mm-453 Subject: Re: Looking for Spare Cycles> MensanatorTry this link and tell me if it works any better:> ---> http://www.gridontap.com/grid.zipI think the other server has had it...Well, I got it downloaded. But I was at my office where we have abroadband connection to the Internet. Even then, there was a log pauseat 30% complete, which is the same symptom I had with the dial-upconnections. The difference here was that the download eventuallyresumed before the download process aborted. So the problem may not befixed.In looking at your readme file below, can you explain what it means to have the latest .Net framework? Such as what the hell it meansand where do I get it? Will having Visual Basic .NET installed on mycomputer suffice?Here's some more advice. Don't assume everyone knows what the hellyou're talking about. Which of the three .exe files is the client?And I was very disappointed that there was no description of what thisparticular client actually does. Ok, it's testing the CollatzConjecture. How? Is it merely checking that each number reaches 1? Isit compiling statistics such as stopping time records? What data isbeing sent back to the host? Will the participants have access to thedata collected? Was that information in one of the pages that timedout when I tried to open it? Maybe this should be included as anothertext file attached to the .zip.You also neglect to explain what happens once you get it running. Doyou have to be connected to the internet while it's running? That's noproblem for broadband users, but dial-up users, like myself, shouldnot even bother because I may have spare cycles, but I don't havespare phone time. What about stopping and restarting?You say that eventually, the grid will handle thousands of projects.Right now it doesn't seem to be handling even one. I realize it's inbeta test, but you might want to think about these issues.Grid on Tap === ========To run the client, you will need to have the latest .Net frameworkinstlled on your computer.Prerequesites: 1) Visit http://distributed.redirectme.net and add a new user account. 2) Run the client, and login. 3) Your system will start processing work if everything went ok.Misc. Notes === ========You should not install this program on a computer in which youdo not have permission. We will have a linux client shortly.If you want to use multiple CPU's, you must create a new directory for each one. We do not run programs that we cannot compile, all code is checked100% and compiled by us. === Subject: Re: Looking for Spare CyclesGood points, I will update the read-me. I'll get back to you soon> MensanatorTry this link and tell me if it works any better:> ---> http://www.gridontap.com/grid.zipI think the other server has had it...> Well, I got it downloaded. But I was at my office where we have a> broadband connection to the Internet. Even then, there was a log pause> at 30% complete, which is the same symptom I had with the dial-up> connections. The difference here was that the download eventually> resumed before the download process aborted. So the problem may not be> fixed.> In looking at your readme file below, can you explain what it means> to have the latest .Net framework? Such as what the hell it means> and where do I get it? Will having Visual Basic .NET installed on my> computer suffice?> Here's some more advice. Don't assume everyone knows what the hell> you're talking about. Which of the three .exe files is the client?> And I was very disappointed that there was no description of what this> particular client actually does. Ok, it's testing the Collatz> Conjecture. How? Is it merely checking that each number reaches 1? Is> it compiling statistics such as stopping time records? What data is> being sent back to the host? Will the participants have access to the> data collected? Was that information in one of the pages that timed> out when I tried to open it? Maybe this should be included as another> text file attached to the .zip.> You also neglect to explain what happens once you get it running. Do> you have to be connected to the internet while it's running? That's no> problem for broadband users, but dial-up users, like myself, should> not even bother because I may have spare cycles, but I don't have> spare phone time. What about stopping and restarting?> You say that eventually, the grid will handle thousands of projects.> Right now it doesn't seem to be handling even one. I realize it's in> beta test, but you might want to think about these issues.> Grid on Tap> === ========> To run the client, you will need to have the latest .Net framework> instlled on your computer.> Prerequesites:> 1) Visit http://distributed.redirectme.net and add a new> user account.> 2) Run the client, and login.> 3) Your system will start processing work if everything> went ok.> Misc. Notes> === ========> You should not install this program on a computer in which you> do not have permission.> We will have a linux client shortly.> If you want to use multiple CPU's, you must create a new directory> for each one.> We do not run programs that we cannot compile, all code is checked> 100% and compiled by us. === Subject: Re: Looking for Spare Cycles>Good points, I will update the read-me. I'll get back to you soonFYI, I tried the grid.exe (it says on the homepage that this is the client)on the computer on which I have installed Visual Basic .NET.It will not run. Says I don't have .NET framework.What, pray tell, exactly_is_ .NET framework? Is this something else I have tobuy? Do I have to re-run the install program and choose that as an option? I'mnot trying to be a dick, I honestly don't know and I'll wager 99% of yourpotential clients don't know either.--Mensanator2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: 3D Math FlashHello everybody ! I am a newbie to 3D Math. I program in actionscript. This is alanguage used for scripting flash graphics. I am doing a project tohelp develop a 3D engine coded in Flash to make a cet game. I want to learn 3D Math. Are there any good resources where ican learn 3D mathematics. I dont get the general idea about rotationand scaling. Please help me friends! I just want to grasp the idea ! === Subject: Re: 3D Math FlashHello everybody ! I am a newbie to 3D Math. I program in actionscript. This is a> language used for scripting flash graphics. I am doing a project to> help develop a 3D engine coded in Flash to make a cet game.> I want to learn 3D Math. Are there any good resources where i> can learn 3D mathematics. I dont get the general idea about rotation> and scaling. Please help me friends! I just want to grasp the idea !A good place to start are Joseph O'Rourke'scomp.graphics.algorithms Frequently Asked Questionsat http://www.faqs.org/faqs/graphics/algorithms-faq/Hugo === Subject: Re: Connecting the Dottiness.> >>.... If you've picked up a textbook on abstract>>algebra, you might notice there are all these little, dry rules.> Well you don't want to both yourself with those little, dry rules. > (Oh... wait a minute... you don't do you?)James Harris has a rule against ad hoc rules. James'srule against ad hoc rules is itself ad hoc. SoJames is not following his own rules. Ha ha.David BernierComments: This message did not originate from the Sender address above. It was remailed automatically by anonymizing remailer software. Please report problems or inappropriate use to the remailer administrator at . === Subject: Comparisons of genius mathematicians with other geniuses.Mail-To-News-Contact: abuse@dizum.comJ. S. Bach - L. EulerW. A. Mozart - C. F. GaussA. Rimbaud - E. Galois === === Subject: Experts Assemble to Test New Theory of Relativity> Exercise 1: Recognizing Absolute Motion Let an observer in the circle universe be at rest in her own inertial> frame of reference. Let her pick a positive and negative direction.> Suppose she has a nice watch on her wrist to note the time. Other than> just looking nice and being able to see how old she's getting, let her> do experiments. Let t1 be the time she measures for a photon to> circumnavigate the universe in the positive direction. Let t2 be the> time she measures for a photon to circumnavigate the universe in the> negative direction. Being only a one-dimensional creature, she is> still smart enough to realize the impossibility of all frames agreeing> on a frame independent law of light propagation. Consequently, if t1> doesn't equal t2, then she is moving at some velocity v with respect> to an absolute frame of reference. Isn't it obvious, based on the> global theorem, that the velocity v is given by the equation:> t1/t2 = (c+v)/(c-v) ?http://www.everythingimportant.org/relativity/ simultaneity.htmEugene ShubertI always wondered what would happen if somebody sent two photons> around the world. Now I know.A new theory must be explained properly, scrutinized and thencarefully tested before it can have any real credibility within thescientific community. Not everyone here respects the peer reviewprocess. All persons interested in participating in respectful peerreview and who are curious about A Viable Alternative to Einstein'sSpecial Relativity Theory are invited to join in a respectfulinterrogation of the author.See http://www.everythingimportant.org/viewtopic.php?t=605Eugene Shuberthttp://www.everythingimportant.orgPS. Registering with a valid email address at the forum is easy butrequired.Starblade, Tom Snyder, Robert Israel: judging from the exceptionalquality and intelligence of your many helpful posts, you areespecially encouraged to sign up. === Subject: Re: Experts Assemble to Test New Theory of Relativity> Here's a brief summary:Where there is empirical data, my theory is the same as SR. > Where there is no empirical data, my theory differs with SR. The logic and math is inescapably trivial yet physicists refuse to> acknowledge that a logically consistent alternative SR type theory> exists. Well, that's because those physicists aren't really physicists, they're really mathematicians in disguise. Who are still having their usual infinite problems with 2, nevermind logical consistently. So, what's typcally done to the solve problem, is to elect a Chemist to Chair the International Philosopher Committee on Communistic Twoness. Someplace in repaved Europe, The Middle East, or Canada is usually chosen as the optimal Center-Of-Flatulence. But, don't tell them that you're charging by the hour for their irrelevent whining about their non-existent Energy Flux-fields saturating the Aether. === Subject: Re: Experts Assemble to Test New Theory of RelativityHere's a brief summary:Where there is empirical data, my theory is the same as SR.> Where there is no empirical data, my theory differs with SR.The logic and math is inescapably trivial yet physicists refuse to> acknowledge that a logically consistent alternative SR type theory> exists.> Show us your derivation using both your alternative and SR for time dilation for say a GPS satellite with observed velocity of 3.9 km/s. Unless your alternative is smoke and dung, it should almost trivially predict the time dilation... lets see the calculation (not just the answer, but the calculation)! === Subject: Re: Gravitons are electrons, Electrons are gravitons>In sci.physics, Satan the Devil>:>> NEWTON'S G EQUATION MAKES A BIG COMEBACK! >> >> Devil Satan,>> >> Whatever you do Lucifer, DO NOT, and I repeat, DO NOT TELL THEM that>> in the central levels, the radius (r), even the Radius (R), IS OF OF>> EACH PARTICLE PARTICLE, not the square of the distance between their>> two centers.The Holy Ghost>Actually, you have a good point. I'm not sure how much difference>it makes to back-of-napkin orbital calculations but it's clear>gravitational attraction. Therefore, a good approximation would>be an integral over the volume of the planet. (An even better>one would be the sum of several integrals, each one modeling>a specific part: one for the crust, one for the mantle, one for>the liquid core, one for the solid core. And even then one>has issues regarding the continents.)F = integral(sphere)( GmM/d^2 dV)where d^2 and dV are linked (e.g., d^2 = x^2+y^2+z^2 and dV = dx dy dz;>however, most would probably want to use polar coordinates :-) ).The Holy Ghost> Aha. And, once you evaluate everything, you get multipole expansion. > The lowest order is simply total mass divided by the square of the > distance from the center of mass (and multiplied by Gm, of course). > There is no dipole moment. So, the next term is proportional to the > quadrupole momement of the mass distribution. Pretty small. And the > contribuion of this one is proportional to 1^r4. Does it make a > measurable difference? For satellites in orbit, yes. For planetary > orbits, no.Mati Meron | When you argue with a fool,> meron@cars.uchicago.edu | chances are he is doing just the same--NEWTON'S FORCE EQUATION MADE A BIG COMEBACK (and you all missed it save and except for)Gravitons are electrons in charged and curved space. Electrons are gravitons in charged and curved space.Offspring of .. F=Gm^2/w^2 for a wave. NEW UNIVERSAL LAWSF=Gm^2/w^2 for a wave. REESTABLISHED UNIVERSAL LAWNEWTON'S RADIUS IS RADIUS OF ATOM PARTICLE INSIDE EACH ATOM, AND NEWTON'S RADIUS IS RADIUS OF TWO ATOMS OUTSIDE EACH ATOM. Therefore in Chronological Discovery Order:3) lin...F=Gm^2/w^2 (for a wave in and out of atom) The four horses, the four forces.Welcome back Newton, farewell to quantum,--Satan the Devil === Subject: The Ghost In The Machine Sayeth;> The Ghost In The Machine Sayeth; === Subject: Re: The Ghost In The Machine Sayeth;Denke - if you start your iterative spamming again, off t.o you willgo, again.You have 24 hours to comply. RSVP.--D.-- david iain greig greig@ediacara.orgmoderator, talk.origins sp4 koxhttp://www.ediacara.org/~greig arbor plena alouattarum === Subject: Re: The Ghost In The Machine Sayeth;Denke - if you start your iterative spamming again, off t.o you will> go, again.You have 24 hours to comply. RSVP.--D.Three (3) more; None (0) t.o.Thank you for writing.Kill fulfiled.--G. === Subject: Re: Gravitons are electrons, Electrons are gravitonsIn sci.physics, Satan the Devil<61ca1ba2.0308241031.44aea8bb@ posting.google.com>:>>In sci.physics, Satan the Devil>><61ca1ba2.0308231250.2c4f6ae9@ posting.google.com>:> NEWTON'S G EQUATION MAKES A BIG COMEBACK! > > Devil Satan,> > Whatever you do Lucifer, DO NOT, and I repeat, DO NOT TELL THEM that> in the central levels, the radius (r), even the Radius (R), IS OF OF> EACH PARTICLE PARTICLE, not the square of the distance between their> two centers.The Holy GhostHoly?That's amusing, considering I'm atheist.> >>Actually, you have a good point. I'm not sure how much difference>>it makes to back-of-napkin orbital calculations but it's clear>>gravitational attraction. Therefore, a good approximation would>>be an integral over the volume of the planet. (An even better>>one would be the sum of several integrals, each one modeling>>a specific part: one for the crust, one for the mantle, one for>>the liquid core, one for the solid core. And even then one>>has issues regarding the continents.)>>F = integral(sphere)( GmM/d^2 dV)>>where d^2 and dV are linked (e.g., d^2 = x^2+y^2+z^2 and dV = dx dy dz;>>however, most would probably want to use polar coordinates :-) ).The Holy Ghost> >> Aha. And, once you evaluate everything, you get multipole expansion. >> The lowest order is simply total mass divided by the square of the >> distance from the center of mass (and multiplied by Gm, of course). >> There is no dipole moment. So, the next term is proportional to the >> quadrupole momement of the mass distribution. Pretty small. And the >> contribuion of this one is proportional to 1^r4. Does it make a >> measurable difference? For satellites in orbit, yes. For planetary >> orbits, no.I think Meron's on the right track here, pending evaluation.Of course it depends on the distance between masses.>> >> Mati Meron | When you argue with a fool,>> meron@cars.uchicago.edu | chances are he is doing just the same--NEWTON'S FORCE EQUATION MADE A BIG COMEBACK > (and you all missed it save and except for)[rest snipped]You're not. :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: Re: Gravitons are electrons, Electrons are gravitons> Of course it depends on the distance between masses.F=Gm1M2/r^2Only if outside of atom, however inside of atom;.~THE NEW UNIVERSAL LAWS OF FORCE3) lin...F=Gm^2/w^2 for each wave in and out of atom. .~ Newton's force equation; For all creatures for all time.> You're not. Charge thou shalt not know;:-)(-:Charge thou shalt not know;:+)(+:No thou shalt not know;Satan the Devil.~Holy are all. === Subject: Re: Gravitons are electrons, Electrons are gravitonsThe Holy Ghost > The four horses,> the four forces.Welcome back Newton,> farewell to quantum,--Satan the DevilHas this e owt to do with maths?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.hHis mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Weakly dense in linear order > >> > Let S be a linear order. Claim: S is order isomorphic to a > subset of reals iff some countable C subset S with for all x,y > in S, if x < y, then some z in C with x <= z <= y. ie., C is > weakly dense in S. >> >> An easy exercise for the weekend: >> >> Is there always a topology on S such that C weakly dense in S <==> >> C dense in the topology. >> > > I fail to see the significance of this ambigously stated problem. > > If S linear order and C subset S, then there's a topology T such that > C is weakly dense in S iff C is topological dense in (S,T) > provided S is a multi-point space, in which case T may be taken as: > indiscrete topology when C is weakly dense in S discrete topolgy when > C not weakly dense in S. > > If S is single point, then C = nulset is counterexample. > > -- If S linear order, then there's a topology T such that for all C > subset S C is weakly dense in S iff C is topological dense in (S,T) > ? > > No, S = {0} and C = nulset counterexample. > > To build another counter example Let S = {0,1,2}. {0,2}, {0,1}, > {1,2} and {1} weakly dense in S {0}, {2} not weakly dense in S If {0} > not dense, some open nonnul U with 0 not in U If {2} not dense, some > open nonnul V with 2 not in V If {1} dense, for all open nonnul W, 1 > in W. Thus 1 in U,V and {1} = U/V is open. Hence {0,2} is not dense. > > > ----The significance is that, as you correctly proved, the concept weakly dense is different from the concept topologically dense. In contrast to the usual order dense concept, which is equivalent to topologically dense for a suitable topology. === Subject: Re: Weakly dense in linear order === Subject: Re: Weakly dense in linear order > >The significance is that, as you correctly proved, the concept >weakly dense is different from the concept topologically dense. >In contrast to the usual order dense concept, which is equivalent >to topologically dense for a suitable topology. >Would you clarify? Order dense is a property of the whole space,topologically dense is a property of a subset of the whole space.S (order) dense when for all x,y in S, x < y ==> some s in S with x < s < yA (order) dense in S when for all x,y in S, x < y ==> some a in A with x < a < yA weakly (order) dense in S when for all x,y in S, x < y ==> some a in A with x <= a <= yD (topologically) dense subset S when cl D = SD subset linear order S ==> D dense in S iff D dense subset S, S order densewhere the topology of S is presumed to be thelinear order topology of the linear order of SA weakly dense in S, S order dense iff A dense in SD subset dense linear order S ==> D weakly dense in S iff D dense subset S iff D dense in SDoes that cover your loosely stated assertion?BTW, nulset dense in {0} but nulset not dense subset {0}so all the above with the caveat S is multi-point or D nonnul---- === Subject: Re: Weakly dense in linear order === > dense in linear order > >> >> The significance is that, as you correctly proved, the concept >> weakly dense is different from the concept topologically dense. >> In contrast to the usual order dense concept, which is >> equivalent to topologically dense for a suitable topology. >> > Would you clarify? Order dense is a property of the whole space, > topologically dense is a property of a subset of the whole space. > > S (order) dense when for all x,y in S, x < y ==> some s in S with x < > s < y > > A (order) dense in S when for all x,y in S, x < y ==> some a in A > with x < a < y A weakly (order) dense in S when for all x,y in S, x < > y ==> some a in A with x <= a <= y > > D (topologically) dense subset S when cl D = S > > D subset linear order S ==> D dense in S iff D dense subset S, S > order dense > > where the topology of S is presumed to be the linear order topology > of the linear order of S > > A weakly dense in S, S order dense iff A dense in S > > D subset dense linear order S ==> D weakly dense in S iff D dense > subset S iff D dense in S > > Does that cover your loosely stated assertion? > > BTW, nulset dense in {0} but nulset not dense subset {0} so all the > above with the caveat S is multi-point or D nonnul > > ---- Clarification:First, definitions:Let S be a (partially) ordered set S. A subset D of S is called (order)dense in S if:for each x,y in S such that x I have a triangle with verts (x1,y1),(x2,y2) and(x3,y3). The center of the > triangle is (x0,y0). Is there an equation which tells me if the point > (x4,y4) lies inside the triangle?The algorithm used in finite element work is: [ 1 1 1] [L1] [ 1]> let [T] = [x1 x2 x3], compute [L2] = [T]^(-1) [x4] > [y1 y2 y3] [L3] [y4]If each Li (i=1,2,3) is between 0 and 1, the point is inside.> (Actualy only 2 tests are required because L1+L2+L3=1.)> The Li are called barycentric coordinates. Note that x0 and y0 > are not used. The algorithm extends trivially to tetrahedra > in 3D, and to simplices in n dimensions.Forgot to add that it is also sufficient to test that thethree coordinates L1,L2,L3 are positive (for strict inclusion)or nonnegative (if side points are accepted as interior).If the procedure is to be repeated for many points and the sametriangle, the inverse of [T] can be precomputed. Its closedform expression is given in equation (15.11) ofhttp://caswww.colorado.edu/courses.d/IFEM.d/IFEM.Ch15.d/ IFEM.Ch15.pdf === Subject: Resolving the classic card paradox!Consider the following problem:> Start with a normal deck of playing cards.> Discard all but the aces and kings. Now you > have eight cards left. Your friend draws > two cards and hides them from you. He truthfully > tells you that at least one of his cards is an ace. > What is the probability that he holds two aces?> > He replaces the cards and you shuffle them well.> Your friend again draws two cards, and truthfully > tells you that one of them is the ace of spades.> What is the probability that he now holds two aces?> (Hint: this isn't the same answer as above!)Okay . . . I don't get it. Why isn't it the same > answer both times?The answer IS the same (3/11) in each case if we set up the problem inthe correct way. Indeed, if the probability in the second case wasactually 3/7 (as some respondents have asserted) then we obtain theapparent paradox: the same conclusion would hold for all four suits.How can this be? The declared ace must have SOME suit, and hence, bysymmetry, the answers for the two parts of the question must beidentical!So, why is the answer 3/7 wrong? Let's think about the secondsituation a little more.Suppose you knew that your friend had a deeply-engrained preferencefor suits, and his policy was to declare the ace of spades whenever hedrew it, to declare the ace of diamonds whenever he drew it (providingthe ace of spades was not also drawn), to declare the ace of clubswhenever he drew it (provided neither the ace of spades nor the ace ofdiamonds was also drawn) and to declare the ace of hearts whenever hedrew it (provided none of the other aces had been drawn). In otherwords, your friend's preference amongst the aces was S - D - C - H.Now, UNDER SUCH A KNOWN PREFERENCE, it is indeed true that the desiredprobability is 3/7. There are three relevant 'double ace' points inthe sample space, since the ace of spades is named precisely when itis drawn, and there are four other points in the sample space(corresponding to drawing the ace of spades and a king).However, is this a natural model? If we know that the ace of spadeswill be declared whenever it is drawn, this knowledge has distortedour set of probabilities. Indeed, under the S - D - C - H preferencedetailed above, if the ace of hearts is named, the probability ofanother ace being held is zero! The existence of a PREFERENCE acrossthe four suits means the probabilities are not all equal (inparticular, not all equal to 3/7) but actually bce out to give3/11 (the value in the absence of suit information).A more natural model exploits the inherent symmetry of the setup. Withno preference between suits, if two aces are drawn then your friendhas a choice as to which ace to declare. If this choice is madearbitrarily (probability 0.5 for each of the two aces) then each pointin the probability space corresponding to 'two aces' is divided intotwo, representing the two possible choices. This may be shown on atree diagram: each of the hands AS/AH, AS/AC and AS/AD branch off intotwo possibilities depending on which card is named. Then, theprobability of your friend's other card being another ace, given thathe has declared an ace of spades, is not 3/7 but 1.5/5.5 = 3/11!In summary, there is no paradox. The (erroneous) calculation whichproduces a value of 3/7 for the second part of the original questionassumes that the friend will always name the ace of spades if he drawsit. However, if this is indeed his preference, then the problem is notsymmetrical with respect to the four suits as it should be! Byassuming that the friend names either ace arbitrarily (uniformly) inthe case where two aces are drawn, the linguistic symmetry in thestatement of the problem (with respect to replacing 'spades' by'hearts' say) is carried over to the mathematical model. In all cases,the probability of the remaining card being an ace is 1.5/5.5 = 3/11.HJThe key point is that the probability in the second case is only 3/7if your friend's protocol is to name the Ace of Spades in everyinstance in which he draws two aces. === Subject: Re: Resolving the classic card paradox!> Consider the following problem:> Start with a normal deck of playing cards.> Discard all but the aces and kings. Now you > have eight cards left. Your friend draws > two cards and hides them from you. He truthfully > tells you that at least one of his cards is an ace. > What is the probability that he holds two aces?> > He replaces the cards and you shuffle them well.> Your friend again draws two cards, and truthfully > tells you that one of them is the ace of spades.> What is the probability that he now holds two aces?> (Hint: this isn't the same answer as above!)Okay . . . I don't get it. Why isn't it the same > answer both times?The answer IS the same (3/11) in each case if we set up the problem in> the correct way. Indeed, if the probability in the second case was> actually 3/7 (as some respondents have asserted) then we obtain the> apparent paradox: the same conclusion would hold for all four suits.> How can this be? The declared ace must have SOME suit, and hence, by> symmetry, the answers for the two parts of the question must be> identical!What symmetry would that be?The answers for the two questions aren't identical becausethe questions aren't identical.You appear to be making some additional assumptions, ashave other people who disagree with the 3/7 answer. Inthe simplest possible interpretation, there are 7 possibleequally likely hands which the second person could be holding, andthree of them contain two aces.> So, why is the answer 3/7 wrong? Let's think about the second> situation a little more.Suppose you knew that your friend had a deeply-engrained preference> for suits,Unwarranted additional assumption.> and his policy was to declare the ace of spades whenever he> drew it, to declare the ace of diamonds whenever he drew it (providing> the ace of spades was not also drawn), to declare the ace of clubs> whenever he drew it (provided neither the ace of spades nor the ace of> diamonds was also drawn) and to declare the ace of hearts whenever he> drew it (provided none of the other aces had been drawn). In other> words, your friend's preference amongst the aces was S - D - C - H.Now, UNDER SUCH A KNOWN PREFERENCE, it is indeed true that the desired> probability is 3/7. There are three relevant 'double ace' points in> the sample space, since the ace of spades is named precisely when it> is drawn, and there are four other points in the sample space> (corresponding to drawing the ace of spades and a king).However, is this a natural model?No. A more natural one would be that you are going toask him two questions and he has agreed to answer truthfully.The first one is: Are you holding the ace of spades?> The key point is that the probability in the second case is only 3/7> if your friend's protocol is to name the Ace of Spades in every> instance in which he draws two aces.Yes, OK, I'll agree with that. But in all such paradoxes Ifind the assumptions needed to get answers different fromthe obvious population conditioning require some rathersevere bending of the interpretation.Obviously, things like natural interpretation aresubjective. All such arguments eventually end up in a disagreement on what is natural, and eventually weall agree that the test maker should pose the questionin an unambiguous way. - Randy === Subject: Metric EngineeringNote that the extrinsic spatial curvature must rapidly fall to zero exponentially away from the exotic vacuum propulsion boundary layers in order that the UFO not wreak havoc on its local environment, i.e. not be a WMD. This is where the art of the metric engineer is crucial. We are basically free to design any local warp drive vacuum propeller metric we like by shaping the /zpf zero point energy density exotic vacuum field appropriately. This is analogous to quantum dots that are designer atoms shaped to a particular technological application in nanotechnology.Go to http://qedcorp.com/APS/warpdrivephysics.pdffor evolving details. === Subject: Re: Metric Engineering(Top of post)>Note that the extrinsic spatial curvature must rapidly fall to zero >exponentially away from the exotic vacuum propulsion boundary layers in >order that the UFO not wreak havoc on its local environment, i.e. not be >a WMD. What UFO? Should this be a Re: post? Did you perhaps have a little too muchcoffee this morning?Ah. Didn't attend chapel, drank coffee all day, had a breakthrough. Or aabout the Bhuddist hotdog vendor?He said:All day long people walk up to me and say One with everything?, and I sayYes, thank you and then they ask me for a hotdog, I make it, they pay, andthey and go off happy. I don't get it.....>We are >basically free to design any local warp drive vacuum propeller metric we >like by shaping the /zpf zero point energy density exotic vacuum field >appropriately. I thought the Casimir and Unruh effects are the only known ways to interactwith the ZPE. >This is analogous to quantum dots that are .89[Thorn]designer >atoms.89 shaped to a particular technological application in nanotechnology.Ok, we already have quantum dot LEDs, so I can follow that, but how could theybe used in propulsion?>Go to http://qedcorp.com/APS/warpdrivephysics.pdf>for evolving details.I am not currently accepting commands from your location, Jack. IT FROM BITTo which I addBIT FROM ITTo get the participatory parallel universes as a kind of .89[Thorn]self-excitedcircuit.89 (Wheeler) or spontaneously self-organizingemergent .89[Thorn]More is different.89 (P.W. Anderson) physicalreality. We have all recently experienced .89[Thorn]bad programming.89on the Internet. J!!!!RE-EDIT!!!! !!DELETE!!I like the part where he admits I'm guessing.I have posted a reply to an old post, a challenge question, in a thread insci.physics.research:If an infinite array of one-gram Pu spheres a meter apart is suddenly broughtinto being, will it end up in a subcritical, critical, or supercritical state?The answer is supercritical because the mean free path before interactionwill be only 5000 meters, and the array is infinite. Now we could reason thata 5000 meter array could be the biggest bomb in history, containing 125 * 10^6grams of Pu, (I do not know the world supply) and I asked about a method ofinitiation that would cause the nucelar reactions to accelerate inward to afocus, suspecting the energy would be of big bang magnitude, an ideaopportunity for study.The post is in transit and will not appear in any newsgroup until, well,moderated! Isn't that wry? Of course, the array spacing could be recalculated to use our existinginventory of pits, but this is at present theoretical.Now the question I have for you is can we access zpe at big bang conditions,and if so, how close do we have to get?Yours,Doug Goncz, Replikon Research, Seven Corners, VA The hormones work at different speeds: In a fight-or-flight scenario, glucocorticoids are the ones drawing up blueprints for new aircraft carriers;epinephrine is the one handing out guns. === Subject: Re: Sides of a right triangle given only area and hypotenuse?Could this be the problem? You only see posts that you haven't seen before. Ifyou want to see all the posts, you have to click on list all when you openthe newsgroup. If you just press enter or double-click it will automaticallylist unread>Is this subject of the newsgroup not working or what? I still see none of my>postmessages about this hypotenuse area problem. In fact I see no posts of>any of its responses except the original post, or the last one I tried to>quote>and comment for. === Subject: Re: Sides of a right triangle given only area and hypotenuse?My only option is LIST; no other button is available for bringing up allposts. When I did this, I was only presented the one very recent post frombsinthewall8@aol.com, and NONE others. About the original subject, I became somewhat tangled in my efforts to examinethe algebra concerning this triangle situation, and hoped somebody would haveposted a more successful result about YES or NO; My attempt gave mecontradictory results, but intuitively, I believe that the sides should befindable based on known hypotenuse and area. G C === Subject: Re: Sides of a right triangle given only area and hypotenuse?>Could this be the problem? You only see posts that you haven't seen before.>If>you want to see all the posts, you have to click on list all when you open>the newsgroup. If you just press enter or double-click it will automaticallylist unread>>Is this subject of the newsgroup not working or what? I still see none of>my>>postmessages about this hypotenuse area problem. In fact I see no posts of>>any of its responses except the original post, or the last one I tried to>>quote>>and comment for.Not sure. Maybe the sequence of choices matters, but still I never saw theother posts, or a couple of mine which I expected. I'll try a differentsequence of LIST ALL or LIST, to see what happens. G C === Subject: probabilityCan some please tell me what is being asked in these problem more clearly.1st triangular means { 0 x < a fx(x) = k{x - a) a ? x < (a+b)/2 K(b - x) (a+b)/2 ? x < b 0 b ? x, }1)Recall that for a normal random variable U with mean ? and variance ?2,the new random variable V = ?U + ? is normal with mean ?? + ? and variance?2?2, as long as ? ? O. Is this property true for triangular randomvariables X? In other words, is the triangular shape of X preserved in Y =?X + ??Do I need to just look at the graph of fx(?X + ?)??2)Recall that any two independent normal random variables U1 and U2 withmeans ?l and ?2 and variances ?12 and ?22, the new random variable W = U1+ U2 is normal with mean (?l + ?2) and variance (?12 + ?22). Is thisproperty true for any two independent triangular random variables Xl and X2?In other words, is Z = Xl + X2 triangular?Again, do I need to just look at the graph of fx(Xl + X2 )and since fx(x) islinear the answer is yes? === Subject: enseignement en facbonjour,je voudrais savoir quel est le cursus requis pour pouvoir enseigner .88 lafac. Y en a t il qu'un seul ou est-il possible d'arriver .88 ce but pardiverses fa.8dons ?Merci === Subject: Re: enseignement en fac|je voudrais savoir quel est le cursus requis pour pouvoir enseigner .88 la|fac. Y en a t il qu'un seul ou est-il possible d'arriver .88 ce but par|diverses fa.8dons ?Je ne sais pas. Mais les gens au fr.sci.maths le savent bien, j'en suissur. Ils sont toujours en train de discuter la fac, etc. Et ils savent.8ecrire bien en francais, aussi. :-)Keith RamsayP.S.|i would like to know what is the required course to be able to teach|at school (the faculty). Is there only one or is it possible to arrive|at this end by various means?I don't know. But the folks at fr.sci.maths know it well,I'm sure of it. They are always in the process of talking aboutthe faculty, etc. And they know how to write in French well,too. :-) === Subject: Re: enseignement en fac> bonjour,> je voudrais savoir quel est le cursus requis pour pouvoir enseigner .88 la> fac. Y en a t il qu'un seul ou est-il possible d'arriver .88 ce but par> diverses fa.8dons ?> Merci s le newsgroup fr.education.entraide.maths ou peut-.90tre fr.sci.maths se trouvent des lecteurs fran.8dais qui sauront r.8epondre. === Subject: partial and totally ordered sets Let P be a partially ordered set. Is there a totally ordered Tand f:P -> T such that for all x, y in P, x < y implies f(x) < f(y)? I don't see any way to do this without AC, or any easy way todo this with AC. === Subject: Re: partial and totally ordered sets> Let P be a partially ordered set. Is there a totally ordered T> and f:P -> T such that for all x, y in P, x < y implies f(x) < f(y)? I don't see any way to do this without AC, or any easy way to> do this with AC.What about using Zorn's Lemma? === Subject: Re: partial and totally ordered sets> > Let P be a partially ordered set. Is there a totally ordered T> and f:P -> T such that for all x, y in P, x < y implies f(x) < f(y)?> > I don't see any way to do this without AC, or any easy way to> do this with AC.>> >What about using Zorn's Lemma?> To be more precise than the OP, given a poset (S,P), does there exist a totally ordered set (S',T) and an injection f: S -> S' preserving order?Halmos has as an exercise that this is true for S' = S. To elaborate on ANN's hint, apply Zorn's lemma to {R in Power(S^2): P is a subset of R and R is a partial order of S}.I believe that AC is needed, but I will allow the masters to confirm this. One guess at a beginning: Order P(N) by inclusion. By the above prepostion, we can extend this to a total order on P(N) which preserves inclusion. Can we play with this to come up with a with a well-ordering?-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: partial and totally ordered sets <3F493496.2020902@rutcor.rutgers.edu> === Subject: Re: partial and totally ordered sets>> Let P be a partially ordered set. Is there a totally ordered T>> and f:P -> T such that for all x, y in P, x < y implies f(x) < f(y)? >To be more precise, given a poset (S,P), does there exist a totally >ordered set (S',T) and an injection f: S -> S' preserving order? >Halmos has as an exercise that this is true for S' = S. >To elaborate, apply Zorn's lemma to {R in Power(S^2): >P is a subset of R and R is a partial order of S}. >Order <= of S extendable to total order let P be all orders of S containing <=C subset P, C totally ordered by set inclusion ==> <= subset Union C, Union C in Psome maximal M in P. Use Zorn's lemmaM total order. Otherwise: some a,b in S with not aMb, not bMa M proper subset M / { x | xMa }x{ y | bMy } in P, which cannot beDetails left to reader.---- === Subject: Re: partial and totally ordered setsOP:| Let P be a partially ordered set. Is there a totally ordered T| and f:P -> T such that for all x, y in P, x < y implies f(x) < f(y)?|To be more precise than the OP, given a poset (S,P), does there exist a |totally ordered set (S',T) and an injection f: S -> S' preserving order?That's a little stronger than the original statement, because in theoriginal, one might have f(x)=f(y) if x != y but x is not < y andy is not < x.[...]|I believe that AC is needed, but I will allow the masters to confirm |this. One guess at a beginning: Order P(N) by inclusion. By the |above prepostion, we can extend this to a total order on P(N) which |preserves inclusion. Can we play with this to come up with a with a |well-ordering?No. We can order P(N) by letting S < T if the smallest n which is amember of one of S and T but not the other one, is a member of T.That doesn't require AC.I had thought the statement implied the existence of a choice setfor a family of finite sets, which I believe is not provable in ZF. I wasgoing to let the partially ordered set be the disjoint union of thesets in the family. Your version of the statement implies that there'sa uniform ordering on all of them, and the first element of each set inthe family under that ordering would form a choice set. But in theoriginal version of the statement, we could map all of the elementsto a single-point set. I still think AC is needed, but it's harder to show.Keith Ramsay === Subject: Re: partial and totally ordered sets Let P be a partially ordered set. Is there a totally ordered T> and f:P -> T such that for all x, y in P, x < y implies f(x) < f(y)? I don't see any way to do this without AC, or any easy way to> do this with AC.What about using Zorn's Lemma?Zorn is equivalent to AC. === Subject: conic sectionsthe graph of E is the intersection of a plane and a cone (with just a handful oftrivial exceptions like xy=yx, or x^2=(-1), or what-have-you). It seems likethere should be some deep reason for this. Does anybody know of one? Can anybodypoint me to a cute proof? TIA.Peace,EJ === Subject: Symmetric designsDoes anyone know of a list of feasible parameter sets for symmetric(v,k,lambda) designs? If not, I'd like to generate my own. Whatrestrictions are known on such parameter sets, in addition to thefollowing?[r is the replication number, b is the number of blocks, n is k-l].(1) r=b. [Obvious](2) b geq v. [Fischer's Theorem](3) v-2k+lambda>0.(4) 4n-1 leq v leq n^2+n+1.(5) The Bruck-Chowla-Ryser theorem.(6) Connor's theorem.Felix. === Subject: Re: Symmetric designsOriginator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow)>Does anyone know of a list of feasible parameter sets for symmetric>(v,k,lambda) designs? If not, I'd like to generate my own. What>restrictions are known on such parameter sets, in addition to the>following?The first place to look is the CRC Handbook of Combinatorial Designs(ed. Colbourn & Dinitz). Did you look there?-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: Symmetric designsSurely there are several such lists published.A classical one is in M.Hall's Combinatorial Theory.There are various tables in Beth, Jungnickel, and Lenz Design theory. > Does anyone know of a list of feasible parameter sets for symmetric> (v,k,lambda) designs? If not, I'd like to generate my own. What> restrictions are known on such parameter sets, in addition to the> following?[r is the replication number, b is the number of blocks, n is k-l].(1) r=b. [Obvious]> (2) b geq v. [Fischer's Theorem]> (3) v-2k+lambda>0.> (4) 4n-1 leq v leq n^2+n+1.> (5) The Bruck-Chowla-Ryser theorem.> (6) Connor's theorem.> HTH,Dmitriihttp://www.thi.informatik.uni-furt.de/~dima/ === Subject: 'INDIA'S SCHOOLBOOK HISTORIES' by Subhash KakX-Url: http://www.mantra.com/jai http://www.flex.com/~jaiX-Warning: Not for commercial use. Views expressed by others not necessarily poster's.India's schoolbook historiesBy Subhash KakRediff On The NetLiving as we do in the multicultural age, it is notfashionable to speak of the soul of a nation. But byexamining a nation's achievements, one can form an ideaof the moving spirit behind its history.A strong ideological position or cultural prejudice maymake a person denigrate achievements in the fields of artor philosophy. Thus early nineteenth century Englishhistorians, unable to judge Indian sculpture usingWestern canon with its notion of progress, dismissedIndian art.India's contributions to science, technology and craftsare well documented, if not widely known. For example,before the British arrived, Indians had a system ofinoculation against smallpox; year-old live smallpoxmatter was used, and it was very effective. Tikadarswould fan out into the country before the smallpox seasonbook in 1767 describing the system and how it was safe.European medicine did not have any treatment against thisdisease at that time.Inoculation against smallpox using cowpox wasdemonstrated by Edward Jenner in 1798 and it became apart of Western medicine by 1840. No sooner did thathappen that the British in India banned the older methodof vaccination, without making certain that sufficientnumber of inoculators in the new technique existed.Smallpox in India became a greater scourge than before. India's technology was flourishing before the British. Ithas been estimated that India's share of world trade in1800 was about 20 percent (equal to America's share ofworld trade in 2000). The historian Ruttonjee Wadia saysthat ships built at Mumbai in its heyday were 'vastlysuperior to anything built anywhere else in the world.'According to Dharampal, there were 10,000 iron and steelfurnaces operating in the eighteenth century India. The story of the destruction of India's textile industryby the British is too well known to need repeating. TheBritish became masters of India at a very opportune time.First, they cut off India's export markets. Soon theinnovations of the dawning industrial revolution gavetheir products a cost advantage that became permanent inthe absence of new investments to upgrade Indianfactories. As India became de-industrialised, it turned into a hugemonopoly market for British products. British Raj madetoken investments in science and technology. In 1920,India's scientific services had a total of 213 scientistsof whom 195 were British!But this story of India's economic decline (and the lossof memory of its previous condition) is a complex one.Here, I would like to speak of scientific progress alone,a unique measure of a culture that can be appreciated byall, even those whose ideology or experience of art andaesthetics is different. Suppose you were offered a history of the English withoutreference to Newton, Faraday, and Maxwell or of theAmericans without mention of Edison, Michelson, orFeynman, you would say it overlooks the real genius ofthese nations. Youth in these countries brought upwithout the stories of these masters would not be quiteEnglish or American in spirit. Given this, why is it that Indian schools leave outmention of India's great scientists from its textbooks?Most readers of this column will have heard only one ortwo names of the greatest Indian scientists andmathematicians: Lagadha, Baudhayana, Panini, Pingala,Aryabhata, Bhaskara, Madhava, Nilakantha, whose ideashave shaped the world.astronomical text in 1300 BC. Baudhayana (800 BC) gavethe 'Pythagoras theorem' centuries before the Greek.Panini (400 century BC) has been called the greatestgenius who ever lived: his grammar of the Sanskritlanguage is exhaustive and yet it uses only 4,000computer program-like rules. Pingala (400 BC) inventedthe binary number system (counting by 0s and 1s) that,2,500 years later, turned out to be basic to computeroperations.The astronomers Aryabhata and Bhaskara may be familiar tosome from the eponymous spacecrafts of the Indian SpaceOrganization. Aryabhata (500 AD) took the earth to spinon its axis and he described the planet periods withreference to the sun. He also took the solar system to beseveral hundred million miles across. In all of thesethings he was ahead of the rest of the world by more thana thousand years. Bhaskara (12th century) was a brilliantmathematician. The last two names belong to the amazing Kerala school ofmathematics and astronomy. M.89dhava (c 1340-1425) andN.94lakantha (c 1444-1545), who made fundamentalcontributions to power series, calculus and astronomy,are amongst the greatest scientists who have ever lived.Their invention of calculus came two hundred years beforeNewton and Leibnitz.Three British historians have recently suggested thatKerala mathematics may have provided key ideas for thescientific revolution in Europe. The need for clocks tokeep accurate time on ships became of critical importanceafter the colonisation of America. There were significantfinancial rewards for new navigation techniques. Thesehistorians argue that information was sought from Indiadue to the prestige of the eleventh century Arabictranslations of Indian navigational methods. They suggestthat Jesuit missionaries were the intermediaries in thediffusion of Kerala mathematical ideas into Europe.The only rational explanation for leaving out mention ofIndia's great scientists from schoolbooks appears eitherto be bureaucratic sloth in the centralised textbookwriting agency or the internalisation by Indians of theideology of British colonialists who justified the Empireon the ground that India had a lot of religion but noscience, thus being incapable of self-rule.Macaulay's programme to estrange Indians from theirculture has been so successful that most textbook authorsare not even aware of the Kerala school, or of Pingalaand Panini's scientific contributions. Many who arepassionate in their love for India are so misguided bythe prestige of the Orientalist narratives that theybelieve that Madhava and Nilakantha are fictionalcharacters, product of a conspiracy to create an imaginedgreatness for ancient India. To me, India represents the spirit of excellence andpursuit of beauty and truth in art and science. There isnothing parochial about this idea; it is inspiring, andit helps us find our common ground with all humanity. Ourschools must celebrate this spirit and remembering greatmasters is a part of the process. Making peace with thepast will make it easier to move on with the business ofcreating a great future for everyone. External Link:Ian Pearce's history of Indian mathematicshttp://www-history.mcs.stws.ac.uk/history/Projects/ Pearce/index.hSubhash Kak, professor at Louisiana State University,will contribute a monthly column to rediff.com Subhash KakRead the complete news at:http://www.rediff.com Jai Maharaj http://www.mantra.com/jai Om Shanti Shubhanu Nama Samvatsare Dakshinaya Jivana Ritau Singh Mase Krishna Pakshe Bhanu Vasara YuktayamPunarvasu-Pushya Nakshatr Vyatipat Yog Taitil-Gar Karan Dvadashi-Trayodashi Yam TithauHindu Holocaust Museumhttp://www.mantra.com/holocaustHindu life, principles, spirituality and philosophyhttp://www.hindu.orghttp://www.hindunet.orgThe truth about Islam and Muslimshttp://www.flex.com/~jai/satyamevajayate o Not for commercial use. Solely to be fairly used for theeducational purposes of research and open discussion. The contents ofthis post may not have been authored by, and do not necessarily representthe opinion of the poster. The contents are protected by copyright lawand the exemption for fair use of copyrighted works. o If you send private e-mail to me, it will likely not be read,considered or answered if it does not contain your full legal name,current e-mail and postal addresses, and live-voice telephone number. o Posted for information and discussion. Views expressed by othersare not necessarily those of the poster. === Subject: Re: 'INDIA'S SCHOOLBOOK HISTORIES' by Subhash KakIndia's schoolbook historiesThe Toilet Papers of IndiaThe intensive historical cataloging of 5000 years of Indian sanitarytissue production and availability has arrived. This is a punctiliousexamination of the sociopolitical, religious, and aesthetic evolutionof the subcontinent's bumpwipes unlike any other. No effort has beenspared to include the most minute pertinent facts over five millenniaof careful exhaustive documentation. 26 pages plus illustrations.--Uncle Alhttp://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Re: What cardinality may a ring have?> In my old-fashioned world, all rings had at least two members, an>> additive identity and a distinct multipicative identity.Wow, you must be *old*. In my 50-year-old copy of Van der Waerden's>_Modern Algebra_ that requirement had already been abolished: The>integers form a ring C with identity; the even numbers a ring without>identity. There are also rings with one or more right identities but>without a left identity, or vice versa.When this came up in another thread, I checked the only two algebra> texts I have (Van der Waerden and Herstein), and couldn't find> anything about a multiplicative identity.A more interesting question to me is whether {0} should be considered> a ring. I can't decide off the top of my head which would cause more> difficulty... in one case one would have to put 'non-trivial' in for> some statements, in the other case, it would sometimes be necessary to> say 'ring or {0}'.Perhaps some insist on a multiplicative identity so they don't have to> deal with {0}. In the ring {0} there's only one way to define +, * and 0; and 0 _is_ amultiplicative identity, look: 0*0 = 0. > If I were concerned about that, I would have> (R,+) is an Abelian group with identity 0.> (R/{0}',*) is a semigroup> and distributivityas the definition. This would guarantee that there are at least two> elements.Larry> (this space unintentially left blank .....> make obvious deletion for email-- G.C. === Subject: Re: 27 straight lines on a cubic surface> ....> I did say that I was reading this in a *history* text so I was hoping> for an simple explanation, not a ticking off....I thought a cubic surface was a 3D object so I couldn't see how it> could necessarily contain any infinite lines.... You've had some technically correct answers, but perhaps they've leftyour intuition lagging behind a bit. If you think about Robin Chapman's example of a cone, you'll see thatinfinitely many straight lines lie on it. The same applies to ahyperboloid of one sheet, and a hyperbolic paraboloid, surfaces which youmay like to look up. Architects sometimes use them to make interestinglycurved roofs, which is not too difficult precisely because they can beframed with straight beams. All these surfaces have second-degreeequations, and have infinitely many straight lines on them. With that background, 19th-century geometers were accustomed to somesurfaces containing infinitely many straight lines, and of course manyothers containing none. It may have been quite a surprise for them tocome upon a third-degree surface having just the finite number 27 ofstraight lines embedded in it here and there. HTH === Subject: Re: Plotting a graphSorry for posting this twice but I didn't get a reponse and wasSuppose I have recorded some distances (they're actually pointer> distances between two objects in memory). Since these distances could be> very large, I have separated them into buckets which are at intervals> of Log base 2. So I then have buckets such as: 0, 2, 4, 16, ... to> 2147483648. Distance data such as: 34, 16, 15, 100 etc will be in> buckets 32, 16, 16 and 64 respectively. I then keep a count for each> bucket to know the number of distances that belong to each one of them> (actual distances are not required).I want to be able to plot this information on a graph. On the y-axis I> need to place these distances and on the x-axis I want to keep the> number of times these distances are sampled (i.e. the total number of> pointer mutations.. this can go to millions!). Now someone adviced me> that I normalize the y-axis (i.e. the distances that should be recored> as percentages) in terms of the largest heap size encountered (i.e. the> denominator in the normalization should be the same for all points on> the x axis). What exactly does this mean? What should I do? Should I> divide the bucket sizes by the largest heap size, and then depending> upon the number of distances that were measured for that particular> bucket I take a percentage of the overall mutations that there were?> Could someone guide me as to how I should do this and use the givenTiffThat's how I would do it:When your chosen bucket centers are at 2^n you should count ameasurementin that bucket i that minimizes the relative distance.100 should therefore be counted at 128 and not at 64, because 100/128< 64/100. The bucket boundaries are at 2^n/sqrt(2) and 2^n*sqrt(2)(90.51,181.02) in the example. The interval length of bucket n isl_n=2^n/sqrt(2). If a uniform distribution of your measurements shallgive a horizontal line in the histogram graph, then you have to dividethe accumulated counts in the buckets by the bucket length.Let's assume you have m=100 measurements x with uniform distribution in0<=x<=16. First and last bucket are special cases. 1 | 2 | 4 | 8 | 16 buckets at 2^n0.0 1.414 2.828 5.657 11.314 | boundaries 9 9 18 35 29 measurements counted 8.839 8.839 17.678 35.355 29.289 l_n*m/(total range) 1.018 1.018 1.018 0.990 0.990 plotted in graphHugo === Subject: ProbabilityCan some please tell me what is being asked in these problem more clearly.1st triangular means { 0 if x < a fx(x) = k{x - a) if a < x < (a+b)/2 K(b - x) if (a+b)/2 < x < b 0 if b < x, }1)Recall that for a normal random variable U with mean mu and variancesigma^2,the new random variable V = alphaU + beta is normal with mean alpha*mu +beta and variancealpha^2*sigma^2, as long as alpha /= O. Is this property true for triangularrandomvariables X? In other words, is the triangular shape of X preserved in Y =alpha*X + beta?Do I need to just look at the graph of fx(alpha*X + beta )?2)Recall that any two independent normal random variables U1 and U2 withmeans mul and mu2 and variances sigma1 and sigma2, the new random variableW = U1+ U2 is normal with mean (mul + mu2) and variance (sigma1^2 + sigma2^2). Isthisproperty true for any two independent triangular random variables Xl and X2?In other words, is Z = Xl + X2 triangular?Again, do I need to just look at the graph of fx(Xl + X2 )and since fx(x) islinear the answer is yes? === Subject: Re: Probability> Can some please tell me what is being asked in these problem more clearly.> 1st triangular means> { 0 if x < a fx(x) = k{x - a) if a < x < (a+b)/2 K(b - x) if (a+b)/2 < x < b 0 if b < x, }I am going to assume that you actually meant to use thesame symbol k, not two different symbols k and K. Thatway fx(x) has the same limit of k(a+b)/2 whetherapproaching (a+b)/2 from the left or the right.Small nit: some of your inequalities should be<=, not strictly less than, or else you have failedto define fx(x) everywhere. What is its value ata for instance?> 1)Recall that for a normal random variable U with mean mu and variance> sigma^2,> the new random variable V = alphaU + beta is normal with mean alpha*mu +> beta and variance> alpha^2*sigma^2, as long as alpha /= O. Is this property true for triangular> random> variables X? In other words, is the triangular shape of X preserved in Y => alpha*X + beta?Do I need to just look at the graph of fx(alpha*X + beta )?That won't answer the question, but it will give you apictorial clue. It will also help you to pick the rightparameters. To really answer the question you need to:1. Find the density fy(y) of the random variable Y. (It'sbecause the height is something different. You should havetransformation rules for finding fy(y) in terms of fx(y)).2. Determine if it follows the triangular form asdefined: - 0 outside some limits [ay, by] - linear with slope k from ay to (ay+by)/2 - linear with slope -k from (ay+by)/2 to by> 2)Recall that any two independent normal random variables U1 and U2 with> means mul and mu2 and variances sigma1 and sigma2, the new random variable> W = U1> + U2 is normal with mean (mul + mu2) and variance (sigma1^2 + sigma2^2). Is> this> property true for any two independent triangular random variables Xl and X2?> In other words, is Z = Xl + X2 triangular?Here you must use the transformation rules again, or answerthe question from first principles (see below).> Again, do I need to just look at the graph of fx(Xl + X2 )and since fx(x) is> linear the answer is yes?No, because again fx(x1+x2) is not the density of thevariable Y = X1 + X2.Here's what I mean by first principles, though not quiterigorous. The density of Y, fy(y) is defined by the probabilitythat Y lies in [y, y+dy] being fy(y) dy.What is the probability that Y lies in a small intervalnear y? Since Y = X1 + X2, you could have X1 = 0 andX2 = y, or X1 = y/2 and X2 = y/2, or X2 = 0 and X1 = y,etc. In general, you could have X1 be any valuex and X2 be in [y-x, y-x + dy].What is the probability of that? The probability thatX1 is in [x, x+dx] is fx(x) dx. Given a value of x,the probability that X2 is in [y-x, y-x + dy] is fx(y-x) dy. The total probability that X1 and X2 addup to some value in [y, y+dy] is integral(x = -inf toinf) fx(y) fy(y) dx dySince this is equal to fy(y) dy, then we can identify fy(y) = integral(x=-inf to inf) fx(x) fx(y-x) dxIn the first example, a similar consideration will leadyou to the correct transformation, but it's much simpler.What values does x take such that y is in a range[y, y+dy]? Keep in mind there that dy has an expressionin terms of dx.The general transformation rules come from working thisout for y being any function of x. - Randy === Subject: Re: a matrix problem related to eigevalues> I have a matrix A(n), > where A(2)= [0 1/2> 1 1/2]> A(3)= [0 0 1/3> 0 1/2 1/3> 1 1/2 1/3]> A(4)= [0 0 0 1/4> 0 0 1/3 1/4> 0 1/2 1/3 1/4> 1 1/2 1/3 1/4]> And so on.> My purpose is to find the eigevalues. Through matlab simulation I> found they should be:> 1, -1/2, 1/3, -1/4,..., (-1)^n/n> But I cannot prove it.ClimberYou can (more or less) explicitly write down the eigenvectors and thenshow that they satisfy the eigenvalue equation with the eigenvaluesyou give. This probably isn't the easiest or best method ofdemonstrating your eigenvalues are correct, but you get theeigenvectors as a bonus which you might have wanted.The kth component of the rth eigenvector for the nxn case isa_k = a(n,r,k) = sum over i of(-1)^(i-1)*(k choose i)*(r+i-1 choose i-1)*(n-i choose r-i)where 1<=r<=n, 1<=k<=n, and the rth eigenvalue is (-1)^(r-1)/r.We also defineb_k = b(n,r,k) = sum over i of(-1)^(r+i-1)*(k choose i)*(r+i-1 choose i)*(n-i choose r-i)(Slight change to middle binomial. It will turn out that a_k=b_{n-k}.)The eigenvalue equations area_k/k+a_{k+1}/(k+1)+...+a_n/n=(lambda)a_{n+1-k} (k=1...n),or subtracting consecutive equations and using lambda=(-1)^(r-1)/r,r*a_k = (-1)^(r-1)*k*(a_{n+1-k}-a_{n-k}) (k=1...n, a_0=0).The above is what we need to show - it will show the eigenvalues andeigenvectors are correct.One way to do this sort of thing is to use generating functions:So multiply a(n,r,k) by X^k*Y^r*Z^n and sum over k,r,n. You getA(X,Y,Z) = -(1-Z-YZ)/[(1-Z)(1-Z-XZ-YZ+XYZ+XZ^2)](Do the sums over k,n first, then r. You repeatedly use that sum overr of (a+r choose b)*x^r = x^(b-a)/(1-x)^(b+1).)Similarly for b(n,r,k):B(X,Y,Z) = YZ/[(1-Z)(1-Z-XZ+YZ-XYZ+XZ^2)]B(X^-1,Y,XZ) will be the generating function for b(n,r,n-k), andA(X,Y,Z)-B(X^-1,Y,XZ) = -1/[(1-XZ)(1-Z)] which is independent of Y, soa(n,r,k)=b(n,r,n-k), except possibly for r=0 which we don't careabout.Also Y(d/dY)A(X,Y,Z) = X(d/dX)((1-X)B(X,-Y,Z))which proves that r*a_k=k*(-1)^r*(b_k-b_{k-1}) i.e.,r*a_k = (-1)^(r-1)*k*(a_{n+1-k}-a_{n-k}), which is what we wanted.Alex === Subject: Re: This Week's Finds in Mathematical Physics (Week 197) This discussion of 24 reminds of Pythagoras in the 7th century B.C. Hestudied mathematics, and found the same numbers popping up over andover again, and people thought this had some deep and profoundmeaning, and eventually the Pythogoreans evolved into a religious cultdevoted to numerology. Something similar seems to be happening here.Jeffery Winkler === Subject: Re: This Week's Finds in Mathematical Physics (Week 197)> But, if we take the same formulas and think of g_1 and g_2> as *formal* variables, we are working not over the complex> numbers but over the field of rational functions in g_1 and g_2. > And, we get a formal group law over this big field. We get> all the *specific* formal group laws just by setting the> variables g_1 and g_2 to equal specific values. That's almost how we stitch together all the formal > group laws for specific elliptic curves into one big fat > formal group law. But instead of working with the field> of rational functions in g_1 and g_2, we should really > work with the field of rational functions on the moduli> space of elliptic curves.Yes, that's obvious. I should have realized that myself, guessmy brain was too tired.> We should, but I don't know if anyone has mastered the> technology to do it this way! There's a technology for> getting complex oriented cobodism theories from formal> group laws over a field, but when the field is actually> the rational functions on a *stack*, it's not really a> field anymore and I don't know if people have generalized> all the necessary theorems to this case. Ha, yeah, I wonder what _would_ it be. This relates to one thingI've been wondering about before:An algebraic stack is in a sense the categorification of the usual notion of (say) an algebraic scheme (it would be moreprecise to say an algebraic space i.e. we usually have etale orother such topology there rather than Zarisky, as far as Iunderstand). Now, a scheme is a ringed space. An algebraic space,I guess, may also be described as a ringed space, just with oneof those fancy Grothendick topologies underlying it? If so, canone describe a stack a categoringed-space i.e. a sheaf of ring(or rig) categories?Best , Squark-------------------------------------------------------- ----------Write to me using the following e-mail:Skvark_Nuclearsto@excite.exeextension in the obvious way) === Subject: Re: Zeno's paradoxes (was Re: reciting N and other sets) >> Actually I would say that many people do not understand Zeno's paradoxes.>> As an example, I offer the hordes of calculus teachers who resolve the>> Achilles paradox by explaining that infinite series can converge. This>> makes little sense when one realizes that the Achilles paradox relies on>> the fact that a particular series converges.I think you're being a little unfair here. If someone> stated the Achilles paradox and asked what is wrong> with such reasoning, is it not sufficient to point out> one step in the argument that is false? In this case,> the false assumption seems to be that an infinite series> of positive numbers cannot converge *because* it is> infinite.You might find an explicit evaluation of the particular> series to be a more satisfying resolution, but I think> that's a quibble over the meaning of resolution, not> evidence of any greater or lesser understanding.I don't really understand the second paragraph here, but that's naturalsince your statement about what the false assumption is in the Achillesparadox is puzzling to me.Let me begin with stating the standard version of the Achilles paradox:We suppose that Achilles and the Tortoise are running on a linearracetrack. Let's say that Achilles can run at twice the speed of theTortoise (the exact numbers don't really matter here). Achilles should beable to catch the Tortoise eventually (even though the Tortoise has aheadstart). But, whenever Achilles gets to where the Tortoise was, theTortoise must necessarily be a little ahead. If Achilles runs to catchup, again the Tortoise will always be a little ahead. The distancebetween Achilles and the Tortoise will diminish, but never vanish. Therefore, Achilles can never catch the Tortoise.For convenience, I'm going to change this relative form of the paradoxto an absolute form. Achilles starts at spot A and is trying to get to spot B. When he ishalfway between A and B, he needs to get to halfway between where he isand B. Even when he gets to that spot, he still has to get to the pointhalfway between where he is and B. Ad infinitum. Therefore, he cannever get to B.The common thing to do at this point is to say, But of course, we knowfrom Calculus that infinite series of positive numbers can add up to somefinite number. How silly of Zeno not to realize this!This kind of thinking misses two major points. The setting of theparadox relies on the fact that infinite series of positive numbers canaddup to some finite number. In particular, it relies on the fact thatyou can divide up a geometric segment in half, and then in half, and thenin half, ad infinitum. Knowing this fact is part of the *setting* of theparadox. It in no way resolves anything.The other point is that Zeno, as a Greek philosopher, was well aware ofthis kind of calculus-type reasoning. He lived during the fifth and fourthcenturies BC, but the Pythagoreans, for example, had been around beforehim. The kinds of things they studied, for example, constructing thegolden rectangle, and creating the spiral formed by it, would have, Ibelieve, been known to Zeno. To better see that calculus does not resolve the Achilles paradox,consider the following addition to the above absolute form of the paradox:as Achilles travels in smaller intervals of space, he alternates openingand closing his mouth, i.e. When he is traveling to the first midpoint, heopens his mouth, when he is traveling from the first midpoint to thesecond midpoint, he closes his mouth, ad infinitum. What is the positionof his mouth when he arrives at his destination? Open, or closed? Orneither?The typical objection to this question is to say, there is no last term ina sequence, so it doesn't make sense to ask this question. You can'tassign some state to each event in this infinite sequence of events, andthen ask what is the final state, since that is contradictory to theidea of an infinite sequence. I answer this objection as follows: you allowed me to describe the actionof getting to some destination as the end result of an infinite sequenceof events. If you allow that, why don't you allow me to ask this questionof Achilles opening and closing his mouth? In fact, I can describeanother variation of the paradox:Achilles can never move. For, before he can get to some spot, he needs tomove through a spot inbetween that and his starting spot. And of course,I can divide even that further. Since there is no last term in asequence, Achilles can never get started. Note this variation explicitly uses the fact that there is no last termin an infinite sequence. Actually, as I hope is clear by now, the basicissue is in describing some event to be the result of an infinite sequenceof events. I believe this kind of description is in itself wrong, but notfor mathematical reasons, but physical ones. Knowing calculus is not asolution to these kind of paradoxes, it is the beginning. Similar kindsof paradoxes come up in half-life decay problems. According to themathematics, your sample should never vanish, but become increasinglysmaller as time goes on. But of course, reality says otherwise.I don't believe there can be a purely mathematical way to resolve Zeno'sparadoxes, as I believe they are paradoxes about the nature of space andtime. Any resolution must rely on physics in some way. === Subject: Re: Zeno's paradoxes (was Re: reciting N and other sets)> >> Actually I would say that many people do not understand Zeno's paradoxes.>> As an example, I offer the hordes of calculus teachers who resolve the>> Achilles paradox by explaining that infinite series can converge. This>> makes little sense when one realizes that the Achilles paradox relies on>> the fact that a particular series converges.I think you're being a little unfair here. If someone> stated the Achilles paradox and asked what is wrong> with such reasoning, is it not sufficient to point out> one step in the argument that is false? In this case,> the false assumption seems to be that an infinite series> of positive numbers cannot converge *because* it is> infinite.You might find an explicit evaluation of the particular> series to be a more satisfying resolution, but I think> that's a quibble over the meaning of resolution, not> evidence of any greater or lesser understanding.> I don't believe there can be a purely mathematical way to resolve Zeno's> paradoxes, as I believe they are paradoxes about the nature of space and> time. Any resolution must rely on physics in some way.If I may, I'd like to share a few things on this subject:The paradox is an expression of the historical potential infinity vs.actual infinity debate, is it not? Many before and some after Cantor,including some noted mathematicians, rejected the actually infiniteand accepted instead only the potentially infinite.It's not strictly necessary to accept the existence of an actualinfinity to accept the existence of the limit of a sequence. Manymathematicians pre-Cantor and some mathematicians post-Cantor rejectedthe former and accepted the latter. (I believe that in a pre-Cantorvein, Zeno accepted the existence of the limit of a sequence withoutaccepting the existence of an actual infinity.)So the paradox seems to be resolved if and only if the existence of anactual infinity is philosophically accepted. If it is, then oneactually reaches the limit of a sequence, and if it is not, then oneactually does not reach the limit of a sequence.Note: The love-knot denotation of infinity, which we see incalculus, etc. associated with limits, is not the actual infinity, thesimplest of which is denoted by lower case omega, the ordinal numberof the natural numbers. A rather nice book by Conway, ON NUMBERScontext of his surreal numbers. It sets the love-knot (potentialinfinity) equal to lower case omega (the simplest actual infinity)rooted to the power of V (the class of all sets, which is ______infinity - Rucker calls it Absolute Infinity, denoted by him as uppercase omega.) I wonder what Zeno would have thought of that!I doubt that physics can resolve Zeno's paradoxes, since the storiesseem to use space and time only as a metaphor for the tension betweenthe two views of infinity. But I also share your doubt thatmathematics can resolve them. Accepting or rejecting actual infinityseems to be a purely a priori philosophical bias that one brings tothe table of mathematics or physics. === Subject: Re: is there any Gram-Schimt techique for integer matrix?> Maybe what you really want to do is to approximate>a real-valued orthogonal matrix by a rational-valued>orthogonal matrix?>That seems harder. The easiest case should be>2x2 matrices.I think every orthogonal matrix can be written asa product of rotations in planes (given by two of thecoordinates) possibly followed by a reflection in onecoordinate. In the 3 x 3 case, these rotations correspondto the Euler angles. So it all reduces to the 2x2 case.Now given any angle theta in (0,Pi/2), approximate cos(theta)/(1+sin(theta)) by a rational number r, andcos(theta) and sin(theta) are approximated by the rationals2 r/(1+r^2) and (1-r^2)/(1+r^2) respectively. So anyangle can be approximated by an angle whose sine and cosineare rational, and any orthogonal matrix can be approximatedby one with rational entries.Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: is there any Gram-Schimt techique for integer matrix?Hi Chip,> Actually the Gram-Schmidt procedure preserves rationality if we leave out> normalizing the output vectors to unit length.> Consider two rational vectors u, v with the residue r when u is projected> onto v:> r = u - (u,v)/||v||^2> Note that r is again a rational vector. I regard this as a widely known> fact> of uncertain provenance. It underlies such specific developments as the> root-free Cholesky decomposition, root-free QR, etc.1) The G-S can maintain rationality, but the resultant matrix R may needdifferent scaling factor for each column in order to scale them to integer.If one common scaling factor cannot be found, then by round(c*R) where c isa scaling factor, error is again introduced into the round-off procedure.2) There is no problem that G-S can definitely lead to a orthogonalmatrix... but sometimes the matrix structure will be destroyed due to theorthogonalization procedure... but here I need the matrix structure to beable to maintain considerably, and under this condition, change/modify alittle bit to make the matrix orthogonalize.For example, a typical transform is DFT, by applying the DFT transform, theoutput coefficients have interpretation of frequency components. People mayneed to post-processing on these transformed frequency components byapplying another matrix multiplication... but now if I scale the DFTtransform matrix and round to integer, the orthogonality is lost, then if Iuse G-S, the whole matrix structure will be destroyed, the post-processingmatrix multiplication with the transformed result will have no meaning atall...Is there any mathematical work for solving the above two issues?Please point me to some resources!-Walala === Subject: Re: is there any Gram-Schimt techique for integer matrix?> For some reason, I want to restrict the construction of an orthogonal> matrix within the domain of integer matrix.Is there a Gram-Schimt kind of technique which can deal with integer> matrix and get the result as near as orthogonal matrix as> possible?I am amazed that it has been a day and no-one has mentioned this yet: There is the Lenstra, Lenstra, and Lovasz basis-reduction algortihm(also called the LLL algorithm or L^3 algorithm). This is one of themost important algorithms of the last 20 years and it has numerousapplications. New applications are being found regularly. It iswidely used in computer-algebra packages. === Subject: Re: is there any Gram-Schimt techique for integer matrix? <3a64b911.0308241216.7cda9635@posting.google.comFor some reason, I want to restrict the construction of an orthogonal> matrix within the domain of integer matrix.Is there a Gram-Schimt kind of technique which can deal with integer> matrix and get the result as near as orthogonal matrix as> possible?An old fact which can be incorporated in your efforts: (First of all, the term orthogonal matrix referstraditionally to a matrix with orthonormal columns, whetherwe like it or not; so an integer orthogonal matrix would haveonly 0's, 1's and -1's as entries in appropriate places. Ipresume you are looking for orthogonal matrices with rationalentries.)The fact is: If U is an orthogonal matrix (U'*U=U*U'=I) suchthat I+U is invertible (e.g. U is sufficiently close to I) then H = (U-I)/(U+I) (matrix division)is skew-symmetric, and elementary inversion tells us that U = (I+H)/(I-H)The nice thing is: You can find rational skew-symmetric matricesK arbitrarily close to H, and then by continuity V = (I+K)/(I-K)are rational orthogonal matrices close to U.This V can be obtained from K using exact rational arithmetic(representing rational numbers as pairs of integers).Of course, beause of rounding errors, H may not come out exactlyskew-symmetric, so you skew-symmetrize by (H-H')/2.Hope it helps, ZVK(Slavek). === Subject: Re: is there any Gram-Schimt techique for integer matrix?> I am amazed that it has been a day and no-one has mentioned this yet:> There is the Lenstra, Lenstra, and Lovasz basis-reduction algortihm> (also called the LLL algorithm or L^3 algorithm). This is one of the> most important algorithms of the last 20 years and it has numerous> applications. New applications are being found regularly. It is> widely used in computer-algebra packages.Can your LLL algorithm work under the following two restrictions?1) The G-S can maintain rationality, but the resultant matrix R may needdifferent scaling factor for each column in order to scale them to integer.If one common scaling factor cannot be found, then by round(c*R) where c isa scaling factor, error is again introduced into the round-off procedure.2) There is no problem that G-S can definitely lead to a orthogonalmatrix... but sometimes the matrix structure will be destroyed due to theorthogonalization procedure... but here I need the matrix structure to beable to maintain considerably, and under this condition, change/modify alittle bit to make the matrix orthogonalize.For example, a typical transform is DFT, by applying the DFT transform, theoutput coefficients have interpretation of frequency components. People mayneed to post-processing on these transformed frequency components byapplying another matrix multiplication... but now if I scale the DFTtransform matrix and round to integer, the orthogonality is lost, then if Iuse G-S, the whole matrix structure will be destroyed, the post-processingmatrix multiplication with the transformed result will have no meaning atall...Is there any mathematical work for solving the above two issues?Please point me to some resources!-Walala === Subject: Re: Mean Reverting ProcessOops... the equation didn't come out right. Let me retype it out in words:dX = Kappa*(Miu - X)dt + sigma*X^gammadW === Subject: Re: Mean Reverting ProcessI'm not sure whether I can help, but I have some thoughts on this. In > the context of mathematical finance mean-reverting stochastic > differential equations appear when modeling interest rate dynamics. For > example the Vasicek model is mean-reverting. It is characterized by the > following SDE:dr(t)=(b-ar(t))dt+sdW(t)Here r is the interest rate you're modeling, in this case the short > rate. b and a are parameters to be estimated from the market, W is a > standard Brownian motion under the martingale measure (which is not > unique) and s is the volatility. This process is now reverting to the > mean level b/a. (Note that you can write the SDE in integral form to be > able to see what kind of process it is. The SDE is solvable and the > solution has a Gaussian distribution.)So sure, it depends on the parameters to which level the process is > reverting. However, where this is explained in a more general context I > don't know.By the way, you speak of *the* mean reverting process. Which process > is that?Cheers, Michamost general one, i.e. dX = κ(μ - X)dt + σX^γdW.To rephrase my question, given the process above and that κ > 0,μ > 0, is 0 a reflecting or adsorbing barrier for X? I alwaysthought it to be always reflecting. But now, I am told it depends onthe parameters, i.e. for certain parameters, it will be adsorbing, forothers, it will be reflecting.I find that hard to visualise because as X -> 0, drift is positive(κ*μ) and volatility -> 0, i.e. it should be reflecting. === Subject: 7x+1Has there been any progress in the 7x+1 problem?. Do all starts end in 1 or19? Are there any conjectured unbounded cases? You start with any integer ofthe form 6*n+1 or 6*n-1. Multiply by seven add one then divide by 2 or 3 asmany times as possible and repeat the process until you get 1 or 19 orinfinity.. Example 7,50,25,176,88,44,22,11,78,39,13,92,46,23,162,81,27,9,3,1. This is form of the Colletz 3x+1 problem === Subject: Re: Name this operation?> Can anyone tell me if the following operation has a special name in math, or> possibly other fields?A(x), B(x) and C(x) are discrete time series. T(a, b) is some function> taking and returning numbers.C(x) = T(A(x), B(x) )I found something similar in fuzzy sets called 'fuzzy conjunction' but that> had limitations on what T could be.> With functions instead of time series, I would call it function composition.> Is there such a thing as time series composition?> I would not use the term Composition since that would be something like C(x)=B(A(x)). Your T is essentially mixing the two outputs, so just use Mixer or some other word with similar meaning.Since most statistical filters tend to do the inverse operation that your T does, you could also call it an inverse Filter. The scenario is given signal a(x) and disturbance b(x) find the random output. So you unfilter them with T.MK === Subject: Re: UFO Warp Drive Metric Engineering> Mr. Sofruity, I know you think your verbal diarrhea impresses the> rubes, but can you please write in English next time? Stringing> together a random assortment of buzzwords does not mean you've> produced a readable sentence.check the picture, move a couple lines of the letters around and theequation is Sta TrkHerc === Subject: Parity of the number of prime factors.This is a very informal post, as is usual for me.It's something I was playing around with recently. I was alwaysstruck by the Hardy result that the very rough number of prime factorsof a random integer is of the order of log log N, and that this was so*regardless* of whether those prime factors were considered in total,or merely the distinct prime factors.So anyway, it struck me recently that the number of prime factors ofa very large random number ought to be equally likely to be oddor even. The randomness in the parity ought to be 50% each way.Once the numbers become big enough, of course. A few numericalchecks seemed to bear this out quite well.But then I had to concern myself with the above matter - distinct primefactors, or total number of same? Either way, the 50-50 predictionstill seemed to come out OK.So far so good. But you know I'm a bi-dichotomy fan; so I checkedthe combinations of the two parities:- total/distinct vs odd/even;expecting them to be more or less independent.But no! The figures were WELL askew....Numbers 1 to 100: distinct Even Odd Even 39 10 total Odd 18 33 ...and then...Numbers 101 to 200: distinct Even Odd Even 35 8 total Odd 17 40 (I did these by hand at first)These struck me as being WAY too far off 25-25-25-25 for even the famedlaw of small numbers to account for; even considering that in mattersof primality the first several numbers are typically very atypical. (Hehe!)So then I thought about this apparent strong correlation betweenthe distinct-parity and the total-parity, and it finally hit me:-of COURSE they must be correlated! - because a very substantialproportion of numbers have to be the same - being square-free!!!parity for both. So about 30% each of all numbers would go into the topleft and bottome right cells, straight off. Then the remainder, thesquareful numbers, (why is this term never used?), would now presumablysplit up into 10% for each cell. Giving a final ratio of 40-10-10-40.This seemed to fit the data so far reasonably well!So now it was time to make use of the indispensible Maple,and check out bigger numbers. Here is what I got...Numbers 1 to 1000: distinct Even Odd Even 379 114 total Odd 158 354 ...and then...Numbers 1001 to 2000: distinct Even Odd Even 368 128 total Odd 140 364 ...and then...Numbers 2001 to 3000: distinct Even Odd Even 361 128 total Odd 127 384Not too bad, but not too good either. There seems to be a consistent*under-appearance* of equi-parity types, the main diagonal, from the newpredicted count. Finally a giant check from among higher numbers still...Numbers 10001 to 20000: distinct Even Odd Even 3659 1361 total Odd 1289 3691Numbers 40001 to 50000: distinct Even Odd Even 3699 1347 total Odd 1269 3685The figures are internally very consistent with one another,(so much for the law of small numbers!), but clearly NOT followingthe predicted 40-10-10-40 pattern.After some more thought it became a little more clear why.My idea that the square-free numbers would be 50-50 was OK; but my glibassumption that the squareful ones would also be, was somewhat off.For example, there is no doubt a fixed fraction of numbers of the typep^2.q.r.s... , those with any number of factors, all single except for ONE double; we might call them the almost square-free numbers.I'm sure the fraction can be worked out in principle, just as forthe square-frees. And this group will ALL contribute toward theoff-diagonal cells in the counts above, re-skewing the data that waya little. These will in turn be countered by the not-quite-so-squarefrees,of type p^2.q^2.r.s.t... , but these latter will be still fewer, over all.So finding the proper counts for the four parity types of numberswould be quite a job:- rather beyond me, I fear, but perhaps someoneelse here would like to have a go?Unless it's all in Hardy already, of course...----------------------------------------------------- ------------------------- Taylor W.Taylor@math.canterbury.ac.nz-------------------------------- ---------------------------------------------- a priori a posteriori .--------------------------. synthetic | math | science | |------------+-------------| The KANT-TAYLOR bi-dichotomy. analytic | logic | sophistry | `--------------------------' === Subject: Re: Parity of the number of prime factors.> This is a very informal post, as is usual for me.It's something I was playing around with recently. I was always> struck by the Hardy result that the very rough number of prime factors> of a random integer is of the order of log log N, and that this was so> *regardless* of whether those prime factors were considered in total,> or merely the distinct prime factors.So anyway, it struck me recently that the number of prime factors of> a very large random number ought to be equally likely to be odd> or even. The randomness in the parity ought to be 50% each way.> Once the numbers become big enough, of course. A few numerical> checks seemed to bear this out quite well.But then I had to concern myself with the above matter - distinct prime> factors, or total number of same? Either way, the 50-50 prediction> still seemed to come out OK.So far so good. But you know I'm a bi-dichotomy fan; so I checked> the combinations of the two parities:- total/distinct vs odd/even;> expecting them to be more or less independent.But no! The figures were WELL askew....> Numbers 1 to 100: distinct Even Odd> Even 39 10> total Odd 18 33 ...and then...Numbers 101 to 200: distinct Even Odd> Even 35 8> total Odd 17 40 (I did these by hand at first)> These struck me as being WAY too far off 25-25-25-25 for even the famed> law of small numbers to account for; even considering that in matters> of primality the first several numbers are typically very atypical. (Hehe!)So then I thought about this apparent strong correlation between> the distinct-parity and the total-parity, and it finally hit me:-> of COURSE they must be correlated! - because a very substantial> proportion of numbers have to be the same - being square-free!!!parity for both. So about 30% each of all numbers would go into the top> left and bottome right cells, straight off. Then the remainder, the> squareful numbers, (why is this term never used?), would now presumably> split up into 10% for each cell. Giving a final ratio of 40-10-10-40.>...I have no idea if this is directly applicable to all of this, butidentity 6 from:http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe= off&threadm=b4be2fdf.0307111419.24913cca%40posting.google.com& rnum=5&prev=is (quote): limit{m->oo} (1/m) *sum{k=1 to m} (M(k) - N(k))= sum{p=primes} 1/(p(p-1)) (which converges),where M(k) is sum of all exponents in prime factorization of k, andN(k) is the number of distinct primes dividing k. I hope the above is true. But whether the above is true or not, andeven if it is not useful to your problem directly, *perhaps* still thegeneral theorem in the link might be of use somehow.These kind of problems can also be usually tackled usingzeta-function-like products.For instance (using my M() and N() notation from above): product{p=primes} 1/(1 -x/p^r) =sum{m=1 to oo} x^M(m)/m^r.Andproduct{p=primes} (1 +x/(p^r -1)) =sum{m=1 to oo} x^N(m)/m^r.(I think)So, assuming one can take derivatives of both sides of each of aboveequations without problems with uniform convergence,zeta(r) sum{p=primes} 1/(p^r -1) =sum{m=1 to oo} M(m)/m^r.And:zeta(r) sum{p=primes} 1/p^r =sum{m=1 to oo} N(m)/m^r.(I think.) And, back to the first product/sum identities (more related to yourproblem):sum{m=1 to oo} (-1)^M(m)/m^r =zeta(2r)/zeta(r).And,(sum{m=1 to oo} (-1)^N(m)/m^r) *(sum{k=1 to oo} 2^M(k)/k^r)= zeta(r).(I think...)This will all not necessarily be useful to the original poster, butmaybe.Leroy Quet === Subject: Re: Parity of the number of prime factors.> Then the remainder, the squareful numbers, (why is this term never > used?)It is used, but with a different meaning; n is squareful means for all primes p if p divides n then p^2 divides n.-- === Subject: Vigier Conference in Paris Sept 15-18Richard et-alre: http://qedcorp.com/APS/Revised Abstract for Vigier Paris ConferenceWhat is the Universe made out of? Jack Sarfatti ISEP, San FranciscoThe Question is: What is The Question? Archibald Wheeler April The new observational data of precision cosmology has drastically changed our conception of the cosmos beyond what anyone imagined even as late as 1999. It is now known with a precision uncertainty of about 2% that only 4% of the stuff of our spatially flat universe we detect with our telescopes on the large scale of 10 megaparsecs and greater is made out of ordinary atoms, radiation, ionized plasma, MHD near fields, neutrinos etc. A whopping 96% of all the stuff of the world is something else! What is it?All the King's Horses and All The King's Men are, I suggest, asking the wrong question of Dame Nature. They are looking for some kind of new of the missing mass as shown from gravitational lensing, COBE, WMAP etc. It started to become clear in 1999 with the type 1a supernovae data that ~ 73% of the missing stuff of the universe in the past light cones of our telescopes is an exotic anti-gravitating zero point energy density phase of vacuum with w = pressure/(energy density) = 1. The zero point energy vacuum fluctuations of all the quantum fields obey w = -1 and it is the quantum zero point pressure that dominates Einstein's generalized Newton-Poisson local field equation for the gravitational properties of the vacuum. Einstein's cosmological constant / is the net large scale effect of the quantum zero point vacuum in the FRW metric of standard cosmology.It has been a big mystery why / is so small when naive arguments suggest it is enormous. I claim to have solved this problem by noticing that the Dirac filled negative energy off-mass-shellelectron vacuum is intrinsically unstable to the non-perturbative ODLRO/ BCS pairing of virtual electrons and positrons near the zero Fermi energy of momentum ~ 10^19Gev/c. This coheres the random zero point fluctuations from all the quantum fields like throwing oil on the turbulent waters, calming them, giving rise to a local scalar vacuum field /zpf(x) that is closely related to the complex macro-quantum scalar field PSI(x) with Higgs amplitude |PSI(x)| and Goldstone phase argPSI(x) of chaotic inflationary cosmology. Indeed Einstein's geometrodynamic field guv(x) emerges out of the modulation of the Goldstone phase of the complex scalar inflation field consistent with Hagen Kleinert's idea of the 4D world crystal lattice with curvature as disclination defect density and torsion as dislocation defect density. This local field /zpf(x) merges to Einstein's cosmological constant for large scales.One uses a wavelet transform generalization of the Wigner phase space density replacing momentum with scale. The result is that /zpf(x) = 0 is the ordinary non-gravitating vacuum of common sense. However, /zpf(x) can be positive and negative as well as zero. /zpf(x) > 0 is dark energy with negative zero point pressure forming an anti-gravitating exotic vacuum that is ~ 73% of the large scale structure of the universe causing the cosmological expansion to accelerate rather than slow down. Similarly /zpf(x) < 0 is the elusive dark matter demystified with positive zero point pressure forming a gravitating exotic vacuum region like the spherical halos holding galaxies together. A large clump of dark matter exotic vacuum /zpf < 0 whose internal structure is w = -1 nevertheless moves as a coherent rigid blob with an apparent w ~ 0.justifying J.P. Vigier's idea of the spatially extended electron for example that arises in his tight atomic state theory for a new form of atomic energy release. The rotating electron is Newton's hard massy sphere of size e^2/mc^2 ~ 1 fermi with a strongly gravitating /zpf ~ - 1/(1fermi)^2 dark matter core preventing the self-charge from exploding and also bcing the centrifugal inertial force from the h/4pi rotation. This real on-mass-shell electron is a Bohm hidden variable surrounded by a plasma cloud of virtual electron-positron pairs extending out to the Compton wave length h/mc ~ 10^-11 cm (more accurately ~ 137 times the classical electron radius). But the electron scattering. That is easily explained as strong micro-curvature excited by deep probe momentum transfers of the Heisenberg scattering microscope. The same idea applies to lepto-quarks in general and is consistent with the QCD Lite bag model in which the hadronic masses are ~ 1Gev mostly from trapped kinetic energy of the light lepto-quark rest masses m ~ e^2|/zpf|^1/2 ~ 1 Mev. The hadronic Regge trajectory data is also almost trivially understood in this picture since the universal microscopic |/zpf| ~ 1/(1fermi)^2 simply determines the measured string tension alpha' ~ 1/(1Gev)^2. Susskind's holographic universe plays a key role in the new model because, the micro |/zpf| = Lp^-4/3(c/Ho)^-2/3 from quantum gravity and cosmology together where Lp^2 = hG/c^3 ~ 10^-66 cm^2, and 1/Ho ~ 13.7 ion years so that c/Ho ~ 10^28 cm.Last, but not least, new vista's for Hal Puthoff's dream of metric engineering space-time open up with a simple example of how these macro-quantum cohered zero point energy ideas apply to Alcubierre's toy model warp drive metric as well as to recent notions of the coupling of rotating superconductors to the space-time continuum for potential application to propellantless space vehicle propulsion.As an epilogue Archibald Wheeler and his students (notably W.H. Zurek) are fascinated with IT FROM BIT, i.e. geometry and matter-energy from information. Wheeler also envisions the participatory universe as a self-excited circuit. What is missing from Wheeler's Vision is the complementary BIT FROM IT forming a two way relation in the sense of Bohm and Hiley in which the parallel universes on a single post-inflationary bubble self-organize and emerge in the entropy-lowering spontaneous breakdown of vacuum symmetry in the BCS pairing of virtual electrons and positrons setting the direction initial arrow of time of the global cosmic clock in our Hubble-horizoned FRW metric large scale universe on the bubble that is one of an infinity of bubbles. A single post-inflationary bubble can be pictured as an expanding and accelerating flat infinite rubber sheet with a lattice of circular Hubble horizons that are Goldstone phase singularities analogous to the lattice of vortex cores in a type II superconductor where the local macro-quantum order parameter PSI (x) vanishes.http://qedcorp.com/APS/Ukraine.doc to be published in Progress in Quantum Physics Research (Nova)http://qedcorp.com/APS/Vigier4.pdfhttp://qedcorp.com/APS /warpdrivephysics.pdfFinal call for papers to 4th Vigier Paris symposium. Details on conference website;www.mindspring.com/~cerebroscopicPlease excuse lack of apparent motion or updates on the website.A number of factors have caused this; I have been in both Europe and Asia for the past month or so and although I can send and receive email; I do not no how to upload remotely, now that I am back in USA I have also had to move my office and thus computer is not yet set up; in spending some hours preparing to use a friends computer at the last minute he changed his mind being fearful of the recent virus and worm attacks.So eventually the site will be updated with abstracts and program; but perhaps not in any timely manner. This symposium will also be smaller than expected due to delays and to 's strokes and travel fears regarding SARS and remnants of 9-11 etc etc.Currently we expect about 30 participants a smaller but goodly number which can also lead to some time for longer talks and round table disdussions which is not so often available.Currently a luncheon is planned with on the Tuesday of the conference; this will probably be his only attendence; although small groups may go to his home for modest visits.I hope this allays some concerns for small actions on website...If you haven't sent abstract please do so immediatley, reg fees can be paid on site in Paris. Get your hotel from searching the Paris-hotels web listed on the conference pageThere will be a wine and cheese gathering on the Saturday before the conference in the Gravitation dept. Tour 22 etage 4 from 7 to 9 PMWe look forward to seeing you soon,R AmorosoChairman (and person to blame for any blame due) === Subject: Re: probability and/or logic puzzle> Start with a normal deck of playing cards.> Discard all but the aces and kings. Now you > have eight cards left. Your friend draws > two cards and hides them from you. He truthfully > tells you that at least one of his cards is an ace. > What is the probability that he holds two aces?Impossible to tell without further information. Scenario 1: my friend always tells me at least one card is an ace whenever both cards are aces, otherwise he tells me nothing. In this case the probability that he holds two aces is 1. Scenario 2: my friend always tells me at least one card is an ace whenever exactly one card is an ace, otherwise he tells me nothing. In this case the probability that he holds two aces is zero. Basically, the answer can be anything from 0 to 1 inclusive, depending on the psychology of my friend.-- === Subject: Re: probability and/or logic puzzle Start with a normal deck of playing cards.> Discard all but the aces and kings. Now you > have eight cards left. Your friend draws > two cards and hides them from you. He truthfully > tells you that at least one of his cards is an ace. > What is the probability that he holds two aces?Impossible to tell without further information. Scenario 1: my friend always tells me at least one card is an ace > whenever both cards are aces, otherwise he tells me nothing. In > this case the probability that he holds two aces is 1. Scenario 2: my friend always tells me at least one card is an ace > whenever exactly one card is an ace, otherwise he tells me nothing. > In this case the probability that he holds two aces is zero. Basically, the answer can be anything from 0 to 1 inclusive, > depending on the psychology of my friend.We had one deal and one statement. I don't think you can assume out ofthis one to one relationship.We can therefore assume that had the deal been two kings, our friendwould have said at least one is a king. How would this change yourthinking?Then, suppose that the friend's statement said, one is the king ofdiamonds.Would that alter your thinking?Eldon Moritz === Subject: Re: probability and/or logic puzzle>Yes, this came out of a textbook . . .>but I passed that class about 25 years ago,>so I'm not cheating on my homework.> Start with a normal deck of playing cards.> Discard all but the aces and kings. Now you > have eight cards left. Your friend draws > two cards and hides them from you. He truthfully > tells you that at least one of his cards is an ace. > What is the probability that he holds two aces?Number of pairs of which one is an ace: A + K = 16 A + A = C(4,2) = 6Fraction of these in which both are aces = 6/16 = 3/8.That's the probability that he holds two aces.> > He replaces the cards and you shuffle them well.> Your friend again draws two cards, and truthfully > tells you that one of them is the ace of spades.> What is the probability that he now holds two aces?> (Hint: this isn't the same answer as above!)Number of pairs of which one is the ace of spaces:7.Number of these in which the second card is an ace:3.Fraction of these hands in which both are aces: 3/7.>Okay . . . I don't get it. Why isn't it the same >answer both times?Because, in the language of probability, both the conditionedpopulation and the event being measured differ in the two cases.In the first case we could have 6 successes out of a total of 16candidates. In the second case both the number of candidates (7) andthe number of successes (3) were different.These are conditional probabilities, the probability that A is truegiven that you know B to be true, written P(A | B). The first one is:P(he holds any two aces | he holds any one ace)The second one is:P(he holds an ace of spaces and another ace | he holds an ace ofspades).You can see that the statements marked A and B are not the same. - Randy>Also, in case it matters, this isn't a math >book. It's a (non-mathematical) logic book.Ted Shoemaker>shoemakerted@yahoo.com === Subject: Re: probability and/or logic puzzle>Now you > have eight cards left. Your friend draws > two cards and hides them from you. He truthfully > tells you that at least one of his cards is an ace. > What is the probability that he holds two aces?3/7And I don't see why it isn't the same answer for both questions, unless its at question somehow.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~ I am by, hence my e-mail. === Subject: Re: Origin of Ben lin's Force EquationsWas Ben lin in France when he discovered these? Thou shalt not know, however: from henceforth;All in all, I'd rather he was in Philadelphia.~.Thou shalt not know this,lin...F=Gm^2/w^2 for a wave in and out of atom Thou shalt not know this,~.No thou shalt not know.Satan the Devil === Subject: Choice and Peano AxiomsThe way the Axiom of Choice reads in my book (paraphrased since I can't dothe symbols).*****Let I and X be non-empty sets, and let A_alpha be a non-empty subset of Xfor each alpha in I. Then, there exists a function f from I to X such thatf(alpha) is in A_alpha.*****Assuming the Peano axioms, and induction in particular, I am looking for thecases where AC follows. One case is when X is countable. That's becauseinduction and the existence of 1 imply that N is well ordered and so anycountable set can also be well ordered, and we can just choose the leastelement for each A_alpha.That one is pretty easy, but I'm having trouble in the case where I iscountable and the A_alphas are possibly uncountable. Does AC follow then?With induction, I can logically define an arbitrarily long -finite-sequence, but how do I go from that point to an -infinite- sequence withoutassuming AC? Maybe it's obvious, but am getting very confused and itshurting my head. :( The problem is that I can't always tell if I amaccidentally assuming the thing I need to prove.Marshall === Subject: Re: Choice and Peano Axioms>The way the Axiom of Choice reads in my book (paraphrased since I can't do>the symbols).>*****>Let I and X be non-empty sets, and let A_alpha be a non-empty subset of X>for each alpha in I. Then, there exists a function f from I to X such that>f(alpha) is in A_alpha.>*****>Assuming the Peano axioms, and induction in particular, I am looking for the>cases where AC follows. One case is when X is countable. That's because>induction and the existence of 1 imply that N is well ordered and so any>countable set can also be well ordered, and we can just choose the least>element for each A_alpha.This will work if X is well-ordered, countable or not.>That one is pretty easy, but I'm having trouble in the case where I is>countable and the A_alphas are possibly uncountable. Does AC follow then?Even if the A_alphas are all countable, it does notfollow. Solovay has a model in which the reals are theunion of a countable number of countable sets. Even ifthe A_alphas are all 2-element sets, one might not beable to choose an element from each; the problem ofchoosing one sock each from a countable family of pairsof socks involves some choice.>With induction, I can logically define an arbitrarily long -finite->sequence, but how do I go from that point to an -infinite- sequence without>assuming AC? Maybe it's obvious, but am getting very confused and its>hurting my head. :( The problem is that I can't always tell if I am>accidentally assuming the thing I need to prove.For the relation of these, and other, properties to theAxiom of Choice, I suggest Howard and Jean Rubin,_Consequences of the Axiom of Choice_. Relative strengthsof various propositions are given.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: an Axiom of Choice question (was: Choice and Peano Axioms> Even if the A_alphas are all countable, it does not> follow. Solovay has a model in which the reals are the> union of a countable number of countable sets. Even if> the A_alphas are all 2-element sets, one might not be> able to choose an element from each; the problem of> choosing one sock each from a countable family of pairs> of socks involves some choice.A question I sometimes ask...Give an example of an object, defined in the language of ZF,such that (1) it is provable in ZF that it is a countably infiniteset of pairs, but that (2) there is no obvious choice function.When you omit countably infinite I can do it: the set ofall pairs of subsets of R. But of course that is not countablyinfinite.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Choice and Peano Axioms> The way the Axiom of Choice reads in my book (paraphrased since I can't do> the symbols).*****> Let I and X be non-empty sets, and let A_alpha be a non-empty subset of X> for each alpha in I. Then, there exists a function f from I to X such that> f(alpha) is in A_alpha.> *****Assuming the Peano axioms, and induction in particular, I am looking for the> cases where AC follows. One case is when X is countable. That's because> induction and the existence of 1 imply that N is well ordered and so any> countable set can also be well ordered, and we can just choose the least> element for each A_alpha.That one is pretty easy, but I'm having trouble in the case where I is> countable and the A_alphas are possibly uncountable. Does AC follow then?No, because if it did you could well-order any uncountable set.-- Aatu Koskensilta (aatu.koskensilta@xortec.fi)Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Choice and Peano AxiomsThat one is pretty easy, but I'm having trouble in the case where I is> countable and the A_alphas are possibly uncountable. Does AC followthen?> No, because if it did you could well-order any uncountable set.Ok. I didn't say clearly what I meant to say. What I meant was does theconclusion of AC follow in this particular case? If I is countable, isthere a function f from I to X so that for each alpha in I, f(alpha) is inA_alpha? electron-dot-cloud are galaxies === === === Subject: Re: normal forms for algebraic numbers>Given the labels:>A: algebraics (roots of univariate polys with rational (wlog integer)>coefficients)>Modulo any qualms you may have for the definitions above, what is the >status of unique normal forms for A, or other subsets between Q and A, >e.g. Q augmented with just sqrt?For algebraic numbers, a unique normal form is given by 1) the minimal polynomial P(x) of which the number is a root (i.e a polynomial with integer coefficients, irreducible over the rationals, leading coefficient positive, gcd of coefficients 1, having this number as a root, and of these polynomials the unique one with minimal degree), and2) an index to specify which root of P(x)For example, in Maple the indexed RootOf notation can be used for such a form. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: normal forms for algebraic numbers> >>Given the labels:> >>A: algebraics (roots of univariate polys with rational (wlog integer)>>coefficients)> >>Modulo any qualms you may have for the definitions above, what is the >>status of unique normal forms for A, or other subsets between Q and A, >>e.g. Q augmented with just sqrt?For algebraic numbers, a unique normal form is given by > 1) the minimal polynomial P(x) of which the number is a root > (i.e a polynomial with integer coefficients, irreducible over the > rationals, leading coefficient positive, gcd of coefficients 1, > having this number as a root, and of these polynomials the unique > one with minimal degree), and> 2) an index to specify which root of P(x)OK, it's easy to see that this is unique. (aren't algebraic -integers-such things with the extra restriction that the leading coeff is 1?)However, I have trouble figuring out how to compute with this, that is:1) given an algebraic number, say sqrt(1+sqrt(17+cuberoot(19)))- cuberoot(5)Oh. OK, that's not necessarily algebraic according to my definition.So...uh... how about I change the set on you but keep the same question.For -forms- involving arbitrary roots, etc, no vars, is there a uniquenormal form? I simplemindedly can't see an obvious way to convert theabove form to a poly of which it is a root. (for some forms yes, but notall (e.g. like the one above)).Likewise is there a way (algorithm) to get a such a form from a polynomial?and2) I realize there exists a way to index (totally order) the roots, butI wonder what is computationally acceptable. That is, I can do itnumerically (Sturm sequence/binary search), but is there a better way?more symbolic? an implicit who cares?> For example, in Maple the indexed RootOf notation can be used for such a > form. Mathematica, too. But MMA documentation is not very expressive about its methods.-- HarrisLehrstuhl fuer Automatentheorie, Fakultaet InformatikTechnische Universitaet Dresden, Deutschlandhttp://tcs.inf.tu-dresden.de/~harris === Subject: Re: normal forms for algebraic numbers|>OK, it's easy to see that this is unique. (aren't algebraic -integers-|>such things with the extra restriction that the leading coeff is 1?)|>However, I have trouble figuring out how to compute with this, that is:|>1) given an algebraic number, say|> sqrt(1+sqrt(17+cuberoot(19)))- cuberoot(5)|>Oh. OK, that's not necessarily algebraic according to my definition.|>So...uh... how about I change the set on you but keep the same question.|>For -forms- involving arbitrary roots, etc, no vars, is there a unique|>normal form? I simplemindedly can't see an obvious way to convert the|>above form to a poly of which it is a root. (for some forms yes, but not|>all (e.g. like the one above)).|>Likewise is there a way (algorithm) to get a such a form from a polynomial?Yes, there is. In Maple:> q:= sqrt(1+sqrt(17+19^(1/3)))-5^(1/3); P:=sort(evala(Norm(convert((q-x),RootOf)))); 36 34 33 31 30 29 P := x - 18 x + 60 x - 720 x + 3282 x + 9720 x 28 27 26 25 - 16146 x - 78820 x + 295956 x - 287820 x 24 23 22 21 - 1792506 x + 13743360 x + 40209912 x - 14096280 x 20 19 18 - 279700236 x + 387516960 x - 523775078 x 17 16 15 + 730492920 x + 11686352778 x + 85080734460 x 14 13 12 + 9488713392 x - 254456753220 x - 616299273531 x 11 10 9 + 857922985800 x + 1211443482834 x + 277908076980 x 8 7 6 - 5266258028172 x + 3345011588160 x + 1484108332884 x 5 4 3 + 4356940483080 x - 14804808163020 x + 15364752902640 x 2 - 7922401473600 x + 2087405575200 x - 225309308400Of course, then you have to figure out which of the 36 roots of P this is,which might not be so simple (requiring high-precision numericalcalculation). === Subject: Re: normal forms for algebraic numbers> |>For -forms- involving arbitrary roots, etc, no vars, is there a unique> |>normal form? I simplemindedly can't see an obvious way to convert the> |>above form to a poly of which it is a root. (for some forms yes, but not> |>all (e.g. like the one above)).> |>Likewise is there a way (algorithm) to get a such a form from a > polynomial?Yes, there is. > In Maple:> >>q:= sqrt(1+sqrt(17+19^(1/3)))-5^(1/3); P:=sort(evala(Norm(convert((q-x),RootOf)))); 36 34 33 31 30 29> P := x - 18 x + 60 x - 720 x + 3282 x + 9720 x...> Of course, then you have to figure out which of the 36 roots of P this is,> which might not be so simple (requiring high-precision numerical> calculation).so to get the polynomial P(x) from the form q, there -seems- to be somekind of elimination going on (take a reasonably high power of q,subtract off a smaller power of q which removes the leading term withrespect to the radicals, continue with smaller and smaller powers).But I don't see the actual details of the process. Is it obvious? Is this just basic algebra?And I don't see how to go the other direction, that is, get some rootbased form from the polynomial.OK, with just a rereading of that, it seems really naive. Slappingmyself in the forehead, that would be equivalent to solving an arbitrary polynomial in radicals which is known to be undecidable. So finding a polynomial, one of whose solutions is a given algebraic, is quite a bit easier than finding the algebraic itself. Sounds like a one-way function to me!!-- HarrisLehrstuhl fuer Automatentheorie, Fakultaet InformatikTechnische Universitaet Dresden, Deutschlandhttp://tcs.inf.tu-dresden.de/~harris === Subject: Re: normal forms for algebraic numbers> OK, with just a rereading of that, it seems really naive. Slapping> myself in the forehead, that would be equivalent to solving an arbitrary> polynomial in radicals which is known to be undecidable.It's not undecidable -- it's impossible.-- Timothy Murphy e-mail: tim@birdsnest.maths.tcd.ietel: +353-86-233 6090s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: normal forms for algebraic numbers> >>OK, with just a rereading of that, it seems really naive. Slapping>>myself in the forehead, that would be equivalent to solving an arbitrary>>polynomial in radicals which is known to be undecidable.It's not undecidable -- it's impossible.Poor usage/memory on my part. so some of what I said is thenmeaningless..Then what is known here?:1) Is it decidable whether or not a particular polynomial has solutionsin radicals? (compute the galois group of a poly, check for beingsolvable; that is are those two procedures decidable) If so, what is thecomplexity?2) given a formula in +,-,*,/, and radicals, and its computed minimalpolynomial (for which the formula is a solution), are the other realsolutions expressible similarly (as formulas in +,-,+,/,radicals)?-- HarrisLehrstuhl fuer Automatentheorie, Fakultaet InformatikTechnische Universitaet Dresden, Deutschlandhttp://tcs.inf.tu-dresden.de/~harris === Subject: Re: normal forms for algebraic numbers> Then what is known here?:1) Is it decidable whether or not a particular polynomial has solutions> in radicals? (compute the galois group of a poly, check for being> solvable; that is are those two procedures decidable) If so, what is the> complexity?It is decidable.There is an algorithm for determining the galois group of a polynomial(ie of the splitting field of the polynomial).IIRC, such an algorithm is given in van der Waerden, Modern Algebra.(I'm not certain of that, as both my volumes have been stolen.However, I am certain there is such an algorithm.)Since the galois group is finite, any question about it is decidable --eg to determine if it is simple one goes through all subsets,and decides if each one is a normal subgroup.[I'm not suggesting that is a sensible thing to do; just that it is a possible procedure.]I don't know the complexity of the algorithm,but I would be slightly surprised if it were not polynomial timein the degree of the polynomial. > 2) given a formula in +,-,*,/, and radicals, and its computed minimal> polynomial (for which the formula is a solution), are the other real> solutions expressible similarly (as formulas in +,-,+,/,radicals)?Yes, the other roots (ie the conjugates of the given root)are obtained by replacing each nth root by the other nth roots,eg replacing a^{1/n} by wa^{1/n} where w is an nth root of 1.-- === Subject: Re: horseshoe mapSmale forever!> How to construct the periodic table> for the horseshoe map for periods up to 6? Thank you very much!:) === Subject: Re: algrithm to list all possible partitioning of a network> answer. It looks the question was not sufficiently clear though. Here> is another way to put it. If a web of nodes is to be cut into pieces> (two or more), is there an algorithm to list all the different ways to> cut it (or all the outcome)? Consider each way of cutting to be the> set of cuts needed to finish the job.Hm...what does finish the job mean? totally disconnect the graph (remove all nodes)?OK, here's an example that you can confirm or disconfirm.Here are the 15 partitions of 4 objects:1:{{a},{b},{c},{d}},2:{{a,b},{c},{d}},3:{{a,c},{b},{d} },4:{{a,d},{b},{c}},5:{{a},{b,c},{d}},6:{{a},{b,d},{c}},7:{{a} ,{b},{c,d}},8:{{a,b},{c,d}},9:{{a,c},{b,d}},10:{{a,d},{b,c}}, 11:{{a,b,c},{d}},12:{{a,b,d},{c}},13:{{a,c,d},{b}},14:{{a},{b, c,d}},15:{{a,b,c,d}},Let a,b,c,d be the nodes of the following graph: c / |a--b | | dThen really you want the partitions with respect to this graph to be what? Here is my guess (another try):The partition {{a,d},{b,c}} is equivalent (in your sense) to {{a},{b,c},{d}} because a and d are already disconnected.Likewise, {{a,c},{b,d}} and {{a,c},{b},{d}}and by cursory inspection those are the only pairs made the same by this graph.So it seems the sparser the graph, the more similarities you'll have and the shorter the list of partitions you'll really have to check.Does that accord with your intuition? That is, as others have suggested, is all you care about that the subgraphs induced by the partition are connected?-- HarrisLehrstuhl fuer Automatentheorie, Fakultaet InformatikTechnische Universitaet Dresden, Deutschlandhttp://tcs.inf.tu-dresden.de/~harris === Subject: Beyond =?iso-8859-1?Q?G=F6del?=Think about TWO axiom systems S1 and S2.How can they be related?- Not at all (different language)- S1 is a subset of S2- Incompatible: a truth of S1 is a falsehood of S2(e.g. Euclidean and non-Euclidean geometry) - Intersecting. Let S3=S1+S2 be a consistent system,but S1 and S2 are not equal.Question: Is there always a G.9adel sentence of S1that is provable in S2?Are there more possibilities how S1 and S2 caninteract? -- Hauke Reddmann <:-EX8 Private email:fc3a501@math.uni-hamburg.deFor our chemistry workgroup,remove math from the addressFor spamming, remove anything else === Subject: Re: Are all mathematicians music lovers?> It seems every math/science type I meet is a Bach lover, having fallen for> the propaganda that music expands the mind. Actually, there are people> who go crazy from repetitive tunes that won't stop inside their head. I've> been able to drive otherwise calm-and-collected math/science types berserk> by saying music is stupid. Are there any mathematicians today who have the> courage to say music is stupid?Even trolls have music (of a kind).Gib === Subject: Re: Are all mathematicians music lovers?> >It seems every math/science type I meet is a Bach lover, having fallen for>the propaganda that music expands the mind. Actually, there are people>who go crazy from repetitive tunes that won't stop inside their head.>> I've>been able to drive otherwise calm-and-collected math/science types berserk>by saying music is stupid. Are there any mathematicians today who have>> the>courage to say music is stupid?>> Define music. Axioms are not necessary.> Music is that which has exactly zero to do with either Goedel, Escher, or Bach.> Since none of them where musicians. Bach was an instrumentalist.Perhaps you meant to say Bach was a composer?Gib === Subject: Re: [Primes|Assymptotics] What is the order of Sum_{p <= n, p prime} Sum_{i=1..n} [n/p^i] pNow I understand all.> And you answered me to the second question: S(n) is c n^2/log n> assymptotically following the link you provide me.I want not to abuse of you ;-), but you know anymore about nologXan. Well I didn't but inhttp://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?Anum=A001414(check the links there too :http://mathworld.wolfram.com/SumofPrimeFactors.h for a picture) onefinds that it is sometimes called sopfr(n) (sopfr(n!) in your case) and is completely additive (sopfr(a*b)= sopfr(a)+sopfr(b) for anyinteger a and b) so that it's a kind of integer logarithm with theadditional convention that sopfr(prime)=prime. For more information you could perhaps contact Benoit Cloitre orWouter L. J. Meeussen (e-mail in my initial link). Enjoy, Raymond === Subject: Re: Asymptotic Formula for Trinomial Coefficients>The coefficients of the expansion of the following expression: (x^2 + x +>1)^n, are what are called 'trinomial coefficients'. For example:> (x^2 + x + 1)^2 = x^4 + 2x^3 + 3x^2 + 2x + 1>so the coefficients are 1 2 3 2 1. I need to find an asymptotic/approximate>formula for arbitrary coefficents of such expressions. I've checked>numerous math sites including Mathworld, but none have any approximation>formulas.Let c(n,k) be the coefficient of x^k.Consider a random walk X_n starting at 0, where each step is 0 with probability 1/(1+x+x^2), 1 with probability x/(1+x+x^2) or 2 with probability x^2/(1+x+x^2), and x > 0.Then P(X_n = k) = c(n,k) x^k/(1+x+x^2)^n. If n -> infinity and k -> infinitywith k/n -> K (a constant with 0 < K < 2), the Central Limit Theoremsays X_n is approximately normal with mean mu n and variance sigma^2 nwhere mu = x (1 + 2 x)/(1 + x + x^2) and sigma^2 = x (1 + 4 x + x^2)/(1 + x + x^2).When k = mu n + O(sqrt(n)), I would expect P(X_n = k) to be approximately1/sqrt(2 pi sigma^2) exp(-(k - mu n)^2/(2 sigma^2))Now K = mu for x = x_K = (K - 1 + sqrt(1 + 6 K - 3 K^2))/(4 - 2K)and so I would expectc(n,Kn) ~= 1/sqrt(2 pi sigma^2 n) (1+x_K+x_K^2)^n/x_K^(K n) For example, x_K = 2 for K = 10/7, with sigma^2 = 26/49, soc(n, 10/7 n) ~= 1/sqrt(52/49 pi n) 7^n/2^(10/7 n)which seems to work pretty well, e.g. for n = 70 we havecf(70, 100) = 7.385565663 * 10^27 and the approximation is7.410297540 * 10^27.Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Asymptotic Formula for Trinomial Coefficients> The coefficients of the expansion of the following expression: (x^2 + x +> 1)^n, are what are called 'trinomial coefficients'....> I need to find an asymptotic/approximate> formula for arbitrary coefficents of such expressions.... Have you searched for the word multinomial? === Subject: Re: Measure question>Say S is an abstract space, G is a subset of S and a Borel field, and>u(X) is a measure on G.>For each n, Q_n is in G, and the union of all Q_n=S. Also for each n,>u(Q_n) is finite. How does one prove that for each A in G:>limsup u(Q_n intersection A)= u(A)>All help appreciated.Haven't we been through this before, in the thread Limit of measurefrom last month? The only change, it seems to me, is that you havelimsup instead of lim. The same easy counterexamples apply. === Subject: Re: Measure questionSay S is an abstract space, G is a subset of S and a Borel field, and>u(X) is a measure on G.>For each n, Q_n is in G, and the union of all Q_n=S. Things are very confused already. First, if G is a subset of S and> a Borel field then S cannot be an abstract space, it must be> a set of subsets of some set. Now when you say that the union of> the Q_n is S it becomes pretty clear that you didn't mean that> G was a subset of S at all.I _think_ that what you mean here is that S is a topological> space, G is the field of Borel subsets of S, and u is a measure> on G (I'm also very confused by what you mean by saying> u(X) is a measure...)> >Also for each n,>u(Q_n) is finite. How does one prove that for each A in G:limsup u(Q_n intersection A)= u(A)You don't prove this because it's not true. (Simple counterexample:> A = S = [1, infinity), u = Lebesgue measure, and Q+n = [n, n+1).)In addition to garbling the hypotheses that you stated you> _omitted_ a crucial hypothesis! I could tell you what the> hypothesis you omitted was, but that seems like a bad> idea - you're trying to learn this stuff, and you're never> going to learn to _solve_ these problems until you can> at least _read_ and _state_ them correctly.So. Exactly what was the problem again?> >All help appreciated.1.- S is an abstract space.2.- T is its total Borel field, and G is an element of T.3.- u is a sigma-finite measure on G.4.- For each n, Q_n is in G, and the union of Q_n's = S.5.- For each n, u(Q_n)is finite.How does one prove that for each A in Gliminf u(A intersection Q_n)= Asee howliminf u(A intersection (Union Q_n))might equal A. But I cannot seehow it can be true in the problem as stated, since nothing is saidabout Q_n's being increasing. Many thanks for your help and forpointing out the incorrect hypotheses. === Subject: Re: Measure question>> >>Say S is an abstract space, G is a subset of S and a Borel field, and>>u(X) is a measure on G.>>For each n, Q_n is in G, and the union of all Q_n=S. >> >> Things are very confused already. First, if G is a subset of S and>> a Borel field then S cannot be an abstract space, it must be>> a set of subsets of some set. Now when you say that the union of>> the Q_n is S it becomes pretty clear that you didn't mean that>> G was a subset of S at all.>> >> I _think_ that what you mean here is that S is a topological>> space, G is the field of Borel subsets of S, and u is a measure>> on G (I'm also very confused by what you mean by saying>> u(X) is a measure...)>>Also for each n,>>u(Q_n) is finite. How does one prove that for each A in G:>>limsup u(Q_n intersection A)= u(A)>> >> You don't prove this because it's not true. (Simple counterexample:>> A = S = [1, infinity), u = Lebesgue measure, and Q+n = [n, n+1).)>> >> In addition to garbling the hypotheses that you stated you>> _omitted_ a crucial hypothesis! I could tell you what the>> hypothesis you omitted was, but that seems like a bad>> idea - you're trying to learn this stuff, and you're never>> going to learn to _solve_ these problems until you can>> at least _read_ and _state_ them correctly.>> >> So. Exactly what was the problem again?>>All help appreciated.>> >> >> >> >1.- S is an abstract space.>2.- T is its total Borel field, and G is an element of T.>3.- u is a sigma-finite measure on G.>4.- For each n, Q_n is in G, Meaning that Q_n is a _subset_ of G?>and the union of Q_n's = S.Was this a typo for union of Q_n's = G?>5.- For each n, u(Q_n)is finite.>How does one prove that for each A in G>liminf u(A intersection Q_n)= AAnd this was a typo forliminf u(A intersection Q_n)= u(A)?Assuming my interpretations of your revised versionare correct, then you cannot prove this - you coulduse _exactly_ the same counterexample as theexample I gave yesterday to show this!What does the problem _really_ ask? You are_still_ leaving out some hypothesis...>see how>liminf u(A intersection (Union Q_n))might equal A. There you said A instead of u(A) again, so maybe there's amisunderstanding somewhere instead of just a typo.liminf u(A intersection (Union Q_n)) is a number andA is a set, so they can't be equal.(You might also note that liminf u(A intersection (Union Q_n))doesn't make much sense, because u(A intersection (Union Q_n))is a single number, not a sequence of numbers...)>But I cannot see>how it can be true in the problem as stated, since nothing is said>about Q_n's being increasing. Many thanks for your help and for>pointing out the incorrect hypotheses.It may be that there is simply an error in the book.Although I'm not convinced of that yet, because you areclearly still not writing exactly what you mean to write. === Subject: Re: Measure question>1.- S is an abstract space.>2.- T is its total Borel field, and G is an element of T.>3.- u is a sigma-finite measure on G.>4.- For each n, Q_n is in G, Meaning that Q_n is a _subset_ of G?Subset or just an element of G.> >and the union of Q_n's = S.Was this a typo for union of Q_n's = G?No, Union of Q_n's = S> >5.- For each n, u(Q_n)is finite.How does one prove that for each A in Gliminf u(A intersection Q_n)= AAnd this was a typo forliminf u(A intersection Q_n)= u(A)?Yes, I'm sorry. One of these days I'll learn to read my stuff beforeposting.Assuming my interpretations of your revised version> are correct, then you cannot prove this - you could> use _exactly_ the same counterexample as the> example I gave yesterday to show this!What does the problem _really_ ask? You are> _still_ leaving out some hypothesis...> >see howliminf u(A intersection (Union Q_n))might equal A. There you said A instead of u(A) again, so maybe there's a> misunderstanding somewhere instead of just a typo.> liminf u(A intersection (Union Q_n)) is a number and> A is a set, so they can't be equal.You are correct. I made the same mistake twice.(You might also note that liminf u(A intersection (Union Q_n))> doesn't make much sense, because u(A intersection (Union Q_n))> is a single number, not a sequence of numbers...)I should have stated:liminf u(A intersection (Union Q_n))= u(A) nmeaning the liminf of the sequence of u's = u(A)This is from Chung's A Course in Probabilility Theory,(3rd. Ed.) inthe Supplement on Measure and Integral.Many thanks for your help. === Subject: Re: Measure question>>1.- S is an abstract space.>>2.- T is its total Borel field, and G is an element of T.>>3.- u is a sigma-finite measure on G.>>4.- For each n, Q_n is in G, >> >> Meaning that Q_n is a _subset_ of G?>Subset or just an element of G.Aargh. Subset and element are two very differentthings - when you say it can be a subset or justan element that just doesn't make any sense.>>and the union of Q_n's = S.>> >> Was this a typo for union of Q_n's = G?>No, Union of Q_n's = S>> >>5.- For each n, u(Q_n)is finite.>>How does one prove that for each A in G>>liminf u(A intersection Q_n)= A>> >> And this was a typo for>> >> liminf u(A intersection Q_n)= u(A)?>Yes, I'm sorry. One of these days I'll learn to read my stuff before>posting.>> >> Assuming my interpretations of your revised version>> are correct, then you cannot prove this - you could>> use _exactly_ the same counterexample as the>> example I gave yesterday to show this!>> >> What does the problem _really_ ask? You are>> _still_ leaving out some hypothesis...>> >>see how>>liminf u(A intersection (Union Q_n))might equal A. >> >> There you said A instead of u(A) again, so maybe there's a>> misunderstanding somewhere instead of just a typo.>> liminf u(A intersection (Union Q_n)) is a number and>> A is a set, so they can't be equal.>You are correct. I made the same mistake twice.>> >> (You might also note that liminf u(A intersection (Union Q_n))>> doesn't make much sense, because u(A intersection (Union Q_n))>> is a single number, not a sequence of numbers...)>I should have stated:>liminf u(A intersection (Union Q_n))= u(A)> n>meaning the liminf of the sequence of u's = u(A)Aargh again. This still makes more or less no sense,because u(A intersection (Union Q_n)) is _one__number_, not a sequence of numbers.>This is from Chung's A Course in Probabilility Theory,(3rd. Ed.) in>the Supplement on Measure and Integral.It's impossible to tell from your posts _what_ is in Chung,because the story keeps changing. Someone elserequested that you post the problem _verbatim_.That means post _exactly_ what the problem in Chungasks. Using _exactly_ the words Chung uses, inexactly the same order.>Many thanks for your help. === Subject: Re: Measure question> This is from Chung's A Course in Probabilility Theory,(3rd. Ed.) in> the Supplement on Measure and Integral.Do us a favor and quote the assertion *verbatim* from Chung's book.-- A. === Subject: Richmond surface relationship to catenoidhttp://www.math.hmc.edu/faculty/gu/curves_and_surfaces /surfaces/richmond.htmlIs a special case isometric to a catenoid? I ask this as it has aplanar circular boundary and is also a minimal surface like thecatenoid. === Subject: Simple question about matricesJust a quick question. This is something that's so blindingly simple I'mhaving difficulty with it. I'm looking at someone's work at the momentand they've got a calculation which I'm suspicious of. Essentially, I thinkthey've made a mistake in their paper which propagates throughout and altersthe main result significantly. I can trace this back to their calculation ofthe variation of a symmetric matrix g_ab.If we let g=det(g_ij) be the metric determinant, then the variation of g andsqrt(g) is given bydelta g = g*g^(ij)*delta(g_ij)delta g^(1/2) = g*g^(ij)*delta(g_ij)I'm certain that both of these results are correct. Now, I think theirmistake occurs when they calculate the variation of 1/sqrt(g). The result Iobtain for this isdelta (1/sqrt(g)) = -(g^ij) / (2*sqrt(g)) delta(g_ij)but in their work, they are missing the minus sign. Can someone verify thislast calculation (it's only two lines long) and tell me if I'm going crazydavidoff === Subject: Re: Simple question about matrices> Just a quick question. ... I can trace this back to their calculation of> the variation of a symmetric matrix g_ab.> If we let g=det(g_ij) be the metric determinant, then the variation of g and> sqrt(g) is given bydelta g = g*g^(ij)*delta(g_ij)delta g^(1/2) = g*g^(ij)*delta(g_ij)I'm certain that both of these results are correct. ...You gave the same formula for delta g and delta g^(1/2). ell === Subject: Re: Simple question about matrices> Just a quick question. ... I can trace this back to their calculation of> the variation of a symmetric matrix g_ab.> If we let g=det(g_ij) be the metric determinant, then the variation of gand> sqrt(g) is given bydelta g = g*g^(ij)*delta(g_ij)delta g^(1/2) = g*g^(ij)*delta(g_ij)I'm certain that both of these results are correct. ...> You gave the same formula for delta g and delta g^(1/2).> ellI know, sorry about that. The correct expressions are in the thread afterDoug Goncz's reply.davidoff === Subject: Re: Simple question about matrices>delta g = g*g^(ij)*delta(g_ij)>delta g^(1/2) = g*g^(ij)*delta(g_ij)I doubt it. The left sides are not the same, while the right sides, as written,are identical.Cutting and pasting: g*g^(ij)*delta(g_ij) = g*g^(ij)*delta(g_ij) identicalbut why should g=sqrt(g)?Let's stop there. The rest is based on this, anyway. Did something go wrong inposting? You do know that Usenet cannot correctly interpret h, don't you?So what is delta, some kind of Tex? Can't post that either.....Usenet is ASCII, a 7-bit code. Not IBM PC and Appple Macintosh 8-bit ASCII.Now, if someone pastes: http colon slash slash something dot com, it may showin your reader as a hyperlink, but please don't put things where they don'tbelong. This particular AOL newsreader won't let me drag a favorite place intothis edit box, but the edit box is too dumb to be immune to all the possiblepastes.http://something.comSee?Is it blue?Yours,Doug Goncz, Replikon Research, Seven Corners, VA The hormones work at different speeds: In a fight-or-flight scenario, glucocorticoids are the ones drawing up blueprints for new aircraft carriers;epinephrine is the one handing out guns. === Subject: Re: Simple question about matrices Sorry about that. What I've written is of course wrong, so thanks forpointing it out. What I meant to say was:delta( g ) = g*g^(ij)*delta( g_(ij) )delta( sqrt(g) ) = (1/2)*sqrt(g)*g^(ij)*delta(g_(ij))And the answer I get for delta( 1/sqrt(g) ) isdelta( 1/sqrt(g) ) = -( 1 / (2*sqrt(g)) )*g^(ij)*delta( g_(ij) )I'm using delta to represent a 'small' variation of the term in bracketsimmediately after. You're right that this is a hangover from me using TeX,but it's not actually meant to be correct TeX code. I've put that at thebottom of this post in case you'd like to look at it. My apologies at makingthis post unnecessarily long, but Usenet is rubbish for trying tocommunicate equations with.davidoff% Begin TeX Sourcebegin{document}Just a quick question. This is something that's so blindingly simple I'mhaving difficulty with it. I'm looking at someone's work at the momentand they've got a calculation which I'm suspicious of. Essentially, I thinkthey've made a mistake in their paper which propagates throughout and altersthe main result significantly. I can trace this back to their calculation ofthe variation of a symmetric matrix g_ab.If we let g=det(g_ij) be the metric determinant, then the variation of g andsqrt(g) is given bybegin{equation}delta g = g g^{ij} delta g_{ij},end{equation}begin{equation}deltasqrt{g} = frac{1}{2}sqrt{g} g^{ij} delta g_{ij}.end{equation}noindent I'm certain that both of these results are correct. Now, I thinktheirmistake occurs when they calculate the variation of 1/sqrt(g). The result Iobtain for this isbegin{equation}delta left( frac{1}{sqrt{g}} right) = -left( frac{1}{2sqrt{g}}right) g^{ij}delta g_{ij}.end{equation}noindent but in their work, they are missing the minus sign. Can someoneverify thislast calculation (it's only two lines long) and tell me if I'm going crazyend{document}% End TeX source === Subject: Re: Factorial/Exponential Identity, InfinityWhat is the name for a set that contains a maximum element of n andeach element of N or Z+ less than n: {1, ..., n}? It's the set ofcounting numbers less than or equal to n, it's the set of contiguouselements of N from 1 to n, etcetera.How is done summing the products of each triple of this set?Summing the products of pairs of the set is as an expression ((sumn)^2 - sum(n squared))/2. Is the sum of products of triples in theform ((sum n)^3 - sum(n cubed))/8?Consider the case with 1, 2, 3. There is only one triple, {1, 2, 3},and the sum of the product of itself is 6.1 1 2 31 1 2 32 2 4 63 3 6 92 1 2 31 2 4 62 4 8 123 6 12 183 1 2 31 3 6 92 6 12 183 9 18 27The diagonal of that, in a way, would be 1, 8, 27, for (1, 1, 1), (2,2, 2), (3, 3, 3). The value of (sum n)^3 is 6^3 = 216. sum (n^3) is36. (sum n)^3 - sum (n^3) = 180. 180 / 30 = 6, the result. Theresult is also the sum and the product of (1, 1, 1), (1, 1, 2), and(1, 1, 3).For n=4, there are these triples:{1, 2, 3}{1, 2, 4}{1, 3, 4}{2, 3, 4}The value of (sum n)^3 is 1000, the value of sum n^3 is 100. The sumof the products of the triples is 6+8+12+24=50. 900/18 = 50.For n=3, the divisor x is 30, for n=4, 18, to get a correct expressionfor the sum of the products of triples being ((sum n)^3 - sum n^3) /x.The idea here is to determine forms for the coefficients of thepolynomial (n+1)...(n+n)/n^n. Determining a function of n for the sumof the product of each triple of {1, ...n} would give an expressionfor the fourth coefficient, a_4. The third coefficient, a_3, has asan expression ((sum n)^2-sum n^2)/2, the second coefficient has theform sum n = (x+1)x/2, the first coefficient is equal to 1. It wouldbe acceptable/preferable to have expressions of the coefficients thatonly apply to an infinite value of n, or fractional expressions of n.For n=5, sum n is 15, (sum n)^3 is 3375, and sum (n^3) is 225, theirdifference is 3150. The triples of {1,...5} are{1, 2, 3}{1, 2, 4}{1, 2, 5}{1, 3, 4}{1, 3, 5}{1, 4, 5}{2, 3, 4}{2, 3, 5}{2, 4, 5}{3, 4, 5}2*(4)3*(2+4+8+12)5*(2+3+4+6+8)The sum of their products is 6+8+10+12+15+20+24+30+40+60 = 225. Thevalue of x for 3150/x=225 is 14. Thus for some f(x), f(3)=30,f(4)=18, f(5)=14.I guess what I should make here is an algorithm to list each subset ofa given number of elements, here 3, of {1, ..., n}. Also I shouldresearch the function that gives the number of subsets of a givenlength of a set of a given length. What is the analysis of a bitsequence of length n with x on bits, and each of those sequences?I look to MathWorld and it notes that subsets of length k of a set arecalled k-subsets.http://mathworld.wolfram.com/k-Subset.hThere are n choice k many k-subsets of a set of length n: n! / (n-k)!k!. What I want to determine here are what their products are for {1,..., n}. In the case of k=3, for n=3, 4, 5 there are 1, 4, and 10permutations, the sum of their products is 6, 50, 225.Earlier, I noted that ((sum n)^2 - sum(n^2)) / 2 = (n^2+n)^2 / 4 - sum(n^3+n^2)/2. (sum n)^2 - sum(n^2) = ((n^2+n)^2) / 2 - sum(n^3+n^2)(sum n)^2 = ((n^2+n)^2) / 2 - sum(n^3+n^2) + sum(n^2)(sum n)^2 = ((n^2+n)^2) / 2 - sum(n^3)Here when I have been writing sum (f(n)) it means sum for i=1 to n off(n).(sum n)^2 = (2 (sum n)^2 - sum(n^3)sum(n^3) = (sum n)^2Hmm, that's kind of interesting. I wasn't aware of that relation. The sum of the cubes of the numbers from 1 to n is equal to the squareof the sum of the numbers from 1 to n.That's saying that taking the sum of the cubes of the numbers from oneto a zillion, that that sum is equal to the square of the sum of thenumbers from one to a zillion.What's up with that? Is that well-known or trivial, and if so, inwhat context? Here a zillion is an arbitrarily large integer, and apower of ten, and is less than a gazillion, another abitrarily largepower of ten.Ross === Subject: Re: Polynomial problem>Hello>I'm trying to prove the following statement, but got really stuck.>Does anyone have a suggestion?>Let P be a polynomial of the n_th degree such that P(x) is in [-1,1]>for every x in [-1, 1]. Then, P'(x) is in [-n^2, n^2] for every x in>[-1, 1].>Apparently, this is not that hard, but I couldn't get trough.Well I thought about it a little yesterday and I don't see how todo it.What material about polynomials was covered recently?(Anything about special polynomials, like Chebyschevpolynomials or whatever?)>Thank you.PS: I think this is true only if the coefficients of P are real. === Subject: Re: Polynomial problem> Hello> I'm trying to prove the following statement, but got really stuck.> Does anyone have a suggestion?> Let P be a polynomial of the n_th degree such that P(x) is in [-1,1]> for every x in [-1, 1]. Then, P'(x) is in [-n^2, n^2] for every x in> [-1, 1].> Apparently, this is not that hard, but I couldn't get trough.> Thank you.PS: I think this is true only if the coefficients of P are real.Since p(x) is a polynomial then p'(x) exists for all x in [-1 , 1]. Nowsince p'(x) is also a polynomial (der. of poly is a poly) it has a max, M,and min, m,on [-1 , 1] Let M* = Max {|M|, |m|}. Then P' (x) is in [-M* , M*]for x in [-1, 1]. Finish up from here. === Subject: Re: Polynomial problem>> Hello>> I'm trying to prove the following statement, but got really stuck.>> Does anyone have a suggestion?>> Let P be a polynomial of the n_th degree such that P(x) is in [-1,1]>> for every x in [-1, 1]. Then, P'(x) is in [-n^2, n^2] for every x in>> [-1, 1].>> Apparently, this is not that hard, but I couldn't get trough.>> Thank you.>> >> PS: I think this is true only if the coefficients of P are real.>Since p(x) is a polynomial then p'(x) exists for all x in [-1 , 1]. Now>since p'(x) is also a polynomial (der. of poly is a poly) it has a max, M,>and min, m,on [-1 , 1] Let M* = Max {|M|, |m|}. Then P' (x) is in [-M* , M*]>for x in [-1, 1]. Finish up from here.Finish up from here? You haven't said anything significant aboutthe problem. (If you want to dispute that you should show us_how_ to finish up from here...) === Subject: Re: Polynomial problemMaybe this will be of some help:If P is monic (leading coefficient is 1), then abs(P')<=n*abs(P)^(1-1/n)everywhere; (to prove: let P be L1...Ln, where each Li is linear; then P' is thesum of all the terms L1...Ln/Li, and the product of all these terms is P^(n-1);the triangle theorem and the amgm inequality will bring you home)Now you just (?) need to extend the result to nonmonic P| Hello| I'm trying to prove the following statement, but got really stuck.| Does anyone have a suggestion?|| Let P be a polynomial of the n_th degree such that P(x) is in [-1,1]| for every x in [-1, 1]. Then, P'(x) is in [-n^2, n^2] for every x in| [-1, 1].|| Apparently, this is not that hard, but I couldn't get trough.| Thank you.| || PS: I think this is true only if the coefficients of P are real. === Subject: Re: Polynomial problem>Maybe this will be of some help:>If P is monic (leading coefficient is 1), then abs(P')<=n*abs(P)^(1-1/n)>everywhere; Wade has pointed out a problem with your proof of this fact,but he didn't mention that the inequality itself is false. It is(if it were true it would follow that any zero of P was a zeroof P'; I'll let you come up with a counterexample to that...)>(to prove: let P be L1...Ln, where each Li is linear; then P' is the>sum of all the terms L1...Ln/Li, and the product of all these terms is P^(n-1);>the triangle theorem and the amgm inequality will bring you home)>Now you just (?) need to extend the result to nonmonic P>| Hello>| I'm trying to prove the following statement, but got really stuck.>| Does anyone have a suggestion?>|>| Let P be a polynomial of the n_th degree such that P(x) is in [-1,1]>| for every x in [-1, 1]. Then, P'(x) is in [-n^2, n^2] for every x in>| [-1, 1].>|>| Apparently, this is not that hard, but I couldn't get trough.>| Thank you.>| >|>| PS: I think this is true only if the coefficients of P are real. === Subject: Re: Polynomial problem>Maybe this will be of some help:>If P is monic (leading coefficient is 1), then abs(P')<=n*abs(P)^(1-1/n)>everywhere; Wade has pointed out a problem with your proof of this fact,> but he didn't mention that the inequality itself is false.Not true, I gave the example x^2 - 1. === Subject: Re: Polynomial problem>>Maybe this will be of some help:>>If P is monic (leading coefficient is 1), then abs(P')<=n*abs(P)^(1-1/n)>>everywhere; >> >> Wade has pointed out a problem with your proof of this fact,>> but he didn't mention that the inequality itself is false.>Not true, I gave the example x^2 - 1.Sorry. === Subject: Re: Polynomial problem> Maybe this will be of some help:> If P is monic (leading coefficient is 1), then abs(P')<=n*abs(P)^(1-1/n)> everywhere; (to prove: let P be L1...Ln, where each Li is linear; then P' is the> sum of all the terms L1...Ln/Li, and the product of all these terms is P^(n-1);> the triangle theorem and the amgm inequality will bring you home)> Now you just (?) need to extend the result to nonmonic P Elaine.But, I didn't get very well the meaning of the Li's . Do you mean theyare linear functions? Could you detail your proof a bit more?Based on your inequality, we conclude that if P is monic then p'(x) isin [-n, n] for every x in {-1, 1], right? === Subject: Re: Polynomial problemHi . Here is the idea in complete detail:Let P be a monic polynomial of degree n. By the Fundamental Theorem of Algebrathere are monic linear polynomials L1...Ln such that P=product_i(Li). (Each Liis x->(x-r) for some root r of P. If r is a root of P with multiplicity m, thenexactly m of the Li's are x->(x-r).) Since Li'=1 for each Li, the Product Ruleand an easy induction show that P'=sum_i[product_{j<>i}(Lj)]. (If n=2 thenP'=L1+L2, if n=3 then P'=L1*L2+L1*L3+L2*L3, and so on.) Henceabs(P')/n<=sum_i[abs[product_{j<>i}(Lj)]]/n (Triangle Inequality)<==abs[product_i[product_{j<>i}(Lj)]]^{1/n}=abs(P^{ n-1})^{1/n}=abs(P)^{1-1/n}Hence abs(P')<=n*abs(P)^{1-1/n}. In particular, if abs(P)<=1 on an interval I,then abs(P')<=n on I. This is what I was saying before. Now let's take it alittle further by dropping the assumption that P is monic. Let's say P=aM, whereM is monic. (In other words, a is the leading coefficient of P). If abs(P)<=1 onan interval I, then abs(M)<=1/abs(a) on I andabs(P')=abs(a)*abs(M')<=n*abs(a)*abs(M)^{1-1/n}<=n*abs(a)*[ 1/abs(a)]^{1-1/n}=n*abs(a)^{1/n}...on I. So you're finished if you can only (?) prove that your hypotheses forcethe leading coefficient of P to be at most n^n. The question mark means I don'tknow how tough or easy it is to prove this conjecture. Maybe the conjecture isfalse. Maybe it's true but you won't be able to prove it with only a reasonableinvestment of effort. There's no guarantee this is the best way to approach yourproblem. Good luck. Let me know how it goes.Peace,EJ| > Maybe this will be of some help:| > If P is monic (leading coefficient is 1), then abs(P')<=n*abs(P)^(1-1/n)| > everywhere; (to prove: let P be L1...Ln, where each Li is linear; then P' isthe| > sum of all the terms L1...Ln/Li, and the product of all these terms isP^(n-1);| > the triangle theorem and the amgm inequality will bring you home)| > Now you just (?) need to extend the result to nonmonic P|| Elaine.| But, I didn't get very well the meaning of the Li's . Do you mean they| are linear functions? Could you detail your proof a bit more?| Based on your inequality, we conclude that if P is monic then p'(x) is| in [-n, n] for every x in {-1, 1], right?| === Subject: Re: Polynomial problem> Let P be a monic polynomial of degree n. By the Fundamental Theorem of > Algebra> there are monic linear polynomials L1...Ln such that P=product_i(Li). (Each > Li> is x->(x-r) for some root r of P. If r is a root of P with multiplicity m, > then> exactly m of the Li's are x->(x-r).) Since Li'=1 for each Li, the Product > Rule> and an easy induction show that P'=sum_i[product_{j<>i}(Lj)]. (If n=2 then> P'=L1+L2, if n=3 then P'=L1*L2+L1*L3+L2*L3, and so on.) Henceabs(P')/n> <=> sum_i[abs[product_{j<>i}(Lj)]]/n (Triangle Inequality)> <=> => abs[product_i[product_{j<>i}(Lj)]]^{1/n}> => abs(P^{n-1})^{1/n}> => abs(P)^{1-1/n}Hence abs(P')<=n*abs(P)^{1-1/n}.x^2 - 1 is a counterexample to the inequality in the last line. === Subject: Re: Polynomial problemQuite right. Sorry, . Just an honest mistake.|| > Let P be a monic polynomial of degree n. By the Fundamental Theorem of| > Algebra| > there are monic linear polynomials L1...Ln such that P=product_i(Li). (Each| > Li| > is x->(x-r) for some root r of P. If r is a root of P with multiplicity m,| > then| > exactly m of the Li's are x->(x-r).) Since Li'=1 for each Li, the Product| > Rule| > and an easy induction show that P'=sum_i[product_{j<>i}(Lj)]. (If n=2 then| > P'=L1+L2, if n=3 then P'=L1*L2+L1*L3+L2*L3, and so on.) Hence| >| > abs(P')/n| > <=| > sum_i[abs[product_{j<>i}(Lj)]]/n (Triangle Inequality)| > <=| > =| > abs[product_i[product_{j<>i}(Lj)]]^{1/n}| > =| > abs(P^{n-1})^{1/n}| > =| > abs(P)^{1-1/n}| >| > Hence abs(P')<=n*abs(P)^{1-1/n}.|| x^2 - 1 is a counterexample to the inequality in the last line. === Subject: Re: FLT and the Barcelona conjecture>In fact, proving the Barcelona conjecture also proves FLT Now, to me, that sounds like some fun.Unfortunately I am working with p=3 right now, and not on FLT.> 1=(x+y+z)^3/(24xyz) is what you are working on? The BarcelonaConjecture states no integer solutions satisfy this equality. Provingthis also demonstrates no integer a,b,c satisfy a^3+b^3=c^3.As many conjectures related to FLT are formulated for positiveintegers, I'd like to mention that x,y and z in this case areintegers, negative being allowed.You will also find that integers of the forms 4k+1,4j+3 and (2^(np))rare requisites for x,y and z ( r odd, by the way).Obviously, xyz not equal to zero.. === Subject: Re: Solving systems of quadratic equalities and inequalities <3F355A66.8030804@comcast.netMathematica can solve systems of polynomial inequalities.I like massima or yacas, they are free soft... === Subject: uniformly convergence??hello....my doctorfn(x) = (1-x) / (1+(x^n)) , x in [1,2]prove fn(x) is uniformly convergence------------------------f(x) = lim fn(x) = 0 (n->00) on 1<= x <= 2|fn(x) - f(x)| = |(1-x) / (1+(x^n))| = (x-1) / (1+(x^n))next, i can't proceedi know that n -> 00 => |fn(x)-f(x)| -> 0but i can't affirm U.C. from this factbecause, if fn(x) = x^n , (0 fn(x) = (1-x) / (1+(x^n)) , x in [1,2]> prove fn(x) is uniformly convergence> f(x) = lim fn(x) = 0 (n->00) on 1<= x <= 2> |fn(x) - f(x)| = |(1-x) / (1+(x^n))| = (x-1) / (1+(x^n))> next, i can't proceed> i know that n -> 00 => |fn(x)-f(x)| -> 0> but i can't affirm U.C. from this factYes, you need yet more.> because, if fn(x) = x^n , (0 i think that i would have to find the concrete N(epsilon)You have to find a N = N(eps) such thatfor all x in [1,2], n > N, (x-1) / (1 + x^n) < epsFor that we do some hair spliting.(x-1) / (1 + x^n) < x-1 < eps when 1 <= x < 1+eps <= 1/(1+x^n) < 1/x^n <= 1/(1+eps)^n when 1+eps <= x <= 2Now we want 1/(1+eps)^n < eps 1/eps < (1+eps)^n - log eps < n log(1+eps) -(log eps)/log(1+eps) < nCheck to see if that gives the right formula for N. === Subject: Re: thank sir. david, williamin fact, i could not understand style of david doctor for my inability.but i could understand by detailed explanation of willam doctor.thanks to have solved my problem. === Subject: Re: uniformly convergence??>hello....my doctor>fn(x) = (1-x) / (1+(x^n)) , x in [1,2]>prove fn(x) is uniformly convergence>------------------------>f(x) = lim fn(x) = 0 (n->00) on 1<= x <= 2>|fn(x) - f(x)| = |(1-x) / (1+(x^n))| = (x-1) / (1+(x^n))>next, i can't proceed>i know that n -> 00 => |fn(x)-f(x)| -> 0>but i can't affirm U.C. from this fact>because, if fn(x) = x^n , (0i think that i would have to find the concrete N(epsilon)Yes you do. Fix epsilon > 0.First note that 0 <= fn(x) <= x - 1 for all x; this givesthe inequality you need for 1 <= x < 1 + epsilon,_regardless_ of what N(epsilon) you choose. Nowfind an N(epsilon) that works for 1 + epsilon <= x <= 2.(Or you could use Dini's theorem, if you've covered that...)>help me for my worry. doctor === Subject: Re: Byron and FullerInjector-Info: news.mailgate.org; posting-host=gatekeeper.tripos.com; posting-account=35661; posting-date=1061774471X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ 3a9d1995cb92e175fe23cc50e94ecf25.35661%40mygate.mailgate.org> |> Are you just typing these posts because you have nothing> |> better to do or what? There _is_ no discussion of the difference> |> between closed and complete above. That's because there> |> is no difference. This has already been made very clear.> No need to get snappy...Thank you! It doesn't take much to upset people round these parts.....-- === Subject: Re: Byron and FullerInjector-Info: news.mailgate.org; posting-host=gatekeeper.tripos.com; posting-account=35661; posting-date=1061774471X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ cdb1806df513decb39a0496c5003e4fa.35661%40mygate.mailgate.org> Theorem 5.2. A set of orthonormal functions in Hilbert space is> complete if and only if it is closed.I simply can't believe that you read that, thought about it,> and didn't realize it was saying that closed and complete> are equivalent.Really? Okay, so I have another book which says (paraphrasing) a metricspace is compact if and only if it is complete.In that case I take compact and complete to be equivalent, do I?The point is that you should _believe_ that I didn't appreciate the fullsignificance of the statement in the above theorem, because there aremany if and only if statements that don't work the other way round.Add to this the unconventional defintion of the term closed and thefact that I was under the misapprehension that the discussion showed upa difference between the two terms closed and complete, and youshould be able to see why I didn't immediately understand your comments.Here's another: water is pure if and only if it is colourless.This is a genuine attempt to explain why I didn't follow your comments -I was not posting because I didn't have anything better do.I have no doubt that you will wish to respond to these comments, and Ilook forward to reading that response, but I will say now that I willresist any wish to respond further as I can't see it achieving anything.-- === Subject: Re: Byron and FullerSupersedes: > Theorem 5.2. A set of orthonormal functions in Hilbert space is>> complete if and only if it is closed.>> >> I simply can't believe that you read that, thought about it,>> and didn't realize it was saying that closed and complete>> are equivalent.Really? Okay, so I have another book which says (paraphrasing) a metric> space is compact if and only if it is complete.Then I suggest casting it into the flames, for thereare complete metric spaces which are not compact.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.hHis mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Byron and Fuller>> Theorem 5.2. A set of orthonormal functions in Hilbert space is>> complete if and only if it is closed.>> >> I simply can't believe that you read that, thought about it,>> and didn't realize it was saying that closed and complete>> are equivalent.>Really? Okay, so I have another book which says (paraphrasing) a metric>space is compact if and only if it is complete.What???????????????????What book is that?>In that case I take compact and complete to be equivalent, do I?If the theorem you cite were true (of course it isn't) then you shouldtake the two to be equivalent _in_ the context of metric spaces, yes.(Not sure whether it's relevant, since I'm not sure what your pointis, but we should perhaps note that this use of the word completeis totally different from the complete in complete orthonormalset.)>The point is that you should _believe_ that I didn't appreciate the full>significance of the statement in the above theorem, because there are>many if and only if statements that don't work the other way round.Huh? I'm not sure what you mean by if and only if statement thatdoesn't work the other way around, so I'm not sure whether I'veseen any. You seem to be taking the above as an example.Supposing for the sake of argument that it _were_ true that ametric space was complete if and only if it was compact - whatdo you mean when you say this does not work the other way around?>Add to this the unconventional defintion of the term closed and the>fact that I was under the misapprehension that the discussion showed up>a difference between the two terms closed and complete, and you>should be able to see why I didn't immediately understand your comments.>Here's another: water is pure if and only if it is colourless.>This is a genuine attempt to explain why I didn't follow your comments ->I was not posting because I didn't have anything better do.>I have no doubt that you will wish to respond to these comments, and I>look forward to reading that response, but I will say now that I will>resist any wish to respond further as I can't see it achieving anything.> === Subject: Re: Byron and FullerInjector-Info: news.mailgate.org; posting-host=gatekeeper.tripos.com; posting-account=35661; posting-date=1061805322X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ 77f39134bcfbe150cb07aa1f9b7a4d36.35661%40mygate.mailgate.org> What book is that?In Methods of Real Analysis by R. R. Goldberg, 2nd. Ed., page 160 itsays 6.5A. DEFINITION. The metric space is said to be compact if is both complete and bounded.So maybe this wasn't a good example. The point I'm trying to make isthis (and I believe very well that you know what point I'm trying tomake): if a metric space is compact it is complete; but if a metricspace is complete, it is NOT necessarily compact, since it might not bebounded. There's probably some mathematical way of describing thissituation.Water is pure if and only if it is colourless. Is this a truestatement?Does it mean that anything which is colourless is pure?You can get water which is colourless but which isn't pure.So whichever way you look at it, the statement doesn't work both ways.So just because a set of orthogonal functions is closed if and only ifit is complete doesn't mean that the two are the same unless you proveit the other way round i.e. that it is complete if and only if it isclosed. This is what Byron and Fuller end up doing, but I didn't learnthis from our earlier discussion.-- === Subject: Re: Byron and Fuller>[...]>So just because a set of orthogonal functions is closed if and only if>it is complete doesn't mean that the two are the same unless you prove>it the other way round i.e. that it is complete if and only if it is>closed.Um, maybe I should add something. I take back the things I'vesaid about how I couldn't believe you even glanced at variousthings here and in the book - when I said those things I wasassuming you understood the meaning of the phrase if andonly if. Sorry. === Subject: Re: Byron and Fuller>[...]>So just because a set of orthogonal functions is closed if and only if>it is complete doesn't mean that the two are the same unless you prove>it the other way round i.e. that it is complete if and only if it is>closed.> Um, maybe I should add something. I take back the things I've> said about how I couldn't believe you even glanced at various> things here and in the book - when I said those things I was> assuming you understood the meaning of the phrase if and> only if. Sorry.A question please: In the phrase if and only if is the first if necessary?IOW, A only if B is not the same as A if and only if B?> > === Subject: Re: Byron and Fuller> A question please: In the phrase if and only if is the first if necessary?> IOW, A only if B is not the same as A if and only if B?Yes, both if's are significant. The following statements all mean exactly the same thing: 1. A only if B. 2. B if A. 3. If A, then B. 4. A is sufficient for B. 5. B is necessary for A.You get an entirely different meaning if you exchange A and Bthroughout those 5 statements.The following statements are all equivalent to each other and are allquite different from the 5 statements above: 6. A if and only if B. 7. A if B, and A only if B. 8. A is logically equivalent to B. 9. A is necessary and sufficient for B. 10. If A then B, and conversely.In statements 6-10, exchanging A and B has no effect on the meaning. === Subject: Re: Byron and FullerInjector-Info: news.mailgate.org; posting-host=gatekeeper.tripos.com; posting-account=35661; posting-date=1061805322X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ 915122c4294129638faeb18a27b3e8be.35661%40mygate.mailgate.org> Um, maybe I should add something. I take back the things I've> said about how I couldn't believe you even glanced at various> things here and in the book - when I said those things I was> assuming you understood the meaning of the phrase if and> only if. Sorry.That's fine. I'm glad we pursued it to the point of finding out what wasgoing wrong. If we hadn't gone as far as we did I would never haveunderstood the meaning of if and only if statements, so I've learntthat at least.-- === Subject: Re: Byron and Fuller>> What book is that?>In Methods of Real Analysis by R. R. Goldberg, 2nd. Ed., page 160 it>says 6.5A. DEFINITION. The metric space is said to be compact if> is both complete and bounded.It actually says exactly that? (It's not talking about subsets of R^n,If that book actually says exactly that then you need a differentbook - a complete and bounded metric space most assuredlyneed _not_ be compact. (Example: The set {x: ||x|| <= 1} ina Hilbert space.)>So maybe this wasn't a good example. The point I'm trying to make is>this (and I believe very well that you know what point I'm trying to>make): if a metric space is compact it is complete; but if a metric>space is complete, it is NOT necessarily compact, since it might not be>bounded. There's probably some mathematical way of describing this>situation.No, it was not clear that that was your point. And (again, supposingfor the sake of argument that the definition above is correct, whichof course it isn't) this is _not_ an example of an if and only ifstatement that only goes one way, because it doesn't saythat it's compact if and only if it's complete.Water is pure if and only if it is colourless. Is this a true>statement?>Does it mean that anything which is colourless is pure?>You can get water which is colourless but which isn't pure.>So whichever way you look at it, the statement doesn't work both ways.The statement Water is pure if and only if it's colorless is simply_false_. When I said there was no such thing as an if and onlyif that doesn't work both ways I was talking about _true_ if andonly if statements...>So just because a set of orthogonal functions is closed if and only if>it is complete doesn't mean that the two are the same unless you prove>it the other way round i.e. that it is complete if and only if it is>closed. Golly. At least we see here where the confusion lies - it's not aproblem with high-powered mathematical terminology, it's aproblem with simple _English_: A if and only if B meansexactly the same thing as B if and only if A.No, the (incorrect) statement about metric spaces aboveis not a counterexample to this. It says that a metric spaceis compact if and only if it's complete and bounded; that'sthe same as saying it's complete and bounded if andonly if it's compact.(I have a hard time believing it really says that. Could be,I suppose.)Seriously: You can take this as a flame or as constructiveadvise, as you prefer, but if you really don't realize thatA if and only if B says the same thing as B if and onlyif A you're not going to have much of a chance tryingto learn about measure theory and Hilbert space andsuch things, because the people writing those booksare assuming you know what phrases like if and onlyif mean. You should start with a book or a courseon basic mathematical reasoning.>This is what Byron and Fuller end up doing, but I didn't learn>this from our earlier discussion.No, Byron and Fuller do _not_ prove complete if and only if closedand then prove closed if and only if complete separately. Whatthey do is prove complete if closed and also prove closedif complete.> === Subject: Re: Byron and FullerInjector-Info: news.mailgate.org; posting-host=gatekeeper.tripos.com; posting-account=35661; posting-date=1061805322X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ 53d90c0fbec6dc41e0fc3426050c8a3c.35661%40mygate.mailgate.org> In Methods of Real Analysis by R. R. Goldberg, 2nd. Ed., page 160 it>says 6.5A. DEFINITION. The metric space is said to be compact if> is both complete and bounded.It actually says exactly that? (It's not talking about subsets of R^n,I've checked the book and you're right - I should have written totallybounded - sorry for any inconvenience.-- === Subject: Re: Byron and FullerInjector-Info: news.mailgate.org; posting-host=gatekeeper.tripos.com; posting-account=35661; posting-date=1061805322X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ 0830c3b2132eb5cc3e3c48d023b7c721.35661%40mygate.mailgate.org> In Methods of Real Analysis by R. R. Goldberg, 2nd. Ed., page 160 it>says 6.5A. DEFINITION. The metric space is said to be compact if> is both complete and bounded.It actually says exactly that? (It's not talking about subsets of R^n,> I've checked the book again and you're right, I did inadvertently missout the word bounded. Sorry for any inconvenience.-- === Subject: Re: Affine Connection & Gauge Theory> >Geometry on the space with affine connection is more general than>Riemanian geometry. I thought the Grassmann algebla on the former>space might be relevant with the gauge transformation. Namely the>structure equation on the Grassmann algebla corresponds to the gauge>field.No, because the gauge group is not limited to the linear> transformations of the tangent space or of the Grassmann algebra on> it. You need to deal with connections on an arbitrary vector bundle,> not necessarily related to the tangent bundle.Sorry, for I should have written correctly.Generally the principal fiber bundle is defined as the quotient spaceof the direct product of local base space and transformation groupbased on an equivalence relations, while the space I've dealt with inthe following webpage:http://139.134.5.123/tiddler2/gauge4/gauge.htmis the direct product of Lorentz space and inner space. There Lorentzspace was supposed as linear space (i.e. ignoring the effect ofgravity) and the inner space was supposed as the space with affineconnection. The transformation group on Lorentz space is of courseLorentz transformation group and that on the inner space is Lie group.Namely the base space is not Lorentz space, but the inner spaceitself. === Subject: Proving that the codebook of an (n,k) RS code is an ideal?How can I prove that the (n,k) RS code book, i.e. all the valid codewords of the specific (n,k) Reed-Solomon code, is an ideal?The (n,k) RS code has a generator g(x), consisting of n-k consecutiveroots in GF(2^m). (I.e. a valid RS codeword is divisible by g(x))Also, g(x), the generator polynomial has order n, i.e. g(x) dividesx^n-1.For an ideal, the following must hold:A nonempty subset G of a ring R is an ideal of R if:1. a - b in G whenever a,b in G.2. da and ad are in G, whenever a in G and d in R. The first condition is met because of the linearity of Reed-Solomoncodes.That is, the addition (or subtraction) of any two elements in (n,k) RS(=> G), will result in another valid codeword (element in G).The second condition I am not sure how to tackle.One way that I thought of was to use the remainder theorem. For anyelement c in the Ring R, we can express c as c = pq + r, where p,q andr are in R, with 0 <= deg (r) < deg(p).If we assume that c = ad, where a in G and d in R, (or c = da, becausethe ring GF(2^m)[x] is commutative) and because RS is a cyclic code,we can divide c by x^n - 1. (The remainder should then give us a validcodeword)Thus c = p(x^n-1) + r, where 0 <= deg(r) < deg(x^n-1)The condition on the degree of r assures that r will have the correctdegree. (p is an arbitrary polynomial)Also, when deg(ad) < deg(x^n-1), r = ad, and because a is in G, wehave a = p(x)g(x), thus r = [d(x)p(x)]g(x), which is in G.How can I prove that when deg(ad) >= deg(x^n-1), that r will have theform g(x)p(x), where g(x) is the generator, and p(x) some polynomial? === Subject: Re: Proving that the codebook of an (n,k) RS code is an ideal?Thus c = p(x^n-1) + r, where 0 <= deg(r) < deg(x^n-1)> The condition on the degree of r assures that r will have the correct> degree. (p is an arbitrary polynomial)Also, when deg(ad) < deg(x^n-1), r = ad, and because a is in G, we> have a = p(x)g(x), thus r = [d(x)p(x)]g(x), which is in G.How can I prove that when deg(ad) >= deg(x^n-1), that r will have the> form g(x)p(x), where g(x) is the generator, and p(x) some polynomial?So let's collect all the things you have:1) the polynomial a(x) (just 'a' above) is a multiple of g(x)2) x^n-1 is a multiple of g(x)3) r(x) = d(x)a(x)-p(x)(x^n-1)and you need to prove that r(x) is also a multiple of g(x).Shouldn't the conclusion be obvious from 1-3 above? If 'a' is a multiple of 'g', and 'b' is a multiple of 'g', thensurely 'da-pb' is a multiple of 'g' no matter what 'd' and 'p'are? === Subject: Re: Proving that the codebook of an (n,k) RS code is an ideal?>How can I prove that the (n,k) RS code book, i.e. all the valid code>words of the specific (n,k) Reed-Solomon code, is an ideal?>The (n,k) RS code has a generator g(x), consisting of n-k consecutive>roots in GF(2^m). (I.e. a valid RS codeword is divisible by g(x))>Also, g(x), the generator polynomial has order n, i.e. g(x) divides>x^n-1.>For an ideal, the following must hold:>A nonempty subset G of a ring R is an ideal of R if:>1. a - b in G whenever a,b in G.>2. da and ad are in G, whenever a in G and d in R. <<<<<<<<<<How can I prove that when deg(ad) >= deg(x^n-1), that r will have the>form g(x)p(x), where g(x) is the generator, and p(x) some polynomial?g(x) is a divisor of x^n - 1, and every codeword a(x) is a multiple of g(x). Thus, for any d(x) such that deg((a(x).d(x)) > n-1,we divide a(x).d(x) by x^n - 1 to get a quotient Q(x) anda remainder r(x) of degree < n. In other words, we have that a(x).d(x) = Q(x).(x^n - 1) + r(x)or equivalently, r(x) = a(x).d(x) - Q(x).(x^n - 1). The rightside is a multiple of g(x).....Hope this helps (and that the query was not from homework....)--