounds a lot like you. Do you think forwarding some of your reprehensible behavior to some > random people there would be a good idea? -------------- > psychopath > ----------------------- whinerpath- Hide quoted text - - Show quoted text - You're the wiener here Eric. === Subject: Re: #348 an argument that only a Well Ordered Field are viable Number Systems; new book 2nd edition: New True Mathematics <87bprhvzht.fsf@phiwumbda.org> <49D4028E.8060208@hotmail.com> <49D45B5A.6070601@hotmail.com> negatives. So the entirety of the AP-Reals as a checkerboard has 0 > as the least element. Then it's not an ordered field at all. Feel free to call it an AP-ordered-field if you like, but it's not an ordered field. Part of the definition of an ordered field is that a < b implies a+c < b+c for all c. In particular for a = 0, b > 0 and c = -b, then -b < 0. - Tim === Subject: Re: #349 why the AP-Reals form a Well Ordered Field yet the Old-Reals cannot and thus flawed; new book 2nd edition: New True Mathematics <49D4028E.8060208@hotmail.com> <49D45B5A.6070601@hotmail.com> of Reals, and in like fashion, keeping the Negative Reals as > *undefined* for < or >, similarly has no harming affect on the Field > properties of Reals so long as the Positive Reals have the above > properties. Refer back to that page again: one of the defining properties of a total order is that *every* pair of elements a, b satisfies exactly one of a = b, a < b, or a > b. If there are any elements for which this is not true, for example by being undefined, then the field fails to be an ordered field. If you mean instead that -a is undefined for any positive a, then it is not even a field: it fails the existence of additive inverse property. Your inability to even recognise the basic definitions of what you claim is why I don't normally bother posting to correct your numerous errors - a state to which I am now returning. - Tim === Subject: Re: String Manipulation Puzzle > Given a string of characters e.g. ABBCBCDAAD what is the (any) largest > segment that has exactly N different characters e.g for N=2 the answer > is BBCBC? What is the tuple of the largest segment that has exactly N > different characters for N equals 1 through the total number of > different characters, 4 in the example? The problem is that of efficiency. How to avoid the square of the > length of the string checks? Define a Perfect Segment to be one such > that there is no adjacent character (to its left or right) that is > contained in the segment. ... I don't follow your Perfect Segment notion -- do you mean the substring is at end of string, or is maximal, or what? Anyway, re the main question, here is a simple algorithm that is O(n) if the alphabet is bounded and a character costs O(1) to access. Let the string be S[0..m-1], that is, is m characters in array S. 1. Initialize P[c]=0 for every character c. Set L=0, R=k, so that S[0..k] contains N different characters.[1] Set WL=L, WM=R-L to record best substring so far, and inRun=N. 2. While R N, subtract 1 from P[S[L]] and if P[S[L]] becomes 0, deduct 1 from inRun; add 1 to L at each pass. [3] 4. If R-L is a new record, set WL = L and WM = R-L. [4] [1] R = -1; L = inRun = 0; while (inRun < N && R N) { c = S[L]; ++L; --P[c]; if (P[c]==0) --inRun;}} else ++P[c]; [4] if (R-L > WM) { WL = L; WM = R-L;}} [See rest of program at http://pat7.com/jp/s/maxNsubstr.c . It takes a couple of milliseconds to find the best substring of any given length in an 80000-character string] > (The answer can help save the country. Really.) How? -- jiw === Subject: Re: String Manipulation Puzzle ... > substring is at end of string, or is maximal, or what? Anyway, re the > main question, here is a simple algorithm that is O(n) if the alphabet O(n) is a typo; should be O(m) > is bounded and a character costs O(1) to access. Let the string be > S[0..m-1], that is, is m characters in array S. ... -- jiw === Subject: Re: #333 *Archimedean Postulate* works just as well with equal as with greater than; new book 2nd edition: New True Mathematics <87bprhvzht.fsf@phiwumbda.org> <87vdpoq6pa.fsf@phiwumbda.org> <87hc17qoum.fsf@phiwumbda.org> <87r60bp8th.fsf@phiwumbda.org> <877i22bfph.fsf@phiwumbda.org> <87zlew1vyn.fsf@phiwumbda.org I'm more doubtful, just because it seems to \ me that analysts like to > do analysis, rather than play with toy theories that are vaguely > similar to R. LW's personal definition of standard analyst appears to be any person who believes that those who repeatedly post mathematical rubbish are cranks. That's why I put the term in quotes. I didn't even require that such people would be actively interested in such a theory, only that they would be happy with it being posted. As opposed to triggering a flood of posts pointing out the numerous errors in the garbage that cranks actually do post. - Tim === Subject: Continuity, integration, and uniform convergence. Given the following: * a sequence g_n of positive continuous functions for which the integral of g_n from -1 to 1 is 1 * g_n converges uniformly on [-1, -c] and [c, 1] for any c > 0 * a function f continuous on [-1, 1] I'm trying to show that the integral of f(x)g_n(x) from -1 to 1 converges to f(0). I'm having trouble showing this, as I don't think I have my approaches straight. First, in the outer pieces of [-1, 1], I can see that f(x)g_n(x) converges uniformly to 0. Which means that I have to show that f(x)g_n(x) converges to f(0) in the middle third. The only thing that I can think of that I can use to introduce f(0) is the fact that f is continuous at 0 (which presumably will also give me how to divide [-1, 1] into the three pieces. However, I fail to see how I can parlay the fact that the integral of g_n is 1 to the desired conclusion. I'm having trouble basically joining these disjointed parts together. Any clues that can help me put the pieces together? === Subject: Re: yet another f(f(x)) = x equation amy666 a .8ecrit : > f(x) =/= x > f(f(x)) = x > exp(f(x)) = f(exp(x)) > give examples of f(x) satisfying all the above 3 > equations at once. its silent again hmm > I gave you full examples 4 days ago. Do you want something more ? === Subject: Re: yet another f(f(x)) = x equation > f(x) =/= x f(f(x)) = x exp(f(x)) = f(exp(x)) > give examples of f(x) satisfying all the above 3 > equations at once. f(x) = complex conjugate of x -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === === Subject: Re: #322 proof outlines that Euclidean Geom. has only finite lines \ and that the Old Reals are not the True Reals (Continuous EveryWhere, Differentiable Nowhere) ; new book 2nd edition: New True Mathematics posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Many (most notably WM) have claimed > that since all (standard) natural numbers are finite, there are > only finitely many natural numbers. Sorry. I have not claimed that. My claim (and the obvious truth) is > that there are not more than X bits available to any mathematician. > [...] > I do not believe that there is a largest number because ingeniousity > may overcome any obstacle to produce larger and larger numbers, if > desired. Wow. So there are an infinite number of representations of > numbers, but there are only a finite number of bits available > to any representation. How does that work? How do you get > an infinite number of combinations to fit in a finite number > of bits? I do not need an infinite number of representations in order to construct a natural surpassing a given natural. I need simply another interpretation or another word in a given language. As the set of all interpretations and of all words invented up to a certain time is finite, there will be another one constructable from the finite set of bits. Further: Given the highly improbable case that all available matter had been used up to construct a larger number, then there is some matter remaining that constitutes the brain recognizing this number. There seems not to be a fixed limit on how much matter the brain ultimately requires. So there is no fixed limit on the number of bits that can be used to represent that large number. But as I said, it is very improbable that an intelligent civilization will attempt to turn the accesible universe into a big computer only in order to construct a large number. === Subject: Re: #322 proof outlines that Euclidean Geom. has only finite lines \ and that the Old Reals are not the True Reals (Continuous EveryWhere, Differentiable Nowhere) ; new book 2nd edition: New True Mathematics posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > Sorry. I have not claimed that. My claim (and the obvious truth) is > that there are not more than X bits available to any mathematician. > [...] > I do not believe that there is a largest number because ingeniousity > may overcome any obstacle to produce larger and larger numbers, if > desired. > Wow. So there are an infinite number of representations of > numbers, but there are only a finite number of bits available > to any representation. How does that work? How do you get > an infinite number of combinations to fit in a finite number > of bits? I do not need an infinite number of representations in order to > construct a natural surpassing a given natural. I need simply another > interpretation or another word in a given language. As the set of all > interpretations and of all words invented up to a certain time is > finite, there will be another one constructable from the finite set of > bits. So you are saying that If there are K representations, then there is always another interpretation possible. In other words, given K interpretations, there are always K+1 interpretations possible. Logically, then, the sequence of all K is infinite. Or did I miss something? I don't see anything in what you upper bound, the sequence of all K is infinite. Yet the number of bits available to represent any given K is finite. So I ask again, how do you fit an infinite number of interpretations into a finite number of bits? If you can, then you may qualify for some kind of prize. If you cannot, then at least one of your statements is false. === Subject: Re: #322 proof outlines that Euclidean Geom. has only finite lines \ and that the Old Reals are not the True Reals (Continuous EveryWhere, Differentiable Nowhere) ; new book 2nd edition: New True Mathematics posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Many (most notably WM) have claimed > that since all (standard) natural numbers are finite, there are > only finitely many natural numbers. > Sorry. I have not claimed that. Hmmm. I know that someone on sci.math has claimed that since > all naturals numbers are finite, there are only finitely many > natural numbers. But I don't recall who it is. Sorry, when read your text I understood that you meant a *fixed* finite set. Perhaps I misinterpreted that. When I write that the natural numbers form a finite set, then it is clear to me that this finity is not a fixed one but can change. Therefore I prefer to call it potential infinity. It is clear however, that potential infinity means always finite (though not fixed). So you were right, but it should be stressed that this finity is not a finite set in the sense of set theory. Therefore she cannot produce a number that requires more than X bits > to define it. In my text book I present a little example for my > students:http://www.hs-augsburg.de/~mueckenh/P5%20Zusfass.pdf > Consider an array that can display 7 letters of a usual keyboard. This > display can show every number between 0 and 9999999. It can also show > 10^1000 or 9^9^9^9 but not 12345678. > I do not believe that there is a largest number because ingeniousity > may overcome any obstacle to produce larger and larger numbers, if > desired. OK. But this does sound a lot like the Berry paradox. A link from > Wikipedia gives: http://en.wikipedia.org/wiki/Berry paradox the smallest possible integer not definable by a given number of > words Replace words with bits and there you go. That phrase has 72 > characters, and at eight bits per byte that's 576 bits. So a number > not > definable by 576 bits suddenly is definable by that many bits. I wonder how WM's theory avoids the Berry paradox. Do you know what number is addressed when I say The smallest possible integer that is not definable with 10^80 bits? Nobody knows it, nobody can define it because nobody has 10^80 bits. Therefore this number does not exist. If it existed, then it would be smaller than 10^10^100. But it does not exist. We will have to learn that this non-existence is as natural as the non-existence of sharp conjugate variables in quantum mechanics. In the display with seven symbols, the number 33643823 does not exist. But there can be remedy by extending the display. The mass of the universe however cannot be extended. === Subject: WM messes up again. Many (most notably WM) have claimed > that since all (standard) natural numbers are finite, there are > only finitely many natural numbers. > Sorry. I have not claimed that. Hmmm. I know that _someone_ on sci.math has claimed that since > all naturals numbers are finite, there are only finitely many > natural numbers. But I don't recall who it is. Sorry, when read your text I understood that you meant a *fixed* > finite set. Perhaps I misinterpreted that. When I write that the > natural numbers form a finite set, then it is clear to me that this > finity is not a fixed one but can change. Therefore I prefer to call > it potential infinity. It is clear however, that potential infinity > means always finite (though not fixed). So you were right, but it > should be stressed that this finity is not a finite set in the sense > of set theory. It is not even a set in the sense of any mathematical set theory. Do you know what number is addressed when I say The smallest possible > integer that is not definable with 10^80 bits? Nobody knows it, > nobody can define it because nobody has 10^80 bits. Therefore this > number does not exist. That description does not define a number, but that is quite different from implying that some number has been described but does not exist. If it existed, then it would be smaller than 10^10^100. But it does > not exist. We will have to learn that this non-existence is as > natural as the non-existence of sharp conjugate variables in quantum > mechanics. What number does WM claim does not exist? Unless it exists, one cannot even speak of it. What WM really means is that certain descriptions do not, in fact, describe a number. We might say that the description is ill-formed, but in any truly mathematical sense it is nonsense to say THAT number does not exist, as it requires the existence of the non-existent. === Subject: Probability problem I have the following problem: Let's assume that we are absolutely sure that a dice we own is unfair - we know that the probability that we will get 6 in a throw is less than 0.1 \ . Can you give any example of probabilistic experiment / process that will ensure us that we know p with error not larger than 0.01 with probability of \ 95%? (Equivalent: How many times should we throw our dice to have 95% of confidence that we can estimate the p with error not larger than 0.01?) I believe that to solve this problem one have to use Chebyshev's inequality \ and Bernoulli's scheme, but I have no idea how to do it... Can I ask you for \ some help? Chris === Subject: Re: Probability problem posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) On Apr 5, 4:19am, Christopher Kolago =95% sure that your estimate lies within 0.01 of the true p value, it is straightforward. The standard estimate of p is the sample average = X/N, where N = number of tosses and X = number of sixes obtained. X has distribution binomial(N,p), so has mean N*p. We have P{p - 0.01 <= est(p) <= p + 0.01} = P{N*p - 0.01*N <= X <= N*p + 0.01*N}, which could be computed from the binomial distribution for given N and p. However, assuming large N, it should be well approximated by a normal distribution, so we want P{N*p-.01*N <= N(mean=N*p,var=N*p*(1-p)) <= N*p + .01*N}/ If Phi is the standard N(0,1) cdf, this probability equates to P = Phi(b)-Phi(a), where a = -0.01*N/sqrt(N*p*(1-p)) and b = +0.01*N/sqrt(N*p*(1-p)). For each p we can solve for N from the equation P = 0.95; then N must be at least as large as the rounded-up value of that solution for that value of p. We can do this for a range of p and take the largest overall value of N. Here is what I get using Maple 9.5: p Nmin 0.020 753.000 0.030 1118.000 0.040 1476.000 0.050 1825.000 0.060 2167.000 0.070 2501.000 0.080 2828.000 0.090 3147.000 0.100 3458.000 So, overall, it suffices to toss the die 3458 times. This will satisfy your requirements for any of the unknown values of p <= 0.10. If you toss the die 3458 times the estimate = (number of 6s)/3458 will be within 0.01 of the true value of p (for p <= 0.10) with probability >= 0.95. In fact, for p less that 0.10 the probability will be > 0.95, because we are using a larger sample than necessary. For instance, if we knew that p = 0.07 we would need only 2501 tosses. R.G. Vickson === Subject: Re: Probability problem >I have the following problem: Let's assume that we are absolutely sure that a dice we own is unfair - we know that the probability that we will get 6 in a throw is less than 0.1 \ . >Can you give any example of probabilistic experiment / process that will ensure us that we know p with error not larger than 0.01 with >probability of 95%? (Equivalent: How many times should we throw our dice to \ have 95% of confidence that we can estimate the p >with error not larger than 0.01?) > For a similar problem, I found Wald's sequential testing and variations thereof to be most effective. http://en.wikipedia.org/wiki/Wald_test Also: http://www.isixsigma.com/library/content/c030811a.asp http://www.measuringusability.com/wald.htm http://en.wikipedia.org/wiki/Sequential_probability_ratio_test === Subject: Re: Probability problem posting-account=a6woBRAAAADpNFZJBA7ZBx35zXaKmaP4 Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) On Apr 5, 7:19am, Christopher Kolago dice we own is unfair - we know that the > probability that we will get 6 in a throw is > less than 0.1 . Can you give any example of > probabilistic experiment / process that will > ensure us that we know p with error not larger > than 0.01 with probability of 95%? (Equivalent: > How many times should we throw our dice to have > 95% of confidence that we can estimate the p > with error not larger than 0.01?) I believe that to solve this problem one have > to use Chebyshev's inequality and Bernoulli's > scheme, but I have no idea how to do it... Can > I ask you for some help? You can use Chebyshev's inequality to get a confidence interval in terms of the standard deviation, but this is a distribution-free result and correspondingly weak in terms of how many dice throws it will require. Think more efficiently in terms of the central limit theorem and the normal approximation of a binomial distribution (dice throw gives 6 or not 6). You'll get a more practical answer... Chris === Subject: Re: Probability problem > I have the following problem: Let's assume that we are absolutely sure that a > dice we own is unfair - we know that the > probability that we will get 6 in a throw is > less than 0.1 . Can you give any example of > probabilistic experiment / process that will > ensure us that we know p with error not larger > than 0.01 with probability of 95%? (Equivalent: > How many times should we throw our dice to have > 95% of confidence that we can estimate the p > with error not larger than 0.01?) I believe that to solve this problem one have > to use Chebyshev's inequality and Bernoulli's > scheme, but I have no idea how to do it... Can > I ask you for some help? You can use Chebyshev's inequality to get a > confidence interval in terms of the standard > deviation, but this is a distribution-free > result and correspondingly weak in terms of > how many dice throws it will require. I know that Chebyshev's inequality will give very thick estemitation (it will be weak ine termin of needed number of throws). I don't need very precise answer. > Think more efficiently in terms of the central > limit theorem and the normal approximation of > a binomial distribution (dice throw gives 6 > or not 6). You'll get a more practical > answer... That is really interesting idea. But there is one, slight problem - I know the CTL theorem, but I have never used it before? Can you show me please, \ how we can apply it to my problem? Chris === Subject: Re: Probability problem <28984745.29994.1238953528963.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=a6woBRAAAADpNFZJBA7ZBx35zXaKmaP4 Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) On Apr 5, 1:44pm, Christopher Kolago dice we own is unfair - we know that the > probability that we will get 6 in a throw is > less than 0.1 . Can you give any example of > probabilistic experiment / process that will > ensure us that we know p with error not larger > than 0.01 with probability of 95%? (Equivalent: > How many times should we throw our dice to have > 95% of confidence that we can estimate the p > with error not larger than 0.01?) I believe that to solve this problem one have > to use Chebyshev's inequality and Bernoulli's > scheme, but I have no idea how to do it... Can > I ask you for some help? You can use Chebyshev's inequality to get a > confidence interval in terms of the standard > deviation, but this is a distribution-free > result and correspondingly weak in terms of > how many dice throws it will require. I know that Chebyshev's inequality will give > very thick estemitation (it will be weak > ine termin of needed number of throws). I don't > need very precise answer. Think more efficiently in terms of the central > limit theorem and the normal approximation of > a binomial distribution (dice throw gives 6 > or not 6). You'll get a more practical > answer... That is really interesting idea. But there is one, > slight problem - I know the CTL theorem, but I have > never used it before? Can you show me please, how > we can apply it to my problem? Chris Let p = Pr(throw of dice (die?) is 6), and q = 1-p = Pr(throw is not 6). Assume repeated throws are independent events, and we keep track of how many throws are 6 (or not). This Bernoulli process has a binomial distribution. You said that p < 0.1 is known already. Apparently we need only an absolute margin of error in estimating p, not precision (relative error), so our goal is to estimate p within 0.01 with 95% confidence. Chebyshev's inequality (I'll adopt your spelling; I'm told he used quite a number of variants himself!) requires some knowledge of the standard deviation of a distribution. It may be best to think in terms of the sample mean of the Bernoulli process, as this will be p (if we take outcome to be 1 when a throw is 6 and otherwise 0). The standard deviation for samples of size N is s N = SQRT(pq/N): Since we don't know p exactly, neither do we know standard deviation s N. But we can give an upper bound: p < 0.1 q <= 1 s N < 1/SQRT(10N) What does Chebyshev's inequality say? http://en.wikipedia.org/wiki/Chebyshev%27s inequality No more than 1/k^2 of the values are more than k standard deviations away from the mean. In algebraic terms let m N be the (random variable) sample mean (for sample of size N), and as before the standard deviation s N: Pr( |m N - p| > k*s N ) <= 1/k^2 You have control over k and N. Keep the goal in mind: no more than 5% of the sample means for your chosen sample size should be more than 0.01 away from the true value p. So here's what you need to work out. How big should k be so the right hand side of the inequality is 5% or less (k can be any positive real number, so you can make 1/k^2 = 5% exactly if you wish)? For your given choice of k, how big does N need to be to guarantee the error bound k*s N is not more than 0.01? (The upper bound we have on s N is sufficient for this purpose. k is fixed by the above, and you are only required to increase N to accomplish the task.) If you work through this, then we can discuss the normal approximation and why it is more efficient when it applies (if at that point you are still interested). === === Subject: Re: Pepin factoring method question This I took from a book. Quote As one of several examples of his method, Pepin considered > N=(5^13 - 1)/4 = 305175781. Since any prime divisor of N is > of the form 26x + 1 and since there is no such prime divisor > less than 1000, he was able to say that if N = (26x+1)(26y+1) > then 1343 < x+y < 5946. Furthermore, 26xy + x + y = (N-1)/26= > 11737530; hence, x + y = 26z + 12 and > 169z^2 + 157z - 451407 = u^2 > with 51 < z < 229. End quote. my question is: 1- Why should a prime divisor of N be of the form 26x + 1? > 2- I can't figure out where the 5946 comes from in > 1343 < x + y < 5946. any help will be appreciated. If an odd prime p is a divisor of N = (5^13 - 1), then 5 has order 13 > in the reduced residue group modulo p, which has order p - 1. Hence > 13 is a divisor of p - 1, or p = 1 (mod 13), or > p = 13a + 1 for some positive integer a. If a is odd, then so is p, > so a = 2b, and p = 26a + 1. Achava thank you very much for your time. I still cannot figure out how he came up with the 5946 in the x + y > inequality. I know 1343 is tied to the square root of N something like > 2*sqrt(N)/26 +1 = 1343. But I fail to find something similar with the > upper > limit 5946. Maybe it goes like this. 26 x + 1 > 1000; x > 999 / 26; > also y > 999 / 26; if x + y > 5946 > then x y > (999 / 26)(5946 - (999 / 26)), > so (26 x + 1)(26 y + 1) > 26 x 26 y > 999 ((5946)(26) - 999) > which is maybe bigger than N. > -- > GM I understand what you did but the choice of x+y > 5946 looks arbitrary to > me. It does not seem to be justified. he could have used any other number. > and this is what I am still missing. How did he end up chosing 5946 in the > first place? The argument I gave - does it work if you use a number smaller than 5946? -- GM cg === === Subject: Has this already been done? Did anybody ever study the function x' = ax^2(1-bx^2), to see if it exhibits bifurcations similar to x' = ax(1-x^2) ? === Subject: Re: Has this already been done? Did anybody ever study the function x' = ax^2(1-bx^2), to see if it exhibits bifurcations similar to x' = ax(1-x^2) ? Yes: http://www.bjoernfredrik.de/bildschirm/hintern.html w. === Subject: Re: Has this already been done? anybody ever study the function >x' = ax^2(1-bx^2), >to see if it exhibits bifurcations similar to >x' = ax(1-x^2) ? > Yes: > http://www.bjoernfredrik.de/bildschirm/hintern.html Please do not do that when I've a mouthful of tea. The result can be spectacular, but not especially neat. [grin] -- RLW === Subject: Re: $50US prize for progression on equivalence between polysign P4 \ and RxC posting-account=n26igQkAAACeF9xA2Ms8cKIdBH40qzwr Gecko/20070505 Iceape/1.0.9 (Debian-1.0.13~pre080614i-0etch1),gzip(gfe),gzip(gfe) > This has been an ongoing debate by a few here. I have decided to reopen the $50 US prize ... This offer stands to anyone who can instantiate an isometric > isomorphism between P4 and RxC which is consistent under product. > The layout of this construction should include a transformation T such > that > y( P4 ) <--> x( R ), z( C ) > e.g. T( y ) = x, z > such that > T( y1 y2 ) = T( y1 ) T( y2 ). > The right hand side arithmetic product should be clearly defined on > RxC which is the part of the puzzle that remains troubling. Hagen already did that for you, you apparently couldn't quite see > that. His work was quite elegant and clear, but since you seem to > desire things spelled out extremely explicitly: Suppose y1 = -a+b*c#d where a,b,c,d are in RR. The Hagen shows the > isomorphism follows via the map T(y1) |-> ( b+d-a-c , d-b + (a-c)i ). > Now, if y2 = -r+s*t#u where r,s,t,u are in RR, then if I follow your > polysign ideas correctly we have: y1y2 = +ar*as#at-au > *br#bs-bt+bu > #cr-cs+ct*cu > -dr+ds*dt#du which if we collect terms (or signs -- whatever the right term is) we > get: y1y2 = -(au + bt + cs + dr)+(ar + bu + ct + ds)*(as + br + cu + dt)# > (at + bs + cr + du) where the '+' signs inside braces is the usual addition of positive > real numbers. By definition of the map we have (using square braces > for clarity): T(y1y2) = ((ar + bu + ct + ds) + (at + bs + cr + du) - (au + bt + cs + > dr) - (as + br + cu + dt) , (at + bs + cr + du) - (ar + bu + ct + ds) > + [(au + bt + cs + dr) - (as + br + cu + dt)]i ) Alternatively we have: T(y1)T(y2) = ( b+d-a-c , d-b + (a-c)i )( s+u-r-t , u-s + (r-t)i ) > = ( (b+d-a-c)(s+u-r-t) , (d-b + (a-c)i)(u-s + (r-t)i) ) Which expands out to (using square braces for clarity): T(y1)T(y2) = ( (ar + bu + ct + ds) + (at + bs + cr + du) - (au + bt + > cs + dr) - (as + br + cu + dt) , (at + bs + cr + du) - (ar + bu + ct + > ds) + [(au + bt + cs + dr) - (as + br + cu + dt)]i ) Thus we have T(y1y2) = T(y1)T(y2) as expected. Hagen showed more than > this, of course, including the isometry using an appropriate metric on > RR x CC, and much more elegantly. In short, you owe Hagen $50. Yes, I agree to what you say here. Hagen has refused the money, but I would still like to pay him. And yes he has gone far above and beyond my request in his pdf http://www.von-eitzen.de/math/PolysignNumbers.pdf much of which I am confused by. I don't understand how a polynomial constructs a multidimensional system. That is what seems to be going on starting at section three. I guess my thinking was partly that the transform phi() which Hagen presented is offensive to the geometry which I have assumed to be Euclidean in my own attempts. In other words since RxC is a 3D space the distance of the transform phi() could maintain the distance presented it in P4. As I process this attachment that I have I see that clearly Hagen's way must be the right way to maintain the simple isolated product in RxC, but that another way might be stated which does maintain the distance of the source space and actually looks very likely to exist. Yet the arithmetic product in RxC will then appear more contorted. This solution would expose a series I believe, as exposed by this analysis: http://bandtechnology.com/PolySigned/Deformation/P4T3Comparison.html which does conserve distance in the transform from P4 to RxC, which I call T3 since RxC matches the tatrix format of the polysign dimensional progression P1,P2,P3,... Taking the freedom to use two distance transforms to solve the problem feels suspicious to me, but it works and I have implemented the P4 version of Hagen's work into my code. Still I am curious what a distance conserving version would do to the arithmetic product definition in RxC so I offer up even another prize to expose this. You seem so bothered Leland, but I am happy that you your own two cents in here. At some level all of this is just a diversion to me, but it seems to be drawing out some sort of result as evidenced by Hagen's excellent work. The polysign numbers are extremely primitive. They build the real numbers and the complex numbers, on which Hagen's work relies. They yield spacetime through their curious distance behaviors in P4+. I don't feel that these features are well born out by Hagen's paper, but that is his agenda and I respect that. Especially to overlook the curious behaviors of P1 and the time correspondence is a missed opportunity to lead into physical phenomena. This is where the polysign paradigm leads and signals of coherence are present, but they are hidden under the shroud of the math that Hagen purveys, which treats the real number as fundamental. Under the polysign paradigm the sign symbols in front of every 'real' number are not fundamental but are evidence of substructure and this theory is born out by the polysign construction. - Tim === Subject: Re: $50US prize for progression on equivalence between polysign P4 \ and RxC posting-account=LhNEWgkAAACbPEcThjkaaetq3IIrFwRO Gecko/20080528 Epiphany/2.22 Firefox/3.0,gzip(gfe),gzip(gfe) This has been an ongoing debate by a few here. I have decided to reopen the $50 US prize ... This offer stands to anyone who can instantiate an isometric > isomorphism between P4 and RxC which is consistent under product. > The layout of this construction should include a transformation T such > that > y( P4 ) <--> x( R ), z( C ) > e.g. T( y ) = x, z > such that > T( y1 y2 ) = T( y1 ) T( y2 ). > The right hand side arithmetic product should be clearly defined on > RxC which is the part of the puzzle that remains troubling. Hagen already did that for you, you apparently couldn't quite see > that. His work was quite elegant and clear, but since you seem to > desire things spelled out extremely explicitly: Suppose y1 = -a+b*c#d where a,b,c,d are in RR. The Hagen shows the > isomorphism follows via the map T(y1) |-> ( b+d-a-c , d-b + (a-c)i ). > Now, if y2 = -r+s*t#u where r,s,t,u are in RR, then if I follow your > polysign ideas correctly we have: y1y2 = +ar*as#at-au > *br#bs-bt+bu > #cr-cs+ct*cu > -dr+ds*dt#du which if we collect terms (or signs -- whatever the right term is) we > get: y1y2 = -(au + bt + cs + dr)+(ar + bu + ct + ds)*(as + br + cu + dt)# > (at + bs + cr + du) where the '+' signs inside braces is the usual addition of positive > real numbers. By definition of the map we have (using square braces > for clarity): T(y1y2) = ((ar + bu + ct + ds) + (at + bs + cr + du) - (au + bt + cs + > dr) - (as + br + cu + dt) , (at + bs + cr + du) - (ar + bu + ct + ds) > + [(au + bt + cs + dr) - (as + br + cu + dt)]i ) Alternatively we have: T(y1)T(y2) = ( b+d-a-c , d-b + (a-c)i )( s+u-r-t , u-s + (r-t)i ) > = ( (b+d-a-c)(s+u-r-t) , (d-b + (a-c)i)(u-s + (r-t)i) ) Which expands out to (using square braces for clarity): T(y1)T(y2) = ( (ar + bu + ct + ds) + (at + bs + cr + du) - (au + bt + > cs + dr) - (as + br + cu + dt) , (at + bs + cr + du) - (ar + bu + ct + > ds) + [(au + bt + cs + dr) - (as + br + cu + dt)]i ) Thus we have T(y1y2) = T(y1)T(y2) as expected. Hagen showed more than > this, of course, including the isometry using an appropriate metric on > RR x CC, and much more elegantly. In short, you owe Hagen $50. Yes, I agree to what you say here. Hagen has refused the money, but I > would still like to pay him. And yes he has gone far above and beyond > my request in his pdf > http://www.von-eitzen.de/math/PolysignNumbers.pdf > much of which I am confused by. I don't understand how a polynomial > constructs a multidimensional system. That is what seems to be going > on starting at section three. I would suggest that this is a point that you need to work on. It's great that you're interested in exploring the properties of polysign numbers. Having ideas and exploring their mathematical consequences is a great thing to do. An important point, however, is that polysign numbers can be described very neatly within a standard commutative algebra framework (or via group algebras). This point is important because it allows for generalisation and more efficient explication -- the existing framework exists for a reason: it is both flexible, and powerful. What seems difficult to discern from polysign numbers (say the isomorphism between P4 and RR x CC) is often straightforward to see and prove in the usual framework. Now, not knowing the usual framework of commutative algebra is not a bad thing: lots of people don't know it, and don't have a need to know it. However, it has been pointed out to you several times that a good understanding of such things (rings, polynomial rings, group rings, etc.), and associated theorems, will go a long way toward allowing you to see and prove more regarding your polysign numbers. You really should be investing the effort to learn exactly what it is people like Hagen are talking about. The best place to start is a decent introductory book on modern algebra (or abstract algebra). I believe Dover publications have a number of offerings at very reasonable prices (ranging from $13 to $35). Alternatively your local library may have some suitable books. There is even a fair range of material on-line for free; you could try: http://www.math.miami.edu/~ec/book/ or for something a little more advanced try: http://www.math.uiuc.edu/~r-ash/Algebra.html followed by http://www.math.uiuc.edu/~r-ash/ComAlg.html or just do some searching for on-line course notes. It will be fun and interesting and it will give you a whole range of far more powerful tools and techniques for you to use in your studies. === === === Subject: Re: Conformal mapping from one annuli to another I thought of something that makes me almost certain > that the argument > the authors had in mind is bogus: Of course it's impossible to show that it does not > follow that fg = g, > because strictly speaking it does follow - we'vc > assumed something > _false_, and _anything_ follows from that, fg = g as > well as 2 + 2 = > 5. But. Consider the case where a_1 = a_2. In that case > there _do_ > exist conformal maps f from A_1 onto A_2. In > particular, f could > be a non-trivial rotation. In that case (g_n) does > not converge. > It does have subsequences converging to g, but fg = g > is clearly > false. So if there _is_ some argument showing that fg = g it > must > depend on the fact that a_2 < a_1; in particular it > can't be > as simple or obvious as the authors seem to suggest. Heh-heh: This is why you should be reading Complex > Made > Simple instead. I just got an email from a guy who's > evidently > read the entire book very carefully. He didn't find > any actual > errors, just 10 or 15 typos (the most serious being > that I omitted > simply connected at one point and I omitted > Hausdorff > from the definition of one-dimensional complex > manifold.) Steve David C. Ullrich Understanding Godel isn't about following his formal > proof. > That would make a mockery of everything Godel was up > to. > (John Jones, My talk about Godel to the post-grads. > in sci.logic.) But I *do* read Complex Made Simple! Perhaps you might remember I posted a question about it not too long ago. I don't even *own* the Greene \ & Krantz book; it is only for class that I have to do exercises out of that book. As someone who has done work in both textbooks, I definitely agree yours suits my style much better. I especially liked Chapter 11, and your explanation of the group of Mobius transformations. [I'm not being sycophantic here, this is an honest, personal opinion.] Steve === Subject: Re: Conformal mapping from one annuli to another This is a question based on an exercise in Greene & > Krantz' book. The idea is to assume there is a > conformal mapping from the annulus >A_1 = { z in C | 1 < |z| < R} >to the annulus >A_2 = { z in C | 1 < |z| < r} with r < R. Call this mapping f. The hint is to > look at the family {g_n}, where g_n = f^n, that is, > the n-th iterate of f. This family is normal, so > some subsequence converges to a holomorphic mapping > g: A_1 -> A_2. >The next part is what is confusing me: it says to > use the fact fg(z) = f(z), where fg means here f > composed with g. >My question is: why is this fact true? I think it > has something to do with the image of f getting > smaller each time, but I don't see how to make this > precise. I am fairly certain g(A_1) is exactly the > fixed set of f, but I don't see how that shows f > fixes g(A_1) pointwise. I'm gonna start over. I don't see why a convergent > subsequence implies > that fg = g, and I suspect that the authors don't > either, although I > could certainly be wrong about that. One _can_ show that the images of g_n are getting > smaller in a > suitable sense, allowing one to complete the argument > the > authors sketch without knowing that fg = g. But when > I > think about how to show this I realize that once > we've > shown what needs to be shown to do that we don't > need most of the argument the authors sketch, we're > already done for simpler reasons. Ok. First, repeating from last night: Since the inverse of f is continuous, it follows that > if z_n tends to the boundary of A_1 then f(z_n) > tends to the boundary of A_2. A topological > argument shows that one of the following must > hold: (i) If |z_n| -> 1 then |f(z_n)| -> 1 while if |z_n| > -> a_1 > then |f(z_n)| -> a_2 or (ii) the other way around: if |z_n| -> 1 then > |f(z_n)| > -> a_2 and if |z_n| -> a_1 then |f(z_n)| -> 1. If (ii) happens then replacing f by f(c/z) makes > (i) happen. So assume (i). Now, last night I said something about using the > fact that |f| is subharmonic. That together with (i) > does give suitable estimates to show that |g_n| > tends to 1, I believe, but come to think of it > it's better to use the fact that log|f| is > subharmonic. > That, with (i), shows that log|f(z)| <= c log|z|, > where c = log(a_2)/log(a_1), and that inequality > shows that |g_n| -> 1. But duh: In fact log|f| is _harmonic_. With (i) > this shows that (*) log|f(z)| = c log|z|, where 0 < c < 1. And there's any number of ways to > see immediately that (*) is impossible. For example, (*) shows that |f(z)/z^c| = 1, > where z^c is defined in some disk contained in A_1. > Hence f(z)/z^c is constant in that disk, hence it can > be analytically continued to all of A_1, so f is a > constant times z^c in A_1, but there's no such thing > as z^c in A_1. Or a little more formally: Say D is some disk > contained > in A_1. There exist functions log(f) and log(z) in D, > and (*) shows that Re(log(f) - c log(z)) = 0 in D. > So log(f(z)) - c log(z) is constant in D; > diifferentiating, > this shows that (**) f'/f = c/z in D. Hence (**) holds in all of A_1. This shows > that 1/(2 pi i)int f'/f = c (if the integral is about > a circle centered at the origin), contradicting the > fact that that integral/(2 pi i) must be an integer. Steve David C. Ullrich Understanding Godel isn't about following his formal > proof. > That would make a mockery of everything Godel was up > to. > (John Jones, My talk about Godel to the post-grads. > in sci.logic.) Wow, I really appreciate that! That is a very nice, and simple, proof. It's also more in line with the theme of the chapter: harmonic functions. I am in complete agreement that Greene & Krantz probably though fg = g was an afterthought; I am more tempted to believe - but by no means convinced - the integral of (g_n)'/g_n is what they say. Steve === Subject: Converse to Riemann-Lebesgue lemma The lemma (for Fourier series) states that if |f(x)| is Lebesgue integrabls and periodic then its Fourier coefficients a_n -> 0. Is there a simple example of a sequence a_n -> 0 which does not arise from an L^1 function in this way? -- Timothy Murphy e-mail: gayleard /at/ eircom.net tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College Dublin === Subject: Re: Converse to Riemann-Lebesgue lemma > The lemma (for Fourier series) states that > if |f(x)| is Lebesgue integrabls and periodic > then its Fourier coefficients a_n -> 0. > Is there a simple example of a sequence a_n -> 0 > which does not arise from an L^1 function in this way? I think there is an exercise in Rudin (PRINCIPLES OF MATHEMATICAL > ANALYSIS) that constructs such a sequence. A softer argument shows such a sequence exists without actually > constructing it: if not, then by the closed graph theorem, Banach > spaces L_1[0,1] and c_0 would be isomorphic, which is false (for > example, one has separable dual but not the other). A slightly different approach is: if the natural map F from L_1([0,1]) into c_0 happened to be surjective then, since it would be bounded bijective between Banach spaces, it would be an isomorphism (by the open mapping theorem). Therefore, there would be some M > 0 such that, for each _c_ in c_0, ||F^{-1}(c)|| <= M.||c||, which is equivalent to the assertion that, when _f_ in L_1([0,1]), ||F(f)|| >= M^{-1}.||f||. It is easy to see that no such M can exist, using the functions _f_ of the type f(x) = sum_{-n <= k <= n}exp(i k x). Jose Carlos Santos === Subject: Re: Converse to Riemann-Lebesgue lemma > The lemma (for Fourier series) states that > if |f(x)| is Lebesgue integrabls and periodic > then its Fourier coefficients a n -> 0. Is there a simple example of a sequence a n -> 0 > which does not arise from an L^1 function in this way? > A slightly different approach is: if the natural map F from L 1([0,1]) > into c 0 happened to be surjective then, since it would be bounded > bijective between Banach spaces, it would be an isomorphism (by the > open mapping theorem). Therefore, there would be some M > 0 such that, > for each c in c 0, ||F^{-1}(c)|| <= M.||c||, which is equivalent to > the assertion that, when f in L 1([0,1]), ||F(f)|| >= M^{-1}.||f||. > It is easy to see that no such M can exist, using the functions f of > the type f(x) = sum {-n <= k <= n}exp(i k x). I have to confess that this was an exam question here a year or two ago, with the above solution suggested. I was convinced that there must be a straightforward counter-example, but it seems my colleague has one up on me. -- Timothy Murphy e-mail: gayleard /at/ eircom.net tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College Dublin === Subject: Re: Converse to Riemann-Lebesgue lemma The lemma (for Fourier series) states that > if |f(x)| is Lebesgue integrabls and periodic > then its Fourier coefficients a_n -> 0. > Is there a simple example of a sequence a_n -> 0 > which does not arise from an L^1 function in this way? > A slightly different approach is: if the natural map F from L_1([0,1]) > into c_0 happened to be surjective then, since it would be bounded > bijective between Banach spaces, it would be an isomorphism (by the > open mapping theorem). Therefore, there would be some M > 0 such that, > for each _c_ in c_0, ||F^{-1}(c)|| <= M.||c||, which is equivalent to > the assertion that, when _f_ in L_1([0,1]), ||F(f)|| >= M^{-1}.||f||. > It is easy to see that no such M can exist, using the functions _f_ of > the type f(x) = sum_{-n <= k <= n}exp(i k x). I have to confess that this was an exam question here a year or two ago, >with the above solution suggested. >I was convinced that there must be a straightforward counter-example, There's probably an explicit counterexample somewhere in Zygmund. Giving an explicit example can't be all that easy - for example you > can't write down a sequence and say that it's not the Fourier > coefficients of an L^1 function because it tends to 0 too slowly, > because of the following curious fact: If c_n >= 0 and c_n -> 0 > then there exists f in L^1 with f^(n) >= c_n for all n. On the other hand, there is this: if a_n is decreasing with limit 0, and sum_{n=1}^infty a_n / n = infty, then sum_n a_n sin(nx), although everywhere convergent, is not the Fourier series of an L^1 function. See Edwards, Fourier Series, A Modern Introduction, sec. 7.3.4. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Converse to Riemann-Lebesgue lemma > The lemma (for Fourier series) states that > if |f(x)| is Lebesgue integrabls and periodic > then its Fourier coefficients a_n -> 0. Is there a simple example of a sequence a_n -> 0 > which does not arise from an L^1 function in this way? > A slightly different approach is: if the natural map F from L_1([0,1]) > into c_0 happened to be surjective then, since it would be bounded > bijective between Banach spaces, it would be an isomorphism (by the > open mapping theorem). Therefore, there would be some M > 0 such that, > for each _c_ in c_0, ||F^{-1}(c)|| <= M.||c||, which is equivalent to > the assertion that, when _f_ in L_1([0,1]), ||F(f)|| >= M^{-1}.||f||. > It is easy to see that no such M can exist, using the functions _f_ of > the type f(x) = sum_{-n <= k <= n}exp(i k x). I have to confess that this was an exam question here a year or two ago, with the above solution suggested. I was convinced that there must be a straightforward counter-example, but it seems my colleague has one up on me. ------------- I think your colleagues are trying to tell you something... === Subject: f continuous, 2pi periodic; s_n(f) does not converge to f uniformly (as n-> infinity) posting-account=JsYpPwoAAABhdgohxXUSVJ-FjzcpTBpO .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) If f continuous and 2pi periodic, How can I show that the nth partial sums of Fourier series of f, s_n (f) , does not converge to f uniformly ? === Subject: Re: f continuous, 2pi periodic; s_n(f) does not converge to f uniformly (as n-> infinity) > If f continuous and 2pi periodic, How can I show that the nth partial sums of Fourier series of f, s_n > (f) , does not converge to f uniformly ? Wrong question. s_n(f) _might_ not converge to f uniformly, but for many f's (and probably for any f that you are likely to write down explicitly) they will converge to f uniformly. Perhaps what you mean is, how can I show that _there exists_ f .... -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: f continuous, 2pi periodic; s_n(f) does not converge to f uniformly (as n-> infinity) posting-account=JsYpPwoAAABhdgohxXUSVJ-FjzcpTBpO .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) On Apr 5, 1:46pm, Robert Israel > If f continuous and 2pi periodic, How can I show that the nth partial sums of Fourier series of f, s n > (f) , does not converge to f uniformly ? Wrong question. s n(f) might not converge to f uniformly, but for > many f's (and probably for any f that you are likely to write down > explicitly) they will converge to f uniformly. Perhaps what you mean is, > how can I show that there exists f .... Just a question from a poorly written review sheet (hint was given as: Growth of Lebesgue constants.). It's okay, I will come back to it later. On the other hand, can you answer this question for me? When we deal with Fourier series and Lp spaces, why do we change the norms on these spaces to have 1/(2pi) in front of integral? Is there a motivation for this? Then shouldn't we also change the norm on L infinity to also have 1/ (2pi) in front? If not the relationship ||f|| infinity = lim(p->infinity) ||f|| p will not hold. Correct? === Subject: Re: f continuous, 2pi periodic; s_n(f) does not converge to f uniformly (as n-> infinity) > On Apr 5, 1:46pm, Robert Israel > If f continuous and 2pi periodic, How can I show that the nth partial sums of Fourier series of f, s_n > (f) , does not converge to f uniformly ? Wrong question. s_n(f) _might_ not converge to f uniformly, but for > many f's (and probably for any f that you are likely to write down > explicitly) they will converge to f uniformly. Perhaps what you mean is, > how can I show that _there exists_ f .... Just a question from a poorly written review sheet (hint was given as: > Growth of Lebesgue constants.). It's okay, I will come back to it > later. On the other hand, can you answer this question for me? When we deal with Fourier series and Lp spaces, why do we change the > norms on these spaces to have 1/(2pi) in front of integral? Is there > a motivation for this? So that the exponentials e^(int) form an orthonormal basis. > Then shouldn't we also change the norm on L_infinity to also have 1/ > (2pi) in front? If not the relationship ||f||_infinity = lim(p->infinity) ||f||_p will not hold. Correct? No, the L^oo norm equals the essential supremum. That doesn't change by normalizing. (Another way to see it: [1/(2pi)]^(1/p) -> 1 as p -> oo.) === === Subject: Re: fourier series of f ' relative to f > What is the theorem that can allow me to justify moving > differentiation operator inside infinite sums (or infinite limits)? I think you need absolute convergence of the original series. No, you need uniform convergence of the original series, and > _also_ uniform convergence of the series of derivatives. If your >periodic function is continuosly differentiable, in other words belong >to C^1, then you have it. Yes, C^1 implies the Fourier series for f converges absolutely. > But that's not enough to say the derivative of the sum is the > sum of the derivatives. C^1 does _not_ imply that the > differentiated series converges uniformly (precisely because > there exist continuous functions with divergent Fourier > series; take f so that f' is continuous with a divergent > Fourier series.) A condition known as C^(1 + epsilon) is enough. But > (i) that doesn't answer the OP's question, and (ii) > the actual solution is very simple, just using the > definition of the Fourier coefficients instead of > trying to differentiate the series. The basic idea is simple, but I don't see how to follow through without measure theory. I think the problem was either misstated by the OP or badly chosen by its author. If the author intended a classical proof using the Riemann integral, then it can't be done as stated. (Maybe the author intended piecewise C^1.) If an appeal to the Lebesgue integral was intended, then why not just assume f is absolutely continuous? You get a stronger and more natural result - and the piecewise business can be left out. > David C. Ullrich Understanding Godel isn't about following his formal proof. > That would make a mockery of everything Godel was up to. > (John Jones, My talk about Godel to the post-grads. > in sci.logic.) === Subject: Re: fourier series of f ' relative to f <9c4ht4pd2op8ib24gbge1mdkslkihj0804@4ax.com> posting-account=iyXB5AkAAABPCtkDKhRJOsNWmafzHSRE Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > What is the theorem that can allow me to justify moving > differentiation operator inside infinite sums (or infinite limits)? I think you need absolute convergence of the original series. No, you need uniform convergence of the original series, and > also uniform convergence of the series of derivatives. If your >periodic function is continuosly differentiable, in other words belong >to C^1, then you have it. Yes, C^1 implies the Fourier series for f converges absolutely. > But that's not enough to say the derivative of the sum is the > sum of the derivatives. C^1 does not imply that the > differentiated series converges uniformly (precisely because > there exist continuous functions with divergent Fourier > series; take f so that f' is continuous with a divergent > Fourier series.) A condition known as C^(1 + epsilon) is enough. But > (i) that doesn't answer the OP's question, and (ii) > the actual solution is very simple, just using the > definition of the Fourier coefficients instead of > trying to differentiate the series. The basic idea is simple, but I don't see how to follow through > without measure theory. Maybe you forgot that the derivative f` was bounded. By mean value theorem for any (a,b) on [-pii,pii], there is a c on (a,b) such that f`(c)(b - a) = f(b) - f(a) so when you take sup f`(c) on [-pii,pii] you obtain bound sup f`(c)(b - a) for all of the variations. >I think the problem was either misstated by > the OP or badly chosen by its author. If the author intended a > classical proof using the Riemann integral, then it can't be done as > stated. (Maybe the author intended piecewise C^1.) If an appeal to the > Lebesgue integral was intended, then why not just assume f is > absolutely continuous? You get a stronger and more natural result - > and the piecewise business can be left out. David C. Ullrich Understanding Godel isn't about following his formal proof. > That would make a mockery of everything Godel was up to. > (John Jones, My talk about Godel to the post-grads. > in sci.logic.) === Subject: Re: fourier series of f ' relative to f > What is the theorem that can allow me to justify moving > differentiation operator inside infinite sums (or infinite limits)? I think you need absolute convergence of the original series. No, you need uniform convergence of the original series, and > _also_ uniform convergence of the series of derivatives. If your >periodic function is continuosly differentiable, in other words belong >to C^1, then you have it. Yes, C^1 implies the Fourier series for f converges absolutely. > But that's not enough to say the derivative of the sum is the > sum of the derivatives. C^1 does _not_ imply that the > differentiated series converges uniformly (precisely because > there exist continuous functions with divergent Fourier > series; take f so that f' is continuous with a divergent > Fourier series.) A condition known as C^(1 + epsilon) is enough. But > (i) that doesn't answer the OP's question, and (ii) > the actual solution is very simple, just using the > definition of the Fourier coefficients instead of > trying to differentiate the series. The basic idea is simple, but I don't see how to follow through > without measure theory. Maybe you forgot that the derivative f` was bounded. By mean value > theorem for any (a,b) on [-pii,pii], there is a c on (a,b) such that > f`(c)(b - a) = f(b) - f(a) so when you take sup f`(c) on [-pii,pii] > you obtain bound sup f`(c)(b - a) for all of the variations. No, I didn't forget that, but it's not going to help with Riemann integrability. >I think the problem was either misstated by > the OP or badly chosen by its author. If the author intended a > classical proof using the Riemann integral, then it can't be done as > stated. (Maybe the author intended piecewise C^1.) If an appeal to the > Lebesgue integral was intended, then why not just assume f is > absolutely continuous? You get a stronger and more natural result - > and the piecewise business can be left out. David C. Ullrich Understanding Godel isn't about following his formal proof. > That would make a mockery of everything Godel was up to. > (John Jones, My talk about Godel to the post-grads. > in sci.logic.) === Subject: Re: fourier series of f ' relative to f <9c4ht4pd2op8ib24gbge1mdkslkihj0804@4ax.com> posting-account=iyXB5AkAAABPCtkDKhRJOsNWmafzHSRE Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > What is the theorem that can allow me to justify moving > differentiation operator inside infinite sums (or infinite limits)? I think you need absolute convergence of the original series. No, you need uniform convergence of the original series, and > also uniform convergence of the series of derivatives. If your >periodic function is continuosly differentiable, in other words belong >to C^1, then you have it. Yes, C^1 implies the Fourier series for f converges absolutely. > But that's not enough to say the derivative of the sum is the > sum of the derivatives. C^1 does not imply that the > differentiated series converges uniformly (precisely because > there exist continuous functions with divergent Fourier > series; take f so that f' is continuous with a divergent > Fourier series.) A condition known as C^(1 + epsilon) is enough. But > (i) that doesn't answer the OP's question, and (ii) > the actual solution is very simple, just using the > definition of the Fourier coefficients instead of > trying to differentiate the series. The basic idea is simple, but I don't see how to follow through > without measure theory. Maybe you forgot that the derivative f` was bounded. By mean value > theorem for any (a,b) on [-pii,pii], there is a c on (a,b) such that f`(c)(b - a) = f(b) - f(a) so when you take sup |f`(c)| on [-pii,pii] you obtain bound sup |f`(c)|(b - a) for all of the variations. So I understand that the riemann integral of f`(x)dx does exist, because the stieljes integral of d(f(x) is the same. But I don`t know how to prove it with elementary methods. === Subject: Re: fourier series of f ' relative to f > What is the theorem that can allow me to justify moving > differentiation operator inside infinite sums (or infinite limits)? I think you need absolute convergence of the original series. No, you need uniform convergence of the original series, and > _also_ uniform convergence of the series of derivatives. If your >periodic function is continuosly differentiable, in other words belong >to C^1, then you have it. Yes, C^1 implies the Fourier series for f converges absolutely. > But that's not enough to say the derivative of the sum is the > sum of the derivatives. C^1 does _not_ imply that the > differentiated series converges uniformly (precisely because > there exist continuous functions with divergent Fourier > series; take f so that f' is continuous with a divergent > Fourier series.) A condition known as C^(1 + epsilon) is enough. But > (i) that doesn't answer the OP's question, and (ii) > the actual solution is very simple, just using the > definition of the Fourier coefficients instead of > trying to differentiate the series. The basic idea is simple, but I don't see how to follow through > without measure theory. Maybe you forgot that the derivative f` was bounded. By mean value > theorem for any (a,b) on [-pii,pii], there is a c on (a,b) such that > f`(c)(b - a) = f(b) - f(a) so when you take sup |f`(c)| on [-pii,pii] > you obtain bound sup |f`(c)|(b - a) for all of the variations. So I understand that the riemann integral of f`(x)dx does exist, > because the stieljes integral of d(f(x) is the same. But I don`t know > how to prove it with elementary methods. No, the Riemann integral of f' need not exist. There exist differentiable functions f on [a, b], with f' bounded on [a, b], such that f' is discontinuous on a set of positive measure, which implies f' is not Riemann integrable on [a, b]. === Subject: Re: fourier series of f ' relative to f <9c4ht4pd2op8ib24gbge1mdkslkihj0804@4ax.com> posting-account=iyXB5AkAAABPCtkDKhRJOsNWmafzHSRE Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > What is the theorem that can allow me to justify moving > differentiation operator inside infinite sums (or infinite limits)? I think you need absolute convergence of the original series. No, you need uniform convergence of the original series, and > also uniform convergence of the series of derivatives. If your >periodic function is continuosly differentiable, in other words belong >to C^1, then you have it. Yes, C^1 implies the Fourier series for f converges absolutely. > But that's not enough to say the derivative of the sum is the > sum of the derivatives. C^1 does not imply that the > differentiated series converges uniformly (precisely because > there exist continuous functions with divergent Fourier > series; take f so that f' is continuous with a divergent > Fourier series.) A condition known as C^(1 + epsilon) is enough. But > (i) that doesn't answer the OP's question, and (ii) > the actual solution is very simple, just using the > definition of the Fourier coefficients instead of > trying to differentiate the series. The basic idea is simple, but I don't see how to follow through > without measure theory. Maybe you forgot that the derivative f` was bounded. By mean value > theorem for any (a,b) on [-pii,pii], there is a c on (a,b) such that > f`(c)(b - a) = f(b) - f(a) so when you take sup |f`(c)| on [-pii,pii] > you obtain bound sup |f`(c)|(b - a) for all of the variations. So I understand that the riemann integral of f`(x)dx does exist, > because the stieljes integral of d(f(x) is the same. But I don`t know > how to prove it with elementary methods. No, the Riemann integral of f' need not exist. There exist > differentiable functions f on [a, b], with f' bounded on [a, b], such > that f' is discontinuous on a set of positive measure, which implies > f' is not Riemann integrable on [a, b]. OK, but I read from here that the stieljes integral is the same than the corresponding riemann integral when the measure is of bounded variation and everywhere differentiable. http://knowledgerush.com/kr/encyclopedia/Riemann-Stieltjes integral/ === Subject: Re: fourier series of f ' relative to f > What is the theorem that can allow me to justify moving > differentiation operator inside infinite sums (or infinite > limits)? I think you need absolute convergence of the original series. No, you need uniform convergence of the original series, and > _also_ uniform convergence of the series of derivatives. If your >periodic function is continuosly differentiable, in other words >belong >to C^1, then you have it. Yes, C^1 implies the Fourier series for f converges absolutely. > But that's not enough to say the derivative of the sum is the > sum of the derivatives. C^1 does _not_ imply that the > differentiated series converges uniformly (precisely because > there exist continuous functions with divergent Fourier > series; take f so that f' is continuous with a divergent > Fourier series.) A condition known as C^(1 + epsilon) is enough. But > (i) that doesn't answer the OP's question, and (ii) > the actual solution is very simple, just using the > definition of the Fourier coefficients instead of > trying to differentiate the series. The basic idea is simple, but I don't see how to follow through > without measure theory. Maybe you forgot that the derivative f` was bounded. By mean value > theorem for any (a,b) on [-pii,pii], there is a c on (a,b) such that > f`(c)(b - a) = f(b) - f(a) so when you take sup |f`(c)| on [-pii,pii] > you obtain bound sup |f`(c)|(b - a) for all of the variations. So I understand that the riemann integral of f`(x)dx does exist, > because the stieljes integral of d(f(x) is the same. But I don`t know > how to prove it with elementary methods. No, the Riemann integral of f' need not exist. There exist > differentiable functions f on [a, b], with f' bounded on [a, b], such > that f' is discontinuous on a set of positive measure, which implies > f' is not Riemann integrable on [a, b]. OK, but I read from here that the stieljes integral is the same than > the > corresponding riemann integral when the measure is of bounded > variation and everywhere differentiable. > http://knowledgerush.com/kr/encyclopedia/Riemann-Stieltjes_integral/ That reference is incorrect. Do you see why? === Subject: Re: fourier series of f ' relative to f <9c4ht4pd2op8ib24gbge1mdkslkihj0804@4ax.com> posting-account=iyXB5AkAAABPCtkDKhRJOsNWmafzHSRE Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > What is the theorem that can allow me to justify moving > differentiation operator inside infinite sums (or infinite > limits)? I think you need absolute convergence of the original series. No, you need uniform convergence of the original series, and > also uniform convergence of the series of derivatives. If your >periodic function is continuosly differentiable, in other words >belong >to C^1, then you have it. Yes, C^1 implies the Fourier series for f converges absolutely. > But that's not enough to say the derivative of the sum is the > sum of the derivatives. C^1 does not imply that the > differentiated series converges uniformly (precisely because > there exist continuous functions with divergent Fourier > series; take f so that f' is continuous with a divergent > Fourier series.) A condition known as C^(1 + epsilon) is enough. But > (i) that doesn't answer the OP's question, and (ii) > the actual solution is very simple, just using the > definition of the Fourier coefficients instead of > trying to differentiate the series. The basic idea is simple, but I don't see how to follow through > without measure theory. Maybe you forgot that the derivative f` was bounded. By mean value > theorem for any (a,b) on [-pii,pii], there is a c on (a,b) such that > f`(c)(b - a) = f(b) - f(a) so when you take sup |f`(c)| on [-pii,pii] > you obtain bound sup |f`(c)|(b - a) for all of the variations. So I understand that the riemann integral of f`(x)dx does exist, > because the stieljes integral of d(f(x) is the same. But I don`t know > how to prove it with elementary methods. No, the Riemann integral of f' need not exist. There exist > differentiable functions f on [a, b], with f' bounded on [a, b], such > that f' is discontinuous on a set of positive measure, which implies > f' is not Riemann integrable on [a, b]. OK, but I read from here that the stieljes integral is the same than > the > corresponding riemann integral when the measure is of bounded > variation and everywhere differentiable. >http://knowledgerush.com/kr/encyclopedia/Riemann-Stieltjes integral/ That reference is incorrect. Do you see why? I understand that there are indeed functions like this http://en.wikipedia.org/wiki/Volterra%27s function === Subject: Re: fourier series of f ' relative to f <9c4ht4pd2op8ib24gbge1mdkslkihj0804@4ax.com> posting-account=iyXB5AkAAABPCtkDKhRJOsNWmafzHSRE Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > What is the theorem that can allow me to justify moving > differentiation operator inside infinite sums (or infinite > limits)? I think you need absolute convergence of the original series. No, you need uniform convergence of the original series, and > also uniform convergence of the series of derivatives. If your >periodic function is continuosly differentiable, in other words >belong >to C^1, then you have it. Yes, C^1 implies the Fourier series for f converges absolutely. > But that's not enough to say the derivative of the sum is the > sum of the derivatives. C^1 does not imply that the > differentiated series converges uniformly (precisely because > there exist continuous functions with divergent Fourier > series; take f so that f' is continuous with a divergent > Fourier series.) A condition known as C^(1 + epsilon) is enough. But > (i) that doesn't answer the OP's question, and (ii) > the actual solution is very simple, just using the > definition of the Fourier coefficients instead of > trying to differentiate the series. The basic idea is simple, but I don't see how to follow through > without measure theory. Maybe you forgot that the derivative f` was bounded. By mean value > theorem for any (a,b) on [-pii,pii], there is a c on (a,b) such that > f`(c)(b - a) = f(b) - f(a) so when you take sup |f`(c)| on [-pii,pii] > you obtain bound sup |f`(c)|(b - a) for all of the variations. So I understand that the riemann integral of f`(x)dx does exist, > because the stieljes integral of d(f(x) is the same. But I don`t know > how to prove it with elementary methods. No, the Riemann integral of f' need not exist. There exist > differentiable functions f on [a, b], with f' bounded on [a, b], such > that f' is discontinuous on a set of positive measure, which implies > f' is not Riemann integrable on [a, b]. OK, but I read from here that the stieljes integral is the same than > the > corresponding riemann integral when the measure is of bounded > variation and everywhere differentiable. >http://knowledgerush.com/kr/encyclopedia/Riemann-Stieltjes integral/ That reference is incorrect. Do you see why? No. === Subject: Re: fourier series of f ' relative to f <9c4ht4pd2op8ib24gbge1mdkslkihj0804@4ax.com> posting-account=iyXB5AkAAABPCtkDKhRJOsNWmafzHSRE Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > What is the theorem that can allow me to justify moving > differentiation operator inside infinite sums (or infinite > limits)? I think you need absolute convergence of the original series. No, you need uniform convergence of the original series, and > also uniform convergence of the series of derivatives. If your >periodic function is continuosly differentiable, in other words >belong >to C^1, then you have it. Yes, C^1 implies the Fourier series for f converges absolutely. > But that's not enough to say the derivative of the sum is the > sum of the derivatives. C^1 does not imply that the > differentiated series converges uniformly (precisely because > there exist continuous functions with divergent Fourier > series; take f so that f' is continuous with a divergent > Fourier series.) A condition known as C^(1 + epsilon) is enough. But > (i) that doesn't answer the OP's question, and (ii) > the actual solution is very simple, just using the > definition of the Fourier coefficients instead of > trying to differentiate the series. The basic idea is simple, but I don't see how to follow through > without measure theory. Maybe you forgot that the derivative f` was bounded. By mean value > theorem for any (a,b) on [-pii,pii], there is a c on (a,b) such that > f`(c)(b - a) = f(b) - f(a) so when you take sup |f`(c)| on [-pii,pii] > you obtain bound sup |f`(c)|(b - a) for all of the variations. So I understand that the riemann integral of f`(x)dx does exist, > because the stieljes integral of d(f(x) is the same. But I don`t know > how to prove it with elementary methods. No, the Riemann integral of f' need not exist. There exist > differentiable functions f on [a, b], with f' bounded on [a, b], such > that f' is discontinuous on a set of positive measure, which implies > f' is not Riemann integrable on [a, b]. OK, but I read from here that the stieljes integral is the same than > the > corresponding riemann integral when the measure is of bounded > variation and everywhere differentiable. >http://knowledgerush.com/kr/encyclopedia/Riemann-Stieltjes integral/ That reference is incorrect. Do you see why? No. Or maybe it is a crappy encyklopedia, anyway I will go to sleep now. === Subject: Re: fourier series of f ' relative to f <9c4ht4pd2op8ib24gbge1mdkslkihj0804@4ax.com> posting-account=iyXB5AkAAABPCtkDKhRJOsNWmafzHSRE Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > What is the theorem that can allow me to justify moving > differentiation operator inside infinite sums (or infinite > limits)? I think you need absolute convergence of the original series. No, you need uniform convergence of the original series, and > also uniform convergence of the series of derivatives. If your >periodic function is continuosly differentiable, in other words >belong >to C^1, then you have it. Yes, C^1 implies the Fourier series for f converges absolutely. > But that's not enough to say the derivative of the sum is the > sum of the derivatives. C^1 does not imply that the > differentiated series converges uniformly (precisely because > there exist continuous functions with divergent Fourier > series; take f so that f' is continuous with a divergent > Fourier series.) A condition known as C^(1 + epsilon) is enough. But > (i) that doesn't answer the OP's question, and (ii) > the actual solution is very simple, just using the > definition of the Fourier coefficients instead of > trying to differentiate the series. The basic idea is simple, but I don't see how to follow through > without measure theory. Maybe you forgot that the derivative f` was bounded. By mean value > theorem for any (a,b) on [-pii,pii], there is a c on (a,b) such that > f`(c)(b - a) = f(b) - f(a) so when you take sup |f`(c)| on [-pii,pii] > you obtain bound sup |f`(c)|(b - a) for all of the variations. So I understand that the riemann integral of f`(x)dx does exist, > because the stieljes integral of d(f(x) is the same. But I don`t know > how to prove it with elementary methods. No, the Riemann integral of f' need not exist. There exist > differentiable functions f on [a, b], with f' bounded on [a, b], such > that f' is discontinuous on a set of positive measure, which implies > f' is not Riemann integrable on [a, b]. OK, but I read from here that the stieljes integral is the same than > the > corresponding riemann integral when the measure is of bounded > variation and everywhere differentiable. >http://knowledgerush.com/kr/encyclopedia/Riemann-Stieltjes integral/ That reference is incorrect. Do you see why? No. === Subject: Re: fourier series of f ' relative to f <9c4ht4pd2op8ib24gbge1mdkslkihj0804@4ax.com> posting-account=iyXB5AkAAABPCtkDKhRJOsNWmafzHSRE Gecko/2009032609 Firefox/3.0.8 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > What is the theorem that can allow me to justify moving > differentiation operator inside infinite sums (or infinite limits)? I think you need absolute convergence of the original series. No, you need uniform convergence of the original series, and > _also_ uniform convergence of the series of derivatives. If your >periodic function is continuosly differentiable, in other words belong >to C^1, then you have it. Yes, C^1 implies the Fourier series for f converges absolutely. > But that's not enough to say the derivative of the sum is the > sum of the derivatives. C^1 does _not_ imply that the > differentiated series converges uniformly (precisely because > there exist continuous functions with divergent Fourier > series; take f so that f' is continuous with a divergent > Fourier series.) A condition known as C^(1 + epsilon) is enough. But > (i) that doesn't answer the OP's question, and (ii) > the actual solution is very simple, just using the > definition of the Fourier coefficients instead of > trying to differentiate the series. OK. I didn`t try to answer the original question. Solution that I now have in mind is: integral -pii^pii (f`(t)e^-itn dt)*e^inx*1/2pii = integral -pii^pii(f`(t)e^in(x-t)*dt)*1/2pii = integral -pii^pii (f`(y) e^-iny dy)*1/2pii by chance of variables and translation invariance. By integration by parts -pii/pii f(y)e^-iny - integral -pii^pii (f`(y) e^-iny dy) = integral -pii^pii - 1/(i*n)*(f(y)e^-iny dy)*1/2pii. Because (-pii/pii f(y)e^-iny) = 0, we have the result. === Subject: Re: fourier series of f ' relative to f posting-account=iyXB5AkAAABPCtkDKhRJOsNWmafzHSRE Gecko/2009032609 Firefox/3.0.8 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > What is the theorem that can allow me to justify moving > differentiation operator inside infinite sums (or infinite limits)? I think you need absolute convergence of the original series. No, you need uniform convergence of the original series, and > also uniform convergence of the series of derivatives. If your >periodic function is continuosly differentiable, in other words belong >to C^1, then you have it. Yes, C^1 implies the Fourier series for f converges absolutely. > But that's not enough to say the derivative of the sum is the > sum of the derivatives. C^1 does not imply that the > differentiated series converges uniformly (precisely because > there exist continuous functions with divergent Fourier > series; take f so that f' is continuous with a divergent > Fourier series.) A condition known as C^(1 + epsilon) is enough. But > (i) that doesn't answer the OP's question, and (ii) > the actual solution is very simple, just using the > definition of the Fourier coefficients instead of > trying to differentiate the series. OK. I didn`t try to answer the original question. Solution that I now > have in mind is: integral -pii^pii (f`(t)e^-itn dt)*e^inx*1/2pii = > integral -pii^pii(f`(t)e^in(x-t)*dt)*1/2pii = integral -pii^pii (f`(y) > e^-iny dy)*1/2pii by chance of variables and translation invariance. > By integration by parts -pii/pii f(y)e^-iny - integral -pii^pii (f`(y) > e^-iny dy) = integral -pii^pii > - 1/(i*n)*(f(y)e^-iny dy)*1/2pii. Because (-pii/pii f(y)e^-iny) = 0, > we have the result. proof. What I mean was just by integration by parts -pii/pii f(y)e^-iny - integral -pii^pii (f`(y) e^-iny dy) = integral -pii^pii -(i*n)*(f(y)e^-iny dy). Because (-pii/ pii f(y)e^-iny) = 0, we have the result when we multiply by - 1/2pii * e^-inx. === Subject: Re: fourier series of f ' relative to f > What is the theorem that can allow me to justify moving > differentiation operator inside infinite sums (or infinite limits)? I think you need absolute convergence of the original series. No, you need uniform convergence of the original series, and > _also_ uniform convergence of the series of derivatives. If your >periodic function is continuosly differentiable, in other words belong >to C^1, then you have it. Yes, C^1 implies the Fourier series for f converges absolutely. > But that's not enough to say the derivative of the sum is the > sum of the derivatives. C^1 does _not_ imply that the > differentiated series converges uniformly (precisely because > there exist continuous functions with divergent Fourier > series; take f so that f' is continuous with a divergent > Fourier series.) A condition known as C^(1 + epsilon) is enough. But > (i) that doesn't answer the OP's question, and (ii) > the actual solution is very simple, just using the > definition of the Fourier coefficients instead of > trying to differentiate the series. OK. I didn`t try to answer the original question. Solution that I now > have in mind is: integral -pii^pii (f`(t)e^-itn dt)*e^inx*1/2pii = > integral -pii^pii(f`(t)e^in(x-t)*dt)*1/2pii = integral -pii^pii (f`(y) > e^-iny dy)*1/2pii by chance of variables and translation invariance. > By integration by parts -pii/pii f(y)e^-iny - integral -pii^pii (f`(y) > e^-iny dy) = integral -pii^pii > - 1/(i*n)*(f(y)e^-iny dy)*1/2pii. Because (-pii/pii f(y)e^-iny) = 0, > we have the result. proof. What I mean was just > by integration by parts -pii/pii f(y)e^-iny - integral -pii^pii (f`(y) > e^-iny dy) = integral -pii^pii -(i*n)*(f(y)e^-iny dy). Because (-pii/ > pii f(y)e^-iny) = 0, > we have the result when we multiply by - 1/2pii * e^-inx. Your notation really sucks. More importantly, why do you think the integration by parts is justified? === Subject: Re: fourier series of f ' relative to f posting-account=iyXB5AkAAABPCtkDKhRJOsNWmafzHSRE Gecko/2009032609 Firefox/3.0.8 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > What is the theorem that can allow me to justify moving > differentiation operator inside infinite sums (or infinite limits)? I think you need absolute convergence of the original series. No, you need uniform convergence of the original series, and > also uniform convergence of the series of derivatives. If your >periodic function is continuosly differentiable, in other words belong >to C^1, then you have it. Yes, C^1 implies the Fourier series for f converges absolutely. > But that's not enough to say the derivative of the sum is the > sum of the derivatives. C^1 does not imply that the > differentiated series converges uniformly (precisely because > there exist continuous functions with divergent Fourier > series; take f so that f' is continuous with a divergent > Fourier series.) A condition known as C^(1 + epsilon) is enough. But > (i) that doesn't answer the OP's question, and (ii) > the actual solution is very simple, just using the > definition of the Fourier coefficients instead of > trying to differentiate the series. OK. I didn`t try to answer the original question. Solution that I now > have in mind is: integral -pii^pii (f`(t)e^-itn dt)*e^inx*1/2pii = > integral -pii^pii(f`(t)e^in(x-t)*dt)*1/2pii = integral -pii^pii (f`(y) > e^-iny dy)*1/2pii by chance of variables and translation invariance. > By integration by parts -pii/pii f(y)e^-iny - integral -pii^pii (f`(y) > e^-iny dy) = integral -pii^pii > - 1/(i*n)*(f(y)e^-iny dy)*1/2pii. Because (-pii/pii f(y)e^-iny) = 0, > we have the result. proof. What I mean was just > by integration by parts -pii/pii f(y)e^-iny - integral -pii^pii (f`(y) > e^-iny dy) = integral -pii^pii -(i*n)*(f(y)e^-iny dy). Because (-pii/ > pii f(y)e^-iny) = 0, > we have the result when we multiply by - 1/2pii * e^-inx. Your notation really sucks. More importantly, why do you think the > integration by parts is justified? I just differentiate f(y)e^-iny by the product rule. So I got d/dy(f (y)e^-iny) = -(i*n)*(f(y)e^-iny) + f`(y)e^-iny. Then I integrate the hole thing from -pii to pii. What is wrong with this? === Subject: Re: fourier series of f ' relative to f posting-account=iyXB5AkAAABPCtkDKhRJOsNWmafzHSRE 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > What is the theorem that can allow me to justify moving > differentiation operator inside infinite sums (or infinite limits)? >I think you need absolute convergence of the original series. > No, you need uniform convergence of the original series, and > also uniform convergence of the series of derivatives. >If your >periodic function is continuosly differentiable, in other words belong >to C^1, then you have it. > Yes, C^1 implies the Fourier series for f converges absolutely. > But that's not enough to say the derivative of the sum is the > sum of the derivatives. C^1 does not imply that the > differentiated series converges uniformly (precisely because > there exist continuous functions with divergent Fourier > series; take f so that f' is continuous with a divergent > Fourier series.) > A condition known as C^(1 + epsilon) is enough. But > (i) that doesn't answer the OP's question, and (ii) > the actual solution is very simple, just using the > definition of the Fourier coefficients instead of > trying to differentiate the series. > OK. I didn`t try to answer the original question. Solution that I now > have in mind is: integral -pii^pii (f`(t)e^-itn dt)*e^inx*1/2pii = > integral -pii^pii(f`(t)e^in(x-t)*dt)*1/2pii = integral -pii^pii (f`(y) > e^-iny dy)*1/2pii by chance of variables and translation invariance. > By integration by parts -pii/pii f(y)e^-iny - integral -pii^pii (f`(y) > e^-iny dy) = integral -pii^pii > - 1/(i*n)*(f(y)e^-iny dy)*1/2pii. Because (-pii/pii f(y)e^-iny) = 0, > we have the result. > proof. What I mean was just > by integration by parts -pii/pii f(y)e^-iny - integral -pii^pii (f`(y) > e^-iny dy) = integral -pii^pii -(i*n)*(f(y)e^-iny dy). Because (-pii/ > pii f(y)e^-iny) = 0, > we have the result when we multiply by - 1/2pii * e^-inx. > Your notation really sucks. More importantly, why do you think the > integration by parts is justified? I just differentiate f(y)e^-iny by the product rule. So I got d/dy(f >(y)e^-iny) = -(i*n)*(f(y)e^-iny) + f`(y)e^-iny. Then I integrate the >hole thing from -pii to pii. What is wrong with this? When you just do that you're using theorems. Those theorems have > hypotheses - the question is why are the hypotheses satisfied. In particular: Integration by parts is just the product rule > for derivatives plus the fundamental theorem of calculus. > But FTC has hypotheses - it's not true, if we're talking > about the Riemann integral, that f differentiable implies > that f(a) - f(b) is the integral of f' from a to b (in fact > f' need not even be Riemann inetegrable). OK. I have used (properly?) the definition of FTC from from Wikipedia. http://en.wikipedia.org/wiki/Fundamental theorem of calculus (Second part)It only assumes real valued function f` on [a,b] and existence of the antiderivative. === Subject: Re: fourier series of f ' relative to f > What is the theorem that can allow me to justify moving > differentiation operator inside infinite sums (or infinite > limits)? >I think you need absolute convergence of the original series. > No, you need uniform convergence of the original series, and > _also_ uniform convergence of the series of derivatives. >If your >periodic function is continuosly differentiable, in other words >belong >to C^1, then you have it. > Yes, C^1 implies the Fourier series for f converges absolutely. > But that's not enough to say the derivative of the sum is the > sum of the derivatives. C^1 does _not_ imply that the > differentiated series converges uniformly (precisely because > there exist continuous functions with divergent Fourier > series; take f so that f' is continuous with a divergent > Fourier series.) > A condition known as C^(1 + epsilon) is enough. But > (i) that doesn't answer the OP's question, and (ii) > the actual solution is very simple, just using the > definition of the Fourier coefficients instead of > trying to differentiate the series. > OK. I didn`t try to answer the original question. Solution that I > now > have in mind is: integral -pii^pii (f`(t)e^-itn dt)*e^inx*1/2pii = > integral -pii^pii(f`(t)e^in(x-t)*dt)*1/2pii = integral -pii^pii > (f`(y) > e^-iny dy)*1/2pii by chance of variables and translation invariance. > By integration by parts -pii/pii f(y)e^-iny - integral -pii^pii > (f`(y) > e^-iny dy) = integral -pii^pii > - 1/(i*n)*(f(y)e^-iny dy)*1/2pii. Because (-pii/pii f(y)e^-iny) = 0, > we have the result. > proof. What I mean was just > by integration by parts -pii/pii f(y)e^-iny - integral -pii^pii (f`(y) > e^-iny dy) = integral -pii^pii -(i*n)*(f(y)e^-iny dy). Because (-pii/ > pii f(y)e^-iny) = 0, > we have the result when we multiply by - 1/2pii * e^-inx. > Your notation really sucks. More importantly, why do you think the > integration by parts is justified? I just differentiate f(y)e^-iny by the product rule. So I got d/dy(f >(y)e^-iny) = -(i*n)*(f(y)e^-iny) + f`(y)e^-iny. Then I integrate the >hole thing from -pii to pii. What is wrong with this? When you just do that you're using theorems. Those theorems have > hypotheses - the question is why are the hypotheses satisfied. In particular: Integration by parts is just the product rule > for derivatives plus the fundamental theorem of calculus. > But FTC has hypotheses - it's not true, if we're talking > about the Riemann integral, that f differentiable implies > that f(a) - f(b) is the integral of f' from a to b (in fact > f' need not even be Riemann inetegrable). OK. I have used (properly?) the definition of FTC from from > Wikipedia. > http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus > (Second part)It only assumes real valued function f` on [a,b] and > existence of the antiderivative. No, read again. It also assumes f is integrable on [a, b]. Assuming integrable means Riemann integrable, that assumption may not be satisfied in our problem. === Subject: Re: fourier series of f ' relative to f posting-account=iyXB5AkAAABPCtkDKhRJOsNWmafzHSRE 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) What is the theorem that can allow me to justify moving > differentiation operator inside infinite sums (or infinite limits)? >I think you need absolute convergence of the original series. > No, you need uniform convergence of the original series, and > also uniform convergence of the series of derivatives. >If your >periodic function is continuosly differentiable, in other words belong >to C^1, then you have it. > Yes, C^1 implies the Fourier series for f converges absolutely. > But that's not enough to say the derivative of the sum is the > sum of the derivatives. C^1 does not imply that the > differentiated series converges uniformly (precisely because > there exist continuous functions with divergent Fourier > series; take f so that f' is continuous with a divergent > Fourier series.) > A condition known as C^(1 + epsilon) is enough. But > (i) that doesn't answer the OP's question, and (ii) > the actual solution is very simple, just using the > definition of the Fourier coefficients instead of > trying to differentiate the series. > OK. I didn`t try to answer the original question. Solution that I now > have in mind is: integral -pii^pii (f`(t)e^-itn dt)*e^inx*1/2pii = > integral -pii^pii(f`(t)e^in(x-t)*dt)*1/2pii = integral -pii^pii (f`(y) > e^-iny dy)*1/2pii by chance of variables and translation invariance. > By integration by parts -pii/pii f(y)e^-iny - integral -pii^pii (f`(y) > e^-iny dy) = integral -pii^pii > - 1/(i*n)*(f(y)e^-iny dy)*1/2pii. Because (-pii/pii f(y)e^-iny) = 0, > we have the result. > proof. What I mean was just > by integration by parts -pii/pii f(y)e^-iny - integral -pii^pii (f`(y) > e^-iny dy) = integral -pii^pii -(i*n)*(f(y)e^-iny dy). Because (-pii/ > pii f(y)e^-iny) = 0, > we have the result when we multiply by - 1/2pii * e^-inx. > Your notation really sucks. More importantly, why do you think the > integration by parts is justified? I just differentiate f(y)e^-iny by the product rule. So I got d/dy(f >(y)e^-iny) = -(i*n)*(f(y)e^-iny) + f`(y)e^-iny. Then I integrate the >hole thing from -pii to pii. What is wrong with this? When you just do that you're using theorems. Those theorems have > hypotheses - the question is why are the hypotheses satisfied. In particular: Integration by parts is just the product rule > for derivatives plus the fundamental theorem of calculus. > But FTC has hypotheses - it's not true, if we're talking > about the Riemann integral, that f differentiable implies > that f(a) - f(b) is the integral of f' from a to b (in fact > f' need not even be Riemann inetegrable). OK. I have used (properly?) the definition of FTC from from > Wikipedia.http://en.wikipedia.org/wiki/Fundamental theorem of calculus > (Second part)It only assumes real valued function f` on [a,b] and > existence of the antiderivative.- Piilota siteerattu teksti - - N.8ayt.8a siteerattu teksti - I haven`t noticed before that this is not the form taught together with the Riemann integral. === Subject: A variant on the Secret Santa I've been given an intriguing puzzle, and it's caught me by surprise. I thought the answer was obvious, but with considerable work I now find it's obviously not obvious. You may disagree with what's obvious and what's not, but here it is. A group of 10 (for some value of 10) people arrange a secret santa. Knowing the usual probability of there being a fixed point in a random permutation, they try to reduce this by using the following procedure: Each of the first nine chooses a name from the hat. If it's their own, they lay it to one side, select another, then replace their own. Given the above procedure, what is the probability that the 10th (for some value of 10) person finds their own name in the hat at the end? I thought this would simply be !9/(!9+!10), where !n represents the number of derangements of n objects. Amusingly, this probability is about 1/(n+1), and I would be interested to hear a heuristic argument as to why this should be so. However, the probability for the given problem is *not* 1/11 because the derangements are not equally likely. So the questions are: * Heuristically, why is !(n-1)/(!(n-1)+!n) = 1/(n+1) * What is the correct answer as 10 approaches infinity? * Is there an exact formula for the given process? === Subject: Re: A variant on the Secret Santa posting-account=oTDIagkAAACTxHurtPutBWvNQS8ZCNO9 Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) On Apr 5, 10:57am, riderofgiraffes me by surprise. I thought the answer was obvious, but > with considerable work I now find it's obviously not > obvious. You may disagree with what's obvious and > what's not, but here it is. A group of 10 (for some value of 10) people arrange a > secret santa. Knowing the usual probability of there > being a fixed point in a random permutation, they try > to reduce this by using the following procedure: Each of the first nine chooses a name from the hat. > If it's their own, they lay it to one side, select > another, then replace their own. Given the above procedure, what is the probability that > the 10th (for some value of 10) person finds their own > name in the hat at the end? I thought this would simply be !9/(!9+!10), where !n > represents the number of derangements of n objects. > Amusingly, this probability is about 1/(n+1), and I > would be interested to hear a heuristic argument as > to why this should be so. However, the probability for the given problem is *not* > 1/11 because the derangements are not equally likely. So the questions are: * Heuristically, why is !(n-1)/(!(n-1)+!n) = 1/(n+1) > * What is the correct answer as 10 approaches infinity? > * Is there an exact formula for the given process? Interesting problem. I also made the same mistake you did at first. All I can say so far is that for large n, !n/n! ~ 1/e implies that for large n, !n/n! ~ !(n-1)/(n-1)! !n/!(n-1) ~ n!/(n-1)! = n which gives the asymptotic result you noticed. === Subject: Re: A variant on the Secret Santa >On Apr 5, 10:57am, riderofgiraffes I've been given an intriguing puzzle, and \ it's caught > me by surprise. I thought the answer was obvious, but > with considerable work I now find it's obviously not > obvious. You may disagree with what's obvious and > what's not, but here it is. > A group of 10 (for some value of 10) people arrange a > secret santa. Knowing the usual probability of there > being a fixed point in a random permutation, they try > to reduce this by using the following procedure: > Each of the first nine chooses a name from the hat. > If it's their own, they lay it to one side, select > another, then replace their own. > Given the above procedure, what is the probability that > the 10th (for some value of 10) person finds their own > name in the hat at the end? > I thought this would simply be !9/(!9+!10), where !n > represents the number of derangements of n objects. > Amusingly, this probability is about 1/(n+1), and I > would be interested to hear a heuristic argument as > to why this should be so. > However, the probability for the given problem is *not* > 1/11 because the derangements are not equally likely. > So the questions are: > * Heuristically, why is !(n-1)/(!(n-1)+!n) = 1/(n+1) > * What is the correct answer as 10 approaches infinity? > * Is there an exact formula for the given process? >Interesting problem. I also made the same mistake you did at first. >All I can say so far is that for large n, !n/n! ~ 1/e implies that for large n, !n/n! ~ !(n-1)/(n-1)! >!n/!(n-1) ~ n!/(n-1)! = n which gives the asymptotic result you noticed. For n = 10, the exact probability is 160127 / 2116800 which is approximately 1/13, not 1/11 as previously suggested. For large n, a little testing suggests that a good asymptotic estimate is might be 1/(n+sqrt(n)). quasi === Subject: Re: A variant on the Secret Santa >On Apr 5, 10:57am, riderofgiraffes me by surprise. I thought the answer was obvious, but > with considerable work I now find it's obviously not > obvious. You may disagree with what's obvious and > what's not, but here it is. A group of 10 (for some value of 10) people arrange a > secret santa. Knowing the usual probability of there > being a fixed point in a random permutation, they try > to reduce this by using the following procedure: Each of the first nine chooses a name from the hat. > If it's their own, they lay it to one side, select > another, then replace their own. Given the above procedure, what is the probability that > the 10th (for some value of 10) person finds their own > name in the hat at the end? I thought this would simply be !9/(!9+!10), where !n > represents the number of derangements of n objects. > Amusingly, this probability is about 1/(n+1), and I > would be interested to hear a heuristic argument as > to why this should be so. However, the probability for the given problem is *not* > 1/11 because the derangements are not equally likely. So the questions are: * Heuristically, why is !(n-1)/(!(n-1)+!n) = 1/(n+1) > * What is the correct answer as 10 approaches infinity? > * Is there an exact formula for the given process? >Interesting problem. I also made the same mistake you did at first. >All I can say so far is that for large n, >!n/n! ~ 1/e >implies that for large n, >!n/n! ~ !(n-1)/(n-1)! >!n/!(n-1) ~ n!/(n-1)! = n >which gives the asymptotic result you noticed. For n = 10, the exact probability is 160127 / 2116800 which is approximately 1/13, not 1/11 as previously suggested. For large n, a little testing suggests that a good asymptotic estimate >is might be 1/(n+sqrt(n)). Correction: For large n, a little testing suggests that a good asymptotic estimate is 1/(n+ln(n)). quasi === Subject: Re: A variant on the Secret Santa <76tkt49un5sf1af2k4bd8m6fbbr71qm1t2@4ax.com> <6fukt4hfrr12u8f1h5ebds5i2d4pm0rvc9@4ax.com> posting-account=U2cMXgoAAADa7uFaEtrTh3s-noWr2c49 Browser; Avant Browser; .NET CLR 2.0.50727; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) [ Original problem clipped to save bandwidth ] For n = 10, the exact probability is 160127 / 2116800 which is approximately 1/13, not 1/11 as previously suggested. For large n, a little testing suggests that a good asymptotic estimate >is might be 1/(n+sqrt(n)). Correction: For large n, a little testing suggests that a good asymptotic estimate > is 1/(n+ln(n)). quasi- Hide quoted text - - Show quoted text - Perhaps an approximation will help. Person 1 will pick number n with probability 1/n-1 . If Person 1 picked i for i < n instead, then Person i will pick n (if n has not already been chosen) with probability 1/(n-i+1). If no one has chosen i before Person i picks, then Person i will pick n with probability 1/(n-i). The desired probability then lies between two expressions, the lower assuming the lesser probability at each stage, and similarly for the higher. This gives the two bounds on the probability for Person n not picking n as: 1/(n-1) + ((n-2)/(n-1))*{1/(n-1) + ((n-2)/(n-1))*[1/(n-2) + ...]} and 1/(n-1) + ((n-2)/(n-1))*{1/(n-2) + ((n-3)/(n-2))*[1/(n-3) + ...]} (I leave filling out each ellipsis and the actual computations to someone with a symbolic computation package.) I guess that both bounds are close to the expected number of 1 - 1/(n+1). Gerhard Ask Me About System Design Paseman, 2009.04.06 === Subject: Re: A variant on the Secret Santa On Apr 5, 10:57am, riderofgiraffes I've been given an intriguing puzzle, and \ it's caught > me by surprise. I thought the answer was obvious, but > with considerable work I now find it's obviously not > obvious. You may disagree with what's obvious and > what's not, but here it is. > A group of 10 (for some value of 10) people arrange a > secret santa. Knowing the usual probability of there > being a fixed point in a random permutation, they try > to reduce this by using the following procedure: > Each of the first nine chooses a name from the hat. > If it's their own, they lay it to one side, select > another, then replace their own. > Given the above procedure, what is the probability that > the 10th (for some value of 10) person finds their own > name in the hat at the end? > I thought this would simply be !9/(!9+!10), where !n > represents the number of derangements of n objects. > Amusingly, this probability is about 1/(n+1), and I > would be interested to hear a heuristic argument as > to why this should be so. > However, the probability for the given problem is *not* > 1/11 because the derangements are not equally likely. > So the questions are: > * Heuristically, why is !(n-1)/(!(n-1)+!n) = 1/(n+1) > * What is the correct answer as 10 approaches infinity? > * Is there an exact formula for the given process? >Interesting problem. I also made the same mistake you did at first. >All I can say so far is that for large n, !n/n! ~ 1/e implies that for large n, !n/n! ~ !(n-1)/(n-1)! >!n/!(n-1) ~ n!/(n-1)! = n which gives the asymptotic result you noticed. >For n = 10, the exact probability is > 160127 / 2116800 >which is approximately 1/13, not 1/11 as previously suggested. >For large n, a little testing suggests that a good asymptotic estimate >is might be 1/(n+sqrt(n)). Correction: For large n, a little testing suggests that a good asymptotic estimate >is 1/(n+ln(n)). Looking at a little more data, now I'm not so sure. It might actually be 1/(n+c) for some constant c. If so, the limiting c appears to be between 3 and 4. quasi === Subject: Random choice of numbers I have problems with writing down a formal solution of the following problem: Two boys - Chris & Michael - are playing the following game: Both of them are selecting randomly numbers from [0,1] interval (with uniform distribution). Chris is taking two numbers (each one is independent of another), and Michael is taking only one number (once again this is independent event). Next, Chris is multiplying both of his numbers, and Michael is calculating a square of his number. The boy with a larger result \ is a winner. Who has more chances of winning this game - Chris or Michael? For me this is obvious, that Michael is the one. (It's important that they are randomly selecting numbers from [0,1]!) But how we can prove this result in a formal way? Dante === Subject: Re: Random choice of numbers As a followup, here are some questions relating to the generalized form ... For n in N, n > 2, let p(n) be the probability that if n numbers are chosen at random from the interval (0,1), the (n-1)'th power of the last number chosen is greater than the product of the first (n-1) numbers. Based on results in this thread, we have p(3) = 5/9. (1) Prove or disprove: p(n) is rational, for all n. Assuming (1) is true, write p(n) = a(n)/b(n), as a fraction in lowest terms, and consider the following two problems ... (2) Prove or disprove: b(n) = n^(n-1), for all n. (3) Prove or disprove: a(n) is always composite, unless n = 3,4,6,8. Also, here's a variant ... For n in N, n > 2, let q(n) be the probability that if n numbers are chosen at random from the interval (0,1), the (n-1)'th power of the median is greater than the product of the other (n-1) numbers. One might intuitively guess that q(3) = p(3) = 5/9. But that's not the case. In fact, it's easy to show that q(3) = 2/3. A few more problems ... (4) Find q(4). (5) Prove or disprove: q(n) is rational, for all n. quasi === Subject: Re: Random choice of numbers > As a followup, here are some questions relating to the generalized > form ... For n in N, n > 2, let p(n) be the probability that if n numbers are > chosen at random from the interval (0,1), the (n-1)'th power of the > last number chosen is greater than the product of the first (n-1) > numbers. Based on results in this thread, we have p(3) = 5/9. (1) Prove or disprove: p(n) is rational, for all n. As geo noted, we can consider X_j = exp(-T_j) for independent exponential random variables T_j with rate 1, and p(n) = P{(n-1) T_n < sum_{j=1}^{n-1} T_j}. Now sum_{j=1}^{n-1} T_j has a Gamma distribution with shape parameter n-1 and scale parameter 1, so p(n) = int_0^infty x^(n-2) exp(-x)/(n-2)! (1 - exp(-x/(n-1))) dx = 1 - ((n-1)/n)^(n-1) > Assuming (1) is true, write p(n) = a(n)/b(n), as a fraction in lowest > terms, and consider the following two problems ... (2) Prove or disprove: b(n) = n^(n-1), for all n. See above. Note gcd(n-1,n) = 1. > (3) Prove or disprove: a(n) is always composite, unless n = 3,4,6,8. a(n) = n^(n-1) - (n-1)^(n-1). This is also prime for n=168. Of course if n-1 is divisible by k, a(n) is divisible by n^k - (n-1)^k. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Random choice of numbers > As a followup, here are some questions relating to the generalized > form ... > For n in N, n > 2, let p(n) be the probability that if n numbers are > chosen at random from the interval (0,1), the (n-1)'th power of the > last number chosen is greater than the product of the first (n-1) > numbers. > Based on results in this thread, we have p(3) = 5/9. > (1) Prove or disprove: p(n) is rational, for all n. As geo noted, we can consider X_j = exp(-T_j) for independent exponential >random variables T_j with rate 1, and >p(n) = P{(n-1) T_n < sum_{j=1}^{n-1} T_j}. >Now sum_{j=1}^{n-1} T_j has a Gamma distribution with shape parameter n-1 and >scale parameter 1, so p(n) = int_0^infty x^(n-2) exp(-x)/(n-2)! (1 - exp(-x/(n-1))) dx > = 1 - ((n-1)/n)^(n-1) Ah, a nice closed form. I was just looking at the data, but still, I should have recogonized that pattern, especially since I already had the denominator. > Assuming (1) is true, write p(n) = a(n)/b(n), as a fraction in lowest > terms, and consider the following two problems ... > (2) Prove or disprove: b(n) = n^(n-1), for all n. See above. Note gcd(n-1,n) = 1. Right -- the closed form makes it instant. > (3) Prove or disprove: a(n) is always composite, unless n = 3,4,6,8. a(n) = n^(n-1) - (n-1)^(n-1). This is also prime for n=168. Of course if n-1 is divisible by k, a(n) is divisible by n^k - (n-1)^k. Right -- so a necessary condition for a(n) to be prime is that n-1 is prime. But given the nature of the closed form, it seems likely that a(n) may in fact be prime infinitely often, though proving or disproving that may be too hard based for today's math. quasi === Subject: Re: Random choice of numbers >As a followup, here are some questions relating to the generalized >form ... For n in N, n > 2, let p(n) be the probability that if n numbers are >chosen at random from the interval (0,1), the (n-1)'th power of the >last number chosen is greater than the product of the first (n-1) >numbers. Based on results in this thread, we have p(3) = 5/9. (1) Prove or disprove: p(n) is rational, for all n. Assuming (1) is true, write p(n) = a(n)/b(n), as a fraction in lowest >terms, and consider the following two problems ... (2) Prove or disprove: b(n) = n^(n-1), for all n. (3) Prove or disprove: a(n) is always composite, unless n = 3,4,6,8. Also, here's a variant ... For n in N, n > 2, let q(n) be the probability that if n numbers are >chosen at random from the interval (0,1), the (n-1)'th power of the >median is greater than the product of the other (n-1) numbers. For this variant, I meant to assume that n is odd. So assume that. >One might intuitively guess that q(3) = p(3) = 5/9. But that's not the case. In fact, it's easy to show that q(3) = 2/3. A few more problems ... (4) Find q(4). Correction (since we are assuming n is odd): (4) Find q(5). >(5) Prove or disprove: q(n) is rational, for all n. quasi === Subject: Re: Random choice of numbers For the variant, here's a natural revision which eliminates the parity restriction on n ... For n in N, n > 2, let q(n) be the probability that if n numbers are chosen at random from the interval (0,1), the n'th power of the median is greater than the product of the n numbers. One might intuitively guess that q(3) = p(3) = 5/9. But that's not the case. In fact, it's easy to show that q(3) = 2/3. A few more problems ... (4) Find q(4) and q(5). (5) Prove or disprove: q(n) is rational, for all n. quasi === Subject: Re: Random choice of numbers <5kujt41hubb3jg1thcu02j0fv8fkmmu4m2@4ax.com> posting-account=SvltewoAAAAi7TTYrD3mAaLUHzDiF2d1 Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > For the variant, here's a natural revision which eliminates the parity > restriction on n ... For n in N, n > 2, let q(n) be the probability that if n numbers are > chosen at random from the interval (0,1), the n'th power of the > median is greater than the product of the n numbers. One might intuitively guess that q(3) = p(3) = 5/9. But that's not the case. In fact, it's easy to show that q(3) = 2/3. A few more problems ... (4) Find q(4) and q(5). (5) Prove or disprove: q(n) is rational, for all n. quasi It seems for odd n = 2k+1, q(n) has the general form (2k+1) C^(2k) k int^{inf} 0 dt exp( -(2k+1)t ) Q(t) where Q(t) is a polynomial in t with rational coefficents. For n = 3, Q(t) is 1 => q(3) = 3 C^2 1 (1/3^2) = 2/3. For n = 5, Q(t) is t^2 (t+1) => q(5) = 5 C^4 2 ( (3!/5^4) + (2!/5^3) ) = 96/125 = 0.768 (q(5) verified by brute force simulation). Still working on the general form of Q(t) for larger k. === Subject: Re: Random choice of numbers > For the variant, here's a natural revision which eliminates the parity > restriction on n ... > For n in N, n > 2, let q(n) be the probability that if n numbers are > chosen at random from the interval (0,1), the n'th power of the > median is greater than the product of the n numbers. > One might intuitively guess that > q(3) = p(3) = 5/9. > But that's not the case. > In fact, it's easy to show that q(3) = 2/3. > A few more problems ... > (4) Find q(4) and q(5). > (5) Prove or disprove: q(n) is rational, for all n. > quasi It seems for odd n = 2k+1, q(n) has the general form (2k+1) C^(2k)_k int^{inf}_0 dt exp( -(2k+1)t ) Q(t) where Q(t) is a polynomial in t with rational coefficents. For n = 3, Q(t) is 1 >=> q(3) = 3 C^2_1 (1/3^2) = 2/3. For n = 5, Q(t) is t^2 (t+1) >=> q(5) = 5 C^4_2 ( (3!/5^4) + (2!/5^3) ) = 96/125 = 0.768 >(q(5) verified by brute force simulation). Nice. >Still working on the general form of Q(t) for larger k. A general form may not be so easily obtained, and if obtainable, may not be so nice. I was just curious about rationality, and about how to actually find an exact answer for a given n. Just a quick toy challenge problem, nothing serious. quasi === Subject: Re: Random choice of numbers <23214088.30045.1238954809767.JavaMail.jakarta@nitrogen.mathforum.org>, > I have problems with writing down a formal solution of the following problem: Two boys - Chris & Michael - are playing the following game: Both of them \ > are selecting randomly numbers from [0,1] interval (with uniform > distribution). Chris is taking two numbers (each one is independent of > another), and Michael is taking only one number (once again this is > independent event). Next, Chris is multiplying both of his numbers, and > Michael is calculating a square of his number. The boy with a larger result > is a winner. Who has more chances of winning this game - Chris or Michael? For me this is obvious, that Michael is the one. (It's important that they \ > are randomly selecting numbers from [0,1]!) But how we can prove this result in a formal way? Isn't it just a question of comparing integral 0 to 1 x^2 dx with integral 0 to 1 integral 0 to 1 x y dy dx? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Random choice of numbers \ , > I have problems with writing down a formal solution of the following problem: > Two boys - Chris & Michael - are playing the following game: Both of them > are selecting randomly numbers from [0,1] interval (with uniform > distribution). Chris is taking two numbers (each one is independent of > another), and Michael is taking only one number (once again this is > independent event). Next, Chris is multiplying both of his numbers, and > Michael is calculating a square of his number. The boy with a larger result > is a winner. Who has more chances of winning this game - Chris or Michael? > For me this is obvious, that Michael is the one. (It's important that they > are randomly selecting numbers from [0,1]!) > But how we can prove this result in a formal way? > Isn't it just a question of comparing integral 0 to 1 x^2 dx > with integral 0 to 1 integral 0 to 1 x y dy dx? No, that's a different question. The question was, who is more likely to win, not whose product has the higher expected value. -- Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. === Subject: Re: Random choice of numbers ><23214088.30045.1238954809767.JavaMail.jakarta@nitrogen.mathforum.org>, > I have problems with writing down a formal solution of the following problem: > Two boys - Chris & Michael - are playing the following game: Both of them > are selecting randomly numbers from [0,1] interval (with uniform > distribution). Chris is taking two numbers (each one is independent of > another), and Michael is taking only one number (once again this is > independent event). Next, Chris is multiplying both of his numbers, and > Michael is calculating a square of his number. The boy with a larger result > is a winner. Who has more chances of winning this game - Chris or Michael? > For me this is obvious, that Michael is the one. (It's important that they > are randomly selecting numbers from [0,1]!) > But how we can prove this result in a formal way? Isn't it just a question of comparing integral 0 to 1 x^2 dx >with integral 0 to 1 integral 0 to 1 x y dy dx? True. Which easily resolves the OP's question. But to calculate the actual probability that Michael wins, that problem is equivalent to the problem of finding the volume in the unit cube above the surface z^2 = xy. quasi === Subject: Re: Random choice of numbers <0lmjt4lbn981iq51fjoskmhp237ssuttp9@4ax.com> posting-account=lMPBJwoAAADEIw-zQg-LUcxd542NXZ4c SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) ><23214088.30045.1238954809767.JavaMail.jaka...@nitrogen.mathforum.org>, I have problems with writing down a formal solution of the following problem: Two boys - Chris & Michael - are playing the following game: Both of them > are selecting randomly numbers from [0,1] interval (with uniform > distribution). Chris is taking two numbers (each one is independent of > another), and Michael is taking only one number (once again this is > independent event). Next, Chris is multiplying both of his numbers, and > Michael is calculating a square of his number. The boy with a larger result > is a winner. Who has more chances of winning this game - Chris or Michael? For me this is obvious, that Michael is the one. (It's important that they > are randomly selecting numbers from [0,1]!) But how we can prove this result in a formal way? >Isn't it just a question of comparing integral 0 to 1 x^2 dx >with integral 0 to 1 integral 0 to 1 x y dy dx? True. Really? How do you show that? Which easily resolves the OP's question. But to calculate the actual probability that Michael wins, that >problem is equivalent to the problem of finding the volume in the unit >cube above the surface z^2 = xy. quasi David C. Ullrich - Show quoted text - If x is standard exponential, then e^(-x) is uniform on [0,1), so the problem may be put in the form of x+y < 2*z for three independent standard exponential variates x,y,z, avoiding the problem of edges in the unit cube. === Subject: Re: Random choice of numbers <23214088.30045.1238954809767.JavaMail.jakarta@nitrogen.mathforum.org>, > I have problems with writing down a formal solution of the following problem: > Two boys - Chris & Michael - are playing the following game: Both of them > are selecting randomly numbers from [0,1] interval (with uniform > distribution). Chris is taking two numbers (each one is independent of > another), and Michael is taking only one number (once again this is > independent event). Next, Chris is multiplying both of his numbers, and > Michael is calculating a square of his number. The boy with a larger result > is a winner. Who has more chances of winning this game - Chris or Michael? > For me this is obvious, that Michael is the one. (It's important that they > are randomly selecting numbers from [0,1]!) > But how we can prove this result in a formal way? Isn't it just a question of comparing integral 0 to 1 x^2 dx >with integral 0 to 1 integral 0 to 1 x y dy dx? >True. Really? How do you show that? I don't. My mistake. As you pointed out in your less sarcastic reply to Gerry, he was comparing expected values. I read it too quickly -- I thought he was comparing the distribution functions. >Which easily resolves the OP's question. >But to calculate the actual probability that Michael wins, that >problem is equivalent to the problem of finding the volume in the unit >cube above the surface z^2 = xy. But I did at least identify a typical method for finding the actual probability. quasi CC: luiroto@yahoo.com, Stephen J. Herschkorn === Subject: Re: Random choice of numbers >I have problems with writing down a formal solution of the following problem: Two boys - Chris & Michael - are playing the following game: Both of them are selecting randomly numbers from [0,1] interval (with uniform distribution). Chris is taking two numbers (each one is independent of another), and Michael is taking only one number (once again this is independent event). Next, Chris is multiplying both of his numbers, and Michael is calculating a square of his number. The boy with a larger result \ is a winner. Who has more chances of winning this game - Chris or Michael? For me this is obvious, that Michael is the one. (It's important that they are randomly selecting numbers from [0,1]!) But how we can prove this result in a formal way? > > Others have shown how to show the probability that Chris wins is 4/9. (Ludovicus's response is incorrect.) Another way to do the computation is via the integral of the uniform density on the unit cube over the region {(x,y,z) in [0,1]^3: xy >= z^2}, which is integral(z = 0..1, y = z^2..1, x = z^2/y..1, 1). -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey === Subject: Re: Random choice of numbers I have problems with writing down a formal solution of the following >problem: Two boys - Chris & Michael - are playing the following game: Both of them >are selecting randomly numbers from [0,1] interval (with uniform >distribution). Chris is taking two numbers (each one is independent of >another), and Michael is taking only one number (once again this is >independent event). Next, Chris is multiplying both of his numbers, and >Michael is calculating a square of his number. The boy with a larger >result is a winner. Who has more chances of winning this game - Chris or >Michael? For me this is obvious, that Michael is the one. (It's important that they >are randomly selecting numbers from [0,1]!) But how we can prove this result in a formal way? > > Others have shown how to show the probability that Chris wins is 4/9. > (Ludovicus's response is incorrect.) Another way to do the > computation is via the integral of the uniform density on the unit cube > over the region {(x,y,z) in [0,1]^3: xy >= z^2}, which is integral(z = > 0..1, y = z^2..1, x = z^2/y..1, 1). ... and more generally, if the random numbers are selected from the interval [a, 1], I get that Chris's probability of winning is 4/9 (1 - 2 a^(3/2) + a^3)/(1-a)^3 which >= 1/2 if a >= 0.0445338838868045196876255090994 approximately. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Random choice of numbers Dante a .8ecrit : > I have problems with writing down a formal solution of the following problem: Two boys - Chris & Michael - are playing the following game: Both of them \ are selecting randomly numbers from [0,1] interval (with uniform distribution). Chris is taking two numbers (each one is independent of another), and Michael is taking only one number (once again this is independent event). Next, Chris is multiplying both of his numbers, and Michael is calculating a square of his number. The boy with a larger result \ is a winner. Who has more chances of winning this game - Chris or Michael? For me this is obvious, that Michael is the one. (It's important that they \ are randomly selecting numbers from [0,1]!) But how we can prove this result in a formal way? Dante Say Michael choosed x and Chris choosed y and z. The probability that Michel wins is the probability that x^2>yz For x and y given, this probability is min(1, x^2/y) For x given, this becomes int{y=0 to 1; min(1, x^2/y)dy} = x^2(1-2ln(x)) So, the given probability is int{x=0 to 1, x^2(1-2ln(x))dx} = 5/9 So Michael has 25% chance more (5/9 versus 4/9) than Chris to win. === Subject: Re: Random choice of numbers > I have problems with writing down a formal solution of the following problem: > Two boys - Chris & Michael - are playing the following game: Both of them \ are selecting randomly numbers from [0,1] interval (with uniform distribution). Chris is taking two numbers (each one is independent of another), and Michael is taking only one number (once again this is independent event). Next, Chris is multiplying both of his numbers, and Michael is calculating a square of his number. The boy with a larger result \ is a winner. Who has more chances of winning this game - Chris or Michael? > For me this is obvious, that Michael is the one. (It's important that they \ are randomly selecting numbers from [0,1]!) > But how we can prove this result in a formal way? Suppose Chris's two numbers are x and y. Find Chris's probability of winning as a function of x and y, and then integrate the result over the unit square. (Yes, it turns out that the integral is less than 1/2, and therefore Michael is more likely to win.) -- Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. === Subject: Re: Random choice of numbers > I have problems with writing down a formal solution of the following > problem: Two boys - Chris & Michael - are playing the following game: Both of them > are selecting randomly numbers from [0,1] interval (with uniform > distribution). Chris is taking two numbers (each one is independent of > another), and Michael is taking only one number (once again this is > independent event). Next, Chris is multiplying both of his numbers, and > Michael is calculating a square of his number. The boy with a larger result > is a winner. Who has more chances of winning this game - Chris or Michael? For me this is obvious, that Michael is the one. (It's important that they > are randomly selecting numbers from [0,1]!) Not so obvious to me! It would be the same, btw, if you changed the [0,1] to [0,L] for any L. But I think Chris would have the advantage if you used the interval [1,2]. > But how we can prove this result in a formal way? Let Chris's and Michael's results be C and M respectively. Condition on M. I find that P{C >= M} = 4/9, so indeed Michael is somewhat more likely to win. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Random choice of numbers posting-account=Z9v8GgkAAAB54xGwWI4tZlTl3U6kju-g Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) > I have problems with writing down a formal solution of the following problem: Two boys - Chris & Michael - are playing the following game: Both of them are selecting randomly numbers from [0,1] interval (with \ uniform distribution). Chris is taking two numbers (each one is independent of another), and Michael is taking only one number (once again this is independent event). Next, Chris is multiplying both of his numbers, and Michael is calculating a square of his number. The boy with a larger result \ is a winner. Who has more chances of winning this game - Chris or Michael? For me this is obvious, that Michael is the one. (It's important that they \ are randomly selecting numbers from [0,1]!) But how we can prove this result in a formal way? Dante Identical. Because the mean of the two numbers of Chris is 0.5, its mean product = 0.25 The mean of the number of Michael is 0.5 and its mean square = 0.25. ?What if the question were : Chris and Michael both take two numbers. Chris sums and divides by 2 , Michael multiply the numbers and takes the square root. Ludovicus === Subject: Re: Random choice of numbers posting-account=Z9v8GgkAAAB54xGwWI4tZlTl3U6kju-g Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) Excuse me. I was wrong. The mean product of Chris is 0.25 But the mean square of Michael is 1/3 Experimentally Michael advantage Chris in proportion 1.25 to 1. Ludovicus === Subject: Addition of rotations posting-account=9GVhJwoAAABWT_ES8RBoMoRidijZAHy9 Gecko/2008120121 GranParadiso/3.0.4,gzip(gfe),gzip(gfe) I was wondering the following: Suppose I have operators which are rotations in d-dimensions in addition to scaling (a mathematician would call that tensor product of rotation group and a 1d real variable?). Concatenation of these operators is written by the multiplication operator. Now I want to create a ring and define the addition of these operators as to get a new rotation with scaling (no shearing, ...). Do rotations have enough properties to conclude that introducing addition is only possible in dimensions 2 and 4 (complex numbers and quaternions)??? === Subject: Re: Addition of rotations- I was wondering the following: Suppose I have operators which are rotations in d-dimensions in > addition to scaling (a mathematician would call that tensor product of > rotation group and a 1d real variable?). Concatenation of these operators is written by the multiplication > operator. Now I want to create a ring and define the addition of these > operators as to get a new rotation with scaling (no shearing, ...). Do rotations have enough properties to conclude that introducing > addition is only possible in dimensions 2 and 4 (complex numbers and > quaternions)??? Combining rotations of Euclidean space about a fixed point with scaling is rather tricky. Let me explain: first of all, throw out scaling by a factor zero. This is metaphorically speaking The Big Crunch. It cannot be undone! so it cannot be an element of any multiplication group. next, what scaling groups do you want to consider? The multiplicative group M+ = (R+, X) of positive reals? The multiplicative group M = (R{0}, X) of all nonzero reals? Discrete subgroups of M+ or M? at last, the combination of such groups with the N-dimensional rotation group SO(N): in odd-dimensional spaces the central inversion, i.e. the scaling by a factor -1, is not a rotation; in even-dimensional spaces the central inversion is a rotation. To be specific, the central inversion in 2N-dimensional space consists of rotations over pi (180 degrees) in each of N mutually orthogonal planes through the rotation centre. In all cases the combination of a rotation group SO(N) and a scaling group SG yields a group of similarity transformations with a fixed point. This group is the direct product SO(N) x SG; it has SO(N) and SG as invariant subgroups. BTW, I guess the term tensor product is rarely used in the context of group theory. Now about addition of rotations: what is this? Perhaps addition of their matrices? Then one observes that in general the sum matrix is not a rotation matrix, even not a multiple of a rotation matrix. So there does not exist such a thing as addition of rotations. Finally the special cases of 2D, 3D and 4D: In 2D we have the 1-1 mapping of similarity transformations with fixed point \ O onto the nonzero complex numbers. Addition of 2D rotations would be interpreted as addition of their representative complex numbers. This results in general in a similarity transformation with a scaling factor different from unity. In 3D and 4D we have by and large the same situation: addition of unit quaternions (which represent 3D and 4D rotations without reduction of magnification) does not yield in general unit quaternions. So in 3D and 4D we have also no addition of rotations. All essentials of representation of 3D and 4D rotations by quaternions may be found in my http://www.xs4all.nl/~jemebius/Contributions-to-arXiv.htm Happy studies: Johan E. Mebius === Subject: Re: Addition of rotations- > I was wondering the following: > Suppose I have operators which are rotations in d-dimensions in > addition to scaling (a mathematician would call that tensor product of > rotation group and a 1d real variable?). > Concatenation of these operators is written by the multiplication > operator. > Now I want to create a ring and define the addition of these > operators as to get a new rotation with scaling (no shearing, ...). > Do rotations have enough properties to conclude that introducing > addition is only possible in dimensions 2 and 4 (complex numbers and > quaternions)??? Combining rotations of Euclidean space about a fixed point with scaling > is rather tricky. > Let me explain: first of all, throw out scaling by a factor zero. This is metaphorically > speaking The Big Crunch. It cannot be undone! so it cannot be an element > of any multiplication group. next, what scaling groups do you want to consider? > The multiplicative group M+ = (R+, X) of positive reals? > The multiplicative group M = (R{0}, X) of all nonzero reals? > Discrete subgroups of M+ or M? at last, the combination of such groups with the N-dimensional rotation > group SO(N): > in odd-dimensional spaces the central inversion, i.e. the scaling by a > factor -1, is not a rotation; in even-dimensional spaces the central > inversion is a rotation. > To be specific, the central inversion in 2N-dimensional space consists > of rotations over pi (180 degrees) in each of N mutually orthogonal > planes through the rotation centre. In all cases the combination of a rotation group SO(N) and a scaling > group SG yields a group of similarity transformations with a fixed > point. This group is the direct product > SO(N) x SG; it has SO(N) and SG as invariant subgroups. > BTW, I guess the term tensor product is rarely used in the context of > group theory. Now about addition of rotations: what is this? Perhaps addition of their > matrices? Then one observes that in general the sum matrix is not a > rotation matrix, even not a multiple of a rotation matrix. So there does > not exist such a thing as addition of rotations. Finally the special cases of 2D, 3D and 4D: > In 2D we have the 1-1 mapping of similarity transformations with fixed > point O onto the nonzero complex numbers. Addition of 2D rotations would > be interpreted as addition of their representative complex numbers. This > results in general in a similarity transformation with a scaling factor > different from unity. In 3D and 4D we have by and large the same situation: addition of unit > quaternions (which represent 3D and 4D rotations without reduction of > magnification) does not yield in general unit quaternions. So in 3D and > 4D we have also no addition of rotations. All essentials of representation of 3D and 4D rotations by quaternions > http://www.xs4all.nl/~jemebius/Contributions-to-arXiv.htm Happy studies: Johan E. Mebius Han de Bruijn === Subject: Re: Addition of rotations- <49D951B1.2090600@xs4all.nl> posting-account=9GVhJwoAAABWT_ES8RBoMoRidijZAHy9 Gecko/2008120121 GranParadiso/3.0.4,gzip(gfe),gzip(gfe) merely a physicist. > first of all, throw out scaling by a factor zero. This is metaphorically speaking The Big > Crunch. It cannot be undone! so it cannot be an element of any multiplication group. I didn't think much about zero scaling. > next, what scaling groups do you want to consider? > The multiplicative group M+ = (R+, X) of positive reals? Yes, I think positive reals. > at last, the combination of such groups with the N-dimensional rotation group SO(N): > in odd-dimensional spaces the central inversion, i.e. the scaling by a factor -1, is not a > rotation; in even-dimensional spaces the central inversion is a rotation. That is interesting to know :) Will it be important? > Now about addition of rotations: what is this? Perhaps addition of their matrices? Then > one observes that in general the sum matrix is not a rotation matrix, even \ not a multiple > of a rotation matrix. So there does not exist such a thing as addition of rotations. It doesn't need to have an easy meaning. I only wonder if it is possible at all to define addition. Well, thinking of quaternions it is possible at least in dimension 4, since quaternions are exactly rotations (unit quaternions) with scalings? > In 3D and 4D we have by and large the same situation: addition of unit quaternions (which > represent 3D and 4D rotations without reduction of magnification) does not \ yield in > general unit quaternions. So in 3D and 4D we have also no addition of rotations. Quaternions are a 4D rotation, too? As I allow to include scaling, I am not restricted to *unit* quaternions?! So addition is possible? The scaling factor is determined by the modulus of the quaternion. > All essentials of representation of 3D and 4D rotations by quaternions may \ be found in my === Subject: Re: Is there a Discrete Exponential Distribution? posting-account=Z9v8GgkAAAB54xGwWI4tZlTl3U6kju-g Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) Working with some problems on primes distribution > I found the necessity of an Exponential Distribution > but applied to integer random variables. > Is there a Discrete Exponential Distribution? If you mean positive integers, yes. Each n gets a weight exp(-n*tau) > where tau is the scale of exponential > you want. You can start at 0 or 1, as you like. Then divide these > weights by the sum to infinity to > make the total probability 1. So starting with 0, > p(n)=exp(-n*tau)/(1-exp(-tau)) If you mean positive and negative integers, yes again if you consider > the absolute value of n. If you > don't, the weights will go to infinity as n->-infinity, so no. What means tau = scale of exponential? > My frequencies are: 5291, 2463,1099,557,240,166,96,40,25,11,5,4,3,0 > n = 0, 1, 2,... And the number of tests = 10000. Essentially, 'tau' is related to the 'p' in the Geometric distribution > P{X=n} = p*(1-p)^n, n=0,1,2,... . Presumably, you want to estimate the > value of 'p' from your data. One way would be to compute the sample > mean and equate it to the formula EX = (1-p)/p. The sample mean = sum > (n*f n,n=0..13)/N = 0.9407, so solving (1-p)/p = 0.9407 gives p = > 0.5152779925. Here the f n are the observed frequencies you have given > above. There are other ways of estimating p that may give better > results: solve the nonlinear optimization problem of minimizing the > squared (or absolute) deviations between the theoretical frequencies > 10000*p*(1-p)^n and the observed frequencies f n. For instance, if we > want to minimize the total squared deviation S = sum [10000*p*(1-p)^n > - f n]^2,n=0..13) we can set the derivative to zero and solve > (numerically); Maple 9.5 gives the optimal value as p = .5305282271, > giving min S = 10231.63924. The frequency fit is as follows: > n fit exact > 0 5305.28227 5291.00000 > 1 2490.68027 2463.00000 > 2 1169.30408 1099.00000 > 3 548.95526 557.00000 > 4 257.71900 240.00000 > 5 120.99180 166.00000 > 6 56.80223 96.00000 > 7 26.66704 40.00000 > 8 12.51942 25.00000 > 9 5.87752 11.00000 > 10 2.75933 5.00000 > 11 1.29543 4.00000 > 12 0.60817 3.00000 > 13 0.28552 0.00000 R.G. Vickson I deduced a simple formula that fit better the empirical data: Taking tau = 1/Mean p(n) = (1 - exp(-tau))*exp(-n*tau) For the given experiment Mean = 1.328 ; tau = 0.753 The adjusted values are: 5257, 2493, 1183, 561, 266, 126, 60, 28, 13, 6, 3, 1, 1, 0. For for those interested in prime numbers this information may be useful: The given statistics corresponds to : Frequency of the first x that makes 6(N - x^2) + 1 a prime when N varies from N1 to N1 + D in the sequence of non negative integers. In this case N1 = 10^7 ; D = 10000 The n of the formula is the order of the increasing equal intervals of x. Ludovicus Ludovicus === Subject: x_n converging to x in normed space (V, || ||) posting-account=2kYL5QoAAAAmH1e_g3WSyZ5Z9zXmRQX3 .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) When we say x_n converges to x in normed space (V, || ||) It means lim ||x_n - x || = 0. Now what if for each n, x_n - x is in V, and lim ||x_n - x || = 0 holds, but we aren't sure if x or even x_n are in V. Does this happen in math? In this case I guess it would be better to say x_n converges to x in the norm || || without mentoin of V. === Subject: Re: x_n converging to x in normed space (V, || ||) When we say x_n converges to x in normed space (V, || ||) It means lim ||x_n - x || = 0. Now what if for each n, x_n - x is in V, and lim ||x_n - x || = 0 > holds, but we aren't sure if x or even x_n are in V. Does this happen in math? In this case I guess it would be better to say x_n converges to x in the norm || || without mentoin of V. If you say x_n converges to x, x_n and x must be in a space on which your norm is defined. If you say x_n converges to x in V, you are implying that this space is V. So if x_n and x are not necessarily in V, I would say x_n - x converges to 0 in V, not x_n converges to x. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: x_n converging to x in normed space (V, || ||) > When we say x_n converges to x in normed space (V, || ||) It means lim ||x_n - x || = 0. Now what if for each n, x_n - x is in V, and lim ||x_n - x || = 0 > holds, but we aren't sure if x or even x_n are in V. Does this happen in math? > Yes, consider x = 2, x_n = 2 + 1/n, V = [0,1]. > In this case I guess it would be better to say x_n converges to x in the norm || || without mentoin of V. > Seemingly V is irrelevant. === Subject: Re: -- Packing unit circles in circles: new results In this post, four new packings are given. (There still seems to be a backlog at Packomania. But once it has been brought up to date, I will post my promised approximation for r in terms of N.) ------------------------------------ N = 327 : r = 19.649180991046..., symmetry group C_3 The best packing previously known has r = 19.6504... and is not symmetric. ------------------------------------ N = 402 : r = 21.733062157664..., symmetry group C_3 The best packing previously known has r = 21.7342... and is not symmetric. ------------------------------------ N = 403 : r = 21.769628263694..., symmetry group C_6 The best packing previously known has r = 21.7731... and is not symmetric. ------------------------------------ N = 565 : r = 25.575806310403..., symmetry group C_6 The best packing previously known has r = 25.575885... and is not symmetric. ------------------------------------ Note: Much earlier in this thread, I had posted packings for N = 402 and 403 which had symmetry groups D_3 and D_6, resp. Since that time, better asymmetric packings were found. And now, I have once again found better symmetric ones, albeit not as highly symmetric as my previous packings. David W. Cantrell === Subject: Re: The complete infinite binary tree has only countably many infinite paths. posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/20080201 Firefox/2.0.0.12 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > It is really so simple that mentioning seemed unnecessary to me. Why > do you think that I chose the tree instad of general sets or lists? In > the tree there are only linear subsets. That is the point. That is NOT the point. Restricting to linear subsets DOES NOT HELP. THERE ARE JUST AS MANY linear subsets as there are subsets. === Subject: Re: JSH: Understanding the 'why' of Pell's Equation > ALL the records of my interactions on the newsgroups including posts I > deleted from Google Groups which are otherwise preserved will be > carefully archived, and studied. Wait. Are you threatening people with _your_ messages that _you_ attempted to delete from history from a well placed sense of shame? -- Michael Press === Subject: Re: JSH: Understanding the 'why' of Pell's Equation <49d75e0f@dnews.tpgi.com.au> posting-account=lj-kRgoAAADbp-o6laOVCIWXjFF53GYz Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > Well I'm just a computer nerd who has interacted with your maths skills Sorry, MichaelW. Well put Me2, Peter === Subject: Re: MathForum I post from MathForum, but that doesn't tell you if you'd lose anyone you'd care about. === Subject: Re: MathForum <15102587.26617.1238869088007.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > However, if by respect', you mean having vast > attention paid to one's least utterance, in > hopes of finding understanding, then I'm afraid > I have to disagree. It looks to me as though > cranks receive much more respect than anyone else > does on sci.math. > that doesnt make any sense. > you want proofs ? > ok > 1) L Renfro posts from MathForum. > is a crank ? > should you killfile ? Renfro is the exception to Carmody's blanket killfile. In fact, this is the reason that Carmody started this thread in the first place. He wanted to know whether there ought to be any exceptions to his blanket killfile of mathforum. And so, Renfro will apparently be the only exception. Is Renfro a so-called crank? Interesting question. Obviously he isn't considered a crank, otherwise he wouldn't be the exception to the mathforum killfile. Yet, as I've pointed out before, Renfro once made a post in which he makes errors confusing cardinality and ordinality and presents as true certain statements whose negations are true in every model of ZFC (including some models in which CH is false). I've shown this to tommy1729 before, but I repeat that post here. The post is nearly a decade old, so I doubt that Renfro has repeated the error since. The context: Renfro was trying to explain CH to a layperson who might not be familiar with ZFC. NOTATION: Let N* be the cardinality of the natural numbers and R* be the cardinality of the reals. First problem -- nonstandard notation. It's been mentioned that cranks often use nonstandard notation. Instead of saying card(N) or, better yet, aleph 0, Renfro invents the new notation N*. Let N** be the smallest uncountable cardinal number (i.e. the next largest infinity after N*). In other words, N** is aleph 1. If CH fails, the situation becomes more complicated. For instance, it is possible for there to be more than N*, more than R*, or even more than 2^R* many distinct cardinal numbers lying between N* and R*, and also between R* and 2^R*. This is invalid in ZFC+~CH. There can't possibly be more than R*, and especially not more than 2^R*, cardinals between N* and R*. Notice that there are only R* ordinals between N* and R*, and most of those ordinals aren't cardinals. What Renfro probably meant was something like: If CH fails, the situation becomes more complicated. For instance, it is possible for c to be greater than aleph omega, greater than aleph aleph 1, even greater than aleph aleph omega, and the same is true for 2^c. But in his zeal to avoid mentioning alephs, Renfro had made statements that are false in every model of ZFC. However, it is known that there cannot be exactly N* distinct cardinals lying between N* and R*. Also invalid in ZFC+~CH. Renfro was trying to state the cofinality rule with the cardinality of the continuum, but since N* means the cardinal aleph 0, what he ended up writing is that c can't equal aleph alpha for any countably infinite alpha! What Renfro probably meant to write was: However, it is known that c can't be aleph omega. but in his zeal to avoid mentioning standard terms that is false in some models of ZFC+~CH. If anyone else started making errors of this type, they'd be labeled crank faster than anyone can blink an eye. Inventing nonstandard terminology is considered one sign of being a crank, especially if one misuses that terminology, as Renfro did. So to answer tommy1729's question, no, Renfro isn't really considered to be a crank. === Subject: Re: MathForum <15102587.26617.1238869088007.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > Yet, as I've pointed out before, Renfro once made a post in > which he makes errors confusing cardinality and ordinality > and presents as true certain statements whose negations are > true in every model of ZFC (including some models in which > CH is false). I've shown this to tommy1729 before, but > I repeat that post here. The post is nearly a decade old, > so I doubt that Renfro has repeated the error since. This got me curious, because I don't recall using the notation in the parts you quoted, so I looked up the original post: sci.math -- combinatorial set theory question (4 August 1999) I suspect the reason I was using nonstandard terminology is because at the time I didn't know the convention for using underlines for subscripts. This was one of my earliest usenet posts, as I think I only began posting a week or two before this. At the time I didn't have internet access at home, so my posts during that period (July-August 1999) were sometimes written at home using Notepad or MicroSoft Word, saved on floppy's, and then posted at a library computer that I later drove to. Short posts were written on the spot, in the library, but for replies that were longer, I took a copy of the post home with me and composed a reply on my computer at home. I note that in this thread (from 1999) there are several posts that use w_1 and/or omega_1, but it's possible that I didn't see those. (I didn't drive the 15 miles to the library every day.) I also note that I didn't copy the relevant part of the post I was replying to, but that was because at the time I thought these posts were all through Math Forum, and so anyone reading a post could clearly see the posting-history tree (which I think Math Forum used to do a better job with back then than now). Shortly after this post (2 weeks? 3 weeks?) I began not only citing the relvant parts of the post I was replying to, but I developed a standard that I used for the next 6 or 7 years, where I included the date of the post and its Math Forum URL, so that anyone coming across the post from a search could easily see the previous post by clicking on the URL. But then a funny thing began happening. Every 2 or 3 years all the Math Forum URLs would be changed, with no redirecting to the changed URLs. I believe this has happend 3 times since I began posting. I've now pretty much discontinued this practice of including the replied-to post's a couple of URL changes, at least links using their old URLs get redirected to the correct place. The errors mentioned all seem to be correctly identified as errors, for what it's worth. They occur in one paragraph of a somewhat tangential remark, but the remark is sufficiently long that this wasn't a case of my quickly adding a few poorly edited sentences and overlooking some things. On the other hand, it seems to me that much of what I did write in that post is correct and correctly written, unlike the case with most crank posts. L. Renfro === Subject: Re: MathForum sha1:4IKxpJ986+E8LtHe9IKQuNm+Cgc= > Is Renfro a so-called crank? Interesting question. Obviously > he isn't considered a crank, otherwise he wouldn't be the > exception to the mathforum killfile. Yet, as I've pointed out before, Renfro once made a post in > which he makes errors confusing cardinality and ordinality > and presents as true certain statements whose negations are > true in every model of ZFC (including some models in which > CH is false). I've shown this to tommy1729 before, but > I repeat that post here. The post is nearly a decade old, so > I doubt that Renfro has repeated the error since. Making an error does not make one a crank. Repeated, obvious errors despite clear corrections and an insistence that one is right and his errors are in fact revolutionary insights. The context: Renfro was trying to explain CH to a layperson > who might not be familiar with ZFC. NOTATION: Let N* be the cardinality of the natural > numbers and R* be the cardinality of the reals. First problem -- nonstandard notation. It's been mentioned > that cranks often use nonstandard notation. Instead of > saying card(N) or, better yet, aleph_0, Renfro invents the > new notation N*. Using N* instead of |N| is perhaps mildly regrettable, but not particularly crankish. (You do know the fallacy of affirming the consequent, don't you? Sometimes, mathematicians use non-standard notation for convenience sake. As long as the notation is explained clearly, no harm is done.) Let N** be the smallest uncountable > cardinal number (i.e. the next largest infinity after N*). In other words, N** is aleph_1. If CH fails, the situation becomes more complicated. > For instance, it is possible for there to be more than > N*, more than R*, or even more than 2^R* many > distinct cardinal numbers lying between N* and R*, > and also between R* and 2^R*. This is invalid in ZFC+~CH. There can't possibly be > more than R*, and especially not more than 2^R*, > cardinals between N* and R*. Notice that there are > only R* _ordinals_ between N* and R*, and most of > those ordinals aren't cardinals. Okay, so he made a mistake. You've shown that. So what? Did he persist in his error when corrected and claim that the error showed in fact more mathematical insight than the bulk of living mathematicians? What Renfro probably meant was something like: If CH fails, the situation becomes more complicated. > For instance, it is possible for c to be greater than > aleph_omega, greater than aleph_aleph_1, even greater > than aleph_aleph_omega, and the same is true for 2^c. But in his zeal to avoid mentioning alephs, Renfro had > made statements that are _false_ in every model of ZFC. > However, it > is known that there cannot be exactly N* distinct > cardinals lying between N* and R*. Also invalid in ZFC+~CH. Renfro was trying to state > the cofinality rule with the cardinality of the > continuum, but since N* means the _cardinal_ aleph_0, > what he ended up writing is that c can't equal > aleph_alpha for _any_ countably infinite alpha! What Renfro probably meant to write was: However, it is known that c can't be aleph_omega. but in his zeal to avoid mentioning standard terms > that is _false_ in some models of ZFC+~CH. If anyone else started making errors of this type, > they'd be labeled crank faster than anyone can > blink an eye. Inventing nonstandard terminology is > considered one sign of being a crank, especially > if one misuses that terminology, as Renfro did. Oh, bull. Everyone makes mistakes. A mistake made once is no sign at all that one is a crank. -- Jesse F. Hughes ... And I'm Michele Norris. -- Quincy P. Hughes === Subject: Re: MathForum sha1:ypi/KbAS51jHyMt/3l46iJP9s9A= > Repeated, obvious errors despite clear corrections and an insistence > that one is right and his errors are in fact revolutionary > insights[...] ... make one a crank. -- Yup, you guessed it. If worse comes to worse, I *will* turn to the Army to help me with mathematicians. And then mathematicians don't think the NSA or CIA can save your asses, as generals LIKE me. -- James Harris's latest foray into mathematical logic. === Subject: Re: MathForum posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Oh yeah, the famous hidden variable example, where there's > a hidden third variable, such as warmer weather (since > people are more likely to eat ice cream when it's not > freezing out, people are more likely to go outside and > commit crimes when it's not freezing out). > I'll mention again that you have not shown there is any > such correlation between money and respect in sci.math. > Still, you do admit that turning your imagined > correlation into a causal relationship was a fallacy > on your part. And yet, it's not a fallacy when Carmody makes the exact same assumption? Carmody killfiles based on where posts come from. > (One irony here is that Phil Carmody probably has > more evidence (crank posters posting from (emphasis mine) And how are those places similar, besides the obvious fact that they are free? If Burns believes that I'm wrong that price is the common factor, then let Burns state what the real similarity is. > I believe that in this case, there may be a hidden third > variable at work too. In this case, it's professorship -- > professors, due to having a steady income, are more likely > to afford full-priced newsreaders, and professors, having > earned Ph.D.'s, have taken classes in standard analysis, > so are less likely to post so-called crank theories. > You are working very hard to maintain the idea that > which newsreader someone uses adds or subtracts from > their crank score. You have only a smattering of > of evidence, only anecdote, really, but if you're right, > you should rejoice. To Carmody, which newsreader someone uses adds or subtracts from their crank score -- otherwise he wouldn't killfile based on which newsreader someone uses. > This is what you've been looking for. Just get the cranks > to change their newsreaders and they'll stop being > regarded as cranks, right? Possibly -- but not for me. Even if I were to change my own newsreader, I'll still be in Carmody's killfile. Even though Carmody can't read what I write, I can read what I also know that lwalke3 is heavily burried in my killfile because of his propensity to spout crap, so the above BS is hardly unexpected. And therefore I'm in Carmody's killfile separately from the Google killfile, because I post crap. So even if I bought a newsreader, I'd still be killfiled. What do I post that Carmody considers crap? Perhaps it's because I defend theories other than ZFC (but this may set off yet another argument that one isn't called a crank merely due to opposition to ZFC). But it is sufficient to say that if I, from the day I first started posting to sci.math, had always sided with the standard analysts and agreed with them that WM, etc., are cranks and spent all my time telling WM, etc., how they are 100% wrong in their beliefs, I wouldn't be in Carmody's killfile today separately from the Google killfile. > That is what it comes down > to, right? And it should be vastly easier for you to > harangue them into using a different newsreader than > to change their methods of argumentation. Newsreaders cost money. It's difficult to harangue anyone into spending money as many a salesperson has learned. So actually it would be easier to change their methods of argumentation, as long as buying a book isn't necessary (or else it would come down to the cost of the book vs. the cost of the newsreader). > Even so amended, I'm not sure I agree with you. It depends > on what you mean by respect. > However, if by respect', you mean having vast > attention paid to one's least utterance, in > hopes of finding understanding, In this context, by respect I mean not being killfiled. So with Carmody at least, being killfiled depends at least on what newsreader one uses due to the blanket killfilings. I'm killfiled separately not based on a newsreader, so Carmody obviously doesn't look only at the newsreader. When one is killfiled, there's no chance for redemption. So even if I bought a newsreader and agreed with the majority of the standard analysts that the crank label is deserved, and posted nothing but ZFC from now on, I'd still be sitting in Carmody's killfile. Many mention zuhair as an example of a reformed crank. At least to Carmody, I can never be reformed. === Subject: Re: MathForum > Newsreaders cost money. Challenge: * Players take turns. * You show us a newsreader and its price. * I show you a free newsreader. * Bundled newsreaders do not qualify for either category; only a newsreader obtainable on its own. * If you pass on a turn, I must show one more free newsreader. * Last man standing wins. -- Michael Press === Subject: Re: MathForum posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) [...] I'll mention again that you have not shown there is any > such correlation between money and respect in sci.math. > Still, you do admit that turning your imagined > correlation into a causal relationship was a fallacy > on your part. And yet, it's not a fallacy when Carmody makes the exact > same assumption? Carmody killfiles based on where posts > come from. So? How does it follow that Carmody is turning a correlation into a causal relationship? Correlation is a perfectly adequate reason for him to killfile posts from GG if he thinks the correlation is strong enough that killfiling GG is in his best interest. > (One irony here is that Phil Carmody probably has > more evidence (crank posters posting from (emphasis mine) And how are those places similar, besides the obvious > fact that they are free? If Burns believes that I'm wrong > that price is the common factor, then let Burns state > what the real similarity is. Both of them are web interfaces, rather than traditional news servers. Since most newsreaders are free and many news servers are too, price obviously has little to do with it. > [...] To Carmody, which newsreader someone uses adds or subtracts > from their crank score -- otherwise he wouldn't killfile > based on which newsreader someone uses. This is what you've been looking for. Just get the cranks > to change their newsreaders and they'll stop being > regarded as cranks, right? Possibly -- but not for me. Even if I were to change my > own newsreader, I'll still be in Carmody's killfile. Even > though Carmody can't read what I write, I can read what I also know that > lwalke3 is heavily burried in my killfile because of his propensity > to spout crap, so the above BS is hardly unexpected. And therefore I'm in Carmody's killfile separately from > the Google killfile, because I post crap. So even if I > bought a newsreader, I'd still be killfiled. What do I post that Carmody considers crap? Perhaps > it's because I defend theories other than ZFC (but this > may set off yet another argument that one isn't called a > crank merely due to opposition to ZFC). But it is > sufficient to say that if I, from the day I first started > posting to sci.math, had always sided with the standard > analysts and agreed with them that WM, etc., are cranks > and spent all my time telling WM, etc., how they are 100% > wrong in their beliefs, I wouldn't be in Carmody's > killfile today separately from the Google killfile. What is your evidence for this? > That is what it comes down > to, right? And it should be vastly easier for you to > harangue them into using a different newsreader than > to change their methods of argumentation. Newsreaders cost money. No, they don't. How many times will you continue to say this after people have pointed out that they don't, even providing examples of free newsreaders and free servers? What on Earth is wrong with you anyway? > It's difficult to harangue anyone > into spending money as many a salesperson has learned. So > actually it would be easier to change their methods of > argumentation, as long as buying a book isn't necessary > (or else it would come down to the cost of the book vs. > the cost of the newsreader). Even so amended, I'm not sure I agree with you. It depends > on what you mean by respect. > However, if by respect', you mean having vast > attention paid to one's least utterance, in > hopes of finding understanding, In this context, by respect I mean not being killfiled. So > with Carmody at least, being killfiled depends at least on > what newsreader one uses due to the blanket killfilings. I'm > killfiled separately not based on a newsreader, so Carmody > obviously doesn't look only at the newsreader. When one is killfiled, there's no chance for redemption. So > even if I bought a newsreader and agreed with the majority > of the standard analysts that the crank label is deserved, > and posted nothing but ZFC from now on, I'd still be sitting > in Carmody's killfile. Many mention zuhair as an example of a reformed crank. At > least to Carmody, I can never be reformed. Unless you change your from header. Though I suspect you'd also have to change your personality, if you wanted to stay out of Phil's killfile for long. (As it happens I'm also in Phil's killfile, since I usually post from GG and am not one of his exceptions. But I don't give a , because Phil has never answered any question I've asked and I don't have the expertise to answer any question he is likely to ask. I imagine the same applies to you, so why does it bother you that you're in his killfile?) === Subject: Re: MathForum > (One irony here is that Phil Carmody probably has > more evidence (crank posters posting from > (emphasis mine) And how are those places similar, besides the obvious > fact that they are free? If Burns believes that I'm wrong > that price is the common factor, then let Burns state > what the real similarity is. Since Walker has asked so nicely, Burns will tell Walker: The real similarity is ... Phil Carmody does not want to read the vast majority of posts from those sites!!!! If Walker really wants to change this dreadful situation, Walker would do better spending Walker's time on Walker's project of reforming the cranks, so that Phil will want to read those posts. The only other option I see is for you to harangue Phil into reading those posts, even though he doesn't want to. As slight as the chances are for you to succeed in that, I still think you should hope that you fail. Whatever patience is shown cranks in sci.math has its origins in the ability of readers to walk away whenever their individual patience (however stout or feeble) has been exhausted. Do you really think cranks are not shown respect now? Imagine what they will be shown if we /must/ read them. [...] > That is what it comes down > to, right? And it should be vastly easier for you to > harangue them into using a different newsreader than > to change their methods of argumentation. Newsreaders cost money. And how do you know this? You've looked, have you? You must have paid the teensiest bit of attention to what others have *told* you, gone and looked, but (alas!) they were wrong. Hee hee. Of course, you did. http://www.mozillamessaging.com/en-US/thunderbird/ Thunderbird 2 * Free Download 2.0.0.21 for Windows, English (US) (6.4MB) You are impervious to reason. *That* is why you are in Phil Carmody's kill-file. Jim Burns === Subject: Re: MathForum sha1:laA5L1TseOhPbWH1+FdzJzIRAzE= > Oh yeah, the famous hidden variable example, where there's > a hidden third variable, such as warmer weather (since > people are more likely to eat ice cream when it's not > freezing out, people are more likely to go outside and > commit crimes when it's not freezing out). > I'll mention again that you have not shown there is any > such correlation between money and respect in sci.math. > Still, you do admit that turning your imagined > correlation into a causal relationship was a fallacy > on your part. And yet, it's not a fallacy when Carmody makes the exact > same assumption? Carmody killfiles based on where posts > come from. Not the same assumption at all. Does Phil killfile all free news servers? No. Does he killfile all free news readers? Certainly not, or he would have to killfile himself. You can't really be so stupid that you fail to see the difference between your description and Phil's real behavior. I think you're being deliberately dishonest here. -- Jesse F. Hughes Would you please stop talking and start talking? -- Vincent Price as the Saint === Subject: Re: MathForum Oh yeah, the famous hidden variable example, where there's > a hidden third variable, such as warmer weather (since > people are more likely to eat ice cream when it's not > freezing out, people are more likely to go outside and > commit crimes when it's not freezing out). > I'll mention again that you have not shown there is any > such correlation between money and respect in sci.math. > Still, you do admit that turning your imagined > correlation into a causal relationship was a fallacy > on your part. > And yet, it's not a fallacy when Carmody makes the _exact > same_ assumption? Carmody killfiles based on where posts > come from. Not the same assumption at all. Does Phil killfile all free news servers? No. Does he killfile all > free news readers? Certainly not, or he would have to killfile > himself. I did killfile myself once at work. Set what should have been an 'AND' filter to 'OR', and magically stopped receiving everything! > You can't really be so stupid that you fail to see the difference > between your description and Phil's real behavior. I think you're > being deliberately dishonest here. We've just seen him be that stupid about a dozen times, I can only conclude that in fact he is that stupid. Anyway, the filter's enabled, I'll see who I need to pluck out of it. Phil -- Marijuana is indeed a dangerous drug. It causes governments to wage war against their own people. -- Seaman (sci.math, 19 Mar 2009) === Subject: Re: MathForum posting-account=U44YcwkAAAAbGXB70Qr7gA3kornmKE4i Gecko/20080922 Ubuntu/7.10 (gutsy) Firefox/2.0.0.17,gzip(gfe),gzip(gfe) It baffles me why you care who or what news server is in someone's killfile. If they don't publicly announce it, how would you even know? What business is it of yours? I have hundreds of entries in my wish my newsreader had the capability of zapping not only its direct targets but any thread started by them. The targets of my killfile don't know they are in it and my usenet browsing is much the more pleasant for it. If someone puts you specifically in their killfile it just means they don't want to talk to you. That's their prerogrative. Much like hanging up on unwanted soliciters on the telephone. --Lynn http://math.asu.edu/~kurtz === Subject: Re: MathForum <823acpaycq.fsf@A166.veli3.tontut.fi> posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > A petty thing indeed. Why do you insist on doing so? > I really don't think LW can see the point about himself that your > posts illustrate. If I had to take a guess what point Aatu is making here, it's that, in trying to avoid judging posters based on the cost of their Usenet access like Carmody, I've overcorrected. Whereas Carmody favors full-priced newsreaders over free access, I've made the opposite mistake, and started favoring free access over full-priced newsreaders. spent on Usenet access. I want to judge posters on what is posted, not how much money is spent. So how can I achieve my goal, then? Certainly not by trying to killfile free access, as Carmody is about to do. And if I had a newsreader with killfiling powers, then I wouldn't killfile based on the price of the access. Instead, I'd killfile based on the use of those two annoying words, crank and troll, designed to belittle the opposition. But what can I do now to avoid making the opposite mistake from Carmody and favor free access? I don't know what I can do now -- especially since I can't distinguish those who use free access from those who pay to post. Not all standard analysts pay to post and not all so-called cranks use free access, so I can't reliably distinguish between the two based on that. (Of course, neither can Carmody.) All I know is that Carmody obviously has a full newsreader with killfiling ability. I want to be neutral towards how one posts on Usenet, unlike Carmody who killfiles based on where one posts from. I don't know what I can do to convince Aatu or Little of this. === Subject: Re: MathForum sha1:iWbPi2Qs+8IQpdkHDsrcnqoAjoU= > If I had to take a guess what point Aatu is making here, it's > that, in trying to avoid judging posters based on the cost of > their Usenet access like Carmody, I've overcorrected. Whereas > Carmody favors full-priced newsreaders over free access, I've > made the _opposite_ mistake, and started favoring free access > over full-priced newsreaders. You're purposely stupid, aren't you? This nonsense about full-priced newsreaders might be ever so slightly more plausible, if only Phil Carmody didn't use a free newsreader. -- Jesse F. Hughes Depression hits more people than thought. --headline in Lexington, KY newspaper, as reported on NPR's Morning Edition === Subject: Re: MathForum > If I had to take a guess what point Aatu is making here, it's that, > in trying to avoid judging posters based on the cost of their Usenet > access like Carmody, I've overcorrected. What is objectionable is that you judge posters based on the thickness of their Cockney. > Whereas Carmody favors full-priced newsreaders over free access, > I've made the _opposite_ mistake, and started favoring free access > over full-priced newsreaders. Why do you favour full-priced newsreaders over free ones? There are many fine free newsreaders, such as Gnus. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: to show that C[a,b] is subspace of Lp[a,b] (p finite) posting-account=2kYL5QoAAAAmH1e_g3WSyZ5Z9zXmRQX3 .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) To show that C[a,b] is subspace of Lp[a,b] (p finite), is it enough to show that p-norm applied to elements of C[a,b] are all finite? Is that a valid proof that C[a,b] is subspace of Lp[a,b]? Basically p-norm(f) <= (b-a)^(1/p) supnorm(f) is immediate because |f (x)|^p <= (supnorm(f))^p for all x And I can use this very same proof to show that Loo is subspace of Lp [a,b] because |f(x)|^p <= (esssupnorm(f))^p for almost everywhere x My reasoning correct? === Subject: Re: to show that C[a,b] is subspace of Lp[a,b] (p finite) To show that C[a,b] is subspace of Lp[a,b] (p finite), is it enough > to show that p-norm applied to elements of C[a,b] are all finite? Say it is a subset ... when you say subspace that may mean it is > closed, which is false. ???? Calling it a subspace is perfectly standard. The only quibbling point is that C[a,b] is usually thought of as a space of functions, while the elements of the metric space L^p[a, b] are equivalence classes of functions. > Is that a valid proof that C[a,b] is subspace of Lp[a,b]? Basically p-norm(f) <= (b-a)^(1/p) supnorm(f) is immediate because |f > (x)|^p <= (supnorm(f))^p for all x > And I can use this very same proof to show that Loo is subspace of Lp > [a,b] because |f(x)|^p <= (esssupnorm(f))^p for almost everywhere x > My reasoning correct? === Subject: Re: to show that C[a,b] is subspace of Lp[a,b] (p finite) To show that C[a,b] is subspace of Lp[a,b] (p finite), is it enough > to show that p-norm applied to elements of C[a,b] are all finite? Say it is a subset ... when you say subspace that may mean it is > closed, which is false. ???? Calling it a subspace is perfectly standard. The only quibbling > point is that C[a,b] is usually thought of as a space of functions, > while the elements of the metric space L^p[a, b] are equivalence > classes of functions. I was thinking more along these lines ... subspace of Lp[a,b] is > often taken to mean closed linear subspace of Lp[a,b] Never heard that. How does one refer to dense subspaces of Lp[a,b] then? === Subject: I am Isaac Newton. My identity had been proven though Esdras II and Nostradamus. I have a maths question. Maths about Nostradamus's prophecies. posting-account=b56JqgoAAACplhzVwdIQXOHh7423vMXz Presto/2.1.1,gzip(gfe),gzip(gfe) 1727+the great seventh number.81fs revolution = the birthday of I returned. 10.74 The great seventh number.81fs revolution, It will appear a time of the year for hecatomb. Not far from the great millennial age, When those can enter will leave the tomb. All English can see what is happening to the whole China airspace in July 1999. A terror of face appear over the whole China airspace. 10.72 The year 1999, seventh month, To bring back to life the great King of Angolmois, Before and after Mars to reign by good luck. According to Saturn will come back again late, you can calculate the year, it is 2008. According to The eye plucked out by the Goshawk at Narbonne, America must be protected Taiwan as soon as the Taiwan Strait War breaks out. According to Empire transferred towards the Dusky nation, a black dog become the American President. 3.92 The world near the last period.81C Saturn will come back again late. Empire transferred towards the Dusky nation.81C The eye plucked out by the Goshawk at Narbonne. ... Just a little. According to Milton William Cooper, Jesus is an alien: http://www.wrongmountain.com/www/cosmia/majority.htm http://en.wikipedia.org/wiki/Milton_William_Cooper http://www.v-j-enterprises.com/cooper.html According to Esdras II, see more: 5.http://epochtimes.com/gb/8/12/12/n2360196.htm 6.http://epochtimes.com/gb/8/12/9/n2356210.htm 7.http://epochtimes.com/gb/8/11/24/n2339729.htm 8.http://epochtimes.com/gb/8/12/15/n2363196.htm 14.NASA: 2012 .81fspace Katrina.81f may cripple U.S. for months http://revolutionradio.org/2009/01/11/nasa-2012-space-katrina-may-cripple-us\ -for-months/ 20.http://www.epochtimes.com/gb/8/1/26/n1991466.htm 21.http://www.epochtimes.com/gb/7/8/7/n1794562.htm 22.http://www.epochtimes.com/gb/7/5/2/n1696812.htm 23.http://www.epochtimes.com/gb/7/4/4/n1668028.htm 24.http://www.epochtimes.com/gb/8/12/3/n2350595.htm 25.http://www.epochtimes.com/gb/8/6/12/n2152318.htm 26.http://www.epochtimes.com/gb/8/5/24/n2129552.htm 28.http://www.epochtimes.com/gb/9/2/10/n2424161.htm 29.http://www.epochtimes.com/gb/9/1/16/n2399273.htm 30.http://news.sina.com.cn/w/2008-02-04/021913379568s.shtml 31.http://blog.cctv.com/html/76/495576-223892.html 32.http://www.epochtimes.com/gb/9/1/14/n2397212.htm 72.http://www.epochtimes.com/gb/8/10/22/n2305396.htm 79.http://epochtimes.com/gb/9/4/1/n2481752.htm 80.http://epochtimes.com/gb/9/3/2/n2447920.htm 81.http://epochtimes.com/gb/9/2/26/n2443892.htm 82.http://epochtimes.com/gb/9/3/22/n2470780.htm 84.http://epochtimes.com/gb/9/2/24/n2441123.htm 104.http://news.sohu.com/20070905/n251977503.shtml 105.http://world.huanqiu.com/roll/2008-11/284494.html 108.http://epochtimes.com/gb/8/12/9/n2357371.htm 109.http://news.163.com/09/0207/19/51ISF7NJ000120GU.html 110.http://epochtimes.com/gb/9/1/8/n2389159.htm 111.http://epochtimes.com/gb/9/1/23/n2406899.htm 112.http://epochtimes.com/gb/9/1/7/n2388403.htm 114.http://news.sina.com.cn/w/p/2009-03-20/081417445950.shtml 115.http://epochtimes.com/gb/9/3/2/n2447541.htm 116.http://epochtimes.com/gb/9/3/2/n2447217.htm 118.http://photo.voloer.com/photo/2008-05/26/content-1183.html 119.http://www.ldnews.cn/news/thenews/todaynews/ 200809/20080912090827.html 120.http://jkl.wjq.gov.cn/E_ReadNews.asp?NewsID=156 121.http://www.chinadaily.com.cn/hqzx/2008-07/30/content_6888623.htm 120.http://www.232100.cn/html/200807/30/124751568.htm 121.http://www.232100.cn/html/200807/29/080415376.htm 122.http://news.yninfo.com/world/hqss/200810/t20081018_708616.htm 123.http://china.huanbohainews.com.cn/system/ 2008/06/19/010077361.shtml 125.http://www.232100.cn/html/200807/24/0813401092.htm 128.http://news.online.tj.cn/news/zggd/ 2008/826/0882691444924AK080826091907K70.html 129.http://www.newsyn.com/html/guonei/20080324/52971.html 130.http://news.nen.com.cn/72340194296070144/20081014/2508657.shtml 131.http://cn.chinareviewnews.com/doc/1006/7/5/1/100675173.html 132.http://news.163.com/08/1122/11/4RBOPRKG0001125G.html 133.http://news.xinhuanet.com/photo/2008-10/09/content_10169523.htm 136.http://bbs.aboluowang.com/thread-10028-1-1.html 142.http://epochtimes.com/gb/7/7/16/n1774628.htm 143.http://epochtimes.com/gb/7/7/19/n1777311.htm 144.http://epochtimes.com/gb/7/7/19/n1777582.htm 145.http://epochtimes.com/gb/7/7/19/n1777604.htm 146.http://www.0086money.com/board,view,62101.html ... Please notice the pictures and the dates. I have added and subtracted some new links. Please notice that they happen to the Earth since the second half of 2006. My first voice appear on Radio Free Asia in Octorber 2006, all English can see what happen to America and Canada in Octorber 2006. The greatest snownstrom happened to America and Canada in Octorber 2006 in 137 years. Read more: Isaac Newton Since 25th December, 2008 === Subject: Re: I am Isaac Newton. My identity had been proven though Esdras II and Nostradamus. I have a maths question. Maths about Nostradamus's \ prophecies. posting-account=b56JqgoAAACplhzVwdIQXOHh7423vMXz Presto/2.1.1,gzip(gfe),gzip(gfe) I am Isaac Newton. I can tell you that all about his prophecies is true. Maths about Nostradamus's prophecies. According to Saturn in Capricorn, Jupiter and Mercury in Taurus: Venus also, Cancer, Mars in Virgo and Sun .81ior Soil.81jtwentieth of Taurus the earth will tremble very mightily, you can calculate the date, it is 12th May 2008. Isaac Newton Since 25th December, 2008 === Subject: Re: I am Isaac Newton. My identity had been proven though Esdras II and Nostradamus. I have a maths question. Maths about Nostradamus's \ prophecies. posting-account=b56JqgoAAACplhzVwdIQXOHh7423vMXz Presto/2.1.1,gzip(gfe),gzip(gfe) I am Isaac Newton. I can tell you that all about his prophecies is true. Maths about Nostradamus's prophecies. According to Saturn in Capricorn, Jupiter and Mercury in Taurus: Venus also, Cancer, Mars in Virgo and Sun .81ior Soil.81jtwentieth of Taurus the earth will tremble very mightily, you can calculate the date, it is 12th May 2008. Isaac Newton Since 25th December, 2008 === Subject: Re: I am Isaac Newton. My identity had been proven though Esdras II and Nostradamus. I have a maths question. Maths about Nostradamus's \ prophecies. posting-account=b56JqgoAAACplhzVwdIQXOHh7423vMXz Presto/2.1.1,gzip(gfe),gzip(gfe) I am Isaac Newton. I can tell you that all about his prophecies is true. Maths about Nostradamus's prophecies. According to Saturn in Capricorn, Jupiter and Mercury in Taurus: Venus also, Cancer, Mars in Virgo and Sun .81ior Soil.81jtwentieth of Taurus the earth will tremble very mightily, you can calculate the date, it is 12 May 2008. Isaac Newton Since 25th December, 2008 === Subject: Re: Wikipedia becoming a search-engine and better than Google, and where sci.newsgroups under the auspices of Wikipedia search It would be nice if someone did a mashup > using Wikimapia and the Microsoft's bird's eye Live Search > so that one could click on a Wikimapia point, > and jump to the bird's eye view. > I'll bet you are one of those Microsoft used, Tom. Perhaps even using (for a newsreader) Microsoft Outlook Express 6.00.2900.2180 === Subject: Interior of a subset Let Y be a subspace of X and let A be a subset of Y. Denote by Int_X(A) the \ interior of A in the topological space X and by Int_Y(A) the interior of A \ in the topological space Y. Prove that Int_X(A) is a subset of Int_Y(A). What \ is an example that shows, generally, that Int_X(A) is not equal to Int_Y(A). === Subject: Re: Interior of a subset > Let Y be a subspace of X and let A be a subset of Y. Denote by Int_X(A) > the interior of A in the topological space X and by Int_Y(A) the > interior of A in the topological space Y. Prove that Int_X(A) is a > subset of Int_Y(A). What is an example that shows, generally, that > Int_X(A) is not equal to Int_Y(A). > If U subset A subset S, then cl_A U = A / cl_S U, int_A U = A / int_S (U / SA) because cl_A U is the smallest A-closed set containing U int_A U is the largest A-open set contained by U. Details are left as an exercise for the reader to demonstrate. === Subject: Re: Interior of a subset posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > Let Y be a subspace of X and let A be a subset of Y. Denote by Int X(A) the interior of A in the topological space X and by Int Y(A) the interior of \ A in the topological space Y. Prove that Int X(A) is a subset of Int Y(A). What is an example that shows, generally, that Int X(A) is not equal to Int \ Y(A). First, recall that the interior of A with respect to some topology is the largest open subset of A, so if V is open in Y and V is a subset of A then V is a subset of Int Y(A). Also, the open subsets of Y are by definition those sets of the form Y n U, where U is an open subset of X (here n denotes the intersection, and c will denote is a subset of in what follows). Now Int X(A) is open in X, and Int X(A) c A c Y, so Int X(A) n Y = Int X(A), and therefore Int X(A) is open in Y. Therefore Int X(A) c Int Y(A). For an example that shows that Int X(A) need not equal Int Y(A), let X be the set of real numbers with the usual topology, let Y be the closed unit interval [0,1], and let A = Y. Then Int X(A) = (0,1), whereas Int Y(A) = [0,1]. === Subject: Re: Interior of a subset posting-account=_l4K0QkAAAC09JhOoK_ZfoJKXOmr_jZf Gecko/2009032609 Firefox/3.0.8 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > Let Y be a subspace of X and let A be a subset of Y. Denote by Int X(A) the interior of A in the topological space X and by Int Y(A) the interior of \ A in the topological space Y. Prove that Int X(A) is a subset of Int Y(A). What is an example that shows, generally, that Int X(A) is not equal to Int \ Y(A). You should really tell us what you have tried, and where you are stuck, rather than just copy the problem onto the newsgroup. Unless, that is, you are trying to assign homework to the group. In that case, I think that my best bet for a solution is to post it somewhere and hope someone will do my homework for me so I don't have to think about it. Do you have any suggestions of a forum or newsgroup or website where I might try that? What are the definitions of interior? When is a point x in the interior of A? When is it in the interior of Y? If you think about it for just a little less time than it took you to post to the newsgroup, you might even figure out an answer. -- Arturo Magidin === Subject: Re: #350 the lackadasical math community with never a definition of \ finite versus infinite; new book 2nd edition: New True Mathematics posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > The Old Reals were designed around absolute continuity > for which there is no evidence of that in Physics. In fact Planck's > Constant contradicts absolute-continuity. > Physics also contradicts infinity. The observable universe > is not infinite. And yet your AP-reals and AP-adics have > infinite numbers of digits. So how can your AP-reals and > AP-adics be based on physics? Tribble is still reeling off the fact that he lost an important > argument: Tribble said words to the effect, that you cannot do physics without > math but that you can do math without physics. Then I pointed out to him the flaw of his reasoning: > Without Physics, well, there is no Universe, no life, no math. Tribble is as shoddy in logic with the above as he is shoddy in logic > with finite versus infinity. He never defines finite from infinite to > tell whether this Peano Number is finite or infinite: > 00000111111....11111 Physics defines it as infinite since at 10^500 there is no more > physical measurement that can > take place via Planck number restraints, but according to Tribble's > use of current definitions > by the math community at large that number is finite. What say you, there, Tribble? Isn't it funny, Tribble, how Peano > crafted axioms for Naturals > but never had the brilliance of defining what is a finite Natural from > an infinite Natural. answering my question. Trying to hurt my feelings? Logically, I should just conclude from your non-response that you agree that the AP-reals cannot be based on Physics. Because it's obvious that you can't provide an example of infinity to support your infinite digits. === 2nd Ed by David Griffiths posting-account=0DE1ygoAAABMcw-D370W5jQ3W0RNK2pW Gecko/2009021910 Firefox/3.0.7,gzip(gfe),gzip(gfe) Do you suffer from a tough class? Are you looking for instructor solutions manual to do your homework? Just send me email with its name and edition and I may be able to help you in low price! It is my list, however if you don't find it here don't give up because it is only a list of some. 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What can be said about partitions of unity using C-infinity functions in relation to paracompactness? Also, does the product of spaces admitting partitions of unity(using C-infinity functions) also admit partitions of unityusing c-infinity functions)? === Subject: Re: Partitions of unity > I have read about partitions of unity in R^n where C-infinity > functions are used. I was wondering what kind of spaces in general > admit the concept of partitions of unity using C-infinity functions. > If we talk about partitions of unity using continuous functions only, > then it is equivalent to paracompactness in hausdorff spaces. What can > be said about partitions of unity using C-infinity functions in > relation to paracompactness? In order to bo able to talk about C-infinity functions you will have to be working with manifolds, that is, about topological spaces locally homeomorphic with R^n such that coordinate changes are differentiable. In general, there are no C-infinity partitions of unity on such a space. In order to assure that they exist, you will have to assume that the space is paracompact. It turns out that this condition is equivalent (for manifolds) to these ones: 1) each connected component is sigma-compact; 2) each connected component is second countable; 3) the manifold is metrizable. Jose Carlos Santos === Subject: Re: Anti-relativists and Judaism >Page 33 Long before Hitler's seizure of power a small group of German >physicists, styling themselves national researchers, had formed >around the Nobel prize winners Lenard and Stark. This group boldly >declared Einstein's theory of relativity to be Jewish world-bluff. >They attempted to dismiss, under the summary heading of Jewish >physics, all studies based on the data of Einstein and Bohr. > It is unfortunate, but history may reveal them to be right. As it > currently stands, relativity has failed two major tests--LIGO and > GPB. And before you lie about a GPB positive result again, relativity > is incompatible with quantum mechanics, and the latter clearly has > more validity. http://www.ligo.caltech.edu/ http://www.nasa.gov/mission_pages/gpb/index.html Explain your abbreviations at least once in a thread, please. Han de Bruijn === Subject: Re: Anti-relativists and Judaism >Besides Einstein the nitwit, the plagiarist, and the liar, the other >Jews were Levi-Civita and Schwarzschild. Karl Schwarzschild (October 9, 1873 [CapitalEth] May 11, 1916) was a German Jewish physicist. ^^^^^^ Han de Bruijn === Subject: Re: Anti-relativists and Judaism [Babbling nonsense snipped] >(As I already observed, the Nazis dismissed relativity >as Jewish science, >Nonsense. Nonsense? Are you sure? Think about your answer carefully. That's assuming you care if you > look like an idiot, though. Which we know to be not the case. and this was not because the Nazis simply latched >onto all fringe ideas. Other anti-semites apparently do the same.) >** The principle of relativity was discovered by Galileo. But applied to E&M by Einstein. >** The invariance in the observed speed of light was first postulated >by Voigt. Incorrect. >** The Lorentz transform was first derived by Voigt and Larmor. Correct. However Voigt/Larmor didn't publish, and Lorentz did. That's > why its' called the LORENTZ transform. >** The concept of spacetime was first discussed by H. G. Wells No, that was the 4th dimension. Try to keep your sci-fi straight. > There's a whole load of context that you will never understand due to > pervasive stupidity. >and >mathematically by Minkowski. Incorrect, but for a different reason. Minkowski recognized the > geometric significance of Einstein's theory. Let us not lose sight of the fact you still can't understand > Minkowski's formalism. >** The principle of equivalence was discovered by Galileo again. It >formed the backbone to the Newtonian law of gravity. Like hell it does. The equivalence principle implies a metric theory > of gravitation. It doesn't. > The energy of gravitation in Newton does not gravitate - Newton does > not work with the EP. > Just because >Einstein the nitwit, the plagiarist, and the liar finally understood >the Newtonian law of gravity, it does not give him as a nitwit, a >plagiarist, and a liar to claim to have rediscovered the principle of >equivalence. And the copy-paste insult comes out. >** The Riemann tensor was derived by Ricci based on his manmade >covariant derivative. Man-made? You mean everything else isn't man made, or are you simply > stating this like how creationists emphasize that evolution is a > theory? Let us not lose sight of the fact that you do not understand the > Riemann tensor. You do not understand the symmetries nor do you > understand the equivalence of the alternative definitions of the > object. >** The Ricci tensor was stated by Levi-Civita (a Jew). Only to you would this be important. >** The idea of linking the Ricci scalar to the field equations was >suggested by Nordstrom. Easy to claim, hard to prove. Just like everything else you do. >** The field equations were finally derived by Hilbert after pulling >out the so-called Lagrangian to the Einstein-Hilbert action (whatever >crap it is) from his own @ss. Yes, the field equations. The derivation of which you still do not > understand. Derivation ? Take any other model with enough free parameters to adjust for a decent match with good old Newtonian gravity. General Relativity is a splendid piece of mathematics, built on quicksand. > Plus the whine shows that you - remarkably - still do not understand > classical mechanics. A system's Lagrangian is postulated - it all > starts from there. That you whine it was picked arbitrarily not only > exhibits great ignorance of concepts like weak coupling, but that you > are a ing idiot as well. >** The first static, spherically symmetric, and asymptotically flat >solution to the field equations was solved by Schwarzschild (a Jew). You don't understand the Schwarzschild solution either. Still think > that the surface area of a sphere in the Schwarzschild metric is equal > to 4 pi r^2 even though a simple calculation proves otherwise? >** The second static, spherically symmetric, and asymptotically flat >solution to the field equations was solved by Hilbert. After >realizing there are infinite such static, spherically symmetric, and >asymptotically flat solutions to the field equations, Hilbert walked >away from the monster he had created and allowed Einstein the nitwit, >the plagiarist, and the liar to take full credit. understand that a coordinate system does not change a tensor. >Besides Einstein the nitwit, the plagiarist, and the liar, the other >Jews were Levi-Civita and Schwarzschild. Do you ever stop whining about the jews? [More babbling nonsense snipped] Han de Bruijn === Subject: Re: Anti-relativists and Judaism posting-account=rIfu6QoAAAD5nXG3h9QEE0J3dZn1U45R Gecko/20090401 Gentoo Firefox/3.1b3,gzip(gfe),gzip(gfe) [...] Einstein Dingleberry [...] lick up each other's droppings [...] > Einstein Dingleberries [...] Einstein Dingleberries [...] Einstein > Dingleberries. [...] eat each other's crap Stay classy. If you have better ideas to describe these Einstein Dingleberries > including yourself, let's hear it. In the meantime, my descriptions > proudly stand. Proud? You are proud of your behavior here? === Subject: Re: Anti-relativists and Judaism <1238662666_15@sicinfo3.epfl.ch> <20090402142538.370$Nr@newsreader.com I can play an acceptable game of Texas \ Hold'em, make cranks weepy little > bitches, and drone on about the world's largest pinecone. Moreover, you're rediscovering your sense of humor. 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Full details: http://jobs.phds.org/job/13322/gqr-global-quant-recruitment/reinsurance-life\ - settlements -------------------------------------------------------------------- Title: Experienced FX, Commodities, Rates Quantitative Analyst Required for Top Tier Investment Bank Employer: GQR | Global Quant Recruitment - Quant Jobs Location: New York, NY, United States Experienced FX, Commodities, Rates Quantitative Analyst Required for Top Tier Investment Bank .89´\.8b New York The Client: Top Tier European Investment Bank Location: New York - USA The Role: My Client is a top tier investment bank currently in the... Full details: http://jobs.phds.org/job/13321/gqr-global-quant-recruitment/experienced-fx-c\ o mmodities -------------------------------------------------------------------- Title: Fixed Income Macro Risk Manager .89´\.8b 2+ Yrs Experience .89´\.8b High Profile Asset Managers - London Employer: GQR | Global Quant Recruitment - Quant Jobs Location: London, United Kingdom Fixed Income Macro Risk Manager .89´\.8b 2+ Yrs Experience .89´\.8b High Profile Asset Managers - London The Client: Asset Managers Location: London, UK The Role: My Client is an investment asset managers with a good reputation in the market place and are... Full details: http://jobs.phds.org/job/13320/gqr-global-quant-recruitment/fixed-income-mac\ r o-risk -------------------------------------------------------------------- Title: Interest Rates Derivatives Quant Employer: Options Group Location: Chicago, IL, United States Our client is looking for an experienced Interest rate derivatives quant. The candidate needs to have strong math skills with a very good knowledge of Stochastic Calculus as well as well as outstanding C++ skills. Background in statistics and some... Full details: http://jobs.phds.org/job/13318/options-group/interest-rates-derivatives -------------------------------------------------------------------- Title: Equity Derivatives Quant Employer: Options Group Location: Chicago, IL, United States Description: Our client is looking for an experienced equity derivatives quant. The candidate needs to have strong math skills with a very good knowledge of Stochastic Calculus as well as well as outstanding C++ skills. Background in statistics and... Full details: http://jobs.phds.org/job/13317/options-group/equity-derivatives-quant -------------------------------------------------------------------- Title: Commodities Quant - Energy Employer: Options Group Location: New York, NY, United States Our global banking client is seeking a commodtities quant to work on their presitigious desk. Requirements: 1-3 years experience working on a commodities trading desk. Energy preferred. Ph.D. required. Full details: http://jobs.phds.org/job/13316/options-group/commodities-quant-energy -------------------------------------------------------------------- Title: High Frequency Quantitative Researcher / strategist with C++ for hedge fund Employer: Hedge Fund Location: New York, NY, United States Hedge Fund in New York is hiring for a Quant Strategist in high frequency to join hedge fund to aid in revenue growth. You will be responsible for managing massive data sets and optimizing trading strategies as well as build your own models. The... Full details: http://jobs.phds.org/job/13345/hedge-fund/high-frequency-quantitative -------------------------------------------------------------------- Title: 2 Senior FX Quantitative Analysts/Developers required with a leading Investment Bank team - London Employer: GQR | Global Quant Recruitment - Quant Jobs Location: London, United Kingdom 2 Senior FX Quantitative Analysts/Developers required with a leading Investment Bank team - London The Client: Leading Investment Bank Location: London, UK The Role: My Client is an investment bank who has had a strong quarter despite of the... Full details: http://jobs.phds.org/job/13319/gqr-global-quant-recruitment/2-senior-fx-quan\ t itative -------------------------------------------------------------------- Title: Software Automation Engineer - QA Employer: Options Group Location: Houston, TX, United States Our client is looking for an experienced Software Engineer with experience of building test automation infrastructure and tools for real-time, production critical applications. Requirements: * 5+ years experience in at least one major scripting or... Full details: http://jobs.phds.org/job/13315/options-group/software-automation-engineer -------------------------------------------------------------------- Title: Quantitative Strategist Employer: Confidential Location: New York, NY, United States The E-Trading Group is looking to hire a Quantitative Strategist to work on High Frequency execution analysis. * An advanced degree in a quantitative field such as: * * Statistics * Mathematics * Physics * Engineering * Strong computer science skills... Full details: http://jobs.phds.org/job/13314/confidential/quantitative-strategist -------------------------------------------------------------------- Title: Quant Researcher / Analyst with Statistics background and high frequency experience Employer: Huxley Associates Location: New York, NY, United States Electronic Market Maker in New York is hiring for a skilled mathematician for a quant analyst to contribute to the statistical modeling of high frequency electronic trading group. You will analyze real data in a quantitative manner applying your... Full details: http://jobs.phds.org/job/13313/huxley-associates/quant-researcher-analyst -------------------------------------------------------------------- Title: Quantitative Model Validation - Commodities Employer: Deutsche Bank Location: New York, NY, United States About Deutsche Bank Deutsche Bank is a leading global investment bank with a strong and profitable private clients franchise. A leader in Germany and Europe, the bank is continuously growing in North America, Asia and key emerging markets. With 80,... Full details: http://jobs.phds.org/job/13312/deutsche-bank/quantitative-model-validation -------------------------------------------------------------------- Title: Portfolio Manager (High Frequency) Employer: Hedge Fund Location: New York, NY, United States The ideal candidate will have a track record of running a book in the asset classes of equities, futures, and currencies, medium or high frequency. Sharpe Ratio MUST be above 4.0 Full details: http://jobs.phds.org/job/12315/hedge-fund/portfolio-manager-high -------------------------------------------------------------------- Title: Market Making Desk seeks C++ Guru Employer: Confidential Location: New York, NY, United States Highly regarded Market Making desk seeks a talented C++ Engineer * An advanced degree in Computer Science * Solid Knowledge of data structures and algorithms (i.e. know the technical tradeoffs of various approaches, performance implications,... Full details: http://jobs.phds.org/job/11675/confidential/market-making-desk-seeks -------------------------------------------------------------------- Title: Automated Market Making Developer Employer: Confidential Location: New York, NY, United States Automated Market Making group within an Investment Bank is looking Market Maker * The role offers the successful candidate... Full details: http://jobs.phds.org/job/11538/confidential/automated-market-making -------------------------------------------------------------------- Title: Network Engineer - Market Making Group Employer: Confidential Location: New York, NY, United States Premier Market Making Group is seeking a true networking visionary. .89´¢ Maintain LAN/WAN network, connectivity between all datacenters and to all financial extranets; monitor network and network equipment for problems, and contact vendors to... Full details: http://jobs.phds.org/job/12270/confidential/network-engineer-market -------------------------------------------------------------------- Award in Bioengineering Employer: Institute for BioNanotechnology in Medicine at Northwestern University Location: Chicago, IL, United States Northwestern University has funding available for young Development Award in Bioengineering provides up to two years of funding... Full details: http://jobs.phds.org/job/13298/institute-for-bionanotechnology/postdoctoral-\ f ellows -------------------------------------------------------------------- Title: RISK QUANT/ REINSURANCE STRATEGIES GROUP/ LONDON Employer: COMPREHENSIVE RECRUITING Location: London, United Kingdom Top tier investment bank seeks quantitative strategist for Reinsurance Strategies Group in London. On a daily basis, the candidate will collaborate with underwriting, origination, marketing and trading teams with regard to valuation and risk analysis... Full details: http://jobs.phds.org/job/13304/comprehensive-recruiting/risk-quant-reinsuran\ c e -------------------------------------------------------------------- Title: RISK MODELING/ ASSOC. VP/ NYC Employer: GLOBAL FINANCIAL MARKETS FIRM Location: New York, NY, United States This person will focus on the development and implementation of Rating Templates, Probability of Default (PD), Loss Given Default (LGD), and Exposure At Default (EAD) modelling methodologies as part of the ongoing enhancement of credit models for... Full details: http://jobs.phds.org/job/13303/global-financial-markets/risk-modeling-assoc-\ v p -------------------------------------------------------------------- Title: Quant Analyst/ Model Validation/ Wall Street Employer: Tier 1 Investment Bank Location: New York, NY, United States Global Investment Bank is looking to add a senior quantitative analyst to their Global Financial Research team in London and New York. This person will be part of a team that is responsible for innovation, model enhancement, quantitative solutions... Full details: http://jobs.phds.org/job/13302/tier-1-investment-bank/quant-analyst-model-va\ l idation -------------------------------------------------------------------- Title: Senior Quantitative Analyst Employer: Large company in energy Location: West Coast, United States Quantitative Analyst / Senior Quantitative Analyst position We are looking for candidates with experience in Quantitative Modeling, Risk management for our clients on the West Coast. Candidates with a Financial Engineering degree or a Ph.D would be... Full details: http://jobs.phds.org/job/13301/large-company-in-energy/senior-quantitative-a\ n alyst -------------------------------------------------------------------- Title: Cash Equity Quantitative Analyst Employer: Integrated Management Resources, Inc. Location: London, United Kingdom Well-known Investment Bank is looking for a PhD Quant to add to their research efforts in London. Dynamic position working directly with the trading desk, developing pricing and risk models. Qualified candidates will have a PhD from a reputable... Full details: http://jobs.phds.org/job/13300/integrated-management/cash-equity-quantitativ\ e -------------------------------------------------------------------- Title: Quantitative Model Developer Department: Financial Engineering Employer: Investment Technology Group Location: Boston, MA, United States ITG's Financial Engineering group is involved in research, development and production of sophisticated, statistical/quantitative tools for investment and trading decisions. The group consists of 20 employees of whom the majority has advanced degrees... Full details: http://jobs.phds.org/job/13299/investment-technology/quantitative-model-deve\ l oper -------------------------------------------------------------------- Title: Energy Risk Manager Employer: Large company Location: West Coast, United States Risk Manager/ Principal Quantitative Analyst We are looking for Risk Managers/ Principal Quantitative Analysts. These positions are full time positions and would be based with our clients on the West coast Please note the skills we are looking for... Full details: http://jobs.phds.org/job/13272/large-company/energy-risk-manager -------------------------------------------------------------------- Title: Quantitative Strategist / PhD Degree Employer: Integrated Management Resources, Inc. Location: New York, NY, United States Top Tier Investment Bank seeks a Strategist possessing a PhD in relevant field, to support strategies group in New York. Candidate will be working on pricing, risk management and modeling analytics and technology. This individual will assist other... Full details: http://jobs.phds.org/job/11714/integrated-management/quantitative-strategist\ -------------------------------------------------------------------- Title: Quantitative Analyst / Oil-Petroleum Background Employer: Integrated Management Resources, Inc. Location: Chicago, IL, United States Major Energy Company is seeing a Quantitative Analyst with a Masters and/or a PhD in Financial Engineering, Mathematics, Physics or other related field, to support commodity desk. Candidate must have an oil/petroleum background. Must have a strong... Full details: http://jobs.phds.org/job/11588/integrated-management/quantitative-analyst -------------------------------------------------------------------- Title: Physicist Postdoc Fellow 22509 Employer: Lawrence Berkeley National Laboratory Location: Berkeley, CA, United States POSITION SUMMARY: The Ion Beam Technology Group of LBNL's Accelerator and Fusion Research Division has an opening for a postdoctoral fellow to study various types of ion sources and low energy beam systems for the purpose of developing neutron and... Full details: http://jobs.phds.org/job/13296/lawrence-berkeley-national/physicist-postdoc-\ f ellow -------------------------------------------------------------------- Title: Physicist Post Doc Fellow 22449 Employer: Lawrence Berkeley National Laboratory Location: Berkeley, CA, United States Lawrence Berkeley National Laboratory (LBNL) is a world leader in science and engineering research, with 11 Nobel Prize recipients over the past 75 years, and 59 present members of the National Academy of Sciences. LBNL conducts unclassified research... Full details: http://jobs.phds.org/job/13295/lawrence-berkeley-national/physicist-post-doc\ - fellow -------------------------------------------------------------------- Title: Computational Sci PD Fellow 21475 Employer: Lawrence Berkeley National Laboratory Location: Berkeley, CA, United States KEY SKILLS: Scientific computing, Applied Mathematics, C++, Fortran POSITION SUMMARY: The Center for Computational Sciences and Engineering has two openings for postdoctoral researchers to work on the development of high-resolution methods for... Full details: http://jobs.phds.org/job/13293/lawrence-berkeley-national/computational-sci-\ p d -------------------------------------------------------------------- Title: Valuation Review Analyst Employer: Comprehensive Recruiting Location: New York, NY, United States Responsibilities for this role include reviewing linear products in one or more commodity portfolio to insure measurement at fair value. Additional responsibilities include: Proactive in evaluating the seasonableness of traders commodity marks;... Full details: http://jobs.phds.org/job/13292/comprehensive-recruiting/valuation-review-ana\ l yst -------------------------------------------------------------------- Title: Quant Analyst - Interest Rates Employer: Comprehensive Recruiting Location: New York, NY, United States Prestigious Global Investment Bank seeks a Senior Quantitative Analyst to join their New York City based Interest Rate Derivative trading team. The ideal candidate will have 5-7 years of financial experience with extensive experience supporting... Full details: http://jobs.phds.org/job/13291/comprehensive-recruiting/quant-analyst-intere\ s t -------------------------------------------------------------------- Title: Quantitative Analyst - Model Validation Employer: Comprehensive Recruiting Location: New York, NY, United States Global Investment Bank is looking to add a senior quantitative analyst to their Global Financial Research team in London and New York. This person will be part of a team that is responsible for innovation, model enhancement, quantitative solutions... Full details: http://jobs.phds.org/job/13290/comprehensive-recruiting/quantitative-analyst\ -------------------------------------------------------------------- Title: Quant Trader with High Frequency Strategy .89´\.8b Prop Group Toronto Employer: Huxley Associates Location: Toronto, ON, Canada Prop trading firm in Toronto hiring high frequency quantitative trader with a proven high frequency model. You will take part in the full lifecycle of quantitative trading: idea generation, research and back-test, optimization and execution. Looking... Full details: http://jobs.phds.org/job/13289/huxley-associates/quant-trader-with-high -------------------------------------------------------------------- Title: Quantitative Engineer - High Frequency Tick Data - Equity Trading - Algorithmic Trading, PhD Employer: Financial Software - leader in the area of high-frequency trading \ and execution optimization Location: New York, NY, United States QUANTITATIVE ENGINEER Research/Quant/Quantitative Engineer COMPANY is seeking new members for our quant team. COMPANY is an industry leader in the area of high-frequency trading and execution optimization. Our product distinguishes itself from... Full details: http://jobs.phds.org/job/13285/financial-software-leader/quantitative-engine\ e r -------------------------------------------------------------------- Title: Quant Modeler to support Proprietary Trading desk Employer: Investment Bank Location: New York, NY, United States Quant modeler is being hired by a mid-sized Investment Bank to work as a dedicated quant to support their proprietary trading desks which trade multiple asset classes. This role will suit a candidate with hands on quant modeling skills in a Front... Full details: http://jobs.phds.org/job/13284/investment-bank/quant-modeler-to-support -------------------------------------------------------------------- Title: Postdoctoral Position in Gas Hydrates Department: Chemical Engineering Department Employer: Colorado School of Mines Location: Golden, CO, United States 1-2 Postdoctoral Fellowship positions are available in the following areas: hydrogen storage in clathrate materials, hydrates in flow assurance (maintaining flow in subsea pipelines). Interested candidates should submit a curriculum vitae and cover... Full details: http://jobs.phds.org/job/13283/colorado-school-of-mines/postdoctoral-positio\ n -------------------------------------------------------------------- Title: Postdoctoral and Senior Research Awards Department: Fellowship Programs Employer: National Research Council of the National Academies Location: Washington, DC, United States Research Associateship Program Research Opportunities. The National Research Council of the National Academies is accepting applications for competitive awards for independent postdoctoral scientific research to be conducted at participating US... Full details: http://jobs.phds.org/job/10274/national-research-council/postdoctoral-and-se\ n ior -------------------------------------------------------------------- Title: Editor- Physics Employer: International Scientific Publisher Location: New York, NY, United States Our client is seeking an Acquisitions Editor of Physics to join this prestigious international scientific publisher of books, journals, and electronic media, with subsidiaries and representatives all over the world. Publications range from medicine... Full details: http://jobs.phds.org/job/13280/international-scientific/editor-physics -------------------------------------------------------------------- Title: C++ Linux/Windows Developer (Server Side) for Futures Prop Trading Firm Employer: Futures Group Location: New York, NY, United States Prop Trading Firm is seeking a C++ Linux/Windows Developer (Server Side) to join their Futures Trading desk. This is a rare opportunity to join an elite team of experienced technologists and Traders, contributing to the development of their Futures... Full details: http://jobs.phds.org/job/13278/futures-group/c-linux-windows-developer -------------------------------------------------------------------- Title: Postdoctoral position in Direct numerical simulation of particulate flows Department: Centre for Material Forming - CEMEF Employer: MINES ParisTech Location: Valbonne, France === Subject: Direct numerical simulation of particulate flows.Application to rheology of suspensions: Dense or highly concentrated particulate medium are very common in several engineering fields such as civil engineering, composite materials, food or... Full details: http://jobs.phds.org/job/13276/mines-paristech/postdoctoral-position -------------------------------------------------------------------- Title: Postdoctoral position in Modelling of the thermo-visco hyper elastic couplings in polymers Department: Centre for Material Forming - CEMEF Employer: MINES ParisTech Location: Valbonne, France Subject : Modelling of the thermo-visco hyper elastic couplings in polymers; Application to the identification of constitutive parameters. Visco elastic behaviour and dissipative effects in polymers make necessary to account for thermo mechanical... Full details: http://jobs.phds.org/job/13277/mines-paristech/postdoctoral-position -------------------------------------------------------------------- Title: Senior High Frequency Quant Developer / Researcher with C++ Employer: Prop Trading Firm Location: Chicago, IL, United States Proprietary trading group in Chicago is hiring for a senior quantitative developer with strong C++ and market knowledge to pioneer growth effort in high frequency trading. You will work with a small group of front office quants reporting directly to... Full details: http://jobs.phds.org/job/13275/prop-trading-firm/senior-high-frequency -------------------------------------------------------------------- Title: Research Associate/Research Fellow (2 positions): Bioinformatics Department: Phenomics and Bioinformatics Research Centre Employer: University of South Australia Location: Adelaide Sa, Australia The Phenomics and Bioinformatics Research Centre (PBRC), a joint initiative of the University of South Australia and the Australian Centre for Plant Functional Genomics (ACPFG), is seeking to appoint Research Associates or Research Fellows to conduct... Full details: http://jobs.phds.org/job/13274/university-of-south-australia/research-associ\ a te-research -------------------------------------------------------------------- Title: Postdoctoral Position in Organic Chemistry and Materials Research Department: Department of Chemistry/Sieburth Group Employer: Temple University Location: Philadelphia, PA, United States Immediate opening for postdoctoral research fellow for interdisciplinary effort in materials research, funded by the NIH. Review of applications will begin immediately. Ideal applicant would have a strong background in organic chemistry, polymer... Full details: http://jobs.phds.org/job/13267/temple-university/postdoctoral-position -------------------------------------------------------------------- Title: Post Doc position in inorganic phosphate chemistry Employer: Ductile Ceramics Location: Wilson, NC, United States Develop acid base cements using phosphates and metal oxides. Use phosphate/metal oxide combinations to create resins for composites. Use acid/base combinations to create paints and coatings. Hands on research working directly with the creator of the... Full details: http://jobs.phds.org/job/13264/ductile-ceramics/post-doc-position-in -------------------------------------------------------------------- Title: Geological Postdoc Fellow Employer: Lawrence Berkeley National Laboratory Location: Berkeley, CA, United States This one-year term post-doctoral researcher position focuses on conducting laboratory experiments to characterize the geophysical, mechanical, and hydrological properties of porous materials at reservoir pressures and temperatures, analyzing the... Full details: http://jobs.phds.org/job/13265/lawrence-berkeley-national/geological-postdoc\ - fellow -------------------------------------------------------------------- Title: Physicist Postdoc Fellow 22599 Employer: Lawrence Berkeley National Laboratory Location: Berkeley, CA, United States Postdoctoral Position in Experimental Nuclear Physics. The Nuclear Science Division of the Lawrence Berkeley National Laboratory invites applications for a postdoctoral position with the Nuclear Structure Group. Our experimental activities use state-... Full details: http://jobs.phds.org/job/13261/lawrence-berkeley-national/physicist-postdoc-\ f ellow -------------------------------------------------------------------- Title: Physicist Postdoc Fellow 22372 Employer: Lawrence Berkeley National Laboratory Location: Berkeley, CA, United States Be part of something great! The scientific focus for this position will be characterization of metalloenzymes such as nitrogenase and hydorgenase by spectroscopic methods. The incumbent will be responsible for the collection and interpretation of... Full details: http://jobs.phds.org/job/13262/lawrence-berkeley-national/physicist-postdoc-\ f ellow -------------------------------------------------------------------- Title: Physicist Postdoc Fellow 22135 Employer: Lawrence Berkeley National Laboratory Location: Berkeley, CA, United States The Physics Division at the Lawrence Berkeley National Laboratory is seeking outstanding candidates for a postdoctoral position in experimental cosmology. We are particulary interested in candidates with interest in the SDSS-III Baryon Oscillation... Full details: http://jobs.phds.org/job/13263/lawrence-berkeley-national/physicist-postdoc-\ f ellow -------------------------------------------------------------------- Title: Derivatives C++ Technical Lead Employer: Clarity Technical Search Location: New York, NY, United States The Derivatives Application group is seeking a hands-on technical lead with in-depth knowledge of interest rate derivatives or structured notes or convertible bonds. The candidate will have significant experience developing systems involving... Full details: http://jobs.phds.org/job/13271/clarity-technical-search/derivatives-c-techni\ c al -------------------------------------------------------------------- Title: Equities C++ Developer Employer: Clarity Technical Search Location: New York, NY, United States As a member of a small team, for this international bank, you will write multithreaded applications on UNIX(Linux)/C/C++/Java platforms to ensure that real-time exchange-distributed pricing is efficiently received and delivered for processing by the... Full details: http://jobs.phds.org/job/13270/clarity-technical-search/equities-c-developer\ -------------------------------------------------------------------- Title: Postdoctoral Fellow in Natural Product Chemistry/Chemical Ecology Department: Botany and Zoology, School of Biology Employer: The Australian National University Location: Canberra Act, Australia A three year fixed-term position of Postdoctoral Fellow (Level A) in natural product chemistry/chemical ecology is available from July 2009. The position is funded by a multidisciplinary Australian Research Council Linkage grant..89´[Hyphen] The position is a... Full details: http://jobs.phds.org/job/13257/the-australian-national/postdoctoral-fellow-i\ n -------------------------------------------------------------------- Title: Deputy Director Employer: Space Telescope Science Institute, Director's Office Location: Baltimore- Johns Hopkins Homewood Campus, United States The Space Telescope Science Institute (STScI), the science center for the Hubble Space Telescope and James Webb Space Telescope, and a premier astronomical research institution, is conducting a search for a Deputy Director. Full details: http://jobs.phds.org/job/13260/space-telescope-science/deputy-director -------------------------------------------------------------------- Title: LiDAR applied to landscape archaeology (France & Slovenia) Employer: MSHE Ledoux Location: BesaníÛon, France Applications are invited to work with the team of Dr Laure Nuninger (archaeologist), Dr FraníÛois-Pierre Tourneux (geographer) and Pr Kristof Ostir (physicist, geodesist) to study landscape management over the long term using LiDAR data, Color /... Full details: http://jobs.phds.org/job/12579/mshe-ledoux/lidar-applied-to-landscape -------------------------------------------------------------------- Title: Senior Biotechnology Research Employer: Cleantech Location: Eastern Provinces, Canada Top Canadian biotechnology firm is seeking a senior biotechnology professional to join their clean energy technology group. Candidate will be responsible for guiding research as well as managing the research team. Candidates should come from a top-... Full details: http://jobs.phds.org/job/13254/cleantech/senior-biotechnology -------------------------------------------------------------------- Title: Statistical Programmer Employer: Top Oncology Pharmaceuticals Company Location: Boston, MA, United States Large pharmaceuticals company is seeking a statistical programmer to join their oncology clinical trials team. Candidates will be responsible for programing and validating statistical analysis of clinical data. In addition to developing new analytics... Full details: http://jobs.phds.org/job/13253/top-oncology-pharmaceuticals/statistical-prog\ r ammer -------------------------------------------------------------------- Title: Bioinformatics / Computational Biology Professional Employer: Mid-Sized Biotechnology Firm Location: Boston, MA, United States Drug discovery platform at top biotechnology firm is seeking a bioinformatics professional to perform research and analysis on large, gene networks in order to identify therapy targets. Candidates will be responsible for developing analytics... Full details: http://jobs.phds.org/job/12623/mid-sized-biotechnology/bioinformatics-comput\ a tional -------------------------------------------------------------------- Title: Materials Project Scientist Employer: Lawrence Berkeley National Laboratory Location: Berkeley, CA, United States The LBNL Windows and Daylighting Group has an opportunity for a Post Doc Researcher to help develop a new generation of energy-saving electrochromic windows. As a Project Scientist, you will: * Characterize thin film components of electrochromic... Full details: http://jobs.phds.org/job/13244/lawrence-berkeley-national/materials-project-\ s cientist -------------------------------------------------------------------- Title: Options Market Making Architect with C++ skills Employer: Futures Group Location: Jersey City, NJ, United States Premier Financial Firm seeks Lead Architect to join an elite team that is building a brand new Options Making system. This candidate will be responsible for formulating and driving the implementation of the initial platform, setting the course for... Full details: http://jobs.phds.org/job/13248/futures-group/options-market-making -------------------------------------------------------------------- Title: Multi-Terabyte Database Developer w/ Java experience Employer: Futures Group Location: New York, NY, United States Global Investment firm is seeking Multi-Terabyte Database Developer w/ Java experience for thier highly visible and profitable High-Frequency Equity Options Trading Desk. Canidates will be responsible for designing, implementing and maintaining... Full details: http://jobs.phds.org/job/13249/futures-group/multi-terabyte-database -------------------------------------------------------------------- Title: Sr. C++/C# Analytics Developer for (MBS/ABS) Employer: Futures Group Location: Greenwich, CT, United States Top tier Global Financial Firm is seeking a Sr. C++/C# Analytics Developer needed to join a highly profitable team that is building a brand new MBS Trading system. Candidates will develop, validate, and implement quantitative and analytical models to... Full details: http://jobs.phds.org/job/13250/futures-group/sr-c-c-analytics-developer -------------------------------------------------------------------- Title: Senior C++ Market Data Engineer Employer: Futures Group Location: New York, NY, United States Top Global Firm is seeking a Senior C++ Market Data Engineer for their New York City based office. Candidates will be responsible for migrating existing client market data applications to various data products, developing new applications (based on... Full details: http://jobs.phds.org/job/13251/futures-group/senior-c-market-data -------------------------------------------------------------------- Title: Sr. quantitative analysts and developers Department: Fixed Income Research and Analytics Employer: Asset Management / Distressed Hedge Fund Location: Miami, FL, United States A well positioned hedge fund located in Miami FL is seeking two senior level quant/developer in its fixed income research and analytics group. This is an outstanding opportunity for a Jr. and Sr. quantitative analysts and developers to land their... Full details: http://jobs.phds.org/job/13255/asset-management-distressed/sr-quantitative-a\ n alysts -------------------------------------------------------------------- Title: Senior Machine Learning Scientist / Bioinformatics Scientist/ Computational Biology Scientist Department: Machine Learning Department Employer: Aureon Laboratories, Inc. Location: Yonkers, NY, United States About the position: Aureon.89´\.9cs Machine Learning Department receives data defining a patient.89´\.9cs disease profile from a variety of domains, including clinical information (i.e., race, age), bio-molecular marker expression, and image analysis of the... Full details: http://jobs.phds.org/job/12457/aureon-laboratories-inc/senior-machine-learni\ n g -------------------------------------------------------------------- Title: Quantitative Analyst Employer: Comprehensive Recruiting Location: New York, NY, United States Global Investment Bank is looking to add a Quantitative Analyst to support their Equity Derivative Prop Trading desk. Candidates should be skilled in developing quantitative models & proprietary trading algorithms for equity derivatives. The... Full details: http://jobs.phds.org/job/13242/comprehensive-recruiting/quantitative-analyst\ -------------------------------------------------------------------- Title: Research associate in applied physics-MEMS Department: Nanofabrication Employer: National Research Council Location: Ottawa, ON, Canada Research associate in applied physics-MEMS-Nanofabrication The major responsibility of this position is to undertake research focused on fabrication of electrostatically actuated curling electrodes for smart windows applications, as part of a small... Full details: http://jobs.phds.org/job/13240/national-research-council/research-associate-\ i n -------------------------------------------------------------------- Title: Junior Quant Analyst Employer: Comprehensive Recruiting Location: London, United Kingdom Global hedge fund seeks junior quantitative analyst for UK equities group. Candidate must have a PH.D in a hard science from a top university and strong quantitative and programming skills. Finance background a plus, but not essential. Candidate must... Full details: http://jobs.phds.org/job/13241/comprehensive-recruiting/junior-quant-analyst\ -------------------------------------------------------------------- Title: Postdoctoral position in fluids mechanics : cavitation with Employer: Laboratoire des Ecoulements Gí©ophysiques et Industriels Location: Grenoble, France A post doctoral position is available immediately for one year at the Geophysical and Industrial Flow Laboratory (LEGI) of the University of Grenoble, France. The aim of the research is to investigate the raise of the temperature within a collapsing... Full details: http://jobs.phds.org/job/13239/laboratoire-des-ecoulements/postdoctoral-posi\ t ion -------------------------------------------------------------------- Post your job (free!): http://jobs.phds.org/job/post PhDs.org: Science, Math, and Engineering Career Resources --------------------------------------------------------- * Job Listings: http://jobs.phds.org/ - Job board with hundreds of listings for Ph.D.s - Reach tens of thousands of Ph.D.s each month * Graduate School Rankings: http://graduate-school.phds.org/ - Comprehensive, customizable rankings of graduate programs * Career Resources: http://www.phds.org/ - Pointers to the best resources on the web for: + getting into graduate school + writing your dissertation + jobs for Ph.D.s in academia and industry * Engineering Science Weblog: http://blog.phds.org/ - Building better scientists and engineers === Subject: Find the area of common region posting-account=em7BlwoAAADOhRXv8AilqN49TjuokE7m Trident/4.0; .NET CLR 2.0.50727; FDM; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) Find the area of common region Pythagorean Triangles share common inradius r where r is a prime number >=5, Find the area of common region shared by those triangles in terms of r. This is not a homework :) === Subject: Re: Find the area of common region posting-account=H1y7YgoAAADzGQwbcYaL9UvwttgsjOjp AppleWebKit/525.27.1 (KHTML, like Gecko) Version/3.2.1 Safari/525.27.1,gzip(gfe),gzip(gfe) > Find the area of common region Pythagorean Triangles share common > inradius r where r is a prime number >=5, Find the area of common > region shared by those triangles in terms of r. Doesn't sound right to me. If, say, two triangles have the same inradius, that's not enough to compute the area of their intersection, you have to know how the triangles are situated. -- GM === Subject: Re: Find the area of common region > Doesn't sound right to me. If, say, two triangles have the same > inradius, that's not enough to compute the area of their > intersection, you have to know how the triangles are situated. Yes, it's a rather poorly stated question. I assumed that the triangles are all supposed to share a common right angle. - Tim === Subject: Re: Find the area of common region posting-account=em7BlwoAAADOhRXv8AilqN49TjuokE7m Trident/4.0; .NET CLR 2.0.50727; FDM; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) Doesn't sound right to me. If, say, two triangles have the same > inradius, that's not enough to compute the area of their > intersection, you have to know how the triangles are situated. Yes, it's a rather poorly stated question. I assumed that the > triangles are all supposed to share a common right angle. - Tim Ok, Let me tell you that my following conjecture have been proved If there exist only THREE Pythagorean Triangles which share same incircle with inradius radius r>=5, r is a prime number. You can have a look at proof http://answers.yahoo.com/question/index?qid=20090330053142AAp0lta So now the question becomes find the common area shared by those THREE Triangles :) === Subject: x^3+y^3+z^3 = 1 and Pell equations posting-account=-ACVjwoAAAAVqSiDl929-Pe1jSK2zs-Q FunWebProducts; GTB5; .NET CLR 2.0.50727; .NET CLR 1.1.4322; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) Hello all, Turns out solving x^3+y^3+z^3 = 1 can entail solving a certain Pell equation p^2-dq^2 = 1. We give the identity, (a+c^2-ac^3)^3 + (1-ac+bc)^3 + (ac^3-b-c^2)^3 = 1 {a,b,c} = {12qrt, 3(q-r)(3q+r)t, 3t^2}, or {a,b,c} = {12qrt, 3(q+r)(3q-r)t, 3t^2}, where r = p-18qt^3 and {p,q} satisfy the Pell equation: p^2 - 3(108t^6-1)q^2 = 1 This has a polynomial fundamental soln: {p 1, q 1} = {216t^6-1, 12t^3} from which one can easily derive other polynomial solns {p n, q n} for higher n. Using trivial {p 0, q 0} = {1,0} on the {a,b,c} gives the soln, (1-9t^3)^3 + (9t^4)^3 + (3t-9t^4)^3 = 1 Using {p 1, q 1} and {p 1, -q 1} gives a 10th and 16th deg soln. Using {p 2, q 2} and {p 2, -q 2} gives a 22th and 28th deg soln, and so on. You can easily verify this with Mathematica (or Maple, etc). (The polynomial solns are of degree 6m+4 and are the same as given by a recursion found by D.H. Lehmer.) P.S. This day of the year is significant for me, so I'm posting this identity today ahead of the others. :-) There's also a Pell equation for x^4+y^4 = z^2+1, but that's another story\.83 [CapitalEth] Titus === Subject: Re: x^3+y^3+z^3 = 1 and Pell equations Hello all, Turns out solving x^3+y^3+z^3 = 1 can entail solving a certain Pell equation p^2-dq^2 = 1. We give the identity, (a+c^2-ac^3)^3 + (1-ac+bc)^3 + (ac^3-b-c^2)^3 = 1 {a,b,c} = {12qrt, 3(q-r)(3q+r)t, 3t^2}, or {a,b,c} = {12qrt, 3(q+r)(3q-r)t, 3t^2}, where r = p-18qt^3 and {p,q} satisfy the Pell equation: p^2 - 3(108t^6-1)q^2 = 1 This has a polynomial fundamental soln: {p_1, q_1} = {216t^6-1, 12t^3} from which one can easily derive _other_ polynomial solns {p_n, q_n} for higher n. Using trivial {p_0, q_0} = {1,0} on the {a,b,c} gives the soln, (1-9t^3)^3 + (9t^4)^3 + (3t-9t^4)^3 = 1 Using {p_1, q_1} and {p_1, -q_1} gives a 10th and 16th deg soln. Using {p_2, q_2} and {p_2, -q_2} gives a 22th and 28th deg soln, and so on. You can easily verify this with Mathematica (or Maple, etc). (The polynomial solns are of degree 6m+4 and are the same as given by a recursion found by D.H. Lehmer.) P.S. This day of the year is significant for me, so I'm posting this identity today ahead of the others. :-) There\.89s also a Pell equation for x^4+y^4 = z^2+1, but that\.89s another storyá \.9a Titus Neat but wait until JSH reads about it. I am expecting his post will start with They stole my work about Pell's equation.. cg === Subject: Re: x^3+y^3+z^3 = 1 and Pell equations Hello all, Turns out solving x^3+y^3+z^3 = 1 can entail solving a certain Pell > equation p^2-dq^2 = 1. We give the identity, (a+c^2-ac^3)^3 + (1-ac+bc)^3 + (ac^3-b-c^2)^3 = 1 {a,b,c} = {12qrt, 3(q-r)(3q+r)t, 3t^2}, or > {a,b,c} = {12qrt, 3(q+r)(3q-r)t, 3t^2}, where r = p-18qt^3 and {p,q} satisfy the Pell equation: p^2 - 3(108t^6-1)q^2 = 1 This has a polynomial fundamental soln: {p_1, q_1} = {216t^6-1, 12t^3} from which one can easily derive _other_ polynomial solns {p_n, q_n} > for higher n. Using trivial {p_0, q_0} = {1,0} on the {a,b,c} gives the soln, (1-9t^3)^3 + (9t^4)^3 + (3t-9t^4)^3 = 1 Using {p_1, q_1} and {p_1, -q_1} gives a 10th and 16th deg soln. > Using {p_2, q_2} and {p_2, -q_2} gives a 22th and 28th deg soln, and > so on. You can easily verify this with Mathematica (or Maple, etc). (The polynomial solns are of degree 6m+4 and are the same as given by > a recursion found by D.H. Lehmer.) P.S. This day of the year is significant for me, so I'm posting this > identity today ahead of the others. :-) There\.89s also a Pell equation > for x^4+y^4 = z^2+1, but that\.89s another storyá \.9a Titus Neat but wait until JSH reads about it. > I am expecting his post will start with They stole my work about Pell's > equation.. cg > JSH could never make it beyond squares, or square roots (that +_ nasty stuff) === Subject: Re: Is relativity I L L O G I C A L? PD, let me be clear with you. I am trying to engage in a scientific discussion here. That is why I use scientifically valid references such as the Handbook of Physics. Your insistence on using an unscientific reference such as wikipedia (only Eric uses it for his thesis, thus his continued repetition of the courses) only adds substance to the claim that your sole purpose here is to obfuscate the issues. As I am sure you have forgotten the definition, according to the technical reference Handbook of Physics, here it is again: Inertial frame - a frame in which Newton's laws hold, in particular the law of inertia. In such a frame, a body that is free of forces remains in its state of motion. Therefore, inertial frames are those that move with uniform speed relative to each other. Note again the specific reference to Newton's law of inertia. Now a photon is in uniform linear motion. Unless acted on by an outside force, such as gravity, it will remain in that uniform linear motion. The photon satisfies all the requirements of an inertial frame, except of course your requirement that it 'simply does not'. Which is about as unscientific as it gets. === Subject: Re: Is relativity I L L O G I C A L? posting-account=-Yr6sAoAAAArJwgdHLA4MBxm4oIdzdtZ InfoPath.1; .NET CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729; MS-RTC EA 2),gzip(gfe),gzip(gfe) (NetCache NetApp/6.0.2P1) > PD, let me be clear with you. I am trying to engage in a scientific > discussion here. That is why I use scientifically valid references > such as the Handbook of Physics. I'm happy to provide you with 5 or 6 more scientifically valid > references. You seem to be remarkably resistant to that idea. Why? Your insistence on using an > unscientific reference such as wikipedia (only Eric uses it for his > thesis, thus his continued repetition of the courses) only adds > substance to the claim that your sole purpose here is to obfuscate the > issues. As I am sure you have forgotten the definition, according to > the technical reference Handbook of Physics, here it is again: Inertial frame - a frame in which Newton's laws hold, in particular > the law of inertia. In such a frame, a body that is free of forces > remains in its state of motion. Therefore, inertial frames are those > that move with uniform speed relative to each other. Note again the specific reference to Newton's law of inertia. Now a > photon is in uniform linear motion. Unless acted on by an outside > force, such as gravity, it will remain in that uniform linear motion. > The photon satisfies all the requirements of an inertial frame, except > of course your requirement that it 'simply does not'. Which is about > as unscientific as it gets.- Hide quoted text - - Show quoted text - What is your resistance to the Handbook of Physics? An inertial frame is a basic concept. What makes you think we need 6 references? === Subject: Re: Is relativity I L L O G I C A L? posting-account=-Yr6sAoAAAArJwgdHLA4MBxm4oIdzdtZ InfoPath.1; .NET CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; MS-RTC LM 8; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) (NetCache NetApp/6.0.2P1) > PD, let me be clear with you. I am trying to engage in a scientific > discussion here. That is why I use scientifically valid references > such as the Handbook of Physics. I'm happy to provide you with 5 or 6 more scientifically valid > references. You seem to be remarkably resistant to that idea. Why? Your insistence on using an > unscientific reference such as wikipedia (only Eric uses it for his > thesis, thus his continued repetition of the courses) only adds > substance to the claim that your sole purpose here is to obfuscate the > issues. As I am sure you have forgotten the definition, according to > the technical reference Handbook of Physics, here it is again: Inertial frame - a frame in which Newton's laws hold, in particular > the law of inertia. In such a frame, a body that is free of forces > remains in its state of motion. Therefore, inertial frames are those > that move with uniform speed relative to each other. Note again the specific reference to Newton's law of inertia. Now a > photon is in uniform linear motion. Unless acted on by an outside > force, such as gravity, it will remain in that uniform linear motion. > The photon satisfies all the requirements of an inertial frame, except > of course your requirement that it 'simply does not'. Which is about > as unscientific as it gets.- .83.9f.83ëfi\.91\.98\.9d[ATi lde]\.8f \.91\.8c\.8e\.92ó\.94\.95[UG rave] \.97\.8c [Eth]\.87\.98Ë'\.8c\.97\.8f - - \.a6\.92\.beË\.94\.8e\.97\.8f \.91\.8c\.8e\.92ó\.94\.95[UG rave] \.97\.8c [Eth]\.87\.98Ë'\.8c\.97\.8f - What is your resisitence to the Handbook of Physics? An inertial frame is a basic concept. What makes you think we need 6 references? === Subject: Re: Cantor's argument is erroneous Nntp-Posting-Host: hera.cwi.nl ... > proposing > alternate theories are the cranks, not non-cranks. As far as I can see there is only one person who tries to actually formulate an alternate theory. Make of it what you wish. > The point is, I want to discuss theories other than standard ZFC. > Nothing prevents you from doing so. =A0You don't need to latch onto a > crank thread for that. > > The vast majority of sci.math threads about alternate systems are > labeled as so-called crank threads. That alone provides evidence of > the general attitude towards alternative theories. Because they do not formulate anything close to a theory. > No, that's a fabrication. =A0The fact that he doesn't know and isn't > willing to learn is why it's pointless to ask him questions about > fields in connection with his infinitesimal. =A0(Singular. =A0MR has made > it quite clear that he considers only one infinitesimal to exist) > > Then I wonder what a better question to ask MR is, such that I can > learn more about MR's proposed infinitesimal. MR is not proposing anything. He is just *stating* things. > I wonder what such an actually mathematical study might entail. Since > many of these cranks embrace a geometry with adjacent _points_, > maybe starting with the geometry might work better. Perhaps. Because I have not seen any formulation that is even close to coherent I do not know. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor's argument is erroneous Nntp-Posting-Host: hera.cwi.nl ... > Can you come up with a theory such that AP's claim is right that he is the > only man ever who has come up with a correct proof of the infinitude of > primes? > > I've only seen AP's comments about the Infinitude of Primes once or > twice, and I'm not completely sure what his objection to Euclid's > proof actually is. > > The only objection I can imagine he'd have is that Euclid's proof is > based on the standard natural numbers, while AP's proof is based > on his own AP-adics. You are wrong. Roughly AP's proof is along the following lines: (1) Given the set of all primes S (2) Calculate p1 * p2 * ... for all primes in S and add 1, call this number N. (3) Because N is not divisible by any of the primes in S, it is prime. (4) Hence there are infinitely many primes. He objects to restating (3) as: (3a) So either N is prime, a contractiction, because S contains all primes. (3b) Or N is composite, but than it must be divisible by a prime not in S, as it is not divisible by a prime in S. And he calls his proof a *direct* proof because it finds a prime not in S or somesuch. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor's argument is erroneous >... > Can you come up with a theory such that AP's claim is right that he is the > only man ever who has come up with a correct proof of the infinitude of > primes? I've only seen AP's comments about the Infinitude of Primes once or > twice, and I'm not completely sure what his objection to Euclid's > proof actually is. The only objection I can imagine he'd have is that Euclid's proof is > based on the standard natural numbers, while AP's proof is based > on his own AP-adics. You are wrong. Roughly AP's proof is along the following lines: >(1) Given the set of all primes S >(2) Calculate p1 * p2 * ... for all primes in S and add 1, call this > number N. >(3) Because N is not divisible by any of the primes in S, it is prime. >(4) Hence there are infinitely many primes. He objects to restating (3) as: >(3a) So either N is prime, a contractiction, because S contains all primes. >(3b) Or N is composite, but than it must be divisible by a prime not in S, \ as > it is not divisible by a prime in S. >And he calls his proof a *direct* proof because it finds a prime not in S >or somesuch. A valid point does come out of this (I won't go so far as to say that AP has a point). When I was younger and dumber I assumed that step #3 meant that if you took all the primes up to a certain number, mutliplied them together, and added one, you got a prime number. Understanding why this is not the case and why the proof does not claim that this is the case is not trivial (if you are ten), so his modification can help in the understanding. Alan -- Defendit numerus === Subject: Re: Cantor's argument is erroneous posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG 1.1.4322; .NET CLR 2.0.50727; MS-RTC LM 8; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > I've only seen AP's comments about the Infinitude of Primes once or > twice, and I'm not completely sure what his objection to Euclid's > proof actually is. > The only objection I can imagine he'd have is that Euclid's proof is > based on the standard natural numbers, while AP's proof is based > on his own AP-adics. You are wrong. Roughly AP's proof is along the following lines: > (1) Given the set of all primes S > (2) Calculate p1 * p2 * ... for all primes in S and add 1, call this > number N. > (3) Because N is not divisible by any of the primes in S, it is prime. > (4) Hence there are infinitely many primes. He objects to restating (3) as: > (3a) So either N is prime, a contradiction, because S contains all primes. > (3b) Or N is composite, but then it must be divisible by a prime not in S, \ as > it is not divisible by a prime in S. > And he calls his proof a *direct* proof because it finds a prime not in S > or somesuch. Clearly AP does not comprehend the point of Euclid's proof, or more specifically, the reason it *is* a proof. AP has stated that axioms are not essential to deriving mathematical truths, which must be founded on geometry and physics. Walker would do well to read some of AP's proofs in detail. === Subject: Re: Cantor's argument is erroneous posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > Get _them_ to change their behavior? As far as I'm concerned, > _both_ sides of the crank debate have to change their > behavior before they'd be able to agree on anything. Yes, that's true. Unfortunately, the mathematicians would have to reject simple logic in order to agree with the cranks. Maybe you should be asking if it's a good thing that the mathematicians don't accept the pronouncements of the cranks, and if it's a bad thing that the cranks don't accept the way the mathematicians use logic. === Subject: Re: Cantor's argument is erroneous posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > To me, the word crank is just a label, and AP and tommy1729 > have already been labeled cranks. So it's too late for me to > make it so that they _never were_ (called) cranks. The best I > can do is make it so that they _stop_ being (called) cranks. No, you can't. Only they can do that. === Subject: Re: Cantor's argument is erroneous posting-account=71XbuAoAAACx3_UV8yBrbgOAHUYjIUR6 Gecko/2009011913 Firefox/3.0.6 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) OK then, so let me try to divide the so-called cranks into > three groups, based on whether their beliefs are closest to > choices (1), (2), or (3). Since not every crank considers > Cantor's theorem in the argument, let me generalize the > three possibilities as: (1) Some unspecified theory proves the negation of some > theorem of ZFC. > (2) ZFC proves the negation of some theorem of ZFC (and > so ZFC is inconsistent). > (3) Some theorem of ZFC is really false, no matter what > ZFC says. > I've come across another categorization scheme which you may find a pleasant pastime to try to fit the various systems (standard or non- standard) into: A Soviet dissident logician Essenin-Volpin divided formal logical systems into two classes: democratic and totalitarian. In a democratic system, the rules tell you what is forbidden. By default, the rest is allowed. In a totalitarian system, the rules tell you what is allowed. By default, the rest is forbidden. (For those of you who understand only the language of categories, democratic and totalitarian systems are the final and initial objects of the appropriate category.) In the September,1993 issue of Ferment, editor Roy Lisker has an which he states: In this history of mathematics, we may perhaps speak of the Russell model as the democratic version, and the Hilbert model as the totalitarian one, and Brouwer, paradoxically as the most totalitarian www.fermentmagazine.org/FermentVIII/Volp1.doc Later Lisker says: He then goes on to state three postulates of ïtotalitarian logic' on which any authoritarian state may be based,( I paraphrase): Postulate I. We can prove premise A by proving premise B which resembles it, then call this a proof of A. Postulate II. A thinking person is someone who understands his personal and public usefulness Corollary: It is impossible that a thinking person would dispute postulate I, for he would therefore not be a thinking person. Postulate III:( in two parts) (a) Persons who do not understand their personal usefulness are to be commited to mental hospitals. (b) Persons who do not understand their public usefulness are to be sent to jail or executed. Finally he listed ñtwo principles of the understandingî. These fall outside of logic proper and can be called ñMeta-Logicalî PRINCIPLE I: The word ñto understandî means either to understand as we understand, or as somebody else understands. PRINCIPLE II: If somebody states that he does not wish to understand something, then he does not understand. Had Volpin's essay been published in Western Europe, it might have merely been considered clever or funny by those who understood it, and as merely ïununderstandable' by those who did not wish to understand it. In the Russia of 1959, it earned its author yet one more year's incarcerated in the Leningrad Special Psychiatric Hospital. === Subject: Re: Cantor's argument is erroneous <87r6081v63.fsf@phiwumbda.org> posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > But, crank or not, it still doesn't give me a rigorous theory > which proves the existence of adjacent infinitesimals. > I posted a toy theory doing just this. Twice. You never commented. The problem is that these threads grow so fast that it's hard for me to keep up with every post in the thread. As we look at Hughes's toy theory, we'll check to see whether it satisfies MR's intutions, since MR is the standard with which we must consider such theories. Recall that this theory was originally posted in the MR thread. > My structure is the set R' of all functions > (w + 1) -> {0,...,9} OK, so this is simply a list of digits isomorphic to omega+1, rather than the standard omega. > There's a natural injection [0,1) -> R' defined as follows. Let x in > [0,1) and let > (x1)(x2)(x3)... > denote the decimal expansion of x. If x has two decimal expansions, > one ending in 0000... and the other in 999..., choose the former. > We map x to R' via > x |-> Okay, so R' is an extension of [0,1), but now we need a *different* > map interpreting digit strings as elements of R'. Note: so far, we > embedded actual real numbers into R' using their decimal expansions, > but *now* we are mapping sequences of {0,...,9} into [0,1), so this is > a somewhat *different* matter. > The map is nonetheless very similar. We map a sequence > |-> (x1)(x2)(x3)... eventually stabilizes to a digit k, then pick z = k. > That is, if there is an n such that for all m > n xm = k, then z = k. > Otherwise, let z = 0. > where z is defined just as above. So in other words, we have: 1/3 = {3, 3, 3, 3, ..., 3} 1/6 = {1, 6, 6, 6, ..., 6} but 2/11 = {1, 8, 1, 8, ..., 0} sqrt(2)-1 = {4, 1, 4, 2, ..., 0} pi-3 = {1, 4, 1, 5, ..., 0}. And of course, for numbers mentioned by MR, we have: 1/infinity = .0r1 = {0, 0, 0, 0, ..., 1} .9r = {9, 9, 9, 9, ..., 9} .9r8 = {9, 9, 9, 9, ..., 8} > Now, could we define addition and so forth? Perhaps. It's a little > hard to know what to do when the w'th position has a carry, but maybe > there's something sensible to be done (I kinda doubt it). I remember how one so-called crank -- not MR, but a crank who doesn't post on sci.math -- once said something to the effect that if there's an (omega)th decimal place, then there must also be an (omega-1)st decimal place since that's where the carry from the (omega)th decimal place goes. Notice the implication being made here is that there must also be an (omega-2)nd place since that's where the carry from the (omega-1)st place goes, and eventually an (omega-n)th place for every natural number n. Thus the digits here would not be wellordered. The resulting system would be end up being more like AP-reals, But none of this is MR, of course. The ultimate test to find out whether the toy theory would satisfy MR's intuitions is, since MR has already declared .9r8 to be the predecessor of .9r, so we can see what MR considers the .9r8 to be. And if he states that it is .9r7, then we can see what the predecessor of .9r7, and keep on going until he reaches .9r0. Then we can ask MR what the predecessor of .9r0 is. Notice that: {9, 9, 9, 9, ..., 0} has no predecessor in the toy theory. Thus, if MR states that .9r0 has no predecessor, then we can be sure that the toy theory accurately matches MR's intuitions. > This is why I think you don't quite express your project well. It's > not enough to find a theory or structure making two or three or n > pronouncements true. The theory has to be useful, since it is > supposed to be analogous to the remarkably useful theory of real > numbers. That word useful can mean different things to different people, so I dislike using the word useful to describe any theory. Naturally every poster, whether a standard theory or a so-called crank, believes that his own theory is useful and that the opponent's theory isn't useful. But still, what may be helpful would be to try coming up with axioms for geometry first. Axiomatically assume that adjacent points exist, and then find out what numbers would describe the geometry. Most cranks who desire adjacent infinitesimals really want there to be adjacent points . === Subject: Re: Cantor's argument is erroneous sha1:gJUBaKeFG6imq4hS8Dhj7WBHF2c= > But, crank or not, it still doesn't give me a rigorous theory > which proves the existence of adjacent infinitesimals. > I posted a toy theory doing just this. Twice. You never commented. The problem is that these threads grow so fast that it's hard for > me to keep up with every post in the thread. As we look at Hughes's toy theory, we'll check to see whether > it satisfies MR's intutions, since MR is the standard with which > we must consider such theories. Recall that this theory was > originally posted in the MR thread. I make absolutely no promises that it satisfies anything other than the two principles we were discussing at the time. Besides, above you said that you wanted a rigorous theory that proves the existence of adjacent infinitesimals. That's what this is. Why are you changing your standards now? If you're really interested just in adjacent infinitesimals, then anything else said by Mitch is irrelevant. And, if you're interested in proving that Mitch is not a crank, then no theory I provide can do the job. You seem to be waffling on your aims. > My structure is the set R' of all functions > (w + 1) -> {0,...,9} OK, so this is simply a list of digits isomorphic to omega+1, > rather than the standard omega. > There's a natural injection [0,1) -> R' defined as follows. Let x in > [0,1) and let > (x1)(x2)(x3)... > denote the decimal expansion of x. If x has two decimal expansions, > one ending in 0000... and the other in 999..., choose the former. > We map x to R' via > x |-> (that is, to the function f such that f(i) = xi \ and f(w) = 0). > Okay, so R' is an extension of [0,1), but now we need a *different* > map interpreting digit strings as elements of R'. Note: so far, we > embedded actual real numbers into R' using their decimal expansions, > but *now* we are mapping sequences of {0,...,9} into [0,1), so this is > a somewhat *different* matter. > The map is nonetheless very similar. We map a sequence > |-> where z is defined as follows: If the \ decimal expansion > (x1)(x2)(x3)... eventually stabilizes to a digit k, then pick z = k. > That is, if there is an n such that for all m > n xm = k, then z = k. > Otherwise, let z = 0. > where z is defined just as above. So in other words, we have: 1/3 = {3, 3, 3, 3, ..., 3} > 1/6 = {1, 6, 6, 6, ..., 6} but 2/11 = {1, 8, 1, 8, ..., 0} > sqrt(2)-1 = {4, 1, 4, 2, ..., 0} > pi-3 = {1, 4, 1, 5, ..., 0}. And of course, for numbers mentioned by MR, we have: 1/infinity = .0r1 = {0, 0, 0, 0, ..., 1} > .9r = {9, 9, 9, 9, ..., 9} > .9r8 = {9, 9, 9, 9, ..., 8} That's really all I was shooting for: that .9r is the greatest number less than 1 and that there is a least positive number. At the time, that was all that you were interested in, as I recall. > Now, could we define addition and so forth? Perhaps. It's a little > hard to know what to do when the w'th position has a carry, but maybe > there's something sensible to be done (I kinda doubt it). I remember how one so-called crank -- not MR, but a crank who > doesn't post on sci.math -- once said something to the effect that > if there's an (omega)th decimal place, then there must also be an > (omega-1)st decimal place since that's where the carry from the > (omega)th decimal place goes. Notice the implication being made > here is that there must also be an (omega-2)nd place since that's > where the carry from the (omega-1)st place goes, and eventually > an (omega-n)th place for every natural number n. Thus the digits > here would not be wellordered. The resulting system would be > end up being more like AP-reals, Yes, that might work. I didn't want to begin with a complicated theory, so I chose omega + 1 sequences rather than omega + omega^op sequences. > But none of this is MR, of course. The ultimate test to find out > whether the toy theory would satisfy MR's intuitions is, since MR > has already declared .9r8 to be the predecessor of .9r, so we can > see what MR considers the .9r8 to be. And if he states that it > is .9r7, then we can see what the predecessor of .9r7, and keep > on going until he reaches .9r0. Then we can ask MR what the > predecessor of .9r0 is. Notice that: {9, 9, 9, 9, ..., 0} has no predecessor in the toy theory. Thus, if MR states that > .9r0 has no predecessor, then we can be sure that the toy theory > accurately matches MR's intuitions. Yes, you're right. That number has no predecessor. Regardless of what Mitch says (and I don't really care about his incoherent utterances), that's not what I wanted. So perhaps you can fix it. Consider omega + omega^op sequences, as you suggested. Get your hands dirty and do something. And stop whining that you want a theory similar to Mitch's proclamations, as if it is the job of others to make it happen. > This is why I think you don't quite express your project well. It's > not enough to find a theory or structure making two or three or n > pronouncements true. The theory has to be useful, since it is > supposed to be analogous to the remarkably useful theory of real > numbers. That word useful can mean different things to different people, so > I dislike using the word useful to describe any theory. Naturally > every poster, whether a standard theory or a so-called crank, > believes that his own theory is useful and that the opponent's > theory isn't useful. But still, what may be helpful would be to try coming up with > axioms for geometry first. Axiomatically assume that adjacent > points exist, and then find out what numbers would describe > the geometry. Most cranks who desire adjacent infinitesimals > really want there to be adjacent points . Doesn't sound like a good approach to me, but feel free to give it a go. -- Jesse F. Hughes Most of my research is irreducibly complex. -- James S. Harris === Subject: Re: Cantor's argument is erroneous You have sketched out a possible model for a number system in which 0.9r <> \ 1.0 I doubt if many people would call this a crank post, any more than mathematicians think Conway was a crank for constructing surreals, or ??? was a crank for constructing hyperReals. The differences between cranks and people legitimately constructing non-standard theories (such as yourself) are: 1. You are not claiming results that overturn a century of intensive mathematical study, which in this case would also be a proof of the inconsistency of ZF. Specifically you are not claiming |N| = |R|. 2. You provide a mathematical argument, not an appeal to intuition. People who are labeled cranks are labeled such because they break one of both of these rules, not because their theories are non-standard. === Subject: Re: Cantor's argument is erroneous posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Get them to change their behavior? As far as I'm concerned, > both sides of the crank debate have to change their > behavior before they'd be able to agree on anything. > As far as you're concerned, yes - that's because you've got a > completely false idea of standard analyst behaviour. Since said > idea persists against all the evidence provided to you, and you have > been completely unable to provide evidence in support of it, I'm > coming to believe that your concerns are delusional. There's one aspect of standard analyst behavior that not even Little can deny -- standard analysts have a tendency to refer to certain posters as cranks. So that's at least one aspect of standard analyst behavior that I'd like to change, but most likely can't -- I want to make it so that they don't call certain posters cranks. > And the standard analysts would have to give such axiomatizations, > once provided, at the very least some reasons that ZFC axioms are > superior. > Many reasons have already been given supporting that in the case of > many alternative systems. Still, research into such theories > proceeds, which is a very good thing. Such research is conducted by > non-cranks of course, because cranks are generally not capable of > research. proposing alternate theories are the cranks, not non-cranks. > The point is, I want to discuss theories other than standard ZFC. > Nothing prevents you from doing so. You don't need to latch onto a > crank thread for that. The vast majority of sci.math threads about alternate systems are labeled as so-called crank threads. That alone provides evidence of the general attitude towards alternative theories. > It's been pointed out in the other thread that MR doesn't know the > difference between an algebraic field and a gravitational field, and > that's why MR is confused about adjacent infinitesimals. > No, that's a fabrication. The fact that he doesn't know and isn't > willing to learn is why it's pointless to ask him questions about > fields in connection with his infinitesimal. (Singular. MR has made > it quite clear that he considers only one infinitesimal to exist) Then I wonder what a better question to ask MR is, such that I can learn more about MR's proposed infinitesimal. > And thus, changing the cranks' attitude isn't sufficient to erase > the concept of adjacent infinitesimals -- which, BTW, have been > proposed by many cranks, not just MR. > Of course. However, such a change may mean that they might conduct an > actually *mathematical* study of systems of adjacent infinitesimals, > which would be a good thing, instead of dribbling disconnected > nonsense filled with undefined terms and non-sequitur arguments as is > currently the case. I wonder what such an actually mathematical study might entail. Since many of these cranks embrace a geometry with adjacent points , maybe starting with the geometry might work better. > But, crank or not, it still doesn't give me a rigorous theory > which proves the existence of adjacent infinitesimals. > There are infinitely many such theories. Three have already been > outlined, and you certainly don't need to go outside ZF to find one. > Coming up with such theories is something most competent > mathematicians can do in at most minutes. What is usually lacking is > any reason why such a system is mathematically interesting. > The only person in this group who thinks that merely claiming > non-standard number properties makes someone a crank, is you. I hope that mathematically interesting isn't implicitly defined in a way such that only standard analysis is interesting, or such that once one lists all the interesting properties a theory must have in order to be considered, all sets with all such interesting properties are isomorphic to standard R. === Subject: Re: Cantor's argument is erroneous posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) Get them to change their behavior? As far as I'm concerned, > both sides of the crank debate have to change their > behavior before they'd be able to agree on anything. > As far as you're concerned, yes - that's because you've got a > completely false idea of standard analyst behaviour. Since said > idea persists against all the evidence provided to you, and you have > been completely unable to provide evidence in support of it, I'm > coming to believe that your concerns are delusional. There's one aspect of standard analyst behavior that not even Little > can deny -- standard analysts have a tendency to refer to certain > posters as cranks. They do? I wonder whether the vast majority of analysts who work in classical analysis (standard analysists) even explicitly know that there are such Internet meeting places as sci.math and sci.logic. Moreoverr, it seems to me that there a many, many standard analysists who post on sci.math who don't bother even to discuss cranks. > So that's at least one aspect of standard > analyst behavior that I'd like to change, but most likely can't -- I > want to make it so that they don't call certain posters cranks. But calling people 'cranks' is not even a given aspect of being a standard analyist. Moreover, calling cranks out as cranks has a good purpose, whereas being a crank has none. > And the standard analysts would have to give such axiomatizations, > once provided, at the very least some reasons that ZFC axioms are > superior. > Many reasons have already been given supporting that in the case of > many alternative systems. Still, research into such theories > proceeds, which is a very good thing. Such research is conducted by > non-cranks of course, because cranks are generally not capable of > research. proposing > alternate theories are the cranks, not non-cranks. Okay. So? > The point is, I want to discuss theories other than standard ZFC. > Nothing prevents you from doing so. You don't need to latch onto a > crank thread for that. The vast majority of sci.math threads about alternate systems are > labeled as so-called crank threads. That alone provides evidence of > the general attitude towards alternative theories. No it doesn't. The majority are called 'crank threads' because they are centered on cranks. Calling a crank 'crank' or a thread dominated by a crank 'a crank thread' doesn't entail that one has any particular attitude about alternative theories. > I hope that mathematically interesting isn't implicitly defined in a > way such that only standard analysis is interesting, or such that > once one lists all the interesting properties a theory must have in > order to be considered, all sets with all such interesting > properties > are isomorphic to standard R. What is interesting is subjective to each person. On the other hand, it is fair to ask of a given arbitrary theory, So what? Why should I care? Most foundational theories are interesting because of their scope - that we can express and prove a great deal of mathematics, especially (as is claimed by many a author) virtually all the mathematics for the sciences. Moreover, certain theories also provide for an axiomatization of results in mathematical logic itself. An alternative theory that gets all heated up with the friction of its own intricacies but doesn't provide for an axiomatization for the mathematics of the sciences is not likely to be taken seriously as a FOUNDATIONAL alternative. MoeBlee === Subject: Re: Cantor's argument is erroneous > Moreover, calling cranks out as cranks has a good purpose, whereas > being a crank has none. Even the worst of us, like cranks, can serve as bad examples! === Subject: Re: Cantor's argument is erroneous posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > proposing alternate theories are the cranks, not non-cranks. Who might those posters be? All I've seen is posters who propose that the standard theories and theorems are wrong. Some examples: Plutonium - Standard reals are flawed, 10^500 is largest finite, Peano axioms produce infinite naturals, irrationals are not complete, pi is imaginary, et al. Mueckenheim - ZFC is inconsistent, there are no actual infinite sets, 10^1000 is the largest countable natural. Orlow - ZFC is inconsistent, unit infinities exist. Harris - Current factoring algorithms are wrong. Petry - Cantor's theorem is wrong. Zick - 'not' is the basic unit of all truth, axioms are not true. Easterly - (something about infinities)? Finlayson - All infinities are equivalent. There are several others I'm not quite as familiar with. I've probably misquoted a few (esp. Ross F), but that's only because I can't tell what the hell they are trying to say. Zuhair is the only success story according to your criteria. He started by posting problems with standard set theory, slowly got more educated, and then tried to invent his own simpler version of set theory. I don't know how far he got, but what made him step away from crankdom was his willingness to learn actual math and use actual logic. === Subject: Re: Cantor's argument is erroneous So that's at least one aspect of standard analyst behavior that I'd > like to change, but most likely can't -- I want to make it so that > they don't call certain posters cranks. Correct, you can't change it. Those posters could in principle change it *themselves* if they don't like it, by ceasing to behave as cranks. > proposing alternate theories are the cranks, not non-cranks. I disagree. Almost all of what the cranks propose doesn't qualify as a mathematical theory, alternate or not. Generally their posts simply don't have the mathematical content to be considered a theory at all. A few non-cranks (yourself included) have posted alternate theories in the follow-up discussions to these cranks. These alternate theories have to the best of my recollection *always* been ignored or rejected by the original posters (cranks). > I wonder what such an actually mathematical study might > entail. Since many of these cranks embrace a geometry with > adjacent _points_, maybe starting with the geometry might work > better. Perhaps an alternative place to start instead of geometry: the *primary* requirement seems to be that Intuition is King: any result that does not accord with their intuition is false in an absolute sense, which drives any system entailing such a result to be rejected. So it would appear that any such system must be logically weak, in that any new axiom must not entail many results, for risk of one or more of those results conflicting with intuition and invalidating the whole system. It may be possible for a logically strong system to accord with a pre-existing intuition in all respects, but it seems overwhelmingly unlikely. > I hope that mathematically interesting isn't implicitly defined in > a way such that only standard analysis is interesting, Not really, but I think it does include logically strong. Empirically, the systems most interesting to mathematicians appear to be those that have relatively simple axioms and definitions generating a *very* rich space of results. If some of those results conflict with intuition, so be it: intuition can be extended. - Tim === Subject: Re: Cantor's argument is erroneous > I hope that mathematically interesting isn't implicitly defined in a > way such that only standard analysis is interesting, Non-standard analysis is certainly of interest to many mathematicians, as are a number of areas in which what mathematicians call analysis does not enter at all. === Subject: Re: Cantor's argument is erroneous The opposite of Cantor's results can be easily obtained by inductive and/or topological means, that is, from the basic properties of numbers and spatial relations. ***************** Cool. ZF is inconsistent. Please show us the easy inductive and topological \ proofs you refer to. === Subject: Re: Cantor's argument is erroneous posting-account=71XbuAoAAACx3_UV8yBrbgOAHUYjIUR6 Gecko/2009011913 Firefox/3.0.6 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > Notice that Yessenin-Volpin's ultrafinitism is far more severe than > even > the sci.math ultrafinitists. The upper limits for the sci.math > ultrafinitists > are usually based on physics. > Indeed. Jean-Yves Girard says: A proposition, not quite serious, by Essenin-Volpin : there would be integers only up to say 19. This want of earnestness is confirmed by his claim that these methods can be used to prove. . . the consistency of set theory ZF ! http://iml.univ-mrs.fr/~girard/coursang/coursang5.pdf.gz Scroll down to the 5th page which is page 387 of these lecture notes. === Subject: Re: Cantor's argument is erroneous > And I really don't know what you mean by we cannot prove nor disprove > Cantor's results with a proof that embeds them in its axioms. In what > sense is the result embedded in the axioms anymore so that any > theorem that follows from some axioms is embedded in those axioms? > You seem to be going about that completely backwards It's you who are getting backwards my usage of embedded. Cantor's > arguments are embedded into the standard axioms in the sense that they > are an inspiration for such axioms. Those axioms were not inspired by Cantor's theorem so much as by the Peano properties and such problems as Russell's paradox, which are, by themselves, sufficient to warrant ZF(C) . > The fact that, *a posteriori*, you > can restate Cantor's theorems in formal terms and show that they are a > consequence of the given axioms is what I consider an incorrect > methodology in the context of an unsound approach to mathematics: you > have can show that such theorems formally derive from those axioms, > but you have not even gone near to show the soundness of Cantor's > arguments per se, because that comes before any formal theories and > their axioms. Then with what absolute truths does LV propose by which to replace axiom systems? Do not they then become an axiom system? -- your argument > is with the axioms, and you should start there. My argument is indeed that one should *not* start there: axioms are > rather a consequence of a series of informal assumptions. Or, as the late C. R. Wylie Jr. put it: PARADOX Not truth, nor certainty. These I forswore In my novitiate, as young men called To holy orders must abjure the world. 'If . . . , then . . . ,' this only I assert; And my successes are but pretty chains Linking twin doubts, for it is vain to ask If what I postulate be justified, Or what I prove possess the stamp of fact. Yet bridges stand, and men no longer crawl In two dimensions. And such triumphs stem In no small measure from the power this game, Played with the thrice-attenuated shades Of things, has over their originals. How frail the wand, but how profound the spell! Cantor's argument, > which is perfectly sound in ZFC, is neither here nor there as far as > your complaints are ultimately concerned; it's no more pertinent than > the Banach-Tarski theorem, or any other theorem derivable from ZFC > that isn't derivable in whatever your preferred axiom system is. And > that's the crux of the matter: what is your preferred axiom system? > That is where you need to begin, because that is where your point of > contention ireally s. It is, in a sense, the point of contention, but only in the sense that > I refuse such an approach. I am using sound vs. invalid as this > informal sense): As in full of sound and fury, signifying nothing? [quote] > A deductive argument is said to be valid if and only if it takes a > form that makes it impossible for the premises to be true and the > conclusion nevertheless to be false. Otherwise, a deductive argument > is said to be invalid. A deductive argument is sound if and only if it is both valid, and all > of its premises are actually true. Otherwise, a deductive argument is > unsound. > [/quote] And which of these two opposites does LV hold to? Mathematicians and logicians hold to the first. Only those who believe that they (and only they) know The Truth, hold the second. The opposite of Cantor's results can be easily obtained by inductive > and/or topological means, that is, from the basic properties of > numbers and spatial relations. I do not see you, or anyone else, obtaining them, at least not without secretly assuming them. > Cantor's results are an unwarranted invention, with disastrous > consequences for the overall soundness of math and logic: despite > that you have made a formal system out of it, whose consistency > cannot be disproved given that the system is indeed unsound and let's > you prove what you like and dispove what you don't... LV is stating his creed, based only on his faith, not on fact. LV claims ZFC is unsound while simultaneously admitting that no one can justify that claim of unsoundness. Faith overriding fact. === Subject: Re: Cantor's argument is erroneous posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) I cannot agree: there is such a thing called soundness, implying > that the relationship of theory to application cannot be neglected. YOU DON'T KNOW what soundness is. Shut up. -LV === Subject: Re: Cantor's argument is erroneous posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/20080201 Firefox/2.0.0.12 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > I cannot agree: there is such a thing called soundness, implying > that the relationship of theory to application cannot be neglected. > YOU DON'T KNOW what soundness is. Shut up. Make me, dip. The tragic part is that EVEN AFTER I GAVE YOU THREE DEFINITIONS for it, YOU STILL don't know what soundness is. Telling you to shut up would be redundant; even when you speak, you are not saying anything. === Subject: Re: Cantor's argument is erroneous I cannot agree: there is such a thing called soundness, implying > that the relationship of theory to application cannot be neglected. YOU DON'T KNOW what soundness is. Shut up. Shut up. does not explain anything. === Subject: Re: Cantor's argument is erroneous posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) I cannot agree: there is such a thing called soundness, implying > that the relationship of theory to application cannot be neglected. No, there isn't. Yes, there is. -LV === Subject: Re: Cantor's argument is erroneous I cannot agree: there is such a thing called soundness, implying > that the relationship of theory to application cannot be neglected. No, there isn't. Yes, there is. There may be in physics or engineering, but soundness in pure math is independent of any applications. Number theory, for example, was perfectly sound when all computers were still human. === Subject: Re: Cantor's argument is erroneous posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) >Cantor's argument is erroneous and its adoption leads to unsound >mathematics. >The basic idea in the argument is that there is no bijection between >the set of counting numbers and the set of infinite binary strings. >But such a bijection exists, it can be expressed in terms of limit >points, or by transfinite induction; informally, it can be defined as >the correspondence between the paths and the leaf (i.e. limit) nodes >in the infinite binary tree. This invalidates all results relating to >Cantor's transfinite. > This is nonsense. Being charitable about the notion of limit node: > Yes, there's a bijection between the limit nodes and the paths. > Do tell -- in the charitable interpretation, what _is_ a limit node? You guys can be sloppy: http://en.wikipedia.org/wiki/Limit_point Please note that I say limit points and only later, qualified as >informal, I say limit nodes. So what? First, _you_ haven't bothered to explain > what those limit points _are_ - pointing to that > wikipedia page doesn't answer that question > until you specify what topological space > you're talking about. But much more important, however you eventually > define them, those limit points are not nodes of the > tree! A limit point of a set need not be an element > of the set. That's what makes a difference between the potential infinity and the actually complete tree. And that's why I am invoking the transfinite. > It's like you're deaf. Deaf is only your dogmatism. -LV === Subject: Re: Cantor's argument is erroneous posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) Cantor's argument is erroneous and its adoption leads to unsound >mathematics. The basic idea in the argument is that there is no bijection between >the set of counting numbers and the set of infinite binary strings. >But such a bijection exists, it can be expressed in terms of limit >points, or by transfinite induction; informally, it can be defined as >the correspondence between the paths and the leaf (i.e. limit) nodes >in the infinite binary tree. This invalidates all results relating to >Cantor's transfinite. This is nonsense. Being charitable about the notion of limit node: > Yes, there's a bijection between the limit nodes and the paths. Do tell -- in the charitable interpretation, what _is_ a limit node? >You guys can be sloppy: >http://en.wikipedia.org/wiki/Limit_point >Please note that I say limit points and only later, qualified as >informal, I say limit nodes. > So what? First, _you_ haven't bothered to explain > what those limit points _are_ - pointing to that > wikipedia page doesn't answer that question > until you specify what topological space > you're talking about. > But much more important, however you eventually > define them, those limit points are not nodes of the > tree! A limit point of a set need not be an element > of the set. That's what makes a difference between the potential infinity and the >actually complete tree. This is simply nonsense. Those limit points are _not_ nodes of the > tree, by the _definition_ of that tree. If your notion of the actually complete tree means that > the actually complete tree does include those limit points > as nodes that's fine. Cool, you got it. > But in that case you're not refuting > the statements you think you're refuting, you're simply > talking about something else. I don't think so: as I see it, Cantor's arguments are all about the transfinite. But I'll be pondering this: at least a serious objection. -LV === Subject: Re: Cantor's argument is erroneous posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) >Cantor's argument is erroneous and its adoption leads to unsound >mathematics. >The basic idea in the argument is that there is no bijection between >the set of counting numbers and the set of infinite binary strings. >But such a bijection exists, it can be expressed in terms of limit >points, or by transfinite induction; informally, it can be defined as >the correspondence between the paths and the leaf (i.e. limit) nodes >in the infinite binary tree. This invalidates all results relating to >Cantor's transfinite. > This is nonsense. Being charitable about the notion of limit node: > Yes, there's a bijection between the limit nodes and the paths. > Do tell -- in the charitable interpretation, what _is_ a limit node? You guys can be sloppy: http://en.wikipedia.org/wiki/Limit_point Please note that I say limit points and only later, qualified as >informal, I say limit nodes. So what? First, _you_ haven't bothered to explain > what those limit points _are_ - pointing to that > wikipedia page doesn't answer that question > until you specify what topological space > you're talking about. But much more important, however you eventually > define them, those limit points are not nodes of the > tree! A limit point of a set need not be an element > of the set. >That's what makes a difference between the potential infinity and the >actually complete tree. > This is simply nonsense. Those limit points are _not_ nodes of the > tree, by the _definition_ of that tree. > If your notion of the actually complete tree means that > the actually complete tree does include those limit points > as nodes that's fine. Cool, you got it. > But in that case you're not refuting > the statements you think you're refuting, you're simply > talking about something else. I don't think so: Well, that's truly bizarre. You're saying that you don't thing > a _definition_ is correct. This makes no sense - definitions > _are_ correct, since they're simply explanations of what we > _mean_ by a word. In particular, when we say that that tree has more paths > than nodes, the tree we're referring to does _not_ include > the things you're calling limit points as nodes. Saying > you don't think so is saying that you don't think we > mean what we say we mean. This is an awesomely > strange argument. If you are talking about some object that includes those > limit nodes then you _are_ talking about something > else, whether you think so or not. You (not only you of course) must have your brain shut, or just be trolling. I do not think so refers to your you're not refuting the statements you think you're refuting. >as I see it, Cantor's arguments are all about the >transfinite. So what? The fact that Cantor's arguments are all about > the transfinite means that somehow I cannot talk about > the set of all finite binary strings? Discussing around here, as said, is an enlightening experience: no wander all those who still can think by themselves do convert to non- standard whatever-but-this-brain-dead-. -LV === Subject: Re: Cantor's argument is erroneous posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG 1.1.4322; .NET CLR 2.0.50727; MS-RTC LM 8; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > Discussing around here, as said, is an enlightening experience: no > wander all those who still can think by themselves do convert to non- > standard whatever-but-this-brain-dead-. But apparently not enlightening enough for some. === Subject: Re: Cantor's argument is erroneous posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I Clearly you think that limit nodes exist and that any tree that does not include limit nodes is not complete OK. Divide the nodes into two sets, kumquat nodes (a kumquat node is any node at a finite level of the tree) and LV nodes (an LV node is any node that is not a kumquat node). Now take any argument about the tree you think is not the complete tree, add the word kumquat in front of node, and you have an argument about the complete binary tree. In particular, we can define a subset, Q, of the paths such that every element of Q correponds to a subset of the kumquat nodes, and there is no bijection between the natural numbers and Q. So, the paths in the complete binary tree are not countable. - William Hughes === Subject: Re: Cantor's argument is erroneous posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) > Clearly you think that limit nodes exist and > that any tree that does not include limit nodes > is not complete OK. Divide the nodes into two > sets, kumquat nodes (a kumquat node is > any node at a finite level of the tree) and > LV nodes (an LV node is any node that is not > a kumquat node). Now take any argument about the tree you > think is not the complete tree, add > the word kumquat in front of node, and > you have an argument about the complete > binary tree. Given *your* definition, this tree is the *in*complete one. > In particular, we can define > a subset, Q, of the paths > such that every element of Q > correponds to a subset of the kumquat nodes, > and there is no bijection between the natural numbers > and Q. So, the paths in the complete binary > tree are not countable. So this is plain nonsense. And I'm depressed. -LV === Subject: Re: Cantor's argument is erroneous posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I Clearly you think that limit nodes exist and > that any tree that does not include limit nodes > is not complete OK. Divide the nodes into two > sets, kumquat nodes (a kumquat node is > any node at a finite level of the tree) and > LV nodes (an LV node is any node that is not > a kumquat node). Now take any argument about the tree you > think is not the complete tree, add > the word kumquat in front of node, and > you have an argument about the complete > binary tree. Given *your* definition, this tree is the *in*complete one. So? Use your definition of the complete binary tree. The argument (with kumquat node in place of node) holds. In particular, we can define > a subset, Q, of the paths > such that every element of Q > correponds to a subset of the kumquat nodes, > and there is no bijection between the natural numbers > and Q. So, the paths in the complete binary > tree are not countable. So this is plain nonsense. Nope, the argument holds for your definition of the complete tree. - William Hughes === Subject: Re: Cantor's argument is erroneous >Cantor's argument is erroneous and its adoption leads to unsound >mathematics. >The basic idea in the argument is that there is no bijection between >the set of counting numbers and the set of infinite binary strings. >But such a bijection exists, it can be expressed in terms of limit >points, or by transfinite induction; informally, it can be defined as >the correspondence between the paths and the leaf (i.e. limit) nodes >in the infinite binary tree. This invalidates all results relating to >Cantor's transfinite. > This is nonsense. Being charitable about the notion of limit node: > Yes, there's a bijection between the limit nodes and the paths. > Do tell -- in the charitable interpretation, what _is_ a limit node? You guys can be sloppy: http://en.wikipedia.org/wiki/Limit_point Please note that I say limit points and only later, qualified as >informal, I say limit nodes. So what? First, _you_ haven't bothered to explain > what those limit points _are_ - pointing to that > wikipedia page doesn't answer that question > until you specify what topological space > you're talking about. But much more important, however you eventually > define them, those limit points are not nodes of the > tree! A limit point of a set need not be an element > of the set. That's what makes a difference between the potential infinity and the > actually complete tree. And that's why I am invoking the transfinite. Invoking the transfinite is meaningless twaddle. Merely LV outing his ignorance. === Subject: Re: Cantor's argument is erroneous posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) > ... >But such a bijection exists, it can be expressed in terms of limit >points, or by transfinite induction; informally, it can be defined as >the correspondence between the paths and the leaf (i.e. limit) nodes >in the infinite binary tree. This invalidates all results relating to >Cantor's transfinite. > ... > Do tell -- in the charitable interpretation, what is a limit node? You guys can be sloppy: http://en.wikipedia.org/wiki/Limit point Please note that I say limit points and only later, qualified as > informal, I say limit nodes. Pray tell, what is the topology? Do you know what a topological space > actually is? No, I think not. Again? And again: just no point in discussing furtherly. -LV === Subject: Re: Cantor's argument is erroneous sha1:Sups0fRMtEgXn7iTP2i061fkFJ4= > ... >But such a bijection exists, it can be expressed in terms of limit >points, or by transfinite induction; informally, it can be defined as >the correspondence between the paths and the leaf (i.e. limit) nodes >in the infinite binary tree. This invalidates all results relating to >Cantor's transfinite. > ... > Do tell -- in the charitable interpretation, what is a limit node? > You guys can be sloppy: >http://en.wikipedia.org/wiki/Limit point > Please note that I say limit points and only later, qualified as > informal, I say limit nodes. > Pray tell, what is the topology? Do you know what a topological space > actually is? No, I think not. Again? And again: just no point in discussing furtherly. You could just as easily show that you *do* know what a topological space is, by giving the topology you have in mind. Why pretend that there is no point? It would be trivial enough to answer the question. -- They are anti-mathematicians, evil incarnate, dedicated to undermining intellectual development in this area. If you never thought such people could actually exist, outside of myths or legends, welcome to the real world. --James S Harris on evil incarnate's Usenet presence === Subject: Re: Cantor's argument is erroneous > ... >But such a bijection exists, it can be expressed in terms of limit >points, or by transfinite induction; informally, it can be defined > as >the correspondence between the paths and the leaf (i.e. limit) > nodes >in the infinite binary tree. This invalidates all results relating > to >Cantor's transfinite. > ... > Do tell -- in the charitable interpretation, what _is_ a limit node? You guys can be sloppy: http://en.wikipedia.org/wiki/Limit_point Please note that I say limit points and only later, qualified as > informal, I say limit nodes. Pray tell, what is the topology? Do you know what a topological space > actually is? No, I think not. Again? And again: just no point in discussing furtherly. Not while LV remains ignorant of the issues. === Subject: Re: Cantor's argument is erroneous <87fxgsrs5w.fsf@phiwumbda.org> posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) > Look, which of the following do you think that LV is trying to show > us? > (1) In some unspecified theory, Cantor's theorem is false. > (2) Cantor's theorem is false in ZFC. > (3) Cantor's theorem is really false, no matter what ZFC says. Number 3 of course! So then you agree that the proof of Cantor's theorem in ZFC is a valid > proof (in ZFC) and instead reject ZFC as unsound? NO! Didn't I just say *3*?! Cantor's theorem is _really_ false, no matter what ZFC says. -LV === Subject: Re: Cantor's argument is erroneous > Cantor's theorem is _really_ false, no matter what ZFC says. ZFC merely says that it follows from the ZFC axioms, which it does, and which has often been shown to be the case. LV ays it Cantor's theorem is _really_ false, but provides no evidence in support of that claim. SO LV's claim is worth no more than the evidence he produces in support of it. To wit, zero! === Subject: Re: Cantor's argument is erroneous sha1:G/acLNyArbLYY6d3tgFhUieYwwI= > Look, which of the following do you think that LV is trying to show > us? > (1) In some unspecified theory, Cantor's theorem is false. > (2) Cantor's theorem is false in ZFC. > (3) Cantor's theorem is really false, no matter what ZFC says. > Number 3 of course! > So then you agree that the proof of Cantor's theorem in ZFC is a valid > proof (in ZFC) and instead reject ZFC as unsound? NO! Didn't I just say *3*?! Cantor's theorem is _really_ false, no matter what ZFC says. Do you not know what valid means? If ZFC is unsound, then Cantor's theorem may be a valid proof in ZFC, even though the result is false. What I am asking is this: Is Cantor's theorem a valid proof in ZFC? That is, is it a valid proof in first-order logic such that every axiom occurring in the proof is an instance of a ZFC axiom? Let's call this question (*). Evidently, you didn't understand my question, since the fact that you said (3) does not imply an answer to (*). Note: If your answer to (*) is no, then you need to show what step in the proof of Cantor's theorem is invalid, that is, either is an axiom that is not justified by the axiom schema of ZFC, or is an invalid inference. -- Jesse F. Hughes I will admit I can get giddy over these forays into ideas at the extreme edge. Being wrong can just add to the fun, oddly enough. --James S. Harris just wants to have fun. === Subject: Re: Cantor's argument is erroneous <87bprc1723.fsf@phiwumbda.org> posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) Look, which of the following do you think that LV is trying to show > us? (1) In some unspecified theory, Cantor's theorem is false. (2) Cantor's theorem is false in ZFC. (3) Cantor's theorem is really false, no matter what ZFC says. > Number 3 of course! > So then you agree that the proof of Cantor's theorem in ZFC is a valid > proof (in ZFC) and instead reject ZFC as unsound? NO! Didn't I just say *3*?! Cantor's theorem is really false, no matter what ZFC says. Do you not know what valid means? If ZFC is unsound, then Cantor's theorem may be a valid proof in ZFC, > even though the result is false. Exactly! Is it so hard to admit that I just said what I meant? > What I am asking is this: Is Cantor's theorem a valid proof in ZFC? > That is, is it a valid proof in first-order logic such that every > axiom occurring in the proof is an instance of a ZFC axiom? Let's > call this question (*). I said and meant *3*. > Evidently, you didn't understand my question, since the fact that you > said (3) does not imply an answer to (*). I've understood your question and even answered. To be extra clear: I am *not* contending the formal proof does not go through: that's not my point of concern. The answer to your (*) is simply *yes*, as far as the proof has been even mechanically checked, but, again: that is just irrelevant to my concerns here. I said and meant *3*! -LV === Subject: Re: Cantor's argument is erroneous sha1:gG4SLZOCVozkDAWmrQd+hbpZdyQ= > Look, which of the following do you think that LV is trying to show > us? > (1) In some unspecified theory, Cantor's theorem is false. > (2) Cantor's theorem is false in ZFC. > (3) Cantor's theorem is really false, no matter what ZFC says. > Number 3 of course! > So then you agree that the proof of Cantor's theorem in ZFC is a valid > proof (in ZFC) and instead reject ZFC as unsound? > NO! Didn't I just say *3*?! > Cantor's theorem is really false, no matter what ZFC says. > Do you not know what valid means? > If ZFC is unsound, then Cantor's theorem may be a valid proof in ZFC, > even though the result is false. Exactly! Is it so hard to admit that I just said what I meant? Evidently so. Read again the above. Pay particular attention to the part where you say NO! > What I am asking is this: Is Cantor's theorem a valid proof in ZFC? > That is, is it a valid proof in first-order logic such that every > axiom occurring in the proof is an instance of a ZFC axiom? Let's > call this question (*). I said and meant *3*. > Evidently, you didn't understand my question, since the fact that you > said (3) does not imply an answer to (*). I've understood your question and even answered. To be extra clear: I > am *not* contending the formal proof does not go through: that's not > my point of concern. The answer to your (*) is simply *yes*, as far as > the proof has been even mechanically checked, but, again: that is just > irrelevant to my concerns here. I said and meant *3*! Fine. So Cantor's theorem is indeed a theorem of ZFC. I don't really see much controversy then. You say that ZFC happens to be unsound (but not inconsistent). Okay, I suppose that you have some sort of mystical insight into what sets really are, and I lack that insight. At least we agree that Cantor's theorem is a theorem of ZFC. But you could be a little clearer in your criticisms. Which axioms of ZFC do you find false? -- Jesse F. Hughes For a gentle introduction to set theory, see Bourbaki (1970). -- Footnote from Transgressing the Boundaries, Alan Sokal === Subject: Re: Cantor's argument is erroneous <87ljqfxb02.fsf@phiwumbda.org> posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) > Look, which of the following do you think that LV is trying to show > us? > (1) In some unspecified theory, Cantor's theorem is false. > (2) Cantor's theorem is false in ZFC. > (3) Cantor's theorem is really false, no matter what ZFC says. Number 3 of course! So then you agree that the proof of Cantor's theorem in ZFC is a valid > proof (in ZFC) and instead reject ZFC as unsound? > NO! Didn't I just say *3*?! > Cantor's theorem is really false, no matter what ZFC says. > Do you not know what valid means? > If ZFC is unsound, then Cantor's theorem may be a valid proof in ZFC, > even though the result is false. Exactly! Is it so hard to admit that I just said what I meant? Evidently so. Read again the above. Pay particular attention to the part where you > say NO! The point is: are you trying to get my point or just cavilling for the sake of it? Below is my last attempt at coming to terms with your requests. > What I am asking is this: Is Cantor's theorem a valid proof in ZFC? > That is, is it a valid proof in first-order logic such that every > axiom occurring in the proof is an instance of a ZFC axiom? Let's > call this question (*). I said and meant *3*. > Evidently, you didn't understand my question, since the fact that you > said (3) does not imply an answer to (*). I've understood your question and even answered. To be extra clear: I > am *not* contending the formal proof does not go through: that's not > my point of concern. The answer to your (*) is simply *yes*, as far as > the proof has been even mechanically checked, but, again: that is just > irrelevant to my concerns here. I said and meant *3*! Fine. So Cantor's theorem is indeed a theorem of ZFC. That's the NO. Not no because I do not agree: no because that's JUST NOT MY CONCERN! What I have presented is a Cantor's counter- argument: this has nothing to do with ZFC or any axiomatic system you can conjure up *a posteriori*. > I don't really see much controversy then. You say that ZFC happens to > be unsound (but not inconsistent). I DON'T say ZFC is not inconsistent. I just say I don't contend that: that is NOT MY CONCERN. The preminence of the formal is YOUR problem. > Okay, I suppose that you have some > sort of mystical insight into what sets really are, and I lack that > insight. At least we agree that Cantor's theorem is a theorem of ZFC. I don't: I just am NOT CONCERNED and so don't contend that. That of course does NOT imply I am agreeing with you. > But you could be a little clearer in your criticisms. Which axioms of > ZFC do you find false? I find -among other things- unsound this preminence of the formal to the informal. That is indeed just the other side of the same problem: unsoundness, in the methodology even before than in the assumptions. And with this I'll close with this thread for now. I'll have to think the details (the specific definitions involved, etc.) before I can argument any further. experience. -LV === Subject: Re: Cantor's argument is erroneous sha1:9LOsLlVCW5DAmNbKAROs8mlSvsg= > Look, which of the following do you think that LV is trying to show > us? > (1) In some unspecified theory, Cantor's theorem is false. > (2) Cantor's theorem is false in ZFC. > (3) Cantor's theorem is really false, no matter what ZFC says. > Number 3 of course! > So then you agree that the proof of Cantor's theorem in ZFC is a valid > proof (in ZFC) and instead reject ZFC as unsound? > NO! Didn't I just say *3*?! > Cantor's theorem is really false, no matter what ZFC says. > Do you not know what valid means? > If ZFC is unsound, then Cantor's theorem may be a valid proof in ZFC, > even though the result is false. > Exactly! Is it so hard to admit that I just said what I meant? > Evidently so. > Read again the above. Pay particular attention to the part where you > say NO! The point is: are you trying to get my point or just cavilling for the > sake of it? Below is my last attempt at coming to terms with your > requests. I'm trying to get your point, but this latest message isn't helping. > What I am asking is this: Is Cantor's theorem a valid proof in ZFC? > That is, is it a valid proof in first-order logic such that every > axiom occurring in the proof is an instance of a ZFC axiom? Let's > call this question (*). > I said and meant *3*. > Evidently, you didn't understand my question, since the fact that you > said (3) does not imply an answer to (*). > I've understood your question and even answered. To be extra clear: I > am *not* contending the formal proof does not go through: that's not > my point of concern. The answer to your (*) is simply *yes*, as far as > the proof has been even mechanically checked, but, again: that is just > irrelevant to my concerns here. I said and meant *3*! > Fine. So Cantor's theorem is indeed a theorem of ZFC. That's the NO. Not no because I do not agree: no because that's > JUST NOT MY CONCERN! What I have presented is a Cantor's counter- > argument: this has nothing to do with ZFC or any axiomatic system you > can conjure up *a posteriori*. Er, so do you really have *no opinion* on whether Cantor's theorem is a theorem of ZFC? After all, most the folks around here haven't really asserted anything more than that. Personally, I don't see how anything you have said really suggests that Cantor's theorem is false in some absolute sense, but I'd rather not discuss that anyway. That sort of question is just too vague and mushy. I'd prefer that we first settle the simpler question: Is Cantor's theorem a theorem of ZFC? > I don't really see much controversy then. You say that ZFC happens to > be unsound (but not inconsistent). I DON'T say ZFC is not inconsistent. I just say I don't contend > that: that is NOT MY CONCERN. The preminence of the formal is YOUR > problem. > Okay, I suppose that you have some sort of mystical insight into > what sets really are, and I lack that insight. At least we agree > that Cantor's theorem is a theorem of ZFC. I don't: I just am NOT CONCERNED and so don't contend that. That of > course does NOT imply I am agreeing with you. > But you could be a little clearer in your criticisms. Which axioms of > ZFC do you find false? I find -among other things- unsound this preminence of the formal to > the informal. That is indeed just the other side of the same problem: > unsoundness, in the methodology even before than in the assumptions. No idea what the heck that means, so I'll ask again. Which axioms of ZFC are dubious? > And with this I'll close with this thread for now. I'll have to think > the details (the specific definitions involved, etc.) before I can > argument any further. Darn shame. -- It's good for the economy to charge for intellectual property, so open source software cannot be good, while Microsoft is the most far-thinking company around and is doing it all for the good of the public. -- Linus Torvalds paraphrases Microsoft VP Craig Mundie === Subject: Re: Cantor's argument is erroneous posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) > Oh, SHUT UP. You shut up: that does not add anything useful. -LV === Subject: Re: Cantor's argument is erroneous Oh, SHUT UP. You shut up: that does not add anything useful. -LV Not even when LV says it. === Subject: Re: Cantor's argument is erroneous posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) > ... > I (and not only I of course) have! We can produce the opposite result, > i.e. eventually the countability of the reals, with something as basic > as transfinite induction. I think you do not know what transfinite induction actually is. Then no point in discussing furtherly. -LV === Subject: Re: Cantor's argument is erroneous > ... > I (and not only I of course) have! We can produce the opposite result, > i.e. eventually the countability of the reals, with something as basic > as transfinite induction. I think you do not know what transfinite induction actually is. Then no point in discussing furtherly. Then LV concedes his ignorance? === Subject: Re: Cantor's argument is erroneous posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) Cantor's arguments and results are the inspiration for some specific > axioms. This is complete and utter bull. It's not, of course. For instance: please tell me who/what inspired the powerset axiom? (Genuine question, I am not an expert in axiomatic systems.) > Cantor's arguments and theorems: but you simply have it upside down. No, we don't. The formal result just is what it is and it in fact does > NOT > depend on a whole LOT of axioms. The existence of the powerset was > not > motivated by this theorem; that motivation was obviously the other way > around. And as for what founds what, the formal DOES NOT NEED OR CARE > about ANY foundation. The informal, by contrast, HAS TO CONFORM > to the formal: There might be some problem with terminology behind this, anyway I cannot but state that I am convinced of the opposite: the informal founds the formal, not the other way round. Maybe note that I am implicitely saying: informal as in constructive, not informal as in sloppy. > a contradiction doesn't stop being contradictory just > because you stated it informally. Absolutely. And an unsound result does not become sound because you put it formally with no apparent inconsistency. -LV === Subject: Re: Cantor's argument is erroneous Cantor's arguments and results are the inspiration for some > specific axioms. This is complete and utter bull. It's not, of course. Claims absent evidence, such as the one just above, are worth no more than the weight of that non-evidence > For instance: please tell me who/what inspired the powerset axiom? > (Genuine question, I am not an expert in axiomatic systems.) Almost certainly a modeling in sets of the Peano postulates. Cantor's arguments and theorems: but you simply have it upside > down. No, we don't. The formal result just is what it is and it in fact > does NOT depend on a whole LOT of axioms. The existence of the > powerset was not motivated by this theorem; that motivation was > obviously the other way around. And as for what founds what, the formal DOES NOT NEED OR CARE about > ANY foundation. The informal, by contrast, HAS TO CONFORM to the > formal: There might be some problem with terminology behind this, anyway I > cannot but state that I am convinced of the opposite: the informal > founds the formal, not the other way round. Maybe note that I am > implicitely saying: _informal_ as in constructive, not informal as > in sloppy. a contradiction doesn't stop being contradictory just because you > stated it informally. Absolutely. And an unsound result does not become sound because you > put it formally with no apparent inconsistency. And a result certainly does not become unsound because you put it formally with no apparent inconsistency. Of the two, soundness as a result of lack of apparent inconsistency seems considerably more likely than unsoundness, particularly of a system as thoroughly vetted as ZFC. === Subject: Re: Cantor's argument is erroneous posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) > Evidently you don't understand the difference between an assumption and a > conclusion. The nonexistence of a bijection between the naturals and the reals > is a conclusion of Cantor's argument, not an assumption. You have not > identified a false assumption. > It's a conclusion for Cantor's argument, it becomes an assumption > with respect to a formal theory that embeds it as one or more axiom. > What particular formal theory embeds the statement there is no > bijection between the naturals and the reals as an axiom? ZFC, for > example, has no such axiom. Cantor's arguments and results are the inspiration for some specific >axioms. Supposing that that were true, so what? The question of what > inspired this or that has no bearing on mathematical correctness. It of course does, as far as there are many kinds and levels of correctness. For instance, in another post in this thread, I have tried to explain how I am using the words sound vs. valid. -LV === Subject: Re: Cantor's argument is erroneous Cantor's arguments and results are the inspiration for some specific >axioms. Supposing that that were true, so what? The question of what > inspired this or that has no bearing on mathematical correctness. It of course does, as far as there are many kinds and levels of > correctness. For instance, in another post in this thread, I have > tried to explain how I am using the words sound vs. valid. Your notions of sound and valid, are mathematically irrelevant unless they agree to a large extent to those of most mathematicians. And they seem not to. === Subject: Re: Cantor's argument is erroneous posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) > Evidently you don't understand the difference between an assumption and a > conclusion. The nonexistence of a bijection between the naturals and the reals > is a conclusion of Cantor's argument, not an assumption. You have not > identified a false assumption. It's a conclusion for Cantor's argument, it becomes an assumption > with respect to a formal theory that embeds it as one or more axiom. What particular formal theory embeds the statement there is no > bijection between the naturals and the reals as an axiom? ZFC, for > example, has no such axiom. Cantor's arguments and results are the inspiration for some specific > Cantor's arguments and theorems: but you simply have it upside down. > The informal founds the formal, not the other way round. So you still have not said what particular theory embeds the > statement there is no bijection between the naturals and the reals > as an axiom. And what is your basis for the claim that Cantors proof of the > uncountability of the reals inspired any particular axioms? WHICH > axioms in particular. WHO exactly was inspired in the way you claim? > And what is your evidence in this regard? I am not and am not interested to be an historician of mathematics. > As to informal preceding the informal, you haven't said which > particular formal theory you have in mind that was not based on > informal notions. ?? What I'd say regarding this is that *all* theories are, explicitly or implicitly, based on informal notions. It's in fact the preminence of the formal that, among other things, I am implicitly criticising here and elsewhere. -LV === Subject: Re: Cantor's argument is erroneous posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) Evidently you don't understand the difference between an assumption and a > conclusion. The nonexistence of a bijection between the naturals and the reals > is a conclusion of Cantor's argument, not an assumption. You have not > identified a false assumption. It's a conclusion for Cantor's argument, it becomes an assumption > with respect to a formal theory that embeds it as one or more axiom. What particular formal theory embeds the statement there is no > bijection between the naturals and the reals as an axiom? ZFC, for > example, has no such axiom. Cantor's arguments and results are the inspiration for some specific > Cantor's arguments and theorems: but you simply have it upside down. > The informal founds the formal, not the other way round. So you still have not said what particular theory embeds the > statement there is no bijection between the naturals and the reals > as an axiom. And what is your basis for the claim that Cantors proof of the > uncountability of the reals inspired any particular axioms? WHICH > axioms in particular. WHO exactly was inspired in the way you claim? > And what is your evidence in this regard? I am not and am not interested to be an historician of mathematics. That's fine. But you made a particular historical claim. I'm just interested in your basis for that claim. If you have no basis, then fine too; I'll just disregard your claim. > As to informal preceding the informal, you haven't said which > particular formal theory you have in mind that was not based on > informal notions. ?? What I'd say regarding this is that *all* theories are, explicitly > or implicitly, based on informal notions. Okay. So in your view ZFC is based on informal notions. > It's in fact the > preminence of the formal that, among other things, I am implicitly > criticising here and elsewhere. What theory, in which formality is not preeminent, do you recommend? And what is it about that theory that makes its formality not preeminent as in ZFC? For that matter, is preeminence merely an informal estimation of yours, or is there some objective standard by which to determine whether formality is or is not preeminent in a given theory? MoeBlee === Subject: Re: Cantor's argument is erroneous posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse MathPlayer 2.10d; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) > Evidently you don't understand the difference between an assumption and a > conclusion. The nonexistence of a bijection between the naturals and the reals > is a conclusion of Cantor's argument, not an assumption. You have not > identified a false assumption. It's a conclusion for Cantor's argument, it becomes an assumption > with respect to a formal theory that embeds it as one or more axiom. What particular formal theory embeds the statement there is no > bijection between the naturals and the reals as an axiom? ZFC, for > example, has no such axiom. Cantor's arguments and results are the inspiration for some specific > Cantor's arguments and theorems: but you simply have it upside down. > The informal founds the formal, not the other way round. So you still have not said what particular theory embeds the > statement there is no bijection between the naturals and the reals > as an axiom. And what is your basis for the claim that Cantors proof of the > uncountability of the reals inspired any particular axioms? WHICH > axioms in particular. WHO exactly was inspired in the way you claim? > And what is your evidence in this regard? I am not and am not interested to be an historician of mathematics. That's fine. But you made a particular historical claim. I didn't and the fact that you cannot get the difference between a theoretical and a historical analysis tells only your restricted understanding of such matters. > I'm just > interested in your basis for that claim. If you have no basis, then > fine too; I'll just disregard your claim. As to informal preceding the informal, you haven't said which > particular formal theory you have in mind that was not based on > informal notions. ?? What I'd say regarding this is that *all* theories are, explicitly > or implicitly, based on informal notions. Okay. So in your view ZFC is based on informal notions. *All* formal approaches are founded on the informal: this is a theoretical point and ZFC is of course no exception here. > It's in fact the > preminence of the formal that, among other things, I am implicitly > criticising here and elsewhere. What theory, in which formality is not preeminent, do you recommend? *All* formal approaches are founded on the informal. The mistake is yours, not in the theory itself. What I'd recommend is that you broaden your perspective: theoretical has to do with the pure concepts, not with an enumeration of specific instances. -LV > And what is it about that theory that makes its formality not > preeminent as in ZFC? For that matter, is preeminence merely an > informal estimation of yours, or is there some objective standard by > which to determine whether formality is or is not preeminent in a > given theory? === Subject: Re: Cantor's argument is erroneous posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > Evidently you don't understand the difference between an assumption and a > conclusion. The nonexistence of a bijection between the naturals and the reals > is a conclusion of Cantor's argument, not an assumption. You have not > identified a false assumption. It's a conclusion for Cantor's argument, it becomes an assumption > with respect to a formal theory that embeds it as one or more axiom. What particular formal theory embeds the statement there is no > bijection between the naturals and the reals as an axiom? ZFC, for > example, has no such axiom. Cantor's arguments and results are the inspiration for some specific > Cantor's arguments and theorems: but you simply have it upside down. > The informal founds the formal, not the other way round. So you still have not said what particular theory embeds the > statement there is no bijection between the naturals and the reals > as an axiom. And what is your basis for the claim that Cantors proof of the > uncountability of the reals inspired any particular axioms? WHICH > axioms in particular. WHO exactly was inspired in the way you claim? > And what is your evidence in this regard? I am not and am not interested to be an historician of mathematics. That's fine. But you made a particular historical claim. I didn't and the fact that you cannot get the difference between a > theoretical and a historical analysis tells only your restricted > understanding of such matters. You said: Cantor's arguments and results are the inspiration for some specific axioms. That's a claim as to what inspired people to adopt certain axioms, which is an historical claim. In fact it was YOU who responded that you did not wish to answer my question because you are not interested in being an historian of mathematics. In any case, whether such a claim is best described as one about history or not, it still stands that you have not given any basis for you claim, whether historical (which was indeed the reason YOU gave for demurring) or theoretical or WHATEVER. > It's in fact the > preminence of the formal that, among other things, I am implicitly > criticising here and elsewhere. What theory, in which formality is not preeminent, do you recommend? *All* formal approaches are founded on the informal. The mistake is > yours, not in the theory itself. What SPECIFIC mistake are you talking about? Anyway, you didn't answer the questions, which were: What theory, in which formality is not preeminent, do you recommend? And what is it about that theory that makes its formality not preeminent as in ZFC? For that matter, is preeminence merely an informal estimation of yours, or is there some objective standard by which to determine whether formality is or is not preeminent in a given theory? > What I'd recommend is that you > broaden your perspective: theoretical has to do with the pure > concepts, not with an enumeration of specific instances. I don't suppose there is anyone who wouldn't benefit from broadening of his or her mind. But whatever merit there is in the sentiment that theoretical has to do with the pure concepts, not with an enumeration of specific instances, you have not shown any basis for, or even explication of your claims mentioned in these latest posts between us; as I surely hope that you don't think that declaring theoretical has to do with the pure concepts, not with an enumeration of specific instances is some kind of catch-all automatic support for whatever you might say. MoeBlee === Subject: Re: Cantor's argument is erroneous > Evidently you don't understand the difference between an assumption > and a > conclusion. The nonexistence of a bijection between the naturals > and the reals > is a conclusion of Cantor's argument, not an assumption. You have > not > identified a false assumption. It's a conclusion for Cantor's argument, it becomes an assumption > with respect to a formal theory that embeds it as one or more axiom. What particular formal theory embeds the statement there is no > bijection between the naturals and the reals as an axiom? ZFC, for > example, has no such axiom. Cantor's arguments and results are the inspiration for some specific > Cantor's arguments and theorems: but you simply have it upside down. > The informal founds the formal, not the other way round. So you still have not said what particular theory embeds the > statement there is no bijection between the naturals and the reals > as an axiom. And what is your basis for the claim that Cantors proof of the > uncountability of the reals inspired any particular axioms? WHICH > axioms in particular. WHO exactly was inspired in the way you claim? > And what is your evidence in this regard? I am not and am not interested to be an historician of mathematics. As to informal preceding the informal, you haven't said which > particular formal theory you have in mind that was not based on > informal notions. ?? What I'd say regarding this is that *all* theories are, explicitly > or implicitly, based on informal notions. It's in fact the > preminence of the formal that, among other things, I am implicitly > criticising here and elsewhere. Mathematics has a long history of lack of formality leading to error and uncertainty, which makes mathematics treat informality with caution. Those who do not treat informality with caution, like WM, often end up being certain of the truth of falsehoods. === === Subject: New Inverse 19 mathematics Time which is a dimension, has come to introduce Inverse Mathematics at 19, discovered by us here a month or so ago, with the recent alignment of the Pells equation to progression of vector 19 inverse ( Y is the C constant +1 of infinity expansion which was discovered by us ) and basically 90(2^2) +1=19^2 =361---says it all Basically inverse 19 mathematics operates at -1 as the maximum least value for which has a specific inverse value, but if you were to measure sugar by granules, rather than by teaspoons , it would be more accurate and proprtionate. We have discovered the value of the granule-1, that is inversly proportional to the infinte expansion value we call C constant 1:360 value. The maximum inverse proportion is -4 which is not reachable as a mathematical fact in our universe, but -1 inverse mathematical proprtion is attainable(currently we are at +1 max if that). Also the median in the vector 19 is the absolute linear and possibly the least dimension, we do not know. Vector spaces can be multiplied with but not with the units that John Baez had prescribed in that prestigious Sc. math. research recently , that was an idea we started a year ago, and I had sent those to John with no response. We have Officially complained about the bias by math reseachers and journals etc in not understanding or us/communicatig , and the matter is submitted to the American Mathematical society Journal, and the news media for review. We will come forth with the final equations within a month to set forth on a 10 ton Mathematical monument in Wisconsin dedicated to lost mathematics. V.C and T.E. === === Subject: Re: .9 repeating Nntp-Posting-Host: hera.cwi.nl ... > So my goal to have a rigorous theory in which 0.999... < 1 is > not just about satisfying MR's intuitions, but satisfy those of > nearly 40% of physicists as well. > Theories where 0.999... < 1 abound. One of those has been offered to > you already in one of these threads. > > But that theory doesn't have adjacent infinitesimals, and so it > still doesn't satisfy MR's intuitions. That one didn't, indeed. I do not think it is difficult to come up with one where it *is* possible. Consider the Carthesian product R * {0, 1}, with lexicographic ordering. That one would satisfy MR's intuition with respect to the finite numbers. > But you will have difficulty in such > theories to prove that algorithms that he physicists use actually do > work. > > In other words, the algorithms that the physicists use are > difficult in a theory in which 0.999... < 1, and are much > easier in a theory in which 0.999... = 1? No, I did *not* state that. Read again: it is about *proving* that a particular algorithm works. > If so, then why did the survey reveal that almost 40% of > physicists believe that 0.999... < 1? If physics is easier > in a theory in which 0.999... = 1, then the percentage of > physicists who believe that 0.999... = 1 ought to be 100%, > not a little more than 60%. The physics is not easier, what is easier is the *proof* that an *algorithm* that he uses actually does work. But there are apparently many physicists that just write an algorithm and than by trial and error come up with an algorithm that appears to work. The only thing is that they never can be certain that it actually does work, and under what circumstances it does not work. But in general physicists use algorithms designed by mathematicians (numerical mathematicians in general) and the designers do *not* work in a system where 0.999... < 1, and they *prove* that algorithms do work under what circumstances. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: .9 repeating <15439582.26675.1238871277190.JavaMail.jakarta@nitrogen.mathforum.org If I \ were the author, then why would I speak of myself in the 3rd person? Gee, why would I think that a post by iamjohngabriel@gmail.com might discussion? I would guess for the same reason that you (Jason Wells) responded to my post very clearly replying to the person using the email address iamjohngabriel@gmail.com (as quoted below, in case your memory doesn't go back that far): you thought you were using your sock-puppet account but forgot to change the address, you moronic crank. Relevant portion of my post: > 1>0.999.. for if he assumed anything else, he ends up with a > contradiction. Now, go and wipe all the egg off your face. Just out of curiosity, are you the author? - Tim === Subject: Re: .9 repeating <49d740ce$0$7879$afc38c87@news.optusnet.com.au> posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) Fields of force are continuous quantities passing through every value > down to the level of the infinitely small. ************************** > No, you don't understand the question. No, he does understand. He's being deliberately > obtuse just to see if he can get you angry. It's > what trolls do . B. -- > Cheerfully resisting change since 1959. For instance If you accelerate you pass through every speed. Mitch Raemsch === Subject: Re: .9 repeating > Fields of force are continuous quantities passing through > every value down to the level of the infinitely small. > ************************** > No, you don't understand the question. > No, he _does_ understand. He's being deliberately > obtuse just to see if he can get you angry. It's > what trolls _do_. > For instance If you accelerate you pass through > every speed. Hey, Mitch! Remember when you claimed that (1 - .999...)/2 did not exist? You asked us to prove you wrong, so I did. If you want the proof, just ask. Jim Burns === Subject: Re: .9 repeating <49d740ce$0$7879$afc38c87@news.optusnet.com.au> posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) Fields of force are continuous quantities passing through > every value down to the level of the infinitely small. > ************************** > No, you don't understand the question. > No, he does understand. He's being deliberately > obtuse just to see if he can get you angry. It's > what trolls do . > For instance If you accelerate you pass through > every speed. Hey, Mitch! Remember when you claimed that > (1 - .999...)/2 did not exist? You asked us to prove you wrong, so I did. > If you want the proof, just ask. Jim Burns No you just need to prove it to yourself jim not me. Mitch Raemsch === Subject: Re: .9 repeating > Fields of force are continuous quantities passing through > every value down to the level of the infinitely small. > ************************** > No, you don't understand the question. > No, he _does_ understand. He's being deliberately > obtuse just to see if he can get you angry. It's > what trolls _do_. > For instance If you accelerate you pass through > every speed. > Hey, Mitch! Remember when you claimed that > (1 - .999...)/2 did not exist? > You asked us to prove you wrong, so I did. > If you want the proof, just ask. > No you just need to prove it to yourself jim not me. Then I have convinced you without the proof? So easy! Maybe you would like to see the proof anyway, in order to see where you went wrong? Jim Burns === Subject: Re: .9 repeating <49d740ce$0$7879$afc38c87@news.optusnet.com.au> posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) > Fields of force are continuous quantities passing through > every value down to the level of the infinitely small. > ************************** > No, you don't understand the question. > No, he does understand. He's being deliberately > obtuse just to see if he can get you angry. It's > what trolls do . > For instance If you accelerate you pass through > every speed. > Hey, Mitch! Remember when you claimed that > (1 - .999...)/2 did not exist? > You asked us to prove you wrong, so I did. > If you want the proof, just ask. > No you just need to prove it to yourself jim not me. Then I have convinced you without the proof? > So easy! Maybe you would like to see the proof anyway, > in order to see where you went wrong? Jim Burns- Hide quoted text - - Show quoted text - I believe the difference between .9 repeating and one is the infinitesimal nonzero. There is also a first quantity above one. Mitch Raemsch === Subject: Re: .9 repeating > On Apr 4, 1:58 pm, Jim Burns > On Apr 4, 1:07 pm, Jim Burns > Hey, Mitch! Remember when you claimed that > (1 - .999...)/2 did not exist? > You asked us to prove you wrong, so I did. > If you want the proof, just ask. No you just need to prove it to yourself jim > not me. > Then I have convinced you without the proof? > So easy! > Maybe you would like to see the proof anyway, > in order to see where you went wrong? > I believe the difference between .9 repeating > and one is the infinitesimal nonzero. There is also a first quantity above one. This is a long post, Mitch, I know, and I didn't mean it to be. There just came up more and more issues that I didn't feel I could edit out. As an enticement to read all the way to the end, there's a long segment where I explain exactly how you're wrong, *BUT* I follow it immediately with a pretty clever way (if I do say so myself) for you to be right. (There's always the possibility that you don't care about being right, that you just want to stir up trouble. In that case you're a troll, and you've been complete waste of my time.) Do you believe that two numbers multiplied together give some value, some existing number (what is called closure)? Then you are wrong about .9r < 1 and the first quantity above one, unless you deny other things that are even more obvious than closure. Do you deny closure? If you deny closure, then you have opened a very large can of worms, indeed. To start with, does pi = 3.14159... exist or not? It is infinitely close to both numbers that exist and numbers that do not exist. Remember, pi (existing or not) and its existing and non-existing neighbors all have the same decimal expansion. Maybe it's obvious to you how all this gets sorted out, but I, at least, need it explained to me. Do you believe in closure? Then do you believe in 0.5? Then what you believe is wrong. Here's why: Suppose that (1 - .9r) is the first quantity above zero. What is (1 - .9r)*.5? It must be some number. Is it zero? Then either 0.5 is zero or (1 - .9r) is zero, which means 1 = .9r. Is (1 - .9r)*.5 = (1 - .9r)? Then (1 - .9r)*.5 - (1 - .9r)*.5 = (1 - .9r) - (1 - .9r)*.5 and 0 = (1 - .9r)*.5 Thus, as we saw already, 1 = 0.9r Whatever you claim (1 - .9r)*.5 is equal to, there is (1 - .9r) times some number other than 0.5 already there. That will always be enough to prove that 1 = .9r. Here's how you re-work your claims, so that you're happy, everybody's happy: What you'll do is analogous to modular arithmetic, where counting cycles through a finite set of numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, ... (mod 10). We introduce new addition (+') and equality (='), similar to the old ones, to make this work. For example, 6 +' 7 =' 3 because 3 =' 13 =' 89,543. 6 *' 7 =' 2 because 2 =' 42 =' 146,782. All the integers that differ by a multiple of 10 are equal (=') modulo 10. It all works out smoothly, and various forms of modular arithmetic even have important practical applications. You should introduce new forms of addition (+'), multiplication (*'), equality (='), and comparison (<'), where any two numbers that differ by an infinitesimal (let us say, 1 and .9r) are held to be equal under this new modular equality, that is, 1 =' .9r, and so on. (I am probably mis-using terminology here. If someone would give me a better way of describing what I'm calling modular equality, I'd be grateful.) All the usual things the rest of us say about real numbers become true under the new forms of +', *', =', and <', so we no longer have a complaint. However, at the same time, you still really have .9r < 1 and other such facts true in the background, let's say (with the original operations), the same way 6 + 7 really is 13, not 3 in the background. Another advantage to you is that all the stuff mathematicians have learned about the foreground operations (+', *', =', <') will form a solid jumping off point in your explorations of the background operations (+, *, =, <), where, for example .9r < 1. Jim Burns === Subject: Re: .9 repeating <49d740ce$0$7879$afc38c87@news.optusnet.com.au> posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) > On Apr 4, 1:58 pm, Jim Burns > On Apr 4, 1:07 pm, Jim Burns > Hey, Mitch! Remember when you claimed that > (1 - .999...)/2 did not exist? > You asked us to prove you wrong, so I did. > If you want the proof, just ask. No you just need to prove it to yourself jim > not me. > Then I have convinced you without the proof? > So easy! > Maybe you would like to see the proof anyway, > in order to see where you went wrong? > I believe the difference between .9 repeating > and one is the infinitesimal nonzero. There is also a first quantity above one. This is a long post, Mitch, I know, and I didn't > mean it to be. There just came up more and more > issues that I didn't feel I could edit out. As an enticement to read all the way to the end, > there's a long segment where I explain exactly > how you're wrong, *BUT* I follow it immediately > with a pretty clever way (if I do say so myself) > for you to be right. (There's always the possibility that you don't > care about being right, that you just want to stir > up trouble. In that case you're a troll, and > you've been complete waste of my time.) Do you believe that two numbers multiplied together > give some value, some existing number (what is > called closure)? Then you are wrong about .9r < 1 > and the first quantity above one, unless you deny > other things that are even more obvious than > closure. Do you deny closure? If you deny closure, then you have opened > a very large can of worms, indeed. To start with, > does pi = 3.14159... exist or not? It is infinitely > close to both numbers that exist and numbers > that do not exist. Remember, pi (existing or not) > and its existing and non-existing neighbors > all have the same decimal expansion. Maybe > it's obvious to you how all this gets sorted > out, but I, at least, need it explained to me. Do you believe in closure? Then do you believe in 0.5? Then what > you believe is wrong. Here's why: Suppose that (1 - .9r) is the first quantity > above zero. What is (1 - .9r)*.5? It must > be some number. Is it zero? Then either 0.5 is zero or > (1 - .9r) is zero, which means 1 = .9r. Is (1 - .9r)*.5 = (1 - .9r)? > Then > (1 - .9r)*.5 - (1 - .9r)*.5 = > (1 - .9r) - (1 - .9r)*.5 > and > 0 = (1 - .9r)*.5 > Thus, as we saw already, > 1 = 0.9r Whatever you claim (1 - .9r)*.5 is equal > to, there is (1 - .9r) times some number > other than 0.5 already there. That will > always be enough to prove that 1 = .9r. Here's how you re-work your claims, > so that you're happy, everybody's happy: What you'll do is analogous to modular arithmetic, > where counting cycles through a finite set > of numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, > 1, 2, 3, 4, ... (mod 10). We introduce new > addition (+') and equality (='), similar to > the old ones, to make this work. For example, > 6 +' 7 =' 3 because 3 =' 13 =' 89,543. > 6 *' 7 =' 2 because 2 =' 42 =' 146,782. > All the integers that differ by a multiple > of 10 are equal (=') modulo 10. It all works > out smoothly, and various forms of modular > arithmetic even have important practical > applications. You should introduce new forms of addition (+'), > multiplication (*'), equality (='), > and comparison (<'), where any two numbers > that differ by an infinitesimal (let us say, > 1 and .9r) are held to be equal under this > new modular equality, that is, 1 =' .9r, > and so on. (I am probably mis-using terminology here. > If someone would give me a better way of > describing what I'm calling modular equality, > I'd be grateful.) All the usual things the rest of us say about > real numbers become true under the new forms > of +', *', =', and <', so we no longer have a > complaint. However, at the same time, you still really have > .9r < 1 and other such facts true in the > background, let's say (with the original > operations), the same way 6 + 7 really is 13, > not 3 in the background. Another advantage to you is that all the stuff > mathematicians have learned about the foreground > operations (+', *', =', <') will form a solid > jumping off point in your explorations of the > background operations (+, *, =, <), where, > for example .9r < 1. Jim Burns- Hide quoted text - - Show quoted text - Physical changes happen fast or slow at the level of the infintely small. Every physical quantity is exact down to the infinitely small. === Subject: Re: .9 repeating sha1:q0mrjQDrDXOHQuWqW1rKdkG83XU= > This is a long post, Mitch, I know, and I didn't > mean it to be. There just came up more and more > issues that I didn't feel I could edit out. Well, let's see how it goes... [...] > Jim Burns- Hide quoted text - > - Show quoted text - Physical changes happen fast or slow at the level of the infintely > small. Every physical quantity is exact down to the infinitely > small. Geez, Jim, tough luck, but no one could have guessed that Mitch would utterly fail to respond to your careful exposition. -- Rob Enderle [predicts] that Longhorn will provide 'vast improvements in security.' We can cheer this happy prospect, but at the same time we must ignore the snide laughs of Macintosh users who have yet to encounter a virus... -- New York Times === Subject: Re: #318 The AP-Reals with AP-adics satisfies the Archimedean Axiom \ ; Nntp-Posting-Host: hera.cwi.nl ... > In his above opinion, Dik would not accept the Maxwell Equations > because the Gilbert definition > of Magnetism or the Ben Franklin definition of magnetism is not the > same as what magnetism > had been understood by the time of Maxwell. This is just one example > of where Dik is out of > place when a discussion of math theory is taking place. > > Neither are actually definitions. They are what was understood about > them, and in empirical sciences that can change over time. Mathematics > is *not* an empirical science. > > Every science is either empirical or theological. Nevertheless, like > in other realms of reality there are mathematical facts that do not > change. Mathematics is the science investiating these facts, but > mathematics is not the facts itself. You are wrong. > Absolute truth is a matter of theology. Mathematics will turn away > from theology in near future. Hrm, it was *you* who was talking about mathematical things that were absolutely true. So you were talking theology? -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: #318 The AP-Reals with AP-adics satisfies the Archimedean Axiom \ ; posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/20080201 Firefox/2.0.0.12 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > Every science is either empirical or theological. That is just ridiculous. Theology IS NOT a science. Science IS NOT theological, PERIOD, according to THE ACTUAL definition of theological. You, personally, however, are hell- bent on CALLING everything that is not empirical theolgoical. THAT is YOUR PERSONAL false dichotomy. It doesn't have anything to do with reality. In reality, science is THEORETICAL, NOT theological. It involves theories ABOUT the empirical. Math is in a privileged position here because the empirical investigations, being calculations and manipulations OF SYMBOLS, are themselves in an abstract realm. There is a sense in which a computation that has never yet been factually carried out, a calculation whose result is not yet known, is like an experiment, and the completion of the calculation (with the concomitant revelation of its result) is like an observation. But you personally are content to oversimplify all this and pontificate as though YOU were, somehow, some authority that other people are bound to respect. The world will be a better place, in at least that 1 respect, after nature takes its course with you. === Subject: Re: The complete infinite binary tree has only countably many Nntp-Posting-Host: hera.cwi.nl ... > ZFC yields uncountably many real numbers. Real numbers can be > identified with paths in the tree. > > Depends on how you define your tree. The last time I saw a definition by > you it stated that it contained all paths that had finitely many 1's (in > your terminology: that terminates with a sequence of 0's). In that case > your statement is false, i.e. there are reals that can *not* be identified > with paths in the tree. There are even rationals that can *not* be > identified with paths in the tree. > > Say: There are many reals that do not fit into the tree. The tree is > already occupied. I have no idea what this means. With your *definition* of a tree, where each path has finitely many 1's, it is clear that 1/3 is *not* represented in the tree. Not because of some vague notion of already being occupied, but because 1/3 does *not* correspond to a path that has finitely may 1's. With the common definition of the complete tree, 1/3 *is* in that tree. > Result: There are not enough infinite sequences of > bits or digits as would be required to distinguish an uncountable set > of reals. In a tree that contains only paths with finitely many 1's only rationals are represented in the tree (and not even all), and so in such a tree there are not sufficient paths to represent all reals. On the other hand, such a tree does *not* contain all possible sequences of 0's and 1's so from such a tree the result you state is not a result. > For once, decide what your complete tree actually is. > (1) It contains all paths that ultimately continue with an infinite > sequence of 0's. > (2) It contains all paths with 1's and 0's in arbitrary positions. > > Neither nor. In that case will you *again* provide a definition of your complete tree? > The tree is a means to prove that there is no all with > respect to infinity. > We start with (1), conclude that the tree, i.e., the structure of > edges and nodes, then must also satisfy (2), and find that it's > impossible. Can you give the mathematical reasoning behind this all? I have not yet seen it. If the tree is the set of all paths that ultimately continue with an infinite sequence of 0's, I do not see how that tree necessarily also contains paths with 0's and 1's in arbitrary positions. Or is your tree not the set of paths? If not, what is it? Pray give a *mathematical* definition. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: The complete infinite binary tree has only countably many posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > ... > ZFC yields uncountably many real numbers. Real numbers can be > identified with paths in the tree. Depends on how you define your tree. The last time I saw a definition by > you it stated that it contained all paths that had finitely many 1's (in > your terminology: that terminates with a sequence of 0's). In that case > your statement is false, i.e. there are reals that can *not* be identified > with paths in the tree. There are even rationals that can *not* be > identified with paths in the tree. Say: There are many reals that do not fit into the tree. The tree is > already occupied. I have no idea what this means. With your *definition* of a tree, where each > path has finitely many 1's, it is clear that 1/3 is *not* represented in the > tree. Not because of some vague notion of already being occupied, but > because 1/3 does *not* correspond to a path that has finitely may 1's. With the common definition of the complete tree, 1/3 *is* in that tree. With the common definition of a tree, the tree is complete if there is no node missing. This is the case with my tree. Result: There are not enough infinite sequences of > bits or digits as would be required to distinguish an uncountable set > of reals. In a tree that contains only paths with finitely many 1's only rationals > are represented in the tree (and not even all), and so in such a tree there > are not sufficient paths to represent all reals. On the other hand, such > a tree does *not* contain all possible sequences of 0's and 1's so from > such a tree the result you state is not a result. This tree contains all nodes. This tree contains all nodes of the path 1/3. Which one is missing? For once, decide what your complete tree actually is. > (1) It contains all paths that ultimately continue with an infinite > sequence of 0's. > (2) It contains all paths with 1's and 0's in arbitrary positions. Neither nor. In that case will you *again* provide a definition of your complete tree? It is a tree that contains all nodes (and edges connecting them) such that you cannot name a node that would be missing. The tree is a means to prove that there is no all with > respect to infinity. > We start with (1), conclude that the tree, i.e., the structure of > edges and nodes, then must also satisfy (2), and find that it's > impossible. Can you give the mathematical reasoning behind this all? I have not yet > seen it. If the tree is the set of all paths that ultimately continue with > an infinite sequence of 0's, I do not see how that tree necessarily also > contains paths with 0's and 1's in arbitrary positions. Because the tree contains all nodes of every path. Or is your tree not the set of paths? If not, what is it? Pray give a > *mathematical* definition. > -- A decimal representation is defined by its digits. The tree is defined by its nodes. Your problem is that logics does not supply unique results in the infinite. Cantor's list could have a label at every line: Attention, this line cannot contain the diagonal number. But Cantor's list should also have a disclaimer at every line: Attention, this line does not show that the diagonal is not in a later part of the list because this line is not the last one of the list. === Subject: Re: The complete infinite binary tree has only countably many > ... > ZFC yields uncountably many real numbers. Real numbers can be > identified with paths in the tree. Depends on how you define your tree. The last time I saw a definition > by > you it stated that it contained all paths that had finitely many 1's > (in > your terminology: that terminates with a sequence of 0's). In that > case > your statement is false, i.e. there are reals that can *not* be > identified > with paths in the tree. There are even rationals that can *not* be > identified with paths in the tree. Say: There are many reals that do not fit into the tree. The tree is > already occupied. I have no idea what this means. With your *definition* of a tree, where > each > path has finitely many 1's, it is clear that 1/3 is *not* represented in > the > tree. Not because of some vague notion of already being occupied, but > because 1/3 does *not* correspond to a path that has finitely may 1's. With the common definition of the complete tree, 1/3 *is* in that tree. With the common definition of a tree, the tree is complete if there is > no node missing. This is the case with my tree. In such a collection of nodes as WM calls a complete infinite binary tree, there are chains of nodes, say pre-paths, which satisfy all the definitions of being a path in such a tree but which WM declares are not paths in that tree. In every complete tree, every maximal chain of parent-to-child linked nodes/edges is a path, as a path in any tree is defined to be any such being such a maximal chain of parent-to-child linked nodes/edges of that tree. Such WM-trees are at best incomplete, or pre-trees, which need completion by requiring all their pre-paths to become paths, in order to become complete trees. Result: There are not enough infinite sequences of > bits or digits as would be required to distinguish an uncountable set > of reals. In a tree that contains only paths with finitely many 1's only rationals > are represented in the tree (and not even all), and so in such a tree there > are not sufficient paths to represent all reals. On the other hand, such > a tree does *not* contain all possible sequences of 0's and 1's so from > such a tree the result you state is not a result. This tree contains all nodes. This tree contains all nodes of the path > 1/3. Which one is missing? Paths are parent-to-child linked chains of nodes, so that a path can be distinguished from others by such a link (an edge) that it does NOT have but they all do. The path in question does not have any left branch link/edge, but every path in WM's tree has infinitely many of them, so it is infinitely often distinguished from each of them and is not in WM's tree. This also distinguishes all infinitely many of those paths having only finitely many left branches, and thus not in WM's tree, from all of the paths in WM's clearly incomplete tree. For once, decide what your complete tree actually is. > (1) It contains all paths that ultimately continue with an infinite > sequence of 0's. > (2) It contains all paths with 1's and 0's in arbitrary positions. Neither nor. In that case will you *again* provide a definition of your complete tree? It is a tree that contains all nodes (and edges connecting them) such > that you cannot name a node that would be missing. But I, and others, can name a lot of things that satisfy all the properties of being a path in that tree which WM claims are not paths. The tree is a means to prove that there is no all with > respect to infinity. > We start with (1), conclude that the tree, i.e., the structure of > edges and nodes, then must also satisfy (2), and find that it's > impossible. Can you give the mathematical reasoning behind this all? I have not yet > seen it. If the tree is the set of all paths that ultimately continue with > an infinite sequence of 0's, I do not see how that tree necessarily also > contains paths with 0's and 1's in arbitrary positions. Because the tree contains all nodes of every path. Or is your tree not the set of paths? If not, what is it? Pray give a > *mathematical* definition. > -- > A decimal representation is defined by its digits. The tree is defined > by its nodes. > Your problem is that logics does not supply unique results in the > infinite. In mathematics, an inability clearly to define what you are talking about means you do not know what you are talking about. Ergo, WM does not know what he is talking about. Cantor's list could have a label at every line: Attention, this line > cannot contain the diagonal number. > But Cantor's list should also have a disclaimer at every line: > Attention, this line does not show that the diagonal is not in a later > part of the list because this line is not the last one of the list. But the rule of construction of the anti-diagonal is such that one CAN say that the diagonal is not in any later part of the list because the rule acts equally on all lines at once. WM's objection above is like objecting to inductive proofs because there is always another natural. === Subject: Re: quantum computing and cryptography posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > I read up on Shor's algorithm a few minutes ago on Wikipedia, and was > surprised to find that the quantum algorithm for solving integer > factorization can also be modified to solve the discrete logarithm > problem as well. (This is really rather trivial, but it was unknown > to me.) This means that if quantum computers ever become easy to build, > standard public key cryptography is essentially useless. Has there been any effort or progress towards designing a public-key > cryptosystem that is probably not breakable in BQP? (I briefly > examinedhttp://en.wikipedia.org/wiki/Quantum cryptography, but saw In other words, is there a problem in NQP that is probably hard for > quantum computers in the same way that the discrete logarithm problem > and integer factorization are probably hard for standard Turing > machines? Could this be used as the basis for a cryptosystem? Finally, since I don't know much about quantum cryptography, I wanted > to ask--does anyone know of any good books on quantum computing that > would be accessible to someone with some theoretical CS background, > but minimal knowledge of physics or quantum computing? Phil It may take longer to build the mythical quantum computer than it does to find a better factoring algorithm. Why bother using a quantum computer if we can just make up any sci-fi invention that won't even break a sweat with anything crypto? === Subject: Re: quantum computing and cryptography posting-account=ZI4QjQoAAAA329OSaIKUN9SLZLEv-XTR AppleWebKit/525.18.1 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) > Has there been any effort or progress towards designing a public-key > cryptosystem that is probably not breakable in BQP? (I briefly > examinedhttp://en.wikipedia.org/wiki/Quantum cryptography, but saw In other words, is there a problem in NQP that is probably hard for > quantum computers in the same way that the discrete logarithm problem > and integer factorization are probably hard for standard Turing > machines? Could this be used as the basis for a cryptosystem? To oversimplify, quantum computers reduce the time complexity of a problem by log base 2. So to maintain hardness, a problem would have to be superexponentially complex by conventional methods. Maybe there is some sort of super-factoring that can be done. === Subject: Re: quantum computing and cryptography Originator: kristiag@math.ntnu.no (Kristian GjÀsteen) >To oversimplify, quantum computers reduce the time complexity of a >problem by log base 2. This is not an oversimplification. This is plain wrong. -- Kristian GjÀsteen === Subject: Re: quantum computing and cryptography posting-account=ZI4QjQoAAAA329OSaIKUN9SLZLEv-XTR AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) On Apr 6, 3:43am, Kristian GjÀsteen problem by log base 2. This is not an oversimplification. This is plain wrong. -- > Kristian GjÀsteen Yeah. I knew it was one of those two. === Subject: Re: quantum computing and cryptography > Has there been any effort or progress towards designing a public-key > cryptosystem that is probably not breakable in BQP? =A0(I briefly > examinedhttp://en.wikipedia.org/wiki/Quantum_cryptography, but saw > In other words, is there a problem in NQP that is probably hard for > quantum computers in the same way that the discrete logarithm problem > and integer factorization are probably hard for standard Turing > machines? =A0Could this be used as the basis for a cryptosystem? >To oversimplify, quantum computers reduce the time complexity of a >problem by log base 2. So to maintain hardness, a problem would have There are very few problems for which a QC is logarithmic of the classical time. Most problems are at best sqrt of the classical which is still polynomial. >to be superexponentially complex by conventional methods. Maybe there >is some sort of super-factoring that can be done. === Subject: Re: quantum computing and cryptography posting-account=ZI4QjQoAAAA329OSaIKUN9SLZLEv-XTR AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) > Has there been any effort or progress towards designing a public-key > cryptosystem that is probably not breakable in BQP? =A0(I briefly > examinedhttp://en.wikipedia.org/wiki/Quantum cryptography, but saw > In other words, is there a problem in NQP that is probably hard for > quantum computers in the same way that the discrete logarithm problem > and integer factorization are probably hard for standard Turing > machines? =A0Could this be used as the basis for a cryptosystem? >To oversimplify, quantum computers reduce the time complexity of a >problem by log base 2. So to maintain hardness, a problem would have There are very few problems for which a QC is logarithmic of the classical > time. > Most problems are at best sqrt of the classical which is still polynomial. to be superexponentially complex by conventional methods. Maybe there >is some sort of super-factoring that can be done. I thought that, since quantum computers can be in 2^n states for n qubits (while normal computers can be in only one), this allows for exponentially more operations. === Subject: Re: quantum computing and cryptography > Has there been any effort or progress towards designing a public-key > cryptosystem that is probably not breakable in BQP? Certainly. There are at least a couple that are not known to be any > easier to break with quantum computation than classical, according to http://www.cdc.informatik.tu-darmstadt.de/research/QPKC.html > http://www.spectrum.ieee.org/jan09/7095 There are apparently also some proposed cryptosystems that require > quantum computing for their feasibility. > - Tim > You think the banks and/or others institutions that depend heavily on RSA are ready to switch from RSA to some other system overnight if one were to come up with a fast factoring method? Or will it be the straw that broke the \ camel's back in the current financial crisis? really curious to get an idea. cg === Subject: Re: quantum computing and cryptography posting-account=kZanLQoAAABvNhBbAlX1SsCxeprjdiHJ GTB5),gzip(gfe),gzip(gfe) When the anthopological origional war of 'ma bitches and resource ug, do fighty fighy' becomes abstracted into a profit and its force, where body parts are not under threat, and resource exceeds personal need, then a funny perverse species evolves. And not funny ha,ha, but funny pants-que-lear. The lack of evapouration of the natural competiton, is contra to the assumption of monopoly. Hence via the escapements of the playboy island, and the selector bitches having a cahoot of a time, fuzz force one lines up for his pie slip a better man than I. Along came the wanton, doubting of the cake, and then the pie slip rummble surely became fake. Was it not for jesus how held us in our moral. The cup cake and the fondu would have never palital. For those who chose to trust, beyond the bounds of logic, faith upon the market man has solute a trojik. cheers jacko === Subject: Re: quantum computing and cryptography > Has there been any effort or progress towards designing a public-key > cryptosystem that is probably not breakable in BQP? > Certainly. There are at least a couple that are not known to be any > easier to break with quantum computation than classical, according to > http://www.cdc.informatik.tu-darmstadt.de/research/QPKC.html > http://www.spectrum.ieee.org/jan09/7095 > There are apparently also some proposed cryptosystems that require > quantum computing for their feasibility. > - Tim >You think the banks and/or others institutions that depend heavily on RSA >are ready to switch from RSA to some other system overnight if one were to \ >come up with a fast factoring method? Or will it be the straw that broke the >camel's back in the current financial crisis? I do not know that banks do depend on RSA. They tend to do key exchange by sending a courier to the bank with the password AFAIK. But if you are worried about QC as a way of defeating RSA, that will take a few years. === Subject: is nth partial fourier sum s_n(f) in L(T), Lp(T) or C(T) posting-account=JTrNRgoAAABuGIzwXYnlAgkRTDpv8_sI .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) T = |R mod 2pi Say f is from L(T), Lp(T) or C(T), is the nth partial fourier sum s_n(f) also in L(T), Lp(T) or C(T) respectively? For C(T) case it's obvious, because the partial sum would be a combination of all continuous functions. Not sure about other two cases. === Subject: Re: is nth partial fourier sum s_n(f) in L(T), Lp(T) or C(T) > T = |R mod 2pi Say f is from L(T), Lp(T) or C(T), is the nth partial fourier sum s_n(f) also in L(T), Lp(T) or C(T) > respectively? For C(T) case it's obvious, because the partial sum would be a > combination of all continuous functions. Not sure about other two cases. Sn(f) is just a finite linear combination of exponentials, so yes. === Subject: Re: Number Superposition Game posting-account=uEOgRAoAAAASMp5oEBx_0d7qfHMcYtcE Gecko/2009032609 Firefox/3.0.8 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > Here is a simple game for any plural number of players. This game is played on an m-by-m grid drawn on paper. The first player to move places a 1 in any square of the grid. Players thereafter take turns, each player placing the integer k in a > square adjacent to (and in the direction of above, below, right of, or > left of -- with restrictions {see below}) the square where the > 1,2,3,4,5,...) A player may put the integer in an empty square OR may write the > integer in a square that already has a single number in it -- but may > not write a number in a square that already has two numbers written in > it. Nice concept. The snake gets a dramatically improved life expectancy (I'm guessing) since it can cross over itself (at least to a height of 2) and avoid dead ending. A player may also not write a number in the square previously written > in by the player who moved before the last player to move. > (In other words: No U-turns. Any integer k cannot be placed in the > same square as the integer (k-2).) difference between the two numbers in the square} added to his/her > score. The game continues until no more moves can be made. Highest score wins, of course. > The anticlimactic, fizzle out game ending: Play ends when there are no more moves and high score wins. Yuck. When you get a good concept, take the next step and make a good game. I think you can do better. -Mark === Subject: Re: Number Superposition Game posting-account=dGiPYgkAAABSJ3xUlNLViQdT0h489hR6 AppleWebKit/523.10.3 (KHTML, like Gecko) Version/3.0.4 Safari/523.10,gzip(gfe),gzip(gfe) > Here is a simple game for any plural number of players. This game is played on an m-by-m grid drawn on paper. The first player to move places a 1 in any square of the grid. Players thereafter take turns, each player placing the integer k in a > square adjacent to (and in the direction of above, below, right of, or > left of -- with restrictions {see below}) the square where the > 1,2,3,4,5,...) A player may put the integer in an empty square OR may write the > integer in a square that already has a single number in it -- but may > not write a number in a square that already has two numbers written in > it. Nice concept. The snake gets a dramatically improved life expectancy > (I'm guessing) since it can cross over itself (at least to a height of > 2) and avoid dead ending. > A player may also not write a number in the square previously written > in by the player who moved before the last player to move. > (In other words: No U-turns. Any integer k cannot be placed in the > same square as the integer (k-2).) difference between the two numbers in the square} added to his/her > score. The game continues until no more moves can be made. Highest score wins, of course. > The anticlimactic, fizzle out game ending: Play ends when there are > no more moves and high score wins. Yuck. When you get a good > concept, take the next step and make a good game. I think you can do > better. -Mark You could simply end the game after the first player reaches a predetermined score. Maybe that is too simple a fix, but I am mentally tired right now. You could, now that I think a little, end the game when the first player reaches a score that is exactly a number that fits into a certain class, such as, the game ends as soon as a player gets a score that is a factorial. The problem with this approach is that neither player may obtain the magic type of number until there are no more moves possible anyway. The trick would be finding the correct class of number. In any case, we can end the game with a whichever comes first -- no more moves possible, OR a score that is a -rule. Leroy Quet === Subject: Re: Ohanian describes Einstein's Mistakes > I have asked you ñif any of the frequencies utilized on board the GPS > satellites are not corrected for the SR or GR effects, would the > system continue to function?î Professor Andersen said yes. Gisse the > college drop-out, the troll, and the liar said no. What is your > answer? We really cannot go on discussing this issue if you behave > like a troll on this one. For Christ's sake, you are a professor of > physics. So, why don't you answer the question? OK. If the clocks on board the GPS satellites where not corrected for > SR and GR effects, there would be a large regular offset introduced > which would have to be corrected by a large ad-hoc correction. You have to identify what this large regular offset is. broadcasts its signal with a 750.000kHz carrier frequency. Your receiver tunes in to 750.001kHz. What type of large regular offset is there? The answer is none. It is part of the tolerance which every engineer has to take into account. Now, with the signal frequency of GPS, if GR clock effect indeed existed, a 450 parts in a trillion of offset in frequencies between the transmitter and the receiver would not matter at all. There is actually no accumulated offset. Professor Andersen explained this a few months. If you didn't understand it, why did you not object? The accumulated offset that you are thinking of exists when you try to synchronize the timer counters in the satellites with the timer counter on the ground. Then, both the satellite timer frequencies and the ground timer frequency that drive these counters have to be all the same. If not, the due to obvious reasons counters would not be synchronized between the satellites and the ground. That is if the GR effect were indeed true. satellites effectively eliminates the necessity to establish a timer counter on the ground. See the post that I have explained how GPS works exactly. GPS has been working for quite sometimes with 4 satellite acquisition while physicists are still baffled by innovative ingenuities among engineers. The claim by Professor Andersen and Roberts is that there exists something else in the GPS designs that require the same frequencies in both the satellites and the ground. What this mysterious functionality is remains illusive from both of the claimants. I am willing to accept if GR is designed into the GPS if this mysterious functionality is explained. Otherwise, the application of GR in GPS remains a lie. certainly technically possible, just like an ad-hoc correction for > long-range ballistics is possible without attributing the correction > to Coriolis effects. The question is, if the sole purpose of doing so > is to avoid attributing the correction to known effects, is this a > worthwhile and practical endeavor? Why don't YOU answer THAT question? Do you now see that your question makes no sense? Once again. === Subject: Re: Ohanian describes Einstein's Mistakes posting-account=-Yr6sAoAAAArJwgdHLA4MBxm4oIdzdtZ InfoPath.1; .NET CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729; MS-RTC EA 2),gzip(gfe),gzip(gfe) (NetCache NetApp/6.0.2P1) I have asked you ñif any of the frequencies utilized on board the GPS > satellites are not corrected for the SR or GR effects, would the > system continue to function?î Professor Andersen said yes. Gisse the > college drop-out, the troll, and the liar said no. What is your > answer? We really cannot go on discussing this issue if you behave > like a troll on this one. For Christ's sake, you are a professor of > physics. So, why don't you answer the question? OK. If the clocks on board the GPS satellites where not corrected for > SR and GR effects, there would be a large regular offset introduced > which would have to be corrected by a large ad-hoc correction. You have to identify what this large regular offset is. broadcasts its signal with a 750.000kHz carrier frequency. Your > receiver tunes in to 750.001kHz. What type of large regular offset is > there? The answer is none. It is part of the tolerance which every > engineer has to take into account. Now, with the signal frequency of GPS, if GR clock effect indeed > existed, a 450 parts in a trillion of offset in frequencies between > the transmitter and the receiver would not matter at all. There is > actually no accumulated offset. Professor Andersen explained this a > few months. If you didn't understand it, why did you not object? The accumulated offset that you are thinking of exists when you try to > synchronize the timer counters in the satellites with the timer > counter on the ground. Then, both the satellite timer frequencies and > the ground timer frequency that drive these counters have to be all > the same. If not, the due to obvious reasons counters would not be > synchronized between the satellites and the ground. That is if the GR > effect were indeed true. satellites effectively eliminates the necessity to establish a timer > counter on the ground. See the post that I have explained how GPS > works exactly. GPS has been working for quite sometimes with 4 > satellite acquisition while physicists are still baffled by innovative > ingenuities among engineers. The claim by Professor Andersen and Roberts is that there exists > something else in the GPS designs that require the same frequencies in > both the satellites and the ground. What this mysterious > functionality is remains illusive from both of the claimants. I am > willing to accept if GR is designed into the GPS if this mysterious > functionality is explained. Otherwise, the application of GR in GPS > remains a lie. This is > certainly technically possible, just like an ad-hoc correction for > long-range ballistics is possible without attributing the correction > to Coriolis effects. The question is, if the sole purpose of doing so > is to avoid attributing the correction to known effects, is this a > worthwhile and practical endeavor? Why don't YOU answer THAT question? Do you now see that your question makes no sense? Once again. These relativist should not be playing chess at all. They keep losing. They should stick with checkers. They have yet to show the equations that generate the corrections. I bet those equations were made long after the offsets were known. It would be interesting to see what those equations yield for say, a GPS system in the moon, Mars or Jupiter? === Subject: Re: Ohanian describes Einstein's Mistakes <49D780C9.9010006@somewhere.no> posting-account=-Yr6sAoAAAArJwgdHLA4MBxm4oIdzdtZ InfoPath.1; .NET CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; MS-RTC LM 8; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) (NetCache NetApp/6.0.2P1) On 4 .83.9f.83ë.83ì, 11:46, Paul B. Andersen Well here's an easy question for you: Is the quantum mechanical > up the atomic clock in the ground? (Hint: The question may sound complex, but it is answerable by a > simple yes/no, and one of these will be correct). (More hints: The answer is no.) Followup question: Since the solutions are different, would that be > sufficient to account for 50,000ns/day differential? I challenge you to show the two different solutions, > and calculate the differential. What, can't you do it? > Thought so. -- > Paul http://home.c2i.net/pb andersen/ I already said the calculations cannot be done due to complexity. Are you deaf or something? === Subject: Re: Ohanian describes Einstein's Mistakes posting-account=-Yr6sAoAAAArJwgdHLA4MBxm4oIdzdtZ InfoPath.1; .NET CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; MS-RTC LM 8; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) (NetCache NetApp/6.0.2P1) > [Note: PD is unable to discuss quantum mechanics at all. [NonBreakingSpace]Even the > basic double slit experiment he is ignorant of the set-up itself. [NonBreakingSpace]He > thinks it can be done at home with a $5 laser. [NonBreakingSpace]Sheesh.] It thinks it can be done because it can be done rather trivially. A > piece of foil, a sharp box cutter, and a cheap laser is all it takes. Is there a reason you don't think it can be done, ?- .83.9f.83ëfi\.91\.98\.9d[ATi lde]\.8f \.91\.8c\.8e\.92ó\.94\.95[UG rave] \.97\.8c [Eth]\.87\.98Ë'\.8c\.97\.8f - - \.a6\.92\.beË\.94\.8e\.97\.8f \.91\.8c\.8e\.92ó\.94\.95[UG rave] \.97\.8c [Eth]\.87\.98Ë'\.8c\.97\.8f - BWAH HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA We're discussing quantum mechanics IDIOT. You need a detector to confound the photon path IDIOT. When the detector is on, the interference pattern disappears IDIOT. BWAH HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA BWAH HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA BWAH HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA The moron Eric is thinking about the classical DSE and not the quantum DSE. BWAH HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA BWAH HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA BWAH HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA === Subject: Re: Ohanian describes Einstein's Mistakes posting-account=-Yr6sAoAAAArJwgdHLA4MBxm4oIdzdtZ InfoPath.1; .NET CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; MS-RTC LM 8; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) (NetCache NetApp/6.0.2P1) > You slept through your high school physics courses You keep forgetting to add that I had the highest test scores. and none of those covered QM. They did. Name the textbook you used. [...] Why? [NonBreakingSpace]It's not gonna be easier than what you have. How do you know, ? Judging from your 'over-staying time', no textbook will make it easier for you :-) === Subject: Re: Ohanian describes Einstein's Mistakes <1238665395_21@sicinfo3.epfl.ch> <49d4e634$0$977$ec3e2dad@news.usenetmonster.com> <49d80ad3$0$9327$ec3e2dad@news.usenetmonster.com> posting-account=PTS84AoAAACr67p51zvy0Hlr3LkoIUcc SLCC1; .NET CLR 2.0.50727; .NET CLR 3.0.04506; Media Center PC 5.0; \ InfoPath.2),gzip(gfe),gzip(gfe) > xxein: I can wholeheartedly disagree with your opinion --- trying to > explain SR without any math.... Yes, trying to explain SR -- or LET -- without the use of math has led to > huge amounts of confusion on the part of people attempting to understand > it. That is correct. Both LET and SR are merely interpretations to the Lorentz transform. can't understand most physics with NO math, either. Without math, no one can prove you right or wrong. At least with math, you can perform experiments to prove you right. Physics without the math is philosophy where there is no right or wrong. Math is everything. If Dr. Ohanian did publish any math, I don't suppose you would understand it anyway since you cannot see the gross error in (E = m c^2) of Einstein the nitwit, the plagiarist, and the liar in his 1905 paper. subjective mode of thought and will be short-circuited from > understanding the objective physic. I feel very sorry for you. Sheesh, throw in an insult, why don't you. Like you have not. After getting your ass kicked by yours truly in discussing SR and GR, you've resorted to call me NAZI, anti-Semitic, and all sorts of ill-words. Sheesh, you are not only a liar but a hypocrite as well. === Subject: Re: Define mass posting-account=8k9z_QoAAABfJ6zct3wDB2k3FKfRerU2 2.0.50727),gzip(gfe),gzip(gfe) > revelations are as close as you will ever get to science truths! NE You express yourself very well! And you say a major truth: ïEinstein > has held back science.' Your Fritz London figured how the jigsaw > puzzle of atoms can fit together. But the change of state from a gas > to a solid relates more to thermal quantities than to some ñrandomî > attractive force. I sense that you are a ïlearned' person. Too often that means someone > who absorbs what others write rather then tries to make sense of the > whole. You are some of both personas. For the past year or so, I > have revealed on sci.physics that gravity is simply ñdownward flowing > etherî. Light is packets of ether (photons if you wish) traveling > faster than the ether ïfluid' is flowing. And light can travel in the > reverse direction to the ether flow by tunnelingmuch like machine gun > bullets can pass up through falling snow. Via the reverse flow, the > ether density near massive objects is maintained, so gravity can > continue. Stop the replenishment, as in Black Holes, and the gravity > will shut off. Because ether density near massive objects conforms > to the inverse square law, all of the purported ñpredictionsî of > Einstein's GR are really just coincidental results from his having > (after five long years) written an empirical equation to define the > precessed orbit of the planet Mercury. The smallest indivisible energy unitwhich I call an IOTAis probably > doughnut shaped, and with a tangential velocity of ïc'. Depending > upon which side one references, IOTAs can be clockwise or > counterclockwise. So they are polar. Align two IOTAs with the same > direction of rotation and you are beginning a line of magnet flux. If > light hits that chain, it can be disrupted. That is also why the > Earth's magnetic flux can be disturbed by solar winds. Magnetism, > long as those don't cross paths too often. Sue, rather than lamenting that gravity is some unknown energy > mechanism, you should rejoice with me that ñvarying ether flow and > densityî not only explain gravity, but everything else in nature! > NoEinstein John certainly is no Einstein. Read the following threads and > laugh at his incompetetence: Where Angels Fear to Fall > Last Nails in Einstein's Coffin > Pop Quiz for Science Buffs! > An Einstein Disproof for Dummies > Another look at Einstein > Three Problems for Math and Science > Matter from Thin Air > Curing Einstein's Disease > Replicating NoEinstein's Invalidation of M-M (at sci.math) > Cleaning Away Einstein's Mishmash > Dropping Einstein Like a Stone > Plotting the Curves of Coriolis, Einstein, and NoEinstein (is > Copyrighted.) > Are Jews Destroying Objectivity in Science? > The Gravity of Masses Doesn't Bend Light. > KE = 1/2mv^2 is disproved in new falling object impact test. > Light rays don't travel on ballistic curves. >I challenge anyone to define mass. ><< Einstein published his theory of gravitation, or >general theory of relativity, in 1916. And so a new >paradigm, or set of beliefs, was established. It was >not until 1930 that Fritz London explained the weak, >attractive dipolar electric bonding force (known as >Van der Waals' dispersion force or the ñLondon forceî) >that causes gas molecules to condense and form liquids >and solids. Like gravity, the London force is always >attractive and operates between electrically neutral >molecules.And that precise property has been the most >puzzling distinction between gravity and the powerful >electromagnetic forces, which may repel as well as >attract. >So it seems the clue about the true nature of gravity >has been available to chemists [CapitalEth] who are not interested >in gravity [CapitalEth] and unavailable to physicists [CapitalEth] who are not >interested in physical chemistry (and view the world >through Einstein's distorting spectacles). Look at any >average general physics textbook and you will find no >reference to Van der Waals' or London forces. What a >different story might have been told if London's insight >had come a few decades earlier? Physics could, by now, >have advanced by a century instead of being bogged in >a mire of metaphysics. >An excellent illustrated lesson on the London force, or >Van der Waals' dispersion force is given at: >http://www.chemguide.co.uk/atoms/bonding/vdw.html >The London force originates in fluctuating electric dipoles >caused by slight distortion of otherwise electrically neutral >atoms and molecules. The tiny electric dipoles arise >because the orbiting electrons, at any given instant, cannot >shield the positive charge of the nucleus equally in all >directions. The result, amongst a group of similar atoms or >molecules is that the electric dipoles tend to resonate and >line up so that they attract each other. >Obviously, gravity is distinct from the London force. >It is much, much weaker. That should be a clue. What >if we are looking at gravity being due to a similar >electrostatic distortion effect in the far smaller constituents >of each atom? Of course, this is heresy because the electron >challenge this belief. What ismore, this model of an electron >offers a simple mechanism to explain quantum theory and >http://en.wikipedia.org/wiki/Induced gravityhttp://en.wikipedia.org/w... >Sakharov's induced gravity: a modern perspectivehttp://arxiv.org/abs/gr-qc/0204062 >The Origin of Gravity >Authors: C. P. Kouropouloshttp://arxiv.org/abs/physics/0107015http://arxiv.org/abs/physics/\ 0 107... >Sue... Mitch Raemsch- Hide quoted text - >- Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - === Subject: Re: Define mass posting-account=c8o1vgkAAAAnBLUfdutzGql6SMmfFJWF Trident/4.0; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) In short, stress is not mass. Yet another example of why it's best to ignore what you think about > this subject. It would be MUCH better if you actually READ what I said. > The inertia of stress, Medina, Rodrigo, American Journal of Physics, > Volume 74, Issue 11, pp. 1031-1034 (2006). You need to read more carefully: I did NOT say stress has no inertia. Had you read my entire post, you would know that the above sentence was just a summary of the entire post, in which I pointed out that stress does not BEHAVE like mass in GR. That is, while stress has inertia, it is not the same as the inertia of mass -- they are different components of the energy-momentum tensor and thus affect the geometry differently. Calling stress mass is just as silly as calling energy mass. And just as wrong. There is a reason we use different names for them.... Tom Roberts === Subject: Re: Define mass posting-account=5GUrzQkAAADun29oaK3p_W_saUVxxHUF Gecko/2009032609 Firefox/3.0.8,gzip(gfe),gzip(gfe) > In short, stress is not mass. Yet another example of why it's best to ignore what you think about > this subject. It would be MUCH better if you actually READ what I said. The inertia of stress, Medina, Rodrigo, American Journal of Physics, > Volume 74, Issue 11, pp. 1031-1034 (2006). You need to read more carefully: I did NOT say stress has no inertia. Had you read my entire post, you would know that the above sentence was > just a summary of the entire post, in which I pointed out that stress > does not BEHAVE like mass in GR. That is, while stress has inertia, it > is not the same as the inertia of mass -- they are different components > of the energy-momentum tensor and thus affect the geometry differently. Calling stress mass is just as silly as calling energy mass. And > just as wrong. There is a reason we use different names for them.... Tom Roberts ----------------- energy is MASS IN MOTION!! exactly as in macrocosm !!! Y.Porat -------------------------- === Subject: Re: Define mass posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) > In short, stress is not mass. Yet another example of why it's best to ignore what you think about > this subject. It would be MUCH better if you actually READ what I said. The inertia of stress, Medina, Rodrigo, American Journal of Physics, > Volume 74, Issue 11, pp. 1031-1034 (2006). You need to read more carefully: I did NOT say stress has no inertia. Had you read my entire post, you would know that the above sentence was > just a summary of the entire post, in which I pointed out that stress > does not BEHAVE like mass in GR. That is, while stress has inertia, it > is not the same as the inertia of mass -- they are different components > of the energy-momentum tensor and thus affect the geometry differently. Calling stress mass is just as silly as calling energy mass. And > just as wrong. There is a reason we use different names for them.... Tom Roberts ----------------- > energy is > MASS IN MOTION!! exactly as in macrocosm !!! > Y.Porat > --------------------------- Hide quoted text - - Show quoted text - I call it energy flow in the immatterial In momentum it is energy flow exchange Mitch Raemsch === Subject: Re: Define mass posting-account=c8o1vgkAAAAnBLUfdutzGql6SMmfFJWF Trident/4.0; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) On Apr 3, 10:56am, stevendaryl3...@yahoo.com (Daryl McCullough) > pmbsays... >I wanted to add that there are a lot of quantities that can be >eliminated from physics. Elimination #1 - We can do away with rest mass and replace it with >proper energy E 0/c^2. Elimination #2 - We can do away with total inertial energy and replace >it with gamma*E 0. Elimination #3 - We can do away with velocity and replace it with dx/ >dt. Elimination #4 - We can do away with momentum by replacing it with >(gamma*E 0/c^2)dx/dt. Elimination #5 - We can do away with conservation of momentum by >postulatkng that (gamma*E 0/c^2) is conserved. Elimination #6 - We can do away with force by replacing it with d/dt >[(gamma*E 0/c^2)dx/dt]. Elimination #7 - We can do away with time intervals by only refering >to proper time recording on clocks. Elimination #8 - We can do away with length by replacing it with >gamma*L 0 replacing it with gamma*dtau My point is that a complete set of physically > meaningful quantities are those that can be expressed > as scalars, vectors, tensors, (maybe spinors). Then > the other quantities can be understood in terms of those. So yes. We can do all that. But do you really think that it'd be wise >or helpful to do so? I.e. please explain why the following E 0 = m 0 c^2 >E = gamma*E 0 >v = dx/dt >p = mv >F = dp/dt >dt = gamma*dtau helpful when we can do without them? I don't consider them particularly helpful. The > more useful quantities are: 1. momentum 4-vector > 2. tangent vector to a path > 3. 4-force > 4. affine parameterization of a path -- > Daryl McCullough > Ithaca, NY- Hide quoted text - - Show quoted text - Actually ... never mind. I don't know how I let myself get involved in yet another discussion about semantics and philosophy simply over the definition of mass === Subject: Re: Define mass posting-account=kZanLQoAAABvNhBbAlX1SsCxeprjdiHJ GTB5),gzip(gfe),gzip(gfe) When the anthopological origional war of 'ma bitches and resource ug, do fighty fighy' becomes abstracted into a profit and its force, where body parts are not under threat, and resource exceeds personal need, then a funny perverse species evolves. And not funny ha,ha, but funny pants-que-lear. The lack of evapouration of the natural competiton, is contra to the assumption of monopoly. Hence via the escapements of the playboy island, and the selector bitches having a cahoot of a time, fuzz force one lines up for his pie slip a better man than I. Along came the wanton, doubting of the cake, and then the pie slip rummble surely became fake. Was it not for jesus how held us in our moral. The cup cake and the fondu would have never palital. For those who chose to trust, beyond the bounds of logic, faith upon the market man has solute a trojik. cheers jacko === Subject: Re: Define mass posting-account=c8o1vgkAAAAnBLUfdutzGql6SMmfFJWF Trident/4.0; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) On Apr 3, 10:56am, stevendaryl3...@yahoo.com (Daryl McCullough) > pmbsays... >I wanted to add that there are a lot of quantities that can be >eliminated from physics. Elimination #1 - We can do away with rest mass and replace it with >proper energy E 0/c^2. Elimination #2 - We can do away with total inertial energy and replace >it with gamma*E 0. Elimination #3 - We can do away with velocity and replace it with dx/ >dt. Elimination #4 - We can do away with momentum by replacing it with >(gamma*E 0/c^2)dx/dt. Elimination #5 - We can do away with conservation of momentum by >postulatkng that (gamma*E 0/c^2) is conserved. Elimination #6 - We can do away with force by replacing it with d/dt >[(gamma*E 0/c^2)dx/dt]. Elimination #7 - We can do away with time intervals by only refering >to proper time recording on clocks. Elimination #8 - We can do away with length by replacing it with >gamma*L 0 replacing it with gamma*dtau My point is that a complete set of physically > meaningful quantities are those that can be expressed > as scalars, vectors, tensors, (maybe spinors). Then > the other quantities can be understood in terms of those. So yes. We can do all that. But do you really think that it'd be wise >or helpful to do so? I.e. please explain why the following E 0 = m 0 c^2 >E = gamma*E 0 >v = dx/dt >p = mv >F = dp/dt >dt = gamma*dtau helpful when we can do without them? I don't consider them particularly helpful. The > more useful quantities are: 1. momentum 4-vector > 2. tangent vector to a path > 3. 4-force Please explain why you consider then useful === Subject: Re: Define mass posting-account=c8o1vgkAAAAnBLUfdutzGql6SMmfFJWF Trident/4.0; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) On Apr 3, 12:11pm, stevendaryl3...@yahoo.com (Daryl McCullough) > pmbsays... On Apr 1, 11:29=A0am, stevendaryl3...@yahoo.com (Daryl McCullough) > pmbsays... >On Mar 31, 10:52 am, stevendaryl3...@yahoo.com (Daryl McCullough) > Personally, I think that there is no benefit to using relativistic mas= >s, > and I think it only causes confusion. In particular, I think that > F =3D3D d(mv)/dt is the wrong generalization of Newton's second law. >Do you really want to discuss this? This is a very complicated issue. > No, I don't think it is. For whatever purpose relativistic mass > serves, you can use E/c^2 instead, where E is the relativistic > energy. No you can't. The relationship E = mc^2 only holds in special cases Why do you say that? Because it's true. > It's just a definition. You can > take E = mc^2 as the implicit definition of relativistic mass, > or you can take p = mv as the implicit definition. That would required changing the definition of relativistic mass to something it isn't, for no apparent reason other than to give E/c^2 a name. > Or > you can take f = ma as the implicit definition (in this > case, m is a 3-tensor, rather than a scalar). Why is one > any better than the other? > Using proper time, rest mass, and velocity 4-vector U > instead of coordinate time, relativistic mass, and ordinary > velocity v makes it much clearer what's a scalar, what's a > 4-vector, which makes it clearer how things transform to different > coordinate systems. > Yes, mass is not useful in other cases. Why would you want to restrict the concept of mass to pertain only to very simple cases? === Subject: Re: What mass isn't relativistic ? posting-account=8k9z_QoAAABfJ6zct3wDB2k3FKfRerU2 2.0.50727),gzip(gfe),gzip(gfe) > When have you done anything for science? > Mass in not relativistic in fasest time. === Subject: Re: What mass isn't relativistic ? posting-account=8k9z_QoAAABfJ6zct3wDB2k3FKfRerU2 2.0.50727),gzip(gfe),gzip(gfe) When have you done anything for science? > === Subject: Introduction to spectral sequences Hi! I need a short and, definitely, easy introduction to spectral sequences. I am not looking for a book, but rather something that is free to download on the web. The topic seems so dense and hard so I am trying to make the introduction as less scary as I possibly can (also I am under time pressure so I need to learn the basic stuff fast, depth is not needed). Galois === Subject: Re: Introduction to spectral sequences > Hi! I need a short and, definitely, easy introduction to spectral sequences. > I am not looking for a book, but rather something that is free to > download on the web. The topic seems so dense and hard so I am trying to make the introduction > as less scary as I possibly can (also I am under time pressure so I need > to learn the basic stuff fast, depth is not needed). Galois I haven't read this (what you get from downloading the thing described in the URL below), but from what I have read of Allen Hatcher's texts, it should be correct, well-written, and perhaps understandable to the novice: http://www.math.cornell.edu/~hatcher/#SSAT === Subject: Find area of two circles posting-account=em7BlwoAAADOhRXv8AilqN49TjuokE7m Trident/4.0; InfoPath.2; FDM; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) Two unit circles, passing through centers of each other, contains two gray circles as shown in the image, can you find the area of those two (gray colored) circles? http://i233.photobucket.com/albums/ee201/vcpandya/CircleQuestion.jpg === Subject: Re: Find area of two circles AI a .8ecrit : > Two unit circles, passing through centers of each other, contains two > gray circles as shown in the image, can you find the area of those two > (gray colored) circles? > http://i233.photobucket.com/albums/ee201/vcpandya/CircleQuestion.jpg For the rightmost circle, it is easy and just uses Pythagora to get (R-r)^2 = R^2/4 + r^2, hence deduce r/R hence area ratio. The leftmost is trickier : My method is to consider the triangle of centers, with sides a = R, b = R-r' and c = R+r', and altitude r' That is area (of this triangle) A = R.r'/2 but also from Heron is A = sqrt(p(p-a)(p-b)(p-c)) with p = (a+b+c)/2 hence a relation in (R,r') : I got r'^2 = 3(R^2/4 - r'^2) hence ratio r'/R and area ratio. -- Philippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) === Subject: Re: Find area of two circles posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > http://i233.photobucket.com/albums/ee201/vcpandya/CircleQuestion.jpg For the rightmost circle, it is easy and just uses Pythagora > to get (R-r)^2 = R^2/4 + r^2, hence deduce r/R hence area > ratio. How do you know that the center of the large circle, the center of the small rightmost circle, and the upper right point of tangency lie on the same line? I'm sure these three points do lie on the same line, but I don't see why. If we assume they lie on the same line, the equation you gave is immediate . . . L. Renfro === Subject: Re: Find area of two circles posting-account=U2cMXgoAAADa7uFaEtrTh3s-noWr2c49 Browser; Avant Browser; .NET CLR 2.0.50727; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) >http://i233.photobucket.com/albums/ee201/vcpandya/CircleQuestion.jpg For the rightmost circle, it is easy and just uses Pythagora > to get (R-r)^2 = R^2/4 + r^2, hence deduce r/R hence area > ratio. How do you know that the center of the large circle, the center > of the small rightmost circle, and the upper right point of > tangency lie on the same line? I'm sure these three points do > lie on the same line, but I don't see why. If we assume they > lie on the same line, the equation you gave is immediate . . . L. Renfro To see colinearity of the tangent point and centers of two circles of unequal size, just let the smaller circle roll to the bottom. Either that, or use perpedicularity of the tangent line with the appropriate radius (and whatever additional postulates are needed). For the larger of the two inscribed circles, consider the distance between the center of the large left circle and the point where the drawn diameter and left inscribed circle meet. Call this distance d. Then d is part of two right triangles, one with sides r, R-r and d, and another with sides R+r, r and R+d. This gives enough information to solve for d in terms of R and r, and then to solve for r/R, the ratio of the radii of the small to the large circle. Gerhard Ask Me About System Design Paseman, 2009.04.06 === Subject: Re: Find area of two circles > http://i233.photobucket.com/albums/ee201/vcpandya/CircleQuestion.jpg For the rightmost circle, it is easy and just uses Pythagora > to get (R-r)^2 = R^2/4 + r^2, hence deduce r/R hence area > ratio. How do you know that the center of the large circle, the center > of the small rightmost circle, and the upper right point of > tangency lie on the same line? I'm sure these three points do > lie on the same line, but I don't see why. If we assume they > lie on the same line, the equation you gave is immediate . . . There line perpendicular to both circles at the common point of tangency must pass through both centers. > L. Renfro === Subject: Re: Find area of two circles posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > How do you know that the center of the large circle, the center > of the small rightmost circle, and the upper right point of > tangency lie on the same line? I'm sure these three points do > lie on the same line, but I don't see why. If we assume they > lie on the same line, the equation you gave is immediate . . . > There line perpendicular to both circles at the common point > of tangency must pass through both centers. areas of math (probably _the_ weakest). L. Renfro === Subject: what does 1/{x} mean? A problem I saw had this notation. {1/{x} x>=0 x not in Z(Integers) f(x)={oo x>=0 x in Z {0 x<0 Surely the notation for a singleton set makes no sense here, does it? Can anyone clarify what this 1/{x} means? === Subject: Re: what does 1/{x} mean? <060420091506187805%edgar@math.ohio-state.edu.invalid> posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > Probably {x} means fractional part of x, that > is x-[x], where [x] is the integer part of x. Ah, that's probably what it means. I've seen this before, but incorrectly guessed it was [x]. L. Renfro === Subject: Re: what does 1/{x} mean? posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > A problem I saw had this notation. > {1/{x} x>=0 x not in Z(Integers) > f(x)={oo x>=0 x in Z > {0 x<0 Surely the notation for a singleton set makes > no sense here, does it? Can anyone clarify what > this 1/{x} means? The expression {x} might mean the greatest integer less than or equal to x (floor function) or the least integer greater than or equal to x (ceiling function). A specific reference to where you saw this could possibly be helpful. L. Renfro === Subject: Generalized eigenvalue problem Importance: Normal For the generalized eigenvalue problem Ax=eBx, where A and B are real symmetric square matrices, x is the eigenvector, and e is the eigenvalue, the eigenvalues are supposed to be real (see proof below). Are there other requirements for e to be real besides A and B's being real and symmetric? For example, if A=[1,0; 0,-1] and B=[0,1; 1,0], e=+-i. However, \ B^{-1}A is skew-symmetric, and skew-symmetric matrices have imaginary eigenvalues. I do not recall ever seeing (and cannot now find) any requirements on B^{-1}A in order for the generalized eigenvalue problem to have real eigenvalues. What am I missing? Appendix. Proof that Ax=eBx has real eigenvalues if A and B are real symmetric matrices: 1. Given Ax=eBx. 2. Take complex conjugate (*) of Eq. 1: Ax*=e*Bx* 3. Multiply Eq. 1 by x*^T: x*^TAx=ex*^TBx 4. Multiply Eq. 2 by x^T: x^TAx*=e*x^TBx* 5. LHS of Eqs. 3 and 4 are equal, since they are the transposes of each other, and both are scalars. 6. Equate RHS of Eqs. 3 and 4, where x*^TBx=x^TBx* implies e*=e. Q.E.D. However, since I have a counter-example, something must be wrong with this proof, but I cannot find it. === Subject: Re: Generalized eigenvalue problem >For the generalized eigenvalue problem Ax=eBx, where A and B are real >symmetric square matrices, x is the eigenvector, and e is the eigenvalue, >the eigenvalues are supposed to be real (see proof below). Are there >other requirements for e to be real besides A and B's being real and >symmetric? For example, if A=[1,0; 0,-1] and B=[0,1; 1,0], e=+-i. However, >B^{-1}A is skew-symmetric, and skew-symmetric matrices have imaginary >eigenvalues. I do not recall ever seeing (and cannot now find) any >requirements on B^{-1}A in order for the generalized eigenvalue problem to \ >have real eigenvalues. What am I missing? Appendix. Proof that Ax=eBx has real eigenvalues if A and B are real >symmetric matrices: >1. Given Ax=eBx. >2. Take complex conjugate (*) of Eq. 1: Ax*=e*Bx* >3. Multiply Eq. 1 by x*^T: x*^TAx=ex*^TBx >4. Multiply Eq. 2 by x^T: x^TAx*=e*x^TBx* >5. LHS of Eqs. 3 and 4 are equal, since they are the transposes of each >other, and both are scalars. >6. Equate RHS of Eqs. 3 and 4, where x*^TBx=x^TBx* implies e*=e. Q.E.D. >However, since I have a counter-example, something must be wrong with this \ >proof, but I cannot find it. > for the point 6 :and if x^TB*=0? (case of counter-exemple) === Subject: Re: JSH: Considering my binary quadratic Diophantine solution Most excellent! Keep it up. === Subject: Re: Considering my binary quadratic Diophantine solution projection, pathetic, corrected; > I, in contrast, are useless to your world. But I pretend otherwise > and God help us, we're screwed until enough people figure that out, > and toss me out. > James Harris === Subject: Re: Considering my binary quadratic Diophantine solution simple math, for simpletons is SQRT(JSH) + or - or both ? > James Harris === Subject: Re: JSH: Considering my binary quadratic Diophantine solution [corrected;] >My society is like Hollywood science. My people don't know the >meaning of doing real research. Just a fantasy. > James Harris === Subject: Re: JSH: Considering my binary quadratic Diophantine solution it was pure? Or how the nail people might decide that nails need > two heads and call you insane if you argued that they were stupid? > James Harris What strange swirly-pies float around in your small brain. Why ? === Subject: Re: Considering my binary quadratic Diophantine solution > One of the things I did months ago was use what I call tautological > spaces--invented to tackle Fermat's Last Theorem by me back December > 1999--against binary quadratic Diophantine equations--kind of > accidentally as I was going after a three variable expression > originally. Have you tried using a faster-than-light USB toaster? === Subject: Re: Considering my binary quadratic Diophantine solution > One of the things I did months ago was use what I call tautological > spaces--invented to tackle Fermat's Last Theorem by me back December > 1999--against binary quadratic Diophantine equations--kind of > accidentally as I was going after a three variable expression > originally. Have you tried using a faster-than-light USB toaster? === Subject: Re: Considering my binary quadratic Diophantine solution > One of the things I did months ago was use what I call tautological > spaces--invented to tackle Fermat's Last Theorem by me back December > 1999--against binary quadratic Diophantine equations--kind of > accidentally as I was going after a three variable expression > originally. Have you tried using a faster-than-light USB toaster? === Subject: Re: Considering my binary quadratic Diophantine solution > Your society is like Hollywood science. You people don't know the > meaning of doing real research. Just a fantasy. And you do? Right. How does it feel to waste 14 years of your life on useless crap? === Subject: Re: JSH: Considering my binary quadratic Diophantine solution posting-account=WlifZwoAAADn4Qc008FhhuRE4Syn8J58 3.011; GTB5; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) So a rational parameterization of Pell's Equation is not of interest > to the number theorists as they focus on integers. While the > algebraic geometrists, don't do number theory. > James: > Every book on algebraic geometry contains the theorem (usually in the first few pages) that every smooth conic is birationally equivalent to a line. People who are not stupid understand that this statement (along with its proof) has exactly the same content as all your recent posts about rational parameterization. > There are only two ways you could believe you've done something new. Either a) you have never picked up a book on algebraic geometry or b) you were too stupid to understand what that book was saying. My guess is that both things are true. > The first *nontrivial* result in algebraic geometry is that a nonsingular cubic is NOT birationally equivalent to a line. The proof is both beautiful and enlightening. Unfortunately, you are too stupid to understand it. === Subject: Re: JSH: Considering my binary quadratic Diophantine solution >There are only two ways you could believe you've done something new. >Either a) you have never picked up a book on algebraic geometry or >b) you were too stupid to understand what that book was saying. My >guess is that both things are true. JSH has proved his lack of knowledge of Math, and proved his is all troll/crackpot. === Subject: Re: JSH: Considering my binary quadratic Diophantine solution >And you get a parameterization for hyperbolas with integer D>0, for >circles with D=-1, and for ellipses with D<-1. That D number though may hold secrets to our reality. Unfortunately the value of D that actually holds the secret is in the range -1 < D < 0 so your work has missed the target once again James. Did it never occur to you to ACTUALLY CHECK YOUR WORK to see if you had covered the complete range of values of D? YOU NEED TO CHECK YOUR WORK MORE THOROUGHLY. Sorry to shout, but it seems neccessary. All too often you make trivial, avoidable, errors because you fail to check your work. You can do better than this James. rossum === Subject: Re: JSH: Considering my binary quadratic Diophantine solution posting-account=sKfmEQkAAAC8kI3Pv6_U_nt9sVsxZ_ou 1.1.4322),gzip(gfe),gzip(gfe) > And I freaking generally solved binary quadratic Diophantine > equations, getting among other things, the result that every such > equation for a circle, ellipse or hyperbola can instead be solved for > integers by solving an equation of the form u^2 + Dy^2 = F. You are only 201 years behind the times on this too. See Theorie des Nombres by A.N. Legendre published in 1808, === Subject: Re: JSH: Considering my binary quadratic Diophantine solution posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.1),gzip(gfe),gzip(gfe) > Your society is like Hollywood science. You people don't know the > meaning of doing real research. Just a fantasy. James Harris Fantasy is seeing that you call jour jibberish (JSH) research. You understnad the concept of research as well as you understand the meaning of truth! scrambled up. You have been at this nonsense for 14 years and no one will ever accept anything you ever do. You are a cheat, a charlatan and a liar. You have your own entry on crank.net. You are jealous of Dr. Wiles as he actually proved FLT. You are jealous at the rest of us who have dozens of 'published' papers in non-defunct journals! Eat crow and eat you delusional narcissist! You have amounted to a Hollywood nothing and are a circus act that continues to mathurbate! Get your head out of your ass! ~A === Subject: Re: JSH: Considering my binary quadratic Diophantine solution > One of the things I did months ago was use what I call tautological > spaces--invented to tackle Fermat's Last Theorem by me back December > 1999--against binary quadratic Diophantine equations--kind of > accidentally as I was going after a three variable expression > originally. And I freaking circus freaking ? > generally solved binary quadratic Diophantine > equations, getting among other things, the result that every such > equation for a circle, ellipse or hyperbola can instead be solved for > integers by solving an equation of the form u^2 + Dy^2 = F. geometry, highschool. Ever take it ? I too at that time was focused on Diophantine only, and didn't care > about rational or other solutions, but I thought that was a nifty > a journal or two and it got rejected by them. discarded monkey math. But now having moved on to considering rational cases it looks like a > lot of people missed the reality that certain conics in rationals are > best contemplated by what mathematicians called the Pell's Equation > and my result now looks like an exclamation point on that reality. Mathematically, u^2 + Dy^2 = F, is the granddaddy equation, granddaddy ? Obviously, you have not been around Math very much. > and its > special case of u^2 - Dy^2 = 1, is more than up to the task of > completely encompassing all that is mathematically of interest with > circles, ellipses and hyperbolas. This is Jr High school stuff, JSH, you are regressing. My take on what mathematicians do is that they specialize. > people specialize on tools--but do not really use them. It's like if carpenters had to deal with hammer specialists, versus > saw specialists, where saw people can't be bothered with areas that > they think involve the hammer people, and God help you if you're > really concerned about nails, as the hammer people know that hammers > need nails, but they can't be bothered in their pure research with > such practical matters!!! Mathematics is a tool. JSH is a tool. We let some people specialize in pure math and these people screw us > over, in some of the most bizarre ways imaginable, and then can't be > trusted to tell the truth about their own areas. People who do only pure math are like tool people who only make > tools but don't think they should be bothered with how the tools are > used. Can you imagine what weird stuff such people would build? How the hammer people might add weird gew-gaws to hammers and proclaim > it was pure? Or how the nail people might decide that nails need > two heads and call you insane if you argued that they were stupid? These people have ignored my solution to binary quadratic Diophantine > equations, and wouldn't you know, using it, I'm finding all this weird > crap that they've also managed to miss or ignore in even their own > prized pure math areas, and you know what? More than likely I'm just talking to myself here for the good it will > do. You people don't deserve the best knowledge. You cannot understand it. You deserve patting each other on the back for bottom level results > when you could have truly magnificent and huge results, and my only > hope then is that history will despise you and the crap you call > research as much as I do. Projecting again. If humanity survives, your tremendous volume of research will be on > the crap heap, worthless. Projecting..... Transcended by real research done by people who care about knowledge, > are willing to do the work, who know how to be real skeptics who > question. Your crapola is not real, troll. You in contrast, are useless to your world. But you pretend otherwise > and God help us, we're screwed until enough people figure that out, > and toss you out. speak for yourself, tossed outted one. Then maybe our students can get taught how to do real research, > instead of playing pretend. you still playing pretend ? Your society is like Hollywood science. You people don't know the > meaning of doing real research. Just a fantasy. > James Harris (D) === Subject: Re: #364 Ring, multiplicative inverses and AP-adics, and why 2, 1/2, 5000....0000 are special; new book 2nd edition: New True Mathematics posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG 1.1.4322; .NET CLR 2.0.50727; MS-RTC LM 8; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > Now, that is reasonable considering that the AP-adics has all > numbers contained within itself as a infinite string such as > 9999....191817161514131211109876543210 Which we'll call L. In order to figure out where L fits into the rest of the AP-numbers, what can you say about the leftmost string of 9's in L? Is it the same as the AP-number 999...999? The first possibility is that L = 999...999, but we know that can't be true because the rightmost digits are different. That leaves two choices: Is L less than 999...999 (because the rightmost digits of L are 3210 and the right digits of 999...999 are 9999)? Or is L greater than 999...999 (because L contains 999...999 in its left side, so L = 999...999 times some power of 10, plus the rest of the rightmost digits of L)? === Subject: Re: Parametric Pell's Equation and circle Well, if your methods are as superior as you say they are, submit your results to a journal in some other country, or to some group, say in industry. Is there any private company willing to hire you to do mathematics?. I doubt it. Sure there must be a conspiracy in industry too, preventing you from getting a job there. Sci.math conspired with them, right?. If you want to see why your sanity is questioned, just read some of your posts. These are not posts of anyone sane. And if you don't believe me, I can look some of them up for you. So don't hide behind that. === Subject: Re: Parametric Pell's Equation and circle > Hey, I've been ripped on by people like you for > years. And you have > no shame, do you? No sense of apology or anything despite years of > insults and degrading > attacks. You are and have been just as insulting. It's either do that or just grin and bear it. Though I will admit that as a physics undergrad I > routinely heard negative things about mathematicians. not cut it in mathematics. And apparently you were not good-enough to even get your physics degree. Hmm..I guess this is also the mathematical community's fault, right?. You are a proffesional whiner, blaming all your \ problems on everybody but yourself. You are just a victim. I didn't come on these newsgroups with high hopes of > seeing rational behavior, and I also noted a weird need by many math people to question the sanity of people who disagreed with > them. Lots of diagnoses of mental illness from people in > your community. Not a good sign. > Yes, and your repeated cries about an apocalyptic future and of how you will single-handedly save humanity do attest to your sanity, right?. Your claims about how your algorithms will destroy the global economy are definitely the sign of a sane mind, right? Even you admitted to your NPD, so I cannot take you seriously. > On top of that, if you are not a social animal-- > as you condescendingly refer to others--this > should > not bother you. You are, by your own belief, > above > and superior to others. And yet you decry the > (supposed) hierarchy of our world. I don't see myself as superior. That's garbage. You have repeatedly specifically stated that you believe you are superior. However, I DO have > superior > techniques as proven by the results I have. > You have nothing. You storm into a group , a community, start spouting garbage, talking about things you know nothing about, claiming to be a superstar. And don't start whining about your two-bit paper again, I am tired of seeing 10 ing posts about your alleged proofs every time I drop by this site, and reading about your whining, about how you are a victim of everything and every one. > But those techniques, like brainstorming, are ones I > LEARNED. In my opinion, your society refuses to learn better > methods. > In my opinion you are here just for self-aggrandizement ( you have often explicitly stated this), and you mostly do not know what you are talking about. You storm into a group that gets along well, start insulting without provocation, start condescending people, and then you whine when you get it back. But you are trapped in a cycle of self-delusion. And you never even consider that you may play _some_ role on how you are trated. You storm into sci.math and start blabbering about things you know nothing about, start accusing without proof, act like you are like a big shot -- with nothing to show for. Listen: NO ONE is willing to pay you to do mathematics not in academe, not in industry. Can you seriously claim that there is a plot against you in industry You never look at the role you may be playing in how you are treated. This is because you are a proffesional victim, and because you are inherently superficial > I say you were working on a glass ceiling: > class > rules where merit > doesn't matter, as you and the others worked to > create a society where > one group is on top, and the other group--is > permanently on the > bottom. But you constantly talk about how you are better > than everyone else, and explicitly condescend > everyone else. You also talk about _your_ proof, how > _you_ I'm not better. I have better research techniques. (believe you) discovered so and so. How is this > _not_ hierarchical. People like you believe in the rulers and the > ruled. So fear is part of your lives as you try to be > the > ruler, believing > that everything depends on you stepping on some > other > person, those you wish to rule. Or worse, > believe you already rule, whether they realize it or > not. You have never been part of the mathematical > world, > and you repeatedly make unfounded statements like > those above that many, including myself, find > insulting. > Do you have any real basis for these > speculations, > I mean _real_ basis?. > I like doing my own mathematical research. In retaliation for me doing what I am free to do, > math people have insulted me, refused to acknowledge mathematical proofs, \ refused to allow that anything I do is of value, and maintained a hostile > attitude for years, and yes, I will be hostile to you > back. > Oh, poor you. I cry thinking about how you are so innocent, you never contribute to your own problems, and yet you get mistreated by the bad, ugly mathematicians. Sob... You are such a poor victim: you are _never_ hostile and aggressive, and yet you get mistreated. But it's never really your fault. You are just a victim. > As far as I'm concerned you're angry nasty people who > pretend to like something you despise: mathematics. > No. We despise assholes like you who don't know what you're talking about and try to condescends us and how to do our job. You have insulted even those that offer you help. You have trashed those who disagree with you, only to \ (at least often) admit you were wrong. But this does not elicit an apology from poor proffesional victim you. You post around SEVEN ING times a day. Do you really believe you have SEVEN brilliant ideas every day?. No, you seem to believe that every single thought that comes to you is brilliant, and trash this site with every stupidity that pops into your head. But that is not insulting to others because her majesty is posting her brilliant thoughts. > And, you insult people a lot. And do a lot of > questioning of other people's sanity. > Take a mirror and look at yourself. Read a few of your posts talking about how you will guide humanity to its future, how you will be the ruler of society in the future, etc. and tell me if those are posts of a sane person. > BUT then you come back, and ask for the world to > treat you like you > are objective and rational people who behave. You are walking contradictions in your own little > wacky worlds. And you demonstrate hatred of mathematics, not love > of it. So why should I be nicer to you? > James Harris If you don't like it, post somewhere else. I am sure no one will ever pay you to do mathematics. Neither as an academic nor in industry You can continue lying to yourself about conspiracies. Is there also a conspiracy in the private sector, blocking you from getting a research job there? === Subject: Re: Parametric Pell's Equation and circle > There never has been a *single* parameterization that handles circles, > ellipses, and hyperbolas just by sign before: How can you make a claim like that reasonably?. (snip) Really: what kind of response do you expect > to get by acting this way?. Are you completely unaware?. Have you ever considered posting in > someone else's thread to offer someone else help > on what _they're_ doing, instead of just being > self-involved 100% of the time?. You come off > as being very selfish in this respect. > the reason he does not help others is because he has nothing to offer and \ he > knows it. it's that simple. > so I agree with the word selfish you used but I will add ignorant to it. > That pretty much covers all the bases. Nasty? Delusional? Megalomaniacal? Isolated? Moribund? -- Michael Press === Subject: Re: Parametric Pell's Equation and circle There never has been a *single* parameterization that handles circles, > ellipses, and hyperbolas just by sign before: How can you make a claim like that reasonably?. (snip) Really: what kind of response do you expect > to get by acting this way?. Are you completely unaware?. Have you ever considered posting in > someone else's thread to offer someone else help > on what _they're_ doing, instead of just being > self-involved 100% of the time?. You come off > as being very selfish in this respect. > the reason he does not help others is because he has nothing to offer and he > knows it. it's that simple. > so I agree with the word selfish you used but I will add ignorant to it. > That pretty much covers all the bases. Nasty? Delusional? Megalomaniacal? Isolated? Moribund? Overweening? Hypocrite? Drama Queen? -- Michael Press === Subject: Re: Parametric Pell's Equation and circle > On Apr 4, 3:43am, David Bernier > Intriguingly, in rationals, my parametric > solution for Pell's > Equation, can be used to give the familiar > parametric equation result > for the circle: With x^2 - Dy^2 = 1 Pffft. x and y can be complex. Consider y pure > imaginary, and > let D = 1. Except I'm considering rationals. I have proven: y = 2[f_2*v - 1]/[D - (f_2*v - 1)^2] and x = [D + (f_2*v - 1)^2]/[D - (f_2*v - 1)^2] where f_2 is an integer factor of D-1, so with > D=-1, you have x^2 + > y^2 = 1, and I'll give the circle result in its > familiar form: x = (1 - t^2)/(1 + t^2) y = 2t/(1 + t^2) See:http://mathworld.wolfram.com/Circle.htmleqns. > 16 & 17 Cool! Isn't mathematical research wonderful? The parametric form for Pell's Equation was > clearly previously unknown > or a lot of textbooks are hiding a beautiful > result. Also I'll remind that I've proven that Pell's > Equation itself is more > easily solved using alternates to it, like the > negative Pell's > Equation: j^2 - Dk^2 = -1, where x = 2Dk^2 - 1 = 2j^2 + 1 or j^2 - Dk^2 = -2, where x = Dk^2 - 1 = j^2 + 1 or j^2 - Dk^2 = 2, where x = Dk^2 + 1 = j^2 - 1 and, of course, x^2 - Dy^2 = 1. One of the alternates is always true if D is a > prime, where the > negative Pell's Equation is the dominant one, and > is true if D is a > prime that is 1 mod 4 or is the product of primes > that all are 1 mod > 4, while the other alternates are true for cases > when D is -1 mod 4, > or is the product of primes where all either have > 2 as a quadratic > residue or have -2 as a quadratic residue, else > there is yet another > alternate equation which is just not as pretty, > so I'm not giving it > here. Because the first x that solves Pell's Equation > is roughly j^2, it is > mathematically naive to solve Pell's Equation > directly, rather than > use the alternates which are also solvable by the > same techniques > available for Pell's Equation, for instance, a > continued fraction > solution must exist. It is amazing that a parametric equation for > Pell's Equation which > gives the circle parametric equation was > previously unknown, but I > think it is a beautiful pure result which maybe > just didn't want to > be born into this world until the 21st century. Why not? What if your ego meets a Black Hole? Hey, I've been ripped on by people like you for > years. And you have > no shame, do you? No sense of apology or anything despite years of > insults and degrading > attacks. You are and have been just as insulting. On top of that, if you are not a social animal-- as you condescendingly refer to others--this should not bother you. You are, by your own belief, above and superior to others. And yet you decry the (supposed) hierarchy of our world. I say you were working on a glass ceiling: class > rules where merit > doesn't matter, as you and the others worked to > create a society where > one group is on top, and the other group--is > permanently on the > bottom. > But you constantly talk about how you are better than everyone else, and explicitly condescend everyone else. You also talk \ about _your_ proof, how _you_ (believe you) discovered so and so. How is this _not_ hierarchical. > People like you believe in the rulers and the ruled. So fear is part of your lives as you try to be the > ruler, believing > that everything depends on you stepping on some other > person, those you wish to rule. Or worse, believe you already rule, whether they realize it or not. > You have never been part of the mathematical world, and you repeatedly make unfounded statements like those above that many, including myself, find insulting. Do you have any real basis for these speculations, I mean _real_ basis?. Again, you keep claiming how better you are than everyone else, how you are at a superior level,so you are no egalitarian yourself. > So for you, the new fear is that instead you will be > moved to the bottom, and be stepped on. > More baseless claims. You have never met any of the people you are referring to, yet you speculate and say frankly offensive things. Face it: you can dislike what others do and disagree with it, but when you speculate about their motivation, you really don't have any basis, you have no idea of what you're talking about. But yet you blabber on. That is offensive to many. If you don't want to be insulted, don't insult others. You _dont_ know our motivations, jus like we don't know yours, your pseudo-sociological rants notwithstanding. So don't tell me why I am doing what I am doing. More simply: don't dish it out if you don't like receiving it. > Your class wars are boring. People like you are > fighting feudal > battles the rest of the world evolved beyond years > ago. You're recidivists lusting for a new royal age--with > you the new > nobility--in a world that has moved on. If humanity is forever split in a war with itself, > then it will split > into two species, but I don't think that will happen. Your vision of the future in all its rage and > darkness, is one I do > not agree with, as I think reality will give us > greater problems to face than ourselves. > You don't know what my vision of the future is. Don't tell me what I believe nor why I believe it. I find this insulting, and if you continue with it, I will insult in return. It is not just you that finds certain things offensive. There is a world outside of your own small bubble. > Human beings will not always be their own greatest > enemies. > James Harris === Subject: Re: Parametric Pell's Equation and circle >Well, Fermat was a lawyer by trade. Mathematics was an amateur >activity for fun. How is that a problem James? You also are an amateur mathematician, unless that is you have recently got a job in a Department of Mathematics at some university. rossum === Subject: Re: Parametric Pell's Equation and circle <31889182.25289.1238822200547.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=lj-kRgoAAADbp-o6laOVCIWXjFF53GYz Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) But if you really doubt me, why don't you go to the library if you > distrust Internet searches? > How would you now what's at the library, have you actually gone to one and done any research on Pell's eqn? Didn't think so. No, the real question is why don't you go to the library? Who knows, you might actually learn something. Or just do a search on: parametric equations circle James Harris and if you do, (with or without quotes), your website does not come up (at least not in the first 5 pages, and then I got bored). One reason why, which apparently you do not understand, is that a site that you have already visited and is in your cache, will come up high on a search list. For the other 6.6 billion, you like to think are viewing your sites, your site does not appear (which gives us all some hope I guess). James, you are truly a hero in your own mind. Peter === Subject: Re: Parametric Pell's Equation and circle posting-account=WlifZwoAAADn4Qc008FhhuRE4Syn8J58 3.011; GTB5; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) Oh, James, it's just astonishing how stupid you are. > Your parameterization comes down to three observations: > 1) The point (1,0) satisfies your Pell equation. > 2) A line meets a conic in exactly two points. > 3) For each v, there exists a line through (1,0) with slope 1/(-1+f2 v) > Which of these three observations did you think had gone unnoticed for the past few millennia? === Subject: Re: Parametric Pell's Equation and circle > On Apr 4, 3:43am, David Bernier > the circle result in its Cool! Isn't mathematical research wonderful? Yes, if you knew how to do it. > think it is a beautiful pure result which maybe > just didn't want to > be born into this world until the 21st century. Why not? What if your ego meets a Black Hole? Hey, I've been ripped on by people like you for > years. And you have > no shame, do you? You have done as much ripping yourself. No sense of apology or anything despite years of > insults and degrading > attacks. See above, hypocrit. I say you were working on a glass ceiling: class > rules where merit > doesn't matter, as you and the others worked to > create a society where > one group is on top, and the other group--is > permanently on the > bottom. People like you believe in the rulers and the ruled. So fear is part of your lives... That is laughable coming from someone who gives repeated predictions of an \ apocalyptic future. as you try to be the > ruler, believing > that everything depends on you stepping on some other > person, those > you wish to rule. Or worse, believe you already > rule, whether they > realize it or not. So for you, the new fear is that instead you will be > moved to the > bottom, and be stepped on. This is why you receive help here, but never offer any to any one. You are a parasite, you want to take from others but never give. Because you think you're at the top and you're better. Hypocrit. Your class wars are boring. People like you are > fighting feudal > battles the rest of the world evolved beyond years > ago. You're recidivists lusting for a new royal age--with > you the new nobility--in a world that has moved on. > This is why you repeatedly call yourself the brightest and the best mathematician, comparing yourself to NEwton, Gauss, etc. and tell us (laughs0 how much better you are than everyone, right?. Hypocrit. > If humanity is forever split in a war with itself, > then it will split into two species, but I don't think that will happen. As long as your not in my side, I don't care. Your vision of the future in all its rage and > darkness, is one I do > not agree with, as I think reality will give us > greater problems to > face than ourselves. Yes, this is why you give your apocalyptic visions of the future, right?. Human beings will not always be their own greatest > enemies. No. We will always have psychopathic parasites like you, Madoff, Limbaugh, etc. to take that job. > James Harris