mm-1026 === Subject: Re: JSH: Weird, eh? > Curse him and his ilk!!! LOL. Im in some mood today. Youre obviously uncontrollably driven by your moods and character ßaws. Too bad youre not driven by love of the truth. By the way, hows *your* ilk. > James A legend in his own mind. Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Reply, reply, reply >Come on, reply to my posts!!! Come why dont you say something else, >huh? >Try to point out some mistake Nora Baron you stupid shit! >Hey, you lapdog David Ullrich, yeah, I called you a LAPDOG Ullrich!!! >Arent you mad now? Hey, you know, you HAVE to reply right, or Ill >call your school and try to get you fired, right? >Ullrich the university professor cursing in the muck, with no shame. >REPLY TO MY POSTS YOU CUR! >Hey Dik Winter, you fucking shit, reply again! Why dont you throw >more of my old arguments on a webpage, you stupid fuck!!! >Yeah, youll all be back want you? Along with C. Bond and all the >rest. >lapdogs >You are curs, my , my little traveling circus of obsessive >repliers. >I spit on you. >But youll be back, now wont you? Totally bizarre. When people reply to your posts your curse them and tell them not to. When you make a post or two and people dont reply immediately, possibly because theyre at work or something, you complain about that. The strangest thing Ive ever seen. And considering that Ive been reading _your_ posts for years thats saying something. >James Harris ************************ David C. Ullrich === Subject: Re: . The hardest of all hard facts . it should be Michelson-Morley-Miller, with Michelson being the pioneer of interferomteric methods for this. D.C.Millers refinement highlighted the anomolies that MM had found, but this seems to have been dysmissed, because of a comment that Eintstein made, on one of his brief forays to his office at CalTech. there is no vacuum, per se, anyway, unless you rely on the formalism of atoms as zero-dimensional points (and renormalization etc.). > The speed of light through vacuum is the same for all observers. > However, when the speed of light is less than c (eg when passing > through air), it would not be the same for all observers. > The explains how the Michelson-Morley experiment gave results that > were non-null, due to the effect of air in the apparatus. > The MMX has been repeated in a vacuum. No change in the result. anyway, the Protocol of the Elders of Kyoto is on, as of Dec.12 (LA(TRIBco)Times in November -- on the Chicago BoT, I think), so that the oilcos free market has a new pricing tool; have you heard any thing about this?... the dystribution is mainly tied-up with Dutch and British interests, where the Common Market is more radical than California on these enviro matters -- except for the bull**** republican policy of not getting oil & gas off of our own coasts! thus quoth: A COORDINATED WALL ST. ASSAULT Who Killed U.S. Nuclear Power? by Marsha Freeman --zero-dimensional holograms. utterly compressed information. DemConK2 cloning malfunction: (Citizen John-mick) Kane/Bore/Gush/Nadir; LaRouche et al versus Fowler et al, Ô96 -- Supreme Decision, March 27, 2000. but, hey; the Voting Rights Act is up for republication in Ô07! --Give the World a Trickier Dick Cheeny -- out of office after GIGA years. http://www.benfranklinbooks.com/ http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: Re: How many ways to put 5 balls into 500 ordered cups? > The short answer, nCr(504,5), or in general for n balls and k cups, > nCr(k+n-1n). Sorry, small typo, should ne nCr(k+n-1,n) Proof is coming... - Tim -- Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University === Subject: Re: How many ways to put 5 balls into 500 ordered cups? >--CELKO-- escribi.97: >> You are given 500 numbered cups and five identical balls. Any cup can >> hold up to five balls. How many ways can you put the five balls into >> the 500 cups? >You need to chose five cups among 500, with possibly repetitions. You need >to count the number of ways to put together 499 identical separators and 5 >identical balls. >The answer has a dozen of decimal digits ... Maybe its 520906250200. Whether thats right or not, and how we came up with that number, is left as an exercise for the reader. === Subject: Re: JSH: Not a solution >Obsessively replying to my posts to insult me or just aggravate me is >not a solution. Are we typing while drunk again or what? Less than a half hour ago you made a post just _begging_ people to reply to your posts - it was really sort of pathetic, the sort of thing that would make one feel sorry for you if you werent such a fucking asshole. And now youre back to complaining about the replies. >Sure, it might make you feel better or like youre accomplishing >something if you can piss me off, but thats just a bad reßection on >you. >Notice, here I am again: posting. >Those of you who make it your business to obsessively reply to my >posts are not doing other people a favor. I post for many reasons, >including just to toss out math ideas that Im considering, and >theres no reason to think that Ill stop. Nobodys stupid enough to think youre going to stop. >Some of you seem to be stupid in a way that defies commonsense as now >Im one of the most read posters on sci.math and it seems to me that >you MUST on some level understand that the obessive posters have >something to do with it. >Yet you continue to support them, and they continue to go at it. >Now Im in a mood so I feel like berating some of you, not because I >think youll go away, but because I feel confident you will not. >Its kind of, sickly, *relaxing* to curse at Nora Baron or David >Ullrich. Sickly indeed. You may be on to something here. >Im starting to worry that Im finding it appealing to have my own >dumpground, where when I get frustrated I can just come out and curse >at them, knowing theyll just be back, for more. >Its not a solution though, just another display of rather absurd and >sick human behavior played out to a worldwide audience. >But hey, were mostly Americans here! Surprise. Surprise. Good point. If you were actually alternately making a fool of yourself mathematically and exhibiting kind of sickly behavior several times a day in front of the entire world that would be something youd want to worry about. But as long as its only the US, what the heck. You should really try to avoid being such a fucking asshole - if you werent so obnoxious then people all around the world wouldnt find it so amusing when you ßip out like youre doing today. >James Harris ************************ David C. Ullrich === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs > Define an actual experiment where you randomly chose a real > between 0 and 1, and Ill show you where you took a shortcut. >> This is sci.math. If you want to talk about actual experiments, >> take it to sci.physics. >> Playing the physics card isnt helpful here. Im claiming that its >> *mathematically* impossible to randomly choose a real from [0,1]. Not >> physically impossible. Im asking you to convince me otherwise, use a >> thought experiment if you want, or not. >> And the fact that you cant specify which real you chose is not the only >> reason you cant choose one. If you find some way of specifying which >> one you chose, Ill find a way to show you where you failed to do what >> you set out to do. > While I originally agreed with Patrick on this point, after giving >it a little thought Im no longer sure. Patrick, how about this >random number? > R = Sum from i=1 to +inf of (2^-i)*(r_i) > where each r_i = a random integer in [0,1] inclusive. ie, generate a random number by ßipping a coin to determine whether its bigger than 0.5, the again to determine whether its bigger than 0.25 (or 0.75), repeat forever. Theres the kicker - forever. When forever comes around, Ill let you know how you went wrong. Or you can say I told you so. Either way, its not going to happen. You never chose your number, youre still choosing it. The existence of a construct like fom i=1 to +inf which allows you to theoretically specify a random real doesnt in my opinion entitle you to claim youve actually found a way to make a probability-zero event happen. >We can certainly pick either 0 or 1 randomly. And this real number >is pretty well specified: I can give you its approximation to any >precision you like; agreed > I can add and subtract it from things in a well- >defined manner; etc. Not so sure about this one, I could be persuaded maybe. If you accept that constructs like infinite sums involving random numbers are well-defined! > So I guess it is possible to choose a random >real number in [0,1). In a sense... but not a convincing enough one yet that Ill accept as an example of zero-probability event. > I havent studied formal probability yet, so perhaps Im talking >nonsense here, but I *would* argue that the probability of choosing >R=x, for any real number x, is not zero but rather infinitesimal. I would argue that R = Sum from i=1 to n of (2^-i)*(r_i) approaches zero as n approaches infinity. But talking about what happens *at* infinity is tricky. I dont think its Ôinfinitesimal, I think its zero. But say you actually got there and Ill ask to see the picture you surely brought back! > Saying that the probability of an event that *has occurred* was >zero seems to me like foolishness, no matter how big a sample space >were talking about. So... youre actually agreeing with me :> Once again, you cant hide behind the Ôreality card. You can say Ill >> pick an integer randomly from a uniform distribution across the whole >> set but thats garbage mathematically speaking, and Im claiming that >> choosing a real from [0,1] is roughly the same ßavour of garbage. > I claim that my example above shows its not. But I think in any case >its true that you cant map [0,1):R onto [0,+inf):N in such a way as to >make the two *equivalent* ßavors of garbage. ;-)) I knew Id get in trouble for the Ôroughly the same ßavour thing. -- Patrick Hamlyn posting from Perth, Western Australia Windsurfing capital of the Southern Hemisphere Moderator: polyforms group (polyforms-subscribe@egroups.com) === Subject: Re: How many ways to put 5 balls into 500 ordered cups? >> You are given 500 numbered cups and five identical balls. Any cup >> can hold up to five balls. How many ways can you put the five balls >> into the 500 cups? > The short answer, nCr(504,5), or in general for n balls and k cups, > nCr(k+n-1n). Check a few of these...using Excel for lack of a better > tool Suppose that the cups are labelled C_1, C_2, ..., C_k. A distribution of balls into cups can be summarized by listing for each ball the cell into which it goes. Then, a distribution corresponds to a collection of n cups with repetition allowed, corresponding to nCr(n+k-1,n). Why does that work? Stars and bars proof: Draw 5 stars, * * * * *. Now you have to partition these stars into 500 cups. However, if you have two cups, you only need one bar | because placing that bar divides it into two sections, one before the bar and one after. If you have three cups, you only need two bars, | |, because placing these bars between the stars divides it into 3 partitions. Now imagine you have n stars and k cups. This corresponds to n stars and k-1 bars. Now you have to put these stars and bars in order. You have (n+k-1) positions to fill. Put the bars out first, then pick a spot for the n stars left. That is, you have (n+k-1) objects (the spaces) and pick n of them (where you put the stars). Hope that helps, if not, let me know and I can try to be more clear. some. - Tim -- Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University === Subject: How to diagonalize a Hermitian matrix Hello group, Could somebody please tell me a numerical algorithm to diagonalize a Hermitian matrix H, so that I end up not only with the eigenvalues perched along the diagonal, but also with the unitary matrix U that conjugates with H to diagonalize it. I should probably say what I really want. I have a positive-definite Hermitian matrix H, and I need its matrix square root. The above is simply the approach that occurred to me for obtaining that square-root. Achava === Subject: Re: How many ways to put 5 balls into 500 ordered cups? > You are given 500 numbered cups and five identical balls. Any cup can > hold up to five balls. How many ways can you put the five balls into > the 500 cups? Oh, and the answer, it is 265661562600, by using the formula nCr(504,5) - Tim -- Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University === Subject: Re: How many ways to put 5 balls into 500 ordered cups? >>--CELKO-- escribi.97: > You are given 500 numbered cups and five identical balls. Any cup can > hold up to five balls. How many ways can you put the five balls into > the 500 cups? >>You need to chose five cups among 500, with possibly repetitions. You need >>to count the number of ways to put together 499 identical separators and 5 >>identical balls. >>The answer has a dozen of decimal digits ... >Maybe its 520906250200. Whether thats right or not, and how we came >up with that number, is left as an exercise for the reader. Or maybe its 265661313100 ... or maybe its 265661562600 ... One of these is right ... why dont you figure it out and tell us? === Subject: Re: How many ways to put 5 balls into 500 ordered cups? >> You are given 500 numbered cups and five identical balls. Any cup can >> hold up to five balls. How many ways can you put the five balls into >> the 500 cups? >Oh, and the answer, it is 265661562600, by using the formula nCr(504,5) > - Tim No fair doing peoples homework for them :-) === Subject: Re: No Set Contains Every Computable Natural (was Church-Turing compared to Zuse-Fredkin thesis) > The set of all natural numbers is not a recursive set. Yes it is. The characteristic function of the set of all natural numbers is the function (lambda x )[1]. (I am using Churchs excellent lambda notation for functions.) All constant functions are recursive. For a trivial proof see: _Computability and Unsolvability_ by Martin Davis or _Theory of Recursive Functions and Effective Computability_ by Rogers. In fact, in the latter source, see page 6 where the author defines the class of primitive recursive functions. This is quite a standard definition (although equivalent alternatives exist in the literature). You will see that (lambda x)[1] is primitive recursive by definition. It is well-known that the primitive recursive functions are recursive (and indeed this is often part of the definition of recursive function). Therefore, the set of natural numbers is not only recursive, it is primitive recursive. This is all quite trivial and should be mastered on the first day of studying recursion theory. -Leonard Blackburn === Subject: Re: 3 Squares Covering 1 Circle you mean, such that the tips of the three tetragona [Skwares] touch at the center, and two of their edges are tangent (at their opposite vertices) ?? sounds like it might be the best, if its not Jims stack oblocks one, because it doesnt seem that letting them overlap would improve it. wait. I can see taht Jims is better, because you can rotate the three tetragona to an L shape, and slide one of them to ... maybe. anyway, try it for a sphere of 4pi/3 unit-hexahedra [Qyoobs]. > Heres a similar situation that does better than the 80.4% > with part (a). Use this 120 deg symmetry, but place the diagonals > of the square along the radii. The optimum placement of the > squares (they can be shifted along the radii) covers 83.0% of > the circles area. This is the best that Ive found, but I dont > know if its the best possible. anyway, the Protocol of the Elders of Kyoto is on, as of Dec.12 (LA(TRIBco)Times in November -- on the Chicago BoT, I think), so that the oilcos free market has a new pricing tool; have you heard any thing about this?... the dystribution is mainly tied-up with Dutch and British interests, where the Common Market is more radical than California on these enviro matters -- except for the bull**** republican policy of not getting oil & gas off of our own coasts! thus quoth: A COORDINATED WALL ST. ASSAULT Who Killed U.S. Nuclear Power? by Marsha Freeman --zero-dimensional holograms. utterly compressed information. DemConK2 cloning malfunction: (Citizen John-mick) Kane/Bore/Gush/Nadir; LaRouche et al versus Fowler et al, Ô96 -- Supreme Decision, March 27, 2000. but, hey; the Voting Rights Act is up for republication in Ô07! --Give the World a Trickier Dick Cheeny -- out of office after GIGA years. http://www.benfranklinbooks.com/ http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: Re: Mathematical induction look for the 2-page proof that mathematical induction (infinite descent, *reductio ad absurdum*) is isomophic with mathematical deduction, in Mathematics Magazine. > I have so far only found pages with 1 or 2 exercises or some with execises > but no solutions... anyway, the Protocol of the Elders of Kyoto is on, as of Dec.12 (LA(TRIBco)Times in November -- on the Chicago BoT, I think), so that the oilcos free market has a new pricing tool; have you heard any thing about this?... the dystribution is mainly tied-up with Dutch and British interests, where the Common Market is more radical than California on these enviro matters -- except for the bull**** republican policy of not getting oil & gas off of our own coasts! thus quoth: A COORDINATED WALL ST. ASSAULT Who Killed U.S. Nuclear Power? by Marsha Freeman --zero-dimensional holograms. utterly compressed information. DemConK2 cloning malfunction: (Citizen John-mick) Kane/Bore/Gush/Nadir; LaRouche et al versus Fowler et al, Ô96 -- Supreme Decision, March 27, 2000. but, hey; the Voting Rights Act is up for republication in Ô07! --Give the World a Trickier Dick Cheeny -- out of office after GIGA years. http://www.benfranklinbooks.com/ http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: Re: JSH: Reply, reply, reply > Come on, reply to my posts!!! Come why dont you say something else, > huh? Hello James my little tickle-puppy. Cootchie cootchie coo! > Try to point out some mistake Nora Baron you stupid shit! Wow, if she replies, then shes slavishly doing your bidding. If she doesnt, then your fiendishly subtle reverse psychology triumphs again. What unfalsifiable cleverness! > Hey, you lapdog David Ullrich, yeah, I called you a LAPDOG Ullrich!!! > Arent you mad now? Hey, you know, you HAVE to reply right, or Ill > call your school and try to get you fired, right? > Ullrich the university professor cursing in the muck, with no shame. > REPLY TO MY POSTS YOU CUR! Amazing! Hell probably reply to you again at some point, and then youll be the puppet master pulling his strings. Or maybe hell never reply again, in which case, oh how deftly done James! > Hey Dik Winter, you fucking shit, reply again! Why dont you throw > more of my old arguments on a webpage, you stupid fuck!!! Whether he does or doesnt, it will be because thats what youve manipulated him into doing. I fear your awesome, unfalsifiable powers. > Yeah, youll all be back want you? Along with C. Bond and all the > rest. I am back right now. Whoa! Just as James predicted! In fact, is James controlling my hands right now as I type? HeLlO! i Am ThE sPiRiT oF jAmEs HaRrIs! MaKe Me A wEbSiTe LiKe ThIs OnE: http://www.williamhung.net/ In less than one minute, William Hung became a star by sucking at something shamelessly and spectacularly. James Harris has sucked at something for eight years, but does he have a website where female fans can write to say how cute they think he is in all of his pathetic glory? Come on, Dik Winter, Erik Max Francis, and Rick Decker: make James a star! > You are curs, my lapdogs, my little traveling circus of obsessive > repliers. Id never considered how powerful it must make a pile of dog feces feel that various dogs come by to sniff at it. > I spit on you. Well, you waft foul aromas at us, but close enough. > But youll be back, now wont you? Its hard to understand what we dogs find so interesting about that smell, but yeah. Well be back. -Jim Ferry === Subject: Re: Got a speeding ticket and need to fight back > The Lord of Chaos (Suresh Devanathan) > The last i checked, I dictate the laws of physics. I am God, to many > people. Your court is beneath me. Your constitution is beneath me. In fact, > you cannot even prosecute me, because the constitution depends on me. I tell > what the laws are. And thats final. > Not another case of Narcissistic Personality Disorder... > btw, does your boss keep your arround because he can use your face to wipe his ass? > -suresh No. Christian is a very good traffic court judge. Small world, eh? ;-) -- Paul Hovnanian mailto:Paul@Hovnanian.com ------------------------------------------------------------- ----- Live Faust, die Jung. === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >> Define an actual experiment where you randomly chose a real between 0 and 1, and >> Ill show you where you took a shortcut. >This is sci.math. If you want to talk about actual experiments, take it >to sci.physics. >> Playing the physics card isnt helpful here. Im claiming that its >> *mathematically* impossible to randomly choose a real from [0,1]. Not physically >> impossible. Im asking you to convince me otherwise, use a thought experiment if >> you want, or not. >The probability space is simply ([0,1],L,m), where L is the set of all >Lebesgue-measurable subsets of [0,1], and m is Lebesgue measure on [0,1]. >See . >For each x in [0,1], the probability of the event X element {x} is the >Lebesgue measure of the set {x}, which is zero. >Done. I trawled up and down those definitions, nothing seemed to jump out as supporting the Ôzero-probability-event actually happened scenario. Not that I pretend to understand very much of those pages. >> And the fact that you cant specify which real you chose is not the only reason >> you cant choose one. If you find some way of specifying which one you chose, >> Ill find a way to show you where you failed to do what you set out to do. >And I suppose you also object to things like geometric constructions, on >the grounds that they cant be carried out exactly in the real world. >Yawn. You can specify your real however you want. Geometric constructions translate just fine into the real world. But what did you have in mind? Choose an angle randomly from 0 to 90 degrees, and then construct a line from the 90 degree vertex of a triangle to the opposite side (which has length 1), at that angle from one of the other sides? Youre just translating the domain, choosing the angle is effectively the same as choosing the real directly. >> Once again, you cant hide behind the Ôreality card. You can say Ill pick an >> integer randomly from a uniform distribution across the whole set but thats >> garbage mathematically speaking, and Im claiming that choosing a real from >> [0,1] is roughly the same ßavour of garbage. >No, there is an enormous difference between the two. There is no such >thing as a uniform probability measure on a countably infinite set, but >uncountable sets are quite a different matter, as the Lebesgue measure >example demonstrates. >The example I gave is correct. >> Its so Ôcos I said its so doesnt wash with me. >Its not my fault if you dont know the definition of a probability >space. No. But its disappointing (to me at least) that my error cant be explained to me in terms I can understand. It seems such a simple concept, and we have to resort to opaque definitions like Probability Spaces defined in terms of a Lebesgue Measure of Lebesgue Measurable subsets. <...Look at and read the >third paragraph down, where the concept of a probability function is >mentioned. Notice that the function f: [0,1] -> R given by f(x) = 1 is >an example of a probability function or probability density function >as described there. The fact that the integral of f over [0,1] = 1 is >what makes the function properly normalized to define a probability >measure, which turns out to be ordinary Lebesgue measure on [0,1]. >Look at the probability axioms that are stated farther down on that page. >A probability measure is something that satisfies those axioms. Notice >that the description of probability says nothing about what you can or >cannot do in the real world; its an abstract mathematical concept. So... youre saying that if we accept some set of definitions and constructs which dont relate to the real world, we can then say Ôa zero-probability-event can happen in terms of those definitions. Well maybe you can, but then youre not saying what I understand by the phrase. -- Patrick Hamlyn posting from Perth, Western Australia Windsurfing capital of the Southern Hemisphere Moderator: polyforms group (polyforms-subscribe@egroups.com) === Subject: Re: No Set Contains Every Computable Natural <3sRWb.24383$cM.3142@newssvr25.news.prodigy.com> be a Turing machine. > Requiring the input tape to be finite is a severe limitation. > For example, no language, L, could contain every binary > representation of rational numbers between 0 and 1. > 1/3 has an infinite binary representation in base 2. Languages are usually defined as finite concatenations of a set of alphabet symbols. For example, the set of natural numbers expressed in unary notation is a perfectly fine language... a mistake you made in an earlier post was to assume that a word infinitely long is in that language; instead, this is an infinite language containing all 1-strings of finite length. For your example above, there are better ways of representing every rational between 0 and 1. > Every definition of TMs that I have seen assumes > that an infinite tape of blanks is allowed. But the input (non-blank symbols) is always bounded in length by some finite amount. M.C: > No natural number requires an infinite unary representation. All may be > represented using a finite number of 1s. R.E: > There is no TM that can decide if the string has an > infinite string of 1s. Russell, I dont see how your response is relevant to what Michael said. His point is that the language {1^n | n is a natural number} is a set of finite-lengthed strings. It is not hard to show that there is a TM that can decide this language. And, yes, if you include an infinitely-long string of 1s, then no TM can decide it (in a finite amount of time, which is a crucial requirement.) This is why infinitely-long strings are not usually considered in languages. M.C: > All Turing machines can be finitely represented. There is no such > thing as a TM transition function that requires an infinite state > automata for representation. R.E: > There is no TM that can decide if the transition table is > infinitely long. Again, your response is kind of irrelevant. His point is that if you have a set representing the transition function, it MUST be finite sized if it is going to represent a TM. > The Halting Problem is ill posed. > There is no set that contains every TM > Sure there is : A = { | M is a valid TM representation} > It is impossible to decide if M is a valid representation. Actually, it is quite simple to decide if M is a valid representation. > It is easy to check, given a data structure, whether the input is a > valid TM or not. > It is impossible for a TM to decide if M is a valid representation. Again, it is quite simple. Look up a definite of a Turing machine... it is some tuple of sets, and it is straightforward to check that, for example, the claimed start state is an element of the state set. And that the transitions form a subset of the cross product of the state set with itself... and so on. > All natural numbers are computable. However, they _all_ cant be > generated in a single run of any algorithm/TM. > The output of a TM cant contain a representation of every natural? That is correct. Since a TM must halt to complete its output, it can only have a finite number of natural number representations. Dont feel that TMs are Ôweak because they cant output ALL the naturals. What TM can do, however, is output all the natural numbers from 1 to k for arbitrarily large k. > I didnt say that a TM cant print out certain natural numbers. > I said that a TM that requires a fixed, finite amount of time to > perform an operation cant print certain natural numbers in a > reasonable amount of time. Taking about withing a reasonable amount of time has very little to do with decidable sets or recursively enumerable ones. These sets do not say anything about the time required to accomplish any computation; they are merely telling you what can and can not be computed. You probably want to investigate the notion of time-bounded (or even space-bounded) turing machines. J === Subject: algorithms for skeleton tracing and loop extraction Do you know of any standard algorithms for skeleton tracing and loop extraction? Here are the details: What I have: I have a binary image that contains a skeleton of a contour. This skeleton can be thought of as a collection of links connecting feature points of three types: end points, T- and X-intersections. This skeleton may contain loops of various degrees of complexity, e.g. a simple loop where a link starts and ends at the same FP or a complex loop where several FPs are connected into a single loop. What I need: I need to trace this contour, i.e. to turn a 2d image into a sequence of skeleton pixels going from one FP to another in a certain order. I also need to detect loops. This implies that when traced, all links belonging to the same loop should be listed successively without any gaps. Any help will be greatly appreciated! Ilya === Subject: Re: the anticlassicalist }{ ii: the spectre continues > There is, after all, a set of first-order axioms defining > lambda-calculus. Do you know of any online presentations? :-) mitch === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs > ie, generate a random number by ßipping a coin to determine whether its > bigger than 0.5, the again to determine whether its bigger than 0.25 (or > 0.75), repeat forever. > Theres the kicker - forever. When forever comes around, Ill let you know how > you went wrong. Or you can say I told you so. Either way, its not going to > happen. You never chose your number, youre still choosing it. That argument leads to a denial of the existence of most real numbers. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: Reply, reply, reply > Come on, reply to my posts!!! Come why dont you say something else, > huh? First of all, you might not have a life, but we do. We all have jobs, you apparently dont > Try to point out some mistake Nora Baron you stupid shit! Real mature, James. Keep it up. > Hey, you lapdog David Ullrich, yeah, I called you a LAPDOG Ullrich!!! > Arent you mad now? Hey, you know, you HAVE to reply right, or Ill > call your school and try to get you fired, right? Why would you try to get someone fired for not replying? You know people do have lives, too bad you dont. > Ullrich the university professor cursing in the muck, with no shame. > REPLY TO MY POSTS YOU CUR! > Hey Dik Winter, you fucking shit, reply again! Why dont you throw > more of my old arguments on a webpage, you stupid fuck!!! > Yeah, youll all be back want you? Along with C. Bond and all the > rest. > You are curs, my lapdogs, my little traveling circus of obsessive > repliers. > I spit on you. > But youll be back, now wont you? > James Harris David Moran === Subject: Re: 3 Squares Covering 1 Circle > squares arranged like: ____ > ! ! > ! ! > --------- > ! ! ! > ! ! ! > --------- However the first picture above (three nonoverlapping > squares, arranged as shown) cant be optimal when overlap > is allowed, because sliding the lower two squares upwards > increases the covered area until the top and bottom edges > of the squares have equal length segments of the boundary. I am very happy to have found a new application for my beloved sliding ladder which gives rise to the function slad(x) = sqrt(x*(2-x)). Think of a ladder of length 1 leaning at the wall of a room with height 1. Let x be the distance between the top of the ladder and the ceiling. And let y be the distance of the foot of the ladder from the wall. Then y = slad(x). 
: +-------------- ceiling
: |x
: o y^2 + (1-x)^2 = 1
: height |
: of wall| ---> y = slad(x)
: is 1 | ladder length 1
: | i.e.
: |
: | y = sqrt(x*(2-x))
: | y
: +-------o-------- ßoor
There is a single point in the configuration above where all three squares meet each other. Because of the T-shape here we may call the point just T. Placing the circle with its center in T covers a lot of its area. But sliding it down a little increases the covered area, since at first we loose only very little at the bottom of the two squares but gain a lot at the top. The maximum is reached as soon as gain and loss are in equilibrium. And that situation corresponds to a double-sliding-ladder configu- ration! 
:
: |
: | The lower right square
: 1/2 | z shown in more detail.
: +---------+--------o+
: |x | Midpoint of the circle
: o | is x below the top.
: | | The circle is cutting
: | | the square at distances
: | | y and z from the corners.
: | |
: | | we have x = slad(z)
: | | and y = slad(x)
: | y |
: +-------o-----------+
The drawing shows that you loose covering at the bottom proportional to y while you gain covering at the top proportional to (1/2 - z). Equilibrium is reached for z + y = 1/2. Solving for z + slad(slad(z)) = 1/2 thus gives the optimal position x of the circles midpoint! I will try to find a nice closed solution involving sqrt(55) as Jim Ferry found out. But for now Ill just show how fine this works and how the ladder is sliding (tiny z, small x, large y): 
: z y=slad(slad(z)) z+y x
: --------------------------------------------------
: 0.000 0.000 0.000 0.000
: 0.001 0.296 0.297 0.045
: 0.002 0.350 0.352 0.063
: 0.003 0.386 0.389 0.077
: 0.004 0.413 0.417 0.089
: 0.005 0.436 0.441 0.100
: 0.006 0.455 0.461 0.109
: 0.007 0.471 0.478 0.118
: 0.008 0.486 0.494 0.126
: 0.00838 0.49162 0.500 0.12919 !!!
: 0.009 0.500 0.509 0.134
: 0.010 0.512 0.522 0.141
:
We see the solution 0.12919 for the best placement of the circle so far. Rainer Rosenthal r.rosenthal@web.de === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >>Playing the physics card isnt helpful here. Im claiming that its >>*mathematically* impossible to randomly choose a real from [0,1]. Not physically >>impossible. Im asking you to convince me otherwise, use a thought experiment if >>you want, or not. >The canonical example is a spinner which can take on a random angle. >This gets you [0,2pi) I think we had this one last time we explored this concept. At some point the quantum state of the universe takes over. There are a countable number of angles your spinner can indicate. Not surprising that they all have non-zero probability, and that one actually happened. -- Patrick Hamlyn posting from Perth, Western Australia Windsurfing capital of the Southern Hemisphere Moderator: polyforms group (polyforms-subscribe@egroups.com) === Subject: Re: PLEASE HELP!!! > FIND EXTREME FUNCTIONS f AND DRAW ITS GRAPH Extreme functions result from things such as skysurfing, mountain rollerblading, and skateboarding. Try some Mountain Dew. Doug === Subject: Mean Value Theorem Two places Ive seen the equivalent statement: If f is continuously differentiable on [0,1], then |f(x)-f(y)|<=sup|f(u)||x-y| where the sup is extended over 0<=u<=1. I cant see why we need f to be continuous. It seems by the MVT we have f(x)-f(y)=f(c)(x-y) for some c in [0,1], so that |f(x)-f(y)| = |f(c)||x-y|, for some c in [0,1], <= sup_u|f(u)||x-y|. === Subject: Re: help with solutions for three questions from the past contest > Id greatly appreciate your help on the solutions for these questions. > 1. Al and Bob are at opposite ends of a diameter of a silo in the > shape of a tall right circular cylinder with radius 150 ft. al is due > west of Bob. Al begins walking along the edge of the silo at 6 ft. per > second at the same moment that Bob begins to walk due east at the same > speed. The value closest to the time in seconds when Al first can see > Bob is what? answer: 48 > 2. if a, b, c, and d are nonzero numbers such that c and d are > solutions of x^2+ax+b=0 and a and b are solutions of x^2+cx+d, find > a+b+c+d. ans: -2 > 3. A boat with an ill passenger is 7.5 mi north of a straight > coastline which runs east and west. A hospital on the coast is 60 > miles from the point on shore south of the boat. If the boat starts > toward shore at 15 mph at the same time an ambulance leaves the > hospital at 60 mph and meets the ambulance, what is the total distance > (to the nearest 0.5 mile) traveled by the boatand the ambulance? ans: > 62.5 > Choogu All right, heres question 2. Remember that the roots of x^2 + ux + v add up to -u and multiply to v. Hence a+b = -c and c+d = -a, so -b = (a+c) = -d or b = d. Also ab = d so a = 1, and cd = b and c = 1 (since b =d and they are not 0). Now b = d = -(a +c) = -2, so a+b+c+d = 1-2+1-2 = -2 as desired. Maybe later for #1, maybe not. Achava === Subject: Re: No Set Contains Every Computable Natural > The set of all natural numbers is not a recursive set. > Yes it is. The characteristic function of the set of all natural numbers > is the function (lambda x )[1]. (I am using Churchs excellent lambda > notation for functions.) All constant functions are recursive. For > a trivial proof see: _Computability and Unsolvability_ by Martin Davis > or _Theory of Recursive Functions and Effective Computability_ by Rogers. > In fact, in the latter source, see page 6 where the author defines the > class of primitive recursive functions. This is quite a standard definition > (although equivalent alternatives exist in the literature). You will > see that (lambda x)[1] is primitive recursive by definition. It is well-known > that the primitive recursive functions are recursive (and indeed this > is often part of the definition of recursive function). Therefore, the > set of natural numbers is not only recursive, it is primitive recursive. > This is all quite trivial and should be mastered on the first day of studying > recursion theory. I havent read Computability and Unsolvability or Theory of Recursive Functions and Effective Computability I did find a reference to Theory of Recursive Functions and Effective Computability at Mathworld: http://mathworld.wolfram.com/GeneralRecursiveFunction.html There are two camps of thought on the meaning of general recursive function. One camp considers general recursive functions to be equivalent to the usual recursive functions. For members of this camp, the word general emphasizes that the class of functions includes all of the specific subclasses, such as the primitive recursive functions (Rogers 1987, p. 27). The other camp considers general recursive functions to be equivalent to total recursive functions. This is Mathwrlds definition of total recursive functions: http://mathworld.wolfram.com/RecursiveFunction.html A recursive function is a function generated by (1) addition, (2) multiplication, (3) selection of an element from a list, and (4) determination of the truth or falsity of the inequality a < b according to the technical rules: 1. If F and the sequence of functions G1,G2,G3, ..., Gn are recursive, then so is F(G1,G2,...Gn). 2. If F is a recursive function such that there is an x for each a with , then the smallest x can be obtained recursively. A Turing machine is capable of computing recursive functions. My approach assumes computable means TM computable. My argument falls into the other camp. I am using the definition of decidable language given at wikipedia: A decidable or recursive language is a formal language that is a recursive set, i.e., for which there exists an algorithm to solve the following decision problem: Given string w, does w belong to the language? The algorithm is not allowed to run into an infinite loop and has to produce a YES/NO answer for any input string after a finite amount of time. To formalize the rather vague term algorithm, one usually employs Turing machines, but several other equivalent approaches are possible. All regular, context-free and context-sensitive languages are recursive, but there exist recursively enumerable languages that are not recursive; one example is given by the halting problem. I have shown why a Turing machine can not decide if a string, w, represents a natural number. wikipedia says several other equivalent approaches are possible. Rogers definition can not be one of these equivalent approaches. The set of all natural numbers is not a recursive set if our definition of recursive set requires there exist a TM that can decide membership in the set. Mathworlds definition indicates there are differences between definitions of general recursive function. This appears to be one of those differences. Russell - 2 many 2 count === Subject: Re: Mean Value Theorem >Two places Ive seen the equivalent statement: If f is continuously >differentiable on [0,1], then |f(x)-f(y)|<=sup|f(u)||x-y| where the >sup is extended over 0<=u<=1. >I cant see why we need f to be continuous. It seems by the MVT we >have > f(x)-f(y)=f(c)(x-y) for some c in [0,1], Right. >so that > |f(x)-f(y)| = |f(c)||x-y|, for some c in [0,1], > <= sup_u|f(u)||x-y|. But without continuity (or something extra beyond differentiability) how can you conclude that the derivative f is bounded on [0, 1]? You have to have boundedness in order to justify your last step... G. Rodrigues === Subject: Re: e is transcendental (was: classes of transcendental numbers ? >>So? The real part of exp(i pi) is cos(pi), and its imaginary part >>is sin(pi), so all you are saying is that cos(pi) = -1 and >>sin(pi) = 0, and we were already aware of these facts. There is >>NO reason to conclude that exp(i pi) = 0. > I,have a reason , with my due respect. > Panagiotis Stefanides >>Yes, but you DONT tell us what your reason is. You cant expect us to >>accept your claims without giving support for those claims. So what >>possible reason could you have for expecting us to agree with your claim >>that exp(i pi) = 0? >The reason is simply that exp(ipi0=-1 should be accompanied >by the statement that this is the real part solution. >Is it fair? >>No, that is not a fair comment. exp(i pi), the complex number, is >>equal to -1, the complex number. There is no need to appeal to the >>real part. Also, what does exp(i pi) = -1 is the real part solution >>mean? You look like you are using terminology in a manner not >>recognized in mathematics. >>David McAnally >>-------------- >e^[i*pi] ,as accepted ,is a phasor. >>In most of the relevant fields of mathematics, e^[i pi] is a complex >>number. >It is only fair to state that its >polar representation is : >e^[i*pi] = MOD 1 , ARG 180 . >>That is Arg 180 degrees, not just Arg 180. And so what? That does >>not lead to your assertion that exp[i pi] = 0, a result for which >>you have given absolutely no support. Why dont you just give up? >>David McAnally >> At the moment, they (the Time Lords) are far from being all-powerful. >> Thats why its been left up to me and me and me. >> quote by: Patrick Troughton in The Three Doctors >>------- >I, have made myself very explicit. >My original question of the implication >of the imaginary component: e^[ipi]=j*0 (to the related proof) Perhaps you mean that (the imaginary part of e^[i pi]) = 0, in which case you are correct, but you have had a lot of problem expressing yourself, especially in view of the way that you initially made the claim by stating that e^[i pi] = 0, which you described as the imaginary part solution, using a terminology that nobody but you knows. Nobody knows what you mean when you make a statement like exp[i pi] = -1 is the real part solution of exp[i pi] = -1, or that exp[i pi] = 0 is the imaginary part solution of exp[i pi] = -1. I asked you to explain your terminology but you havent bothered. When I take the imaginary part of the equation exp[i pi] = -1, I get sin(pi) = 0, a fact which is already known to us. Have you thought also of the fact that if you have exp[i pi] = -1 (your real part solution) and exp[i pi] = 0 (your imaginary part solution), then you could conclude that 0 = exp[i pi] = -1, and get a contradiction? >was not answered and still it is open. Because under no circumstances can you ever claim that exp[i pi] = 0. David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. === Subject: Re: JSH: Weird, eh? > Im curious to see if any of you will actually follow through my > previous post to its logical conclusion, and especially if *any* of > you can resolve the apparent contradiction. Interesting. As you start a new thread here, what *is* your previous post? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: No Set Contains Every Computable Natural A decidable or recursive language is a formal language that is a recursive > set, i.e., for which there exists an algorithm to solve the following > decision problem: Given string w, does w belong to the language? The > algorithm is not allowed to run into an infinite loop and has to produce a > YES/NO answer for any input string after a finite amount of time. To > formalize the rather vague term algorithm, one usually employs Turing > machines, but several other equivalent approaches are possible. Note that it says a finite amount of time. > All regular, context-free and context-sensitive languages are recursive, > but there exist recursively enumerable languages that are not recursive; > one example is given by the halting problem. Did you realize that the natural numbers in unary (or binary, or any standard base) is a *regular* language, and hence context-free, and recursive (i.e. decidable.). You reference this site, and then you try to make claims contradicting them. > I have shown why a Turing machine can not decide if a string, w, > represents a natural number. Thats because you think the string of an infinite number of 1s is a natural number. Once again, you have to understand that every natural number is finite, even in unary representation. > The set of all natural numbers is not a recursive set if our definition > of recursive set requires there exist a TM that can decide membership > in the set. Mathworlds definition indicates there are differences > between definitions of general recursive function. > This appears to be one of those differences. Did you even notice that your definitions of these sets are based on *formal languages* ? Did you even look up what a formal language is? Here you go, from Wikipedia: > In mathematics, logic and computer science, a formal language is a > set of finite-length words (or strings) over some finite alphabet. Note that a formal language is a set of FINITE-LENGTHED strings. J === Subject: Re: JSH: Weird, eh? ... > Im here to help guide you along, if you manage to get past your > mental roadblocks, and past posters like Dik Winter. > Curse him and his ilk!!! LOL. Im in some mood today. Yes, it appears like that. A remark though. Somewhere you noted that for b + sqrt(b^2 + 28) to be a factor of 1 + sqrt(-167) sqrt(b^2 + 28) should be of the form -167.n^2 for some integer n. This is wrong. Take u some factor of 7 and v = 7/u, and b = u - v, we now have: b + sqrt(b^2 + 28) = u. Now take: u = (44444 - 111.sqrt(-167))^(1/11) v = (44444 + 111.sqrt(-167))^(1/11) w = ([- 27781 - 2017.sqrt(-167)]/2)^(1/11) all three algebraic integers. You may verify: u*v = (1977326743)^(1/11) = (7^11)^(1/11) and u*w = ([- 1272087893 - 86559857.sqrt(-167)]/2)^(1/11) = = ( ([1 + sqrt(-167)]/2)^11 )^(1/11) what of that? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: the anticlassicalist }{ i: linguistic negation > BTW, what are linguistics > triumphs? You are my good Sir, and so is the Universe and everything, even 42. Allow me to prove it... shit, thats the Book of Mormon... Urantia Book? no.. where are my Gideons Bibles when I need them? This will have to do: L.8evangile selon Saint Jean Au commencement le Verbe .8etait et le Verbe .8etait avec Dieu et le Verbe .8etait Dieu. Il .8etait au commencement avec Dieu. Tout fut par lui et sans lui rien ne fut. Voil.88. On ne saurait .90tre plus clair: in the beginning was the Word and the Word was with God and the Word was God. The Word was in the beginning with God, and by Him/It all was made, and without Him/It nothing. (or something to that effect). There, Sir, lies the Triumph of Linguistics. Hors la linguistique, point de salut, cest clair. Beat that! === Subject: Re: Calculating fraction period lengths in different bases Xns94937D4FC49Dcparkesactewaglnetau@203.194.27.1, Carl Parkes > Can anyone prove these statements > My interest is recreational. > > > Statement 1 > > If u=2n^2-n+2 and v=2n^2+n+2 > > Then u^2 mod (u+v-1) =u-1 and v^2 mod (u+v-1) = u-1 > > explicidly: > (1) u+v-1 = 4n^2+3 > (2) u^2 = 4n^4-4n^3+9n^2-4n+4 > = (4n^2+3)(n^2) - 4n^3 + 6n^2-4n + 4 > = (4n^2+3)(n^2-n) + 6n^2 - n + 4 > = (4n^2+3)(n^2-n + 1) + 2n^2 - n + 1 > ^^^^^^^^^^^^ > u-1 > (3) v^2 = 4n^4+4n^3+9n^2+4n+4 > = (4n^2+3)(n^2) + 4n^3 + 6n^2+4n + 4 > = (4n^2+3)(n^2+n) + 6n^2 + n + 4 > = (4n^2+3)(n^2+n + 1) + 2n^2 + n + 1 > ^^^^^^^^^^^^ > v-1 > (4) u^2 mod (u+v-1) = u-1 > (5) v^2 mod (u+v-1) = v-1 > * (5) contradiction of above > (6a)v = u + 2n > (6) v^2 = (u + 2n)^2 = u^2 + 2.u.2n + 4n^2 > =u^2 + 8n^3 + 8n > =u^2 + 2(4n^2+3) + 2n > For all n>0 > > Length of fraction period in base u and v =6 >> I think you have to tell us just what fraction you have in mind. I made a typo in the above statement, mistyping a u for a v, oh well, the thing that interests me is that there are groups or families of equations that will calculate the residues but also will immediately tell you what length the fraction periods will be for the different bases of value u or v. I have read that the factoring of numbers is helped if the fraction period length can be calculated. Presumably this information from numerous bases would also help. It is not surprising that period length of 1/x in a non quadratic base will be twice that of its residue and so on until an odd divisor of the number of relatively prime numbers less than x is reached. For an odd prime number p this will be p-1 eg 7-1=6 the only lengths available are 1,2,3,6. For a composite such as 49 the count of relatively prime numbers is 42 The only lengths available are 1,2,3,6,7,14,21,42 What are the trees, graphs of the non residues to residues called. At some stage the residues form a ring and the length of these show interesting features, has anyone studied these? If so where have they published? Would be quite interested in further information if possible === Subject: Re: Reply, reply, reply James Harris > Come on, reply to my posts!!! Come why dont you say something else, > huh? [infantile tantrum snipped] As always, you are talking about yourself. You live in dread of losing your audience on usenet. Pathetic but true. If one of your worthless threads goes unanswered for even 12 hours, you panic, and put up three more within one hour. If you behave like this and you expect to be admired, or even tolerated, then you are insane. Bad news, Harris: I will not respond to you again. === Subject: Re: Metamath Axiom of Choice > ................... >As I stated above, the Metamath formulation of the Axiom of Choice is >effectively the statement that given any set x, there exists a set y >such that for all nonempty elements w of x, w n (Us) has exactly one >element, where s = {t in y : w in t}. >As stated, it is false. Add the elements of x are disjoint >and it is one of the standard forms. >To see that it is false as stated, let x have three elements, >{1, 2}, {1, 3}, {2, 3}. Any set intersecting all of these in >at least one element has to contain both elements of one. But in the statement above, s is explicitly dependent on w (the is not the existence of a single set with one element in common with every nonempty element. If we take x = {{1,2},{1,3},{2,3}}, then take y = {{{1,2},1}, {{1,3},1},{{2,3},2}}. For the element w = {1,2}, s = {t in y : w in t} = {{{1,2},1}}, so that Us = {{1,2},1}, and so w n (Us) = {1} has exactly one element. For the element w = {1,3}, s = {t in y : w in t} = {{{1,3},1}}, so that Us = {{1,3},1}, and so w n (Us) = {1} has exactly one element. For the element w = {2,3}, s = {t in y : w in t} = {{{2,3},2}}, so that Us = {{2,3},2}, and so w n (Us) = {2} has exactly one element. David McAnally At the moment, they (the Time Lords) are far from being all-powerful. Thats why its been left up to me and me and me. quote by: Patrick Troughton in The Three Doctors ------- === Subject: Re: the anticlassicalist }{ i: linguistic negation : > Formal linguistics is founded on the same rigour : > that one finds in quantum mechanics, so dont : > give me your bullshit. : : Bullshit again. Linguistics is a loud empty hole that has pulled a : donut about itself. It proposes to follow scientific method on : databases that are at best weakly statistical and that have no : empirical falsification possible, inputs or outputs. Like psychology, : it shouts its self-percieved triumphs and whimpers : heteroskedasticity at its failures. BTW, what are linguistics : triumphs? Well, the first and most apparent of all triumphs has been the ability to formalise and rigorise communication to avoid the ambiguity of everyday talk and create that useful thing called science. Logical linguistics and semiotics has abstracted in many different directions, including describing the foundations of mathematics and the logical structure of scientific models. It is also able to describe a lot of structure in natural language, but I dont want to distract you. There is even a lot of research mapping neural connections, cramming electrodes into the brains of various eumetazoan critters, determining glucose consumption through radiological methods, or nicely detailing RNA expression or neurotransmitter response to freshly sliced brain sections of various experimental subjects. You can block your eyes and plug your ears all you want, but there is already a quite large and developing community studying the origin of auditory and visual symbolic processing in the brain, and strangely enough, we already have the beginnings of linguistic expression forming in the research. And they match quite nicely with logical linguistic research that formal languages are built on. So where do you cut the line, mister scientist? Where do models stop, through whatever mystical explanation you care to provide, being able to accurately represent reality? Have you not seen the quite accurate models that fill research proceedings every year on modeling various brain regions? computatrix. Of course, you also snipped all the references to math and physics, perhaps indicative that you never wanted to engage in meaningful dialectic? That is typical for one such as yourself who considers their daily minesweeper practise infinitely more useful to society than the farm workers whos citizenship may be questioned. -- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: No Set Contains Every Computable Natural > You cant input an infinite string to the tape. If you could, it wouldnt > be a Turing machine. > Requiring the input tape to be finite is a severe limitation. > For example, no language, L, could contain every binary > representation of rational numbers between 0 and 1. > 1/3 has an infinite binary representation in base 2. > Languages are usually defined as finite concatenations of a set of > alphabet symbols. An outright admission that a TM cant decide if a string is infinite. > For example, the set of natural numbers expressed in > unary notation is a perfectly fine language... a mistake you made in an > earlier post was to assume that a word infinitely long is in that > language; instead, this is an infinite language containing all 1-strings > of finite length. Finite length must be decided by a human operator. > For your example above, there are better ways of representing every > rational between 0 and 1. Yes, I have presented such representations. You still need a human to decide if a representation is finite. No TM can make this determination. > Every definition of TMs that I have seen assumes > that an infinite tape of blanks is allowed. > But the input (non-blank symbols) is always bounded in length by some > finite amount. A blank is just another symbol. Why would we allow input tapes with an infinite string of blanks and not allow inputs with an infinite number of 1s. I can easily represent natural numbers in unary using blanks. The input tape would be a string of blanks followed by a 0. b0=1, bb0=2, bbb0=3, etc. A TM that decided this language would not halt on a blank tape. > M.C: > No natural number requires an infinite unary representation. All may be > represented using a finite number of 1s. > R.E: > There is no TM that can decide if the string has an > infinite string of 1s. > Russell, I dont see how your response is relevant to what Michael said. > His point is that the language {1^n | n is a natural number} is a set of > finite-lengthed strings. It is not hard to show that there is a TM that > can decide this language. Only if we assume that a human operator decided the iinput string was finite. > And, yes, if you include an infinitely-long string of 1s, then no TM can > decide it (in a finite amount of time, which is a crucial requirement.) > This is why infinitely-long strings are not usually considered in > languages. Then, no decidable language can contain an irrational number. > M.C: > All Turing machines can be finitely represented. There is no such > thing as a TM transition function that requires an infinite state > automata for representation. > R.E: > There is no TM that can decide if the transition table is > infinitely long. > Again, your response is kind of irrelevant. His point is that if you > have a set representing the transition function, it MUST be finite sized > if it is going to represent a TM. Again, a human operator would have to decide the function is finite. There is no automated way to make this detemination. > The Halting Problem is ill posed. > There is no set that contains every TM Sure there is : A = { | M is a valid TM representation} > It is impossible to decide if M is a valid representation. > Actually, it is quite simple to decide if M is a valid representation. Not for a TM. It is impossible for a TM to decide the representation is finite. I dont think it is that simple for human operators, either. > It is easy to check, given a data structure, whether the input is a > valid TM or not. Only if the human operator lives forever. > It is impossible for a TM to decide if M is a valid representation. > Again, it is quite simple. Look up a definite of a Turing machine... it > is some tuple of sets, and it is straightforward to check that, for > example, the claimed start state is an element of the state set. And > that the transitions form a subset of the cross product of the state set > with itself... and so on. > All natural numbers are computable. However, they _all_ cant be > generated in a single run of any algorithm/TM. > The output of a TM cant contain a representation of every natural? > That is correct. Since a TM must halt to complete its output, it can > only have a finite number of natural number representations. > Dont feel that TMs are Ôweak because they cant output ALL the > naturals. What TM can do, however, is output all the natural numbers from > 1 to k for arbitrarily large k. Turing gives an example of such a TM. He doesnt explicitely claim this TM produces every natural number. But, why would he give it as an example unless he was trying to show a TM CAN produce a tape with every natural number? > I didnt say that a TM cant print out certain natural numbers. > I said that a TM that requires a fixed, finite amount of time to > perform an operation cant print certain natural numbers in a > reasonable amount of time. > Taking about withing a reasonable amount of time has very little to do > with decidable sets or recursively enumerable ones. I think it has a lot to do with decidable sets. > These sets do not say > anything about the time required to accomplish any computation; they are > merely telling you what can and can not be computed. Implicit in these definitions is the assumption that an infinite number of computations can be done in a finite amount of time. Russell - 2 many 2 count === Subject: Re: No Set Contains Every Computable Natural > I am using the definition of decidable language given at wikipedia: > A decidable or recursive language is a formal language that is a recursive > set, i.e., for which there exists an algorithm to solve the following > decision problem: Given string w, does w belong to the language? The > algorithm is not allowed to run into an infinite loop and has to produce a > YES/NO answer for any input string after a finite amount of time. To > formalize the rather vague term algorithm, one usually employs Turing > machines, but several other equivalent approaches are possible. > Note that it says a finite amount of time. I ponted out in an earlier post that this should say in a finite number of steps. Hypercomputers can perform an infinite number of operations in a finite amount of time. > All regular, context-free and context-sensitive languages are recursive, > but there exist recursively enumerable languages that are not recursive; > one example is given by the halting problem. > Did you realize that the natural numbers in unary (or binary, or > any standard base) is a *regular* language, and hence context-free, and > recursive (i.e. decidable.). Recursively enumerable, but not recursive. > You reference this site, and then you try to make claims contradicting > them. I contradict lots of people. I even contradict myself, sometimes. > I have shown why a Turing machine can not decide if a string, w, > represents a natural number. > Thats because you think the string of an infinite number of 1s is a > natural number. Once again, you have to understand that every natural > number is finite, even in unary representation. I never said an infinite string of 1s represents a natural number. I said such a string does NOT represent a natural number. No TM can decide that an infinite string of 1s does not represent a natural number in a finite number of steps. > The set of all natural numbers is not a recursive set if our definition > of recursive set requires there exist a TM that can decide membership > in the set. Mathworlds definition indicates there are differences > between definitions of general recursive function. > This appears to be one of those differences. > Did you even notice that your definitions of these sets are based on > *formal languages* ? Did you even look up what a formal language is? > Here you go, from Wikipedia: > In mathematics, logic and computer science, a formal language is a > set of finite-length words (or strings) over some finite alphabet. > Note that a formal language is a set of FINITE-LENGTHED strings. No TM can decide whether an arbitrary set of strings is a formal language. Russell - 2 many 2 count === Subject: Re: How to diagonalize a Hermitian matrix >Hello group, > Could somebody please tell me a numerical algorithm to >diagonalize a Hermitian matrix H, so that I end up not only with the >eigenvalues perched along the diagonal, but also with the unitary >matrix U that conjugates with H to diagonalize it. > I should probably say what I really want. I have a >positive-definite Hermitian matrix H, and I need its matrix square >root. The above is simply the approach that occurred to me for >obtaining that square-root. There are available algorithms to do this diagonalization. Any good library should have such. I am assuming that by square root you mean the Hermitian square root. If you just want a matrix A with H=A*A~, the conjugate transpose of A, the Cholesky decomposition is the fast way to do it. One can use Newtons method to obtain the square root of a matrix; the rate of convergence is not bad. There are other ways as well. I suggest you ask the question in sci.math.num-analysis. -- 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: the anticlassicalist }{ ii: the spectre continues : : -=-=-=-=-= the spectre continues =-=-=-=-=-=- : : : : Now, with this long history of logical analyses of language, : : I always find it strange : : Well, it is not strange. : You are ßaunting ignorance here. It is strange, since it is unnecessary. I am ßaunting ignorance there. : : that there are comments like: : : : : Attention being devoted here entirely to the classical : : two-valued theories of truth-functions, quantifiers, and : : identity, with syntax, semantics, and pragmatics built : : upon their basis, there will be no concern with alternative : : forms of logic, so-called three-valued (or, more generally, : : n-valued) logics, modal logics, intuitionist logics, and : : the like. The view is that whatever is valuable in these : : alternatives can be achieved more readily within the : : classical framework by suitable extensions. : : : : in the introduction to Semiotics and linguistic structure : : by R. M. Martin, where such a representation is not : : faithful to actual usage or expression. : : Actual usage or expression is just irrelevant. Unless you are a linguist : whose object of study is how real people use natural language, natural : language IS JUST STUPID, because PEOPLE are just stupid. There is a formalisation to logic. We can formalise many forms of reasoning. There is no necessity to choose one logical form over another except for representing the model faithfully. So unfornutately, your comment doesnt help resolve my ignorance. : : Now, I just recently posed some questions on the : : newsgroups concerning nonclassical logic, and certain : : linguists and physicists were actually quite : : confrontational about the idea of educating about these : : nonclassical logics. Expressibility was always proffered : : as a reason, even though no one can even claim that : : (lambda calulus is rich enough). : : The parenthetical statement is incredibly stupid. Lambda calculus is : BIGGER AND BADDER than classical boolean anything; it is MORE : complex. SO OF COURSE it is rich enough to express : whatever. That is not even the question. The question : is whether you can achieve similar expressiveness with : SIMPLER machinery (like classical boolean algebra). : There is, after all, a set of first-order axioms defining : lambda-calculus. So it is complexity, then? Because I know an orthomodular logic to describe quantum propositions that is simpler than any embedding could be into Boolean. I know many models whose Heyting structure is far more simplistic than the corresponding Boolean embedding. And since Heyting algebras have a potential universality hinted by the Curry-Howard isomorphism, why is it so necessary to fall back on the classical approach. I still do not see from where this desire arises at forcing the ontology of a model... -- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: Reply, reply, reply > James Harris > Come on, reply to my posts!!! Come why dont you say something else, > huh? > [infantile tantrum snipped] > As always, you are talking about yourself. You live in dread of losing your > audience on usenet. Pathetic but true. If one of your worthless threads goes > unanswered for even 12 hours, you panic, and put up three more within one > hour. If you behave like this and you expect to be admired, or even > tolerated, then you are insane. > Bad news, Harris: I will not respond to you again. Evidently, JSH thrives on attention. Sad that a grown man requires that much attention. David Moran === Subject: Re: What is a number?/What is not a number? ... > The question is, What is it we really mean by the label number? The answer is, there is no clear-cut definition in use. > There are two ways to define the set of number sets: One way is to > simply list all number sets, the other is to give a rule (given by a > definition) that takes the place of the listing of number sets. My > question is, Is there a definition that provides a rule that creates > the same set of number sets as there are in the set of number sets > created as a listing by fiat? Again, the term number is erratic. Rarely do mathematicians use the term number without qualification. And if they do it is (or should be) in a well understood context. > So, we have the set of number sets S = {N, Z, Rationals, Reals, > Complex, Cayley, ...}. So, what definition of number gives us the > set S? All elements of sets obtained by the standard Cayley-Dickson doubling from the real numbers. This will also give quaternions, sedenions, etc.. But I would think that most would consider high order numbers in this sequence not really numbers. And it excludes the cardinals and ordinals. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Dik Winters claims revisited, dependency issue Decker put forward the quadratic > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his as are roots of > a^2 - (x - 1)a + 7(x^2 + x). > Dik Winter has repeatedly asserted that there exists varying algebraic > integer functions w_1(x) and w_2(x), such that > w_1(x) w_2(x) = 7 > and that the factors (5a_1(x) + 7) and (5a_2(x) + 7) have w_1(x) and > w_2(x) as factors, respectively. > Now then, introducing f_1(x) and f_2(x) as the other factors of the > as, you have > w_1(x) f_1(x) = 5a_1(x) + 7, and w_2(x) f_2(x) = 5a_2(x) + 7, > so > a_1(x) = (w_1(x) f_1(x) - 7)/5, and a_2(x) = (w_2(x) f_2(x) - 7)/5. > Now it matters to use the fact that the as are roots of > a^2 - (x - 1)a + 7(x^2 + x), > as solving that quadratic, and picking a_1(x) for the positive sign > root gives > a_1(x) = ((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2, so Typo (or thinko) here and below. ... + 28(x^2 + x)))/2 should be ... - 28(x^2 + x)))/2 Rick === Subject: Re: principal ideals in F[x,y] Arturos proof is certainly correct, but I think that after the line > h(x,y)*f(x,y) = x and h(x,y)*g(x,y) = y. its easier to simply plug in some values. Putting x=0 into the first identity gives h(0,y)*f(0,y)=0, so either h(0,y)=0 or f(0,y)=0. Now if f(0,y)=0, then f(x,y)=x*F(x,y) for some polynomial F, and hence h(x,y)*x*F(x,y)=x. This implies that h is a constant. But h(0,0)=0, since h is in the ideal, so h(x,y)=0. Contradiction. Hence f(0,y) is not 0. This proves that h(0,y)=0, so h(x,y)=x*H(x,y) for some polynomial H. Substituting into the second equation gives x*H(x,y)*g(x,y)=y, and putting x=0 into this identity yields the absurdity 0 = y. Hence the ideal is not principal. BTW, a similar argument shows that if p is a prime number, then the ideal { f(y) in Z[y] such that f(0) = 0 (mod p) } is not a principal ideal. (Notice that this is the ideal generated by p and y, so its very similar to the ideal of F[x,y] generated by x and y.) Joe Silverman >If F is a field, and I is the ideal {fin F[x,y] : f(0,0)=0} in >F[x,y], then why is I not principal? >Im trying to show by contradiction using the fact that > Both x and y are in I. > So if I is principal, say I=(h(x,y)), then there must be polynomials > f(x,y) and g(x,y) such that > h(x,y)*f(x,y) = x and h(x,y)*g(x,y) = y. > of any monomial in h(x,y) is 1; that is, h(x,y) must be of the form > h(x,y)=a+bx+cy. > The same is true for both f and g. Say f(x,y) = r+sx+ty. > Then ct=0, (bt+cs)=0, ra=0, (at+rc)=0, and as+br=1. > If g(x,y)=u+vx+wy, then > ua = 0, vb=0, cw=0, (bw+cv)=0, (av+bu)=0, and (aw+uc)=1. > So either a=0, or both r and u are 0. > If a is zero, then either b=0 or u=0 (from av+bu)=0; but you cannot > have both a and b zero, since as+br=1. So a=0 implies u=0. But then > aw+uc=0, and it is supposed to be 1. So a is not zero. > Therefore, r=u=0. Since a cannot be zero and at+rc=0, then t=0. Which > means that cs=0; If s=0, then as+br=0, which is impossible. So > c=0. Since bw+cv=0, then b=0 or w=0. But w cannot be zero (since > aw+uc=1), so b=0. Now we have a=t=u=c=b=0. But then as+br=1 is > impossible. > Thus, no such h exists. (Other, simpler arguments, are of course > possible). Therefore, I is not principal. > -- > Its not denial. Im just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > Arturo Magidin > magidin@math.berkeley.edu === Subject: Re: 3 Squares Covering 1 Circle > For b, what does a C3v (invariant under 120 deg rotation) > symmetric arrangement yield? (Place the outermost edge of > the squares tangent to the circle.) > Heres a similar situation that does better than the 80.4% > with part (a). Use this 120 deg symmetry, but place the diagonals > of the square along the radii. The optimum placement of the > squares (they can be shifted along the radii) covers 83.0% of > the circles area. This is the best that Ive found, but I dont > know if its the best possible. To be a little more precise, the area is 2.60808705... I have an exact form, but its pretty messy. This happens when the diagonal of each square extends .13980931 past the center of the circle (along the diameter). === Subject: Re: . The hardest of all hard facts . > it should be Michelson-Morley-Miller, > with Michelson being the pioneer of interferomteric methods > for this. D.C.Millers refinement highlighted the anomolies > that MM had found They found none. Millers results could not be duplicated by anyone but Miller with his apparatus. > but this seems to have been dysmissed, because > of a comment that Eintstein made, It was dismissed because it could not be repeated. It was dismissed because it was not verfiable. > on one of his brief forays to his office at CalTech. > there is no vacuum, per se, anyway, unless > you rely on the formalism of atoms as zero-dimensional points > (and renormalization etc.). David A. Smith === Subject: Re: No Set Contains Every Computable Natural > There is no TM that can decide if the string has an > infinite string of 1s. > Has anyone claimed otherwise? People have claimed the set of all natural numbers is a recusive set. This is equivalent to claiming there is a TM that can decide, in a finite number of steps, that an infinite string of 1s does not represent a natural number. > Turings TM doesnt produce a unary representation of every natural? > Which ones does it leave out? I describe a three state TM that calculates which one. Are you sure you want me to post it? How fast is your modem? > I can define a three state TM that will find a natural number > that is not on this tape: 1) Read right until there is a 0. > 2) Read right until a second 0 is found. > 3) Backup and write a 1 on the previous 0. > Repeat steps (1) through (3). This TM will always produce a tape that has exactly one 0. > This 0 will be at a finite position on the tape and the string of 1s > that preceds this 0 represents a natural number that was not on > the original input tape. > This only holds if the initial tape is finite and your last zero comes > before the end of the tape. This certainly applies if the input tape is finite. It also applies if the input tape is infinitely long. Russell - 2 many 2 count === Subject: Re: TSP and the Bell curve What i m trying to get, is very interesting fundamentally. got an interesting insight, by working with the Traveling Sales Man problem, which is known to be NP-Hard. It seems that distribution of paths are gaussian. And based on this assumption, it is possible to construct an Suboptimal alogorithm finder that finds a solution in O(N^2). The solution however, heuristically seems to be acccurate, to only about sqrt(N!)/N! = 1/sqrt(N!) 1/sqrt(N!) of all total paths are missed by this algorithm. Ofcourse there many other assumptions, the algorithm makes. One is that costs are symmetric and that costs are unique. Now , if N-> infty, 1/sqrt(N!) would goto 0. In a way, may be quantum computers, work because of this property? -suresh === Subject: Re: How many ways to put 5 balls into 500 ordered cups? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1I2QCr06788; >You are given 500 numbered cups and five identical balls. Any cup can >hold up to five balls. How many ways can you put the five balls into >the 500 cups? >As a warm-up, lets try smaller sets of ordered cups: >Cups Arrangements >1 1 >2 6 >3 21 >4 56 >5 126 >6 252 >7 462 >8 792 >9 1287 >10 2002 >This is not Eulers partition function because the cups are ordered; >thus, the two-cup case is: >(0, 5) >(5, 0) >(1, 4) >(4, 1) >(2, 3) >(3, 2) >My first thought was that if I throw the first ball into the cups, it >can land #1, thru #500; likewise the second, third, fourth and fifth >balls. So the answer would be (500^5) = 31, 250, 000, 000, 000 ways. >This is wrong, but it sounds good. >What Id like to get is a function (generator, recurrence relation or >closed form) for this that I can explain to someone relatively easily. >I have a brute force solution using an SQL query from hell (my >specialty), which is not quite the same thing as a formula and proof. > I cannot remember enough Combinatorics and Finite Differences to make >the next step. ARRRGH! >It is bothering me enough that I will offer a copy of my next book >(TREES & HIERARCHIES IN SQL) as a prize for the best solution. Why not break it into cases? CASE 1: All 5 balls in 1 cup 500-Choose-1 gives 500 ways CASE 2: 4 balls in 1 cup and 1 ball in another cup There are 500-Choose-2 ways to choose which two cups are used. Then there are 2! ways to permute which cup has 4 balls and which cup has 1 ball So you have 500-Choose-2 * 2! CASE 3: 3 balls in 1 cup and two balls in another cup There are 500-Choose-2 ways to choose which two cups are used. Then there are 2! ways to permute which cup has 3 balls and which cup has 2 balls So you have 500-Choose-2 * 2! CASE 4: 3 balls in 1 cup, 1 ball in another cup, 1 ball in another cup There are 500-Choose-3 ways to choose which three cups are used. Then there are 3!/2! ways to permute which cup has 3 balls and which 2 cups have 1 ball each. So you have 500-Choose-2 * 3!/2! CASE 5: 2 balls in 1 cup, 2 balls in another cup, 1 ball in another cup There are 500-Choose-3 ways to choose which three cups are used. Then there are 3!/2! ways to permute which cup has 1 ball and which 2 cups have 2 balls each. So you have 500-Choose-3 * 3!/2! CASE 6: 2 balls in 1 cup, 1 ball in another cup, 1 ball in another cup, 1 ball in another cup There are 500-Choose-4 ways to choose which four cups are used. Then there are 4!/3! ways to permute which cup has 2 ball and which 3 cups have 1 ball each. So you have 500-Choose-4 * 4!/3! CASE 7: 5 cups with 1 ball each There are 500-Choose-5 ways to choose which three cups are used. Then there is 5!/5! = 1 way to permute which cups have 1 ball each. So you have 500-Choose-5 SUMMARY: (1) Find the ways to partition the number 5: 5 4 1 3 2 3 1 1 2 2 1 2 1 1 1 1 1 1 1 1 (2) For a given case, if there are n numbers that add to 5, then 500-Choose-n cups are used. Then there are n!/c! ways to permute which of the n cups have which number of balls, where c is the number of repititions of the same number of balls in a cup. Multiply these two values to get the total for the case. (3) Sum over all the cases. Suppose that you were partitioning 19 balls into 500 cups. One of the ways to partition 19 is 4 3 3 3 2 2 1 1 Here you are choosing 8 cups, so you have 500-Choose-8. There are 8!/(1! 3! 2! 2!) ways to permute the numbers of balls among the cups. So, for this case, you have 500-Choose-8 * 8!/(1! 3! 2! 2!). === Subject: Re: Dik Winters claims revisited, dependency issue by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1I2QCc06781; Small mistake below, at [*****]. NB >> I let it get personal, and made an earlier post with a silly sign >> mistake. Then I just got really pissed, and made some posts. But >> heres what I feel is the proper approach, with a focus on an implied >> dependency that Dik Winter never bothered to handle. >> Consider the Decker quadratic example (reference at bottom). >> Decker put forward the quadratic >> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) >> where his as are roots of >> a^2 - (x - 1)a + 7(x^2 + x). >> Dik Winter has repeatedly asserted that there exists varying algebraic >> integer functions w_1(x) and w_2(x), such that >> w_1(x) w_2(x) = 7 >> and that the factors (5a_1(x) + 7) and (5a_2(x) + 7) have w_1(x) and >> w_2(x) as factors, respectively. > Your description is accurate. I agree with Dik on this. We have >both given explicit definitions of w_1(x) and w_2(x). See below. >> Now then, introducing f_1(x) and f_2(x) as the other factors of the >> as, you have >> w_1(x) f_1(x) = 5a_1(x) + 7, and w_2(x) f_2(x) = 5a_2(x) + 7, >> so >> a_1(x) = (w_1(x) f_1(x) - 7)/5, and a_2(x) = (w_2(x) f_2(x) - 7)/5. >> Now it matters to use the fact that the as are roots of >> a^2 - (x - 1)a + 7(x^2 + x), >> as solving that quadratic, and picking a_1(x) for the positive sign >> root gives >> a_1(x) = ((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2, so >> (w_1(x) f_1(x) - 7)/5 = ((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 >> so >> w_1(x) f_1(x) - 14 = 5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2, so >> f_1(x) = (5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 + 14)/w_1(x), >> which implies a relationship between f_1(x) and w_1(x), but >> f_1(x) f_2(x) = 25x^2 + 30x + 2 > Correct. First, a_1(x) is a root of >[*] a^2 - (x - 1)*a + 7*(x^2 + x). >Thus a_1(x) is clearly dependent on x. Note however that it >is also dependent on 7: if you replace 7 by, say, 17, then >then a_1(x) is not a root of [*] but instead is a root of > a^2 - (x - 3)*a + 17. [*****} This should have been a^2 - (x - 3) + 17*(x^2 + x). > Now, back to a_1(x) when 7 is used: w_1(x) is also a function >of both x and 7, because when 7 is used, you get > w_1(x) = GCD(a_1(x), 7). >and when 17 is used, you would get > w_1(x) = GCD(a_1(x), 17). > Note that when you replace 7 by 17, a_1(x) also changes; >therefore w_1(x) does also. > What you call f_1(x) is: > f_1(x) = (5 a_1(x) + 7)/w_1(x), >which is ALSO clearly dependent on x, but also clearly dependent >on 7: again, if you replace 7 by 17, you would get a different >value for a_1(x), w_1(x), AND f_1(x). If you want all these >functions to be distinguished for situations other than 7, you >really should write them as a_1(x, 7), a_1(x, 17), f_1(x, 7), >f_1(x, 17), etc.. >> and it is arbitrary that 7 was the multiple before as it could have >> been 11 or 13 or any of an infinity of other numbers, and w_1(x) and >> w_2(x) are themselves not determined, so its a spurious appearance of >> dependency. > No - its not spurious at all. If you replace 7 by something >else, everything changes. For example, with 17, the original >assumption would be >[**] (5 a_1(x) + 17)*(5 a_2(x) + 17) = 17*(25*x^2 + 30*x + 2). >There is no reason to think that the same a_1(x) will work for >17 as for 7; the factorizations are quite different. [For one >thing, the CONSTANT terms are different.] Similarly >for w_1(x) and f_1(x): they will be different for 17 than they >are for 7, etc.. >> Thats the little detail that Dik Winter never bothered to address. > Why should he? He was considering only the 7 factorization. >You appear to think that a_1(x), etc., will not change when you >replace 7 by other numbers. Clearly from the left side of >[**] above, that is not going to be true. >> Now then, if f_1(x) is in fact independent of w_1(x), then how do you >> account for the appearance of a dependency in > f_1(x) is NOT independent of w_1(x). You *defined* it as > f_1(x) = (5 a_1(x) + 7)/w_1(x). >> f_1(x) = (5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 + 14)/w_1(x)? >> If you try to push the issue that it isnt so required consider what >> happens if you try to divide through by w_1(x), as then you have >> f_1(x) = >> (5((x-1)/w_1(x)+sqrt((x-1)^2 + 28(x^2 + x)))/(2w_1(x) + 2w_2(x)) >> and the problem is that all of the ws need to go away. So assuming >> that a_1(x) has w_1(x) as a factor means, introducing g(x), that you >> have >> a_1(x) = w_1(x) g(x), so >> f_1(x) = (5g(x)/2+ 2 w_2(x)) >> which indicates that f_1(x) is dependent on w_2(x). > It is, of course. w_1(x) and w_2(x) are highly dependent >also since their product is 7. >> However, as I pointed out w_1(x) w_2(x) = 7 does NOT determine them as >> an infinity of functions will work, > Not for fixed values like 7. There are a lot of constraints. >For example, for 7, a_1(x) is a root of a^2 - (x - 1) + 7*(x^2 + x), >so a_1(x) is restricted to only being one of two possible numbers. >Then since the right definition of w_1(x) is > w_1(x) = GCD(a_1(x), 7), >w_1(x) is determined once you have chosen a_1(x). Why? Because >the GCD function is well-defined (to within units) in the ring >of algebraic integers. >> while f_1(x) is independent of >> w_1(x) and w_2(x) since >> f_1(x) f_2(x) = 25x^2 + 30x + 2. >> The point is you can multiply some polynomial like 25x^2 + 30x + 2, by >> anything you choose. Theres no way that its locking into it that >> functions that are factors of 7 are required. > Not right - see above. If you replace 7 by something else, like 17, >all of a_1(x), w_1(x), and f_1(x) change. They are all linked >together by the equation > f_1(x) = (5 a_1(x) + p)/w_1(x), >where p could be 7 or 17, etc. >> The simplest answer is that w_1(x) either equals 7, or it equals 1, >> and checking with the first gives >> f_1(x) = (5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/14 + 2)/w_1(x) > That simplest answer however does not give algebraic integer >coefficients as you well know, and that is what you wanted in >the first place. Its simple, but it is the wrong factorization. >Things are not always simple! >> implying that 5((x-1)+sqrt((x-1)^2 + 28(x^2 + x))) has a factor that >> is 14, which is false in the ring of algebraic integers, so Dik Winter >> and his cohorts falsely seize on that to hide the issue with their own >> claims. >> Theyve gone so far that Keith Ramsay even claimed to have posted a >> solution for the ws, but how did he pick ws from infinity? > He didnt. There is a well-defined algorithm for computing >the ws, starting with the number (1 + sqrt(-167))/2. If you >had replaced 7 with 17, the starting number would have been >(-1 + sqrt(-407))/2 and the result would be different. >> Given that his polynomial is of degree 22, its quite possible that he >> just chose a really big polynomial! > Actually I think it had degree 11. I think he used a program >to do the computation, and I think it finds the lowest-degree >polynomial that works. I believe it computes powers of an >ideal, ^M, until it finds an M where that ideal >is principal. It is not a matter of human choice, just a >programmable algorithm. >> Remember the issue here is f_1(x) and its *lack* of a dependency with >> w_1(x). > See above. They are quite dependent, and both change >if you replace 7 by other numbers. >> The problem for Dik Winter and his buddies is that you have two >> independent equations: >> w_1(x) w_2(x) = 7 >> and >> f_1(x) f_2(x) = 25x^2 + 30x + 2 >> To try and dispute my results, posters like Dik Winter or Nora Baron >> simply skip past mathematical consequences of their claims, like the >> independence between those equations. > We focussed on the 7 case because that was Deckers original >example and your own main interest. Other cases are similar >*but not the same*. We never claimed they were. >> Im still pissed. > Too bad. > In fact, seeing the number of threads you have started today >on this, and their content, it looks to me like you are frantic, >perhaps even dangerously upset. You need to calm down about >this. We are not evil conspirators trying to rob you of your >rightful credit. We are just ordinary people trying to show you >where you are making mistakes. > Nora B. >> But theres not a lot I can do when people like Dik >> Winter can so easily get away with basic problems in their claims, on >> a newsgroup that doesnt seem to give a damn about mathematical truth. >> James Harris === Subject: Re: x^2 + y^4 = z^4 > it is known that the equation x^2 > + y^4 = z^4 has no positive-integer solutions. Is the proof of this > result short enough for some kind soul to post it, or need I make a trip > to the library? Depends on what youre willing to accept as proved. (1) (z^4 - y^4)(y^2)(z^2) = (x^2)(y^2)(z^2). But we also (always) have (2) (2(y^2)(z^2))^2 + (z^4 - y^4)^2 = (z^4 + y^4)^2. So, by (1), the Pythagorean triangle with sides 2(y^2)(z^2) and z^4 - y^4 would have an area that is a square. Since 1 is not a congruent number, this is impossible. John Robertson === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >>The probability space is simply ([0,1],L,m), where L is the set of all >>Lebesgue-measurable subsets of [0,1], and m is Lebesgue measure on [0,1]. >>See . >>For each x in [0,1], the probability of the event X element {x} is the >>Lebesgue measure of the set {x}, which is zero. >>Done. > I trawled up and down those definitions, nothing seemed to jump out as > supporting the Ôzero-probability-event actually happened scenario. Not that I > pretend to understand very much of those pages. I dont even know what it means to say that an event actually happened. I know what it means to say that an event is nonempty. An event is really a set, and the probability of an event is its measure. Look at and notice where it says that the Cantor set (an uncountable set) has measure zero. Thus, if we transfer that language to the probability space we have been discussing, then the probability of the event (X in Cantor Set) is 0, despite the fact that the Cantor set contains uncountably many points. Its a very long way from being empty. > And the fact that you cant specify which real you chose is not the only reason > you cant choose one. If you find some way of specifying which one you chose, > Ill find a way to show you where you failed to do what you set out to do. >>And I suppose you also object to things like geometric constructions, on >>the grounds that they cant be carried out exactly in the real world. >>Yawn. > You can specify your real however you want. Geometric constructions translate > just fine into the real world. But what did you have in mind? Choose an angle > randomly from 0 to 90 degrees, and then construct a line from the 90 degree > vertex of a triangle to the opposite side (which has length 1), at that angle > from one of the other sides? > Youre just translating the domain, choosing the angle is effectively the same > as choosing the real directly. I didnt say anything about angles. I didnt say anything about choosing a number at all. I gave you the definition of a particular probability space and I pointed out that in that space, there are nonempty events that have probability zero. > Once again, you cant hide behind the Ôreality card. You can say Ill pick an > integer randomly from a uniform distribution across the whole set but thats > garbage mathematically speaking, and Im claiming that choosing a real from > [0,1] is roughly the same ßavour of garbage. >>No, there is an enormous difference between the two. There is no such >>thing as a uniform probability measure on a countably infinite set, but >>uncountable sets are quite a different matter, as the Lebesgue measure >>example demonstrates. >>The example I gave is correct. > Its so Ôcos I said its so doesnt wash with me. >>Its not my fault if you dont know the definition of a probability >>space. > No. But its disappointing (to me at least) that my error cant be explained to > me in terms I can understand. It seems such a simple concept, and we have to > resort to opaque definitions like Probability Spaces defined in terms of a > Lebesgue Measure of Lebesgue Measurable subsets. What is the length of a point? Zero, right? What is the length of the unit interval [0,1]? Its 1, right? How many points are there in [0,1]? Uncountably many. This shows that you cant find the length of the interval by adding up the lengths of all the consitituent points. There are too many of them. Probability (like measure in general) is countably additive, which means that if you have a *sequence* of pairwise disjoint events, then the probability of the union is the sum of the probabilities. That property doesnt hold for the [0,1] model because there are too many points to be contained in a sequence. The basic problem is that there are exactly the same number of points in the interval [0,1/2] as there are in [0,1]. If you could find the total probability by just adding up the constituent parts, then every nondegenerate interval would have the same probability, which doesnt fit very well with our notion of how probability ought to behave. ><...>Look at and read the >>third paragraph down, where the concept of a probability function is >>mentioned. Notice that the function f: [0,1] -> R given by f(x) = 1 is >>an example of a probability function or probability density function >>as described there. The fact that the integral of f over [0,1] = 1 is >>what makes the function properly normalized to define a probability >>measure, which turns out to be ordinary Lebesgue measure on [0,1]. >>Look at the probability axioms that are stated farther down on that page. >>A probability measure is something that satisfies those axioms. Notice >>that the description of probability says nothing about what you can or >>cannot do in the real world; its an abstract mathematical concept. > So... youre saying that if we accept some set of definitions and constructs > which dont relate to the real world, we can then say Ôa zero-probability-event > can happen in terms of those definitions. > Well maybe you can, but then youre not saying what I understand by the phrase. I am saying that events are sets, and a zero-probability event may be a nonempty set. If an event is represented by a nonempty subset of the sample space, then we can say that it represents a possible outcome. -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: looking for an eltry soln to an old problem A long time ago, J.J. Sylvester posed the problem: if I have arbitrarily many 5 cent or 17 cent stamps, what is the largest denomination I cannot make? In general, if we have p and q cent stamps, it turns out the answer is pq-p-q (granted p and q are coprime). I have derived a solution to the problem, but Id like to teach this to my undergraduates, some of whom have a limited background. So my question is: is there a very nice & friendly proof of this fact? (For example, which avoids any nonobvious facts from number theory.) In case youre interested, the class Im teaching is linear algebra, but as you can see I like to give puzzles to the class which are not necessarily related to the material (in an obvious way). -Tyler === Subject: Re: No Set Contains Every Computable Natural <3sRWb.24383$cM.3142@newssvr25.news.prodigy.com> You still need a human to decide if a representation is finite. > No TM can make this determination. Same comment as above. > A blank is just another symbol. > Why would we allow input tapes with an infinite string of blanks > and not allow inputs with an infinite number of 1s. Because then the input is not *finite-sized* . A TM is essentially an algorithm. The definition of an algorithm requires that it terminates in a finite number of steps. If a TM (an algorithm) is required to read its input (which is a usual requirement) then it will never read its input in a finite number of steps if the input is infinite. This is a good reason why such inputs are not allowed. > I can easily represent natural numbers in unary using blanks. > The input tape would be a string of blanks followed by a 0. > b0=1, bb0=2, bbb0=3, etc. > A TM that decided this language would not halt on a blank tape. You cant ask a TM to decide whether an infinitely long string has any properties. You have to ask it something like does this tape have the first 100 numbers? the first 1000000? the first k? > Only if we assume that a human operator decided the iinput string was > finite. Again, if the human is not working with a finite string, then the human should not be feeding it to a TM. > And, yes, if you include an infinitely-long string of 1s, then no TM can > decide it (in a finite amount of time, which is a crucial requirement.) > This is why infinitely-long strings are not usually considered in > languages. > Then, no decidable language can contain an irrational number. Depends on how you represent them. Here is a set of all possible solutions to quadratic equations with integer coefficients: { (a,b,c,d) such that a,b,c,d are integers } An element of this set (a,b,c,d) should be interpreted to mean (a+b*sqrt(c))/d . > Actually, it is quite simple to decide if M is a valid representation. > Not for a TM. It is impossible for a TM to decide the representation > is finite. I dont think it is that simple for human operators, either. Again, this depends on the representation. I can make a TM that requires an input to be (q_0, Q, T, F, S) where Q,T,F,S are specified by a list of their elements. Then any finite input will have to be describing finite sets. If the data structure, given in this specified way, is not finite, then you cant make it an input to a TM in the first place. > It is easy to check, given a data structure, whether the input is a > valid TM or not. > Only if the human operator lives forever. Again, it depends on the input format. If I ask for a list of all the elements, then I can be sure it is finite. If you want to specify a set of states having a state for every even number representable as a sum of two prime numbers, then no one knows if it is finite. The point, again, is that such claims are representation-dependent. > Turing gives an example of such a TM. > He doesnt explicitely claim this TM produces every natural number. > But, why would he give it as an example unless he was trying to show > a TM CAN produce a tape with every natural number? I dont know what Turings intentions were and I dont know the example you are referring to. > Taking about withing a reasonable amount of time has very little to do > with decidable sets or recursively enumerable ones. > I think it has a lot to do with decidable sets. No, time and number-of-steps have nothing to do with decidable sets (besides, of course, the fact that time and number-of-steps are supposed to be finite.) Talking about time and related properties of TMs (or algorithms) essentially leads to algorithm runtime, NP-completeness, and complexity theory on that sense, but computability theory is independent of time. > These sets do not say anything about the time required to accomplish > any computation; they are merely telling you what can and can not be > computed. > Implicit in these definitions is the assumption that an infinite number > of computations can be done in a finite amount of time. An infinite number of computations is never done in finite time. Yes, it might be considered implicit or hidden assumptions to you, but finiteness of algorithms, turing machine sizes, and input sizes are all *explicitly* defined for all these concepts. J === Subject: Recursion and the Quadratic Function If Recursion is related to Summation then If Summation is a Quadratic Function then Perhaps recursion routines can be solved by finding the correct Quadratic Function. Dick Saunders Jr. End If End If === Subject: Re: branch of log z >... stop throwing stones, were both in the glass house. > I dont want to throw stones, i dont even know, am i inside or > outside. It took me 25 years to find out, what is behind this veil of: > imaginary axis, argand graph, complex plane, imaginary part. Caspar > Wessel and Hamilton didnt use or invented these, as i know now. These > words are still used in teaching people, who are interested in math: > Ask Dr.math > Mathworld.wolfram.com > Complex plane > These two dont mention, that they are talking in their websites about > the R2, just the simple R2 with + and *. > At another level - You mentioned Rudin,Marsden, Royden (books for > 100$): i only found some lecture notes to Rudin. Following these, he > starts with C=RxR, but then for example calling a rectangular > coordinate system an Argand Graph making this special. But you are > not told, what is the difference - it cant be the multiplication, can > it? Im sure you could find any of these at your local library. > So really i am thankfull, that You told me,that i got this - no > differennce between i-axis and y-axis - right. Just one similar > opinion was recently written to sci.math from Lynn Kurtz. It seems to > me, that other people try to avoid a clear statement. So many > mathematicians are reading this and no confirmation or opposition from > them can be found here in sci.math about it. > In R3 you can write a position or an arrow with (x,y,z) and sometimes > its written as x*i+y*j+z*k or as x*e1+y*e2+z*e3. (linear combination > of three basis-vectors). Different notations for the same thing (and > multiplication of vectors is not a necessary condition for it). Good point. I was a bit hasty in suggesting that this would be impractical. > When you introduce the dot-product in R3 you call it euclidian > vectorspace. But you wont say its isomorph to the first, the dot > ignored. It stays the same, with extra dot-multiplication. > So i think (a,b) =a+i*b. a is called the Real part and b the > Imaginary part. a is the first component, the first coordinate and > you ommit 1 or k or e3, as R is embedded ( In R4 they do it too) .And > b is the second component, second coordinate. > Are You going here a step backwards : R^2 and the Complex plane are > isomorphic (if..... ) ? The only strict definition of C is > (R^2,+,*) , as far as i could find. Do You have another one? Well, not of C, but of R^2 as a vector space. To be a bit more precise, R^2 is a vector space with + and scalar * and C is a field with elements from R^2 and the operations +, scalar * and vector *. In the sense you described, I am going backwards. > looks like: complex analysis and R2 analysis -one must be part of the > other. Is this so? Yes, complex analysis is a part of real analysis. The problem with thinking that R^2 analysis and complex analysis are the same is that R^2 analysis usually works on the vector space R^2, whereas if one works with C, one usually works with the complex field. That multiplication issue is very important, since it puts constraints on which functions are differentiable (for instance). Have you heard of the Cauchy Riemann equations? Let f(z) be written as f(x,y) = u(x,y) + iv(x,y). f is complex differentiable iff it is R^2 differentiable (that is, component-wise) and the following equations hold: du/dx = dv/dy and du/dy = -dv/dx. Check out http://mathworld.wolfram.com/Cauchy-RiemannEquations.html for a full exposition. Its a very elegant subject--my favorite thus far. > Feeling like a snail, unable to throw stones, creeping up an parking > ramp to an understanding of branches of log, > Hero === Subject: . Very revealing . Hi Bilge, Why do spend so much time in alt.morons ? The fact that you ignore the way Usenet actually works, while blabbering on about useless documentation, is very revealing. You cant ever win, can you Bilge ? === Subject: Re: No Set Contains Every Computable Natural I ponted out in an earlier post that this should say in a finite number of > steps. > Hypercomputers can perform an infinite number of operations > in a finite amount of time. Time for a turing machine is the number of steps. We are talking about turing machines here, not Ôhypercomputers or other constructs. > Did you realize that the natural numbers in unary (or binary, or > any standard base) is a *regular* language, and hence context-free, and > recursive (i.e. decidable.). > Recursively enumerable, but not recursive. Any regular set is recursive. Any context-free set is recursive. Perhaps you should learn about these two classes of sets before moving up on the Chomsky Hierarchy. > You reference this site, and then you try to make claims contradicting > them. > I contradict lots of people. > I even contradict myself, sometimes. And everyone knows one can deduce anything from a contradiction. I suppose thats how you prove a lot of your statements. > Thats because you think the string of an infinite number of 1s is a > natural number. Once again, you have to understand that every natural > number is finite, even in unary representation. > I never said an infinite string of 1s represents a natural number. > I said such a string does NOT represent a natural number. > No TM can decide that an infinite string of 1s does not represent a > natural number in a finite number of steps. And this is essentially why an input to a turing machine MUST be finite. The turing machine must be guaranteed that it will eventually hit the end of input if it continues to read along its tape. You need to start accepting the fact that input sizes are finite-sized, and a lot of these issues of yours will disappear. > In mathematics, logic and computer science, a formal language is a > set of finite-length words (or strings) over some finite alphabet. > Note that a formal language is a set of FINITE-LENGTHED strings. > No TM can decide whether an arbitrary set of strings is a formal language. TMs are not supposed to reason about anything that is NOT (as a precondition) a formal language. Dont expect any further responses from me in this thread because this is essentially arguing something similar to the existence of God. You seem to be accepting a different set of axioms and definitions from standard ones. J === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs Originator: tchow@markov.mit.edu.mit.edu (Timothy Chow) >Theres the kicker - forever. When forever comes around, Ill let you know >how you went wrong. Or you can say I told you so. Either way, its not >going to happen. You never chose your number, youre still choosing it. Forever isnt going to come around physically, but how do you know that it is impossible for it to come around mathematically? Suppose I choose the first number at time t = 1/2, the second number at time t = 3/4, the third number at time t = 7/8, and so on. (There are physical problems with accelerating the choosing process without limit, but were not talking physics.) At time t = 1, what are you going to tell me went wrong? -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: Lebesgue decomposition > U,v finite measures on (X,A). Suppose v has Lebesgue decomposition (d,N) with > respect to a measure u. How do you determine the Lebesgue decomposition of u > with respect to v? I am not inclined to check out the details. But suppose that the absolutely continuous part has Radon-Nikodym derivative f w.r.t. u. Then I would say that the singular part of u with respect to v is u restricted to the set {x:f(x)=0}. Well I am guessing that this might be a homework problem, so I am leaving you with the problem of (a) checking whether I am correct and (b) providing the details if I am correct. === Subject: Re: CA & Numerical Analysis? > Hi! > Can anyone help me to find a paper about the application of cellular > automata in numerical analysis? > Behrang. http://madeira.cc.hokudai.ac.jp/RD/takai/automa.html http://cafaq.com/apps/index.php?open=complexity http://www.ericweisstein.com/encyclopedias/books/ CellularAutomata.html http://mathforum.org/library/topics/cellular_auto/ http://ieeexplore.ieee.org/xpl/abs_free.jsp?arNumber=338094 http://www.stephenwolfram.com/publications/books/ca-wspc.html http://portal.acm.org/citation.cfm?id=966317.966338&coll= GUIDE&dl=GUIDE [ comp.ai is moderated. To submit, just post and be patient, or if ] [ ask your news administrator to fix the problems with your system. ] === Subject: Re: Metamath Axiom of Choice > > > Perhaps he decided to patronize you because, as *everyone* parsed your > post, you are incapable of reading plain English but you pretend to > have the expertise to respond to questions anyway? > Well, I guess you must be right, then. Since *everyone* says so. By > the way, you should note that what I said was _accurate_. I only > responded to the wrong question. I really dont see why my gaffe has > inspired you to post two insulting messages about me. Correction: One insulting message. Your first post was quite cordial--and correct, not the ascerbic dribble you posted later. > > It shouldnt seem patronizing after all, since as he (and other > respondents) read your post, it was an answer an ignoramus might > write. > Im not exactly sure why youd want to get into a pissing contest with > me. But your reply is completely pointless, since it is in response > to a post that wasnt addressed to you. > Ôcid Ôooh > PS. Chill out. This has nothing to do with you. === Subject: . Earthlink . Hi Ha Ha Hanson, You incorrectly made the following equation, hanson@quick.net = !ed99b2be , While hanson@quick.net is sure to be invalid, identify you every time you use Earthlink to post. I was just using your headers in a vain attempt to explain something to Bilge, Im not verifying you. === Subject: Re: No Set Contains Every Computable Natural > Finite length must be decided by a human operator. > The point here is that a TM can never be given an infinitely sized > input. A TM cant be given an infinitely long blank tape as input. > You still need a human to decide if a representation is finite. > No TM can make this determination. > Same comment as above. See mine. > A blank is just another symbol. > Why would we allow input tapes with an infinite string of blanks > and not allow inputs with an infinite number of 1s. > Because then the input is not *finite-sized* . > A TM is essentially an algorithm. The definition of an algorithm > requires that it terminates in a finite number of steps. If a TM (an > algorithm) is required to read its input (which is a usual requirement) > then it will never read its input in a finite number of steps if the input > is infinite. This is a good reason why such inputs are not allowed. Infinitely long blank tapes are part of the definition of a TM. > I can easily represent natural numbers in unary using blanks. > The input tape would be a string of blanks followed by a 0. > b0=1, bb0=2, bbb0=3, etc. > A TM that decided this language would not halt on a blank tape. > You cant ask a TM to decide whether an infinitely long string has any > properties. You have to ask it something like does this tape have the > first 100 numbers? the first 1000000? the first k? > Only if we assume that a human operator decided the iinput string was > finite. > Again, if the human is not working with a finite string, then the human > should not be feeding it to a TM. A language is decidable because a human decided the input is finite. > And, yes, if you include an infinitely-long string of 1s, then no TM can > decide it (in a finite amount of time, which is a crucial requirement.) > This is why infinitely-long strings are not usually considered in > languages. Instead, we rely on a human operator to decide in advance whether all of the elements of the language are finite. > Then, no decidable language can contain an irrational number. > Depends on how you represent them. Here is a set of all possible > solutions to quadratic equations with integer coefficients: > { (a,b,c,d) such that a,b,c,d are integers } > An element of this set (a,b,c,d) should be interpreted to mean > (a+b*sqrt(c))/d . Is this supposed to represent a decidable language? > Actually, it is quite simple to decide if M is a valid representation. > Not for a TM. It is impossible for a TM to decide the representation > is finite. I dont think it is that simple for human operators, either. > Again, this depends on the representation. I can make a TM that requires > an input to be (q_0, Q, T, F, S) where Q,T,F,S are specified by a list of > their elements. Then any finite input will have to be describing finite > sets. If the data structure, given in this specified way, is not finite, > then you cant make it an input to a TM in the first place. A human operator will decide in advance if the data structure is finite? > It is easy to check, given a data structure, whether the input is a > valid TM or not. > Only if the human operator lives forever. > Again, it depends on the input format. If I ask for a list of all the > elements, then I can be sure it is finite. You will examine every element yourself? I think it will require your children and your childrens children. Even then, I doubt they can examine every finite element. > If you want to specify a set of > states having a state for every even number representable as a sum of two > prime numbers, then no one knows if it is finite. The point, again, is > that such claims are representation-dependent. Whether or not an arbitrary irrational has a finite number of digits depends on the representation? Why not just assign a single symbol to every real number? > Turing gives an example of such a TM. > He doesnt explicitely claim this TM produces every natural number. > But, why would he give it as an example unless he was trying to show > a TM CAN produce a tape with every natural number? > I dont know what Turings intentions were and I dont know the example > you are referring to. > Taking about withing a reasonable amount of time has very little to do > with decidable sets or recursively enumerable ones. > I think it has a lot to do with decidable sets. > No, time and number-of-steps have nothing to do with decidable sets > (besides, of course, the fact that time and number-of-steps are supposed > to be finite.) Except for the fact that a human has decide if the decidable set is actually decidable. There is no automated way to make this determination. > Talking about time and related properties of TMs (or > algorithms) essentially leads to algorithm runtime, NP-completeness, and > complexity theory on that sense, but computability theory is independent > of time. No, it is not. Maybe people wish it was. > These sets do not say anything about the time required to accomplish > any computation; they are merely telling you what can and can not be > computed. > Implicit in these definitions is the assumption that an infinite number > of computations can be done in a finite amount of time. > An infinite number of computations is never done in finite time. Yes, it > might be considered implicit or hidden assumptions to you, but > finiteness of algorithms, turing machine sizes, and input sizes are all > *explicitly* defined for all these concepts. The definition explicitly states that a language is decidable if there exists a TM that can decide if a string, x, is a member of the language. Now, it appears that a human must make this decision in advance before the string can be given to the TM. Russell - 2 many 2 count === Subject: Re: No Set Contains Every Computable Natural Russell Easterly says... >> There is no TM that can decide if the string has an >> infinite string of 1s. >> Has anyone claimed otherwise? >People have claimed the set of all natural numbers >is a recusive set. The usual notion of recursive set is that a set S is recursive as a *subset* of the naturals if there is a Turing machine T such that for any *natural* number n (in unary notation, to be specific) T(n) halts and outputs 1 <-> n is an element of S T(n) halts and outputs 0 <-> n is not an element of S If x is some input that is not a representation of a natural number, then there is no constraint on what T(n) does. -- Daryl McCullough Ithaca, NY === Subject: Re: No Set Contains Every Computable Natural > Russell Easterly says... >> There is no TM that can decide if the string has an >> infinite string of 1s. >> Has anyone claimed otherwise? >People have claimed the set of all natural numbers >is a recusive set. > The usual notion of recursive set is that a set S is > recursive as a *subset* of the naturals if there is a > Turing machine T such that for any *natural* number n > (in unary notation, to be specific) The set of all natural numbers is not a proper subset of the set of all natural numbers. > T(n) halts and outputs 1 <-> n is an element of S > T(n) halts and outputs 0 <-> n is not an element of S > If x is some input that is not a representation of a natural > number, then there is no constraint on what T(n) does. According to this definition, the set of all natural numbers is recursive because there exists a TM that doesnt even read the input and always outputs a 1. This is a one state TM that decides the set of all natural numbers: 1) Write a 1. Russell - 2 many 2 count === Subject: Re: help with solutions for three questions from the past contest > Id greatly appreciate your help on the solutions for these questions. > 1. Al and Bob are at opposite ends of a diameter of a silo in the > shape of a tall right circular cylinder with radius 150 ft. al is due > west of Bob. Al begins walking along the edge of the silo at 6 ft. per > second at the same moment that Bob begins to walk due east at the same > speed. The value closest to the time in seconds when Al first can see > Bob is what? answer: 48 > 2. if a, b, c, and d are nonzero numbers such that c and d are > solutions of x^2+ax+b=0 and a and b are solutions of x^2+cx+d, find > a+b+c+d. ans: -2 > 3. A boat with an ill passenger is 7.5 mi north of a straight > coastline which runs east and west. A hospital on the coast is 60 > miles from the point on shore south of the boat. If the boat starts > toward shore at 15 mph at the same time an ambulance leaves the > hospital at 60 mph and meets the ambulance, what is the total distance > (to the nearest 0.5 mile) traveled by the boatand the ambulance? ans: > 62.5 > Choogu I did figure out the solution for no. 2 so only help needed for no.1 choogu === Subject: Need Code for Fractional Brownian Motion Analysis and Simulation I have looked everywhere for Matlab code for fractional Brownian motion analysis and simulation. The only thing I found after searching with both Yahoo and Googe was a dead link. Im posting because I have no other recourse. Please sent me any info you can. Or post it to the newsgroup. === Subject: covering compact set w/squares Given a compact set K in the plane s.t. each pt x is the center of a square Q_x, prove that you can find a subsequence Q_x_i of squares s.t. K is covered by the Q_x_i and the sum of the areas of each of the Q_x_i is no more than 4 times the area of the union of the Q_x_i. I think this is a totally geometric thing, and I know it suffices to show that in such a minimal cover no point is contained in more than 4 squares, but how do i show this? I know how to do this with intervals on the real line (there the answer is 2 times the area....), but i get lost when going up a dimension. === Subject: Re: tensors for tots > Anyway, to take one more baby step, if I have not exhausted my > credits, am I correct in thinking that a reasonable way to introduce a > metric tensor into babyland would be to define the inner product of > a row vector and a column vector, in a particular coordinate system, > to be a bilinear form with a matrix sandwiched in the middle, that > matrix our metric? Urg, its been a while since Ive done this. This sounds about right, but you should definitely check out Michael Spivaks book on Differential Geometry. . Most of the work he does is in terms of linear algebra and differential forms. Granted, the latter are a tensor analysis concept, but if you understand the Generalized Stokes Theorem, you probably know enough about them to get a good start. Ôcid Ôooh === Subject: How to choose a matrix P Now I am thinking of this question. Suppose Y1=D - PEP Y2=D-E where D,E,P : nxn matrix ||P||=1 How to choose P, so that rank Y1 >= rank Y2 and ||Y1||>= ||Y2|| === Subject: Re: No Set Contains Every Computable Natural > Finite length must be decided by a human operator. > The point here is that a TM can never be given an infinitely sized > input. > A TM cant be given an infinitely long blank tape as input. > Infinitely long blank tapes are part of the definition of a TM. And is black the same as white? today ?? > Russell > - 2 nany 2 count === Subject: Re: proving integrability of binomial numbers > Ignacio Larrosa Ca.96estro escribi.97: > Augustin Des escribi.97: >> I know this is a well known integrability. But what would a formal >> proof of it look like ? > Viewing that Comb(n, k) is the number, necessary integer, of ways to > choose k elemnts from n, without order. > But if you insist in a arthmetic prove, show that Comb(n, 0) = > Comb(n, 1) for n >= 0 and Comb(n, k) = Comb(n-1, k - 1) + Comb(n - 1, > k) (Tartaglias or Pascals triangle) > It must be: > But if you insist in a arthmetic prove, show that Comb(n, 0) = Comb(n, n) = > 1 for n >= 0 > and Comb(n, k) = Comb(n-1, k - 1) + Comb(n - 1, k) (Tartaglias or Pascals > triangle) Of course the proof using recursion is the standard proof and the easiest proof. There is another arithmetic proof in Gausss Disquisitionae Arithmeticae, although I cant remember where it is or get my hands on my packed-away copy of it. He actually considered the prime factors of the numerator and denominator. I suppose you could use the formula for the number of powers of a prime p that divide n!, namely [n/p]+[n/p^2]+[n/p^3]+..., ([x] is the greatest integer function), where this is really a finite sum since all terms are 0 once p^a > n. A small language note. The condition of something being a whole number is its integrality, not its integrability. We talk about the integrability of a function to ask if there is an integral as in f(x)dx of this function. Must come from the same root, but the use is different. Achava PS - I have never heard of Pascals triangle referred to as Tartaglias triangle. I only know of Tartaglia as a contributor to solving the general cubic. Is this because I mostly have access to American and other English language sources? === Subject: How to choose 3 unity vectors from 1000 unity vectors, the volume of the parallelepiped formed by the 3 unity vectors are the largest? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1I4taK18046; Given 1000 three dimensional unity vectors, one can form a parallelepiped using arbitrary 3 of them. I want to choose the 3 unity vectors from the 1000 unity vectors, the volume of the parallelepiped spanned by these 3 unity vectors should be the largest amongst all the parallelepiped formed by any possible 3 unity vectors in the 1000 unity vectors. I cannot choose the specified 3 unity vectors by calculate the volumes of all parallelepiped formed by all possible combination of 3 unity vectors from the 1000 unity vectors, because the combination number is too large: about 1.6e8. Is there any easier way to choose the specified 3 unity vectors? Kenki The Hong Kong Polytechnic University. === Subject: Is the Div Theorem well behaved if the surface isnt closed? Hello The subject says it all? Every reference I have found describing the Divergence Theorem always considers a closed surface for the surface of integration. What happens if the surface isnt closed? Bob === Subject: Re: help with solutions for three questions from the past contest > 3. A boat with an ill passenger is 7.5 mi north of a straight > coastline which runs east and west. A hospital on the coast is 60 > miles from the point on shore south of the boat. If the boat starts > toward shore at 15 mph at the same time an ambulance leaves the > hospital at 60 mph and meets the ambulance, what is the total distance > (to the nearest 0.5 mile) traveled by the boatand the ambulance? ans: > 62.5 > > Choogu Boat . | | 7.5| z | Shore |___________________________________.Hospital x 60-x Hours along distance z at 15 mph must equal hours along distance 60-x at 60 mph. And at constant speeds, distance = speed * time. === Subject: Re: Math Too Advanced For Mainstream Economists The pitiful child who posts below is teaches economics at a leading second-tier department of economics in the U.S.A. Hard to believe. > Youve done enough. Have you no sense of decency, sir? At long last, > have you left no sense of decency? > -- Joseph Welch to Sen. Joseph McCarthy, 1954. Army-McCarthy > Hearings. > also > to the sci.math newsgroup for no reason. Stop this. > Thats the longest answer I ever heard to a yes or no question. > Presidential Debate > also > non-answers about whether supply-and-demand is ever valid. > Samuel Johnson observed that that patriotism is the last refuge of > scoundrels, and evidently aggressive editing is the last refuge of > those losing newsgroup slanging matches. Anyone who is interested in > a strange Sraffarian modus operandi is invited to read up on this sad > thread. > %3DUTF-8%26group%3Dsci.econ > So much for the theory that wages and employment are determined > by the intersection of well-behaved supply and demand curves in > the labor market. > and > Poor Mark Witte seems to believe that the supply and demand model > is sometimes reasonable. > I tired to explain that his linked page doesnt really critique the > competitive labor market supply and demand model, where quantities are > related to prices, and other factors shift said curves. It evidently > didnt take, probably owing to confusion on Mr. Vienneaus part on the > difference between partial and general equilbrium. Its a common > intermediate mistake. > I then asked whether Robert had the courage of his convictions to say > whether he felt that supply and demand was ever a reasonable framework > for explaining our world. I made it easy, but in trying to bring the > mountain to Mohammed somehow only led Mohammed to run away. > Roberts citing of empirically based papers on other topics in support > of a model that endeavors to address labor market issues is shameful, > and he does it here yet again. > There are many models of labor market behavior, supply-and-demand > being a very useful on among them. In the tradition of the following > book, > http://www.amazon.com/exec/obidos/tg/detail/-/0226458083/qid= 1076997593//r > ef=sr_8_xs_ap_i1_xgl14/103-6894108-3696656?v=glance&s=books&n =507846 > this thread will gain my reply when the approach Robert Vienneau > advances is shown to explain something, anything, observable about > labor markets that other approaches dont. > To beat a model, one must either show it is invalid, or that another > model is superior empirically. If supply and demand is not invalid, > then for some other model to be preferred, that model must dominate > supply and demand on some empirical dimension. Its that simple, and > many models meet this criterion. Roberts evidently does not. > Until some real evidence is offered in support of the attempted labor > model under discussion, this thread will just be a weird polemic, the > sound of one hand clapping. -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: No Set Contains Every Computable Natural Russell Easterly says... >> The usual notion of recursive set is that a set S is >> recursive as a *subset* of the naturals if there is a >> Turing machine T such that for any *natural* number n >> (in unary notation, to be specific) >The set of all natural numbers is not a proper subset of >the set of all natural numbers. But it *is* a subset. A recursive subset. >> T(n) halts and outputs 1 <-> n is an element of S >> T(n) halts and outputs 0 <-> n is not an element of S >> If x is some input that is not a representation of a natural >> number, then there is no constraint on what T(n) does. >According to this definition, the set of all natural numbers >is recursive because there exists a TM that doesnt even >read the input and always outputs a 1. Right. -- Daryl McCullough Ithaca, NY === Subject: any good book on general optimization, continuous multivariate optimization? I am studying general optimization, particularly continuous multivariate optimization, something like First Order Nessecery Condition, Second Order Nessecary Condition, Second Order Sufficient Condition for local minimizers... etc. I got really confused. The reasons are 1) Multivariate derivatives, gradient, etc. are what I am not very familiar with, so I got confused; 2) Some boundary condition and degenerate cases got me confused: for example, it is easy to have FONC, SONC, SOSC for interior points, thats easy, but what if the FONC, SONC, SOSC on boundary point, involving the directional derivative vector d, etc. And some degenerate cases such as the Hessian is indefinite... etc. And the relationship between FONC, SONC, SOSC, etc. these all got me confused... Please point me to some good books/online lecture notes, etc. that can help me clearify these concepts? -Walala === Subject: Is there a Second Order Sufficient Condtion(SOSC) for boundary and non-interior points? Hi all, I only learned SOSC for interior points, but I find no where about SOSC for boundary and all(boundary, noninterior) points... and how about First Order Sufficient Condition for local minimizer? -WAlala === Subject: Re: Is the Div Theorem well behaved if the surface isnt closed? > Hello > The subject says it all? Every reference I have found describing the > Divergence Theorem always considers a closed > surface for the surface of integration. What happens if the surface isnt > closed? > Bob If the surface isnt closed, then it doesnt really bound a region, then, does it? However, if the two surfaces S1 & S2 share the same boundary, then the Divergence Theorem relates the integrals of V.dN over S1 and S2, via the integral of div(V) over the volume bounded by the union of S1 and S2. Dale. === Subject: how to understand this statement of Second order Necessary Condition? I read the statement of SONC for local minimizer as follows: Let x* a local minimizer of function f over some constraint set O, and d a feasible direction at this point x*. If d*gradient=0, then d*H*d>=0. where d is the transpose of d column vector... I got confused here because this seems to me is a half statement: If d*gradient=0, then d*H*d>=0. but what if d*gradient>0 but not =0 at that point x*? Is this a possible case in the SONC statment? Is that true that when d*gradient>0 but not =0, there is no statement about the SONC? -Walalal === Subject: Re: . Very revealing . Jeff Relf: >Hi Bilge, Why do spend so much time in alt.morons ? You must be even stupider than you appear from your posts. === Subject: Re: No Set Contains Every Computable Natural > Russell Easterly says... >> The usual notion of recursive set is that a set S is >> recursive as a *subset* of the naturals if there is a >> Turing machine T such that for any *natural* number n >> (in unary notation, to be specific) >The set of all natural numbers is not a proper subset of >the set of all natural numbers. > But it *is* a subset. A recursive subset. >> T(n) halts and outputs 1 <-> n is an element of S >> T(n) halts and outputs 0 <-> n is not an element of S >> If x is some input that is not a representation of a natural >> number, then there is no constraint on what T(n) does. >According to this definition, the set of all natural numbers >is recursive because there exists a TM that doesnt even >read the input and always outputs a 1. > Right. The set of all natural numbers is recursive by fiat. Why bother to include computability in the definition? A human must decide if x is a member of this set before x can be given to the TM. Russell - 2 many 2 count === Subject: Re: Mean Value Theorem >Two places Ive seen the equivalent statement: If f is continuously >differentiable on [0,1], then |f(x)-f(y)|<=sup|f(u)||x-y| where the >sup is extended over 0<=u<=1. >I cant see why we need f to be continuous. It seems by the MVT we >have > f(x)-f(y)=f(c)(x-y) for some c in [0,1], > Right. >so that > |f(x)-f(y)| = |f(c)||x-y|, for some c in [0,1], > <= sup_u|f(u)||x-y|. > But without continuity (or something extra beyond differentiability) > how can you conclude that the derivative f is bounded on [0, 1]? You > have to have boundedness in order to justify your last step... If f is not bounded, then we can assign the value Infinity to sup_u|f(u)| and then the relation still holds. So it seems that the differentiability condition is there purely for technical reasons (read: rigor), to avoid having to deal with Infinity because it does not belong to R. > G. Rodrigues === Subject: Collatz Conjecture : Symmetry question. Every time I post on this problem I get beat up so please be kind. I have found many properties in the landscape of this problem and the one that has most of my interest the gross amount of non-regularity tightly coupled to large a domain of symmetry. I have noticed 2 forms of symmetry in this problem which have me thoroughly intrigued, but do not know where to go with these notions at this point. Symmetry 1; I have noticed that given any tree pattern of any size, it is exactly duplicated infinitely many times. For example, take the entire tree starting at A and extend out all branches to a depth of any N. Then that pattern taken generically is repeated at an offset of A + n*(constant value) with (n = 1, 2, 3, ... infinity), with the constant value is a function of N. Symmetry 2; I have also noticed that all values at N (here I include all rationales and integers by taking all possible paths from A to N, keeping NON_INTEGER branches in this process) produces N sets of a collective 2^(N-1) number of values where the number of items in each set follows the Binomial Coefficients or Pascals Triangle (which ever you normally call it). If you take each of these N sets of values and calculate the ordered set difference (N) that the difference sets (N) are constant and equal for all N independent of A. This means that using only the N sets, all sets at N in groups of N, thus all values at N, for any A, can be calculated without extending the tree from A to N through all the intermediate Ns. In other words, all values of absolute stopping distance of N, for any A, can be calculated directly with out traversing the tree to find each of those values. Is this interesting to anyone? Or better yet, Is this anything of interest for this problem. === Subject: Re: Infinite polynomial product >I recently came across the following infinite product: >(1-x)(1-x^2)(1-x^3)(1-x^4)..., >where abs(x) < 1. > Not elementary, but its related to the Dedekind eta function and > Eulers Pentagonal Number Theorem. > See e.g. > . > and the many references given at the last URL. Theres a proof of the Pentagonal Number Theorem on http://planetmath.org/encyclopedia/ PentagonalNumbersTheorem.html LH === Subject: Re: How many ways to put 5 balls into 500 ordered cups? > You are given 500 numbered cups and five identical balls. Any cup can > hold up to five balls. How many ways can you put the five balls into > the 500 cups? Lets make it more interesting: How about *ten* balls (but each cup can still only hold up to five balls)? -Michael. === Subject: Re: JSH: Not a solution >But hey, were mostly Americans here! Surprise. Surprise. Not true. Youve kept me amused here in Oz for the last three years, and the best part of five years before that in the UK. Woof! Woof! === Subject: Re: the anticlassicalist }{ ii: the spectre continues > : : -=-=-=-=-= the spectre continues =-=-=-=-=-=- > : : > : : Now, with this long history of logical analyses of language, > : : I always find it strange > : Well, it is not strange. > : You are ßaunting ignorance here. > It is strange, since it is unnecessary. > I am ßaunting ignorance there. > : : that there are comments like: > : : > : : Attention being devoted here entirely to the classical > : : two-valued theories of truth-functions, quantifiers, and > : : identity, with syntax, semantics, and pragmatics built > : : upon their basis, there will be no concern with alternative > : : forms of logic, so-called three-valued (or, more generally, > : : n-valued) logics, modal logics, intuitionist logics, and > : : the like. The view is that whatever is valuable in these > : : alternatives can be achieved more readily within the > : : classical framework by suitable extensions. > : : > : : in the introduction to Semiotics and linguistic structure > : : by R. M. Martin, where such a representation is not > : : faithful to actual usage or expression. > : Actual usage or expression is just irrelevant. Unless you are a > linguist > : whose object of study is how real people use natural language, natural > : language IS JUST STUPID, because PEOPLE are just stupid. > There is a formalisation to logic. We can formalise many forms of > reasoning. There is no necessity to choose one logical form over another > except for representing the model faithfully. > So unfornutately, your comment doesnt help resolve my ignorance. > : : Now, I just recently posed some questions on the > : : newsgroups concerning nonclassical logic, and certain > : : linguists and physicists were actually quite > : : confrontational about the idea of educating about these > : : nonclassical logics. Expressibility was always proffered > : : as a reason, even though no one can even claim that > : : (lambda calulus is rich enough). > : The parenthetical statement is incredibly stupid. Lambda calculus is > : BIGGER AND BADDER than classical boolean anything; it is MORE > : complex. SO OF COURSE it is rich enough to express > : whatever. That is not even the question. The question > : is whether you can achieve similar expressiveness with > : SIMPLER machinery (like classical boolean algebra). > : There is, after all, a set of first-order axioms defining > : lambda-calculus. > So it is complexity, then? Because I know an orthomodular logic to describe > quantum propositions that is simpler than any embedding could be into > Boolean. Just FYI: If by Ôsimpler, you mean Ôfewer defining properties, then thats not the same as the complexity that was probably meant, here. I know many models whose Heyting structure is far more simplistic > than the corresponding Boolean embedding. Can you name them? Heyting algebras are always infinite, afaik. > And since Heyting algebras have a potential universality hinted by the > Curry-Howard isomorphism, why is it so necessary to fall back on the > classical approach. I still do not see from where this desire arises at > forcing the ontology of a model... Perhaps the following may help (and perhaps not :-) ). A Heyting algebra is a mathematical structure of some kind. Its defined as an infinite set, together with some operators and relations, that satisfies certain (first-order logic) conditions. So even to understand the notion of a Heyting algebra, most people require a good intuitive picture of the Tarski semantics of first order (classical) predicate logic. Constructivism has so many variants that, in order to study them well, most systems are defined and studied using classical means and classical thinking habits. In more vague terms: Ôreasoning on the meta level is still classical. You may find that awful, but thats the way it is. One reason for this is that there is no reason to prefer one form of constructivism over another. You like Heyting algebras, someone else likes some other system. And unlike the several formalizations of classical logic, these variants are far from being mutually equivalent or translatable. Using classical definitions of the lot allows us at least to understand all variants at the same time, and to make mutual comparisons. There is plenty of room for more research and other points of view, but as something that is supposed to give students a good basis for further research, classical logic is still essential, constructivism an extra. That may serve as an explanation why not many have responded very enthousiastic to your pamßet. :-) Herman Jurjus === Subject: Re: JSH: Not a solution >But hey, were mostly Americans here! Surprise. Surprise. > Not true. Youve kept me amused here in Oz for the last three years, > and the best part of five years before that in the UK. He was referring to his multiple personalities, most of whom are American. Doug === Subject: Re: countable sets >> Let E1, E2, E3, ... be a sequence of pairwise disjoint countable sets. Prove >> that union of Ek as k = 1 to oo is countable. I do not need help proving >> this. I need someone to explain exactly what is being asked. In Let E1, E2, >> E3,... be a sequence of pairwise disjoint countable sets is that saying >> that the numbers of Eis are countable or is it saying that each Ei has a >> countable number of elements?? > Others have explained what is being asked, but I was intrigued by your > statement that you dont need help proving it. Now that the question has > been clarified, you might look again at proving it. > Hint: If your proof does not invoke the axiom of choice, then your proof > is wrong. Not so. The countable union of disjoint countable sets is in 1-1 correspondence with NxN -- the pair (n, k) maps to the the k-th element of the n-th set (the set of sets and each set is alreadyknown to be in 1-1- correspodence with N). This is clearly surjctive, and the disjointness implies that its injective. And it does not require AC to show that NxN is countable. === Subject: two other sides of triangle How we can find two sides of triangle if the third side, the radius of inscribed circle and the angle opposite to the third side is given? === Subject: Re: No Set Contains Every Computable Natural <3_OdnTxoJNyqVK_dRVn-hA@comcast.com> Discussion, linux) > The set of all natural numbers is recursive by fiat. > Why bother to include computability in the definition? > A human must decide if x is a member of this set > before x can be given to the TM. Really? And if we give a tape with an infinite number of 1s on it, how does the human decide whether theres a 1 on every square (to the right of the start square) or just very many 1s to the right? What wonderful faculty have humans for determining the contents of an infinitely long tape? -- All intelligent men are cowards. The Chinese are the worlds worst fighters because they are an intelligent race[...] An average Chinese child knows what the European gray-haired statesmen do not know, that by fighting one gets killed or maimed. -- Lin Yutang === Subject: Re: the anticlassicalist }{ i: linguistic negation > They dont know nothing! How bland. Lets really mean it: They dont know nothing about nothing. > The not-house will be built Is this to mean the not-house wont be not-built? Do even number of negations affirm and odd number of negations deny? If he didnt not double negate, didnt he affirm? If he didnt not double negate, didnt he not affirm? How about double affirmatives implying negation? Arent my ideas the greatest? Oh yea, yea. === Subject: Re: No Set Contains Every Computable Natural > The set of all natural numbers is recursive by fiat. > Why bother to include computability in the definition? > A human must decide if x is a member of this set > before x can be given to the TM. > Really? > And if we give a tape with an infinite number of 1s on it, how does > the human decide whether theres a 1 on every square (to the right of > the start square) or just very many 1s to the right? What wonderful > faculty have humans for determining the contents of an infinitely long > tape? Hey, thats my argument. Using the logic OP have presented, I can choose the set of all natural numbers to be any set I want. (1,2,3) is the recursive set of all natural numbers. Proof: This is the TM that decides if x is a member of (1,2,3). 1) Print 1 2) Halt You might think 1111 represents a natural number, but you are wrong. 1111 is too large to be the input for a TM. Russell - Solution to halting problem: Ctrl Alt Del === Subject: Re: looking for an eltry soln to an old problem Tyler Neylon > A long time ago, J.J. Sylvester posed the problem: > if I have arbitrarily many 5 cent or 17 cent stamps, > what is the largest denomination I cannot make? > In general, if we have p and q cent stamps, it turns > out the answer is pq-p-q (granted p and q are coprime). > I have derived a solution > to the problem, but Id like to teach this to my > undergraduates, some of whom have a limited background. > So my question is: is there a very nice & friendly > proof of this fact? (For example, which avoids any > nonobvious facts from number theory.) 1. If an amount x in [1,pq-1] is not doable, then x is not a multiple of p or q. 2. If y is any element of [1,pq-1] other than a multiple of p or q, then y or pq-y is doable, but not both. 3. So, since p+q is the smallest doable number not a multiple of p or q, the largest non-doable number is pq-p-q. But how to show (2) in a sufficiently digestible fashion? LH === Subject: Re: countable sets > Let E1, E2, E3, ... be a sequence of pairwise disjoint countable > sets. Prove that union of Ek as k = 1 to oo is countable. I do not > need help proving this. I need someone to explain exactly what is > being asked. In Let E1, E2, E3,... be a sequence of pairwise > disjoint countable sets is that saying that the numbers of Eis > are countable or is it saying that each Ei has a countable number > of elements?? >> Others have explained what is being asked, but I was intrigued by >> your statement that you dont need help proving it. Now that the >> question has been clarified, you might look again at proving it. >> Hint: If your proof does not invoke the axiom of choice, then your >> proof is wrong. > Not so. The countable union of disjoint countable sets > is in 1-1 correspondence with NxN -- the pair (n, k) maps to > the the k-th element of the n-th set (the set of sets and each > set is alreadyknown to be in 1-1- correspodence with N). AC is quite tricky. Think again: all you know is that there exist a bijection between the n-th set and IN; the k-th object of the n-th set isz ill-defined, since you dont know explicitly the n-th bijection ;-( This > is clearly surjctive, and the disjointness implies that its > injective. > And it does not require AC to show that NxN is countable. === Subject: Re: Dik Winters claims revisited, dependency issue James Harris: |> Theyve gone so far that Keith Ramsay even claimed to have posted a |> solution for the ws, but how did he pick ws from infinity? Im not sure what you mean here by pick ws from infinity. | He didnt. There is a well-defined algorithm for computing |the ws, starting with the number (1 + sqrt(-167))/2. If you |had replaced 7 with 17, the starting number would have been |(-1 + sqrt(-407))/2 and the result would be different. Lets recall what I posted. |It appears that Q(sqrt(-167)) has a class group of order 11. |Let r stand for (1+sqrt(-167))/2 (which is an algebraic integer, |in spite of the 2 in the denominator). | |r^11 = (-592764018-86559857*r) | = (44555-222*r) (-12882-2017*r) | |and | |7^11 = (44555-222*r)(44555+222*r). As an intermediate step, I found the GCD of ((1+sqrt(-167))/2)^11 and 7^11, which is an 11-th power of the GCD of (1+sqrt(-167))/2 and 7. Harris to check the result using a calculator or something like that. |Also (-12882-2017*r)(-12882-2017*(1-r)) = 2^11 * 3^11, which is |relatively prime to 7^11. | |The factor 44555-222*r = 44444 - 111 sqrt(-167) and its conjugate |44444 + 111 sqrt(-167) are roots of x^2-88888*x+7^11=0. One GCD of |r with 7 is then (44444-111 sqrt(-167))^(1/11), which is a root of |x^22 - 88888 * x^11 + 7^11 = 0. The only point of giving the degree 22 polynomial was to show that the GCD (44444-111*sqrt(-167))^(1/11) actually is an algebraic integer, by giving a monic polynomial with integer coefficients of which its a root. James Harris: |> Given that his polynomial is of degree 22, its quite possible that he |> just chose a really big polynomial! Thats just silly. Its easy to find a quadratic that 44444-111*sqrt(-167) is a root of. Getting the polynomial of degree 22 is just a matter of substituting x^11 for x. Its a very simple polynomial. Nora Baron: | Actually I think it had degree 11. I think he used a program |to do the computation, and I think it finds the lowest-degree |polynomial that works. I believe it computes powers of an |ideal, ^M, until it finds an M where that ideal |is principal. It is not a matter of human choice, just a |programmable algorithm. The package Pari-GP has a way of representing ideals in the ring of integers of a number field. There are functions for multiplying or taking powers of ideals, as well as one for determining whether or not an ideal is principal. It also computes the class group of the number field. The class group of Q(sqrt(-167)) has order 11, and the ideal generated by 7 and (1+sqrt(-167))/2 isnt principal. So it must be that the 11-th power of the ideal is principal. I then used the ideal power function and the one for finding the principal generator of an ideal to get the 11-th power of the GCD. I dont know whether you can multiply the GCD by a unit to get one whose minimal polynomial has degree less than 22. That might be a cute exercise for someone. I would guess not, but I havent given it much thought. Keith Ramsay === Subject: Re: Id like to join this group! > Does the Equation below really prove that 2=1 ?? Y N > 1) X=Y ; Given > 2) X^2=XY ; Multiply both sides by X > 3) X^2-Y^2=XY-Y^2 ; Subtract Y^2 from both sides > 4) (X+Y)(X-Y)=Y(X-Y) ; Factor > 5) X+Y=Y ; Cancel out (X-Y) term > 6) 2Y=Y ; Substitute X for Y, by equation 1 > 7) 2=1 ; Divide both sides by Y Go back to step 4. What it really proves is 2x0=1x0. Now try this: 1) I am nothing without her. (Shakespeare) 2) ME - SHE = 0 (Algebraic representation) 3) ME = SHE (Added SHE to both sides) 4) M = SH (Divided by E) 5) MIT = SHIT (Multiplied by IT) So there you have the mathematical proof of what Shakespeare thought of the future American techincal education. :-) === Subject: Re: countable sets >> >Let E1, E2, E3, ... be a sequence of pairwise disjoint countable sets. Prove >that union of Ek as k = 1 to oo is countable. I do not need help proving >this. I need someone to explain exactly what is being asked. In Let E1, E2, >E3,... be a sequence of pairwise disjoint countable sets is that saying >that the numbers of Eis are countable or is it saying that each Ei has a >countable number of elements?? > >>Others have explained what is being asked, but I was intrigued by your >>statement that you dont need help proving it. Now that the question has >>been clarified, you might look again at proving it. >>Hint: If your proof does not invoke the axiom of choice, then your proof >>is wrong. >> >Not so. The countable union of disjoint countable sets >is in 1-1 correspondence with NxN -- the pair (n, k) maps to >the the k-th element of the n-th set (the set of sets and each >set is alreadyknown to be in 1-1- correspodence with N). This >is clearly surjctive, and the disjointness implies that its >injective. >And it does not require AC to show that NxN is countable. But you *have* used the axiom of choice in selecting infinitely many enumerations (one for each E_i). -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >>The canonical example is a spinner which can take on a random angle. >>This gets you [0,2pi) >I think we had this one last time we explored this concept. At some point the >quantum state of the universe takes over. There are a countable number of angles >your spinner can indicate. Not surprising that they all have non-zero >probability, and that one actually happened. Pray tell, what is the non-zero real number that sums countably to 1? And what does the quantum state of the universe have to do with rigorous mathematical definitions? At first I thought you were a finitist but now you turn out to be just another crank. Oh well. -- Im not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: Re: looking for an eltry soln to an old problem >A long time ago, J.J. Sylvester posed the problem: >if I have arbitrarily many 5 cent or 17 cent stamps, >what is the largest denomination I cannot make? >In general, if we have p and q cent stamps, it turns >out the answer is pq-p-q (granted p and q are coprime). >I have derived a solution >to the problem, but Id like to teach this to my >undergraduates, some of whom have a limited background. >So my question is: is there a very nice & friendly >proof of this fact? (For example, which avoids any >nonobvious facts from number theory.) You need a couple of not-quite-obvious facts. Fact 1: if x = a p + b q, then x = (a + n q) p + (b - n p) q for any integer n, and these are all the ways of writing x as a multiple of p plus a multiple of q. Fact 2: if p and q are coprime, every integer x can be written as x = a p + b q for some (not necessarily positive) integers a,b. This can be done using the Euclidean algorithm; or the fact that t -> t p mod q is one-to-one on the integers in {1,...,q-1} coprime to q, plus the pigeonhole principle. Suppose for some x it cant be done with a, b >= 0. If x = a p + b q is one representation with integers a,b, then for each n we have either a + n q < 0 or b - n p < 0. Take a to be the largest a + n q < 0, so -q <= a < 0 and x = a p + b q with b - p < 0. But then x <= -p + (p-1) q = pq - p - q. On the other hand, pq - p - q = ap + bq with a=q-1 and b=-1; a+nq < 0 if n < 0 and b-np < 0 if n >=0. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: How to choose a matrix P >Now I am thinking of this question. >Suppose >Y1=D - PEP >Y2=D-E >where D,E,P : nxn matrix >||P||=1 >How to choose P, so that >rank Y1 >= rank Y2 >and >||Y1||>= ||Y2|| The identity matrix would seem to work... Maybe you have some more requirements youre not telling us. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: two other sides of triangle >How we can find two sides of triangle if the third side, the radius of >inscribed circle and the angle opposite to the third side is given? In case this is homework, I give only a brief outline: Draw the angle bisectors from the endpoints of the known side. Drop a perpendicular to that side. Let x and y be the lengths of the two subsegments of the side so determined. We can now use trigonometry to determine the tangents of half of each of the unknown angles in terms of x and y. Use the tangent addition formula to get an equation expressing the tangent of the known angle in terms of x and y. Combine this with the fact that the length of the known side is x+y sines of the unknown angles. If this is a school assignment, please cite assistance received in submitted work. If not and you would like more details, ask again and I will gladly provide them. By the way, I am not sure this is the most elegant method. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Number property >I am searching for the most general set of numbers {a_i} (i=1...n) >, 0<=a_i<=2pi , such that : > Better to say < 2 pi, because 0 and 2 pi are equivalent in respect > to addition mod (2 pi). >a_i + a_j = a_t mod(2pi) for every i,j=1...n >and such that the set of all {a_t} obtained in this way is equal to >the set {a_i} we started with. > In other words, you want a finite sub-semigroup of T = R/(2 pi Z) > under addition. > Any finite sub-semigroup of a group is a subgroup (in fact any > finite semigroup such that t->a*t and t->t*a are one-to-one for all > a in the semigroup is a group). > So youre looking for the finite subgroups of T. >And these are all >of the form {2pi k/n: k=0...n-1} for positive integers n. Where can I find a demonstration of : And these are all of the form {2pi k/n: k=0...n-1} for positive integers n. ? Fabio. === Subject: re:How big can a manifold be? Only works for aleph1. It gives an example of a really big manifold, but still of cardinality 2^aleph0. My conjecture is that its as big as you get, but I dont know how to prove it. http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: No Set Contains Every Computable Natural <3_OdnTxoJNyqVK_dRVn-hA@comcast.com> <8765e4q2fe.fsf@phiwumbda.org> <8JCdncK7SMPAiK7dRVn-vw@comcast.com> Discussion, linux) >> The set of all natural numbers is recursive by fiat. >> Why bother to include computability in the definition? >> A human must decide if x is a member of this set >> before x can be given to the TM. >> Really? >> And if we give a tape with an infinite number of 1s on it, how does >> the human decide whether theres a 1 on every square (to the right of >> the start square) or just very many 1s to the right? What wonderful >> faculty have humans for determining the contents of an infinitely long >> tape? > Hey, thats my argument. > Using the logic OP have presented, I can choose the set of > all natural numbers to be any set I want. > (1,2,3) is the recursive set of all natural numbers. No, its not your argument. We have a well-defined notion of N, and also of functions N -> N. We have a well-defined notion of algorithms (in the form of Turing machines). We define what it means for an algorithm to represent a Turing machine (by defining which tapes represent which sequences of natural numbers) and hence derive a notion of computability, recursive sets, etc. These notions as stated make the claim that N is recursive trivial. Thats all there is to it. nonsense. It does not. > Proof: > This is the TM that decides if x is a member of (1,2,3). > 1) Print 1 > 2) Halt > You might think 1111 represents a natural number, > but you are wrong. 1111 is too large to be the > input for a TM. This is just more errant nonsense. -- These mathematicians are worse than communists, as how do you explain their behavior? I *am* the American Dream, fighting for what should be mine, having to get past weak-minded academics who are fighting to block my success. But I shall prevail!!! -- James S. Harris === Subject: Re: countable sets >And it does not require AC to show that NxN is countable. > But you *have* used the axiom of choice in selecting infinitely many > enumerations (one for each E_i). But if memory does not betray me, countable choice is a lot weaker than fullblown AC, right? Herman Jurjus === Subject: Re: How big can a manifold be? what about using whitneys embedding theorem? a jump in dimension will have no effect on the cardinality. === Subject: Re: the anticlassicalist }{ i: linguistic negation <103513r3gafocff@corp.supernews.com> <1035esdt30h4d38@corp.supernews.com> In message <1035esdt30h4d38@corp.supernews.com>, galathaea >Of course, you also snipped all the references to math and physics, Did I blink? I havent seen any physics yet. -- Richard Herring === Subject: Re: No Set Contains Every Computable Natural Discussion, linux) > An outright admission that a TM cant decide if a string is infinite. Duh. -- Run mathematicians, RUN!!! Im coming for you. It may take a few months, but Ill get [computer verification of my proof] and then your lives will be ended as you previously knew it. -- JSH meets PVS === Subject: Re: Bound of a sum >I admit I am no math genius, but I need to find some way to bound >k-1 n >sum sum 2/(j-i+1) >i=1 j=k+1 >1 <= k<= n. >Any upper bound that is some constant by n is OK. >>Write the sum as >>S = sum_{r=3}^n f(r,k)/r >>where f(r,k) = min(k-1,n-r+1) - max(0,k-r+1). >Oops, somehow I lost the 2. So my bound would be 2n. I dont know if if matters to the original poster, but I think I can improve the constant to log(4). Let us first reindex and then count how many times 2/j appears for each value of j to get k-1 n --- --- 2 > > ----- --- --- j-i+1 i=1 j=k+1 k-2 n-i --- --- 2 = > > - --- --- j i=0 j=k-i+1 n --- 2 = > - (min(n-j,k-2) - max(k-j+1,0) + 1) --- j j=3 n --- n-1 - |n-j-k+2| - |k-j+1| = > ------------------------- [1] --- j j=3 When j is near 3, the numerator of the summand in [1] is 2(j-2). When j is near n, the numerator of the summand is 2(n-j+1). The numerator reaches a plateau of 2 min(k-1,n-k). ____________ <--- 2 min(k-1,n-k) / / / / <--- 2 3 n Fig. 1: Numerator of the summand in [1] Therefore, when k = [(n+1)/2], the summation in [1] is maximum for the given n. For that k, when j <= k+1 the numerator is 2(j-2) and when j >= k+2, the numerator is 2(n-j+1). Computing the sum for k = [(n+1)/2], we get [1] k+1 n --- 2(j-2) --- 2(n-j+1) = > ------ + > -------- --- j --- j j=3 j=k+2 = n log(4) - 4 log(n) + 6 log(2) - 4 gamma + 1 - (16+(-1)^n)/2n + O(1/n^2) [2] Although a O(1/n) estimate would have probably been enough, I computed the 1/n term since I had expected the even n case to be about 1/n less than the odd n case since the triangle in Fig. 1 is truncated for even n. Estimate [2] verifies this. In any case, both n log(4) - 4 log(n) + 6 log(2) - 4 gamma + 1 and n log(4) are overestimates for the sum when n >= 3. Rob Johnson take out the trash before replying === Subject: Re: No Set Contains Every Computable Natural >> The set of all natural numbers is recursive by fiat. >> Why bother to include computability in the definition? >> A human must decide if x is a member of this set >> before x can be given to the TM. >> Really? >> And if we give a tape with an infinite number of 1s on it, how does >> the human decide whether theres a 1 on every square (to the right of >> the start square) or just very many 1s to the right? What wonderful >> faculty have humans for determining the contents of an infinitely long >> tape? > Hey, thats my argument. > Using the logic OP have presented, I can choose the set of > all natural numbers to be any set I want. > (1,2,3) is the recursive set of all natural numbers. > No, its not your argument. > We have a well-defined notion of N, and also of functions N -> N. Then why pretend that there is an algorithm that can decide them? > We > have a well-defined notion of algorithms (in the form of Turing > machines). We define what it means for an algorithm to represent a > Turing machine (by defining which tapes represent which sequences of > natural numbers) Suddenly, TMs only write sequences of natural nuumbers. > and hence derive a notion of computability, recursive > sets, etc. These notions as stated make the claim that N is recursive > trivial. N is a recursive set by fiat. Why bother to pretend N is computable? Obviously, a human has to decide if a symbol represents a natural number. No TM can do this. A TM doesnt have a well-defined notion of N. Russell - 2 many 2 count === Subject: Re: No Set Contains Every Computable Natural > An outright admission that a TM cant decide if a string is infinite. > Duh. But a human can, right? === Subject: Re: Ôerf function in C > Let phi(t) be the normal density, phi(t) = exp(-t^2/2)/sqrt(2*Pi), > so that > Phi(x) = integral phi(t), t = -infinity to x. > and > cPhi(x) = 1-Phi(x) = integral phi(t), t = x to infinity. > If you define > R(x) = cPhi(x)/phi(x) > then the function R is a well behaved function that has > a rapidly convergent Taylor expansion, an expansion > with terms easily determined by a two-step recursion. Very nice. But can someone explain why you need R = cPhi / phi rather than just S = Phi/phi ? The maths seems to work out just as well in the second case, and you dont need to keep on writing (1-) during it. Thus Phi = phi phi = -x phi therefore S = (phi Phi - Phi phi) / phi^2 = (phi^2 + x Phi phi) / phi^2 = 1 + x S giving you the recurrence in the terms of the power series for S. === Subject: Re: . 53ab2750 is an anonymous user . Yeah. My niece used to laugh like you ... when she was 2 years old === Subject: Re: x^2 + y^4 = z^4 === Subject: Re: x^2 + y^4 = z^4 >it is known that the equation x^2 + y^4 = z^4 has no >positive-integer solutions. >Depends on what youre willing to accept as proved. >(1) (z^4 - y^4)(y^2)(z^2) = (x^2)(y^2)(z^2). >But we also (always) have >(2) (2(y^2)(z^2))^2 + (z^4 - y^4)^2 = (z^4 + y^4)^2. >So, by (1), the Pythagorean triangle with sides 2(y^2)(z^2) and >z^4 - y^4 would have an area that is a square. Since 1 is not a >congruent number, this is impossible. Would you explain why? ---- === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >> ie, generate a random number by ßipping a coin to determine whether its >> bigger than 0.5, the again to determine whether its bigger than 0.25 (or >> 0.75), repeat forever. >> Theres the kicker - forever. When forever comes around, Ill let you know how >> you went wrong. Or you can say I told you so. Either way, its not going to >> happen. You never chose your number, youre still choosing it. >That argument leads to a denial of the existence of most real numbers. mmmm... in a sense it does. But then again in a sense, most real numbers dont have a proper existence. In terms of proportions, 100% of real numbers are not thought of, used, specified, and dont occur naturally as a measure of something. Im excluding extremely vague usage such as you specify Ôall reals larger than x. -- Patrick Hamlyn posting from Perth, Western Australia Windsurfing capital of the Southern Hemisphere Moderator: polyforms group (polyforms-subscribe@egroups.com) === Subject: Re: How fast is the Confinued Fraction factorization algorithm? >>My impression, based on not very scientific >>(or knowledgeable) experiment >>is that the Quadratic Sieve (with many polynomials) >>is unlikely to factorize number with more than 80 digits >>in a reasonable time on a reasonable computer >>(say 1 day on my 667MHz laptop). > What do you mean by unlikely? It makes no sense in this > context. QS succeeds with virtual certainty in time that depends > only on the size of n (with some small statistical variation). Thats just not true in my experience. And my comment makes perfect sense. The quadratic sieve completes in one or two minutes with some 70-digit numbers, and takes hours with others. Which is exactly what I would expect. What reason do you have to suppose the time has only small statistical variation? Have you actually tried it? What if the number is prime? (I gave random 60-digit numbers for factorisation as an assignment to a class of 30. One was prime, and there was a huge variation in the time taken - using quadratic sieve - to factorise the others.) -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: x^2 + y^4 = z^4 > + y^4 = z^4 has no positive-integer solutions. Is the proof of this > result short enough for some kind soul to post it, or need I make a trip > to the library? (I have citations.) Its relatively easy to prove there are no solutions using elliptic curves. Firstly, the given equation has no non-trivial integer solutions if and only if the equation y^2 = x^4 - 1 has no non-trivial rational solutions. The transformation u = 1/(x-1), v = y/(x-1)^2 brings this to standard elliptic form v^2 = cubic in u, and a little manipulation brings this to y^2 = x^3 + 2x^2 + 4x. The standard method (as described in Silverman & Tate) with the associated curve y^2 = x^3 - 4x^2 - 12x shows that this curve has rank 0, and it is easy to verify that there are no points of finite order except for the point (0,0) of order 2 and the zero point. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: the anticlassicalist }{ i: linguistic negation > -=-=-=-=-= linguistic negation =-=-=-=-=-=- > The logic of natural language has been studied by many different schools of > thought throughout history. > That they have one and all avoided coming to any unique conclusion > emphasizes the bankruptcy of the Liberal Arts. Its a hen party. > Anybody can gossip and all are appreciated. Its bullshit. How would > you like to be doing 80 mph in a car that was built by mob rule > instead of cold, hard, dry engineering? > Oh, but that example is different! It certainly is. Liberal Arts are > bullshit and engineering is real world. > Everything in the Liberal Arts and Social Sciences is both right and > wrong. No conclusion at all can be drawn except the need for more > funding. Say what you want - it makes no difference at all (except > for grant funding - then you must agree with the mob or be destroyed). > Klingon was specifically created to be the worst language possible by > folks who knew linguistics. It was enthusiastically embraced by the > mob and it is as good as Korean or Chinese for transferring content. Well, actually, Uncle Al, youre wrong about this. But then Im just a linguist so what the fuck would I know? And youre a fucking chemist or some sucha whizpoop hot-lead scientologician, which means that your opinions about linguistics are worth-- what, 3.978654549 kazillion gold dollars to the peso? -- over anything that somebody who knows something about the subject might say. Okrand says he violated a few human language universals in inventing Klingon, since of course it wasnt meant to be a human language. Nothing about worst language possible. Studies of how Klingon-users actually speak might be interesting. Do you know of any? And anyhow what was this meant to prove, even if it were all true? Sure as hell wouldnt prove whatever you were salivating about in the previous paragraph anyway. Ross Clark === Subject: Re: How many ways to put 5 balls into 500 ordered cups? >> You are given 500 numbered cups and five identical balls. Any cup >> can hold up to five balls. How many ways can you put the five balls >> into the 500 cups? > Lets make it more interesting: How about *ten* balls (but each cup can > still only hold up to five balls)? > -Michael. Ah, generating functions. Now youve given me something to think about all day. - Tim -- Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University === Subject: Re: looking for an eltry soln to an old problem >A long time ago, J.J. Sylvester posed the problem: >if I have arbitrarily many 5 cent or 17 cent stamps, >what is the largest denomination I cannot make? >In general, if we have p and q cent stamps, it turns >out the answer is pq-p-q (granted p and q are coprime). >I have derived a solution >to the problem, but Id like to teach this to my >undergraduates, some of whom have a limited background. >So my question is: is there a very nice & friendly >proof of this fact? (For example, which avoids any >nonobvious facts from number theory.) I answered this question recently. Take a look at Rob Johnson take out the trash before replying === Subject: Re: any good book on general optimization, continuous multivariate optimization? >I am studying general optimization, particularly continuous multivariate >optimization, something like First Order Nessecery Condition, Second Order >Nessecary Condition, Second Order Sufficient Condition for local >minimizers... etc. I got really confused. The reasons are >1) Multivariate derivatives, gradient, etc. are what I am not very familiar >with, so I got confused; >2) Some boundary condition and degenerate cases got me confused: for >example, it is easy to have FONC, SONC, SOSC for interior points, thats >easy, but what if the FONC, SONC, SOSC on boundary point, involving the >directional derivative vector d, etc. And some degenerate cases such as the >Hessian is indefinite... etc. And the relationship between FONC, SONC, SOSC, >etc. these all got me confused... >Please point me to some good books/online lecture notes, etc. that can help >me clearify these concepts? >-Walala http://plato.la.asu.edu/topics/tutorials.html hth peter === Subject: Re: two other sides of triangle > How we can find two sides of triangle if the third side, the radius of > inscribed circle and the angle opposite to the third side is given? Let c be the known side, C the corresponding angle and r the inradius. Then you have the equations (a, b being the unknown sides) : ab sinC = r (a + b + c) --> a + b = ab/r sin C - c (1) (Formula for inradius) c^2 = (a + b)^2 - (2 + 2cosC) ab (2) (Cosine rule) Substitute in (2) the formula for a+b given in (1) and you have a (remarkably simple !) quadratic in ab. Solve, evaluate a+b using (1) and then solve x^2 - (a+b)x + ab = 0 whose two roots will give you the sides. Hope this helps, Dimitris === Subject: Re: No Set Contains Every Computable Natural <871xosptgg.fsf@phiwumbda.org> Discussion, linux) >> An outright admission that a TM cant decide if a string is infinite. >> Duh. > But a human can, right? I cant imagine how. Tell you what, though. Lets test it. Why dont you create a few input tapes, some of which contain an infinite number of 1s? Then traipse down to your local research institute and ask the good fellas if they can determine which of your tapes have an infinite number of 1s and which do not. Clearly, whether a string is infinite is easily refutable but not decidable. Also clearly, all this talk about infinite strings is mere idealization. -- Sure, [my Usenet presence is] like Shaq playing against you in your backyard, but that has its perks, as I find ways to have my fun *and* I can send messages to certain people in the United States Government without concern that the rest of you understand them. -- James Harris === Subject: [PhD-position] Statistics. University of Twente. Netherlands Announcement of PhD-position Statistical analysis of insurance portfolios University of Twente, The Netherlands The Department of Applied Mathematics of the Faculty of Electrical Engineering, Mathematics and Computer Science at the University of Twente, Enschede, The Netherlands has available a PhD-position For information contact: prof.dr. W. Albers (w.albers@math.utwente.nl, tel. 053-4893816/3434). Title: Statistical analysis of dependence effects on insurance portfolios Research Group: The project will be performed in the chair Statistics and Probability. It fits in the research programme Financial Engineering from the Centre for Telematics and Information Technology (CTIT), one of the key research institutes of the University of Twente. The Financial Engineering Laboratory (FELab) is a joint initiative of researchers in this field within the School of Mathematical Sciences and the Department of Finance and Accounting (FMBE) at the School of Technology & Management. At present, 13 full- and 6 part-time faculty members, as well as 6 PhD-students, participate in this laboratory. Supervision: prof.dr. W. Albers (UT; email: w.albers@math.utwente.nl) dr. W.C.M. Kallenberg (UT; email: w.c.m.kallenberg@math.utwente.nl) Funding source: This position is funded by the Technology Foundation STW, applied science division of NWO and the technology programme of the Ministry of Economic Affairs. Period: The project can start immediately; it will take 4 years. Required: The candidate should have a MSc in mathematical statistics, probability or econometrics, and a keen interest in applications Description: Quantities of interest for insurance portfolios are among others the sum S of the claims of the individual risks during a reference period and the stop-loss premiums E{max(0,(S-a))} for various retentions a. Typically it is assumed that the summands in S are independent, although it is clear that in practice dependencies will occur. Implicit in this assumption apparently is the hope that the effects of such dependencies will be negligible, or at most small for practical purposes. Unfortunately, however, this is often not the case: introducing a small dose of dependence may already lead to gross deviations of the stop-loss premiums. To deal with the problem, families of models incorporating dependencies need to be proposed and studied, with an emphasis on practical usefulness. Typically, a balance will have to be found between degree of complication and realistic content. Next, techniques have to be found to obtain adequate approximations to the above mentioned quantities of interest. This is also a nontrivial part, as already in the independent case exact results are generally intractable. Finally, estimation methods have to be developed to fit the models obtained to data occurring in practice. As concerns the first two of these three stages in the above program, some progress has already been made in [1]. Here a simple stochastic mixture model is proposed, which is demonstrated to incorporate to first order many models in common practical use Second order Edgeworth expansions turn out to be quite successful for obtaining adequate approximations to stop-loss premiums, as long as normally distributed claims are used. In [2], the scope is widened considerably by including various other approximations from the literature. For a wide variety of claim size distributions and retention levels, such approximations are compared in this paper to each other, as well as to a quantitative criterion. The third step, fitting such a model to real data, still has to be taken. It is expected that this will generate a new iteration of the program by suggesting directions in to which the basic model should be altered or generalized. [1] Albers, W. (1999). Stop-loss premiums under dependence. Insurance: Math. & Econ. 24, 173-185. for stop-loss reinsurance premiums. TW-report 1695. === Subject: Re: No Set Contains Every Computable Natural <3_OdnTxoJNyqVK_dRVn-hA@comcast.com> <8765e4q2fe.fsf@phiwumbda.org> <8JCdncK7SMPAiK7dRVn-vw@comcast.com> <87hdxovh3h.fsf@phiwumbda.org> Discussion, linux) >> No, its not your argument. >> We have a well-defined notion of N, and also of functions N -> N. > Then why pretend that there is an algorithm that can decide them? >> We >> have a well-defined notion of algorithms (in the form of Turing >> machines). We define what it means for an algorithm to represent a >> Turing machine (by defining which tapes represent which sequences of >> natural numbers) > Suddenly, TMs only write sequences of natural nuumbers. Who the fuck said that? Read, son, read. We adopt a convention that certain of the tapes produced represent natural numbers. Other tapes do not represent natural numbers. If we want to talk about representations of functions N -> N, then we ignore those outputs which dont represent natural numbers (typically, we adopt the convention that function defined by a Turing machine converges iff the machine halts and the tape thus produced represents a natural number). >> and hence derive a notion of computability, recursive >> sets, etc. These notions as stated make the claim that N is recursive >> trivial. > N is a recursive set by fiat. I think thats a fair assessment. We have the mathematical object N. We have the (also mathematical) notion of Turing machines which produce marks on tapes. If we want to reason about computability of functions involving the former using the latter, then we must choose a representation of natural numbers by tapes. If we ask whether N is recursive, then it depends on what context we mean. We may mean whether there is a computable function N -> N which returns 1 iff its input is in N and 0 else. To prove that there is such a function, we need to reason about Turing machines, but only about their behavior on input tapes which represent natural numbers. On the other hand, we might ask whether there is a total function which, on *any* input tape whatsoever, returns 1 iff the tape represents a natural number and 0 else. Now, if any includes tapes which have infinitely many ones at the start, then clearly the answer is no. In *this* sense, N is not recursive. Again, there is a difference between the following two questions: (1) Does there exist a TM t such that, for every input tape i which represents some natural number n, t(i) = 1? (2) Does there exist a TM t such that, on every input tape (including those with an infinite number of ones) i, t halts and outputs 1 iff i represents some natural number? > Why bother to pretend N is computable? I dont know what it means for N to be computable. I think its pretty clear what I mean when I say N is a recursive set. Evidently, its not clear to you, but maybe the above helps. > Obviously, a human has to decide if a symbol represents > a natural number. No TM can do this. No TM can devise a convention, if thats what you mean. At least, no TM can devise a convention in the sense that matters here. > A TM doesnt have a well-defined notion of N. I wouldnt want to claim that a TM has any notions at all. -- But he himself was not to blame for his vices. They grew out of a personal defect in his mother. She did her best in the way of ßogging him while an infant... but, poor woman! she had the misfortune to be left-handed, and a child ßogged left-handedly had better be left unßogged. -- E.A. Poe === Subject: Re: Is there a Second Order Sufficient Condtion(SOSC) for boundary and non-interior points? >Hi all, >I only learned SOSC for interior points, but I find no where about SOSC for >boundary and all(boundary, noninterior) points... and how about First Order >Sufficient Condition for local minimizer? first order sufficient? -> convex case plus slaters condition >-WAlala if you assume a constraint regular constraint point it is easy: let the constraints be h(x)=0 h has p components and g(x) <=0 g has m components assume that the gradients of the h_j and those g_i with g_i(x*)=0 are linearly independent . assume there are a components g then you can elimininate from the system h(x)=0 g_i(x)=0 for i such that g_i(x*)=0 p+a variables , with n-p-a remaining free. (if you have problems with the implicit function theorem which covers this: think of those equations as being linear... ) then you can express the objective function as one depending only on the free variables. apply the criteria for the unconstrained case to this so called reduced function. write down the result. then translate everything back in terms of the original functions. you obtain the first order necessary, the second order necessary and the second order sufficient conditions this way considering the active inequality constraints as equality constraints. it remmains to consider strictly feasible directions with respect to the active inequality constraints. if you assume strict complementarity, then applying taylors theorm once more you see that in the case considered here these apply also to the inequality constrained case. hth peter === Subject: Re: Mean Value Theorem >Two places Ive seen the equivalent statement: If f is continuously >differentiable on [0,1], then |f(x)-f(y)|<=sup|f(u)||x-y| where the >sup is extended over 0<=u<=1. >I cant see why we need f to be continuous. It seems by the MVT we >have > f(x)-f(y)=f(c)(x-y) for some c in [0,1], >so that > |f(x)-f(y)| = |f(c)||x-y|, for some c in [0,1], > <= sup_u|f(u)||x-y|. Thats correct (noting that if f is not continuous the expression on the right may be infinite; the reason people state the thing the way they do is so they dont have to consider the question of what the sup of an unbounded function is, also because saying that something is <= infinity doesnt seem all that interesting.) ************************ David C. Ullrich === Subject: Re: how to understand this statement of Second order Necessary Condition? >I read the statement of SONC for local minimizer as follows: >Let x* a local minimizer of function f over some constraint set O, and d a >feasible direction at this point x*. If d*gradient=0, then d*H*d>=0. >where d is the transpose of d column vector... >I got confused here because this seems to me is a half statement: >If d*gradient=0, then d*H*d>=0. >but what if d*gradient>0 but not =0 at that point x*? >Is this a possible case in the SONC statment? Is that true that when > d*gradient>0 but not =0, there is no statement about the SONC? d is a feasible direction if for some t0 and 0<= t <= t0 x*+t*d is feasible. now consider f(x*)<= f(x*+t*d) (by assumption) = f(x*)+g*d*t + d*H*d*t^2 + o(t^2) by Taylors theorem for two times continuously differentiable t. let t>0, subtract f(x*) from both sides, divide by t and let t tend to zero... done hth peter === Subject: Re: countable sets > Let E1, E2, E3, ... be a sequence of pairwise disjoint countable sets. Prove > that union of Ek as k = 1 to oo is countable. I do not need help proving > this. I need someone to explain exactly what is being asked. In Let E1, E2, > E3,... be a sequence of pairwise disjoint countable sets is that saying > that the numbers of Eis are countable or is it saying that each Ei has a > countable number of elements?? >> Others have explained what is being asked, but I was intrigued by your >> statement that you dont need help proving it. Now that the question has >> been clarified, you might look again at proving it. >> Hint: If your proof does not invoke the axiom of choice, then your proof >> is wrong. > Not so. The countable union of disjoint countable sets > is in 1-1 correspondence with NxN -- the pair (n, k) maps to > the the k-th element of the n-th set (the set of sets and each > set is alreadyknown to be in 1-1- correspodence with N). This > is clearly surjctive, and the disjointness implies that its > injective. See my other post. You used AC when you selected for each i a particular bijection between E_i and N. > And it does not require AC to show that NxN is countable. Agreed. The problem lies elsewhere. -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: covering compact set w/squares >Given a compact set K in the plane s.t. each pt x is the center of a square >Q_x, prove that you can find a subsequence Q_x_i of squares s.t. K is >covered by the Q_x_i and the sum of the areas of each of the Q_x_i is no >more than 4 times the area of the union of the Q_x_i. Are you sure that this is exactly what youre supposed to show? The reason I ask is that its not entirely clear to me that this statement is true, but I know a true fact that I can imagine one could possibly misread as the statement above. >I think this is a totally geometric thing, and I know it suffices to show >that in such a minimal cover no point is contained in more than 4 squares, >but how do i show this? Im not sure whether what you say youre trying to prove is true, but I can tell you that its _not_ true that in a minimal cover by squares no point is in more than four squares! For example, let Q_1, ... Q_5 be squares of side length 1, such that the lower left corner of Q_j is at the point (j/100, j/100), and let K be the union of the Q_j. Then the Q_j form a minimal cover of K, but there are points of K contained in all five of the Q_j. >I know how to do this with intervals on the real >line (there the answer is 2 times the area....), but i get lost when going >up a dimension. Re-read the statement of what youre supposed to prove and let us know whether its really exactly the above. ************************ David C. Ullrich === Subject: Re: Mean Value Theorem >>Two places Ive seen the equivalent statement: If f is continuously >>differentiable on [0,1], then |f(x)-f(y)|<=sup|f(u)||x-y| where the >>sup is extended over 0<=u<=1. >>I cant see why we need f to be continuous. It seems by the MVT we >>have >> f(x)-f(y)=f(c)(x-y) for some c in [0,1], >> Right. >>so that >> |f(x)-f(y)| = |f(c)||x-y|, for some c in [0,1], >> <= sup_u|f(u)||x-y|. >> But without continuity (or something extra beyond differentiability) >> how can you conclude that the derivative f is bounded on [0, 1]? You >> have to have boundedness in order to justify your last step... >If f is not bounded, then we can assign the value Infinity to sup_u|f(u)| >and then the relation still holds. >So it seems that the differentiability condition is there purely for >technical reasons (read: rigor), to avoid having to deal with Infinity >because it does not belong to R. [Continuing the rambling] Well we can always, compactify R by adjoining -infty and infty, so that should not be a problem. But what if x = y? We get 0 <= infty 0 What to make of that? [Stopped the rambling] G. Rodrigues === Subject: Sets That Resemble Derivatives Somewhat by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1IDNQn23184; Define: 1) (A-->B) = (AB) = A U B for any two sets A, B, where Ô is complement and Adjacent letters are intersected. This is the set analog of logical a-->b for a, b propositions, the latter defined as ~(a ^ ~b) = ~a V b with ~ being not (negation), ^ and (conjunction), V or (disjunction including the possibility of conjunction). Then it is fairly easy to prove that: 2) [(A-->B)(A-->B)] = AB U AB 3) [(A-->B)(A-->B)] = AB U AB Equation (2) resembles the derivative or derivation if Ô (comple- ment) takes the place of derivative and the set on the left hand side inside brackets replaces AB, somewhat like (fg) = fg + fg for differentiable functions f, g. Equation (3) puts an even number of primes on expressions that look like AB on the right hand side, that is to say even from 0 to 2. The latter generalizes one idea of the derivative. Derivations are a topic in the literature on Lie Algebras and other structures, but (2) and (3) have a rather different ßavor. Notice on the left-hand-side inside the brackets there are 0 or 2 primes in one pair of parentheses in (2), while in (3) there is one prime in one pair of parentheses, a sort of reverse of the right hand side situation for (2) and (3). Could the complement of a set be a generalized derivative? Well, f(x + h) and f(x - h) get compared with f(x) in a derivative for real functions, and the complement of a set is certainly outside the set in a sense similar to the way x + h and x - h are outside x (to left and right). On the other hand, since the complement of a set is the universe minus the set, we dont really have to compare the complement with the original set since in a sense the latter is determined once the former is, so we can just look at f(x + h) and not f(x) figuratively speaking. To see this more explicitly, notice that: 4) AB U AB U AB U AB = universe which you can visualize rapidly by using Venn diagrams or prove by elementary set theory, for any two different sets A, B. Osher Doctorow === Subject: Re: Perplexing Patterns of Square Numbers by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1IDNQK23203; I tried to follow these messages about perplex patterns but I am a little lost. Could someone summarize what has already been done about this subject, that is how many patterns have been found for n from 1 to 20 (or something like that) and also how many one can expect to discover. Joaquim Nogueira === Subject: Re: e is transcendental (was: classes of transcendental numbers ? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1IDNRr23256; So? The real part of exp(i pi) is cos(pi), and its imaginary part >is sin(pi), so all you are saying is that cos(pi) = -1 and >sin(pi) = 0, and we were already aware of these facts. There is >NO reason to conclude that exp(i pi) = 0. > I,have a reason , with my due respect. >> Panagiotis Stefanides Yes, but you DONT tell us what your reason is. You cant expect us to >accept your claims without giving support for those claims. So what >possible reason could you have for expecting us to agree with your claim >that exp(i pi) = 0? >The reason is simply that exp(ipi0=-1 should be accompanied >>by the statement that this is the real part solution. >>Is it fair? No, that is not a fair comment. exp(i pi), the complex number, is >equal to -1, the complex number. There is no need to appeal to the >real part. Also, what does exp(i pi) = -1 is the real part solution >mean? You look like you are using terminology in a manner not >recognized in mathematics. David McAnally -------------- >>e^[i*pi] ,as accepted ,is a phasor. >In most of the relevant fields of mathematics, e^[i pi] is a complex >number. >>It is only fair to state that its >>polar representation is : >>e^[i*pi] = MOD 1 , ARG 180 . >That is Arg 180 degrees, not just Arg 180. And so what? That does >not lead to your assertion that exp[i pi] = 0, a result for which >you have given absolutely no support. Why dont you just give up? >David McAnally > At the moment, they (the Time Lords) are far from being all-powerful. > Thats why its been left up to me and me and me. > quote by: Patrick Troughton in The Three Doctors >------- >>I, have made myself very explicit. >>My original question of the implication >>of the imaginary component: e^[ipi]=j*0 (to the related proof) Complex Notation, chapter 12 ,Electrical Technology 3RD ED. Edward Hughes Longmans ,page 338: Stares: OA*=OB+jOC=OA(COStheta+jSINtheta) * Symbols representing phasors are printed in bold face italics, while those representing only magnitudes are printed in ordinary italics,.. Here is very clear the difference between PHASOR and MAGNITUDES [which(MAGNITUDES) I, referred to as REAL PART or IMMAGINAR PART, I should have stated COMPONENTS ]. >Perhaps you mean that (the imaginary part of e^[i pi]) = 0, in which >case you are correct, but you have had a lot of problem expressing >yourself, especially in view of the way that you initially made >the claim by stating that e^[i pi] = 0, which you described as the >imaginary part solution, using a terminology that nobody but you >knows. e[i*theta]=COStheta+iSINtheta thetas could be given and calculations could be performed for numerical evaluations. In books is stated that it is FORMULA and also terms such as evaluate:(-1+i*sqrt[3])^10 Are these not solutions to problems? >Nobody knows what you mean when you make a statement like exp[i pi] = -1 >is the real part solution of exp[i pi] = -1, or that exp[i pi] = 0 >is the imaginary part solution of exp[i pi] = -1. I asked you to >explain your terminology but you havent bothered. I, doubt it but ,still I, exlpained it anyway. >As I, exlained earlier, but I, am not the kind of not being bothered >When I take the imaginary part of the equation exp[i pi] = -1, I get >sin(pi) = 0, a fact which is already known to us. >Have you thought also of the fact that if you have exp[i pi] = -1 (your >real part solution) and exp[i pi] = 0 (your imaginary part solution), >then you could conclude that 0 = exp[i pi] = -1, and get a contradiction? Well tthere is always the possibility of an answer such as MULTIVALUED , examplum gratias: e^[2i*pi]=1 ln[1]=0=2i*pi I, make use of the comlex notation , but still ,I, have my natural questions ,and do not accept everything for granted. I, give an example in the form of question: It is required that COS[-i]+i*COS[-i] be evaluated so that is its Modulus and Argument be evaluated (NUMERICALLY]. Panagiotis Stefanides http://www.stefanides.gr/why_logarithm.htm >>was not answered and still it is open. >Because under no circumstances can you ever claim that exp[i pi] = 0. >David McAnally === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs > ie, generate a random number by ßipping a coin to determine whether its > bigger than 0.5, the again to determine whether its bigger than 0.25 (or > 0.75), repeat forever. > > Theres the kicker - forever. When forever comes around, Ill let you know how > you went wrong. Or you can say I told you so. Either way, its not going to > happen. You never chose your number, youre still choosing it. >>That argument leads to a denial of the existence of most real numbers. > mmmm... in a sense it does. But then again in a sense, most real numbers dont > have a proper existence. In terms of proportions, 100% of real numbers are not > thought of, used, specified, and dont occur naturally as a measure of > something. Im excluding extremely vague usage such as you specify Ôall reals > larger than x. Do you agree that the intermediate value property holds? That is, if f: [a,b] -> R is continuous, f(a) < 0, and f(b) > 0, then do you agree that there is a real number x in [a,b] such that f(x) = 0? -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: We come from your future > The Question is: What is The Question? Wheeler 43 === Subject: Re: covering compact set w/squares >Given a compact set K in the plane s.t. each pt x is the center of a square >Q_x, prove that you can find a subsequence Q_x_i of squares s.t. K is >covered by the Q_x_i and the sum of the areas of each of the Q_x_i is no >more than 4 times the area of the union of the Q_x_i. Ah. This is not true (hence the question: what you were really asked to show?). A minute ago I gave a counterexample to the statement below; a slight modification gives a counterexample to the theorem: Let Q^1, ... Q^5 be squares of side length 1, such that the lower left corner of Q^j is the point (j/100, j/100). Let K be the union of Q^1, ... Q^5. Now if x is in K and x happens to be the center of one of the Q^j define Q_x = Q^j. If x is in K and x is not the center of one of the Q^j let Q_x be the square with center x and side length 1,000. Its easy to see that there is no subcollection of the Q_x that covers K and has sum of the areas less than four times the area of K: We certainly cant include any of the Q_x with side length 1000, but if we dont use any of them we need to use all five of the Q^j to cover K, and the sum of the areas is almost five times the area of K. >I think this is a totally geometric thing, and I know it suffices to show >that in such a minimal cover no point is contained in more than 4 squares, >but how do i show this? I know how to do this with intervals on the real >line (there the answer is 2 times the area....), but i get lost when going >up a dimension. ************************ David C. Ullrich === Subject: Re: covering compact set w/squares >Given a compact set K in the plane s.t. each pt x is the center of a square >Q_x, prove that you can find a subsequence Q_x_i of squares s.t. K is >covered by the Q_x_i and the sum of the areas of each of the Q_x_i is no >more than 4 times the area of the union of the Q_x_i. Sorry to reply three times, but the counterexample I gave was utterly stupid, since it can be hugely simplfied: Let K be any compact set with area = 1, and define each Q_x to have side length 2. Duh. >I think this is a totally geometric thing, and I know it suffices to show >that in such a minimal cover no point is contained in more than 4 squares, >but how do i show this? I know how to do this with intervals on the real >line (there the answer is 2 times the area....), but i get lost when going >up a dimension. ************************ David C. Ullrich === Subject: re:puzzle: GCDs of Infinite Set of Integer Pairs My initial guess is 1 - pi*sin(sqrt(6)) / sqrt(6). Based on a heuristic argument. http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: Sets That Resemble Derivatives Somewhat > Define: > 1) (A-->B) = (AB) = A U B > for any two sets A, B, where Ô is complement in what? > and Adjacent letters > are intersected. This is the set analog of logical a-->b for a, > b propositions, the latter defined as ~(a ^ ~b) = ~a V b with > ~ being not (negation), ^ and (conjunction), V or (disjunction > including the possibility of conjunction). > Then it is fairly easy to prove that: > 2) [(A-->B)(A-->B)] = AB U AB > 3) [(A-->B)(A-->B)] = AB U AB > Equation (2) resembles the derivative or derivation if Ô (comple- > ment) takes the place of derivative and the set on the left No. The right side looks similar, Ôcos of your denoting complement by the same notation as derivative, but the right side is completely different! > hand side inside brackets replaces AB, somewhat like (fg) = > fg + fg for differentiable functions f, g. Equation (3) puts > an even number of primes on expressions that look like AB on > the right hand side, that is to say even from 0 to 2. The > latter generalizes one idea of the derivative. One idea? Not the idea of it being a limit of difference quotients methinks. > Derivations are a topic in the literature on Lie Algebras and other > structures, but (2) and (3) have a rather different ßavor. I.e., by not resembling derivation at all? > Notice on the left-hand-side inside the brackets there are 0 or > 2 primes in one pair of parentheses in (2), while in (3) there > is one prime in one pair of parentheses, a sort of reverse of the > right hand side situation for (2) and (3). applause! > Could the complement of a set be a generalized derivative? Well, > f(x + h) and f(x - h) get compared with f(x) in a derivative for > real functions, and the complement of a set is certainly outside > the set in a sense similar to the way x + h and x - h are outside > x (to left and right). Like er yeah! I mean yeah if x + h is different to x then er yeah then x + h is outside x. Wow, groovy! > On the other hand, since the complement > of a set is the universe minus the set, we dont really have to > compare the complement with the original set since in a sense the > latter is determined once the former is, so we can just look at > f(x + h) and not f(x) figuratively speaking. To see this more > explicitly, notice that: > 4) AB U AB U AB U AB = universe Aha! The universe itslef. Next youll be giving us the answer to life the universe and everything. > which you can visualize rapidly by using Venn diagrams or prove by > elementary set theory, for any two different sets A, B. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Help needed - primes program Hallo, I need help with the following : I made a program which gives all the primes from selected area. I made it in Pascal. The only problem is that it hasnt a very good algorithm which I need. Here is the code: ------------------------------------------------------------- --------------- --- Program Primes; Var x,y:word; f:word; i,n:word; ok:boolean; Begin Write(ÔEnter X and Y -> Ô); Readln(x,y); For i:=x to y do Begin f:=2; ok:=true; While f<=i-1 do Begin If i mod f = 0 then Begin f:=i; ok:=false; End else f:=f+1; End; If ok=true then Write(i, is a prime. Ô); End; Readln; End. ------------------------------------------------------------- --------------- --- I wanted to give you attachment with the *.EXE file but I cant so if ypu need to compile this code download Free Pascal (current version is 1.0.10) form www.freepascal.org . Any suggestion and similar is welcome. === Subject: Re: covering compact set w/squares >Given a compact set K in the plane s.t. each pt x is the center of a square >Q_x, prove that you can find a subsequence Q_x_i of squares s.t. K is >covered by the Q_x_i and the sum of the areas of each of the Q_x_i is no >more than 4 times the area of the union of the Q_x_i. Aargh. I think Ill just spend the rest of my life posting replies to this message. I have to say that the two counterexamples Ive given were wrong (I was reading the statement of the problem Note however that the example showing that its not true that in a minimal cover no point is in more than four squares was correct. (Im still not sure whether the result itself is true and I still wonder whether its really what you wanted to prove.) >I think this is a totally geometric thing, and I know it suffices to show >that in such a minimal cover no point is contained in more than 4 squares, >but how do i show this? I know how to do this with intervals on the real >line (there the answer is 2 times the area....), but i get lost when going >up a dimension. ************************ David C. Ullrich === Subject: Re: How fast is the Confinued Fraction factorization algorithm? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1IDjpt25287; >My impression, based on not very scientific >(or knowledgeable) experiment >is that the Quadratic Sieve (with many polynomials) >is unlikely to factorize number with more than 80 digits >in a reasonable time on a reasonable computer >(say 1 day on my 667MHz laptop). >> What do you mean by unlikely? It makes no sense in this >> context. QS succeeds with virtual certainty in time that depends >> only on the size of n (with some small statistical variation). >Thats just not true in my experience. And how much is that? >And my comment makes perfect sense. Sorry, but it makes ZERO sense. >The quadratic sieve completes in one or two minutes >with some 70-digit numbers, and takes hours with others. This is grossly false unless there is a major problem with your implementation. While there is some variation, depending on the quality of the multiplier, that variation is at worst a factor of 2 to 3. On a 1 GHz Pentium, I expect 70 digits to take about 20 minutes to an hour. Your claim of 1-2 minutes *seems* ludicrous, as does the claim of hours. Are you sure you are not talking about ECM? Such variation with ECM would not surprise me at all. In fact, I would expect it. >Which is exactly what I would expect. Huh? QS runs in time that depends on the size of the number being factored. Why would you expect such a large variation? >What reason do you have to suppose the time has only >small statistical variation? >Have you actually tried it? ROTFL. I have factored more numbers with QS than anyone else in the world to date. I have more experience with the method than anyone else. I suggest you read my paper R. Silverman The Multiple Polynomial Quadratic Sieve Math Comp 1987 In particular, read the section discussing the Knuth-Schroeppel function and its meaning/implications. >What if the number is prime? Then the final congruences that get constructed, i.e. x^2 = y^2 mod N will ALWAYS have x = y or x = -y mod N. When N is prime, there are only two solutions to x^2 = a mod N for any a. If N =pq there will be 4 solutions. Two of these will be x = y or x=-y mod N. The other two yield factors. [hint: think about the Chinese Remainder Theorem] >(I gave random 60-digit numbers for factorisation >as an assignment to a class of 30. >One was prime, and there was a huge variation >in the time taken - using quadratic sieve - >to factorise the others.) Which only shows that there was wide variation in the quality of the implementations. QS does have some variation, depending on the value of the Knuth-Schroeppel function. But it does not vary by more than a factor of 3 for numbers of fixed size. === Subject: Who can tell me the recent develepoment on the GCH? W. Huge Woodin said he gave a solution that yield CH is false. What was it gonging on? === Subject: Re: Simple numbers Hallo; Well, p(n) is a n-th prime but if n=123456 how do I know what is the 123456-th prime? I need the correct number (for exampel 1-st prime is -> 2)! > Hallo, > I know that every simple number (beside 2 and 3) has its one formula : > S.N = > 6*n +/- 1 for some n , but not any n={1,2,3,...+} in formula 6*n +/- 1 > gives > the simple number (for example: n=6 => 6*6-1=35 , 35 is not simple , > 5,7|35) > By simple number I think you must mean prime. > so my question is: > If n={1,2,3,...} what is the formula which gives the simple numbers > (beside > 2 and 3) for any n? > Evidently you dont consider p(n) is the nth prime to be a formula. > Any > particular reason? > Computational complexity? > Jon Miller === Subject: Re: Help needed - primes program > Hallo, > I need help with the following : > I made a program which gives all the primes from selected area. I made it in > Pascal. The only problem is that it hasnt a very good algorithm which I > need. I am not going to do your homework for you, but I marked below some obvious areas where you can improve your algorithm. > Here is the code: > ------------------------------------------------------------- --------------- > --- > Program Primes; > Var > x,y:word; > f:word; > i,n:word; > ok:boolean; > Begin > Write(ÔEnter X and Y -> Ô); > Readln(x,y); > For i:=x to y do > Begin > f:=2; > ok:=true; > While f<=i-1 do You can break off the search much before you get to i-1. > Begin > If i mod f = 0 then > Begin > f:=i; > ok:=false; Once you get here, why do you continue to iterate? > End > else > f:=f+1; Why do you bother with even numbers and multiples of 3 at all? > End; > If ok=true then Write(i, is a prime. Ô); > End; > Readln; > End. === Subject: re:How big can a manifold be? Im mainly interested in 1-dimentional manifolds anyway. I dont see how Whitneys theorem helps here. http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: Goldberg dual Re:http://www.geocities.com/nicholasshea1/index.html and the radius for any number of points? on whether the convex hull is allowed to have square faces or whether > those faces are triangulated. There is a broader generalization to be made about this. The ratio between an average edge length of a tesselated sphere and the radius of the sphere, call it beta, will vary between a maximum value and a minimum values. The maximum value will be come from the case of a completely triangulated arrangement of edges such as in a typical geodesic sphere. The minimum value for beta will be for a tesselation with all 3-way vertexes such as in a Goldberg-like structure. The first case has the greatest edge density(edges/area) and the least volume. The second case has the least edge density(edges/area) with the greatest volume. Beta is always somewhere between these extremes. For any number of points distribute on a sphere, beta will be within a finite range. This range depends on the arrangement of the edge. Dick === Subject: Re: How fast is the Confinued Fraction factorization algorithm? >>My impression, based on not very scientific >>(or knowledgeable) experiment >>is that the Quadratic Sieve (with many polynomials) >>is unlikely to factorize number with more than 80 digits >>in a reasonable time on a reasonable computer >>(say 1 day on my 667MHz laptop). > > What do you mean by unlikely? It makes no sense in this > context. QS succeeds with virtual certainty in time that depends > only on the size of n (with some small statistical variation). > Thats just not true in my experience. > And my comment makes perfect sense. > The quadratic sieve completes in one or two minutes > with some 70-digit numbers, and takes hours with others. > Which is exactly what I would expect. > What reason do you have to suppose the time has only > small statistical variation? > Have you actually tried it? > What if the number is prime? > (I gave random 60-digit numbers for factorisation > as an assignment to a class of 30. > One was prime, and there was a huge variation > in the time taken - using quadratic sieve - > to factorise the others.) My experience is mainly with 50-70 digit numbers, and I find the variation to be minimal. A factor of less than 2 differentiates the slow ones from that fast ones, absolute max. Never a factor of over 30 like you suggest. And certainly I see no basis for you to /expect/ such variation. On what are your expectations founded? It just so happens that just now my Lucas project has dumped on me 2 similar sized C65s, that I wasnt going to bother factoring, but just for the sake of this thread I can crack them as anecdotal evidence: <<< C65= 69204974562927543079855580502240033129063083500852882042929615 601 #FactorBase 4480 max FactorBase 93493 max LargePrime 5703070 Upper Bound 1061386330019 1933(4513)/4512 search for cycles 713 duplicate data merged in Large Prime 67 duplicate data merged (f) type 1933 total 4517 (p) type 23965( p_p 2474, p_p_pp 153, p_p_p_p_pp_pp 0) (pp) type 903( pp_pp 17, pp_pp_pp 7, others 0) block Lanczos method #singleton 186 : zero-weight columns 44 4250*4326 matrix 22.7489 entries / row nonzero entry 96683 memory 378K #iteration 137 gained 32 solutions 0:00:04:00 CPU time: 3m45s <<< C65= 47170085404946399943217136513044414839873437700586846592588641 201 #FactorBase 4416 max FactorBase 90499 max LargePrime 5506560 Upper Bound 997165096816 2007(4448)/4448 search for cycles 696 duplicate data merged in Large Prime 80 duplicate data merged (f) type 2007 total 4468 (p) type 22773( p_p 2376, p_p_pp 148, p_p_p_p_pp_pp 0) (pp) type 714( pp_pp 10, pp_pp_pp 7, others 0) block Lanczos method #singleton 242 : zero-weight columns 72 4102*4206 matrix 23.029 entries / row nonzero entry 94465 memory 370K #iteration 132 gained 31 solutions 0:00:04:00 CPU time: 2m38s (Satoshi Tomabechis PPSIQS) Thats pretty close, and about what Ive come to expect for C65s. However, dont just take my word for it - my view is entirely suported by data from other factoring projects: http://www.asahi-net.or.jp/~KC2H-MSM/mathland/part/ppsiqs.htm The times in each column have no more than a factor of 2 spread, and almost always the highest time will correspond to a number a couple of bits larger than the smallest time. And primes are a red herring, it takes a fraction of a second to do a PSW test on numbers that size, or even an APRCL test. Phil -- Unpatched IE vulnerability: NavigateAndFind protocol history Description: cross-domain scripting, cookie/data/identity theft, command execution Reference: http://safecenter.net/liudieyu/NAFjpuInHistory/ NAFjpuInHistory-Content.HTM Exploit: http://safecenter.net/liudieyu/NAFjpuInHistory/ NAFjpuInHistory-MyPage.HTM