mm-4779 === Subject: Re: bounded monotonic sequence proof HELP > If u_n+1 = sqrt ((u_n) + 1), u_1 = 1 > prove that lim { n to infinity} u_n = 1/2 (1 + sqrt 5) > > u_n denoted u subscript n. > > i thank you for any hints and tips you can throw my way. > > cheers Where did you get the subject title from? Can you see that if the limit L exists, that it has to be the above? === Subject: Re: Show that if a_n < 0 and lim a_n = 0, then lim 1/a_n = -oo > >> Show that if a_n < 0 and lim a_n = 0, then lim 1/a_n = -oo > > Not sufficient. For that to be true, it's necessary that > for all n, a_n < 0 and not for just some n. Well, that is exactly the statement of the problem as assigned. I guess I can only assume that the professor intended a_n < 0 for all n? > >> Proof: >> Since a_n < 0 and lim a_n = 0 we have |a_n - 0| < h with |a_n| > M. > > More careless disregard for ranges of variables. > >> Select M=1/h. > > What do you mean select M = 1/h? You've already determined M above. > > >> Since |a_n| > M and a_n < 0 then 1/a_n < -M for n > N and M=1/h. > > Too quick and slick to mean anything rabbit, > other than the tortoise will out do you. > >> Thus, lim 1/a_n = -oo. > > I don't see how you come to that conclusion. > What I see you doing is to select M = 1/h and then come > to the conclusion M = 1/h. No, it's all too sloppy to be > an acceptable proof. Try slowing down to rabbit7.7. Proof: Suppose that lim a_n = 0. Then we have for some arbitrary h > 0 |a_n| < h for all n > N. Given any h > 0, let N=1/h. Then there exists N such that n > N implies that |a_n - 0| = |a_n| < h or |1/a_n| < 1/h. To have lim 1/a_n then we must have 1/a_n < -M for arbitary M > 0 and n > N. Choose M=1/h and then we have 1/a_n < -M and thus lim 1/a_n = -oo. Is that more careful and/or correct? === Subject: #15 Proving that derivatives tacked-onto StockMarket is a Ponzi Scheme; new book: StockMarket Functioning Fairly > good luck; firstly, define what you mean by proof. > > >>All these bells and whistles of new instruments added onto the >>StockMarket are merely Ponzi Pyramid schemes that cheats the people >>who do only buy and sell in cash. > The proof reasoning follows these lines: Chess and checkers are VonNeumann saddle point games. They have an Optimal Strategy and they are fair games. Now, to add on more squares and to add on more pieces such as pawns to those new squares is like adding on derivatives to the StockMarket of buy and sell in cash only with productive companies. In chess, we can either add more squares and more pieces or start decreasing squares and pieces, say we eliminate one of white and black rooks or pawns is like a derivative of chess. This derivative of chess still is a VonNeumann game with a saddle point. But now, we can add new pieces and new squares that changes the rules and affects the old game so much that it no longer is a VonNeumann fair game, nor zero sum, nor does it have a saddle point. If we eliminate a white and black rook, it is still a VonNeumann fair game with saddle point. I am going to call that a VonNeumann Fair Game Commensalism Game where derivatives end up being as fair as the original game. Now I am going to call a VonNeumann Game that has been altered with new rules -- a derivative game that is Parasitic Game. Notice in the 2008 October Crash, that the derivatives that were destroying the StockMarket were Credit Default Swaps coupled with Securitization of mortgages where the aim was to get money payed to the Investment Banks up front and let the buyers of these instruments worry about them getting their money in the future. So this was really a Ponzi scheme of a chain letter where Merrill Lynch and Lehman to name two expected a huge payment up front and let everyone else down the line worry about getting their money as the future rolled by. So now, in chess or checkers, how do we add a Ponzi move into chess and checkers? How do we add a move to chess and checkers that matches the parasitic move of the Investment Banks with regard to cheating the overall system? Securitizing Mortgages was not a company that produces goods and services but merely another packet of money in a different form than cash. This would be like adding a pair of dice to the game of chess and checkers. Once you add dice to the rules of chess and checkers, it no longer remains a VonNeumann saddle point game. VonNeumann Game Theory has ample room to expand. So that we can include derivatives, include what is a Parasitic versus Commensalism game. I need to modify the title because some derivatives are acceptable whereas others are unacceptable. Sometimes a derivative can facilitate the Stockmarket and make it work better such as the new rule that eliminated the 1/8 or 1/64 to that of decimal change in price. The change from fractions to decimal was a commensalism derivative. But the change of allowing securitization of mortgages was a parasitism of the overall StockMarket and amazingly such a derivative took only a single decade to place the entire stockmarket on the verge of collapse. But when you compare that with a game of chess where the dice roll plays an important role in the game, well, that change alters chess so much that it is no longer chess. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: defining a closed set http://mathforum.org/kb/message.jspa?messageID=6477306 > Here is footnote 1 on page 61 of volume I of Kuratowski's > _Topology_ (new edition, revised and augmented 1966): > > The notion of boundary can be taken too as primitive > term for a topological space. See M. Zarycki, _op cit_. > et J. Albuquerque, _La notion de frontiere en topologie_, > Portug. Math. 2 (1941) pp 280-289. > > The _op_ being _cit_ed is > > M. Zarycki, _Quelques notions fondamentales de l'Analysis > Situs au point de vue de l'Algebre de la Logique, Fund. > Math. 9 (1927) pp 3-15. It's worth pointing out that both of these papers are freely available on the internet: Zarycki's 1927 paper is at http://matwbn.icm.edu.pl/spis.php?wyd=1&jez=pl [Fund. Math.] http://matwbn.icm.edu.pl/ksiazki/fm/fm9/fm912.pdf and Zarycki's 1941 paper is at http://purl.pt/index/pmath/vol/PT/ [Port. Math.] http://purl.pt/2101 http://tinyurl.com/6b6ueq [.pdf file of paper] Another paper that shows the boundary operator (with some assumed properties) characterizes the topology of a space is Hyman Gabai, The exterior operator and boundary operator, American Mathematical Monthly 71 #9 (November 1964), 1029-1031. The following may also characterize the boundary operator, but the only description I can find of this paper (I only have access to freely available internet material where I'm at) is in German at the Zbl (Zentralblatt MATH) site. http://www.zentralblatt-math.org/zmath/en/advanced/ Florin Radulescu, On the frontier operator, Bulletin Math.8ematique de la Soci.8et.8e des Sciences Math.8ematiques de la R.8epublique Socialiste de Roumanie (N.S.) 24(72) (1980), 401-406. Dave L. Renfro === Subject: Re: Dr. David Berlinski and Like-minded Mathematicians posting-account=SWr-zgoAAABXP_I1KRp8SBR28UDaox2f Gecko/2008092510 Ubuntu/8.04 (hardy) Firefox/3.0.3,gzip(gfe),gzip(gfe) Hi this is Conrad Countess Unity of the Constants I find it interesting that both -1 and pi donÍt seem to have a square root until you consider the natural unite c. Just as c x c or c^2 as I c in a sense the square root of -1, (h/2pi) which is a measure of the measure of its wavelength, makes (c x c) = (cx2pi) which make (c = 2pi) in this special case. I find it quite interesting that so many constants in nature, and I suspect all of them, are somehow unified around (the constant speed of light or c). Just as c = h = square root of -1 = 2pi, in this special case, c^2 = h/2pi = G. G, h/2pi and c, not c^2 are used to derive the Planck scale which according to http://www.phys.unsw.edu.au/einsteinlight/jw/module6 Planck.htm is said to be (1.6x10-35 metres), and is said to be the scale at which (Quantum, and General Relativity Theory), will be unified, and also the basic natural unites of, E (energy), M( mass), T (time), Q (charge) and G (gravity). But I think that I have sufficient evidence that they are all unified at the critical frequency/wavelength of (c^2) . And this is a good thing because it is within our reach which the Planck scale is not. Conrad Countess === Subject: Re: Two More New Math Ideas posting-account=SWr-zgoAAABXP_I1KRp8SBR28UDaox2f Gecko/2008092510 Ubuntu/8.04 (hardy) Firefox/3.0.3,gzip(gfe),gzip(gfe) Unity of the Constants I find it interesting that both -1 and pi donÍt seem to have a square root until you consider the natural unite c. Just as c x c or c^2 as I c in a sense the square root of -1, (h/2pi) which is a measure of the measure of its wavelength, makes (c x c) = (cx2pi) which make (c = 2pi) in this special case. I find it quite interesting that so many constants in nature, and I suspect all of them, are somehow unified around (the constant speed of light or c). Just as c = h = square root of -1 = 2pi, in this special case, c^2 = h/2pi = G. G, h/2pi and c, not c^2 are used to derive the Planck scale which according to http://www.phys.unsw.edu.au/einsteinlight/jw/module6 Planck.htm is said to be (1.6x10-35 metres), and is said to be the scale at which (Quantum, and General Relativity Theory), will be unified, and also the basic natural unites of, E (energy), M( mass), T (time), Q (charge) and G (gravity). But I think that I have sufficient evidence that they are all unified at the critical frequency/wavelength of (c^2) . And this is a good thing because it is within our reach which the Planck scale is not. Conrad Countess === Subject: Re: nude preteen models, illegal, underage nudity, ls magazine, lolita art posting-account=S2WdxQoAAACdRGHtDU3qqc8bnXhX0lot rv:1.9.0.3) Gecko/2008092414 Firefox/3.0.3,gzip(gfe),gzip(gfe) That's funny) By the way, Ifyou are really wanna find some nude preteens - watch this video - it's full of them) http://vids-online.net/video.php?id=Preteen === Subject: Re: Initial ordinal, forming omega 1 <4905ee2d$0$28668$7a628cd7@news.club-internet.fr> <4906c1dc$0$28672$7a628cd7@news.club-internet.fr> <490732b6$0$28668$7a628cd7@news.club-internet.fr> In message , Dave Seaman >> David Hartley a ?crit : ... > Obviously you don't claim aleph_1 can be embedded in any non-denumerable > set (without AC) - you can't even fit aleph_0 into a Dedekind-finite > infinite set - but do you still think aleph_1 can always be embedded > into P(N)? (I'd be surprised if can , but I don't know the answer.) > I dont know either, but I would be quite surprised :-) According to this post by Fred Galvin in 1999, c may be incomparable >with every uncountable aleph. 7568b1/a0a4cc4b2425c1ba?lnk=gst&q=galvin+continuum+hypothesis#a0a4cc4b24 >25c1ba> It's a bit ambiguous, he may have meant only that he knew of no proof that c must be comparable with some uncountable aleph, rather than that there is a model where it isn't. (If c is comparable with any uncountable aleph, then c >= aleph_1.) -- David Hartley === Subject: Re: Initial ordinal, forming omega 1 <4905ee2d$0$28668$7a628cd7@news.club-internet.fr> <4906c1dc$0$28672$7a628cd7@news.club-internet.fr> <490732b6$0$28668$7a628cd7@news.club-internet.fr> , Dave Seaman > David Hartley a ?crit : > ... >> Obviously you don't claim aleph_1 can be embedded in any non-denumerable >> set (without AC) - you can't even fit aleph_0 into a Dedekind-finite >> infinite set - but do you still think aleph_1 can always be embedded >> into P(N)? (I'd be surprised if can , but I don't know the answer.) > I dont know either, but I would be quite surprised :-) According to this post by Fred Galvin in 1999, c may be incomparable >>with every uncountable aleph. 7568b1/a0a4cc4b2425c1ba?lnk=gst&q=galvin+continuum+hypothesis#a0a4cc4b24 >>25c1ba It's a bit ambiguous, he may have meant only that he knew of no proof > that c must be comparable with some uncountable aleph, rather than that > there is a model where it isn't. > (If c is comparable with any uncountable aleph, then c >= aleph_1.) Looking at it again, I think your interpretation is the right one. -- Dave Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <7450e$490026a4$82a1e228$12813@news1.tudelft.nl> posting-account=6xUtvgkAAAD_jypmLa2oo2HnrV0e8X9q rv:1.7.13) Gecko/20060414,gzip(gfe),gzip(gfe) Inside the sure part of the mathematics constructive methods may be > not as fruitful as methods including the idea of actual infinity. But > the fact, that the concept of actual infinity is incoherent remains. Constructive methods have resulted in: Computer Programming. Which makes > them a successful form of mathematics in practice. There also does exist > a purely theoretical (proper) construcivism, which is to be considered > IMO as far less relevant, as well as unsuccessful, indeed. I hadn't talked about the applications of math, I'd talked about the considerations about math. In this concern one may assume the existence of actual infinity. But always with the background that this is only a 'facon de parler'. There is no actual infinity. Incoherent not inside the mathematics, as it seems, but incoherent in > our reality. Incoherent even _in_ mathematics. All kind of artificialities are needed > to prevent the building from collapsing. Examples are: -- Wrong limits do not commute -- Natural Densities are Probabilities > -- Calculus XOR Probability I think, there could be a branch of math in which the consequences of the assumptions of an actual infinity may considered. I agree with the majority here that math could investigate any thinkable systeme. But all this considerations are underlayed with an unchangeable truth. And a part of this truth is the antinomy of an actual infinity. The question is: should mathematics reflect something (reality) or > should it only be coherent intrinsically. Only a tiny pinch of salt is needed to make mathematics consistent with > the rest of the world and that is: the acceptance of potential infinity > as the _only_ form of infinity. Hence accept that there is an incredible > small error in any of the real numbers. And accept that the naturals are > not a set (i.e. an unfinished set at best) but, instead, a proper class. Han de Bruijn I agree: the potential infinity is the only consistent infinity. I disagree: not confirming with you, any real number is exact. Some questions remain. Albrecht S. Storz === Subject: Re: A consideration concerning the diagonal argument of G. Cantor There is no actual infinity. > How would you KNOW? Did you check it? :-) > > I agree: the potential infinity is the only consistent infinity. > Oh, oh. Have you found any inconsistency in any axiomatic set theory? Please tell us!!! > > Some questions remain. > Indeed. :-) Herb === Subject: Re: Hypergeometrical Universe Predicts both Mercury Perihelion Precession and Gravitational Lensing > Hypergeometrical Universe Predicts both Mercury Perihelion Precession > and Gravitational Lensing My theory is a 5D Spacetime theory Hahahahaha! Do you want to make some easy money, Androcles... No, I have enough, I don't need any more. --------------------------------------------------- Find a fatal flaw in my theory 5D does it, crank. --------------------------------------------- === Subject: Re: Integrate using complex analysis <6iudg41ni99efu1doormnghirahl6olukc@4ax.com> posting-account=ncfGGAoAAACF2UeTuTcJ6PAyMvLk0mth Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) I am having trouble integrating the following: sin( sqrt(a^2+x^2) ) / sqrt(a^2+x^2) from -Inf to +Inf. I read about the Cauchy integral theorem but the >examples I've found have a polynomial denominator whereas this has a >sqrt, so I don't really know how to apply it here. Do you have any good reason to think that this integral can > be evaluated this way? A few comments: 1. It seems unlikely that you can use the standard complex > methods with the first integral you asked about, > 1 / sqrt(a^2+x^2), because that function is not > defined in the upper half-plane, and in particular > it does not have poles where you think it does. > (and that integral doesn't converge). 2. On the other hand, the function > sin( sqrt(a^2+x^2) ) / sqrt(a^2+x^2) is defined > in the upper half-plane, and does have poles > in the obvious places. (and this integral does > converge). So maybe this one can be done this way. 3. On the other other hand, the usual way to do it > would be to use sin(t) = Im(exp(it)) (for real t), > but if you do that you're back to the problem in (1). > (the problem with sin is that it blows up on the > imaginary axis.) > posted my questions and they didn't appear after a day, so I tried again. The integral of interest is no. 2 in your list above, i.e. +Inf / | sin( sqrt(a^2+x^2) ) / sqrt(a^2+x^2) . dx / -Inf It arose from trying to sum a certain type of signal over all frequencies. As you said, it looks well behaved and I can obtain numerical solutions however a closed form would be much preferred (one less numerical integral to perform). Regarding the poles, I naively thought a pole was any 1/0 occurrence, but after reading up on definitions I see that a pole is of the form 1/ z^n as z->0. Multiplying top and bottom by sqrt(a^2+x^2) moves the sqrt to the numerator although I guess that is just window dressing. For a=0 the function is sin(|x|) / |x| which MATLAB tells me has a closed form integral sinint(), which I know comes from complex analysis. I also have a text book which has as a homework problem to integrate sin(x)/sqrt(x) from 0 to Inf. Does this sound like an avenue to pursue? === Subject: Re: Passing through every quantity posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) > When there is a continuous change there is a passage through every > quantity. This applies in the calculus of physics. There can pass through different sizes of infinity of the infinitely > small at the same rate. > This is speed through distance. The limit for speed is light speed. > You can pass through the same size infinity of the infinitely small > slower or faster. > This is the same distance covered at either slower or faster rate. > With the fastest rate being reaching light speed in an infinitely > small interval. This is acceleration rate infinity. MitchRaemsch There is such a thing as an instantaneous jump where you do not pass > through what is inbetween. The primary example is the speed of light. > There is no acceleration. It simply starts moving at that speed. > Matter at time zero might have had an original speed also. MitchRaemsch Anyone have some primary examples of a jumping curve? MitchRaemsch- Hide quoted text - - Show quoted text - Momentum is exchange of motion of bodies through space. > There is a kinetic energy jump down for one up for another. The first > body gives its motion to the second. If the second body was relatively > still then it gains the motion of the first leaving the first still > like the second orginally was. MitchRaemsch- Hide quoted text - - Show quoted text - Passing through every speed at different faster slower rates gives > the magnitude of acceleration. MitchRaemsch- Hide quoted text - - Show quoted text - Weight fluctuation in temporary acceleration causes an object to pass through every quantity of weight inbetween. Mitch Raemsch === Subject: Re: -- finitely presented groups on n generators <20684031.1225205349878.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=-PngCgkAAAD2yUjosqWv1Nf1lkqWP4lp rv:1.8.1.16) Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > I believe it is however recursively presented--but I don't know the reference. Your remark has no context! What do you believe is recursively presented? Derek Holt. === Subject: Re: -- finitely presented groups on n generators <20684031.1225205349878.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008092318 Fedora/3.0.2-1.fc9 Firefox/3.0.2,gzip(gfe),gzip(gfe) > I believe it is however recursively presented--but I don't know the reference. One can trivially go through an enumeration of 4-tuples (a,b,c,d) of elements in a free group on the generatos and write the corresponding identity [[a,b],[c,d]]. That's a recursive presentation, IIUC. -- m === Subject: Re: The highest limit of acceleration/deceleration posting-account=504E-QkAAAA2v90r8nGnJKpfySa_yBSU 5.1),gzip(gfe),gzip(gfe) On Oct 26, 6:03pm, Dirk Van de moortel ge23g5$c7...@registered.motzarella.org How do you manage to get it wrong all the time? How does he manage to get people reply to him all the time? Dirk Vdm Good question. Harry C. === Subject: Re: The highest limit of acceleration/deceleration posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) > On Oct 26, 6:03pm, Dirk Van de moortel > ge23g5$c7...@registered.motzarella.org How do you manage to get it wrong all the time? How does he manage to get people reply to him all the time? Dirk Vdm Good question. Harry C. An accelerated rocket is relegated to low acceleration in every practical sense. It (acceleration/deceleration) never leaves the realm of the small but is potentially infinite. Mitch Raemsch === Subject: Re: algebraic topology problem <20081028041836.U8566@agora.rdrop.com> posting-account=YeC6dQoAAAD4oLD_rhSm45JYeEZE06ik Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) Well, this forum works badly... yours posts (all of which I found very why I've removed my second plea for help. Going back to the discussion. It follows that f*(pi(RP^n))=0, therefore (as far as I remember), f may be lifted by the exponential map exp: R^1-->S^1 And similarly NotP: Therefore, by the theory of covering spaces, the map f factors through the universal covering space p: R -> S^1 As my contribution into the matter I cite a theorem which I think is relevant in this case: -- Let p: (Z, z_0)-->(X, x_0) be a fibration with unique path lifting. Let Y be a connected locally path-connected space. A necessary and sufficient condition that a map f:(Y,y_0)-->(X, x_0) have a lifting (Y,y_0) --> (Z,z_0) is that in pi(X, x_0) f*[pi(Y, y_0)] belongs to p*[pi(Z, z_0)] where f*: pi(Y, y_0) --> pi(X, x_0) is a group homomorphism induced by f -- We have Z=R X=S^1 Y=RP^n p = exp Covering projection exp: R --> S^1 is a fibration with unique path lifting. f*[pi(RP^n)] belongs to exp*[pi(R)](which is also a constant map as pi(R) is just ({0},+)) === Subject: Re: Why meta diagonals are irrelevant posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) >{1,2,3,...,n+1} U {1,2,3,...,n+2} U {1,2,3,...,n+3} U ... = >>{1,2,3,...} = N (U) >You can drop every finite initial segment that ends with a natural >>number, because we can prove that it is not required to form the set >>N. > Obviously, that's incorrect. No, obviously that is correct for every natural number. Look, the unambiguous, correct statement is this: If we let U_n = the set of all finite initial segments S of the > naturals such that size(S) >= n, then Claim 1: forall n, union(U_n) = N That's all there is to say about it. It isn't contradictory, > it is just true. After all that time you don't know what is true? You are one-eyed. For every n: {1,2,3,...,n} can be deleted from the union. forall n, the union of all {1,2,3,...,n} that can be deleted is N. > What you always like to do is to start with a precise > mathematical statement such as Claim 1, and then make > it ambiguous by rendering it into natural language, then > reinterpret the ambiguous statement so that it means > something different from the original statement, and > something provably false. That's the way you go about > proving contradictions. Claim 1 is true. You paraphrase Claim 1 as > You can drop every finite initial segment that ends with > a natural number, and the union is still N. That is ambiguous. It either means Interpretation 1: Forall natural numbers n, you can drop > the first n initial segments, and the union of the remaining > sets is N. That's true, and is equivalent to Claim 1. Alternatively, it means Interpretation 2: You can remove *all* initial segments > and the union of the remaining sets is N. That's clearly false, and is not equivalent to Claim 1. > You would *know* that if you actually used mathematical > notation, instead of sloppy natural language. Look at a simple, finite example. Look at the simple infinite example: Claim A: For all n: n is divisible by n. The union of all n satisfying claim A is what? === Subject: Re: Why meta diagonals are irrelevant posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > To claim that this sequence has limit 0 is of same impertinence as the > claim > > LIM{n --> oo} (1 + 2 + 3 + ... + n) =3D 0 > > In fact, Dik T. Winter did so. But that is not what I call mathematics > (notwithstanding Euler's erroneous results). > > It was a *definition*. With > standard definitions that limit does not exist. So in mathematics it > is possible to *provide* a definition. > > You gave this definition. It is unimportant whether this limit does > exist or does not exist in standard definitions. Your claim that this > definition is in agreement with mathematics is contradicting > mathematics. What part of mathematics does it contradict? BTW, I just remember that I > sum{n = 1...oo} (1 + 2 + 3 + ... + n) = 0 > As that sum is not defined within standard mathematics, that definition is > not in disagreement with standard mathematics. That is no excuse and no reason! A sum of positive natural numbers and without negative numbers cannot be less than 1. That is so in standard mathematics. (There is no negative positive number, and you cannot define such a number, even if it is not defined in standard mathematics.) === Subject: Re: Agreed Question about whether you can construct L posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Given that the rules of the game do not talk about an anti-diagonal of > L to be put into the set H, I don't see why then one would say that > such anti-diagonal is in fact in H. It isn't, until you put H in a list. Then you construct an > antidiagonal by a fixed rule. Then merely constructing a set of infinite binary strings does not > automatically create any anti-diagonals at all. In fact, none can exist > until one has a completed infinite list of infinite binary strings. Except that the list need not be infinite (it can comprise of n numbers), that is right. === Subject: Re: Agreed Question about whether you can construct L <490710ba$0$5483$9a566e8b@news.aliant.net> posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) >Proposed question #2 > Let H be the set of numbers constructed between >> midnight and t 2 (where t 2 is included). >> Let L be a list containing all the elements of H >> Let t 2 be the time at which you try to present L. >> Can you construct L? Let H be the set of numbers constructed after midnight. Can you >construct L after midnight? >Rejected. In this question there is a start time >>for constructing H, but no end time. So H is something that >>can change. Every infinite set is something that can change, Hotly disputed. However, we do agree that some sets > do not change. Call these unchangable sets, u-sets for short. > Note that H is a u-set if and only if an end time is given. What I understand by H is the set of all anti-diagonal numbers constructed by you (or alternatively by mankind or who aver can construct anti-diagonal numbers). There is no time limit, H is simply all. Like the set of all American people is simply all people in America. Proposed question #2 Let H be the u-set of numbers constructed between > midnight and t 2 (where t 2 is included). > Let an H-list be a list containing all the elements of H > Let t 2 be the last time at which you try to construct an H-list. > Is it possible to contruct an H-list? It is not possible to construct a H-list, but it is possible to construct the set H. Therefore there are uncountable and finite sets. === Subject: Re: Q on Riemann Hypothesis > Why is the following not a correct proof of the Riemann Hypothesis? Because it's too short & simple. If a proof were that easy, Riemann would have found it in his sleep. If what you mean is, please find the error(s) in this proof for me, well, finding the errors in your proof is your job, not ours. But when you have found them, please do report back to us, so we can learn from them. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Q on Riemann Hypothesis In message >Why is the following not a correct proof of the Riemann Hypothesis? Definitions: >z(s)=Riemann zeta function=sum_n=1_to_inf{(-1)^(n+1)*(1/n^s)}, valid >for Re(s)>0 Shouldn't that be Re(s) > 1? Might make a difference. -- David Hartley === Subject: Re: Q on Riemann Hypothesis posting-account=yXs7PwoAAABxalAQoV3duo5HJUDgYk1_ MathPlayer 2.10d; .NET CLR 2.0.50727; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > In message Why is the following not a correct proof of the Riemann Hypothesis? Definitions: >z(s)=Riemann zeta function=sum n=1 to inf{(-1)^(n+1)*(1/n^s)}, valid >for Re(s)>0 Shouldn't that be Re(s) > 1? Might make a difference. > -- > David Hartley Sorry - I meant to include the prefactor for the zeta function: z(s)=1/[1-2^(1-s)]*sum n=1 to inf{(-1)^(n+1)*(1/n^s)} which is defined for Re(s)>0 and illustrates the simple pole at s=1. The rest of the proof should not be affected by this since we are looking at the zeros of the function and this prefactor is never 0 in the range 0 http://www.fredericknewspost.com/sections/news/display.htm?StoryID=81729 Reformer wants changes in local schools > Originally published October 22, 2008 By Marge Neal, News-Post Staff reform to today's education system, however, i do not agree with all thinking that basically anyone with a bachelor's degree and is able to pass two tests should be allowed to be a teacher. I feel that it is essential to go through a teacher-education program in order to get certified to be a teacher. In a teacher-education program, as a student you have classes on inclusive education, teaching methods, and stated would miss out on. The most important part of a teacher education program is that as a student you spend hours upon hours in a classroom practicing teaching with a supervisor and a cooperating teacher observing you teach. The supervisor and cooperating teacher are able to help you develop, execute, and reflect on lessons which will help you develop into a great teacher. But these people who take 2 tests and have a bachelor's degree miss out on this critical practice of being a teacher. Additionally, how does having any sort of bachelor's degree and taking two tests show that someone is a great teacher? I feel that taking tests don't truly showcase a person's teaching abilities. In order for reform to take place there is a need for good teachers which I feel are developed through teacher-education certification. === Subject: Re: Reformer wants changes in local schools >> http://www.fredericknewspost.com/sections/news/display.htm?StoryID=81729 >> Reformer wants changes in local schools >> Originally published October 22, 2008 >> By Marge Neal, News-Post Staff >reform to today's education system, however, i do not agree with all >thinking that basically anyone with a bachelor's degree and is able to >pass two tests should be allowed to be a teacher. I feel that it is >essential to go through a teacher-education program in order to get >certified to be a teacher. At least a large number of my colleagues would prefer to see the schools of education abolished. One poster on the sci.math newsgroup posted that, from observation of his fellow students taking education courses, that it seemed that a lobotomy was required. In a teacher-education program, as a >student you have classes on inclusive education, teaching methods, and >stated would miss out on. Much of the current problem is the use of those methods and the content of those courses. .................. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: FW: Fear of Numbers posting-account=doZR-AoAAAB6mxNSgbGzeqKaEDQsZFQH CLR 1.1.4322),gzip(gfe),gzip(gfe) > don't understand or know how to do basic math. So that means > something needs to be changed in the math curriculum right away in > elementary school. I believe that math should be taught using both > traditional and reform traits. I agree that combining traditional and reform math teaching strategies is the way to go. One book to look at would be Children's Mathematics by Carpenter, Fennema, Freanke, Levi and Empson. It focuses on a teaching strategy called Cognitively Guided Instruction (CGI). This book and teaching strategy demonstrates how to take what a child already intuitively knows about how math works and use that to motivate the student to learn math with understanding. Rather than introducing algorithms that students are required to memorize and use, this technique starts with informal math strategies that students invent and feel comfortable with. The teacher's role is to scaffold the students into learning how to represent their thinking more clearly and eventually guide them into more traditional algorithms, all the while focusing intently on understanding as the main goal. Many times a student will feel that math is boring because there is no true understanding, and typically memorizing traditional algorithms can be boring for almost anyone. However, when we can encourage students to use creativity in how they approach math sparks might begin to fly, and oftentimes we find that they already have successful strategies for the problems we're asking them to do, they just don't know how to represent their thinking on paper. This book has changed the way I will approach math in my classroom, and it has helped me motivate many students to look at math in a different way as well. === Subject: Re: FW: Fear of Numbers posting-account=doZR-AoAAAB6mxNSgbGzeqKaEDQsZFQH CLR 1.1.4322),gzip(gfe),gzip(gfe) But I do feel that > combining the strengths and similarities of the traditional and reform > based math would be a good start to success. I agree that combining traditional and reform math teaching strategies is the way to go. One book to look at would be Children's Mathematics by Carpenter, Fennema, Freanke, Levi and Empson. It focuses on a teaching strategy called Cognitively Guided Instruction (CGI). This book and teaching strategy demonstrates how to take what a child already intuitively knows about how math works and use that to motivate the student to learn math with understanding. Rather than introducing algorithms that students are required to memorize and use, this technique starts with informal math strategies that students invent and feel comfortable with. The teacher's role is to scaffold the students into learning how to represent their thinking more clearly and eventually guide them into more traditional algorithms, all the while focusing intently on understanding as the main goal. Many times a student will feel that math is boring because there is no true understanding, and typically memorizing traditional algorithms can be boring for almost anyone. However, when we can encourage students to use creativity in how they approach math sparks might begin to fly, and oftentimes we find that they already have successful strategies for the problems we're asking them to do, they just don't know how to represent their thinking on paper. This book has changed the way I will approach math in my classroom, and it has helped me motivate many students to look at math in a different way as well. === Subject: Iteration Formula? I'm looking for a formula that will allow me to pick the top 10 selections from a list. The formula would look at the first 10 values and check to see if those summed values were less or equal to another value. If the summed values are less or equal to the comparison value, the formula would stop; else it would then look at the top 9 values and the 11th value and again compare. This would proceed through all the selections in place of the 10th value, then the 9th value, etc... until it met the comparison. Know that sounds a bit confusing but I will clarify further if anyone is willing to assist me. Jack (I got my butt kicked by Calc II). === Subject: Re: Iteration Formula? > I'm looking for a formula that will allow me to pick the top 10 > selections from a list. No, you're not looking for a formula. You're looking for an algorithm. > The formula would look at the first 10 values and check to see if those > summed values were less or equal to another value. > If the summed values are less or equal to the comparison value, the > formula would stop; else it would then look at the top 9 values and the No, you don't mean 'the top 9'. You mean the first 9. > 11th value and again compare. This would proceed through all the > selections in place of the 10th value, then the 9th value, etc... until > it met the comparison. > You're being too slick and glib. Thus I'll decide that the algrorithm will then look at 1-8, 10-11; 1-7, 9-11; 1-6, 8-11, etc. > Know that sounds a bit confusing but I will clarify further if anyone is > willing to assist me. > I am not a mind reader. What do you want? An algorithm that'll pick the 10 highest scores out of a list? > Jack (I got my butt kicked by Calc II). > I can beleive it. === Subject: Re: Iteration Formula? posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > I'm looking for a formula that will allow me to pick the top 10 > selections from a list. The formula would look at the first 10 values > and check to see if those summed values were less or equal to another > value. If the summed values are less or equal to the comparison value, > the formula would stop; else it would then look at the top 9 values and > the 11th value and again compare. This would proceed through all the > selections in place of the 10th value, then the 9th value, etc... until > it met the comparison. And if you can't find it, do you then want to try numbers 2nd through 11th then 2nd through 10th + 12th, etc.? Do you want to try every combination of n items taken 10 at a time until you find your target sum? If so, Python 2.6 can easily handle that (previous versions don't have combinations built in). > import itertools as it > a = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15] > b = [i for i in it.combinations(a,10)] This will give you 3003 combinations. A couple of which are: > b[0] (1, 2, 3, 4, 5, 6, 7, 8, 9, 10) > b[100] (1, 2, 3, 4, 5, 6, 9, 10, 14, 15) > b[666] (1, 2, 3, 5, 9, 10, 11, 12, 13, 15) > b[3002] (6, 7, 8, 9, 10, 11, 12, 13, 14, 15) Once you have that, you can make a histogram of how these combinations add up. > c = {} > for i in b: s = sum(i) if s in c: c[s] += 1 else: c[s] = 1 Of the 3003 combinations, only > len(c) 51 different sums are possible. They break down as: > c { 55: 1, 56: 1, 57: 2, 58: 3, 59: 5, 60: 7, 61: 10, 62: 13, 63: 18, 64: 23, 65: 30, 66: 36, 67: 45, 68: 53, 69: 63, 70: 72, 71: 83, 72: 92, 73: 103, 74: 111, 75: 121, 76: 127, 77: 134, 78: 137, 79: 141, 80: 141, 81: 141, 82: 137, 83: 134, 84: 127, 85: 121, 86: 111, 87: 103, 88: 92, 89: 83, 90: 72, 91: 63, 92: 53, 93: 45, 94: 36, 95: 30, 96: 23, 97: 18, 98: 13, 99: 10, 100: 7, 101: 5, 102: 3, 103: 2, 104: 1, 105: 1 } Obviously, some values are more common than others and target sums less than 55 or greater than 105 don't exist. Is this kind of what you want? Know that sounds a bit confusing but I will clarify further if anyone is > willing to assist me. Jack (I got my butt kicked by Calc II). === Subject: elementary school probability It's been a long time since I've had to deal w/ any kind of probability. I feel bad that I can't remember the steps to help out my 4th grader and this type of problem. Any help/advice would be greatful. The problem was. Levi has 3 blue shirts and 1 white shirt. He has 1 pair of blue pants and 1 pair of white pants. What is the probability that Levi ends up with a blue shirt and white pants? === Subject: Re: elementary school probability > > It's been a long time since I've had to deal w/ any kind of probability. I > feel bad that I can't remember the steps to help out my 4th grader and this > type of problem. Any help/advice would be greatful. > > The problem was. > > Levi has 3 blue shirts and 1 white shirt. He has 1 pair of blue pants and 1 > pair of white pants. > What is the probability that Levi ends up with a blue shirt and white pants? > The question is unanswerable. -- He is not here; but far away The noise of life begins again And ghastly thro' the drizzling rain On the bald street breaks the blank day. === Subject: Re: elementary school probability >It's been a long time since I've had to deal w/ any kind of probability. I >feel bad that I can't remember the steps to help out my 4th grader and this >type of problem. Any help/advice would be greatful. The problem was. Levi has 3 blue shirts and 1 white shirt. He has 1 pair of blue pants and 1 >pair of white pants. >What is the probability that Levi ends up with a blue shirt and white pants? Assuming he picks randomly and independently, the probability of picking a blue shirt is 3/4 and the probability he picks white pants is 1/2 so the probability of both is 3/4 * 1/2 = 3/8. --Lynn === Subject: Re: elementary school probability Probably one of the best ways to go through this with a 4th grader is to set it out visually. A table can work pretty well. The below shows the sample space...given the table, you can see that 8 things can ever happen. You are interested in the chance that he ends up with the blue shirt and white pants. In general, for probability of anything, you're finding: Prob of x = the number of ways that x can happen ------------------------------------ the number of ways that anything can happen So, you can see from the table that he ends up with a blue shirt and white pants 3 times out of a possible 8, so your probability would be 3/8. Pants Blue White Shirts Blue BB BW Blue BB BW Blue BB BW White WB WW I'm dreading this sort of thing in a few years when my 1 yr old is asking me questions...just started back to school after a 10-year hiatus, and it's been fun working through some of the cranial cobwebs. Good luck! Mac > It's been a long time since I've had to deal w/ any kind of probability. > I feel bad that I can't remember the steps to help out my 4th grader > and this type of problem. Any help/advice would be greatful. > > The problem was. > > Levi has 3 blue shirts and 1 white shirt. He has 1 pair of blue pants > and 1 pair of white pants. > What is the probability that Levi ends up with a blue shirt and white > pants? > === Subject: Re: How to plot a variable >When I iterate a variable in a do-loop it actually will not print >out. Is it undefined? In Fortran this would never be a problem, but >in mathematica it seems that it must be done different: >Do[ GA = GA + .5*(G[J] + G[J + 1])*DELT ; Amp = Exp[2.*GA] ; >JJ = J + 1; ABB[1] = 1; ABB[JJ] = Amp, {J, 1, N1}] ; >pl1 = Plot[ABB, {n, 1, NAX}, PlotLabel -> Impedans, Filling -Axis, >PlotRange -> All] >ABB gives no result.. And it should not. I believe you are attempting to use the notation f[n] to mean the nth element of an array f. Mathematica sees this notation as the function f to be evaluated at n. Note the difference between doing In[1]:= x = Table[n^2, {n, 5}]; x Out[2]= {1,4,9,16,25} Out[2]= {1,4,9,16,25} and In[3]:= Do[y[n] = n^2, {n, 5}] y Out[4]= y In[5]:= y[3] Out[5]= 9 I've defined x to be an array which Mathematica evaluates when I type x to output the list of elements. But y is defined as a function. So, typing y results in it being unevaluated since there have been no arguments supplied. Typing y[3] evaluates the function y at 3 and returns the 9 since that has been defined as the value of y at 3. If I correctly understand what you intended, that is for ABB to be an array of numeric values rather than a function and your code defines numeric values for n running from 1 to N1, then ListPlot[Table[ABB[n],{n,N1},PlotLabel->Impedans, Filling ->Axis,PlotRange -> All] should produce the plot you want === Subject: Re: Function pure for Select > Hi al, How can I to write a function pure to extract all the first rows of > collection of data, applied to a list?. For example, Select[data,First[#]==30&] This function extracts all rows which first element is equal to 30. > Well, I want to extend this function to a list of values > {30,45,50,66}. I think this is what you want Select[data,#[[;;4]]=={30,45,50,66}&] Ssezi === Subject: Re: Function pure for Select > How can I to write a function pure to extract all the first rows of > collection of data, applied to a list?. > > For example, > > Select[data,First[#]==30&] > > This function extracts all rows which first element is equal to 30. > Well, I want to extend this function to a list of values > {30,45,50,66}. You could use the built-in function *MemberQ*. For example, In[101]:= data = Table[{n, RandomInteger[]}, {n, 10}] Select[data, MemberQ[{3, 5, 7}, First[#]] &] Out[101]= {{1, 0}, {2, 1}, {3, 0}, {4, 0}, {5, 1}, {6, 0}, {7, 0}, {8, 0}, {9, 1}, {10, 0}} Out[102]= {{3, 0}, {5, 1}, {7, 0}} -- Jean-Marc === Subject: Re: How does Mathematica know whether a number is real or AA = {{0.45236, -0.241371}, {-0.241371, 0.173649}}; BB = {-0.500298, 0.141347}; CC = -0.773749; Select[Solve[{{x, y}.AA.{x, y} + BB.{x, y} + CC == 0, x^2 + y^2 == 1}, {x, y}], Element[{{x, y} /. #}, Reals] &] {{x->-0.970229,y->-0.242189},{x->-0.448307,y->0.89388}} Any non-zero imaginary element will result in the number being complex Element[3. + 10^-20*I, Reals] False Element[3. + 10^-20*I, Complexes] True To eliminate very small parts use Chop Chop[3. + 10^-10*I, 10^-6] 3. Bob Hanlon The solution of the equation at the end of this post is {{x -> -0.9702291184141268, y -> -0.24218888864137905}, {x -> -0.44830718122625973, y -> 0.8938795619438703}, {x -> 1.3776284630851778 - 0.7866963118457873*I, y -> -0.9763348915886718 - 1.1100445557562666*I}, {x -> 1.3776284630851778 + 0.7866963118457873*I, y -> -0.9763348915886718 + 1.1100445557562666*I}} Two of the solutions are real. The other two are complex. I'm wondering how Mathematica know whether a solution is real or complex. If a solution actually has a close-to-zero imaginary part, will mathematica incorrect consider it as a real number? Peng AA = {{0.45236, -0.241371}, {-0.241371, 0.173649}}; BB = {-0.500298, 0.141347}; CC = -0.773749; Solve[{{x, y}.AA.{x, y} + BB.{x, y} + CC == 0, x^2 + y^2 == 1}, {x, y}] -- Bob Hanlon === Subject: Re: Speeding up a list construction >that function, but it does exist in 4.1. I don't understand why you think FoldList doesn't exist in version 4.1. The documentation for FoldList in version 6.03 indicates this function was introduced in version 2 of Mathematica. A quick check of the 4th edition of the Mathematica book does show an entry for FoldList on page 1137. === Subject: Diferential eq Hi! I want to compute a partial derivative of the equation (please find it attached) with respect to the variable x: http://img222.imageshack.us/my.php?image=diffeqyi0.png === Subject: Re: Diferential eq > Hi! I want to compute a partial derivative of the equation (please find it > attached) with respect to the variable x: http://img222.imageshack.us/my.php?image=diffeqyi0.png > === Subject: Re: Diferential eq > Hi! I want to compute a partial derivative of the equation (please find it > attached) with respect to the variable x: > > http://img222.imageshack.us/my.php?image=diffeqyi0.png > The command to use is *DSolve* [1] (you will find many examples here and plenty of pointers to relevant section of the documentation). When writing the equation in Mathematica syntax, keep in mind that Mathematica is case sensitive and that built-in commands start with a capital letter, and that square brackets are used to indicate function arguments (so sin^2 phi should be written Sin[phi]^2). Also, the dependence of time must be made explicit (so x[t], x'[t], x''[t]) and equality is denoted by a double equal sign (==). HTH, - Jean-Marc [1] DSolve http://reference.wolfram.com/mathematica/ref/DSolve.html?q=DSolve&lang=en === Subject: Re: Expressions with ellipsis (...) Saying to use a syntax that Mathematica understands is one, entirely appropriate, answer. Another answer is that Mathematica *should* understand the syntax with ellipsis -- and indeed it can with the facilities of David Park's Presentations application package. > > write it in a syntax that Mathematica understand, i.e., > > Limit[Sum[k^2, {k, 1, n}]/n^3, n -> Infinity] > > ?? > > Jens > >> Hi Mathematica Friends, I want to do this: Limit[(n^2 + (n - 1)^2 + (n - 2)^2 + ... + 1)/n^3, n -> Infinity] But Mathematica barfs: expression cannot be followed by .... Searching the help for 'ellipsis' gives me the Unicode 2026 character >> (i.e. ...). How can I get Mathematica to eval this limit? David. > -- Murray Eisenberg murray@math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2859 (W) 710 North Pleasant Street fax 413 545-1801 Amherst, MA 01003-9305 === Subject: Two questions about DSolve My goal is to solve DSolve[r^2 y''[r] + r y'[r] + (r^2 - m^2) y[r] == BesselJ[m, r], y[r], r] Question 1. Is there a way to tell Mathematica that I want the solutions that are finite at r=0? Question 2. I get answers in terms of MeijerG. How does one obtain the special form of this function from the special combination of arguments. For example, I'm would like to learn what function MeijerG[{{1/2}, {-(1/2), 1}}, {{0, 0, 0}, {-(1/2), 0}}, r, 1/2] is in terms of more elementary functions. Aaron === Subject: Re: Two questions about DSolve I get {y[r] -> BesselY[m, r]*C[2] + (r^(2*m)*BesselJ[-m, r]*Gamma[-m]* HypergeometricPFQ[{m, 1/2 + m}, {1 + m, 1 + m, 1 + 2*m}, -r^2])/(2^(2*(1 + m))*m^2*Gamma[m]) - (Pi*r^2*BesselJ[m, r]*Csc[m*Pi]*HypergeometricPFQRegularized[ {1, 1, 3/2}, {2, 2, 2 - m, 2 + m}, -r^2])/8 + BesselJ[m, r]*(C[1] + Log[r]/(2*m))} with Mathematica 6 -- an there i no MeijerG[] in it. And no, Mathematica can't know what m is and that you you wish to take the limit r->0 Jens > > My goal is to solve > > DSolve[r^2 y''[r] + r y'[r] + (r^2 - m^2) y[r] == BesselJ[m, r], > y[r], r] > > Question 1. Is there a way to tell Mathematica that I want the > solutions that are finite at r=0? > > Question 2. I get answers in terms of MeijerG. How does one obtain the > special form of this function from the special combination of > arguments. For example, I'm would like to learn what function > > MeijerG[{{1/2}, {-(1/2), 1}}, {{0, 0, 0}, {-(1/2), 0}}, r, 1/2] > > is in terms of more elementary functions. > > > Aaron > === Subject: Re: Is there a simple way to transform 1.1 to > Of course you could start by inputing strings, which may not be much >> more work. But really it all depends on what sort of use you wan to >> make of this. If the numbers really arise out of some computation you >> could try doing the computations with high precision to get answers >> which look like the above. I don't know how to do that in general, since my understanding is that > if you increase the precision you might end up with a really different > looking number. You can actually perform numerical computations in Mathematica with fixed precision. As other people have explained, the main problem arises not from variable precision but from the fact that Mathematica performs all computations using binary representation and you want a decimal one. >> One more (I hope final) remark. Another, perhaps simpler way, is to >> include just one extra zero (that should not be very hard): If you are talking about inputs, I agree; in this case it appears to > me similar to inputting strings. I meant input. I do not really understand what you are doing and why writing 1.00000000000000010 is much harder than writing 1.0000000000000001. Is this entire discussion motivated by your desire to avoid pressing keys on your keyboard? Andrzej Kozlowski === Subject: Re: Is there a simple way to transform 1.1 to 11/10 > One more (I hope final) remark. Another, perhaps simpler way, is to > include just one extra zero (that should not be very hard): If you are talking about inputs, I agree; in this case it appears to >> me similar to inputting strings. I meant input. I do not really understand what you are doing and why > writing 1.00000000000000010 is much harder than writing > 1.0000000000000001. No, it is not. I just wanted to be sure you were talking about input. I'll prefer the string approach since I can see by myself why it will never fail. I need to think more about the case of numbers coming out from Mathematica computations. So I consider my initial problem --that of inputs-- solved, and I warmly thank everybody. > Is this entire discussion motivated by your desire to avoid pressing > keys on your keyboard? That's one way to put it. I'd rather say that I like elegant solutions. Anyway, transforming a lot of such numbers is error prone and way too time consuming, to the extent that it simply cannot be done. a. === Subject: Re: Is there a simple way to transform 1.1 to > Of course you could start by inputing strings, which may not be much > more work. But really it all depends on what sort of use you wan to > make of this. If the numbers really arise out of some computation you > could try doing the computations with high precision to get answers > which look like the above. I don't know how to do that in general, since my understanding is that if you increase the precision you might end up with a really different looking number. > One more (I hope final) remark. Another, perhaps simpler way, is to > include just one extra zero (that should not be very hard): If you are talking about inputs, I agree; in this case it appears to me similar to inputting strings. If you are talking about results of a computation, then I don't see how to do it. Anyway, that you again for your help. a. === Subject: Re: Is there a simple way to transform 1.1 to OK. I was wrong. It's even harder than I thought. How about this (using strings) ff[z_] := With[{p = ToExpression[StringJoin[DeleteCases[Characters[ToString[z]], .]]]}, p/Round[p/z]] ff[1.001] 1001/1000 Andrzej Kozlowski > So Andrzej Kozlowski proposed: In[1]:= rAndrzej[z_] := With[{w = > {x, a} &, w, 1]]] Someone else proposed in private: RealDigits[x] Bill Rowe proposed: In[3]:= rBill[x_] := Module[{r = Rationalize[x, 0], p}, p = > Round@Log[10, Denominator[r]]; Round[r 10^p]/10^p] They all performed well on early isolated trials, but more systematic > tests show failures: In[4]:= x=1.001;{rAndrzej[x],rSH[x],rBill[x]} 1000999999999999 1000999999999999 1001 > Out[4]= {----------------, ----------------, ----} > 1000000000000000 1000000000000000 1000 In[5]:= x=1.0000000000000001;{rAndrzej[x],rSH[x],rBill[x]} Out[5]= {1, 1, 1} > Then, when getting closer to 1, the first 2 give the expected result, > while the 3rd one gives sometimes the expected result, sometimes 1. I found a way to do what I want, but only by starting from a string, > i.e., I have to enclose the number between double quotes, which is > impractical in general. I could not find a general way to transform a number to a > string. Specifically, is there a way to transform the above number > 1.0000000000000001 (fifteen zeros), which seems to be the most > problematic, to the corresponding string 1.0000000000000001 ? Alain === Subject: Re: Is there a simple way to transform 1.1 to > OK. I was wrong. It's even harder than I thought. How about this > (using strings) > > ff[z_] := > With[{p = > ToExpression[StringJoin[DeleteCases[Characters[ToString[z]], > .]]]}, > p/Round[p/z]] > > ff[1.001] > 1001/1000 In[106]:= ff[1.0000000000000001] Out[106]= 1 Like I said, I couldn't find a way to transform the number 1.0000000000000001 to the corresponding string. I have tried all of $OutputFormats and $ExportFormats without success yet. The most promising is MathMLForm but it fails at least for that number. alain > Andrzej Kozlowski > > > > > > So Andrzej Kozlowski proposed: > > In[1]:= rAndrzej[z_] := With[{w = > {x, a} &, w, 1]]] > > Someone else proposed in private: > > RealDigits[x] > > Bill Rowe proposed: > > In[3]:= rBill[x_] := Module[{r = Rationalize[x, 0], p}, p = > Round@Log[10, Denominator[r]]; Round[r 10^p]/10^p] > > They all performed well on early isolated trials, but more systematic > tests show failures: > > In[4]:= x=1.001;{rAndrzej[x],rSH[x],rBill[x]} > > 1000999999999999 1000999999999999 1001 > Out[4]= {----------------, ----------------, ----} > 1000000000000000 1000000000000000 1000 > > In[5]:= x=1.0000000000000001;{rAndrzej[x],rSH[x],rBill[x]} > > Out[5]= {1, 1, 1} > > > Then, when getting closer to 1, the first 2 give the expected result, > while the 3rd one gives sometimes the expected result, sometimes 1. > > I found a way to do what I want, but only by starting from a string, > i.e., I have to enclose the number between double quotes, which is > impractical in general. > > I could not find a general way to transform a number to a > string. Specifically, is there a way to transform the above number > 1.0000000000000001 (fifteen zeros), which seems to be the most > problematic, to the corresponding string 1.0000000000000001 ? > > Alain > > === Subject: Re: Is there a simple way to transform 1.1 to I should have, perhaps, noted the obvious: that this problem will only occur if you try to write down MachinePrecision numbers with more digits than In[25]:= $MachinePrecision Out[25]= 15.9546 (that's hardware dependent, by the way). CLearly, doing such a thing does not make sense. So only when your numbers are very long will you encounter the problem with below, and then, I think there is no alternative to the ones already mentioned. Andrzej Kozlowski > Well, obviously this isn't going to work since Mathematica considers In[209]:= 1.0000000000000001 > Out[209]= 1. and there is no way to do anything about is since even Hold will not > help: Hold[1.0000000000000001] > Hold[1.] So you really need to input your numbers like this w = 1.0000000000000001`17 > 1.0000000000000001 then, of course, > ff[w] > 10000000000000001/10000000000000000 > Actually, in this case my old ff, which you called rAndrzej, also > will work: In[10]:= rAndrzej[w] > Out[10]= 10000000000000001/10000000000000000 and of course, also in the example you posted earlier. Note that you > do not actually need to count the number of digits in yoru number, > you can just use a very large precision, as in this case: w = 1.00000000000000000000001`100; ff[w] > 100000000000000000000001/100000000000000000000000 rAndrzej[w] > 100000000000000000000001/100000000000000000000000 > Of course you could start by inputing strings, which may not be much > more work. But really it all depends on what sort of use you wan to > make of this. If the numbers really arise out of some computation > you could try doing the computations with high precision to get > answers which look like the above. If you are just typing them in, > then you might as well type strings to begin with or even fractions > (which really is what Mathematica designers meant you to do in the > first place). Andrzej Kozlowski > OK. I was wrong. It's even harder than I thought. How about this > (using strings) ff[z_] := > With[{p = > ToExpression[StringJoin[DeleteCases[Characters[ToString[z]], > .]]]}, > p/Round[p/z]] ff[1.001] > 1001/1000 >> In[106]:= ff[1.0000000000000001] Out[106]= 1 Like I said, I couldn't find a way to transform the number >> 1.0000000000000001 to the corresponding string. I have tried all of >> $OutputFormats and $ExportFormats without success yet. The most >> promising is MathMLForm but it fails at least for that number. alain > Andrzej Kozlowski > So Andrzej Kozlowski proposed: In[1]:= rAndrzej[z_] := With[{w = >> 0 ...} :> {x, a} &, w, 1]]] Someone else proposed in private: RealDigits[x] Bill Rowe proposed: In[3]:= rBill[x_] := Module[{r = Rationalize[x, 0], p}, p = >> Round@Log[10, Denominator[r]]; Round[r 10^p]/10^p] They all performed well on early isolated trials, but more >> systematic >> tests show failures: In[4]:= x=1.001;{rAndrzej[x],rSH[x],rBill[x]} 1000999999999999 1000999999999999 1001 >> Out[4]= {----------------, ----------------, ----} >> 1000000000000000 1000000000000000 1000 In[5]:= x=1.0000000000000001;{rAndrzej[x],rSH[x],rBill[x]} Out[5]= {1, 1, 1} >> Then, when getting closer to 1, the first 2 give the expected >> result, >> while the 3rd one gives sometimes the expected result, sometimes 1. I found a way to do what I want, but only by starting from a >> string, >> i.e., I have to enclose the number between double quotes, which is >> impractical in general. I could not find a general way to transform a number to a >> string. Specifically, is there a way to transform the above number >> 1.0000000000000001 (fifteen zeros), which seems to be the most >> problematic, to the corresponding string 1.0000000000000001 ? Alain > === Subject: Re: Is there a simple way to transform 1.1 to Well, obviously this isn't going to work since Mathematica considers In[209]:= 1.0000000000000001 Out[209]= 1. and there is no way to do anything about is since even Hold will not help: Hold[1.0000000000000001] Hold[1.] So you really need to input your numbers like this w = 1.0000000000000001`17 1.0000000000000001 then, of course, ff[w] 10000000000000001/10000000000000000 Actually, in this case my old ff, which you called rAndrzej, also will work: In[10]:= rAndrzej[w] Out[10]= 10000000000000001/10000000000000000 and of course, also in the example you posted earlier. Note that you do not actually need to count the number of digits in yoru number, you can just use a very large precision, as in this case: w = 1.00000000000000000000001`100; ff[w] 100000000000000000000001/100000000000000000000000 rAndrzej[w] 100000000000000000000001/100000000000000000000000 Of course you could start by inputing strings, which may not be much more work. But really it all depends on what sort of use you wan to make of this. If the numbers really arise out of some computation you could try doing the computations with high precision to get answers which look like the above. If you are just typing them in, then you might as well type strings to begin with or even fractions (which really is what Mathematica designers meant you to do in the first place). Andrzej Kozlowski >> OK. I was wrong. It's even harder than I thought. How about this >> (using strings) ff[z_] := >> With[{p = >> ToExpression[StringJoin[DeleteCases[Characters[ToString[z]], >> .]]]}, >> p/Round[p/z]] ff[1.001] >> 1001/1000 > In[106]:= ff[1.0000000000000001] Out[106]= 1 Like I said, I couldn't find a way to transform the number > 1.0000000000000001 to the corresponding string. I have tried all of > $OutputFormats and $ExportFormats without success yet. The most > promising is MathMLForm but it fails at least for that number. alain > Andrzej Kozlowski > So Andrzej Kozlowski proposed: In[1]:= rAndrzej[z_] := With[{w = > {x, a} &, w, 1]]] Someone else proposed in private: RealDigits[x] Bill Rowe proposed: In[3]:= rBill[x_] := Module[{r = Rationalize[x, 0], p}, p = > Round@Log[10, Denominator[r]]; Round[r 10^p]/10^p] They all performed well on early isolated trials, but more > systematic > tests show failures: In[4]:= x=1.001;{rAndrzej[x],rSH[x],rBill[x]} 1000999999999999 1000999999999999 1001 > Out[4]= {----------------, ----------------, ----} > 1000000000000000 1000000000000000 1000 In[5]:= x=1.0000000000000001;{rAndrzej[x],rSH[x],rBill[x]} Out[5]= {1, 1, 1} > Then, when getting closer to 1, the first 2 give the expected > result, > while the 3rd one gives sometimes the expected result, sometimes 1. I found a way to do what I want, but only by starting from a string, > i.e., I have to enclose the number between double quotes, which is > impractical in general. I could not find a general way to transform a number to a > string. Specifically, is there a way to transform the above number > 1.0000000000000001 (fifteen zeros), which seems to be the most > problematic, to the corresponding string 1.0000000000000001 ? Alain === Subject: Re: Is there a simple way to transform 1.1 to One more (I hope final) remark. Another, perhaps simpler way, is to include just one extra zero (that should not be very hard): w = 1.00000000000000010 1.0000000000000001 ff[w] 10000000000000001/10000000000000000 Andrzej Kozlowski > I should have, perhaps, noted the obvious: that this problem will > only occur if you try to write down MachinePrecision numbers with > more digits than In[25]:= $MachinePrecision > Out[25]= 15.9546 (that's hardware dependent, by the way). CLearly, doing such a thing does not make sense. So only when your > numbers are very long will you encounter the problem with below, and > then, I think there is no alternative to the ones already mentioned. Andrzej Kozlowski >> Well, obviously this isn't going to work since Mathematica considers In[209]:= 1.0000000000000001 >> Out[209]= 1. and there is no way to do anything about is since even Hold will >> not help: Hold[1.0000000000000001] >> Hold[1.] So you really need to input your numbers like this w = 1.0000000000000001`17 >> 1.0000000000000001 then, of course, >> ff[w] >> 10000000000000001/10000000000000000 >> Actually, in this case my old ff, which you called rAndrzej, also >> will work: In[10]:= rAndrzej[w] >> Out[10]= 10000000000000001/10000000000000000 and of course, also in the example you posted earlier. Note that >> you do not actually need to count the number of digits in yoru >> number, you can just use a very large precision, as in this case: w = 1.00000000000000000000001`100; ff[w] >> 100000000000000000000001/100000000000000000000000 rAndrzej[w] >> 100000000000000000000001/100000000000000000000000 >> Of course you could start by inputing strings, which may not be >> much more work. But really it all depends on what sort of use you >> wan to make of this. If the numbers really arise out of some >> computation you could try doing the computations with high >> precision to get answers which look like the above. If you are just >> typing them in, then you might as well type strings to begin with >> or even fractions (which really is what Mathematica designers meant >> you to do in the first place). Andrzej Kozlowski >> OK. I was wrong. It's even harder than I thought. How about this >> (using strings) ff[z_] := >> With[{p = >> ToExpression[StringJoin[DeleteCases[Characters[ToString[z]], >> .]]]}, >> p/Round[p/z]] ff[1.001] >> 1001/1000 > In[106]:= ff[1.0000000000000001] Out[106]= 1 Like I said, I couldn't find a way to transform the number > 1.0000000000000001 to the corresponding string. I have tried all of > $OutputFormats and $ExportFormats without success yet. The most > promising is MathMLForm but it fails at least for that number. alain > Andrzej Kozlowski > So Andrzej Kozlowski proposed: In[1]:= rAndrzej[z_] := With[{w = > 0 ...} : {x, a} &, w, 1]]] Someone else proposed in private: RealDigits[x] Bill Rowe proposed: In[3]:= rBill[x_] := Module[{r = Rationalize[x, 0], p}, p = > Round@Log[10, Denominator[r]]; Round[r 10^p]/10^p] They all performed well on early isolated trials, but more > systematic > tests show failures: In[4]:= x=1.001;{rAndrzej[x],rSH[x],rBill[x]} 1000999999999999 1000999999999999 1001 > Out[4]= {----------------, ----------------, ----} > 1000000000000000 1000000000000000 1000 In[5]:= x=1.0000000000000001;{rAndrzej[x],rSH[x],rBill[x]} Out[5]= {1, 1, 1} > Then, when getting closer to 1, the first 2 give the expected > result, > while the 3rd one gives sometimes the expected result, sometimes > 1. I found a way to do what I want, but only by starting from a > string, > i.e., I have to enclose the number between double quotes, which is > impractical in general. I could not find a general way to transform a number to a > string. Specifically, is there a way to transform the above number > 1.0000000000000001 (fifteen zeros), which seems to be the most > problematic, to the corresponding string 1.0000000000000001 ? Alain > === Subject: Re: Is there a simple way to transform 1.1 to *snip* > I found a way to do what I want, but only by starting from a string, > i.e., I have to enclose the number between double quotes, which is > impractical in general. > > I could not find a general way to transform a number to a > string. Specifically, is there a way to transform the above number > 1.0000000000000001 (fifteen zeros), which seems to be the most > problematic, to the corresponding string 1.0000000000000001 ? One possible way would be using *arbitrary-precision* numbers. For instance, In[1]:= ToString[1.0000000000000001`17] %//FullForm Out[1]= 1.0000000000000001 Out[2]//FullForm= 1.0000000000000001 -- jean-Marc === Subject: Re: Plotting an Implicit Plot with ParametricPlot Here is one way to do it. Let eq = (5*x^19)/(3*(y^20/3 + (2*x^20)/3)^(19/20)) == 4*x be your equation. We will use the substitution y -> v x, where v will be our parameter. We split the equation into two cases, one for x>0 and one for x<0. Then FullSimplify[eq /. y -> v*x, x > 0] 5/(3^(1/20)*(v^20 + 2)^(19/20)) == 4*x and FullSimplify[eq /. y -> v*x, x < 0] -(5/(3^(1/20)*(v^20 + 2)^(19/20))) == 4*x So we have two expressions for x in terms of v and can make two parametric plots. The only problem is to choose a siutable range of values of v. It helps to look at In[79]:= N[Maximize[(5*v)/(4*3^(1/20)*(v^20 + 2)^(19/20)), v]] Out[79]= {0.5212622662745989, {v -> 0.8959584598407622}} You can also see that as v becomes large, y will go to zero very fast, so this will give you a good graph: q1 = ParametricPlot[{5/(4*3^(1/20)*(v^20 + 2)^(19/20)), (v*5)/(4*3^(1/20)*(v^20 + 2)^(19/20))}, {v, -2, 2}]; q2 = ParametricPlot[{-5/(4 Power[3, (20)^-1] (v^20 + 2)^(19/20)), (v 5)/( 4 Power[3, (20)^-1] (v^20 + 2)^(19/20))}, {v, -2, 2}]; Show[{q1, q2}, PlotRange -> All] If you choose a larger range for v you will actually get the same looking picture. Andrzej Kozlowski > hi everyone.I need to plot functions that look like this: -4 x + (5 x^19)/(3 ((2 x^20)/3 + y^20/3)^(19/20)) == 0 > I was using the old function ImplicitPlot to do this, but it doesn't > mix > well with other functions I'm aplying (such as Manipulate) ParametricPlot does work, but I have no idea how to convert this > equation to > a parametric one. Is there a systematic thing I can do or do I have to convert them by > hand to > parametric equations? > As always, thank you very much for your time! cd -- > Por favor eviten enviarme archivos adjuntos de Word o Powerpoint ( > http://www.gnu.org/philosophy/no-word-attachments.es.html ) === Subject: Re: Plotting an Implicit Plot with ParametricPlot Perhaps ContourPlot[-4 x + (5 x^19)/(3 ((2 x^20)/3 + y^20/3)^(19/20)) == 0, {x, -2, 2}, {y, -2, 2}] will accomplish what you want. > hi everyone.I need to plot functions that look like this: > > -4 x + (5 x^19)/(3 ((2 x^20)/3 + y^20/3)^(19/20)) == 0 > > > I was using the old function ImplicitPlot to do this, but it doesn't mix > well with other functions I'm aplying (such as Manipulate) > > ParametricPlot does work, but I have no idea how to convert this equation to > a parametric one. > > Is there a systematic thing I can do or do I have to convert them by hand to > parametric equations? > > > As always, thank you very much for your time! > > > > cd > -- Murray Eisenberg murray@math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2859 (W) 710 North Pleasant Street fax 413 545-1801 Amherst, MA 01003-9305 === Subject: Re: Plotting an Implicit Plot with ParametricPlot ContourPlot has replaced ImplicitPlot but I could not get it to plot this equation. eqn = Simplify[-4 x + (5 x^19)/(3 ((2 x^20)/3 + y^20/3)^(19/20)) == 0]; You can Plot the result of using Solve. I have filtered out the numerous non-real solutions to speed up the Plot function. p = Select[y /. Solve[eqn, y], Im[# /. x -> 1/2] == 0 &]; Plot[p, {x, -.65, .65}] Bob Hanlon hi everyone.I need to plot functions that look like this: -4 x + (5 x^19)/(3 ((2 x^20)/3 + y^20/3)^(19/20)) == 0 I was using the old function ImplicitPlot to do this, but it doesn't mix well with other functions I'm aplying (such as Manipulate) ParametricPlot does work, but I have no idea how to convert this equation to a parametric one. Is there a systematic thing I can do or do I have to convert them by hand to parametric equations? As always, thank you very much for your time! cd -- Por favor eviten enviarme archivos adjuntos de Word o Powerpoint ( http://www.gnu.org/philosophy/no-word-attachments.es.html ) -- Bob Hanlon === Subject: Re: Function pure for Select Split[Sort[Select[data, MemberQ[{30, 45, 50, 66}, #[[1]]] &]], #1[[1]] == #2[[1]] &] However, it is much easier to use a helper function: f[x_] := Select[data, #[[1]] == x &] f /@ {30, 45, 50, 66} Bob Hanlon Hi al, How can I to write a function pure to extract all the first rows of collection of data, applied to a list?. For example, Select[data,First[#]==30&] This function extracts all rows which first element is equal to 30. Well, I want to extend this function to a list of values {30,45,50,66}. -- Bob Hanlon === Subject: Re: Function pure for Select > Split[Sort[Select[data, > MemberQ[{30, 45, 50, 66}, #[[1]]] &]], > #1[[1]] == #2[[1]] &] However, it is much easier to use a helper function: f[x_] := Select[data, #[[1]] == x &] f /@ {30, 45, 50, 66} Bob Hanlon > Hi al, How can I to write a function pure to extract all the first rows of > collection of data, applied to a list?. For example, Select[data,First[#]==30&] This function extracts all rows which first element is equal to 30. > Well, I want to extend this function to a list of values > {30,45,50,66}. > -- Bob Hanlon Here's a look at a few approaches, with timings on my machine: data = Table[{RandomInteger[1000], RandomInteger[1000]}, {100000}]; targets = Table[i, {i, 250}]; Select[data, MemberQ[targets, First[#]] &] ...takes 2.886 seconds f[x_] := Select[data, First[#] == x &]; f /@ targets ...takes 93.616 seconds Cases[data, {Alternatives@@targets, _}] ...takes 1.061 seconds IsTarget[_] := False; Scan[(IsTarget[#] = True) &, targets]; Select[data, IsTarget[First[#]] &] ...takes 0.452 seconds === Subject: Looking for Stop Mathematica has two termination commands: Abort[] interrupts a computation Exit[] or Quit[] returns control to OS I use Exit[] in some course-distributed application programs that detect an irrecoverable error requiring user intervention and program changes. This has an undesirable side effect: all cells containing support modules have to be reinitialized. Abort[] does not have that effect; however it doesnt stop execution of whatever follows. Is there a command in between that stops kernel execution but leaves cells initialized? Constraint: it has to work with versions >=4.1. I looked for Stop[] but there isnt such a thing under versions 4 or 5. Note: played with Throw and Catch, but this only terninates the evaluation of whatever is inside Catch[...]. Not too useful for big programs. I also tought to place a forward Goto to a Label: at the end of the main program; that doesnt work either since a label is not visible until it is reached. What happens if the code hits Break[] and there is no enclosing Do, For or While? Need to check that possibility. === Subject: Re: Looking for Stop > Mathematica has two termination commands: > > Abort[] interrupts a computation > Exit[] or Quit[] returns control to OS > > I use Exit[] in some course-distributed application programs > that detect an irrecoverable error requiring user intervention > and program changes. This has an undesirable side effect: > all cells containing support modules have to be reinitialized. > Abort[] does not have that effect; however it doesnt stop > execution of whatever follows. I think you should understand that Quit or Exit do not only _stop_ the kernel in what it does, they exit/quit the kernel process. When you start the next evaluation, the frontend will start a new kernel and that of course needs to be initialized. > Is there a command in between that stops kernel > execution but leaves cells initialized? Constraint: > it has to work with versions >=4.1. I looked for > Stop[] but there isnt such a thing under versions 4 or 5. Actually from the kernel point of view Abort[] is doing exactly that. Your problem seems to be that once the kernel aborts the evaluation of one cell, the frontend will send it the next cell in its evaluation queue for evaluation. The kernel presumably doesn't know that this already was in the frontend's queue before the Abort[], so it will just treat it as any other evaluation that it will get from the frontend after an Abort[]. I think this makes clear that it is not the kernel that you need to manipulate, it is the frontend that needs to be informed that it should abort the command _and_ clear its evaluation queue. I just played around a little bit and think what you want can be done with: FrontEndTokenExecute[EvaluatorAbort] but note that the frontend token EvaluatorAbort is marked as not fully integrated yet, so it may change or go away. Also I have found some oddities after using it: the very next evaluation sometimes (always?) still returns $Aborted and only the second next will behave as usual. So you need to check whether that is o.k. for you. If not, you could go the hard way and do some FrontEnd-Programming to first remove all cells from the evaluation queue (select all cells, call That said, I share the opinion of Jens that to produce robust code you really need to start early, and the decision to supply code in form of a Notebook that has to be evaluated from top to bottom is probably where you should start, because that approach has several problems when the notebooks is given away to users not familiar with your code and/or mathematica. You can do everything that can be done that way with one function definition, that will produce the various output that otherwise the cells in your notebook would create. When calling just one function from one cell an Abort[] works as you want, and several other possibilities like Catch/Throw give you even more control about how to react to unexpected results. And when the desire arises, you can build a gui for that function with Manipulate or the other Dynamic stuff... hth, albert === Subject: Re: Looking for Stop Quit[] and Exit[] terminate the Kernel .. it is not a good idea to use it in a application. If you need a fool proof program you must write it in that way form the very first line. Return[]/Return[$Failed] or something like that are the functions in you program that terminate a function in case of detect an irrecoverable error requiring user intervention Jens > Mathematica has two termination commands: > > Abort[] interrupts a computation > Exit[] or Quit[] returns control to OS > > I use Exit[] in some course-distributed application programs > that detect an irrecoverable error requiring user intervention > and program changes. This has an undesirable side effect: > all cells containing support modules have to be reinitialized. > Abort[] does not have that effect; however it doesnt stop > execution of whatever follows. > > Is there a command in between that stops kernel > execution but leaves cells initialized? Constraint: > it has to work with versions >=4.1. I looked for > Stop[] but there isnt such a thing under versions 4 or 5. > > Note: played with Throw and Catch, but this only terninates > the evaluation of whatever is inside Catch[...]. Not too > useful for big programs. I also tought to place a forward > Goto to a Label: at the end of the main program; > that doesnt work either since a label is not visible > until it is reached. > > What happens if the code hits Break[] and there is no enclosing > Do, For or While? Need to check that possibility. > === Subject: SpeechNavigation In the option Inspector there is an option called SpeechNavigation. Does anybody know what is it and how to use it? === Subject: Re: Function pure for Select >How can I to write a function pure to extract all the first rows of >collection of data, applied to a list?. >For example, >Select[data,First[#]==30&] >This function extracts all rows which first element is equal to 30. >Well, I want to extend this function to a list of values >{30,45,50,66}. Here are a couple of ways to do this. First using Select and letting testValues be the list of values above, i.e., testValues={30,45,50,66}; then Select[data, MemberQ[testValues, First@#]&] will work. Or this could be done using Cases as: Cases[data,{Alternatives@@testValues,__}] === Subject: Export ListPlots using PlotMarkers to .pdf A nice ListPlot may be made using various PlotMarkers; it's especially easy for me to use symbols like [GrayCircle]: ListPlot[{#, Sin[13 #]} & /@ Range[0, 1, 0.01], PlotMarkers -> [GrayCircle]] However, if I want to save such a plot as a .pdf (whether using Export, SaveSelectionAs, or using print-to-pdf-file), the symbols are screwed up, with the filling and the outlines offset from each other. Is there a way to get better results using such PlotMarkers? ( This is using 6.0 for Linux x86 (32-bit) (June 2, 2008)) C.O. -- Curtis Osterhoudt cfo@remove_this.lanl.and_this.gov PGP Key ID: 0x4DCA2A10 Please avoid sending me Word or PowerPoint attachments See http://www.gnu.org/philosophy/no-word-attachments.html === Subject: Re: Export ListPlots using PlotMarkers to .pdf The same thing with 6.0 for Microsoft Windows (32-bit) (February 7, 2008) Why not use [FilledCircle]? It works for me On 10=D4=C225=C8=D5, =CF=C2=CE=E73=CA=B103=B7=D6, Curtis Osterhoudt A nice ListPlot may be made using various PlotMarkers; it's especially ea= sy for me to use symbols like [GrayCircle]: ListPlot[{#, Sin[13 #]} & /@ Range[0, 1, 0.01], > PlotMarkers -> [GrayCircle]] However, if I want to save such a plot as a .pdf (whether using Export,= SaveSelectionAs, or using print-to-pdf-file), the symbols are screwed up,= with the filling and the outlines offset from each other. Is there a way to get better results using such PlotMarkers? ( This is using 6.0 for Linux x86 (32-bit) (June 2, 2008)) C.O. -- > Curtis Osterhoudt > cfo@remove_this.lanl.and_this.gov > PGP Key ID: 0x4DCA2A10 > Please avoid sending me Word or PowerPoint attachments > Seehttp://www.gnu.org/philosophy/no-word-attachments.html === Subject: Re: Export ListPlots using PlotMarkers to .pdf Unfortunately, no matter what I do (in linux), the produced .pdf files have the same problem, whether I use a third-party program or not. In Windows, the same problem exists if I use Mathematica's save as .pdf option; if I use a program like PDFCreator, then things look OK. I don't have the tremendously nice chance to use OS X's built-in pdf writer. I haven't yet tried all of the permutations of print-to-various-ps-formats-then-convert. Hopefully, the direct-to-pdf option will work in Mathematica in some future versions. C.O. > > A nice ListPlot may be made using various PlotMarkers; it's especially easy for me to use symbols like [GrayCircle]: > > ListPlot[{#, Sin[13 #]} & /@ Range[0, 1, 0.01], > PlotMarkers -> [GrayCircle]] > > However, if I want to save such a plot as a .pdf (whether using Export, SaveSelectionAs, or using print-to-pdf-file), the symbols are screwed up, with the filling and the outlines offset from each other. > > Is there a way to get better results using such PlotMarkers? > > ( This is using 6.0 for Linux x86 (32-bit) (June 2, 2008)) > > You may want to use a third-party program to print/save PDF files > (GostScript maybe). On Mac OS X 10.5.5, if I use the built-in Apple's > PDF writer in the dialog box for Print or Print Selection, the resulting > PDF file displays correctly the custom markers. (I have sent you such a > PDF file by email.) > > -- Jean-Marc > -- Curtis Osterhoudt cfo@remove_this.lanl.and_this.gov PGP Key ID: 0x4DCA2A10 Please avoid sending me Word or PowerPoint attachments See http://www.gnu.org/philosophy/no-word-attachments.html === Subject: Re: Export ListPlots using PlotMarkers to .pdf as to see from plt = ListPlot[{#, Sin[13 #]} & /@ Range[0, 1, 0.01], PlotMarkers -> [GrayCircle]]; ImportString[ExportString[plt, PDF], PDF] this is a bug. Jens > A nice ListPlot may be made using various PlotMarkers; it's especially easy for me to use symbols like [GrayCircle]: > > ListPlot[{#, Sin[13 #]} & /@ Range[0, 1, 0.01], > PlotMarkers -> [GrayCircle]] > > However, if I want to save such a plot as a .pdf (whether using Export, SaveSelectionAs, or using print-to-pdf-file), the symbols are screwed up, with the filling and the outlines offset from each other. > > Is there a way to get better results using such PlotMarkers? > > ( This is using 6.0 for Linux x86 (32-bit) (June 2, 2008)) > > C.O. > > === Subject: Re: Export ListPlots using PlotMarkers to .pdf > A nice ListPlot may be made using various PlotMarkers; it's especially easy for me to use symbols like [GrayCircle]: > > ListPlot[{#, Sin[13 #]} & /@ Range[0, 1, 0.01], > PlotMarkers -> [GrayCircle]] > > However, if I want to save such a plot as a .pdf (whether using Export, SaveSelectionAs, or using print-to-pdf-file), the symbols are screwed up, with the filling and the outlines offset from each other. > > Is there a way to get better results using such PlotMarkers? > > ( This is using 6.0 for Linux x86 (32-bit) (June 2, 2008)) You may want to use a third-party program to print/save PDF files (GostScript maybe). On Mac OS X 10.5.5, if I use the built-in Apple's PDF writer in the dialog box for Print or Print Selection, the resulting PDF file displays correctly the custom markers. (I have sent you such a PDF file by email.) -- Jean-Marc === Subject: Nonlinear Regression Oddities and Questions I am fitting a function with known errors to determine the Confidence Intervals of the fitted parameters. For each data point I have an associated error, that is, the Standard Deviation of the measurement. I can send the data to anyone interested. In Mathematica I use the function NonlinearRegress with Weights->Standard Deviation^(-2). I get an answer for the Confidence Interval. I then tested the data using a plotting program, pro Fit. In pro Fit the Confidence Interval is determined by a Monte Carlo bootstrapping method. Random data is generated within the error bars and a fit is made; this is done 500 times. The confidence interval gives a range of values from the Monte Carlo result. The two programs give much different answers. Do anyone has experience with the validity of the Mathematica results. I do not know which program is correct. Andrew === Subject: Re: Good computer needed > I would like to ask person which have access to quick computer with big > memory with installed Mathematica to run bellow procedure (I was stop my > old PC computer after 7 days of run this proceduer, Mathematica was used > Hard Drive because memory was not sufficient). I hope that strong > machine count these few hours. > I will be greatfull for help because I don't have access to any better > computer as myself. > Best wishes > Artur > > NSolve[{(3 c^2 d + 3 b d^2 + 6 b c e + 3 d e^2 + e^3 + 3 b^2 f + > 3 d^2 f + 6 c e f + 6 d e f + 3 b f^2 + 3 c f^2 + > f^3) == -960 I Sqrt[3/ > 2869], (c^3 + 6 b c d + 3 b^2 e + 3 d^2 e + 3 c e^2 + 3 d e^2 + > 6 c d f + 3 d^2 f + 6 b e f + 6 c e f + 3 b f^2 + 3 e f^2 + > 2 f^3) == > 1200 I Sqrt[3/ > 2869], (20 - 1500 I Sqrt[3/2869]) == (3 b c^2 + 3 b^2 d + d^3 + > 6 c d e + 3 d^2 e + 3 b e^2 + 3 c e^2 + 3 c^2 f + 6 b d f + > 6 c d f + 6 b e f + 3 e^2 f + 3 d f^2 + 6 e f^2 + > f^3), (3 b^2 c + 3 c d^2 + d^3 + 3 c^2 e + 6 b d e + 6 c d e + > 3 b e^2 + e^3 + 6 b c f + 3 c^2 f + 6 b d f + 6 d e f + > 6 e^2 f + 3 c f^2 + 6 d f^2 + 3 e f^2) == (-25 + > 1875 I Sqrt[3/2869]) , > b^3 + 3 c d^2 + 3 c^2 e + 6 b d e + e^3 + 6 b c f + 6 d e f + > 3 e^2 f + 3 c f^2 + 3 d f^2 == 768 I Sqrt[3/2869]}, {b, c, d, e, > f}, WorkingPrecision -> 2000]//Timing Artur, By trial and error, I have found a full set of solutions (243) using a working precision of only 500. (The memory consumption was about 240MB.) In[16]:= Timing[ sols5 = NSolve[eqns, vars, WorkingPrecision -> 500];][[1]] Length[sols5] Count[eqns /. sols5, ConstantArray[True, {Length[eqns]}]] Out[16]= 4258.3 Out[17]= 243 Out[18]= 243 I have sent you the corresponding notebook by email. Let me know if the precision is enough or if you really need a precision of 2000. Hope this helps, -- Jean-Marc === Subject: Re: Good computer needed You could try using FindRoot with suitable initial guesses. Here is an example of this approach: Define the equations. eqns = {(3 c^2 d + 3 b d^2 + 6 b c e + 3 d e^2 + e^3 + 3 b^2 f + 3 d^2 f + 6 c e f + 6 d e f + 3 b f^2 + 3 c f^2 + f^3) == -960 I Sqrt[3/2869], (c^3 + 6 b c d + 3 b^2 e + 3 d^2 e + 3 c e^2 + 3 d e^2 + 6 c d f + 3 d^2 f + 6 b e f + 6 c e f + 3 b f^2 + 3 e f^2 + 2 f^3) == 1200 I Sqrt[3/2869], (20 - 1500 I Sqrt[3/2869]) == (3 b c^2 + 3 b^2 d + d^3 + 6 c d e + 3 d^2 e + 3 b e^2 + 3 c e^2 + 3 c^2 f + 6 b d f + 6 c d f + 6 b e f + 3 e^2 f + 3 d f^2 + 6 e f^2 + f^3), (3 b^2 c + 3 c d^2 + d^3 + 3 c^2 e + 6 b d e + 6 c d e + 3 b e^2 + e^3 + 6 b c f + 3 c^2 f + 6 b d f + 6 d e f + 6 e^2 f + 3 c f^2 + 6 d f^2 + 3 e f^2) == (-25 + 1875 I Sqrt[3/2869]), b^3 + 3 c d^2 + 3 c^2 e + 6 b d e + e^3 + 6 b c f + 6 d e f + 3 e^2 f + 3 c f^2 + 3 d f^2 == 768 I Sqrt[3/2869]}; Find a root starting at the point {b,c,d,e,f} = {1,1,1,1,1}. root = FindRoot[eqns, {{b, 1}, {c, 1}, {d, 1}, {e, 1}, {f, 1}}] {b -> 2.47109- 0.647912 [ImaginaryI], c -> -1.84539 + 1.28647 [ImaginaryI], d -> 0.544229- 0.0101677 [ImaginaryI], e -> 0.0693024+ 0.582924 [ImaginaryI], f -> 0.859564- 0.053643 [ImaginaryI]} Verify that this is indeed a solution. I turn each equation x=y into the expression x-y to avoid some of the equality tests failing due to rounding errors, and then use Chop to force small absolute values to zero. eqns /. Equal[x_, y_] -> x - y /. root // Chop {0, 0, 0, 0, 0} -- Stephen Luttrell West Malvern, UK > I would like to ask person which have access to quick computer with big > memory with installed Mathematica to run bellow procedure (I was stop my > old PC computer after 7 days of run this proceduer, Mathematica was used > Hard Drive because memory was not sufficient). I hope that strong > machine count these few hours. > I will be greatfull for help because I don't have access to any better > computer as myself. > Best wishes > Artur NSolve[{(3 c^2 d + 3 b d^2 + 6 b c e + 3 d e^2 + e^3 + 3 b^2 f + > 3 d^2 f + 6 c e f + 6 d e f + 3 b f^2 + 3 c f^2 + > f^3) == -960 I Sqrt[3/ > 2869], (c^3 + 6 b c d + 3 b^2 e + 3 d^2 e + 3 c e^2 + 3 d e^2 + > 6 c d f + 3 d^2 f + 6 b e f + 6 c e f + 3 b f^2 + 3 e f^2 + > 2 f^3) == > 1200 I Sqrt[3/ > 2869], (20 - 1500 I Sqrt[3/2869]) == (3 b c^2 + 3 b^2 d + d^3 + > 6 c d e + 3 d^2 e + 3 b e^2 + 3 c e^2 + 3 c^2 f + 6 b d f + > 6 c d f + 6 b e f + 3 e^2 f + 3 d f^2 + 6 e f^2 + > f^3), (3 b^2 c + 3 c d^2 + d^3 + 3 c^2 e + 6 b d e + 6 c d e + > 3 b e^2 + e^3 + 6 b c f + 3 c^2 f + 6 b d f + 6 d e f + > 6 e^2 f + 3 c f^2 + 6 d f^2 + 3 e f^2) == (-25 + > 1875 I Sqrt[3/2869]) , > b^3 + 3 c d^2 + 3 c^2 e + 6 b d e + e^3 + 6 b c f + 6 d e f + > 3 e^2 f + 3 c f^2 + 3 d f^2 == 768 I Sqrt[3/2869]}, {b, c, d, e, > f}, WorkingPrecision -> 2000]//Timing === Subject: Re: Graphic Entries in a Grid > I want to create a grid with some graphic entries. The graphics were > created as explained below: GraphicsGrid[{{, , }, {, Graphics[Disk[], ImageSize -> 6], > }, {, , }}, Frame -> True, Spacings -> {3, 3}, > Background -> Red] By rearranging the disks in the above, it is possible to emulate all > six faces of a die. I want the die faces to be entered in the left column and the top row > of the grid. All other entries are numbers. My attempts to enter the die faces failed. Error message: not a list > of lists. John If I understand what you mean, then try the following. First I created code for each of the six normal die faces using your technique. Then I named them: one,two, etc. I have used four in the code below. four = GraphicsGrid[{{Graphics[Disk[], ImageSize -> 6], , Graphics[Disk[], ImageSize -> 6]}, {, , }, {Graphics[Disk[], ImageSize -> 6], , Graphics[Disk[], ImageSize -> 6]}}, Frame -> True, Spacings -> {3, 3}, Background -> Red] GraphicsRow[{four, four, 4, 4}] I hope this helps. Gary === Subject: Re: Graphic Entries in a Grid I want to create a grid with some graphic entries. The graphics were > created as explained below: GraphicsGrid[{{, , }, {, Graphics[Disk[], ImageSize -> 6= ], > }, {, , }}, Frame -> True, Spacings -> {3, 3}, > Background -> Red] By rearranging the disks in the above, it is possible to emulate all > six faces of a die. I want the die faces to be entered in the left column and the top row > of the grid. All other entries are numbers. My attempts to enter the die faces failed. Error message: not a list > of lists. John If I understand what you mean, then try the following. First I created > code for each of the six normal die faces using your technique. Then I > named them: one,two, etc. I have used four in the code below. four = GraphicsGrid[{{Graphics[Disk[], ImageSize -> 6], , > Graphics[Disk[], ImageSize -> 6]}, {, , > }, {Graphics[Disk[], ImageSize -> 6], , > Graphics[Disk[], ImageSize -> 6]}}, Frame -> True, > Spacings -> {3, 3}, Background -> Red] GraphicsRow[{four, four, 4, 4}] I hope this helps. Gary Gary, John === Subject: Re: Is there a simple way to transform 1.1 to >I found a way to do what I want, but only by starting from a string, >i.e., I have to enclose the number between double quotes, which is >impractical in general. >I could not find a general way to transform a number to a string. >Specifically, is there a way to transform the above number >1.0000000000000001 (fifteen zeros), which seems to be the most >problematic, to the corresponding string 1.0000000000000001 ? Given In[10]:= 1.0000000000000001`20 - 1 Out[10]= 1.000*10^-16 In[11]:= $MachinePrecision Out[11]= 15.9546 It is clear there is no way to distinguish 1.0000000000000001 from 1. with machine precision. The only way to achieve what you are attempting is to use Mathematica's variable precision machinery and enter the data with higher than machine precision. Once you do this Rationalize works just fine In[13]:= Rationalize[1.0000000000000001`20, 0] Out[13]= 10000000000000001/10000000000000000 Quite simply, there are clear limits to machine precision. Mathematica supports higher precision. And if you really need something better than machine precision you will need to enter the data as either exact values or with higher precision. Using something other than machine precision will have a cost both in terms of data entry and the speed of the computations. You will have to decide whether the cost is warranted or not. =46inally, this issue is not unique to Mathematica. It is inherent in the way computers work. Other systems can make different trades between performance and precision and get different results. But the basic issue remains. === Subject: Re: How does Mathematica know whether a number is real or complex? > The solution of the equation at the end of this post is > > {{x -> -0.9702291184141268, y -> -0.24218888864137905}, > {x -> -0.44830718122625973, y -> 0.8938795619438703}, > {x -> 1.3776284630851778 - 0.7866963118457873*I, > y -> -0.9763348915886718 - 1.1100445557562666*I}, > {x -> 1.3776284630851778 + 0.7866963118457873*I, > y -> -0.9763348915886718 + 1.1100445557562666*I}} > > Two of the solutions are real. The other two are complex. > > I'm wondering how Mathematica know whether a solution is real or complex. Their heads are different. Floating-point numbers have head *Real* while complex numbers have head *Complex*. In[1]:= AA = {{0.45236, -0.241371}, {-0.241371, 0.173649}}; BB = {-0.500298, 0.141347}; CC = -0.773749; sols = NSolve[{{x, y}.AA.{x, y} + BB.{x, y} + CC == 0, x^2 + y^2 == 1}, {x, y}] {x, y} /. sols Map[Head, %, {2}] Out[4]= {{x -> 1.3776284630851803 - 0.7866963118456223*I, y -> -0.9763348915886673 - 1.1100445557560326*I}, {x -> 1.3776284630851803 + 0.7866963118456223*I, y -> -0.9763348915886673 + 1.1100445557560326*I}, {x -> -0.44830718122626145, y -> 0.8938795619438661}, {x -> -0.9702291184141273, y -> -0.2421888886413809}} Out[5]= {{1.3776284630851803 - 0.7866963118456223*I, -0.9763348915886673 - 1.1100445557560326*I}, {1.3776284630851803 + 0.7866963118456223*I, -0.9763348915886673 + 1.1100445557560326*I}, {-0.44830718122626145, 0.8938795619438661}, {-0.9702291184141273, -0.2421888886413809}} Out[6]= {{Complex, Complex}, {Complex, Complex}, {Real, Real}, {Real, Real}} > If a solution actually has a close-to-zero imaginary part, will > mathematica incorrect consider it as a real number? No. Whenever one uses floating-point arithmetic (i.e. inexact or machine-precision numbers) tiny part, regardless of how close to zero they are and regardless whether they come from rounding errors are kept. Usually one wants to remove such very small numbers and uses *Chop* for this purpose). For instance, In[7]:= NSolve[x^4 - x^2 - 5 == 0, x] % // Chop Out[7]= {{x -> -1.6707147714310542}, {x -> 4.472654307192293*^-17 - 1.3383900206882595*I}, {x -> 4.472654307192293*^-17 + 1.3383900206882595*I}, {x -> 1.6707147714310542}} Out[8]= {{x -> -1.6707147714310542}, {x -> -1.3383900206882595*I}, {x -> 1.3383900206882595*I}, {x -> 1.6707147714310542}} *snip* -- Jean-Marc === Subject: Re: How does Mathematica know whether a number is real or complex? >The solution of the equation at the end of this post is >{{x -> -0.9702291184141268, y -> -0.24218888864137905}, >{x -> -0.44830718122625973, y -> 0.8938795619438703}, {x -1.3776284630851778 - 0.7866963118457873*I, y -> -0.9763348915886718 >- 1.1100445557562666*I}, {x -> 1.3776284630851778 + >0.7866963118457873*I, y -> -0.9763348915886718 + >1.1100445557562666*I}} >Two of the solutions are real. The other two are complex. >I'm wondering how Mathematica know whether a solution is real or >complex. By default, Mathematica is programmed to treat all expressions as having complex values. Solve simply returns all values whether complex or not. >If a solution actually has a close-to-zero imaginary part, >will mathematica incorrect consider it as a real number? No. In fact, since you set up your equations with machine precision numbers it is entirely possible do to round off error Mathematica will report real solutions as complex with a small imaginary part. The solution to this issue is to use Chop to force all small values to be zero or to use exact numbers. In fact, when using machine precision coefficients in equations it is probably better to use NSolve rather than Solve. === Subject: Re: Plotting an Implicit Plot with ParametricPlot > hi everyone.I need to plot functions that look like this: > > -4 x + (5 x^19)/(3 ((2 x^20)/3 + y^20/3)^(19/20)) == 0 > > > I was using the old function ImplicitPlot to do this, but it doesn't mix > well with other functions I'm aplying (such as Manipulate) > > ParametricPlot does work, but I have no idea how to convert this equation to > a parametric one. > > Is there a systematic thing I can do or do I have to convert them by hand to > parametric equations? No need to do that. You can use *ContourPlot* (v6.0+). For example, ContourPlot[-4 x + (5 x^19)/(3 ((2 x^20)/3 + y^20/3)^(19/20)) == 0, {x, -1, 1}, {y, -1, 1}] -- Jean-Marc === Subject: Re: Plotting an Implicit Plot with ParametricPlot One form of ContourPlot now replaces the old ImplicitPlot. ContourPlot[-4 x + (5 x^19)/(3 ((2 x^20)/3 + y^20/3)^(19/20)) == 0, {x, -1, 1}, {y, -1, 1}, MaxRecursion -> 3, Axes -> False] -- David Park djmpark@comcast.net http://home.comcast.net/~djmpark/ > hi everyone.I need to plot functions that look like this: -4 x + (5 x^19)/(3 ((2 x^20)/3 + y^20/3)^(19/20)) == 0 > I was using the old function ImplicitPlot to do this, but it doesn't mix > well with other functions I'm aplying (such as Manipulate) ParametricPlot does work, but I have no idea how to convert this equation > to > a parametric one. Is there a systematic thing I can do or do I have to convert them by hand > to > parametric equations? > As always, thank you very much for your time! cd -- > Por favor eviten enviarme archivos adjuntos de Word o Powerpoint ( > http://www.gnu.org/philosophy/no-word-attachments.es.html ) === Subject: Re: Plotting an Implicit Plot with ParametricPlot > hi everyone.I need to plot functions that look like this: > > -4 x + (5 x^19)/(3 ((2 x^20)/3 + y^20/3)^(19/20)) == 0 > > > I was using the old function ImplicitPlot to do this, but it doesn't mix > well with other functions I'm aplying (such as Manipulate) > > ParametricPlot does work, but I have no idea how to convert this equation to > a parametric one. > > Is there a systematic thing I can do or do I have to convert them by hand to > parametric equations? Use ContourPlot -- Helen Read University of Vermont === Subject: Re: How does Mathematica know whether a number is real or complex? > Solve uses exact algebraic methods to solve equations so actually it does > not solve the approximate equation that you gave it but some rationalized > rationalize an equation, but probably what it does is that it replaces your > equations by: eq = Rationalize[{{x, y} . AA . {x, y} + BB . {x, y} + CC == 0, > x^2 + y^2 == 1}, 0] {((11309*x)/25000 - (241371*y)/1000000)*x - (250149*x)/500000 + > ((173649*y)/1000000 - (241371*x)/1000000)*y + (141347*y)/1000000 - > 773749/1000000 == 0, x^2 + y^2 == 1} I see. Mathematica essentially change the floating point formula into a formula that uses only integers which can be represented without any error. Right? > then sols = Solve[eq, {x, y}]; is the (very complicated) set of exact solutions and N[sols] > {{x -> 1.3776284630851758 - 0.7866963118457833*I, > y -> -0.9763348915886709 - 1.1100445557562664*I}, > {x -> 1.3776284630851758 + 0.7866963118457833*I, > y -> -0.9763348915886709 + 1.1100445557562664*I}, > {x -> -0.9702291184141258, y -> -0.24218888864138183}, > {x -> -0.44830718122625757, y -> 0.8938795619438715}} look like the same solutions as you got except for order (they may be > slightly different because rationalization is not quite unique). Anyway, the point is that for exact algebraic numbers that sols consists of > one can decide whether they are real or not. > In fact, I understand that the determinate for a quadratic equation can be expressed in integers resulted from the rationalization of the equation. But does the determinant of a quartic (or higher or) equation can be expressed in integers resulted from the rationalization of the equation? Since there is no general analytical solution to polynomials with order higher than 4, how to exactly decide if the roots of the following equation are real or not? Solve[-6. + 5 x - 4 x^2 + 3 x^3 - 2 x^4 + x^5 + x^6 == 0, x] Peng === Subject: Re: How does Mathematica know whether a number is real or complex? Solve uses exact algebraic methods to solve equations so actually it does not solve the approximate equation that you gave it but some a unique way to rationalize an equation, but probably what it does is that it replaces your equations by: eq = Rationalize[{{x, y} . AA . {x, y} + BB . {x, y} + CC == 0, x^2 + y^2 == 1}, 0] {((11309*x)/25000 - (241371*y)/1000000)*x - (250149*x)/500000 + ((173649*y)/1000000 - (241371*x)/1000000)*y + (141347*y)/ 1000000 - 773749/1000000 == 0, x^2 + y^2 == 1} then sols = Solve[eq, {x, y}]; is the (very complicated) set of exact solutions and N[sols] {{x -> 1.3776284630851758 - 0.7866963118457833*I, y -> -0.9763348915886709 - 1.1100445557562664*I}, {x -> 1.3776284630851758 + 0.7866963118457833*I, y -> -0.9763348915886709 + 1.1100445557562664*I}, {x -> -0.9702291184141258, y -> -0.24218888864138183}, {x -> -0.44830718122625757, y -> 0.8938795619438715}} look like the same solutions as you got except for order (they may be slightly different because rationalization is not quite unique). Anyway, the point is that for exact algebraic numbers that sols consists of one can decide whether they are real or not. In fact, roots = Flatten[{x, y} /. sols]; Map[Element[#, Reals] &, roots] {False, False, False, False, True, True, True, True} If, on the other hand, you use NSolve, which relies on approximate methods, you cannot be always sure that this question is determined correctly, although in your particular case NSolve also gets the right answer: {x, y} /. NSolve[{{x, y} . AA . {x, y} + BB . {x, y} + CC == 0, x^2 + y^2 == 1}, {x, y}] {{1.3776284630851803 - 0.7866963118456223*I, -0.9763348915886673 - 1.1100445557560326*I}, {1.3776284630851803 + 0.7866963118456223*I, -0.9763348915886673 + 1.1100445557560326*I}, {-0.44830718122626145, 0.8938795619438661}, {-0.9702291184141273, -0.2421888886413809}} Andrzej Kozlowski The solution of the equation at the end of this post is {{x -> -0.9702291184141268, y -> -0.24218888864137905}, > {x -> -0.44830718122625973, y -> 0.8938795619438703}, > {x -> 1.3776284630851778 - 0.7866963118457873*I, > y -> -0.9763348915886718 - 1.1100445557562666*I}, > {x -> 1.3776284630851778 + 0.7866963118457873*I, > y -> -0.9763348915886718 + 1.1100445557562666*I}} Two of the solutions are real. The other two are complex. I'm wondering how Mathematica know whether a solution is real or > complex. > If a solution actually has a close-to-zero imaginary part, will > mathematica incorrect consider it as a real number? Peng AA = {{0.45236, -0.241371}, {-0.241371, 0.173649}}; > BB = {-0.500298, 0.141347}; > CC = -0.773749; > Solve[{{x, y}.AA.{x, y} + BB.{x, y} + CC == 0, x^2 + y^2 == 1}, {x, > y}] > === Subject: Re: How does Mathematica know whether a number is real or complex? > Since there is no general analytical solution to polynomials with > order higher than 4, how to exactly decide if the roots of the > following equation are real or not? There is no formula for the solution of a general polynomial equation in radicals, but of course that does not mean there are no other exact formulas. In fact there are many kind. Methematica uses so caled root objects: roots = x /. Solve[x^7 + x^6 - 3 x^5 + x^2 - 2 x + 1 == 0, x] {Root[#1^7 + #1^6 - 3 #1^5 + #1^2 - 2 #1 + 1 &, 1], Root[#1^7 + #1^6 - 3 #1^5 + #1^2 - 2 #1 + 1 &, 2], Root[#1^7 + #1^6 - 3 #1^5 + #1^2 - 2 #1 + 1 &, 3], Root[#1^7 + #1^6 - 3 #1^5 + #1^2 - 2 #1 + 1 &, 4], Root[#1^7 + #1^6 - 3 #1^5 + #1^2 - 2 #1 + 1 &, 5], Root[#1^7 + #1^6 - 3 #1^5 + #1^2 - 2 #1 + 1 &, 6], Root[#1^7 + #1^6 - 3 #1^5 + #1^2 - 2 #1 + 1 &, 7]} each of them is an exact quantity, can be computed to arbitrary precision: N[roots[[3]], 20] 1.2961147185902588044 and one can perform algebraic operations on them: p=RootReduce[roots[[3]] + 2 roots[[5]]] Root[#1^42 + 18 #1^41 + 69 #1^40 - 540 #1^39 - 4140 #1^38 + 4056 #1^37 + 88385 #1^36 + 52680 #1^35 - 1044698 #1^34 - 1403304 #1^33 + 7480992 #1^32 + 14724900 #1^31 - 29435735 #1^30 - 97086078 #1^29 + 16526534 #1^28 + 427408728 #1^27 + 406645548 #1^26 - 900737748 #1^25 - 1894388578 #1^24 - 2739622320 #1^23 + 3009136239 #1^22 + 28100399562 #1^21 + 17016274110 #1^20 - 115826796636 #1^19 - 119305659793 #1^18 + 246084263286 #1^17 + 434975355518 #1^16 - 178588275984 #1^15 - 727330538673 #1^14 - 1165988887194 #1^13 + 1953899024842 #1^12 - 60734067324 #1^11 + 3954530782343 #1^10 - 4146422391306 #1^9 - 5161380615111 #1^8 + 3915047501976 #1^7 + 7994697374154 #1^6 - 6436046783448 #1^5 - 4271654676433 #1^4 + 14651765344044 #1^3 - 9702852386442 #1^2 + 5559481408020 #1 - 979887667301 &, 24] For each of them Mathematica computes an exact isolating interval, which contains only one root and actually tells you if it is real or complex: IsolatingInterval[roots[[1]]] {-6, 0} hence roots[[1]] is real IsolatingInterval[roots[[5]]] {-4, 4 I} hence this is complex IsolatingInterval[p] {-(37/64) + (37*I)/32, (37*I)/16} this is also complex. These are all exact answers, just as exact as if with radicals but much more convenient both for numerical and algebraic operations. Andrzej Kozlowski === Subject: Re: How does Mathematica know whether a number is real or complex? > Solve uses exact algebraic methods to solve equations so actually it does > not solve the approximate equation that you gave it but some rationalized > rationalize an equation, but probably what it does is that it replaces your > equations by: eq = Rationalize[{{x, y} . AA . {x, y} + BB . {x, y} + CC == 0, > x^2 + y^2 == 1}, 0] {((11309*x)/25000 - (241371*y)/1000000)*x - (250149*x)/500000 + > ((173649*y)/1000000 - (241371*x)/1000000)*y + (141347*y)/1000000 - > 773749/1000000 == 0, x^2 + y^2 == 1} I see. Mathematica essentially change the floating point formula into a formula that uses only integers which can be represented without any error. Right? > then sols = Solve[eq, {x, y}]; is the (very complicated) set of exact solutions and N[sols] > {{x -> 1.3776284630851758 - 0.7866963118457833*I, > y -> -0.9763348915886709 - 1.1100445557562664*I}, > {x -> 1.3776284630851758 + 0.7866963118457833*I, > y -> -0.9763348915886709 + 1.1100445557562664*I}, > {x -> -0.9702291184141258, y -> -0.24218888864138183}, > {x -> -0.44830718122625757, y -> 0.8938795619438715}} look like the same solutions as you got except for order (they may be > slightly different because rationalization is not quite unique). Anyway, the point is that for exact algebraic numbers that sols consists of > one can decide whether they are real or not. > In fact, I understand that the determinate for a quadratic equation can be expressed in integers resulted from the rationalization of the equation. But does the determinant of a quartic (or higher or) equation can be expressed in integers resulted from the rationalization of the equation? Since there is no general analytical solution to polynomials with order higher than 4, how to exactly decide if the roots of the following equation are real or not? Solve[-6. + 5 x - 4 x^2 + 3 x^3 - 2 x^4 + x^5 + x^6 == 0, x] Peng === Subject: PrintTemporary Hello all, PrintTemporary[..] messages which are in fact notebook interface objects are not found by Cases[Notebooks[], _NotebookInterfaceObject]. I would like to remove all of them at certain points as too many temporary messages are generated during the evaluation of several packages. Of course, it would be possible to code temporaryMessages = {}; ... AppendTo[temporaryMessages, PrintTemporary[text 1]; ... AppendTo[temporaryMessages, PrintTemporary[text 2]; ... NotebookDelete /@ temporaryMessages Is there another possibility ? Hannes Kessler === Subject: Re: PrintTemporary > PrintTemporary[..] messages which are in fact notebook interface > objects are not found by Cases[Notebooks[], _NotebookInterfaceObject]. I don't fully understand how NotebookInterfaceObjects are part of a Notebook or can be accessed/found, but as long as they are seen, the with style PrintTemporary, so you can get rid of them at any time with something like the following: PrintTemporary[test]; Pause[1]; NotebookFind[EvaluationNotebook[], PrintTemporary, All, CellStyle]; NotebookDelete[]; Pause[1]; 1 + 1 hth, albert === Subject: Re: PrintTemporary > Hello all, > > PrintTemporary[..] messages which are in fact notebook interface > objects are not found by Cases[Notebooks[], _NotebookInterfaceObject]. > I would like to remove all of them at certain points as too many > temporary messages are generated during the evaluation of several > packages. > > Of course, it would be possible to code > > temporaryMessages = {}; > ... > AppendTo[temporaryMessages, PrintTemporary[text 1]; > ... > AppendTo[temporaryMessages, PrintTemporary[text 2]; > ... > NotebookDelete /@ temporaryMessages > > Is there another possibility ? > > Hannes Kessler > > The idea of PrintTemporary is, of course to print information that vanishes when an evaluation completes - typically some sort of progress information. The whole idea is that you don't have to clean the information out of your notebook! executing something equivalent to: Print[PrintTemporary[Something]] This would print the temporary message and also produce a NotebookInterfaceObject[...]. The temporary messages would vanish after execution but the NotebookInterfaceObject's will remain. Make sure that each call to PrintTemporary in your package is followed by a semicolon. David Bailey http://www.dbaileyconsultancy.co.uk === Subject: ODE solution problem I do not understand why the following code (using Mathematica vers. 5.2) does not give the correct solution. That is, the variable swt in the equation module is never set to 1, although the value of y(t) exceeds the value of yhi. The correct solution should have the following appearance, * * * * * * * * where it should reach a maximum value of 0.1 (yhi) at about 15 seconds, and then start to decrease, reaching a minimum of 0.025 (ylo) at about 30 seconds, etc. That is, the solution should oscillate between ylo and yhi. The following code however, when plotted shows a linear increase in y(t). I would appreciate it greatly if someone can point out my error(s) in the code, since I have not been able to identify why this code does not work properly. ClassicalRungeKutta /: NDSolve`InitializeMethod[ClassicalRungeKutta, __] := ClassicalRungeKutta[]; ClassicalRungeKutta[___][Step[f_, t_, h_, y_, yp_]]:= Block[{deltay, k1, k2, k3, k4}, k1 = yp; k2 = f[t + 1/2 h, y + 1/2 h k1]; k3 = f[t + 1/2 h, y + 1/2 h k2]; k4 = f[t + h, y + h k3]; deltay = h (1/6 k1 + 1/3 k2 + 1/3 k3 + 1/6 k4); {h, deltay} ]; c = 0.6; g = 9.81; R = 0.05; r = 0.007; yhi = 0.1; ylo = 0.025; Qin = 0.00005; eq[y_,swtch_] := (Qin - swtch*c*Sqrt[2*g*y]*Pi*r^2)/(Pi*R^2); Remove[equation]; swt = 0; equation[y_] := Module[{}, If[y <= ylo,swt=0;ex = eq[y,swt]]; If[y >= yhi,swt=1;ex = eq[y,swt]]; ex]; sol = NDSolve[{y'[t]== equation[y[t]],y[0]==0},y[t],{t,0,35},Method->ClassicalRungeKutta,StartingSt epSize->0.02]; --V. Stokes === Subject: Re: ODE solution problem how do you think should Mathematica find out: Remove[equation]; swt = 0; equation[y_] := Module[{}, If[y <= ylo,swt=0;ex = eq[y,swt]]; If[y >= yhi,swt=1;ex = eq[y,swt]]; ex]; that y[t] is smaller or larger than ylo/yhi y[t] is a symbolic expression as to see from here sol = NDSolve[{y'[t]== equation[y[t]],y[0]==0},y[t],{t,0,35},Method->ClassicalRungeKutta,StartingSt epSize->0.02]; Do you mean equation[y_?NumericQ] := Module[{}, If[y <= ylo,swt=0;ex = eq[y,swt]]; If[y >= yhi,swt=1;ex = eq[y,swt]]; ex]; ?? and for what is Module[{},..] good for ? and is ex not always eq[y,swt] and is eq[y,Which[yyhi,swt=1,True,swt]] not shorter? Jens > I do not understand why the following code (using Mathematica vers. 5.2) > does not give the correct solution. That is, the variable swt in the > equation module is never set to 1, although the value of y(t) exceeds > the value of yhi. The correct solution should have the following appearance, > > * > * * * > * * > * > * > where it should reach a maximum value of 0.1 (yhi) at about 15 seconds, > and then start to decrease, reaching a minimum of 0.025 (ylo) at about > 30 seconds, etc. That is, the solution should oscillate between ylo and > yhi. The following code however, when plotted shows a linear increase in > y(t). I would appreciate it greatly if someone can point out my error(s) > in the code, since I have not been able to identify why this code does > not work properly. > > > ClassicalRungeKutta /: > NDSolve`InitializeMethod[ClassicalRungeKutta, __] := > ClassicalRungeKutta[]; > ClassicalRungeKutta[___][Step[f_, t_, h_, y_, yp_]]:= > Block[{deltay, k1, k2, k3, k4}, > k1 = yp; > k2 = f[t + 1/2 h, y + 1/2 h k1]; > k3 = f[t + 1/2 h, y + 1/2 h k2]; > k4 = f[t + h, y + h k3]; > deltay = h (1/6 k1 + 1/3 k2 + 1/3 k3 + 1/6 k4); > {h, deltay} > ]; > c = 0.6; g = 9.81; > R = 0.05; r = 0.007; yhi = 0.1; ylo = 0.025; > Qin = 0.00005; > > eq[y_,swtch_] := (Qin - swtch*c*Sqrt[2*g*y]*Pi*r^2)/(Pi*R^2); > > Remove[equation]; > swt = 0; > equation[y_] := Module[{}, > If[y <= ylo,swt=0;ex = eq[y,swt]]; > If[y >= yhi,swt=1;ex = eq[y,swt]]; > ex]; > > sol = NDSolve[{y'[t]== > equation[y[t]],y[0]==0},y[t],{t,0,35},Method->ClassicalRungeKutta,StartingSt e pSize->0.02]; > > > --V. Stokes > === Subject: Re: Simulation for probability of the roots of a quadratic equation > Hi All, I saw this recent thread on sci.math. 23ceb5aa5d883cbe/24d7ac4ba82b532f#24d7ac4ba82b532f > How can one do that simple simulation in Mathematica? Also, here is an analysis if what the probabililty should be. Do the > results above match this? 0], {a, -n, n}, {b, -n, n}, {c, -n, n}, Assumptions -> n > 0]; vol = Integrate[1, {a, -n, n}, {b, -n, n}, {c, -n, n}, Assumptions -> n > 0]; frac = Simplify[boole/vol] Compare with the simulation: ranfun = Compile[{n}, Module[{a, b, c}, Plus @@ Table[ {a, b, c} = RandomReal[{-1, 1}, 3]; If[b^2 - 4 a*c > 0, 1, 0], {n}]/n]]; Log[ranfun[10^6]/frac] --Mark === Subject: Re: Simulation for probability of the roots of a quadratic equation > Hi All, I saw this recent thread on sci.math. 23ceb5aa5d883cbe/24d7ac4ba82b532f#24d7ac4ba82b532f > How can one do that simple simulation in Mathematica? Also, here is an analysis if what the probabililty should be. Do the > results above match this? 0, count++], {100000}]; count max = 1000000; r := RandomReal[{-max, max}] count = 1; Do[If[r^2 - 4*r*r > 0, count++], {100000}]; count will give approx. 63% probability. If you restrict the a,b and c to positive numbers max = 1000000; r := RandomReal[{0, max}] count = 1; Do[If[r^2 - 4*r*r > 0, count++], {100000}]; count then you will get approx. 25.5%. Norbert Marxer === Subject: My talk The Joy of Tagging at this year's International Hello All: I gave a talk at this year's International Mathematica User Conference titled: The Joy of Tagging: Manipulating and Mining Notebooks in Mathematica. You can get the notebook version of this talk from http://scientificarts.com/worklife/notebooks/ I think that there is some useful stuff here that can be helpful for folks who want to create Notebook oriented Mathematica tools. Here is the abstract: One of the key elements of a Mathematica application that deals with a variety of documents is tagging. Mathematica's user interface for the tagging of notebooks is minimal, even though the functional interface is very powerful. This talk is about the tricks of the trade for tagging and manipulating notebooks that I have learned and discovered through creating a very large document-oriented Mathematica add-on called A WorkLife FrameWork. A great deal of tagging of notebooks is done in A WorkLife FrameWork, and this is used to create a very wide range of functionalities to automatically navigate among content, as well as to nonlinearly extract desired content from notebooks that have a random assortment of material in them=96the wheat from the chaff. Also, A WorkLife FrameWork provides user interface tools to enhance and make much more useful Mathematica's tagging potential. I will describe how these tools were created and provide useful examples. The tricks that I will describe will allow you to design your own tagging tools whether or not you have a license to A WorkLife FrameWork. I hope that this is helpful. --David === Subject: Re: Mathematica Equivalent of Excel VLOOKUP function > Hi - I have two large csv files.The first is the data: mydata=Table[{{jones, 34124324, 605, 54161, C33}, {smith, > 343214234, 550.92, 49505, V55}, {andrews, 54554543, 550.92, 49505, > G55}, {robinson, 565654565, 550.90, 49505, K77}, {burt, 9898966, > 752.61, 54324, U88}}] the second file is the 'lookup table': > mylookup=Table[{{54161, 3.2}, {54324, 16.3}, {49505, 12.3}}] MS Excel has a function called VLOOKUP - it finds a column of your > data (eg, the 4th column in my example (54161, 49505, 49505, > 49505,54324)), and sequentially looks up each value in the lookup > table (which has unique values only), and then grabs' the adjacent > column ([key]value pairs), and adds the looked up value as a final > column in the original data: result = Table[{{jones, 34124324, 605, 54161, C33, 3.2}, {smith, > 343214234, 550.92, 49505, V55, 12.3}, {andrews, 54554543, 550.92, > 49505, G55,12.3}, {robinson, 565654565, 550.90, 49505, K77,12.3}, > {burt, 9898966, 752.61, 54324, U88,16.3}}] > Not elegant -- but try: Table[Flatten[{mydata[[i]],Select[mylookup, #[[1]] == mydata[[i, 4]] &] [[1, 2]]}], {i,Length[mydata]}] === Subject: Re: Mathematica Equivalent of Excel VLOOKUP function You could try something along the lines of: mydata={{jones,34124324,605,54161,C33},{smith, 343214234,550.92,49505,V55},{andrews,54554543,550.92,49505,G55}, {robinson,565654565,550.90,49505,K77},{burt, 9898966,752.61,54324,U88}}; mylookup={{54161,3.2},{54324,16.3},{49505,12.3}}; vlookup[record_, lookupTable_, recordIndex_, matchColumn_: 1, resultColumn_: - 1] := Module[{nTab, key, i, tabResult}, key = record[[recordIndex]]; nTab = Length[lookupTable]; i = 1; While[ i <= nTab && lookupTable[[i, matchColumn]] != key, i++]; If[i <= nTab, tabResult = lookupTable[[i, resultColumn]], tabResult = No Match ]; If[MatchQ[ tabResult, _List], Join[record, tabResult], Append[record, tabResult] ] ] mydata2=vlookup[#, mylookup, 4, 1, 2]&/@mydata; mydata2//TableForm Note that this is written so you can grab more than just a single column from your lookup table. Also, you can compare more than just a single column value. You can use any expression that Part will accept for the recordIndex, matchColumn and resultColumn. Also note that you need to be careful if your lookup table does not contain a matching entry or if it contains more than one matching entry. I hope that helps. Januk > Hi - I have two large csv files.The first is the data: mydata=Table[{{jones, 34124324, 605, 54161, C33}, {smith, > 343214234, 550.92, 49505, V55}, {andrews, 54554543, 550.92, 49505, > G55}, {robinson, 565654565, 550.90, 49505, K77}, {burt, 9898966, > 752.61, 54324, U88}}] the second file is the 'lookup table': > mylookup=Table[{{54161, 3.2}, {54324, 16.3}, {49505, 12.3}}] MS Excel has a function called VLOOKUP - it finds a column of your > data (eg, the 4th column in my example (54161, 49505, 49505, > 49505,54324)), and sequentially looks up each value in the lookup > table (which has unique values only), and then grabs' the adjacent > column ([key]value pairs), and adds the looked up value as a final > column in the original data: result = Table[{{jones, 34124324, 605, 54161, C33, 3.2}, {smith, > 343214234, 550.92, 49505, V55, 12.3}, {andrews, 54554543, 550.92, > 49505, G55,12.3}, {robinson, 565654565, 550.90, 49505, K77,12.3}, > {burt, 9898966, 752.61, 54324, U88,16.3}}] I have found some old posts that come close to a solution, but aren't > quite right: makeLookupTable[symbol_, list_, > column_] := (symbol[___] = Table[0., {Length[list[[1]] - 1]}]; > Scan[With[{key = #[[column]]}, symbol[key] = Delete[#, column]= ] &, > list]); > m1 = Table[{i, Random[], Random[]}, {i, 10}]; > m2 = Table[{Random[], Random[], Random[], i}, {i, 2, 12}]; > allkeys = Union[m1[[All, 1]], m2[[All, 4]]] > Join[{#}, lt1[#], lt2[#]] & /@ allkeys > clsnyder === Subject: Re: Mathematica Equivalent of Excel VLOOKUP function > Hi - > > I have two large csv files.The first is the data: > > mydata=Table[{{jones, 34124324, 605, 54161, C33}, {smith, > 343214234, 550.92, 49505, V55}, {andrews, 54554543, 550.92, 49505, > G55}, {robinson, 565654565, 550.90, 49505, K77}, {burt, 9898966, > 752.61, 54324, U88}}] > > the second file is the 'lookup table': > mylookup=Table[{{54161, 3.2}, {54324, 16.3}, {49505, 12.3}}] > > MS Excel has a function called VLOOKUP - it finds a column of your > data (eg, the 4th column in my example (54161, 49505, 49505, > 49505,54324)), and sequentially looks up each value in the lookup > table (which has unique values only), and then grabs' the adjacent > column ([key]value pairs), and adds the looked up value as a final > column in the original data: > > result = Table[{{jones, 34124324, 605, 54161, C33, 3.2}, {smith, > 343214234, 550.92, 49505, V55, 12.3}, {andrews, 54554543, 550.92, > 49505, G55,12.3}, {robinson, 565654565, 550.90, 49505, K77,12.3}, > {burt, 9898966, 752.61, 54324, U88,16.3}}] > If you remove the syntactically incorrect and useless Table-wrappers from your array definitions, the following does what I think you want: Flatten /@ Transpose[{mydata, mydata[[All, 4]] /. Rule @@@ mylookup}] when coming form excel, you definitly should spend some time learning the basics of mathematica, the two programs are very different. As a starting point, you might search the documentation for ReplaceAll, Rule and Apply, which I have used to create a list of replacement rules from your lookup table. The rest is then only rearranging the lists to append the result of the lookup. There are other possibilities and with very large tables you might need to use another approach, so I expect that you will get many other answers to that question... hth, albert === Subject: Re: Mathematica Equivalent of Excel VLOOKUP function ...prblm solved! clsnyder === Subject: Re: Mathematica Equivalent of Excel VLOOKUP function > Hi - I have two large csv files.The first is the data: mydata=Table[{{jones, 34124324, 605, 54161, C33}, {smith, > 343214234, 550.92, 49505, V55}, {andrews, 54554543, 550.92, 49505, > G55}, {robinson, 565654565, 550.90, 49505, K77}, {burt, 9898966, > 752.61, 54324, U88}}] the second file is the 'lookup table': > mylookup=Table[{{54161, 3.2}, {54324, 16.3}, {49505, 12.3}}] MS Excel has a function called VLOOKUP - it finds a column of your > data (eg, the 4th column in my example (54161, 49505, 49505, > 49505,54324)), and sequentially looks up each value in the lookup > table (which has unique values only), and then grabs' the adjacent > column ([key]value pairs), and adds the looked up value as a final > column in the original data: result = Table[{{jones, 34124324, 605, 54161, C33, 3.2}, {smith, > 343214234, 550.92, 49505, V55, 12.3}, {andrews, 54554543, 550.92, > 49505, G55,12.3}, {robinson, 565654565, 550.90, 49505, K77,12.3}, > {burt, 9898966, 752.61, 54324, U88,16.3}}] I have found some old posts that come close to a solution, but aren't > quite right: makeLookupTable[symbol_, list_, > column_] := (symbol[___] = Table[0., {Length[list[[1]] - 1]}]; > Scan[With[{key = #[[column]]}, symbol[key] = Delete[#, column]= ] &, > list]); > m1 = Table[{i, Random[], Random[]}, {i, 10}]; > m2 = Table[{Random[], Random[], Random[], i}, {i, 2, 12}]; > allkeys = Union[m1[[All, 1]], m2[[All, 4]]] > Join[{#}, lt1[#], lt2[#]] & /@ allkeys > clsnyder Hello One solution to your problem: If I take your data: Clear[mydata, lookup, f]; mydata = {{jones, 34124324, 605, 54161, C33}, {smith, 343214234, 550.92, 49505, V55}, {andrews, 54554543, 550.92, 49505, G55}, {robinson, 565654565, 550.9, 49505, K77}, {burt, 9898966, 752.61, 54324, U88}}; Grid[mydata] lookup = {{54161, 3.2}, {54324, 16.3}, {49505, 12.3}} And define the function f: f[{a_, b_, c_, d_, e_}] := {a, b, c, d, e, Switch[d, Evaluate[lookup /.{List -> Sequence}]]} Then this function will give the desired output: Grid[f /@ mydata] This is not a general solution (where you can select the column etc), but could be used as a starting point for the more general solution. Norbert Marxer === Subject: Re: partial Dividers? I like the Grid of Grids solutions proposed by Bob Hanlon and David Park, which I did not know about. Another more awkward way of achieving the same effect, but which (seemingly) avoids the 'background' problem outlined by David Park, is it create a separate headings list and to use Item and Frame to suppress the unwanted lines. For Example (modifying the example of David P): headings = {{Item[a, Frame -> { True, True, True, False}], Item[b, Frame -> { True, False, True, True}], Item[c, Frame -> { True, False, True, True}], Item[d, Frame -> { True, False, True, True}], Item[e, Frame -> { True, False, True, True}]}}; table = Table[i + j, {i, 1, 5}, {j, 1, 5}]; The grid may now be generated from the following: Grid[Join[headings, table], Frame -> True, FrameStyle -> Orange, Dividers -> {All -> True, {None}}, Background -> {None, {1 -> GrayLevel[.8]}}, ItemSize -> {2, 1.5}] Or using your own example: m = Partition[Range[15], 3]; m[[1, 1]] = Item[1, Frame -> { True, True, True, False}]; m[[1, 2]] = Item[10^6, Frame -> { True, False, True, False}]; m[[1, 3]] = Item[10^6, Frame -> { True, False, True, True}]; And: Grid[m, Frame -> True, FrameStyle -> Orange, Dividers -> {All -> True, {None}}, Background -> {None, {1 -> GrayLevel[.8]}}] I use this approach as it does what I want, and I too like the completely filled background, but there must be a 'neater' way? Tom Dowling. > m = Partition[Range[9], 3]; Grid[{ > {Grid[{First[m]}]}, > {Grid[Rest[m], Dividers -> {{False, True}, False}]}}] Bob Hanlon > Hi all, In a Grid, I'd like to have a divider between columns 1 and 2 (for > example) that goes from the 2nd row to the last row. But I can't > figure out a way to make a divider that doesn't span the entire grid. > Is there a way? --Mark -- Bob Hanlon Being inspired by the suggestions of both Bob and David, I have come > up with the following, which does pretty much what I want (I think), > but it's not fully automatic in the sense that one needs to know the > proper width for the first column (which appears as the first > numerical value in the specification for size). m = Partition[Range[15], 3]; > m[[1, 2]] = 10^5; > m[[1, 3]] = 10^5; With[{size = {{1, {Automatic}}, {1}}, align = Right}, > Grid[{{ > Grid[{Most[First[m]]}, ItemSize -> size, Alignment -> align], > Last[First[m]] > }, > Sequence @@ Rest[m]}, > Dividers -> {{True, True, {False}, True}, {True, True, {False}, > True}}, ItemSize -> size, Alignment -> align] > ] > --Mark === Subject: Re: Controlling the order of evaluations f[x_?NumericQ, y_?NumericQ] := NIntegrate[2^(s*x), {s, 0, y}] Plot[f[x, #] & /@ {2, 3, 4, 5}, {x, 0.5, 1.5}, Evaluated -> True] ?? Jens > to increase my knowledge about Mathematica I looked in the documentation to find information about the Evaluated option. I seems it is nowhere described. The only thing I found out is, that usually it is set to Automatic by default. Can you tell me what it does exactly? Is setting it to true the same as writing Plot[Evaluate[...],{...}]? Gruss Peter -- ==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-== Peter Breitfeld, Bad Saulgau, Germany -- http://www.pBreitfeld.de === Subject: nested derivatives How do I enter the following recursive function in Mathematica: t[n]=(1/y')*t'[n-1], with n the index t[0]=x y is some function of x Huub Mooren === Subject: Re: nested derivatives Hi Huub, trythe following: t[0]=x; t[n_]:=1/D[y[x],x]D[ t[n-1],x]; t[3] hope this helps, Daniel > How do I enter the following recursive function in Mathematica: > t[n]=(1/y')*t'[n-1], > with n the index > t[0]=x > y is some function of x > > > Huub Mooren > -- Daniel Huber Metrohm Ltd. Oberdorfstr. 68 CH-9100 Herisau Tel. +41 71 353 8585, Fax +41 71 353 8907 E-Mail: Internet: === Subject: Re: nested derivatives > How do I enter the following recursive function in Mathematica: > t[n]=(1/y')*t'[n-1], > with n the index > t[0]=x > y is some function of x Since t[n] does not explicitly depends on x, we must write t'[n-1] as D[t[n-1],x]. Having said that, we can define our function either using an explicit recursive definition (see In[1], note that the definition also includes some test on the argument and use memoization [1] for computational efficiency [2]), or we can use higher-level functions such as *Nest* [3] (see In[4], which use a pure function for its recursive call). In[1]:= t[0] = x; t[n_Integer?Positive] := t[n] = 1/y'[x] D[t[n - 1], x] Table[{n, t[n]}, {n, 4}] // TableForm[#, TableHeadings -> {None, {n, t[n]}}] & // OutputForm Out[3]//OutputForm= n t[n] 1 1 ----- y'[x] 2 y''[x] -(------) 3 y'[x] 3 2 (3) 3 y''[x] y [x] --------- - ------- 4 3 y'[x] y'[x] ------------------- y'[x] 2 (3) 3 y''[x] y [x] y''[x] (--------- - -------) 4 3 y'[x] y'[x] (-(----------------------------) + 2 y'[x] 3 (3) (4) -12 y''[x] 9 y''[x] y [x] y [x] (----------- + ---------------- - -------) / y'[x]) / y'[x] 5 4 3 4 y'[x] y'[x] y'[x] In[4]:= tt[n_Integer?Positive] := Nest[1/y'[x] D[#, x] &, x, n] Table[tt[n] == t[n], {n, 4}] Out[5]= {True, True, True, True} - Jean-Marc http://en.wikipedia.org/wiki/Memoization [2] Functions That Remember Values They Have Found http://reference.wolfram.com/mathematica/tutorial/FunctionsThatRememberValue sTheyHaveFound.html [3] Applying Functions Repeatedly http://reference.wolfram.com/mathematica/tutorial/ApplyingFunctionsRepeatedl y.html === Subject: Re: nested derivatives > How do I enter the following recursive function in Mathematica: > t[n]=(1/y')*t'[n-1], > with n the index > t[0]=x > y is some function of x > > > Huub Mooren > I would try: Subscript[t, 0][x_] = x; Subscript[t, n_][x_] := Derivative[1][Subscript[t, n - 1]][x] / Derivative[1][y][x] Block[{y = Exp}, Table[Subscript[t, n][x], {n, 0, 5}]] yields {x, E^(-x), -E^(-2*x), 2/E^(3*x), -6/E^(4*x), 24/E^(5*x)} Peter === Subject: mathlink and complex number Hi all, I'm using mathlink to communicate between C and Mathematica. I wish to know how to get the complex number output(in C) from a mathematica function. The mathematica function SpheroidalS1 gets integer and float inputs, i wish to know how to retrieve the complex number output of this function in my external C program. === Subject: Re: mathlink and complex number int n; complex c; If(MLGetFunction(stdlink,Complex,&n) && 2==n){ double re,im; MLGetDouble(stdlink,&re); MLGetDouble(stdlink,&im); c=complex(re,im); } ?? Jens > Hi all, > I'm using mathlink to communicate between C and Mathematica. > > I wish to know how to get the complex number output(in C) from a mathematica function. > > The mathematica function SpheroidalS1 gets integer and float inputs, > i wish to know how to retrieve the complex number output of this function in my external C program. > > === Subject: Solving systems of eqs involving Integrations over product of hi, here's my integration function: IntegrateOverIndependentDistributions/: IntegrateOverIndependentDistributions::usage=IntegrateOverIndependentDistr ibutions[fun, {d1,d2,...},method:NIntegrate] returns method[fun[x[1],x[2],...]PDF[d1,x1]PDF[d2,x2]...,{x1,-inf,inf}, {x2,-inf,inf}]; IntegrateOverIndependentDistributions[expr_, dists_,method_: NIntegrate] := With[{len = Length[dists]}, Module[{x}, With[{vars = x /@ Range[1, len]}, With[{integrand = Times[expr @@ vars, Times @@ MapThread[PDF, {dists, vars}]], ranges= With[{lst={#, -[Infinity], [Infinity]} & /@ vars},Sequence @@ lst] }, method[integrand, ranges] ] ] ] ]; I now have an integrand of the form say f[g1_][x1_,x2_]=... So I define F[g1_]:=IntegrateOverIndependentDistributions[f[g1], {NormalDistribution[0,1],NormalDistribution[0,1]}]; and do FindRoot[ { F[g1] == 0.0 }, {{g1, -10, 10}} ] ; but i get the error message: FindRoot::nlnum The function value {NIntegrate[...],{($$24632, $$24657, $ $24658)}] is not a list of numbers with dimensions {1} at {[Gamma]1} = {10.}. It seems that $$24632 is not what I should have, I should have g1... By the way evaluation f[g1] on a grid works... Please help. === Subject: Re: reversing the direction of the X axis in Plot and ListPlot Hi Michael, You could use a ParametricPlot to do the reversal for you. It would then simply be a matter of reversing the tick mark labels. You might start with something like the following: SetAttributes[PlotReverse, HoldAll] PlotReverse[func_, {t_, tmin_, tmax_}, opts___] := Module[{pl1, ticks, t1}, pl1 = ParametricPlot[ Evaluate[{tmax - t1, func /. t -> t1}], {t1, tmin, tmax}, opts]; ticks = Ticks /. AbsoluteOptions[pl1]; ticks[[1, All, 2]] = ticks[[1, All, 2]] /. {x_?NumericQ :> tmax - x}; Show[pl1, Ticks -> ticks] ] f[x_] := Exp[x/40] PlotReverse[f[x], {x, 0, 180}, AspectRatio -> 1/GoldenRatio] Plot[f[x], {x, 0, 180}, AspectRatio -> 1/GoldenRatio] I hope that helps, Januk > I'm graphing some bond values where the natural flow is for the x axis, > which represents days to maturity, to run from from big numbers to small > ones. I can achieve this by expressing my function in terms of days past > (which run small to big) and then using Ticks to label the axis the other > direction (as though it were days to maturity), but this feels like a > complex solution to a simple problem. Is there an easy way to Plot a > function from 180 to 0, rather than 0 to 180? > Michael === Subject: Re: reversing the direction of the X axis in Plot and ListPlot Yes, you do have to write a Ticks specification. I don't know if this makes it a lot easier, but the Presentation package has a CustomTicks command that allows you to specify any tick value transformation that is a 1-1 map to the underlying coordinate. Needs[Presentations`Master`] xticks = CustomTicks[180 - # &, {0, 180, 30, 3}]; Plot[100 Exp[-x/90], {x, 0, 180}, PlotRange -> {0, 110}, Frame -> True, FrameTicks -> {{Automatic, Automatic}, {xticks, xticks // NoTickLabels}}, FrameLabel -> {Days to Maturity, Value $}, PlotLabel -> Value of U.S. Treasuries After 10th Trillon Dollar Bailout, BaseStyle -> {FontSize -> 14}, ImageSize -> 600] -- David Park djmpark@comcast.net http://home.comcast.net/~djmpark/ > I'm graphing some bond values where the natural flow is for the x axis, > which represents days to maturity, to run from from big numbers to small > ones. I can achieve this by expressing my function in terms of days past > (which run small to big) and then using Ticks to label the axis the other > direction (as though it were days to maturity), but this feels like a > complex solution to a simple problem. Is there an easy way to Plot a > function from 180 to 0, rather than 0 to 180? > Michael > === Subject: Re: notation using # with exponents and & > A pure function (or anonymous function in some other programming > languages) Anonymous function might in fact seem to be a more meaningful name for the Mathematic construct, especially since what Mathematica refers to as a pure function seems to be significantly different from what Wikipedia refers to as a pure function in mathematics. Or am I misunderstanding the Wikipedia description? (I have no prior experience with either of these concepts.) === Subject: Re: How to plot a variable Hi Knut, you must distinguish between functions and lists: ABB[1] is a function call with the parameter 1, but: ABB[[1]] is a list element. Further, Plot will take a function that is defined for a continuous variable, otherwise use ListPlot. hope this helps, Daniel > When I iterate a variable in a do-loop it actually will not print out. Is it undefined? In Fortran this would never be a problem, but in mathematica it seems that it must be done different: > > Do[ GA = GA + .5*(G[J] + G[J + 1])*DELT ; Amp = Exp[2.*GA] ; > JJ = J + 1; ABB[1] = 1; ABB[JJ] = Amp, {J, 1, N1}] ; > pl1 = Plot[ABB, {n, 1, NAX}, PlotLabel -> Impedans, Filling -> Axis, > PlotRange -> All] > > ABB gives no result.. > -- Daniel Huber Metrohm Ltd. Oberdorfstr. 68 CH-9100 Herisau Tel. +41 71 353 8585, Fax +41 71 353 8907 E-Mail: Internet: === Subject: Re: How to plot a variable > When I iterate a variable in a do-loop it actually will not print out. = Is it undefined? In Fortran this would never be a problem, but in mathema= tica it seems that it must be done different: Do[ GA = GA + .5*(G[J] + G[J + 1])*DELT ; Amp = Exp[2.*GA] ; > JJ = J + 1; ABB[1] = 1; ABB[JJ] = Amp, {J, 1, N1}] ; > pl1 = Plot[ABB, {n, 1, NAX}, PlotLabel -> Impedans, Filling -> Axis= , > PlotRange -> All] ABB gives no result.. > You have not supplied enough of your code to actually run it, but I am guessing that ABB is supposed to be an array. The following piece of code does something similar and illustrates how to set up an array and how to index it (note the double square brackets). Since your resulting data is discrete, ListPlot is the appropriate plotting command. ABB = Array[0 &, {100}]; Do[ ABB[[k]] = Sin[k*0.1], {k, 1, 100} ]; ListPlot[ABB] I would not write this code with a Do loop, but I realise that this is a convenient way of moving from Fortran. Once you have your code working, you may want to look at the Table function. David Bailey http://www.dbaileyconsultancy.co.uk === Subject: Re: How to plot a variable > When I iterate a variable in a do-loop it actually will not print out. Is it undefined? In Fortran this would never be a problem, but in mathematica it seems that it must be done different: > > Do[ GA = GA + .5*(G[J] + G[J + 1])*DELT ; Amp = Exp[2.*GA] ; > JJ = J + 1; ABB[1] = 1; ABB[JJ] = Amp, {J, 1, N1}] ; > pl1 = Plot[ABB, {n, 1, NAX}, PlotLabel -> Impedans, Filling -> Axis, > PlotRange -> All] > > ABB gives no result.. Assuming I have correctly understood your question, you must define ABB *Table*), then use ListPlot to get a graph. Something along the lines, ABB = Table[GA = GA + .5*(G[J] + G[J + 1])*DELT; Amp = Exp[2.*GA]; Amp, {J, 1, N1}]; ABB = Join[{1}, ABB] pl1 = ListPlot[ABB, {n, 1, NAX}, PlotLabel -> Impedans, Filling -> Axis, PlotRange -> All] (Note that *Table* crates a list with the values of amp, then we add the value 1 in front of this list, which is same under the name ABB.) -- Jean-Marc === Subject: Re: Function pure for Select One of the reasons for the difference in timings is that the different approaches are not all solving the same problem, that is, we interpretted the original posters request (extend this function to a list of values) differently as to the manner of extension. The first and third method select all rows that start with any of the target values. The second approach sequentially selects the rows that start with each of the target values. The results select all of the same rows; however, in the second case the result is grouped by first value. Whether this grouping is desired or intended by the original poster I do not know. A shorter example is shown to view the differences. data = Table[{RandomInteger[100], RandomInteger[100]}, {75}]; targets = Table[i, {i, 10}]; t1 = Timing[s1 = Select[data, MemberQ[targets, First[#]] &];] {0.00021,Null} f[x_] := Select[data, First[#] == x &]; t2 = Timing[s2 = f /@ targets;] {0.000762,Null} t3 = Timing[s3 = Cases[data, {Alternatives @@ targets, _}];] {0.000075,Null} s1 == s3 True s1 {{2, 13}, {5, 49}, {9, 6}, {1, 49}, {3, 84}, {4, 68}, {2, 91}, {1, 92}, {2, 9}, {3, 67}, {4, 91}, {4, 64}, {5, 63}} s2 {{{1, 49}, {1, 92}}, {{2, 13}, {2, 91}, {2, 9}}, {{3, 84}, {3, 67}}, {{4, 68}, {4, 91}, {4, 64}}, {{5, 49}, {5, 63}}, {}, {}, {}, {{9, 6}}, {}} Bob Hanlon > Split[Sort[Select[data, > MemberQ[{30, 45, 50, 66}, #[[1]]] &]], > #1[[1]] == #2[[1]] &] However, it is much easier to use a helper function: f[x_] := Select[data, #[[1]] == x &] f /@ {30, 45, 50, 66} Bob Hanlon > Hi al, How can I to write a function pure to extract all the first rows of > collection of data, applied to a list?. For example, Select[data,First[#]==30&] This function extracts all rows which first element is equal to 30. > Well, I want to extend this function to a list of values > {30,45,50,66}. > -- Bob Hanlon Here's a look at a few approaches, with timings on my machine: data = Table[{RandomInteger[1000], RandomInteger[1000]}, {100000}]; targets = Table[i, {i, 250}]; Select[data, MemberQ[targets, First[#]] &] ...takes 2.886 seconds f[x_] := Select[data, First[#] == x &]; f /@ targets ...takes 93.616 seconds Cases[data, {Alternatives@@targets, _}] ...takes 1.061 seconds IsTarget[_] := False; Scan[(IsTarget[#] = True) &, targets]; Select[data, IsTarget[First[#]] &] ...takes 0.452 seconds === Subject: Associate cycles of roots and irreducible factors of a cyclotomic polynomialmath. Importance: Normal Hello everybody, I have this problem: taken n=32; - (done) create a complete arithmetic system of Galois Field GF(n) - (todo) associate cycles of roots (?) and irreducible factors of X^(n-1) - 1 in Zp[X]; where p is the characteristic of GF(n) googling haven't produced any useful result yet, infact I found some texts where cycles of roots are mentioned, but none of them explains what they are and how to calculate them. Any help would be appreciated. --Marco Marsala === Subject: Re: Integrate using complex analysis > I am having trouble integrating the following: > > sin( sqrt(a^2+x^2) ) / sqrt(a^2+x^2) > > from -Inf to +Inf. I read about the Cauchy integral theorem but the > examples I've found have a polynomial denominator whereas this has a > sqrt, so I don't really know how to apply it here. Any help > appreciated. A couple of days ago, you posted, *I want to integrate f(x) = 1/sqrt(a^2+x^2) from -Inf to +Inf using *Cauchy's integral theorem. I know the function can be integrated using *elementary calculus but I want to do it both ways. *The problem I get is: *f(z) = 1/sqrt[ (z+i.a)(z-i.a) ] *Obviously there are poles z = +/-ia but they are inside the sqrt. How *can I proceed from here? And also, *I posted a question earlier but it appears to have got lost in the *ether. Apologies for this (repeat) post on the same theme. *I want to integrate from -Inf to +Inf *f(x) = sin(sqrt(1+x^2)) / sqrt(1+x^2). *I tried using Cauchy's integral theorem but I'm not sure my integral *is of the right form. First I switched to the complex variable x *f(z) = sin(sqrt(1+z^2)) / sqrt( (1+iz)(1-iz) ). *In the examples I've seen, the denominator does not have the sqrt term *so I don't know how to proceed in this case. Numerical calculations *show there is a well behaved solution but I want the analytical form. *Suggestions would be appreciated :) And you got answers to both of those posts. So was there something you didn't understand in the answers you got? and, if so, wouldn't it be better to reply to those answers, to show what you understood and what you needed help with? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Moving to real fun <49009c5b$1@news.x-privat.org> posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) a) the free market is a visciously oligarchical & British lie, per petrated primarily through Adam Smith's second (1776) hoax, The Wealth of Nations , ever since taken as The Holy Economist writ -- you should see the 2-vol. *abridged* Princeton edition; b) don't nuke any more fresh-faced, volunteer-manned e-math-journals; c) they mean, well! > All scientists should learn to live in the free market. Do their thus: yeah, minus one; the smiley was implicated. see, there is a practice in giving math tests, where no answer is zero points, and a wrong answer is minus one; actually, that's for multipe choice -- NOTA!... however, you'd get extra credit for three wrong choices, perhaps. hey, but I didn't check, since I prefer some sort of rationale to be stated, before the incipiently slippery slope of Fermat's so-called last theorem, which was probaably one of his very first. after all, NO OTHER MISTAKES are known. (like, why do you care, that we should care ?-) thus quoth: With three different versions, one is bound to be correct. > You forgot the smiley thus: here is an etymology of THE PHOTON: photoelectrical effect hits measuring device, ballistically imparting its momentum of one quantum (per Planck h-bar) at some frequency; the bug is a feature, proven statistically by Bose (lies, polls, statistics !-) > where did you ever refute Uncle Al's experiments? --USA out of Darfur Cruizade! http://larouchepub.com/other/2008/3537willl zardari survive.html --ROTC, your summer vacation in the Sahara Desert ( S u d a n ) ; presage the Draft for your middleschool class of '12 -- brought to you by Allstate (tm) and Oxford U.Press! http://larouchepub.com/pr/2008/080813moloch brown.html http://wlym.com --Wikipedia deletes notice of nullification of preclearance rule of Voting Rights Act in LaRouche v. Fowler, March 27, 2000; is the VRAo1965 a dead letter? http://en.wikipedia.org/wiki/Voting Rights Act#Pre-Clearance 2 === Subject: Re: #12 Proving that any derivatives tacked-onto StockMarket is a Ponzi Scheme; new book: StockMarket Functioning Fairly posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) good luck; firstly, define what you mean by proof. > All these bells and whistles of new instruments added onto the > StockMarket are merely Ponzi Pyramid schemes that cheats the people > who do only buy and sell in cash. thus: a) the free market is a visciously oligarchical & British lie, perpetrated primarily through Adam Smith's second (1776) hoax, _The Wealth of Nations_, ever since taken as The Holy Economist writ -- you should see the 2-vol. *abridged* Princeton edition; b) don't nuke any more fresh-faced, volunteer-manned e-math-journals; c) they mean, well! > All scientists should learn to live in the free market. Do their thus: yeah, minus one; the smiley was implicated. see, there is a practice in giving math tests, where no answer is zero points, and a wrong answer is minus one; actually, that's for multipe choice -- NOTA!... however, you'd get extra credit for three wrong choices, perhaps. hey, but I didn't check, since I prefer some sort of rationale to be stated, before the incipiently slippery slope of Fermat's so-called last theorem, which was probaably one of his very first. after all, NO OTHER MISTAKES are known. (like, why do you care, that we should care ?-) thus quoth: With three different versions, one is bound to be correct. > You forgot the smiley thus: here is an etymology of THE PHOTON: photoelectrical effect hits measuring device, ballistically imparting its momentum of one quantum (per Planck h-bar) at some frequency; the bug is a feature, proven statistically by Bose (lies, polls, statistics !-) > where did you ever refute Uncle Al's experiments? --USA out of Darfur Cruizade! http://larouchepub.com/other/2008/3537willl_zardari_survive.html --ROTC, your summer vacation in the Sahara Desert ( S u d a n ) ; presage the Draft for your middleschool class of '12 -- brought to you by Allstate (tm) and Oxford U.Press! http://larouchepub.com/pr/2008/080813moloch_brown.html http://wlym.com --Wikipedia deletes notice of nullification of preclearance rule of Voting Rights Act in LaRouche v. Fowler, March 27, 2000; is the VRAo1965 a dead letter? http://en.wikipedia.org/wiki/Voting_Rights_Act#Pre-Clearance_2 === Subject: Within The Curve: Game posting-account=dGiPYgkAAABSJ3xUlNLViQdT0h489hR6 AppleWebKit/523.10.3 (KHTML, like Gecko) Version/3.0.4 Safari/523.10,gzip(gfe),gzip(gfe) Here is a game for any number of players. Each player has m (n-by-n) grids, where m is the number of players. (So there are m^2 grids all together.) I suggest an n of at least 12. On one of their grids each player secretly draws a closed non-self- intersecting curve. (The curve is bounded within the n-by-n grid.) Each player's curve does not go through any intersections of the grid- lines. Next, on one of each of the other player's blank n-by-n grids each player copies his/her curve over. The copies of each curve must go through the same respective squares of each grid as the original curve did. So, there are m copies each of m curves, each player in possession of one copy of each curve. Next, secretly and simultaneously, each player fills in the squares each curve goes through on any particular grid with 1,2,3,...., the integers placed in order and next to each other along the curve. The numbers can start anywhere along a curve, and can go either clockwise or counterclockwise. Next, each player secretly fills in the squares within each curve's interior with 1,2,3,..., the numbers placed in order, each number placed in any empty interior square such that all other numbers (including possibly numbers along the curve) above, right of, left of, or below the number are coprime to that number. (Any pair of adjacent numbers that are both in squares a curve passes through don't have to be coprime. Only interior numbers have to be coprime to adjacent numbers along the curve, or to adjacent numbers that are also on the curve's interior.) Players continue to fill the interior of each curve with numbers until the players can't fill in any more numbers under the rules. When each player has filled in each curve as far as they can, the score for each player is the sum of the top numbers in the interior squares of each of the m curves the player filled (partially) in. Highest score wins. Players may check their opponents' grids after the game is over to make sure that all applicable pairs of adjacent numbers are actually coprime. If a player made a mistake, that player automatically loses the game. Leroy Quet === Subject: Re: Math outside of Time posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) >>The Absolute beginning is an empty timeless hypersphere. One point >>becomes infinity of points of which then expand 4 dimensionally. After >>inflation of the hypersphere Time and Energy are created on its 3d >>surface all at once. >The interior of the hypersphere is the timeless space element that the >>math of growing infinities of the infinitely small applies to. It is >>math for the interior of the hypersphere or the 4th dimension of >>space. >Mitch Raemsch Why are you repeating their idiocy? None of that is true. Neither you nor BURT can say with any certainty whether > or not time exists extrauniversally.- Hide quoted text - - Show quoted text - xxein: becomes, then, growing??? Hmm. Time must not exist?- Hide quoted text - - Show quoted text - Empty hypersphere. Mitch Raemsch === Subject: Re: Infinities of different sizes posting-account=QBdF7AoAAAB1rnNtnpKM2ANKD1Zv7ChU Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > Are there smaller infinities inbetween the larger ones? Mitch Raemsch First answer me this: How many podges does it make to have a hodge? === Subject: Why aren't my posts appearing? posting-account=QBdF7AoAAAB1rnNtnpKM2ANKD1Zv7ChU Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) Testing testing 1 2 3... === Subject: Re: Why aren't my posts appearing? posting-account=cvz5-QoAAABVNzogw177Plx_25TguPUZ CLR 1.0.3705; .NET CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1; MS-RTC LM 8),gzip(gfe),gzip(gfe) > Testing testing 1 2 3... You are fading in and out. Try defibulating your X-band inter- modulator. - MO === Subject: Re: is there exist 3-rd degree real polynomial contains all Fibonacci > > Gerry Myerson a í©crit : > , > > > >> How about the sequence S = {1, 2, 4, 8, 16, ...} of powers of 2. Does there exist a polynomial f in Z[x], with deg(f) > 1, such that S >> is a subset of f(Z)? > > I think there's a theorem to the effect that if f is as above > then the biggest prime dividing f(n) goes to infinity with n. > > f(x) - f(y) is divisible by x-y ; could this help? > > I think it's a lot deeper than that; I think it came out of > work done on extending Baker's stuff with linear forms > in logarithms. > > I don't think you've quoted the theorem right. Note that f(x) = x^2 > would work for the even powers of 2. The biggest prime dividing f(2^n) > is 2. Yes, there have to be more hypotheses. There's an old theorem that y^2 = cubic in x can only have finitely many integer solutions; the cubic must give rise to an elliptic curve, e.g. it can't be of the form (x + a)^3, but that's not very restrictive. So such a cubic can't take on all the values 1, 4, 16, 64, 256, ..., and a fortiori can't take on all the powers of 2. I think the theorem extends to y^2 = f(x) with the restriction that it give a curve of genus at least one, which is a very mild restriction. The Baker-style results I had in mind are like this one of Shorey and Tijdeman, Compositio Math 1976; let f(x) be a polynomial with integer coefficients and at least two distinct zeros, let B be an arbitrary positive number; then for any x > e^(e^e) there is an effectively computable positive number r depending only on B and f such that the greatest prime factor of f(x) exceeds r log log x. Also there's a paper of Schinzel and Tijdeman, Acta Arithmetica 1976, proving that a polynomial with integer coefficients and at least two distinct roots can assume only finitely many perfect powers. So that much was known in 1976. I suspect more is known now. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: inverse AGM <14938393.1225105469514.JavaMail.jakarta@nitrogen.mathforum.org>, > Ah, perhaps now I see. Your question is not what > I thought. You appear not to be using AGM in its > usual sense, but to be using only a special case. > > To see if I understand, can your question be phrased > more accurately like this: > > * Define f(x) = agm(1,x), where agm is the usual > arithmetic-geometric mean. > * Can you compute an inverse of f? > > Is that it? Are you asking specifically about the > case when one of the parameters is 1? I hesitate to speak for tommy, but that is certainly how I interpreted the original question. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: inverse AGM >> You appear not to be using AGM in its usual >> sense, but to be using only a special case. > ... that is certainly how I interpreted the > original question. Fair enough. He communicated effectively with you, but not with me. I looked up the definitions and was left confused, hence my questions. This serves to remind yet again that precision in communication can avoid confusion, a lesson I take from this exchange. === Subject: Re: Modern Aether Properties posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) oops; Bucky did make the bubble metaphor, but he always hankered to triangling for secondpowering, when obviously it should be (superficially) tetrahedroning; however, this is based upon an overbearing bathos for The Tetrahedron, whichis a great point of pedagogy, but clearly wrong for physics: sphering! > certainly, it pertains as well to the sphere, so that > we could say, we are sphering the variable, or > soapbubbling it, which is the metaphor > that Bucky useth'd in S . to wit, > the c-sphered -- superficially, not volumetrically -- > is just the area of the sphere mod expansion > of the wavefront. thus: good luck; firstly, define what you mean by proof. > All these bells and whistles of new instruments added onto the > StockMarket are merely Ponzi Pyramid schemes that cheats the people > who do only buy and sell in cash. thus: a) the free market is a visciously oligarchical & British lie, perpetrated primarily through Adam Smith's second (1776) hoax, The Wealth of Nations , ever since taken as The Holy Economist writ -- you should see the 2-vol. *abridged* Princeton edition; b) don't nuke any more fresh-faced, volunteer-manned e-math-journals; c) they mean, well! > All scientists should learn to live in the free market. Do their thus: yeah, minus one; the smiley was implicated. see, there is a practice in giving math tests, where no answer is zero points, and a wrong answer is minus one; actually, that's for multipe choice -- NOTA!... however, you'd get extra credit for three wrong choices, perhaps. hey, but I didn't check, since I prefer some sort of rationale to be stated, before the incipiently slippery slope of Fermat's so-called last theorem, which was probaably one of his very first. after all, NO OTHER MISTAKES are known. (like, why do you care, that we should care ?-) thus quoth: With three different versions, one is bound to be correct. > You forgot the smiley thus: here is an etymology of THE PHOTON: photoelectrical effect hits measuring device, ballistically imparting its momentum of one quantum (per Planck h-bar) at some frequency; the bug is a feature, proven statistically by Bose (lies, polls, statistics !-) > where did you ever refute Uncle Al's experiments? --USA out of Darfur Cruizade! http://larouchepub.com/other/2008/3537willl zardari survive.html --ROTC, your summer vacation in the Sahara Desert ( S u d a n ) ; presage the Draft for your middleschool class of '12 -- brought to you by Allstate (tm) and Oxford U.Press! http://larouchepub.com/pr/2008/080813moloch brown.html http://wlym.com --Wikipedia deletes notice of nullification of preclearance rule of Voting Rights Act in LaRouche v. Fowler, March 27, 2000; is the VRAo1965 a dead letter? http://en.wikipedia.org/wiki/Voting Rights Act#Pre-Clearance 2 === Subject: =?windows-1252?Q?Re=3A_E=3Dmc=28squared=29_=3D_E=3Dmc=28circled=29=2C_and_ce l estial_orb?= =?windows-1252?Q?its=2C_may_indeed_be=2C_=93Natures_Squaring_of_the_Circle=9 4 =2E?= posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) corrigendum: oops; Bucky did make the bubble metaphor, but he always hankered to triangling for secondpowering, when obviously it should be (superficially) tetrahedroning; however, this is based upon an overbearing bathos for The Tetrahedron, which is a great point of pedagogy, but clearly wrong for physics: sphering! > certainly, it pertains as well to the sphere, so that > we could say, we are sphering the variable, or > soapbubbling it, which is the metaphor > that Bucky useth'd in _S_. to wit, > the c-sphered -- superficially, not volumetrically -- > is just the area of the sphere mod expansion > of the wavefront. thus: good luck; firstly, define what you mean by proof. > All these bells and whistles of new instruments added onto the > StockMarket are merely Ponzi Pyramid schemes that cheats the people > who do only buy and sell in cash. thus: a) the free market is a visciously oligarchical & British lie, perpetrated primarily through Adam Smith's second (1776) hoax, _The Wealth of Nations_, ever since taken as The Holy Economist writ -- you should see the 2-vol. *abridged* Princeton edition; b) don't nuke any more fresh-faced, volunteer-manned e-math-journals; c) they mean, well! > All scientists should learn to live in the free market. Do their thus: yeah, minus one; the smiley was implicated. see, there is a practice in giving math tests, where no answer is zero points, and a wrong answer is minus one; actually, that's for multipe choice -- NOTA!... however, you'd get extra credit for three wrong choices, perhaps. hey, but I didn't check, since I prefer some sort of rationale to be stated, before the incipiently slippery slope of Fermat's so-called last theorem, which was probaably one of his very first. after all, NO OTHER MISTAKES are known. (like, why do you care, that we should care ?-) thus quoth: With three different versions, one is bound to be correct. > You forgot the smiley thus: here is an etymology of THE PHOTON: photoelectrical effect hits measuring device, ballistically imparting its momentum of one quantum (per Planck h-bar) at some frequency; the bug is a feature, proven statistically by Bose (lies, polls, statistics !-) > where did you ever refute Uncle Al's experiments? --USA out of Darfur Cruizade! http://larouchepub.com/other/2008/3537willl_zardari_survive.html --ROTC, your summer vacation in the Sahara Desert ( S u d a n ) ; presage the Draft for your middleschool class of '12 -- brought to you by Allstate (tm) and Oxford U.Press! http://larouchepub.com/pr/2008/080813moloch_brown.html http://wlym.com --Wikipedia deletes notice of nullification of preclearance rule of Voting Rights Act in LaRouche v. Fowler, March 27, 2000; is the VRAo1965 a dead letter? http://en.wikipedia.org/wiki/Voting_Rights_Act#Pre-Clearance_2 === Subject: Re: example of R not integral domain, R[x] P.I.D. posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008092318 Fedora/3.0.2-1.fc9 Firefox/3.0.2,gzip(gfe),gzip(gfe) On Oct 25, 10:25pm, Jack Schmidt Is there a ring R other than a field such that > R[x] is a PID? > Previous discussion ruled out integral domains, > and so this thread is presumably about more general > rings. > ] You mean PIR (principal ideal ring ) since > zero-divisors of R persist in R[x]. The answer > is no, with the same proof as when R is a domain. > Why would you think otherwise? I think if you only want two-sided ideals to be > principal there are more such rings. If R is the > ring of two by two matrices over a field, I think > R[x] has only principal two-sided ideals. The k-algebra (M 2(k))[X] is naturally isomorphic to M 2(k[X]), which in turn is Morita equivalent (as a k-algebra) to k[X]. Now, it is easy to see that the property of a k-algebra A that every bilateral ideal of A is isomorphic, as an A-bimodule, to A itself is invariant under Morita equivalence. Therefore the algebra (M 2(k))[X] has that property, and your claim follows from this. -- m > Of course for one-sided ideals, I think R must be > a skew field. Rings in which every left ideal is > free are well-studied and interesting. They are > the typical generalizations of PIDs to the noncomm > setting. === Subject: Re: maximum of function abs(ai - aj) > > I have a function f(i,j) = | a(i) - a(j) | + | b(i) - b(j)| + |c(i) - > c(j)| + | d(i) - d(j) | for (1<=j<=i<=N) where a ,b,c,d are sequences > of real numbers with length 10^5. The | x| is absolute value of x. > What is the maximum of f(i,j) ? > What is the minimum of f(i,j) ? > How do we interpret this geometrically ? > > Had it been only 1 sequence ie max of f(i,j)=|a(i) - a(j)| i could > sort and find the a(max) - a(min) which would be the max of f(i,j) Think of the 4-tuples [a(i), b(i), c(i), d(i)] as points in R^4. So you have a set of N points. The minimum is obviously 0. The maximum is the diameter of the set, where we use the taxicab or Manhattan metric. Note that f(i,j) = e_1 (a(i) - a(j)) + e_2 (b(i) - b(j)) + e_3 (c(i)-c(j)) + e_4 (d(i) - d(j)) for some [e_1, ..., e_4] in {+1,-1}^4. For each such e, let A(e) = min_i (e_1 a(i) + ... + e_4 d(i)) and B(e) = max_i (e_1 a(i) + ... + e_4 d(i)), and the diameter is the maximum over all e of B(e) - A(e). -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Z[sqrt -2] posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008092318 Fedora/3.0.2-1.fc9 Firefox/3.0.2,gzip(gfe),gzip(gfe) On Oct 26, 8:03am, gaya.pa...@gmail.com Try harder? -- m === Subject: Re: Z[sqrt -2] posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008092318 Fedora/3.0.2-1.fc9 Firefox/3.0.2,gzip(gfe),gzip(gfe) > Z[sqrt(-2)] Is this a Euclidean domain? Yes it is. > If so, is f(a+ib) = a^2 + b^2 the euclidean function? What have you tried doing to see if this function works? -- m === Subject: Re: Z[sqrt -2] posting-account=JTrNRgoAAABuGIzwXYnlAgkRTDpv8_sI 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) On Oct 27, 6:40pm, Mariano Su.87rez-Alvarez Z[sqrt(-2)] Is this a Euclidean domain? Yes it is. If so, is f(a+ib) = a^2 + b^2 the euclidean function? What have you tried doing to see if this function works? > Using the modulus squared function (over complex field), it is a Euclidean function. I mean this function (where f is modulus squared frunction from complex): let u = sqrt(-2) f(x+yu) = f(x+i(sqrt2)y) = x^2 + 2y^2 Using the multiplicative nature of modulus squared (or equivalently, f), can show that it is a Euclidean function (boils down to mimicing the well known proof for Gaussian integers). But in general, this method can't be extended to anything more than -1, -2. So there must be a different way to get -3, -5, and -7. I will try -3 case some other time (going to put this on TO DO LIST). === Subject: Re: Not a popularity contest > As I rant and rave, continually antagonize my audience, and talk about > cutting public funding for basic research, or well, any research at > all, I can imagine mathematicians smiling to themselves. No way > people will ever listen to such nuttiness and they may smugly feel > they can rest secure that I'll never be taken seriously--except I DO > have major mathematical research results. you gotta no results, you gotta no respect, cause you gotta earn it. you sound like that asshole House on the idiot tube. So their satisfaction in my mind is a dismissal of their own field. your mind is a dismissal of rationality. These modern mathematicians clearly despise mathematics. you dispise yourself, because you are not smart enough to learn even simple math. As if, as if a really unlikable guy could possibly get listened to > just because he has mega mathematical discoveries of the type > that come around once in a century, or more. No way, right? Not in > our popularity dominated world! Yes, you are a moron. Don't you know the lesson of MTV? Of The HIlls? lesson of The Partirdge Family. Popularity is all that matters, right? Substance means nothing, > right? That is what YOU DO. And if you believe them you can sit back and relax, but if they're > wrong then you know that an increasingly angry and frustrated major > discoverer is figuring out ways to convince policymakers someday that > sharp reduction in public funding for basic research is the ONLY way > to bring back integrity to the system and insure that people are > actually doing real research of some value. People doing valuable > research will find a way. The cons will leave for easy money > elsewhere. You are a squeek in the wilderness. A unseen leaf, now fallen and eaten by worms. And besides, I figured out my mathematical results without public > funding and with mathematicians from around the world mostly making > fun of me, the nasty bastards. HS Algebra is difficult isn't it ? > IF I can do it, others can. but you can't, and they can. So what good are you academics anyway? Professors teach. Discoverers do. Wrong, discoverers discover. Albert Einstein of course worked in a patent office. What if society > missed the clue? The reality that maybe academia INHIBITS major > discovery. AE *hates* you. What if we got it wrong? What if our universities around the world > while playing a crucial role of teaching are the last places to go to > for innovation? What if ? What if ? blah... blah.... black sheep What if our world has created a massive system for BLOCKING major > advances, where now even the would-be Einsteins of today cannot get > past the professors of today? We block people that are under preformers, like you. The message I'm giving to you if you haven't picked it up, if you are > an academic, is that you are probably a failure at your job, but > surrounded by other failures, you all call each other successes, and > condemn the human race to the muddled mess you call research. JSH and his muddled mess he calls research > Publish or perish. Yeah, you publish, and let the human race perish > for people who are second-rate researchers at best and outright frauds > in many cases in a system that does not work. you have failed to publish, therefore you parish. And my story proves that beyond a doubt. You block major mathematical > discoveries because you are incompetent boobs living off the public in > a white collar welfare system and you legitimately see me as a threat > to your free money. And, you JSH will never get a single dime. Rather you will be sued for all the money you have or ever will have, for threatening people and being a insignificant tick on the side of humanity. Yeah, you con artists. I am a threat to your free money. And you are > a threat to human civilization itself. I Piss on you........sssssssssssssss..ssssssss.....ssss....ss...ss......s.....s.. You are the snakes in the grass. (dude, take yr meds!) > James Harris === Subject: Re: Discrete math with number 5. >mina_world a ňcrit : >> Hello teacher~ >> >> 1, 2, 3, ... , 100000 >> >> Find the number of 5. >> >> -------------------------------------------------------- >> (5)(*)(*)(*)(*) ==> 10^4 >> (*)(5)(*)(*)(*) ==> 10^4 >> (*)(*)(5)(*)(*) ==> 10^4 >> (*)(*)(*)(5)(*) ==> 10^4 >> (*)(*)(*)(*)(5) ==> 10^4 >> >> so, answer is 5*(10^4). How about (5)(5)(*)(*)(*) etc. ? How about it? There are two fives in 55000. >>One of them got counted in the 5**** category >>and the other got counted in the *5*** category. Didn't 55000 get counted in both categories, so counted twice, and >then counted some more in the other categories? There 10^4 5's in the first digit, 10^4 5's in the second digit, 10^4 5's in the third digit, 10^4 5's in the fourth digit and 10^4 5's in the fifth digit. That looks like 5*10^4 5's to me. If we want to check this looking at it a different way, The count of numbers with k 5's is C(5,k) 9^{5-k}. Therefore, the count of 5's is 5 --- 5-k > k C(5,k) 9 --- k=0 5 --- 5-k = > 5 C(4,k-1) 9 --- k=0 5 --- 5-k = > 5 C(4,5-k) 9 --- k=0 4 = 5 (9 + 1) 4 = 5 10 To count how many numbers have a 5 in them, notice that there are 9^5 numbers without them. Thus, there are 10^5 - 9^5 = 40951 numbers with a 5 in them. We could of course check this in the same manner as above. Rob Johnson take out the trash before replying === Subject: Re: Discrete math with number 5. >mina_world a ?crit : > Hello teacher~ > > 1, 2, 3, ... , 100000 > > Find the number of 5. > > -------------------------------------------------------- > (5)(*)(*)(*)(*) ==> 10^4 > (*)(5)(*)(*)(*) ==> 10^4 > (*)(*)(5)(*)(*) ==> 10^4 > (*)(*)(*)(5)(*) ==> 10^4 > (*)(*)(*)(*)(5) ==> 10^4 > > so, answer is 5*(10^4). How about (5)(5)(*)(*)(*) etc. ? How about it? There are two fives in 55000. >One of them got counted in the 5**** category >and the other got counted in the *5*** category. Didn't 55000 get counted in both categories, so counted twice, and >>then counted some more in the other categories? There 10^4 5's in the first digit, 10^4 5's in the second digit, >10^4 5's in the third digit, 10^4 5's in the fourth digit and >10^4 5's in the fifth digit. That looks like 5*10^4 5's to me. If we want to check this looking at it a different way, The count of >numbers with k 5's is C(5,k) 9^{5-k}. Therefore, the count of 5's is 5 > --- 5-k > > k C(5,k) 9 > --- > k=0 5 > --- 5-k > = > 5 C(4,k-1) 9 > --- > k=0 5 > --- 5-k > = > 5 C(4,5-k) 9 > --- > k=0 4 > = 5 (9 + 1) 4 > = 5 10 To count how many numbers have a 5 in them, notice that there are >9^5 numbers without them. Thus, there are 10^5 - 9^5 = 40951 numbers >with a 5 in them. We could of course check this in the same manner as above. Rob Johnson Integrate cos^2(t/2)???? >I used a trig substitution to make it become the integral of 1/2+1/2 >cos(t) dt...which is 1/2t+1/2 sin (t) +C...however, the answer I'm >supposed to get is 1/2 t +cos(t/2) sin (t/2)...how is 1/2 sin (t) >equal to cos(t/2) sin (t/2)? Use sin(2x) = 2 sin(x) cos(x) with x = t/2. --Lynn === Subject: Re: --- --- --- Solutions of an equation <7s47g4l3tlc0fgqu7i3m08l6k958tru6m2@4ax.com> posting-account=iJfBPwgAAACaBAH7DreA6VfAOYvL5VDz Trident/4.0; .NET CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; IEMB3; IEMB3),gzip(gfe),gzip(gfe) >Assertion: >The following equation (1) cannot be satisfied if all of x, y, z are >>integers each > 1. > x^2 + 8y^2 = z^2 (1) Your assertion is false. Try, x=2, y=2, x=6. Another solution: x=41, y=15, z=50. Typo: In the above line, it should be z = 59. >My approach: x, (sqrt (8)y), z represent the three sides of a >>Pythegorian triangle. Given y is an integer sqrt(8)y can never be an >>integer. Right -- since y is a positive integer, sqrt(8)y is irrational, hence >is not an integer. So what? >Therefore, it can be argued that (1) has no integer solutions. No, the logic is flawed -- you haven't shown a contradiction. Moreover, since solutions actually exist, a contradiction is not >achievable. quasi- Hide quoted text - - Show quoted text - **** A general solution of x^2 + 8y^2 = z^2 (1) can be obtained as (2*sqrt(2)y)^2 + (y^2 - 2)^2 = (y^2 + 2)^2 (2) 4 (3) Apparently, (3) does not satisfy all the above examples. ***** === Subject: Re: --- --- --- Solutions of an equation >> x^2 + 8y^2 = z^2 (1) Try, x=2, y=2, x=6. [presumably, that should be z = 6] >Another solution: x=41, y=15, z=50. Typo: In the above line, it should be z = 59. > **** A general solution of x^2 + 8y^2 = z^2 (1) can be obtained as > (2*sqrt(2)y)^2 + (y^2 - 2)^2 = (y^2 + 2)^2 (2) > 4 (3) > Apparently, (3) does not satisfy all the above examples. > ***** 8 y^2 = z^2 - x^2 = (z - x) (z + x) z - x = 2 m^2, z + x = 4 n^2 x = 2 n^2 - m^2, y = m n, z = 2 n^2 + m^2 gives a two-parameter family of solutions. There may be other, similar, families, based on small variations of the calculations above. m = 3, n = 5 gives x = 41, y = 15, z = 59. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: --- --- --- Solutions of an equation x^2 + 8y^2 = z^2 (1) >> Try, x=2, y=2, z=6. >> Another solution: x=41, y=15, z=59. >> >> **** A general solution of x^2 + 8y^2 = z^2 (1) can be obtained as >> (2*sqrt(2)y)^2 + (y^2 - 2)^2 = (y^2 + 2)^2 (2) >> z - x = 4 (3) >> Apparently, (3) does not satisfy all the above examples. >> ***** Still insist with your quadratic equation in Real numbers ? As already said, 2*sqrt(2)y is not an integer for y integer ! However, you get a few solutions here (see below, y=m, n=1). 8 y^2 = z^2 - x^2 = (z - x) (z + x) z - x = 2 m^2, z + x = 4 n^2 x = 2 n^2 - m^2, y = m n, z = 2 n^2 + m^2 > gives a two-parameter family of solutions. There may > be other, similar, families, based on small variations > of the calculations above. m = 3, n = 5 gives x = 41, y = 15, z = 59. That's the process I already gave here two days ago... with just a variation. Your formula doesn't give all of them, in the same manner as x = p^2 - q^2, y = 2pq, z = p^2 + q^2 does not give all pythagorian triples... For instance z = 15 (x=5, y=5) is not given by your formula as 15 can't be a z = 2*p^2 + m^2 number (because of prime factor 5 = 5 modulo 8). This solution being trivially 5*(1,1,3). All are given here by : x = k*abs(2n^2 - m^2), y = k*m*n, z = k*(2n^2 + m^2) The abs allows to consider not only > z - x = 2 m^2, z + x = 4 n^2 but also the case z - x = 4 m^2, z + x = 2 n^2 The cases z - x = 8 m^2, z + x = n^2 and z - x = m^2, z + x = 8 n^2 should also be considered, but as z - x and z + x have same parity, this implies n resp. m even, hence a common factor 2, and these cases are included in the general z - x = 2k m^2, z + x = 4k n^2 resp. z - x = 4k m^2, z + x = 2k n^2 Of course if you consider only *primitive* solutions, that is adding a GCD(x,y,z)=1 condition to the original question, k=1 and your formula suffice (gives all primitive solutions). But it also then gives non primitive ones ! You should add GCD(m,n) = 1 and m odd to get all primitive solutions, and only these. BTW the (2,2,6) solution is not primitive, it is 2*(1,1,3), but as the OP asked for x,y,z > 1, it is the smallest one. The smallest primitive is for m = 1, n = 2 x = 7, y = 2, z = 9 (49 + 32 = 81) The smallest z which gives several primitive solutions is 2601 = 51^2 = 49^2 + 8*5^2 = 47^2 + 8*7^2 These advanced results come from studies of the equation written as x^2 + 2*(2y)^2 = z^2 that is X^2 + 2*Y^2 = Z^2, restricted to Y even. -- Philippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) === Subject: Re: --- --- --- Solutions of an equation <7s47g4l3tlc0fgqu7i3m08l6k958tru6m2@4ax.com> posting-account=iJfBPwgAAACaBAH7DreA6VfAOYvL5VDz Trident/4.0; .NET CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; IEMB3; IEMB3),gzip(gfe),gzip(gfe) >Assertion: >The following equation (1) cannot be satisfied if all of x, y, z are >>integers each > 1. > x^2 + 8y^2 = z^2 (1) Your assertion is false. Try, x=2, y=2, x=6. Another solution: x=41, y=15, z=50. Typo: In the above line, it should be z = 59. >My approach: x, (sqrt (8)y), z represent the three sides of a >>Pythegorian triangle. Given y is an integer sqrt(8)y can never be an >>integer. Right -- since y is a positive integer, sqrt(8)y is irrational, hence >is not an integer. So what? >Therefore, it can be argued that (1) has no integer solutions. No, the logic is flawed -- you haven't shown a contradiction. Moreover, since solutions actually exist, a contradiction is not >achievable. quasi- Hide quoted text - - Show quoted text - + 8y^2 = z^2 (1) as === Subject: Part of gravity motion is straight posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) Riemannian space curvature leads to curved or Riemannian motion. Circular orbits, swivelling eliptical orbits and one time parabolas are gravity motion curvature. Also if you do not follow the curve you can move straight as in drop: straight toward center of mass. Mitch Raemsch === Subject: Re: Show that if lim a_n = -00, then lim 1/a_n = 0 >> and thus we have the desired |1/a_n| < h and >> for all n > N > > That reminds me of an EE student I knew when I was a student at Caltech. > He just cared about the result, not the proof, in any math class he > took. When he did math homework that asked for a proof, his method was > to write out the problem statement, and then skip down and write down > the conclusion. He'd then start working forward from the problem > statement, and backward from the conclusion. Where the downward stream > of forward reasoning and the upward stream of backward reasoning > collided, he'd connect them with we see that or thus we have or > something like that. Really, I am just trying to practice my mathematical writing. I suppose such formal language might look a little silly in a half baked proof posted to usenet though. ;) > You need to fill in that we have the desired part. For a given choice > of M (1/h is the one you've picked), you have some N (dependent on M) > such that a_n < -M for all n > M. Then |a_n| > M = 1/h, and |1/a_n| < > h. That M then works as L for the 0 limit, and you are done. === Subject: suggestions on how to prove the limit problem posting-account=twH9PwoAAABVFxHWRVsv2WYMPEStgjmE AppleWebKit/525.13 (KHTML, like Gecko) Chrome/0.2.149.30 Safari/525.13,gzip(gfe),gzip(gfe) please give me hints at how to solve the following (giving a formal proof) how to prove that: lim {n-> infinity} (sin n)/ n = 0 thank you all === Subject: limit proof HELP posting-account=twH9PwoAAABVFxHWRVsv2WYMPEStgjmE AppleWebKit/525.13 (KHTML, like Gecko) Chrome/0.2.149.30 Safari/525.13,gzip(gfe),gzip(gfe) how to prove that: lim {n-> infinity} (sin n)/ n = 0 please give advice (need to construct a formal proof) === Subject: Re: limit proof HELP posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > how to prove that: > lim {n-> infinity} (sin n)/ n = 0 please give advice (need to construct a formal proof) http://answers.yahoo.com/question/index?qid=20080623061638AAyILy6 === Subject: Re: limit proof HELP posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008072820 Firefox/3.0.1,gzip(gfe),gzip(gfe) > how to prove that: > lim {n-> infinity} (sin n)/ n = 0 please give advice (need to construct a formal proof) Given n>0, how big can sin(n)/n be at most? === Subject: Re: limit proof HELP > how to prove that: > lim {n-> infinity} (sin n)/ n = 0 > > please give advice (need to construct a formal proof) Hint: Use the Squeeze Theorem. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: limit proof HELP posting-account=twH9PwoAAABVFxHWRVsv2WYMPEStgjmE AppleWebKit/525.13 (KHTML, like Gecko) Chrome/0.2.149.30 Safari/525.13,gzip(gfe),gzip(gfe) On Oct 28, 12:46am, Robert Israel > how to prove that: > lim {n-> infinity} (sin n)/ n = 0 please give advice (need to construct a formal proof) Hint: Use the Squeeze Theorem. > -- > Robert Israel isr...@math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada I appreciate it === Subject: Re: limit proof HELP > how to prove that: > lim {n-> infinity} (sin n)/ n = 0 > > please give advice (need to construct a formal proof) So what do you need to show? === Subject: simple extension of subfield posting-account=ks93ywoAAAD0utXjkmfCRXzosT3mNG-2 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Hello Sir, Help pls 1) Every extension of prime degree is a simple extension. 2) Let F= Z_2(x^2, y^2), G=Z_2(x,y). Show [G : F] = 4 and G not simple extension of F. === Subject: Re: simple extension of subfield days. My association with the Department is that of an alumnus. >Hello Sir, Help pls >1) Every extension of prime degree is a simple extension. Hint: Dedekind's Product Formula. >2) >Let F= Z_2(x^2, y^2), G=Z_2(x,y). Show [G : F] = 4 and G not simple >extension of F. Find K, F < K < G, with [G:K]=2 and [K:F]=2 (easy). As to showing that it is not a simple extension, how about showing that any element of G satisfies a polynomial of degree 2 over G? -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Discrete math with subset. Hello teacher~ A = {1, 2, 3, ... , n} B is a subset with |B| = k. (n(B) = k) B does not contain adjacent numbers. (Namely, B = {3, 5, 6, 9} (no!)) Find the number of possible subsets as B. ------------------------------------------------------- Let B = {x_1, x_2, ... , x_k} so, 1 <= x_1 < x_2 < ... < x_k <= n. and x_1 - 1 >= 0 x_2 - x_1 >= 2 x_3 - x_2 >= 2 ... x_k - x_(k-1) >= 2 n - x_k >= 0 Let y_1 = x_1 - 1 y_2 = x_2 - x_1 y_3 = x_3 - x_2 ... y_k = x_k - x_(k-1) y_(k+1) = n - x_k and then y_1 + y_2 + ... + y_k + y_(k+1) = [x_1 - 1] + [x_2 - x_1] + [x_3 - x_2] + ... + [n - x_k] = n - 1 so, y_1 + y_2 + y_3 + ... + y_k + y_(k+1) = n - 1 y_1 >= 0 y_2, y_3, ... , y_k >= 2 y_(k+1) >= 0 so, y_1 + z_2 + z_3 + ... + z_k + y_(k+1) = (n-1) - 2(k-1) y_1 >= 0 z_2, y_3, ... , z_k >= 0 y_(k+1) >= 0 so, y_1 + z_2 + z_3 + ... + z_k + y_(k+1) = n - 2k + 1 so, answer is (k+1)H(n-2k+1) = (n-k+1)C(k) === Subject: Re: Discrete math with subset. posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008072820 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Hello teacher~ A = {1, 2, 3, ... , n} B is a subset with |B| = k. (n(B) = k) B does not contain adjacent numbers. > (Namely, B = {3, 5, 6, 9} (no!)) Find the number of possible subsets as B. I'd denote the number of such subsets as f(n,k). Clearly, forall n>=0: f(n,0) = 1 as we only have the empty set. Also, forall k>=1: f(2k-1,k) = 1 as we only have B={1,3,5,...,2k-1} And we have the recursion f(n,k) = f(n-1,k) + f(n-2,k-1) whenever n>=2, k>=1 (B is a subset of {1,...,n-1} or B is {n} union a smaller subset of {1,...,n-2}) Writing g(n,k) := f(n+k-1,k), we see that the recursion amounts to g(n,k) = f(n+k-1,k) = f(n+k-2,k) + f(n+k-3,k-1) = g(n-1,k) + g(n-1,k-1) and the initial conditions to g(n,0) = f(n,0) = 1 and g(n,n) = f(2n-1,n) = 1 It follows that g(n,k) = nCk, hence f(n,k) = g(n-k+1,k) = (n-k+1)C(k) > ------------------------------------------------------- > Let B = {x_1, x_2, ... , x_k} so, 1 <= x_1 < x_2 < ... < x_k <= n. > and > x_1 - 1 >= 0 > x_2 - x_1 >= 2 > x_3 - x_2 >= 2 > ... > x_k - x_(k-1) >= 2 > n - x_k >= 0 Let > y_1 = x_1 - 1 > y_2 = x_2 - x_1 > y_3 = x_3 - x_2 > ... > y_k = x_k - x_(k-1) > y_(k+1) = n - x_k and then > y_1 + y_2 + ... + y_k + y_(k+1) > = [x_1 - 1] + [x_2 - x_1] + [x_3 - x_2] + ... + [n - x_k] > = n - 1 so, y_1 + y_2 + y_3 + ... + y_k + y_(k+1) = n - 1 > y_1 >= 0 > y_2, y_3, ... , y_k >= 2 > y_(k+1) >= 0 so, y_1 + z_2 + z_3 + ... + z_k + y_(k+1) = (n-1) - 2(k-1) > y_1 >= 0 > z_2, y_3, ... , z_k >= 0 > y_(k+1) >= 0 so, y_1 + z_2 + z_3 + ... + z_k + y_(k+1) = n - 2k + 1 so, answer is (k+1)H(n-2k+1) = (n-k+1)C(k) === Subject: Re: Discrete math with subset. >A = {1, 2, 3, ... , n} B is a subset with |B| = k. (n(B) = k) B does not contain adjacent numbers. >(Namely, B = {3, 5, 6, 9} (no!)) Find the number of possible subsets as B. Let me suggest a different way to look at this problem. Consider bit strings of length n with k bits set (equal to 1) and n-k bits clear (equal to 0). Each bit that is set represents an element of B. We want to count the number of n-bit bit strings with k bits set in which no adjacent bits are set. We can break these down into two cases, those with the terminial bit set and those with the terminal bit clear. The bit strings with the terminal bit clear can be represented by pairing each set bit with a clear bit, like so n=16, k=4: 00(10)0(10)000(10)(10)00 The count of these is C(n-k,k) The bit strings with the terminal bit set are bit strings, like those above, with one fewer total bits and one fewer bits set, followed by a set bit, like so n=16, k=4: 00(10)00000(10)(10)00(1) ___________________/ n=15, k=3 The count of these is C(n-k,k-1). Thus, the total number of such bit strings is C(n-k,k) + C(n-k,k-1) = C(n-k+1,k). Rob Johnson take out the trash before replying === Subject: Re: Discrete math with subset. >>A = {1, 2, 3, ... , n} B is a subset with |B| = k. (n(B) = k) B does not contain adjacent numbers. >>(Namely, B = {3, 5, 6, 9} (no!)) Find the number of possible subsets as B. Let me suggest a different way to look at this problem. Consider bit strings of length n with k bits set (equal to 1) and n-k > bits clear (equal to 0). Each bit that is set represents an element > of B. We want to count the number of n-bit bit strings with k bits > set in which no adjacent bits are set. We can break these down into > two cases, those with the terminial bit set and those with the > terminal bit clear. The bit strings with the terminal bit clear can > be represented by pairing each set bit with a clear bit, like so n=16, k=4: 00(10)0(10)000(10)(10)00 The count of these is C(n-k,k) The bit strings with the terminal bit set are bit strings, like those > above, with one fewer total bits and one fewer bits set, followed by > a set bit, like so n=16, k=4: 00(10)00000(10)(10)00(1) > ___________________/ > n=15, k=3 The count of these is C(n-k,k-1). Thus, the total number of such > bit strings is C(n-k,k) + C(n-k,k-1) = C(n-k+1,k). Yes, good remodeling. More simplier... n = 5, k = 2 as (ooooo) ==> (oxoxo), (xooox), etc... Fix the two x. ==> (*)x(*)x(*) Put o between two x. ==> (*)x(*)ox(*) for not adjacent. Put in (*) remainder two o ==> 3H2 = 4C2. so, we can expand generally. === Subject: Re: How many infinities are there? posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > There are smaller infinities inbetween larger infinities. Such as? === Subject: Re: How many infinities are there? > >>There are smaller infinities inbetween larger infinities. > > Such as? Q: How many infinities are there? A: Just _one_: potential infinity. Han de Bruijn === Subject: Re: JSH: Moving to real fun posting-account=3WPJYgoAAAA55VjhzK9i07RN8h8u8eEs Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) Oh yeah, part of your job is recruitment. I need a lot of people so I > can fire a lot of them, to whittle down to a team of 4. Or 3. Or > maybe 12. I don't know, but I'm sure I'll figure it out. > Let's see, I guess the ad would go something like this... Wanted: Super-geniuses to work with mega-super-genius. Pay: Nothing Benefits: None (except you'd be saving the world and be in the presence of greatness) Track record of recruiter : No publications, zero revenue. And you wonder if you're living in a fantasy world? Go crawl back into your couch-like womb and watch another episode of House, dumbo. M === Subject: Re: JSH: Moving to real fun posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) This Dr. House thing is started to worry me though, is it possible > that I really need to start thinking about teamwork, and getting super > talents from around the world to help solve mega math problems? Or am I drifting into really weird annoying fantasies? James Harris Too late! I think you have been down that road for maybe 12 years now! You are in a dream world that can be best be described as delusional! Seek help! === Subject: Re: Interesting Math Fact? > Have you come up with any interesing math facts that would be a good > addition to www.odd-info.com ? > There are no uninteresting integers. Almost all real numbers don't have a formula or any other way of computing them. === Subject: Re: Interesting Math Fact? <20081027211513.Y58285@agora.rdrop.com> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008092318 Fedora/3.0.2-1.fc9 Firefox/3.0.2,gzip(gfe),gzip(gfe) > Have you come up with any interesing math facts that would be a good > addition towww.odd-info.com? There are no uninteresting integers. Except 14285435. -- m === Subject: Re: Interesting Math Fact? > Have you come up with any interesing math facts that would be a good > addition towww.odd-info.com? There are no uninteresting integers. Except 14285435. Really? That's very interesting! === Subject: Probability density function question This should be easy, but I'm not quite sure how to answer it. Suppose I have some probability density function f(y) for 0 <= y <= 1. I understand how to figure, for example, P(.6 <= Y <= .8), simply integrate the function from .6 to .8. What if we want to know P(Y <= .7 | Y <= .9) ? Does the fact that P(Y<=.9) matter in this case? === Subject: Re: Probability density function question > This should be easy, but I'm not quite sure how to answer it. Suppose I > have > some probability density function f(y) for 0 <= y <= 1. I understand how to > > figure, for example, P(.6 <= Y <= .8), simply integrate the function from > .6 > to .8. > > What if we want to know P(Y <= .7 | Y <= .9) ? Does the fact that P(Y<=.9) > matter in this case? That last sentence doesn't parse. P(Y<=.9) is a number, not a fact. Hint: use the definition of conditional probability. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Probability density function question > This should be easy, but I'm not quite sure how to answer it. Suppose I >> have >> some probability density function f(y) for 0 <= y <= 1. I understand how >> to figure, for example, P(.6 <= Y <= .8), simply integrate the function from >> .6 >> to .8. What if we want to know P(Y <= .7 | Y <= .9) ? Does the fact that >> P(Y<=.9) >> matter in this case? That last sentence doesn't parse. P(Y<=.9) is a number, not a fact. Hint: use the definition of conditional probability. Heya, Prof. Israel :) To be honest I'm not quite sure what they are asking. Are they asking for P(Y <= .7 ) given that P(Y <= .9)? But as you say, P(Y <= .9) is a number, not a logical, so how can you have given that P(Y <= .9)? I could observe that P(Y <= .7) has nothing to do with P(Y <= .9), consider them independent, and simply figure P(Y <= .7). This is the point where I stop and ask for a nudge in the right direction because I don't quite understand what they are asking. === Subject: Re: Probability density function question > This should be easy, but I'm not quite sure how to answer > it. Suppose I have some probability density function f(y) for 0 > <= y <= 1. I understand how to figure, for example, P(.6 <= Y <= .8), simply integrate the > function from .6 to .8. What if we want to know P(Y <= .7 | Y <= .9) ? Does the fact that > P(Y<=.9) matter in this case? That last sentence doesn't parse. P(Y<=.9) is a number, not a >> fact. Hint: use the definition of conditional probability. ... > To be honest I'm not quite sure what they are asking. Are they > asking for P(Y <= .7 ) given that P(Y <= .9)? But as you say, P(Y <= > .9) is a number, not a logical, so how can you have given that P(Y > <= .9)? given that Y <= .9 the probability that Y <= .7 given that Y <= .9 this, you get P(A | B) = P(A and B) / P(B), which you probably know as the definition of conditional probability. In your case, A and B is easy to simplify. > I could observe that P(Y <= .7) has nothing to do with P(Y <= .9), > consider them independent, and simply figure P(Y <= .7). This is the > point where I stop and ask for a nudge in the right direction > because I don't quite understand what they are asking. What about P(Y <= .9 | Y <= .7)? What about P(Y <= .7 | Y >= .9)? Because Y <= .7 has something to do with Y <= .9. It implies it. === Subject: Re: Statistics with confidence interval. posting-account=xM691AkAAACRhg2rzU0Kd6hZjdls4krD rv:1.8.1.17) Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > Hello teacher~ The concept of a confidence interval is quite difficult for > beginning statistics students, > and sometimes for beginning statistics teachers! For example, assume that our population parameter of interest > is the population mean. > What is the meaning of a 95% confidence interval in this situation? Many students want to say that a 95% confidence interval means > that there is a 95% chance that the confidence interval > contains the population mean. > But any particular confidence interval either contains > the population mean, or it doesn't. > The confidence interval shouldn't be interpreted as a probability. The correct interpretation is based on repeated sampling. > If samples of the same size are drawn repeatedly from a population, > and a confidence interval is calculated from each sample, > then 95% of these intervals should contain the population mean. A 1-alpha confidence interval is the set of potential values of the parameter (mean, or whatever) that would not be rejected by an alpha- level test. === Subject: Re: Statistics with confidence interval. > >Hello teacher~ The concept of a confidence interval is quite difficult for >beginning statistics students, >and sometimes for beginning statistics teachers! > > Yes, it does confuse students sometimes, as you demonstrate. > >For example, assume that our population parameter of interest >is the population mean. >What is the meaning of a 95% confidence interval in this situation? Many students want to say that a 95% confidence interval means >that there is a 95% chance that the confidence interval >contains the population mean. >But any particular confidence interval either contains >the population mean, or it doesn't. >The confidence interval shouldn't be interpreted as a probability. > Well, intervals aren't probabilities, so your last sentence is true, > but your point isn't. You don't *know* the population mean when you > sample the voters. So, if you randomly sample enough voters, you get > this confidence interval within which it is appropriate to say there > is a .95 probability that the [*unknown*] population mean is in that > interval. Whether you know it or not, the population mean is a definite number. It is not a random variable, but the endpoints of the confidence interval are. After you produce the confidence interval, it's not appropriate to talk about the probability that the mean is in this particular interval. What is true is that _before_ you take the sample you can say the probability that the mean will be in the confidence interval I am about to produce is .95. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Statistics with confidence interval. Whether you know it or not, the population mean is a definite number. >It is not a random variable, but the endpoints of the confidence interval >are. After you produce the confidence interval, it's not appropriate to >talk about the probability that the mean is in this particular interval. What >is true is that _before_ you take the sample you can say the >probability that the mean will be in the confidence interval I am about to >produce is .95. OK, hard to argue with that, especially since it's been 35 years since I had anything to do with statistics. So here's my question. Say you were hired to take the sample, so you do and you come up with your confidence interval. So you take your interval to your client to collect your paycheck. Here's the confidence interval you paid me to find. OK, here's your check. What do I say in my press release when I announce this? --Lynn === Subject: Show that if a_n < 0 and lim a_n = 0, then lim 1/a_n = -oo Similar to another recent post from me. I will try and use what I learned from that post here. Show that if a_n < 0 and lim a_n = 0, then lim 1/a_n = -oo Proof: Since a_n < 0 and lim a_n = 0 we have |a_n - 0| < h with |a_n| > M. Select M=1/h. Since |a_n| > M and a_n < 0 then 1/a_n < -M for n > N and M=1/h. Thus, lim 1/a_n = -oo. === Subject: Re: Show that if a_n < 0 and lim a_n = 0, then lim 1/a_n = -oo posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008072820 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Similar to another recent post from me. > I will try and use what I learned from that post here. Show that if a_n < 0 and lim a_n = 0, then lim 1/a_n = -oo > Proof: > Since a_n < 0 and lim a_n = 0 we have |a_n - 0| < h with |a_n| > M. > Select M=1/h. > Since |a_n| > M and a_n < 0 then 1/a_n < -M for n > N and M=1/h. > Thus, lim 1/a_n = -oo. Write down again, what does lim a_n = 0 mean: (1) For every h>0, there exists M such that for every n>M: | a_n - 0 | < h Then write down, what does lim 1/a_n = -oo mean: (2) For every L>0 there exists M such that for every n>M: 1/a_n < -L. To prove the result, you start with an arbitrary L>0 as in the claim (1), ponder which h you wish to plug into (2); this will provide you with an M such that ... And if your choice of h was good, for every n>M: | a_n -0 | < h will imply for every n>M: 1/a_n < -L === Subject: Re: Show that if a_n < 0 and lim a_n = 0, then lim 1/a_n = -oo > Show that if a_n < 0 and lim a_n = 0, then lim 1/a_n = -oo Not sufficient. For that to be true, it's necessary that for all n, a_n < 0 and not for just some n. > Proof: > Since a_n < 0 and lim a_n = 0 we have |a_n - 0| < h with |a_n| > M. More careless disregard for ranges of variables. > Select M=1/h. What do you mean select M = 1/h? You've already determined M above. > Since |a_n| > M and a_n < 0 then 1/a_n < -M for n > N and M=1/h. Too quick and slick to mean anything rabbit, other than the tortoise will out do you. > Thus, lim 1/a_n = -oo. I don't see how you come to that conclusion. What I see you doing is to select M = 1/h and then come to the conclusion M = 1/h. No, it's all too sloppy to be an acceptable proof. Try slowing down to rabbit7.7. === Subject: Re: Show that if a_n < 0 and lim a_n = 0, then lim 1/a_n = -oo > Similar to another recent post from me. > I will try and use what I learned from that post here. > > Show that if a_n < 0 and lim a_n = 0, then lim 1/a_n = -oo > Proof: > Since a_n < 0 and lim a_n = 0 we have |a_n - 0| < h with |a_n| > M. What is _h_? What is M? Where did you use the fact that a_n < 0? > Select M=1/h. > Since |a_n| > M and a_n < 0 then 1/a_n < -M for n > N and M=1/h. Why? Jose Carlos Santos === Subject: Re: Posts not appearing? <9g9Nk.4215$%%2.3001@edtnps82> posting-account=JhJDHwoAAADTQCB3r1hoQX-DHI-aT2tr (KHTML, like Gecko, Safari/525.1+) midori,gzip(gfe),gzip(gfe) > Have I been banned? If so, why not ban the actual cranks out there who > are not interested in learning anything? Nobody gets banned from sci.math, trust me. :D It's some software issue. i tried posting 3 times from web interface. None of them appeared in This is my 3rd post to this thread. === Subject: Re: Posts not appearing? <9g9Nk.4215$%%2.3001@edtnps82> posting-account=JhJDHwoAAADTQCB3r1hoQX-DHI-aT2tr (KHTML, like Gecko, Safari/525.1+) midori,gzip(gfe),gzip(gfe) > Have I been banned? If so, why not ban the actual cranks out there who > are not interested in learning anything? Nobody gets banned from sci.math, trust me. :D It's some software issue. yesterday as i didn't find my first post. Neither of them appeared in === Subject: Re: Math and the beginning of a a concept bigger than the universe posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) The beginning of the hypersphere was a single 3d point that > immediately expanded and became an infinity of points. The hypersphere > grew. Then the Time and energy of the universe was created on its > timeless surface. The timeless hypersphere began. Mitch Raemsch xxein: Then time and timeless. Yeah. We can believe you. Hah!!! Timeless hypersphere began before time. Mitch Raemsch === Subject: Re: Math and the beginning of a a concept bigger than the universe posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) The beginning of the hypersphere was a single 3d point that > immediately expanded and became an infinity of points. The hypersphere > grew. Then the Time and energy of the universe was created on its > timeless surface. The timeless hypersphere began. Mitch Raemsch xxein: It's the mushrooms. Follow the mushrooms. It explains all in > little chunks of wisdom like a barf. The beginning of the hypersphere was one zero dimensional point that became an infinity. Mitch Raemsch === Subject: Re: Sizes of infinity jump posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) Finite quantity of speed can exchange in momentum. Mitch Raemsch xxein: Welcome to gravity. Now take your seat and pay attention to > the instructor. Gravity is how God controls. Mitch Raemsch === Subject: maximum of f (i,j) posting-account=K9O-DgoAAACfJNeR4QSYgwIBRN_weVgi Gecko/20080311 (Debian-1.8.1.13+nobinonly-0ubuntu1) Galeon/2.0.4 (Ubuntu 2.0.4-1ubuntu1),gzip(gfe),gzip(gfe) If i have a function defined by f(i,j) = | ai - aj | + | bi -bj | + |ci -cj| + | di -dj | where |x| is absolute value of x. and (1<=i<=j<=N) and a , b, c ,d are sequences of real numbers of length 10^5 . How do i find the maximum of f(i,j) ? Had it been only |ai - aj| the maximum could be got by sorting the sequence a and then taking a(max) - a(min) , that would be the maximum. How do i find the minimum nonzero value of f(i,j) ? (minimum is for i=j but non zero value ) How can we interpret this equation geometrically / all of them failed.Hope this one appears. === Subject: Re: maximum of f (i,j) posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008072820 Firefox/3.0.1,gzip(gfe),gzip(gfe) > If i have a function defined by f(i,j) = | ai - aj | + | bi -bj | + |ci -cj| + | di -dj | where |x| > is absolute value of x. and (1<=i<=j<=N) and a , b, c ,d are sequences > of real numbers of length 10^5 . Can be done in O(N^2) steps. This is sci.math - we have plenty of time. :) > How do i find the maximum of f(i,j) ? > Had it been only |ai - aj| the maximum could be got by sorting > the sequence a and then taking a(max) - a(min) , that would be the > maximum. That would be O(N log(N)) But simply scanning for min and max in one pass takes only O(N). How do i find the minimum nonzero value of f(i,j) ? (minimum is for > i=j but non zero value ) How can we interpret this equation geometrically / all of them failed.Hope this one appears. === Subject: Maximal Torsion-free quotient Let M be a finitely generated R-module. Is it true that quotients M/N of M are also finitely generated? === Subject: Re: Maximal Torsion-free quotient posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008072820 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Let M be a finitely generated R-module. Is it true that quotients M/N of M are also finitely generated? Given a family of generators of M, can you imagine a family of generators of M/N? === Subject: Re: Maximal Torsion-free quotient posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008092318 Fedora/3.0.2-1.fc9 Firefox/3.0.2,gzip(gfe),gzip(gfe) > Let M be a finitely generated R-module. Is it true that quotients M/N of M are also finitely generated? Each quotient M/N is generated by the image of a generating set for M in M/N under the projection M --> M/N. -- m === Subject: Re: Maximal Torsion-free quotient posting-account=IBUqVwoAAADepmzxVr9iEYD5Z0A483SY Gecko/20070530 Fedora/1.5.0.12-1.fc5 Firefox/1.5.0.12,gzip(gfe),gzip(gfe) > Let M be a finitely generated R-module. Is it true that quotients M/N of M are also finitely generated? Yes Victor Meldrew I don't believe it! === Subject: Re: Maximal Torsion-free quotient > Let M be a finitely generated R-module. > > Is it true that quotients M/N of M are also finitely > generated? *********************************************** Yes. If m_1,...,m_n generate M, then m_1 + N,...,m_n + N generated M/N (all taken as R-module) Tonio === Subject: Re: Initial ordinal, forming omega 1 David Hartley a .8ecrit : > In message <4905ee2d$0$28668$7a628cd7@news.club-internet.fr>, Denis > seems to me that you are constructing a set with order type 2^aleph- > not. Read again. It is a subset of P(NxN) (which, by the way, has no order >> type) What has been explained below seems to be The existence of a set > with cardinality of the continuum. >> Which has an ordcerable subset of type w_1 >> Where as w1 would be the first > uncountable ordinal. (Am I missing something here). > Looks like it > The first initial ordinal we can get by axiom infinity. The way I > understand w1 would be the limit of all countable ordinals. >> Correct, and power set ensures this limit can be constructed *in* P(NxN) > > My turn to be confused. > > Surely it isn't being constructed in P(NxN). If it were, we'd have > aleph_1 <= 2^aleph_0, (without assuming AC). > Sorry, you are right. I was not really awake, and read your answer without understanding exactly your objection. As for my suggestion of well ordering the equivalence classes, lets be charitable... > W is a subset of P(NxN), but the set of equivalence classes, which has > order-type omega_1, is a subset of P(W). So we've constructed the limit > in P(P(NxN)). > > So now we have aleph_1 <= 2^(2^aleph_0)) > > Which is neat, and I haven't seen it before. Is it right? === Subject: Re: Initial ordinal, forming omega 1 <4905ee2d$0$28668$7a628cd7@news.club-internet.fr> <4906c1dc$0$28672$7a628cd7@news.club-internet.fr> In message <4906c1dc$0$28672$7a628cd7@news.club-internet.fr>, Denis >David Hartley a .8ecrit : >> Surely it isn't being constructed in P(NxN). If it were, we'd have >>aleph_1 <= 2^aleph_0, (without assuming AC). >Sorry, you are right. I was not really awake, and read your answer >without understanding exactly your objection. As for my suggestion of >well ordering the equivalence classes, lets be charitable... Luckily I read this before I tried to reply to your first post. Obviously you don't claim aleph_1 can be embedded in any non-denumerable set (without AC) - you can't even fit aleph_0 into a Dedekind-finite infinite set - but do you still think aleph_1 can always be embedded into P(N)? (I'd be surprised if can , but I don't know the answer.) -- David Hartley === Subject: Re: Initial ordinal, forming omega 1 David Hartley a .8ecrit : > In message <4906c1dc$0$28672$7a628cd7@news.club-internet.fr>, Denis >> David Hartley a .8ecrit : > Surely it isn't being constructed in P(NxN). If it were, we'd have > aleph_1 <= 2^aleph_0, (without assuming AC). > Sorry, you are right. I was not really awake, and read your answer >> without understanding exactly your objection. As for my suggestion of >> well ordering the equivalence classes, lets be charitable... > > Luckily I read this before I tried to reply to your first post. > > Obviously you don't claim aleph_1 can be embedded in any non-denumerable > set (without AC) - you can't even fit aleph_0 into a Dedekind-finite > infinite set - but do you still think aleph_1 can always be embedded > into P(N)? (I'd be surprised if can , but I don't know the answer.) I dont know either, but I would be quite surprised :-) What is a little more surprising is that, for any explicit denumerable well-ordering, there exists an explicit *increasing * injection in R (or even in Q), yet there is apparently no way to define an injection (not monotonous of course) of aleph_1 in R not using AC... === Subject: Re: Initial ordinal, forming omega 1 <4905ee2d$0$28668$7a628cd7@news.club-internet.fr> <4906c1dc$0$28672$7a628cd7@news.club-internet.fr> <490732b6$0$28668$7a628cd7@news.club-internet.fr David Hartley a ?crit : >> In message <4906c1dc$0$28672$7a628cd7@news.club-internet.fr>, Denis > David Hartley a ?crit : >> Surely it isn't being constructed in P(NxN). If it were, we'd have >> aleph_1 <= 2^aleph_0, (without assuming AC). > Sorry, you are right. I was not really awake, and read your answer > without understanding exactly your objection. As for my suggestion of > well ordering the equivalence classes, lets be charitable... >> >> Luckily I read this before I tried to reply to your first post. >> >> Obviously you don't claim aleph_1 can be embedded in any non-denumerable >> set (without AC) - you can't even fit aleph_0 into a Dedekind-finite >> infinite set - but do you still think aleph_1 can always be embedded >> into P(N)? (I'd be surprised if can , but I don't know the answer.) > I dont know either, but I would be quite surprised :-) According to this post by Fred Galvin in 1999, c may be incomparable with every uncountable aleph. > What is a little more surprising is that, for any explicit denumerable > well-ordering, there exists an explicit *increasing * injection in R (or > even in Q), yet there is apparently no way to define an injection (not > monotonous of course) of aleph_1 in R not using AC... -- Dave Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. === Subject: Re: Initial ordinal, forming omega 1 <4905ee2d$0$28668$7a628cd7@news.club-internet.fr> <4906c1dc$0$28672$7a628cd7@news.club-internet.fr> , Denis >>David Hartley a ?crit : > Surely it isn't being constructed in P(NxN). If it were, we'd have >aleph_1 <= 2^aleph_0, (without assuming AC). >Sorry, you are right. I was not really awake, and read your answer >>without understanding exactly your objection. As for my suggestion of >>well ordering the equivalence classes, lets be charitable... > Luckily I read this before I tried to reply to your first post. > Obviously you don't claim aleph_1 can be embedded in any non-denumerable > set (without AC) - you can't even fit aleph_0 into a Dedekind-finite > infinite set - but do you still think aleph_1 can always be embedded > into P(N)? (I'd be surprised if can , but I don't know the answer.) P(NxN) is large enough in the sense that for every countable ordinal alpha, there is a member of P(NxN) that has order type alpha. However, there may not be a subset having order type w1 in the absence of AC. I suppose it could be called a Dedekind countable set, analogous to Dedekind finite sets. -- Dave Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. === Subject: Re: Initial ordinal, forming omega 1 David Hartley a .8ecrit : > In message <4905ee2d$0$28668$7a628cd7@news.club-internet.fr>, Denis > seems to me that you are constructing a set with order type 2^aleph- > not. Read again. It is a subset of P(NxN) (which, by the way, has no order >> type) What has been explained below seems to be The existence of a set > with cardinality of the continuum. >> Which has an ordcerable subset of type w_1 >> Where as w1 would be the first > uncountable ordinal. (Am I missing something here). > Looks like it > The first initial ordinal we can get by axiom infinity. The way I > understand w1 would be the limit of all countable ordinals. >> Correct, and power set ensures this limit can be constructed *in* P(NxN) > > My turn to be confused. > > Surely it isn't being constructed in P(NxN). If it were, we'd have > aleph_1 <= 2^aleph_0, (without assuming AC). > No (if you mean it implies 2^aleph_0 is an aleph) as such a well ordering cannot be extended to the whole of 2^aleph_0 (of course, the fact that there is an injection from aleph_1 to 2^aleph_0 is true in ZF with non AC) > W is a subset of P(NxN), but the set of equivalence classes, which has > order-type omega_1, is a subset of P(W). So we've constructed the limit > in P(P(NxN)). Isn't there a canonical representant in P(NxN) ? Use the smallest order (in lecxicographical comparison) > > So now we have aleph_1 <= 2^(2^aleph_0)) > > Which is neat, and I haven't seen it before. Is it right? Anyway, aleph_1 is smaller or equal to any nondenumerable set, but in some cases, the injectuion needs AC . Not here, though... === Subject: Re: Small Neutron star in vicinity of the outside solar system > Supernova explosions might leave small imploded cores of neutronium > behind. I believe the outer planetery motion anomaly could be being > created by such a small neutron star left over from the prior > supernova. > What planetary motion anomaly would that be? Cite? How small would the supposed neutron star be? Where would it be located? You do realize that it would have to be more than a solar mass or it could not be neutronium? And something that big would have a major effect on the Oort cloud? > For every explosion there is an implosion. Not for type Ia supernovae. AFAIK the entire star (a white dwarf) is disrupted. > I want to start the search for Nemesis. Feel free: go right ahead on your own time. Mitch Raemsch -- Mike Dworetsky (Remove pants sp*mbl*ck to reply) === Subject: Re: JSH: Not a popularity contest posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e InfoPath.1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > As I rant and rave, continually antagonize my audience, and talk about > cutting public funding for basic research, or well, any research at > all, I can imagine mathematicians smiling to themselves. No way > people will ever listen to such nuttiness and they may smugly feel > they can rest secure that I'll never be taken seriously--except I DO > have major mathematical research results. > ************************************************************ You're way too modest, and you shouldn't shy away from the fitting adjective for your mathematical results: they are HUMONGOUS. Tonio === Subject: Re: JSH: Not a popularity contest posting-account=LTABjAkAAAByRGPhUAD_hXXBIRzgOoy6 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > cutting public funding for basic research, or well, any research at > all, I can imagine mathematicians smiling to themselves. No way > people will ever listen to such nuttiness and they may smugly feel > they can rest secure that I'll never be taken seriously--except I DO > have major mathematical research results. ************************************************************ You're way too modest, and you shouldn't shy away from the fitting > adjective for your mathematical results: they are HUMONGOUS. HUMONGOUS and GINORMOUS! With BANGS!!!! Jim Deutch (JimboCat) -- I have discovered a truly marvelous method of finding the location of any desired data in pi including, of course, the digits representing the ASCII representation of the C source code for the method itself; unfortunately the hard drive of my ISP's news server is too small to contain its starting index. [Bill Snyder] === Subject: Re: JSH: Not a popularity contest > As I rant and rave, continually antagonize my audience, and talk about > cutting public funding for basic research, or well, any research at > all, I can imagine mathematicians smiling to themselves. No way > people will ever listen to such nuttiness and they may smugly feel > they can rest secure that I'll never be taken seriously-- Yes, yes, yes. > except I DO > have major mathematical research results. Oh! And you were doing so well until this. -- Michael Press === Subject: Supersoluble groups posting-account=5i_ghQoAAABbiBbHHRb14_4kV5Y2uwJY Gecko/2008092417 Firefox/2.0;MEGAUPLOAD 1.0,gzip(gfe),gzip(gfe) Hi everybody! If H is a subgroup of a finite group G then |H| divides |G|. This is Lagrange's theorem. It's well known that the alternating group of degree 4, A_4, has no subgroup of order 6; therefore, in general the converse of Lagrange's theorem does not hold. Note that A_4 is soluble but not a supersoluble group. ( http://en.wikipedia.org/wiki/Supersoluble_group ) My question is: is every supersoluble group a converse of Lagrange's theorem group? I was able to prove that every finite abelian group is a converse of Lagrange's theorem group, though. Best wishes, L.B. === Subject: Re: Supersoluble groups > Hi everybody! > > If H is a subgroup of a finite group G then |H| > divides |G|. > This is Lagrange's theorem. > It's well known that the alternating group of degree > 4, > A_4, has no subgroup of order 6; therefore, in > general > the converse of Lagrange's theorem does not hold. > Note that A_4 is soluble but not a supersoluble > group. > ( http://en.wikipedia.org/wiki/Supersoluble_group ) > > My question is: is every supersoluble group a > converse of Lagrange's theorem group? > > I was able to prove that every finite abelian group > is a > converse of Lagrange's theorem group, though. > > Best wishes, > L.B. ************************************************* The answer seems to be yes: check http://www.ria.ie/cgi-bin/ria/papers/100586.pdf , lemma 3.10, page 5. There they send you to a paper from 1971... Tonio === Subject: Re: Supersoluble groups >> Hi everybody! >> >> If H is a subgroup of a finite group G then |H| >> divides |G|. >> This is Lagrange's theorem. >> It's well known that the alternating group of degree >> 4, >> A_4, has no subgroup of order 6; therefore, in >> general >> the converse of Lagrange's theorem does not hold. >> Note that A_4 is soluble but not a supersoluble >> group. >> ( http://en.wikipedia.org/wiki/Supersoluble_group ) >> >> My question is: is every supersoluble group a >> converse of Lagrange's theorem group? >> >> I was able to prove that every finite abelian group >> is a >> converse of Lagrange's theorem group, though. >> >> Best wishes, >> L.B. ************************************************* The answer seems to be yes: check http://www.ria.ie/cgi-bin/ria/papers/100586.pdf , lemma 3.10, page 5. There they send you to a paper from 1971... It is not hard to prove this by induction. A finite supersoluble group has a normal subgroup N of prime order p. Applying induction to G/N and to proper subgroups of G gets you subgroups of G of all orders dividing |G| except for order |G/N| when |G/N| is not divisible by p, but in that case the required subgroup exists by the Schur-Zassenhaus Theorem. Derek Holt. === Subject: Re: Arguing against negative quantities by themselves as being real posting-account=W7RQ6gkAAACLDC5JWapQU1UV8ot8KkAS .NET CLR 1.0.3705; Dealio Toolbar 3.1.1),gzip(gfe),gzip(gfe) > Negative quantities don't exist by themselves. They exist as > subtraction of their absolute value from some equal or greater > positive value. It doesn't matter, since both positive and negative numbers were invented by people who can't multiply. That's why they also invented zero as a permanent monument to their recursive dumbness, robots to place the computeroid dip switches with CD+rw, Optical Computers to place the GM Doffi with life, and On-Line-Publishing, HDTV, Laser-Guide Lasers, and USB to replace the IBM Clones with the 21st Century. They are a negative operator attached to a postive quantity. Their > only reality is in the operation of subtraction. Mitch Raemsch === Subject: Re: mathematic material about genetic algorithms posting-account=06p6LgoAAADUrhhnbrhCAOeCiEdoGHDV Gecko/2008092417 Firefox/3.0.3 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) Now I have to search for a library to get this book, because it is so expensive ^^ but I'll, see thank you > I am a german student and I am going to write my bachelor thesis. > Because I want to write about genetic algorithms (respectively > like that, which deals with genetic algorithms in a mathematical way. > genetic algorithms but I want to learn something about how fast > genetic algorithms are, how fast they lead to a result and how > reliable they are. Perhaps someone can help me with that. I would be > very glad to see some answers here Greets Sebastian Chapter 4 of the 1998 book An Introduction to Genetic Algorithms, by > Melanie Mitchell, has some theoretical material on convergence and the > like. You can see a bit of it for free by doing a Google search on > genetic algorithms and then going to the Google books entry. A more > and Perspectives: a Guide to GA Theory probably has everything you > need. It seems that almost the whole book deals with GA theory, and > includes dynamical system and markov models of GA algorithms. You can > see some of it in Google books. R.G. Vickson R.G. Vickson === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > There are plenty of non-crackpot alternatives to standard ZFC, such > as constructive logic, Aczel's non-well-founded set theory, Quine's > New Foundations, Martin Lof's type theory, category theory, topos > theory, nonstandard analysis, etc. But those alternatives have > something in common with ZFC: they all require mathematical rigor to > be used correctly. In what sense are Aczel's ill-founded set theory or non-standard analysis alternatives to standard ZFC? -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Aatu Koskensilta says... >> There are plenty of non-crackpot alternatives to standard ZFC, such >> as constructive logic, Aczel's non-well-founded set theory, Quine's >> New Foundations, Martin Lof's type theory, category theory, topos >> theory, nonstandard analysis, etc. But those alternatives have >> something in common with ZFC: they all require mathematical rigor to >> be used correctly. In what sense are Aczel's ill-founded set theory or non-standard >analysis alternatives to standard ZFC? In the sense that they are theories that are useful and have different (that is, alternative) axiomatizations from standard ZFC. Of course, in these two cases, one can interpret nonstandard analysis or Aczel theory inside standard set theory, but it isn't *necessary* to do so (unless you are worried about relative consistency). -- Daryl McCullough Ithaca, NY === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Am 28 Oct 2008 11:56:39 +0200 schrieb Aatu Koskensilta: In what sense [is] Aczel's ill-founded set theory [an] alternative[] to > standard ZFC? > Well, since Aczel's ill-founded set theory is ZFC -Foundation + AFA, and hence NOT standard ZFC (but very similar to it, except for one axiom), wouldn't you call it an alternative[] to standard ZFC? After all, it allows for hypersets, while ZFC does not. Herb === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > If I, WM, and all the other so-called cranks would simply learn > some math, then we would also accept ZF(C) as the only theory worth > discussing, and all of the threads against ZF(C) would disappear. This remark is somwhat puzzling in light of the fact you then go on to write > I _do_ want to learn some math. And I feel that I _am_ > learning math through these discussions. But don't think > that I, or anyone else, will suddenly accept ZF(C) as > the only theory worth discussing simply because I've > learned more about it. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > The only reason that I defend WM, that I talk on his > behalf, is because I don't accept that ZF(C) is the > only set theory worth discussing. WM, and the other > so-called cranks, discuss theories other than ZF(C) > and I find such discussions interesting. Your taste in discussions is somewhat peculiar, but each to their own. Should we take it you believe that when M.9fckenheim asserts ZFC is inconsistent he is in fact discussing some theory other than ZFC? > Surely, Countablist!WM opposes Cantor, since Cantor > proved that the Powerset of a countable set is an > uncountable set. One can happily accept the diagonal argument while rejecting the existence of uncountable sets. > Like Newman, I wish that WM would be more consistent, > but still, I would rather support WM and his theories > than simply declare him to be 100% wrong and accept > ZF(C) as the only set theories worth discussing. Accepting that there are many interesting theories and alternative conceptions worth discussing has nothing to do with M.9fckenheim's blather. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > And you say w is a proper class in your theory. What is your proof in > your theory that there exists a proper class whose members are all and > only the natural numbers? In NBG minus infinity it's trivially provable there's a class of the naturals. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <87ej21hzlt.fsf@alatheia.dsl.inet.fi> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > And you say w is a proper class in your theory. What is your proof in > your theory that there exists a proper class whose members are all and > only the natural numbers? In NBG minus infinity it's trivially provable there's a class of the > naturals. But his claim is that NBG w/o axiom of infinity proves that the class of natural numbers is a PROPER class. MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > >>And you say w is a proper class in your theory. What is your proof in >>your theory that there exists a proper class whose members are all and >>only the natural numbers? > > In NBG minus infinity it's trivially provable there's a class of the > naturals. Proper classes are equivalent with predicates. It's easy to say that n is a natural number. Name that predicate N(n) and you're done. Han de Bruijn === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <87ej21hzlt.fsf@alatheia.dsl.inet.fi> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) >And you say w is a proper class in your theory. What is your proof in >>your theory that there exists a proper class whose members are all and >>only the natural numbers? In NBG minus infinity it's trivially provable there's a class of the > naturals. Proper classes are equivalent with predicates. It's easy to say that n > is a natural number. Name that predicate N(n) and you're done. lwal's claim is that, in NBG w/o the axiom of infinity, the class of natural numbers is a proper class. That requires not just proving that the natural numbers provide a class but ALSO that the class is not a set, i.e., that it is not a member of a class. MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=6xUtvgkAAAD_jypmLa2oo2HnrV0e8X9q 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) Maybe the fact, that e.g. the the material helium exists is due to the > gathering of two neutrons and two protons is a hint that counting is > matter of the reality of counting. Ah, come on. Man's role is too strong here. Whatever counting means in > nature, it will be completely different from our usual notion of counting. > I'd rather say, it's a natural law, that with counting, a counting creature > is necessarily involved. You are totally ignoreing my argument. The first act of counting is > defining the domain. Usually men is defining and counting. But what is > with flowerleaves, legs, amounts of protons, neutrons, electrons of > substances? This domains are natural given. Who counts them that they > are like they are? God, the nature? No matter. But the fact, that a > special number of protons, neutrons and electrons gives a special, > unique, distinct element is a strong argument that counting is, in the > first stage, unindepentend of men. It is only human created distinctions/boundaries that differentiate > between one object and another. If you want to think one moment about this assertion of you, you can find easyly that it is wrong. So that absent those distinctions/boundaries which we have ourselves > created, there are no distinct objects to count. It's hopeless with you. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Maybe the fact, that e.g. the the material helium exists is due to > the > gathering of two neutrons and two protons is a hint that counting is > matter of the reality of counting. Ah, come on. Man's role is too strong here. Whatever counting means in > nature, it will be completely different from our usual notion of > counting. > I'd rather say, it's a natural law, that with counting, a counting > creature > is necessarily involved. You are totally ignoreing my argument. The first act of counting is > defining the domain. Usually men is defining and counting. But what is > with flowerleaves, legs, amounts of protons, neutrons, electrons of > substances? This domains are natural given. Who counts them that they > are like they are? God, the nature? No matter. But the fact, that a > special number of protons, neutrons and electrons gives a special, > unique, distinct element is a strong argument that counting is, in the > first stage, unindepentend of men. It is only human created distinctions/boundaries that differentiate > between one object and another. > > If you want to think one moment about this assertion of you, you can > find easyly that it is wrong. Since all 'boundaries' are defined to be boundaries by humans, without humans creating those definitions, there would be no boundaries. There is nothing in the physical world that is a boundary until some human calls it a boundary. And the more I think about that the truer it gets. > So that absent those distinctions/boundaries which we have ourselves > created, there are no distinct objects to count. > > It's hopeless with you. It is certainly hopeless for you to try to convince me that your peculiar world view is the valid one. Particlarly when no one else seems to share it. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <7450e$490026a4$82a1e228$12813@news1.tudelft.nl> posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > Only a tiny pinch of salt is needed to make mathematics consistent with > the rest of the world ... And why do we need to do that? === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > >>Only a tiny pinch of salt is needed to make mathematics consistent with >>the rest of the world ... > > And why do we need to do that? To survive, perhaps ? Han de Bruijn === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > >>Only a tiny pinch of salt is needed to make mathematics consistent with >>the rest of the world ... > > And why do we need to do that? > > To survive, perhaps ? The weight of evidence is that our survival does not need your sort of salt. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Only a tiny pinch of salt is needed to make mathematics consistent with > the rest of the world ... > > And why do we need to do that? A huge pinch of salt, and more, is needed to make HdB's attitudes at all relevant to mathematics. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > >Only a tiny pinch of salt is needed to make mathematics consistent with >the rest of the world ... And why do we need to do that? > > A huge pinch of salt, and more, is needed to make HdB's attitudes at all > relevant to mathematics. Not. _All_ of my postings evolve around one and the same thing: infinity must be _potential_ and not actual. Making the difference between a real number with an incredibly small error xor a real number with error zero. Making the difference between a ZFC without the axiom of infinity (where the naturals are a proper class) xor ZFC with the axiom of infinity plus a completed _set_ of naturals. Making the difference between limits that have a numerical counterpart xor limits that have not. The difference is a unified Probability = Natural Density xor two distinct notions. What a tiny pinch of salt can do .. Han de Bruijn === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > >Only a tiny pinch of salt is needed to make mathematics consistent with >the rest of the world ... And why do we need to do that? > > A huge pinch of salt, and more, is needed to make HdB's attitudes at all > relevant to mathematics. > > Not. _All_ of my postings evolve around one and the same thing: infinity > must be _potential_ and not actual. Making the difference between a real > number with an incredibly small error xor a real number with error zero. > Making the difference between a ZFC without the axiom of infinity (where > the naturals are a proper class) xor ZFC with the axiom of infinity plus > a completed _set_ of naturals. Making the difference between limits that > have a numerical counterpart xor limits that have not. The difference is > a unified Probability = Natural Density xor two distinct notions. What a > tiny pinch of salt can do .. Those anti-mathematical slugs can all be killed by one pinch of salt. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor _All_ of my postings evolve around one and the same thing: infinity > must be _potential_ and not actual. > Why? (Actually, this seems to be a DOGMA of Anti-Cantorism. Note the usage of the word must.) After all, dealing with actual infinity in the context of modern set theory works _quite well_. (Didn't you realize that?) If you don't think so: please, get real! Herb === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > >>_All_ of my postings evolve around one and the same thing: infinity >>must be _potential_ and not actual. > > Why? (Actually, this seems to be a DOGMA of Anti-Cantorism. Note the usage > of the word must.) > > After all, dealing with actual infinity in the context of modern set > theory works _quite well_. (Didn't you realize that?) If you don't think > so: please, get real! The point is that it _doesn't_ work well. In fact, it doesn't even work. (Note: as a decent computer programmer, I might know what work means.) Han de Bruijn === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > >>_All_ of my postings evolve around one and the same thing: infinity >>must be _potential_ and not actual. > > Why? (Actually, this seems to be a DOGMA of Anti-Cantorism. Note the usage > of the word must.) > > After all, dealing with actual infinity in the context of modern set > theory works _quite well_. (Didn't you realize that?) If you don't think > so: please, get real! > > The point is that it _doesn't_ work well. In fact, it doesn't even work. > (Note: as a decent computer programmer, I might know what work means.) Those who equate computer programming with mathematics are unlikely to be decent at either. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <97505$4906e737$82a1e228$12691@news1.tudelft.nl > _All_ of my postings evolve around one and the same thing: infinity > must be _potential_ and not actual. [Han the crank] > Why? (Actually, this seems to be a DOGMA of Anti-Cantorism. Note the >> usage of the word must.) >> >> After all, dealing with actual infinity in the context of modern set >> theory works _quite well_. (Didn't you realize that?) If you don't think >> so: please, get real! > The point is that it _doesn't_ work well. > Ah? In which mathematical field? Could you please be more specific? :-o Don't we USE quite successfully _set theory_ (or at least some fragments of _set theory_) in (almost) ALL mathematical fields? Even in computer science. (Or did you never study computer science?) Again: Get real, man! (You are completely screwed up in some delusion.) Herb === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > >>_All_ of my postings evolve around one and the same thing: infinity >>must be _potential_ and not actual. [Han the crank] Why? (Actually, this seems to be a DOGMA of Anti-Cantorism. Note the >usage of the word must.) After all, dealing with actual infinity in the context of modern set >theory works _quite well_. (Didn't you realize that?) If you don't think >so: please, get real! The point is that it _doesn't_ work well. > > Ah? In which mathematical field? Could you please be more specific? :-o > > Don't we USE quite successfully _set theory_ (or at least some fragments of > _set theory_) in (almost) ALL mathematical fields? Even in computer > science. (Or did you never study computer science?) > > Again: Get real, man! (You are completely screwed up in some delusion.) _Who_ is the deluded person here ? Han de Bruijn === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > >>_All_ of my postings evolve around one and the same thing: infinity >>must be _potential_ and not actual. [Han the crank] Why? (Actually, this seems to be a DOGMA of Anti-Cantorism. Note the >usage of the word must.) After all, dealing with actual infinity in the context of modern set >theory works _quite well_. (Didn't you realize that?) If you don't think >so: please, get real! The point is that it _doesn't_ work well. > > Ah? In which mathematical field? Could you please be more specific? :-o > > Don't we USE quite successfully _set theory_ (or at least some fragments of > _set theory_) in (almost) ALL mathematical fields? Even in computer > science. (Or did you never study computer science?) > > Again: Get real, man! (You are completely screwed up in some delusion.) > > _Who_ is the deluded person here ? > > Han de Bruijn You named him! === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Han de Bruijn says... >_Who_ is the deluded person here ? Is this a trick question? You, of course. -- Daryl McCullough Ithaca, NY === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Han de Bruijn says... > >>_Who_ is the deluded person here ? > > Is this a trick question? You, of course. Wrong answer. Try again. Han de Bruijn === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > Han de Bruijn says... > >>_Who_ is the deluded person here ? > > Is this a trick question? You, of course. > > Wrong answer. Try again. That HdB does not like that answer does not mean it is wrong. And I, for one, agree with it. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor says... >> Han de Bruijn says... >> >_Who_ is the deluded person here ? >> >> Is this a trick question? You, of course. Wrong answer. Try again. Okay, you and WM and georgie and John Jones. -- Daryl McCullough Ithaca, NY === Subject: Re: A consideration concerning the diagonal argument of G. Cantor _All_ of my postings evolve around one and the same thing: infinity > must be _potential_ and not actual. [Han the crank] > Why? (Actually, this seems to be a DOGMA of Anti-Cantorism. Note the >> usage of the word must.) After all, dealing with actual infinity in the context of modern set >> theory works _quite well_. (Didn't you realize that?) If you don't think >> so: please, get real! > The point is that it _doesn't_ work well. > >> Ah? In which mathematical field? Could you please be more specific? :-o >> >> Don't we USE quite successfully _set theory_ (or at least some fragments of >> _set theory_) in (almost) ALL mathematical fields? Even in computer >> science. (Or did you never study computer science?) >> >> Again: Get real, man! (You are completely screwed up in some delusion.) >> > _Who_ is the deluded person here? > Good question. The crank of us two? Why don't you just ANSWER my question(s)? Don't we USE quite successfully _set theory_ (or at least some fragments of _set theory_) in (almost) ALL mathematical fields? Even in computer science. (Or did you never study computer science?) Herb === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > >>_All_ of my postings evolve around one and the same thing: infinity >>must be _potential_ and not actual. [Han the crank] Why? (Actually, this seems to be a DOGMA of Anti-Cantorism. Note the >usage of the word must.) After all, dealing with actual infinity in the context of modern set >theory works _quite well_. (Didn't you realize that?) If you don't think >so: please, get real! The point is that it _doesn't_ work well. Ah? In which mathematical field? Could you please be more specific? :-o Don't we USE quite successfully _set theory_ (or at least some fragments of >_set theory_) in (almost) ALL mathematical fields? Even in computer >science. (Or did you never study computer science?) Again: Get real, man! (You are completely screwed up in some delusion.) >_Who_ is the deluded person here? > Good question. The crank of us two? > > Why don't you just ANSWER my question(s)? Guess why. > Don't we USE quite successfully _set theory_ (or at least some fragments > of _set theory_) in (almost) ALL mathematical fields? Even in computer > science. (Or did you never study computer science?) Han de Bruijn === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Supersedes: <1rjlvnsvf2pen.vpzwejheklas.dlg@40tude.net> _All_ of my postings evolve around one and the same thing: infinity > must be _potential_ and not actual. [Han the Crank] > Why? (Actually, this seems to be a DOGMA of Anti-Cantorism. Note the >> usage of the word must.) After all, dealing with actual infinity in the context of modern set >> theory works _quite well_. (Didn't you realize that?) If you don't think >> so: please, get real! > The point is that it _doesn't_ work well. > Ah? In which mathematical field? Could you please be more specific? :-o Don't we USE quite successfully _set theory_ (or at least some fragments >> of _set theory_) in (_almost_) ALL mathematical fields? Even in computer >> science. (Or did you never study computer science?) Again: Get real, man! (You are completely screwed up in some delusion.) > _Who_ is the deluded person here? > Good question. The crank of us two? >> >> Why don't you just ANSWER my question(s)? > Guess why. > Because you are a crank? And/or because your claim was nonsensical (or at least _false_)? Again, here's the question: Don't we USE quite successfully _set theory_ (or at least some fragments of _set theory_) in (_almost_) ALL mathematical fields? Even in computer science. (Or did you never study computer science?) Hint: You might open any modern mathematical textbook to find the answer. Herb === Subject: Re: A consideration concerning the diagonal argument of G. Cantor >> Only a tiny pinch of salt is needed to make mathematics consistent with >> the rest of the world ... > > And why do we need to do that? Of course, then you'd have everything around it. It's convenient to bring the system of the Earth around the pinch of sand. By itself, really it depends where it was, the pinch of sand, which is conveniently treated as the Sorites or heap. Who pinched them? Ross F. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > here is an > argument that many ZF(C) defenders make. If I, WM, and > all the other so-called cranks would simply learn > some math, then we would also accept ZF(C) as the > only theory worth discussing, and all of the threads > against ZF(C) would disappear. Who EVER made such an argument? Please name just one person who ever posted such a thing. You continually make strawmen. > The only reason we attack > ZF(C) is due to ignorance, and the more we learn about it, > the more we would accept it. No. Rather, the more one learns about ZFC the more one is in a position to avoid INCORRECT claims about it. > Here's an analogy -- suppose you read the Holy Qu'ran > from cover to cover. Would that automatically mean that > you accept Islam as the one true faith? That's an analogy that doesn't apply here. No one claims that reading a set theory textbook intellectually obligates one to think that ZFC is the only viable foundational theory. > Of course not. One can be an expert on a subject without > accepting it fully. Even if I were to read every single > book in ZF(C) ever published in the world, and became > more of an expert of ZF(C) than even the most brilliant > member of sci.math, I still wouldn't accept it as the > only set theory worthy of discussion! Wonderful! But who EVER said that ZFC is the only set theory worthy of discussion? > Now I have the Levy text, and now I have seen several > sites on the Internet as well. All of them use AC in > the proof. Would you tell me the theorem number and page number you're referring to? 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Turner, read more é... === Subject: Anyone who have solution of electric machinery fundamentals stehen j.chapman pdf download 4th edition posting-account=TdfQkAoAAAC4meFkLHtv9GqGv5skmnhx InfoPath.2),gzip(gfe),gzip(gfe) hello any one who have the solution of electric machinery fundamentals stehen j.chapman pdf download 4th edition. please send me on my email tahirimran62@gmail.com === Subject: Anyone who have solution of electric machinery fundamentals stehen j.chapman pdf download 4th edition posting-account=TdfQkAoAAAC4meFkLHtv9GqGv5skmnhx InfoPath.2),gzip(gfe),gzip(gfe) hello any one who have the solution of electric machinery fundamentals stehen j.chapman pdf download 4th edition. please send me on my email tahirimran62@gmail.com === Subject: Re: -- finitely presented groups on n generators posting-account=-PngCgkAAAD2yUjosqWv1Nf1lkqWP4lp Gecko/20080921 SUSE/2.0.0.17-1.3 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > Here's a followup question ... For a fixed, nontrivial group-theoretic identity with > m variables, let G be the freest n-generator group > satisfying the given identity. The normal subgroup you quotient out by is called a > verbal subgroup, and the class of all groups obeying > the identity is called a variety of groups. Prove or disprove: G can be expressed as a finitely presented group on n > generators. Remarks: By an identity on G, we mean a relation that holds > for arbitrary elements of G. not just for the > generators. I think it's obvious that the claim holds if n=1. As a test case, how about the case m=1, n=2 ? I think it is false for m=1, n=2, and the identity > x -> x^p where p is some large specific prime. The > free group is called B(2,p), the Burnside group of > exponent p on two generators. It is not finitely > presented for large primes p, I believe. It has been proved that for n>8000 and r>1, any group defined by a finite presentation with r generators and n relators that are n-th powers has an element of infinite order. So B(r,n) cannot be defined by a presentation whose relators are all n-th powers. But I think it follows easily from that, that B(r,n) cannot have any finite presentation, because if it did, then each of its relators would be in the normal closure of the set of all n-th powers of elements, and hence would be in the normal closure of finitely many such n-th powers, in which case there would be a finite presentation whose relators were n-th powers. But nothing of this kind is known about B(2,5). So it is conceivable (but very unlikely) that it could be infinite but finitely presented. Of course these examples rely on horrendously long and difficult proofs. I think there should be more elementary example to settle quasi's question. For example, taking n=2 and m=4, consider the law [[w,x], [y,z]]. Then the group defined by this law is the free metabelian group on 2 generators, which I am highly confident is not finitely presentable, although the proof is still not totally straightforward. Derek Holt. === Subject: Re: -- finitely presented groups on n generators I believe it is however recursively presented--but I don't know the reference. === Subject: Re: -- finitely presented groups on n generators > For a fixed, nontrivial group-theoretic identity > with m variables, let G be the freest n-generator > group satisfying the given identity. .. > Prove or disprove: G can be expressed as a finitely presented group on > n generators. .. > I think it's obvious that the claim holds if n=1. As a test case, how about the case m=1, n=2 ? I think it is false for m=1, n=2, and the identity x > -> x^p where p is some large specific prime. The free > group is called B(2,p), the Burnside group of > exponent p on two generators. It is not finitely > presented for large primes p, I believe. It has been proved that for n>8000 and r>1, any group > defined by a finite presentation with r generators and > n relators that are n-th powers has an element of > infinite order. So B(r,n) cannot be defined by a > presentation whose relators are all n-th powers. But I > think it follows easily from that, that B(r,n) cannot > have any finite presentation, because if it did, then > each of its relators would be in the normal closure of > the set of all n-th powers of elements, and hence would > be in the normal closure of finitely many such n-th > powers, in which case there would be a finite > presentation whose relators were n-th powers. Excellent. So it is not large p, but rather large n that is the key. But nothing of this kind is known about B(2,5). So it > is conceivable (but very unlikely) that it could be > infinite but finitely presented. Of course these examples rely on horrendously long and > difficult proofs. I think there should be more > elementary example to settle quasi's question. For > example, taking n=2 and m=4, consider the law [[w,x], [y,z]]. Then the group defined by this law is the free > metabelian group on 2 generators, which I am highly > confident is not finitely presentable, although the > proof is still not totally straightforward. I found a reference for this: Baumslag, Gilbert. Finitely presented metabelian groups. Proceedings of the Second International Conference on the Theory of Groups (Australian Nat. Univ., Canberra, 1973), pp. 65--74. Lecture Notes in Math., Vol. 372, Springer, Berlin, 1974. http://www.ams.org/mathscinet-getitem?mr=404462 http://dx.doi.org/10.1007/BFb0065159 http://www.ams.org/mathscinet-getitem?mr=184993 and that it proves as a corollary that the free group of derived length d on n generators for n,d>1 is not finitely presented. It also confirms Arturuo's comment that finite presentation is not (or at least was not) a very common topic of study, at least for metabelian groups. === Subject: Re: -- finitely presented groups on n generators > It has been proved that for n>8000 and r>1, any group > defined by a finite presentation with r generators and > n relators that are n-th powers has an element of > infinite order. So B(r,n) cannot be defined by a > presentation whose relators are all n-th powers. > Excellent. So it is not large p, but rather large n > that is the key. Strike this comment of mine. It was large p, it was just called n. I realized something was wrong since an elementary abelian group of exponent 2 is defined by 2nd powers, and definitely finitely presented. === Subject: Re: -- finitely presented groups on n generators posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008092318 Fedora/3.0.2-1.fc9 Firefox/3.0.2,gzip(gfe),gzip(gfe) > Here's a followup question ... For a fixed, nontrivial group-theoretic identity with m variables, let > G be the freest n-generator group satisfying the given identity. Prove or disprove: G can be expressed as a finitely presented group on n generators. This is not true. For prime p, the freeest group such that the identity x^p = e holds identically for all p is not finitely presented for large p. IIRC, Derek Holt answered my question about its being finitely presented a while ago on sci.math. -- m === Subject: Re: -- finitely presented groups on n generators days. My association with the Department is that of an alumnus. >> Here's a followup question ... For a fixed, nontrivial group-theoretic identity with m variables, let >> G be the freest n-generator group satisfying the given identity. Prove or disprove: G can be expressed as a finitely presented group on n generators. This is not true. For prime p, the freeest group >such that the identity x^p =3D e holds identically for all p is not finitely presented >for large p. IIRC, Derek Holt answered my question about >its being finitely presented a while ago on sci.math. Here's the messagae: It has been proved (by a combination of work of Ol'shanskii, Ivanov and Lysinick and possibly Novikov/Adian) that for all k>=8000, any group defined by a finite presentation with at least two generators, in which all relators are k-th powers, is infinite - and has elements of infinite order. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: algebraic topology problem posting-account=YeC6dQoAAAD4oLD_rhSm45JYeEZE06ik Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) I have no new ideas whatsoever by now, so I really wish someone answered! === Subject: algebraic topology problem > I have no new ideas whatsoever by now, so I really wish someone > answered! > Neither do I. In fact, I deleted the problem statement yesterday and I don't remember what it was other than beyond my ken. It can happen that a post gets no reply. It's happen to me and it's happen to others and now it's happen to you. You could try posting in the algebraic topology forum found at http://at.yorku.ca . There can also be posts there that go without reply. ---- === Subject: Re: algebraic topology problem posting-account=y4pv3gkAAACfNDHjZnyaLwbK4kSH338B 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) On 26 .8e.98, 23:00, katastrofa nadfioletu > I'm taking first steps in algebraic topology and I became stuck at the > problem: > Prove that any map of the real projective space RP^n for n>=2 to S^1 > is null homotopic (Algebraic topology by E. Spanier. Ex.4 chapter2) > I've already proved that any two maps from a simply connected locally > path-connected space to S^1 are homotopic (Ex.3 from same chapter). > Below I sketch what I have done. (I'm afraid this path isn't homotopic > with a solution...). Pleas help :)! > S^n is (simply connected) double cover of RP^n so from the above we > have that any two maps from S^n to S^1 are homotopic. In particular, > for an arbitrary function g:S^n -> S^1 there exists a homotopy > H: S^nxI -> S^1 such that: > H(x,0)=g(x) for all x in S^n > H(x,1)=p for all x in S^n (p is some fixed point from S^1) > Now, for an arbitrary function f: RP^n -> S^1 I define a homotopy > F: RP^nxI -> S^1 by: > F(x,t)=H(y,t) where: > H is a homotopy given above where g is a composition of a covering > projection c: S^n -> RP^n with function f. As c is a double cover I > need to specify which point y choose to compute H(y,t).(for RP^2 it is > quite easy, but what for n greater than 2?) > Looking forward for any clues and comments > magda I suppose, the path to the solution has to be a little bit longer...:) Here is another attempt: Let a be the generator of pi(RP^n), then 2a=0, so 0=f*(2a)=2f*(a) in pi(S^1)=Z, thus f*(a)=0. (Here f* is the induced homomorphism between fundamental groups.) It follows that f*(pi(RP^n))=0, therefore (as far as I remember), f may be lifted by the exponential map exp: R^1-->S^1, so there is a map F: RP^n-->R^1 such that f=expF. But then it follows immediately that f is null homotopic, since F is so (R^1 is contractible). hth, Simeon === Subject: algebraic topology problem posting-account=YeC6dQoAAAD4oLD_rhSm45JYeEZE06ik Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) I'm taking first steps in algebraic topology and I became stuck at the problem: Prove that any map of the real projective space RP^n for n>=2 to S^1 is null homotopic (Algebraic topology by E. Spanier. Ex.4 chapter2) I've already proved that any two maps from a simply connected locally path-connected space to S^1 are homotopic (Ex.3 from same chapter). Below I sketch what I have done. (I'm afraid this path isn't homotopic with a solution...). Pleas help :)! S^n is (simply connected) double cover of RP^n so from the above we have that any two maps from S^n to S^1 are homotopic. In particular, for an arbitrary function g:S^n -> S^1 there exists a homotopy H: S^nxI -> S^1 such that: H(x,0)=g(x) for all x in S^n H(x,1)=p for all x in S^n (p is some fixed point from S^1) Now, for an arbitrary function f: RP^n -> S^1 I define a homotopy F: RP^nxI -> S^1 by: F(x,t)=H(y,t) where: H is a homotopy given above where g is a composition of a covering projection c: S^n -> RP^n with function f. As c is a double cover I need to specify which point y choose to compute H(y,t).(for RP^2 it is quite easy, but what for n greater than 2?) Looking forward for any clues and comments magda === Subject: Analysis book posting-account=Ye_FRAoAAACXS5t7hbLDH-vHrNDDc870 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Anyone know an Introduction to Analysis book that covers basics of metric spaces and topological spaces in general (not just for Euclidean)? The book I started reading (by Marsden and Hoffman) doesn't cover general topological spaces and it does things mainly in Euclidean metric spaces. And I would prefer it if the book covers metric spaces based on the closed if complement is open definition to match topo. space definitions and _then_ introduces accumulation points (instead of limit points). I don't mind limits point beings mentioned, but I prefer if all the theorems are stated in terms of accumulation points when possible (even though there is a simple relationship between the two). === Subject: Re: JSH: Contradictory data posting-account=_k7REQoAAACDJL2M6OKCfSBj5_wTcvrO Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > I've actually been waiting for some indication of pick-up on my ideas > for managing copy of digital media, which I call Digital Media > Equipment Self-Encryption or DMESE for short (some of you may notice > that can sound a lot like dummies). And I put the idea out there, of course, to reality test, as reality > testing is a critical part of modern problem solving which I gleefully > endorse, and it took over a year and a half--I first put out DMESE > January 29, 2007--but here we are with RealNetworks using the self- > encryption part, for DVD copying software, and promptly being sued > into stopping by Hollywood. Bizarre, or is it? Previously with more reality testing when I pulled out and expanded > one piece of my proof of Fermat's Last Theorem, and shifted things > slightly to make a paper, it got published, after quite a bit of > effort on my part, sending it to journals and such, only to be pulled > when sci.math'ers mounted a crafty email assault against the paper to > the editors, and the wankers yanked it after publication, as always Later the entire mathematical journal died. Cameron University wiped > its existence from its websites and some freaking European group EMIS > saved the American journal when its own country didn't want it any > more. 10 years of math papers into the crapper just like that as if were > nothing. So now we have new and ever more dramatic evidence that the world is > stranger than you previously thought, or even I previously, thought > with some illuminating events indicating human interference, and > interference in bizarre ways indicating strange powers to do things > not previously thought possible: 1. An entire mathematical journal can keel over and die after yanking > a paper by a supposed crackpot, and not a single reporter worldwide > can be motivated to give a damn (believe me, I tried). 2. Citizens around the world including supposedly savvy and very > demanding American consumers cannot legally copy DVD's they buy, > despite a FREE IDEA having been given out over a year and a half ago > by the world's benefactor, me, in a wonderful gesture of charity, as > the idea is conservatively worth, oh, about $100 million over its > patentable lifetime, if I could have gotten a patent, of course, as > patent examiners are kind of bonkers, so that's not a certainty. But > I digress. So I give an idea worth hundreds of millions of dollars to test the > ability of the world to remain in control of some rather powerful > people who have revealed themselves by some amazing blocks requiring > extraordinary political ju-jitsu, and massive balls. Or have they? > Or are they a figment of my fevered imagination? But they haven't simply killed me. (Can you imagine walking around > every day wondering if you're going to be assassinated suddenly? > That's my life.) And they left me to do more mathematical research and I obliged by > solving binary quadratic Diophantine equations (I know, quite a > mouthful) and I'm monitoring to see how long people use the old ways > which are less efficient, versus the new way which leads to a more > direct, general way of solving. Or maybe there is no weird group of strange but powerful men capable > of destroying a math journal at a whim and blocking consumer products > without much effort, which means a new theory is needed. And that's where you people come in, as some of you seem to think > you're very intelligent. I need answers people. What does the data mean? Is there some dark cabal working to keep people from knowing about > advanced mathematical techniques and prevent them from legally copying > their bought DVD's? Or is it all just bizarre coincidences where the information isn't > traveling? So no one KNOWS there is DMESE? So what about its rankings in search results? Why does DMESE dominate > with Google, even trouncing the stock symbol of some company with the > same initials? How about the binary quadratic Diophantine solution? How is it possible for a better way to do still popular math with the > people who do that stuff of course sit versus being picked up in our > very connected world? And if there is a conspiracy, why haven't they killed me yet? Ok, so get to it. I need answers in 48 hours, just because. Better > yet, 24 hours. The hecklers will make their nonsense replies. Ignore them. Part of the conspiracy theory is that they are paid agents, so their > PURPOSE is to distract you. Ok then, let's see how brilliant you are. Resolve the conflicting data, answer the questions. The clock is ticking... James Harris Hey House/Sherlock/Miss Marple, all great investigators know that isolated data points can be difficult to understand until other data points are considered. No good detective would willfully ignore important clues just because they didn't fit their preconceived 'solution' to the mystery. reviewers comments. That fact sheds a lot of light on the other clues in the mystery, but you ignore it. Sorry buddy, you're no Quincy. === Subject: Re: JSH: Contradictory data posting-account=_k7REQoAAACDJL2M6OKCfSBj5_wTcvrO Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > I've actually been waiting for some indication of pick-up on my ideas > for managing copy of digital media, which I call Digital Media > Equipment Self-Encryption or DMESE for short (some of you may notice > that can sound a lot like dummies). And I put the idea out there, of course, to reality test, as reality > testing is a critical part of modern problem solving which I gleefully > endorse, and it took over a year and a half--I first put out DMESE > January 29, 2007--but here we are with RealNetworks using the self- > encryption part, for DVD copying software, and promptly being sued > into stopping by Hollywood. Bizarre, or is it? Previously with more reality testing when I pulled out and expanded > one piece of my proof of Fermat's Last Theorem, and shifted things > slightly to make a paper, it got published, after quite a bit of > effort on my part, sending it to journals and such, only to be pulled > when sci.math'ers mounted a crafty email assault against the paper to > the editors, and the wankers yanked it after publication, as always Later the entire mathematical journal died. Cameron University wiped > its existence from its websites and some freaking European group EMIS > saved the American journal when its own country didn't want it any > more. 10 years of math papers into the crapper just like that as if were > nothing. So now we have new and ever more dramatic evidence that the world is > stranger than you previously thought, or even I previously, thought > with some illuminating events indicating human interference, and > interference in bizarre ways indicating strange powers to do things > not previously thought possible: 1. An entire mathematical journal can keel over and die after yanking > a paper by a supposed crackpot, and not a single reporter worldwide > can be motivated to give a damn (believe me, I tried). 2. Citizens around the world including supposedly savvy and very > demanding American consumers cannot legally copy DVD's they buy, > despite a FREE IDEA having been given out over a year and a half ago > by the world's benefactor, me, in a wonderful gesture of charity, as > the idea is conservatively worth, oh, about $100 million over its > patentable lifetime, if I could have gotten a patent, of course, as > patent examiners are kind of bonkers, so that's not a certainty. But > I digress. So I give an idea worth hundreds of millions of dollars to test the > ability of the world to remain in control of some rather powerful > people who have revealed themselves by some amazing blocks requiring > extraordinary political ju-jitsu, and massive balls. Or have they? > Or are they a figment of my fevered imagination? But they haven't simply killed me. (Can you imagine walking around > every day wondering if you're going to be assassinated suddenly? > That's my life.) And they left me to do more mathematical research and I obliged by > solving binary quadratic Diophantine equations (I know, quite a > mouthful) and I'm monitoring to see how long people use the old ways > which are less efficient, versus the new way which leads to a more > direct, general way of solving. Or maybe there is no weird group of strange but powerful men capable > of destroying a math journal at a whim and blocking consumer products > without much effort, which means a new theory is needed. And that's where you people come in, as some of you seem to think > you're very intelligent. I need answers people. What does the data mean? Is there some dark cabal working to keep people from knowing about > advanced mathematical techniques and prevent them from legally copying > their bought DVD's? Or is it all just bizarre coincidences where the information isn't > traveling? So no one KNOWS there is DMESE? So what about its rankings in search results? Why does DMESE dominate > with Google, even trouncing the stock symbol of some company with the > same initials? How about the binary quadratic Diophantine solution? How is it possible for a better way to do still popular math with the > people who do that stuff of course sit versus being picked up in our > very connected world? And if there is a conspiracy, why haven't they killed me yet? Ok, so get to it. I need answers in 48 hours, just because. Better > yet, 24 hours. The hecklers will make their nonsense replies. Ignore them. Part of the conspiracy theory is that they are paid agents, so their > PURPOSE is to distract you. Ok then, let's see how brilliant you are. Resolve the conflicting data, answer the questions. The clock is ticking... James Harris Hey Sherlock/House/Poirot/Miss Marple, all good investigators know that sometimes isolated data points make no sense until more data is found to put things in context. No great detective would willfully ignore important clues. For example, here's a key clue to unlock the *you didn't receive any reveiwer comments* Now think about it, why might that be? Things making more sense now? === Subject: Re: JSH: Contradictory data posting-account=_k7REQoAAACDJL2M6OKCfSBj5_wTcvrO Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > I've actually been waiting for some indication of pick-up on my ideas > for managing copy of digital media, which I call Digital Media > Equipment Self-Encryption or DMESE for short (some of you may notice > that can sound a lot like dummies). And I put the idea out there, of course, to reality test, as reality > testing is a critical part of modern problem solving which I gleefully > endorse, and it took over a year and a half--I first put out DMESE > January 29, 2007--but here we are with RealNetworks using the self- > encryption part, for DVD copying software, and promptly being sued > into stopping by Hollywood. Bizarre, or is it? Previously with more reality testing when I pulled out and expanded > one piece of my proof of Fermat's Last Theorem, and shifted things > slightly to make a paper, it got published, after quite a bit of > effort on my part, sending it to journals and such, only to be pulled > when sci.math'ers mounted a crafty email assault against the paper to > the editors, and the wankers yanked it after publication, as always Later the entire mathematical journal died. Cameron University wiped > its existence from its websites and some freaking European group EMIS > saved the American journal when its own country didn't want it any > more. 10 years of math papers into the crapper just like that as if were > nothing. So now we have new and ever more dramatic evidence that the world is > stranger than you previously thought, or even I previously, thought > with some illuminating events indicating human interference, and > interference in bizarre ways indicating strange powers to do things > not previously thought possible: 1. An entire mathematical journal can keel over and die after yanking > a paper by a supposed crackpot, and not a single reporter worldwide > can be motivated to give a damn (believe me, I tried). 2. Citizens around the world including supposedly savvy and very > demanding American consumers cannot legally copy DVD's they buy, > despite a FREE IDEA having been given out over a year and a half ago > by the world's benefactor, me, in a wonderful gesture of charity, as > the idea is conservatively worth, oh, about $100 million over its > patentable lifetime, if I could have gotten a patent, of course, as > patent examiners are kind of bonkers, so that's not a certainty. But > I digress. So I give an idea worth hundreds of millions of dollars to test the > ability of the world to remain in control of some rather powerful > people who have revealed themselves by some amazing blocks requiring > extraordinary political ju-jitsu, and massive balls. Or have they? > Or are they a figment of my fevered imagination? But they haven't simply killed me. (Can you imagine walking around > every day wondering if you're going to be assassinated suddenly? > That's my life.) And they left me to do more mathematical research and I obliged by > solving binary quadratic Diophantine equations (I know, quite a > mouthful) and I'm monitoring to see how long people use the old ways > which are less efficient, versus the new way which leads to a more > direct, general way of solving. Or maybe there is no weird group of strange but powerful men capable > of destroying a math journal at a whim and blocking consumer products > without much effort, which means a new theory is needed. And that's where you people come in, as some of you seem to think > you're very intelligent. I need answers people. What does the data mean? Is there some dark cabal working to keep people from knowing about > advanced mathematical techniques and prevent them from legally copying > their bought DVD's? Or is it all just bizarre coincidences where the information isn't > traveling? So no one KNOWS there is DMESE? So what about its rankings in search results? Why does DMESE dominate > with Google, even trouncing the stock symbol of some company with the > same initials? How about the binary quadratic Diophantine solution? How is it possible for a better way to do still popular math with the > people who do that stuff of course sit versus being picked up in our > very connected world? And if there is a conspiracy, why haven't they killed me yet? Ok, so get to it. I need answers in 48 hours, just because. Better > yet, 24 hours. The hecklers will make their nonsense replies. Ignore them. Part of the conspiracy theory is that they are paid agents, so their > PURPOSE is to distract you. Ok then, let's see how brilliant you are. Resolve the conflicting data, answer the questions. The clock is ticking... James Harris Post the reviewers comments, that way you can prove the paper was reviewed, rahter than just accepted by mistake by a very poor journal that deserved to close. === Subject: Re: JSH: Contradictory data posting-account=XH7n1goAAAC_c_M3JtSDv3QFzlbwvvMO Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) James, try making a list of things that a person can do which many, many other people will see as being very significant--significant enough that people will take an interest, and word will spread. The things that a person does in their day-to-day life are, for the most part, ignored by others and quickly forgotten; the acts which are not ignored are those which are truly exceptional. Unfortunately, many of the things that we do which we seem very important to us are really of small significance to others, and it takes a great deal of work, dedication, and time to produce something positive in the world which is remembered and celebrated by others. I have this idea that you will say that some of your mathematical projects have been that significant. Before you say that, though, let me ask: how much of the really significant mathematics of the past few years are you familiar with? A lot of people are producing results, some are even producing results which solve very old and difficult problems in novel and powerful ways; and they are proving their results, filling in the gaps themselves and on their own. This is very different from throwing out an idea into a group of people without checking the details to make sure it solves the problem you want it to solve, and hoping that others will do the nitty-gritty work! I think it is quite natural that you would not know of recent, important results in mathematics, as it takes a lot of time and access to a University library to become familiar with such results; but I also think it is quite natural that others would not know of your mathematical projects. === Subject: Re: JSH: Contradictory data > James, try making a list of things that a person can do which many, > many other people will see as being very significant--significant > enough that people will take an interest, and word will spread. The > things that a person does in their day-to-day life are, for the most > part, ignored by others and quickly forgotten; the acts which are not > ignored are those which are truly exceptional. Unfortunately, many of > the things that we do which we seem very important to us are really of > small significance to others, and it takes a great deal of work, > dedication, and time to produce something positive in the world which > is remembered and celebrated by others. > > I have this idea that you will say that some of your mathematical > projects have been that significant. Before you say that, though, let > me ask: how much of the really significant mathematics of the past few > years are you familiar with? A lot of people are producing results, > some are even producing results which solve very old and difficult > problems in novel and powerful ways; and they are proving their > results, filling in the gaps themselves and on their own. This is very > different from throwing out an idea into a group of people without > checking the details to make sure it solves the problem you want it to > solve, and hoping that others will do the nitty-gritty work! I think > it is quite natural that you would not know of recent, important > results in mathematics, as it takes a lot of time and access to a > University library to become familiar with such results; but I also > think it is quite natural that others would not know of your > mathematical projects. James has postulated a breakdown in the math. establishment sometime between the days of Gauss and the days of Dedekind, at least in his pet area, algebra and number theory. So that's more or less 130 years of suspect number theory, according to James; I think. David Bernier === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? Nntp-Posting-Host: hera.cwi.nl ... > Countables and uncountables already existed in common language. > > Outside of set theories, analysis, and other mathematical uses, none of > which are common language, countable and uncountable have > different meanings, so that those different meanings are irrelevant. Eckard Blumschein wants that terms in mathematics have the same meaning as in commong language. With this he ignores the ambiguity that does exist in common languages. Here he is tripped by such an ambiguity. In English countable has a number of meanings, for which two can be translated to the German: (1) zahlbar: the possibility to terminate the counting is present (2) abzahlbar: the possibility to terminate the counting is not necessarily present Blumschein focuses on the first meaning while in mathematics the second meaning is used. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Decoding a date & time format Not sure if this is a good place to post, but we're trying to reverse engineer an Oracle database which has two fields for date & time. They don't seem to be encoded using the standard Oracle date/time fields so we suspect the developer has used some bespoke encoding system. So I thought it would be a good challenge for somebody mathematically minded. The example I currently have is: 17th October 2008 encoded as 131598865 13:54:49 encoded as 221655296 I'll try and get some more examples. Any help appreciated. === Subject: Re: Decoding a date & time format > Not sure if this is a good place to post, but we're trying to reverse > engineer an Oracle database which has two fields for date & time. They don't > seem to be encoded using the standard Oracle date/time fields so we suspect > the developer has used some bespoke encoding system. So I thought it would be a good challenge for somebody mathematically > minded. The example I currently have is: 17th October 2008 encoded as 131598865 Hexadecimal: 0x07D80A11 And: 07D8 = 2008 0A = 10 11 = 17 > 13:54:49 encoded as 221655296 Hexadecimal: 0x0D363100 And: 0D = 13 36 = 54 31 = 49 Phil -- Christianity has such a contemptible opinion of human nature that it does not believe a man can tell the truth unless frightened by a belief in God. No lower opinion of the human race has ever been expressed. -- Robert Green Ingersoll (1833-1899), American politician and scientist === Subject: Re: Decoding a date & time format >> 17th October 2008 encoded as 131598865 Hexadecimal: 0x07D80A11 > And: > 07D8 = 2008 > 0A = 10 > 11 = 17 === Subject: Re: Decoding a date & time format === Subject: Re: Decoding a date & time format >> Not sure if this is a good place to post, but we're trying to reverse >> engineer an Oracle database which has two fields for date & time. They >> don't >> seem to be encoded using the standard Oracle date/time fields so we >> suspect >> the developer has used some bespoke encoding system. So I thought it would be a good challenge for somebody mathematically >> minded. The example I currently have is: 17th October 2008 encoded as 131598865 Hexadecimal: 0x07D80A11 > And: > 07D8 = 2008 > 0A = 10 > 11 = 17 > 13:54:49 encoded as 221655296 Hexadecimal: 0x0D363100 > And: > 0D = 13 > 36 = 54 > 31 = 49 > Phil Which means it's simply byte byte word format. i.e., day + month << 8 + year << 16 decoding is in reverse: day = date & 0xFF; month = date >> 8 & 0xFF year = date >> 16 === Subject: Re: Decoding a date & time format > Not sure if this is a good place to post, but we're trying to reverse > engineer an Oracle database which has two fields for date & time. They > don't > seem to be encoded using the standard Oracle date/time fields so we > suspect > the developer has used some bespoke encoding system. So I thought it would be a good challenge for somebody mathematically > minded. The example I currently have is: 17th October 2008 encoded as 131598865 Hexadecimal: 0x07D80A11 >> And: >> 07D8 = 2008 >> 0A = 10 >> 11 = 17 > 13:54:49 encoded as 221655296 Hexadecimal: 0x0D363100 >> And: >> 0D = 13 >> 36 = 54 >> 31 = 49 Which means it's simply byte byte word format. i.e., day + month << 8 + year << 16 Not in C, it's not. + has higher precedence than <<. Phil -- Christianity has such a contemptible opinion of human nature that it does not believe a man can tell the truth unless frightened by a belief in God. No lower opinion of the human race has ever been expressed. -- Robert Green Ingersoll (1833-1899), American politician and scientist === Subject: Are these two paths linked? Place two lasers in S^3 so they are near each other but at right angles to each other. Turn lasers on. Assume beams of light are idealized lines. After some time the beams of light will illuminate the back end of the lasers. Let these two light paths define two paths in S^3. Is there a simple way to show the two paths are or are not linked? === Subject: Re: Agreed Question about whether you can construct L posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > The agreed question is: Let H be the set of numbers constructed between > midnight and t 2 (where t 2 is included). > Let L be a list containing all the elements of H. > Let t 2 be the time at which L is constructed. > Can you construct L? What is your answer? (my answer is no). Agreed answer. Sorry, I don't get why that should be the case. What about indexing the elements of H as long as you construct them? > That is, construct first element and give it index 1; construct second > element and give it index 2; and so on. Then, if I have not > misunderstood, at time t 2 both H and L have been constructed and are > finite. Both you and William Hughes have agreed that such a list cannot be > constructed, but I would disagree. Yes, both are finite, but the list L is constructed and with it an > antidiagonal is constructed that is a member of the set H but not a > member of the list L. Given that the rules of the game do not talk about an anti-diagonal of > L to be put into the set H, I don't see why then one would say that > such anti-diagonal is in fact in H. It isn't, until you put H in a list. Then you construct an antidiagonal by a fixed rule. OTOH, should one want to list the anti-diagonal, one could -say- give > it index 0 and put it on top of the list by convention. But then you have another list with another antidiagonal that is not in the list. === Subject: Re: Agreed Question about whether you can construct L > Given that the rules of the game do not talk about an anti-diagonal of > L to be put into the set H, I don't see why then one would say that > such anti-diagonal is in fact in H. > > It isn't, until you put H in a list. Then you construct an > antidiagonal by a fixed rule. Then merely constructing a set of infinite binary strings does not automatically create any anti-diagonals at all. In fact, none can exist until one has a completed infinite list of infinite binary strings. === Subject: Re: Agreed Question about whether you can construct L <4906279f$0$5480$9a566e8b@news.aliant.net> posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) >>Proposed question #2 > Let H be the set of numbers constructed between >> midnight and t 2 (where t 2 is included). >> Let L be a list containing all the elements of H >> Let t 2 be the time at which you try to present L. >> Can you construct L? Let H be the set of numbers constructed after midnight. Can you > construct L after midnight? Rejected. In this question there is a start time > for constructing H, but no end time. So H is something that > can change. Every infinite set is something that can change, in particular if you construct it. Otherwise it would be finished which is the opposite of infinite. === Subject: Re: Agreed Question about whether you can construct L >>Proposed question #2 > Let H be the set of numbers constructed between >> midnight and t_2 (where t_2 is included). >> Let L be a list containing all the elements of H >> Let t_2 be the time at which you try to present L. >> Can you construct L? Let H be the set of numbers constructed after midnight. Can you > construct L after midnight? Rejected. In this question there is a start time > for constructing H, but no end time. So H is something that > can change. > > Every infinite set is something that can change, in particular if you > construct it. Otherwise it would be finished which is the opposite of > infinite. A set is something that does not and cannot change. WM is confusing sets with something like processes. They are not at all the same thing. === Subject: Re: Agreed Question about whether you can construct L > >Proposed question #2 Let H be the set of numbers constructed between >> midnight and t_2 (where t_2 is included). >> Let L be a list containing all the elements of H >> Let t_2 be the time at which you try to present L. >> Can you construct L? >Let H be the set of numbers constructed after midnight. Can you >construct L after midnight? Rejected. In this question there is a start time >>for constructing H, but no end time. So H is something that >>can change. > > > Every infinite set is something that can change, Hotly disputed. However, we do agree that some sets do not change. Call these unchangable sets, u-sets for short. Note that H is a u-set if and only if an end time is given. Proposed question #2 Let H be the u-set of numbers constructed between midnight and t_2 (where t_2 is included). Let an H-list be a list containing all the elements of H Let t_2 be the last time at which you try to construct an H-list. Is it possible to contruct an H-list? -William Hughes === Subject: Re: inverse AGM > however here is an example showing you were wrong > when you said agm(a,b)=agm(c,d) => a=c and b=d >> >> I didn't say that ... I expressed doubt that it >> was true ... >> > doubt is not certain enough to consider > yourself understanding something... I was being polite. My actual position was much stronger, but I can see now that being polite simply led to confusion. I won't bother again. > now we can concentrate on the OP > any ideas ?? Now that I know what you want I've been thinking about it a little, but have nothing of value to offer yet. Sorry. I'll keep looking at it on and off. === Subject: Re: Without a unit, can we construct the prodcut of two given lengths a, b? > When we have positive lengths a, b, I know how to construct the length > ab. > But in that case, it is required we already know how to consturct the > unit length 1. > > Without a unit, can we constuct ab? > That is, suppose that in a plane there are segments with lengths a, b. > Then can you construct ab or find the unit length? The product of two lengths is an area, and you can construct that area without a unit. But (as you know) to make areas correspond to lengths, we need a unit length. === Subject: Re: Without a unit, can we construct the prodcut of two given lengths a, b? posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008092318 Fedora/3.0.2-1.fc9 Firefox/3.0.2,gzip(gfe),gzip(gfe) > When we have positive lengths a, b, I know how to construct the length > ab. > But in that case, it is required we already know how to consturct the > unit length 1. Without a unit, can we constuct ab? Carry out the construction for two given segments a and b using two different units? Do youget the same result? What does that imply with respect to this question? > That is, suppose that in a plane there are segments with lengths a, b. > Then can you construct ab or find the unit length? The question is less trivial if you are given a, b *and* ab... -- m === Subject: T or F, every pair of distinct geodesics on S^3 is linked? True or false, every pair of distinct geodesics on S^3 is linked? Is it easy to show either case? === Subject: Re: defining a closed set > > ... When topology was first invented, > there were many alternative axiomatizations considered; does anyone > know if bdry was a primitive in any of them? Here is footnote 1 on page 61 of volume I of Kuratowski's _Topology_ (new edition, revised and augmented 1966): The notion of boundary can be taken too as primitive term for a topological space. See M. Zarycki, _op cit_. et J. Albuquerque, _La notion de frontiere en topologie_, Portug. Math. 2 (1941) pp 280-289. The _op_ being _cit_ed is M. Zarycki, _Quelques notions fondamentales de l'Analysis Situs au point de vue de l'Algebre de la Logique, Fund. Math. 9 (1927) pp 3-15. I have not tried to reproduce a couple of diacritic marks. -- He is not here; but far away The noise of life begins again And ghastly thro' the drizzling rain On the bald street breaks the blank day. === Subject: Re: defining a closed set posting-account=x2WXVAkAAACheXC-5ndnEdz_vL9CA75q Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) >A closed set is defined as the complement of >an open set. ok, I accept that as rigorous. But intuitively, does >not a closed set consist of an open set, plus its >boundary? Is there any problem with defining a >closed set this way, given a suitable concept of boundary? The problem is it's not right. d(x) = distance from origin A = {x | d(x) < 1} A is an open set. B = {x | d(x) = 1} B is the boundary of A. C = A U B C is a closed set. QED >I'm thinking from a pedagogical viewpoint, it >might be an improvement - You think intuition is a bad thing? -- Rich === Subject: Re: defining a closed set posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > d(x) = distance from origin A = {x | d(x) < 1} > A is an open set. B = {x | d(x) = 1} > B is the boundary of A. C = A U B > C is a closed set. QED QED? What is the what proposition do you think you've proven? Look at my post. I anticipated what you're driving at: closure(S) = interior(S) u boundary(S). And, of course, we know that the closure of any set is a closed set. But that doesn't entail that C is closed iff there is an open set S such that C = S u boundary(S). You seem to be confusing two related, but different things: (1) The defintion of a 'closed set'. and (2) The fact that the closure of any set is the union of the interior and boundary of that set and that the closure of any set is a closed set. MoeBlee === Subject: Re: defining a closed set posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > A closed set is defined as the complement of > an open set. ok, I accept that as rigorous. But intuitively, does > not a closed set consist of an open set, plus its > boundary? I take it that you are asking whether 'closed' could be defined (where 'u' stands for binary union): C is closed iff there is an open set S such that C = S u boundary(S). No. A point can't be both in the complement of S and in S. Maybe what you're thinking of is this theorem: closure(S) = interior(S) u boundary(S). MoeBlee === Subject: Convex hull intersecting lattice in 2-dim posting-account=qzUc6goAAACUhmVDc3RDNQi8M7nqqZja Gecko/2008092318 Fedora/3.0.2-1.fc9 Firefox/3.0.2,gzip(gfe),gzip(gfe) I encountered the following 2-dim problem: 1) The given a set of in-equalities a_i x + b_i y <= c_i defines a convex subset C of R^2 . (a_i, b_i , c_i integral). 2) Given an integral lattice L in Z^2 generated by k integral basis- vectors (e_i, f_i), i.e. the set of points (x, y) = sum( ( e_i , f_i) * n_i ) ( with n_i integer , i = 1...k ) ?) Find the convex hull H (=convex polygon) of the intersection of C and L. I know that I can use standard Gomory cutting plane algorithms. However, is there a smarter way to solve the problem 2 dims? I only need the the vertices of H. Any hints appreciated Herbert === Subject: Re: -- closure of a subset of N under some operations <22967695.1225117403129.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008072820 Firefox/3.0.1,gzip(gfe),gzip(gfe) >Here's an alternate version of one of my >>previous conjectures ... >Conjecture: >If S is a nonempty subset of {n in N | n > 2} such that > x,y in S => xy, (x-1)(y-1), xy-x+1 in S >then S contains at least one pair of consecutive positive integers. Remark: If the above conjecture is true, then it can be easily shown >that the set S, as specified above, must in fact contain >infinitely many pairs of consecutive positive integers. >Remark: >Call the above conjecture Conjecture (1). >Then it's not hard to show that Conjecture (1) implies the >>following group-theoretic conjecture: >Conjecture (2): >If G is a group satisfying an identity of the form > (xy)^n = x^n y^n >for some integer n other than 0 or 1, then there exists a >>nonzero integer m such that, for all x in G, x^m is in the >>center of G. >Remarks: >Although I can prove Conjecture (1) implies Conjecture (2), >>I don't see any obvious way to prove either one. Moreover, >>I can't prove the converse, so it's conceivable that >>Conjecture (1) is false while Conjecture (2) is true. >Trivially, Conjecture (2) implies the following >>(apparently weaker) claim. >Conjecture (3): >If G is a group satisfying an identity of the form > Correction: The above line should be > If G is a nontrivial group satisfying an identity Revised Correction: The above line should be >If G is a group, without finite exponent, >>satisfying an identity of the form > (xy)^n = x^n y^n >for some integer n other than 0 or 1, then G has >>a nontrivial center. >Remarks: >As noted, Conjecture (2) trivially implies Conjecture (3), >>however the converse is not so clear. Hence, it might >>be the case that Conjectures (1) and (2) are false while >>Conjecture (3) is true. >Proofs or disproofs for any of Conjectures (1),(2),(3) >>are welcome. >Re. conjecture (3): Let G be the alternating group A_5. >>This group is simple, so has trivial center. And every finite >>group has an exponent, i.e. a positive n such that x^n=1 >>for all x in G. So if n is the exponent of A_5 then the identity >>(xy)^n = x^n y^n holds, even though the center is trivial. >> Am I missing something? No -- I was missing something. But the fix is simple: For Conjecture (3), simply bar groups of finite exponent. For a smaller example let G = S_3 , the permutations of {1,2,3}. >Then G has trivial center. The nonidentity elements of G are the >three 2-cycles and two 3-cycles. So every element of G satisfies >(xy)^6 = x^6 y^6 holds, even though S_3 has trivial center. the flaw. As noted above, Conjecture (3) can be repaired as follows: Conjecture (3): If G is a group, without finite exponent, satisfying an identity of > the form (xy)^n = x^n y^n for some integer n other than 0 or 1, > then G has a nontrivial center. quasi Hm, what if G may have finite exponent, but we exclude n from being an exponent? In other words, is there a counterexample to Conjecture (3'): If G is a group and n an integer >1 and (xy)^n = x^n y^n for all x,y in G and x^n != 1 for some x in G, then G has nontrivial center ? hagman === Subject: Re: -- closure of a subset of N under some operations >Here's an alternate version of one of my >previous conjectures ... >Conjecture: >If S is a nonempty subset of {n in N | n > 2} such that > x,y in S => xy, (x-1)(y-1), xy-x+1 in S >then S contains at least one pair of consecutive positive integers. Remark: If the above conjecture is true, then it can be easily shown >>that the set S, as specified above, must in fact contain >>infinitely many pairs of consecutive positive integers. >Remark: >Call the above conjecture Conjecture (1). >Then it's not hard to show that Conjecture (1) implies the >following group-theoretic conjecture: >Conjecture (2): >If G is a group satisfying an identity of the form > (xy)^n = x^n y^n >for some integer n other than 0 or 1, then there exists a >nonzero integer m such that, for all x in G, x^m is in the >center of G. >Remarks: >Although I can prove Conjecture (1) implies Conjecture (2), >I don't see any obvious way to prove either one. Moreover, >I can't prove the converse, so it's conceivable that >Conjecture (1) is false while Conjecture (2) is true. >Trivially, Conjecture (2) implies the following >(apparently weaker) claim. >Conjecture (3): >If G is a group satisfying an identity of the form > Correction: The above line should be > If G is a nontrivial group satisfying an identity Revised Correction: The above line should be >If G is a group, without finite exponent, >satisfying an identity of the form > (xy)^n = x^n y^n >for some integer n other than 0 or 1, then G has >a nontrivial center. >Remarks: >As noted, Conjecture (2) trivially implies Conjecture (3), >however the converse is not so clear. Hence, it might >be the case that Conjectures (1) and (2) are false while >Conjecture (3) is true. >Proofs or disproofs for any of Conjectures (1),(2),(3) >are welcome. >Re. conjecture (3): Let G be the alternating group A_5. >This group is simple, so has trivial center. And every finite >group has an exponent, i.e. a positive n such that x^n=1 >for all x in G. So if n is the exponent of A_5 then the identity >(xy)^n = x^n y^n holds, even though the center is trivial. > Am I missing something? No -- I was missing something. But the fix is simple: For Conjecture (3), simply bar groups of finite exponent. For a smaller example let G = S_3 , the permutations of {1,2,3}. >>Then G has trivial center. The nonidentity elements of G are the >>three 2-cycles and two 3-cycles. So every element of G satisfies >>(xy)^6 = x^6 y^6 holds, even though S_3 has trivial center. the flaw. As noted above, Conjecture (3) can be repaired as follows: Conjecture (3): If G is a group, without finite exponent, satisfying an identity of >> the form (xy)^n = x^n y^n for some integer n other than 0 or 1, >> then G has a nontrivial center. quasi Hm, what if G may have finite exponent, but we exclude n from being an >exponent? In other words, is there a counterexample to Conjecture (3'): If G is a group and n an integer >1 and (xy)^n = x^n y^n for all x,y >in G and x^n != 1 for some x in G, then G has nontrivial center ? I don't think Conjecture (3') will hold. As a possible counterexample, try this ... Let G be freest 2-generator group of exponent 12 such that (xy)^4 = x^4 y^4 for all x,y in G. I suspect that G has trivial center. quasi === Subject: Re: -- closure of a subset of N under some operations <18u2g41j4psv83v19dfu5rnktrk52hqmr7@4ax.com> posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > Conjecture: > If S is a nonempty subset of {n in N | n > 2} such that > x,y in S => xy, (x-1)(y-1), xy-x+1 in S > then S contains at least one pair of consecutive positive integers >> (and hence, infinitely many such pairs). Assuming the quasi conjecture, we can define for each n>=3 >q(n) := min { m e N | forall S: S as above, n e S -> {m,m+1} c S } If I'm not wrong, q(3)=3, q(4)=120, q(5)=1560 and so on >How far did you actually compute the q sequence that starts >3, 120, 1560, 27090, 439110, 468386688, ... ? I only tested the cases n = 3, 4, 5, 6. For those cases, I got q(n) = 3, 120, 1560, 27090 matching your results. To me, the next term 468386688 looked somewhat discouraging ... hagman === Subject: Re: -- closure of a subset of N under some operations >Here's an alternate version of one of my previous > conjectures ... Conjecture: If S is a nonempty subset of {n in N | n > 2} such > that x,y in S => xy, (x-1)(y-1), xy-x+1 in S >> >>then S contains at least one pair of consecutive > positive integers. Remark: If the above conjecture is true, then it can be > easily shown that the >set S, as specified above, must in fact contain > infinitely many pairs >of consecutive positive integers. > > Remark: > > Call the above conjecture Conjecture (1). > > Then it's not hard to show that Conjecture (1) > implies the following > group-theoretic conjecture: > > Conjecture (2): > > If G is a group satisfying an identity of the form > > (xy)^n = x^n y^n > > for some integer n other than 0 or 1, then there > exists a nonzero > integer m such that, for all x in G, x^m is in the > center of G. > > Remarks: > > Although I can prove Conjecture (1) implies > Conjecture (2), I don't > see any obvious way to prove either one. Moreover, I > can't prove the > converse, so it's conceivable that Conjecture (1) is > false while > Conjecture (2) is true. > > Trivially, Conjecture (2) implies the following > (apparently weaker) > claim. > > Conjecture (3): > > If G is a group satisfying an identity of the form > > (xy)^n = x^n y^n > > for some integer n other than 0 or 1, then G has a > nontrivial center. > > Remarks: > > As noted, Conjecture (2) trivially implies Conjecture > (3), however the > converse is not so clear. Hence, it might be the case > that Conjectures > (1) and (2) are false while Conjecture (3) is true. > > Proofs or disproofs for any of Conjectures > (1),(2),(3) are welcome. > > quasi I'd like to see the details on showing how conjecture (3) [with the added assumption that G does not have finite exponent] follows from conjecture(1). I can't see how to really use the negative assumption >>that G does not have finite exponent. Nor can I see how the existence of pairs n,n+1 in the generated set give rise to anything about centers of groups in general... At least, can someone give a hint about what the >>pairs n,n+1 in such generated sets have to do with centers of groups? I'll post proofs of the implications Conjecture (1) => Conjecture (2) => Conjecture (3) as soon as I get a chance -- probably tonight. I've written up part of it, but the proofs, though mostly routine, are somewhat long and tedious. I won't be able to finish it tonight, and perhaps not until the weekend. However, in the thread -- group identities of the form (xy)^k = x^k y^k I outline a chain of results, stated without proof (but the proofs are easy), which capture some of the key ideas relating to the conjectures in this thread. Here's a link: quasi === Subject: Re: FLT: a short amateur proof for case 2: where is the mistake? <26881667.1222547742794.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008072820 Firefox/3.0.1,gzip(gfe),gzip(gfe) And still another version is at > Re: My copying and pasting errors: Maybe I should mention that I only have limited time at a library > computer. I've spent this morning's time editing out typos in my > _latest_ version that makes the proof gibberish. The corrected version > is at > For some reason that version also had a load of unwanted page from one site to another, I guess. 1) Please check your formulation. The way you write things down, you seem to use: If p>4 is prime and x,y,z are arbitrary positive integers such that x^p + y^p = z^p (esp. not necessarily pairwise coprime), then there exists a positive integer a such that x+y = a^p. I doubt that this is the correct formulation because if x^p + y^p = z^p then (2x)^p + (2y)^p = (2z)^p, hence there exists a positive integer a' such that 2 a^p = 2x+2y = (a')^p, contradicting the irrationality of p-th root of 2 -- an even shorter proof of FLT! ;) 2) How do you know that y has prime factors that do not divide z- x ? === Subject: trvial Hi what is the definition of trivial central extension? (and its equivalents ) I know the definition of cental extension.what does mean Trival. === Subject: The unimaginable mathematics of Torkel Franzen posting-account=5dWnDwkAAAAJX56s4Iiwi0owHzqwvK-v NokiaE51-1/100.34.20; Profile/MIDP-2.0 Configuration/CLDC-1.1 ) AppleWebKit/413 (KHTML, like Gecko) Safari/413,gzip(gfe),gzip(gfe) I have been reading today William Bloch's entertaining new book _The Unimaginable Mathematics of Borges' Library of Babel_, an exploration of the mathematical ideas in Borges' short story. On page 179 he recommends a book by the late Torkel Franzen, an active sci.math poster. Let me quote Bloch: Goedel's Theorem: An Incomplete Guide to Its Use and Abuse, by Torkel Franzen Franzen's book concisely achieves its goal of clearly demarcating the extent and applicability of Goedel's theorem. Along the way of accomplishing that, he shows its power and majesty in the fields of set theory and foundations, and brings into sharp focus many amusing nuances. === Subject: y = x^p; p = ? Help please... Given y = x^p, how do I solve for p? TIA -- John === Subject: Re: y = x^p; p = ? John a .8ecrit : > Given y = x^p, how do I solve for p? What are those datas? Elements of a field in general, reals, complexes, ...? -- Fatal === Subject: Re: y = x^p; p = ? > John a .8ecrit : > >> Given y = x^p, how do I solve for p? > > What are those datas? Elements of a field in general, reals, complexes, > ...? from Greg: p = ln(y)/ln(x) -- John === Subject: Re: y = x^p; p = ? John a .8ecrit : >> John a .8ecrit : > Given y = x^p, how do I solve for p? What are those datas? Elements of a field in general, reals, >> complexes, ...? > > from Greg: p = ln(y)/ln(x) > It's OK if you assume x>0 and y>0. There could be other solutions with integer exponent and negative x, even for some rational exponents such as 1/3. -- Fatal === Subject: Re: y = x^p; p = ? >> John a .8ecrit : > Given y = x^p, how do I solve for p? What are those datas? Elements of a field in general, reals, complexes, >> ...? Greg: p = ln(y)/ln(x) -- > John actually, p = ln (y/x) or log(y/x) === Subject: Re: y = x^p; p = ? > John a .8ecrit : > Given y = x^p, how do I solve for p? What are those datas? Elements of a field in general, reals, complexes, > ...? Greg: p = ln(y)/ln(x) -- >> John actually, p = ln (y/x) or log(y/x) I think not. Example: 125 = 5^3 y = 125 x = 5 p = 3 ln(y)/ln(x) = 3 = p log(y)/log(x) = 3 = p ln(y/x) = 3.21887... NOT equal to p log(y/x) = 1.39794... NOT equal to p In general, ln(z) <> log(z) for real z > 0, and z <> 1. === Subject: Re: y = x^p; p = ? >> Help please... Given y = x^p, how do I solve for p? > > Have you studied logarithms? More than 30 years ago :-) -- John === Subject: Re: y = x^p; p = ? > Help please... Given y = x^p, how do I solve for p? Have you studied logarithms? More than 30 years ago :-) Ah. So a refresher is in order! You might want to Google properties of logarithms. In the meantime, take the log of each side of your equation: ln(y) = ln(x^p) = p*ln(x) So p = ln(y)/ln(x) You can use logs to any base; I just happened to use the natural log, base e, denoted ln. === Subject: Re: y = x^p; p = ? >> Help please... Given y = x^p, how do I solve for p? > Have you studied logarithms? >> More than 30 years ago :-) > > Ah. So a refresher is in order! You might want to Google > properties of logarithms. > > In the meantime, take the log of each side of your equation: > > ln(y) = ln(x^p) > = p*ln(x) > > So p = ln(y)/ln(x) > > You can use logs to any base; I just happened to use the > natural log, base e, denoted ln. Greg, thank you very much. Your help is very much appreciated! -- John === Subject: Re: -- closure of a subset of N under some operations > >>Conjecture: If S is a nonempty subset of {n in N | n > 2} > such > that x,y in S => xy, (x-1)(y-1), xy-x+1 in S then S contains at least one pair of consecutive > positive integers >>(and hence, infinitely many such pairs). Observation: The following is easily proved ... Fix a in N, a > 2. If S is a minimal nonempty subset of {n in N | n 2} such that a in S x,y in S => xy, (x-1)(y-1), xy-x+1 in S then all elements of S are congruent to 0 or 1 > (mod > a). > > Let denote the closure of {a} under the > specified > operations. > > The case a = 3 appears to be special, in the > following sense ... > > As already observed, <3> must be a subset of > > {n in N | n > 2 and n = 0 or 1 (mod 3)} > > However, as noted by LudovicoVan, it appears that, > in > fact, > > <3> = {n in N | n > 2 and n = 0 or 1 (mod 3)} > > How to prove (or disprove) this? > > quasi > > The following gives a proof that <3> is as stated > above: > ... <3> = {n in N | n > 2 and n = 0 or 1 (mod 3)} > As noted, clear that the left side is contained in > the right. > For positive integer k let P(k) be the statement > that, > > for 3 < n < 9k, if n = 0 or 1 mod 3 then n in <3>. > > We give names to the productions namely > a(x,y) = x*y > b(x,y) = (x-1)*(y-1) > c(x,y) = x(y-1)+1. > [Note c(x,y) is same as xy-x+1] > > We need not go through any base case; just > look at the initial part of the sequence of <3>... > > To show P(k+1) we need only show > (under assumption that P(k) holds) > that 9k+r is in <3> for k in {0,1,3,4,6,7}. > We do this by writing each such 9k+r > as one of a,b,or c evaluated at appropriate > pairs (x,y), where each of x,y is less than 9k. > Here goes: > 9k = a(3,3k) > 9k+1 = c(3k,4) > 9k+3 = a(3,3k+1) > 9k+4 = c(3k+1,4) > 9k+6 = b(4,3k+3) > 9k+7 = c(3,3k+3) > Thus we have arrived at the statement P(k+1), i.e. > now know all n less that 9(k+1) which are 0 or 1 mod > 3 > lie in the set <3>. > > Of course the union of the statements P(k) for all k > > now shows that <3> is as stated all integers >= 3 > which are 0 or 1 mod 3. Just a picky correction: P(k) is the statement for 3 <= n < 9k, if n = 0 or 1 mod 3 then n in <3>. [In above post I had the inequality 3 < n < 9k] === Subject: Re: This Week's Finds in Mathematical Physics (Week 271) [..] > Sulfur melts around 115 Celsius. But when you heat it above 160 > Celsius, something weird happens: contrary to the usual pattern for > liquids, it gets more viscous as it gets hotter! The reason: the > atoms start forming long chain polymers, called catena sulfur. If you don't have a chemistry set, you can always watch it on Youtube: http://www.youtube.com/watch?v=JtPbHL5gFKw&NR=1 This one is also fun: http://www.youtube.com/watch?v=mGMR72X8V-U Gerard === Subject: bounded monotonic sequence proof HELP posting-account=twH9PwoAAABVFxHWRVsv2WYMPEStgjmE AppleWebKit/525.13 (KHTML, like Gecko) Chrome/0.2.149.30 Safari/525.13,gzip(gfe),gzip(gfe) If u_n+1 = sqrt ((u_n) + 1), u_1 = 1 prove that lim { n to infinity} u_n = 1/2 (1 + sqrt 5) u_n denoted u subscript n. i thank you for any hints and tips you can throw my way. cheers === Subject: Re: proving not(c = aleph_aleph_null) diagonally without choice posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) how does a free variable work out here? Unless I missed some detail, it's just application of the rule of > existential instantiation (in whatever formulation of the rule you > prefer). Are you asking for a formulation of the rule of existential > instantiation? I was thinking of Gentzen's sequent calculus rule, sometimes > called existential elimination: I think that's included among what I called 'formulations of the rule of existential instantiation'. I.e., I think you and I are talking about basically the same thing. MoeBlee === Subject: Re: proving not(c = aleph_aleph_null) diagonally without choice > How do we know that it is countable? That is the way > I first see it. No need to rummage around for an injection into N. > It comes with the assertion that A is countable. I see you're a closet constructivist. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: proving not(c = aleph_aleph_null) diagonally without choice > David Hartley a .8ecrit :[cut] > maybe if someone showed me > using one of these models where every cardinal is > a countable union of strictly smaller cardinals > how even then c^(aleph_0) = c > maybe i'd finally get it and stop being such a nuisance? I don't have access to their papers, or any knowledge of forcing and >> model theory, so I can't help you there. > > At this stage, this trickstger is saying something like I need a > proof of 2+2=4 valid even in Moskowki's model. What ? You dont have > heard of it? Only Moskowki is an expert on that model? Then I dont > really believe 2+2=4, and will keep being a nuisance about it... you know when i entered this series of questions i mentioned: i shouldn't avoid finding out even if it means learning something in front of others i've been as upfront as possible that spotting choice was something i never trained and that i was approaching this as a new field of which i have only had passing acquaintance through other interests along the way i tried to ask questions ( you didn't answer any of them ) i tried to bring in resources ( you supplied none ) i tried to break the question apart into manageable pieces ... i tried to do the things that i think students should do when they are confronted with something they do not know the process has brought me some new insights some new confusions and a load of new material it will take me some time to get through but because i have a motivating case i am interested and excited to get through the material in the process i have found some more fundamental things that i need to restudy i think through this process i end up understanding more than i did when i entered your actions on the other hand only tend to polarise and promote a process of choosing sides you added nothing to your initial arguments (though you often attack others of not contributing) and you simply threw out little snide remarks if i was like i was when i was 8 years old i would have fled the discussion in panic if i was like i was when i was 14 years old i would have clamped down like your standard crank and shut off any ability to learn it's only because i am mature enough not to fear looking foolish that i stuck it out until my initial questions were resolved and now i leave with more questions and an interest in the field you on the other hand simply leave letting everyone know you got the smallest dick in town (even compared to me and mina!) -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: proving not(c = aleph_aleph_null) diagonally without choice > In message <_umdnWQiVvopRpnUnZ2dnUVZ_qDinZ2d@trueband.net>, galathaea >> that was my initial thought too >> and i didn't really look at it deeply because it seemed so obvious >> but now it doesn't seem so easy in particular >> we aren't given a space of functions we aren't given (2^N)^N > > But we are. c is defined as |2^N|, and c^aleph_0 as |(2^N)^N|. > >> a set of cardinality c^(aleph_0) >> has the same cardinality as the set of functions >> from a set of cardinality aleph_0 >> to a set of cardinality c to choose a particular function >> for each member of aleph_0 >> seems equivalent to providing a choice function on c > > You don't have to choose such functions. All you need is a specification > of a bijection so that, *given* a function N -> 2^N you can determine > the corresponding function N x N -> 2. It wouldn't even matter if there > were no such function, the proof didn't assume (2^N)^N is non-empty. > (Although we know of course that |(2^N)^N| >= c.) > > There's a similarity here with your qualms over aleph_0 x aleph_0 = > aleph_0. It's well-known that AC is equivalent to the product of a > family of non-empty sets is non-empty. In general in ZF a countable > product of non-empty sets can be empty. But (2^N)^N is not a product of > arbitrary sets, it's (equivalent to) a countable product of copies of > one set and provably non-empty. the choiceless formula would seem to actually need to be something like U aleph >> aleph aleph aleph 0 >> 0 0 0 >> ( 2 ) = 2 if we aren't going to assume a choice is made > > I can't understand this. Presumably the second exponent means the > cardinality of some countable union of countable sets, but we know this > depends on the particular sets. The only countably infinite set we've > got to play with is N, but in that case its aleph_0 x aleph_0 anyway. i must be looking at things from the wrong direction i think it may be a fundamental problem i've been trying hard not to assume choices and i think i've been trying to making unclear distinctions along the way ignore cardinalities for a second this is one way i learned currying: Y Z is the space of functions from Y to Z where Y and Z are some given set by the graph theorem for every element y of Y a function f: Y -> Z produces f(y) e Z so the space of functions is bijective with (x) Z Y so for instance 3 Z = Z x Z x Z now when one has nested exponents Y S ( Z ) one can prove currying holds by noting Y S Y ( Z ) = (x) ( Z ) (by the graph theorem) S (+) Y S = Z (by the product rule) S x Y = Z (disjoint sum of identical sets as cartesian product) writing it out this way uses only elementary relations of exponents in the proof and directly proves the currying relationship when writing out the currying operation as successive applications e.g. f(a)(b) = g(a, b) i saw a shorthand that explicitly follows elements and so was a place that i thought could be hiding a choice by looking at the proof that goes through explicitly i saw steps that introduced the forms which are questionably reducible when choice is not available in particular the infinite product and sum forms so in my admittedly confused mind i thought that this proof of curryability either through the demonstration of repeated application which required member placeholders or the path of direct derivation which went through these questionable forms could be indicating hiding the use of choice >> or >> maybe to make my thinking clearer >> i'll just ask some questions - isn't the existence of a function that chooses an element of c >> for each of the sets of cardinality (2^(aleph_0))^(aleph_0) >> and each element of the set of cardinality aleph_0 >> and produces the appropriate function >> equivalent to a choice function on (2^(aleph_0))^(aleph_0)? > What sets of cardinality (2^(aleph_0))^(aleph_0)? Again we've only got > one. I think you mean something like ... No. I just can't work out what > you mean. I hope I answered it above anyway. > (Incidentally I don't think c is necessarily a set of cardinality c (if > we're using Scott's trick), so you need to use 2^N.) now let me describe again in my own confused manner here the way i have been viewing cardinals and choice and where things may be going wrong in my understanding cardinality has meant to me the existence of a bijection e.g. iff card(X) = card(Y) then thereExists bijection f: X <-> Y choice has meant if a collection C has card(C) infinite choice shows thereExists function f suchThat forAll sets S e C f(S) e S in my unclear state i've been thinking something along: let T = some collection suchThat card(T) = card((2^N)^N) look at functions N -> 2^N the existence of the cardinality bijection implies that forEach t e T thereExists a function f_t: N -> 2^N the reason i have become suspicious is that once you can talk about applying elements n of N you are talking about a function apply_n suchThat forEach t e T apply_n(f_t) is an element of 2^N and if you view each function N -> 2^N as it's graph in N x 2^N apply_n(f_t) is producing an element of f_t ( even more explicit example: if every set S in some collection C is of the form {(1, a1), (2, a2), ...} then apply_1 could select out the element (1, a1) e S forEach S in C ) and i thought isn't that choice? avoiding the element-wise solution seemed to bring me back to countable unions and such again choicy areas to be in >> maybe if someone showed me >> using one of these models where every cardinal is >> a countable union of strictly smaller cardinals >> how even then c^(aleph_0) = c >> maybe i'd finally get it and stop being such a nuisance? > > I don't have access to their papers, or any knowledge of forcing and > model theory, so I can't help you there. it doesn't matter in my hunting expeditions trying to understand this i've just found http://aps.arxiv.org/abs/0704.3998 which explicitly states that this whole thing with c^(aleph_0) = c is true in ZF for the reasons given (see second part to theorem 13) it's clear my concerns were ill-founded i'm sorry to have wasted anyone's time -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: proving not(c = aleph_aleph_null) diagonally without choice >> In message <_umdnWQiVvopRpnUnZ2dnUVZ_qDinZ2d@trueband.net>, galathaea > that was my initial thought too > and i didn't really look at it deeply because it seemed so obvious > but now it doesn't seem so easy in particular > we aren't given a space of functions we aren't given (2^N)^N But we are. c is defined as |2^N|, and c^aleph_0 as |(2^N)^N|. > a set of cardinality c^(aleph_0) > has the same cardinality as the set of functions > from a set of cardinality aleph_0 > to a set of cardinality c to choose a particular function > for each member of aleph_0 > seems equivalent to providing a choice function on c You don't have to choose such functions. All you need is a >> specification of a bijection so that, *given* a function N -> 2^N you >> can determine the corresponding function N x N -> 2. It wouldn't even >> matter if there were no such function, the proof didn't assume (2^N)^N >> is non-empty. (Although we know of course that |(2^N)^N| >= c.) There's a similarity here with your qualms over aleph_0 x aleph_0 = >> aleph_0. It's well-known that AC is equivalent to the product of a >> family of non-empty sets is non-empty. In general in ZF a countable >> product of non-empty sets can be empty. But (2^N)^N is not a product >> of arbitrary sets, it's (equivalent to) a countable product of copies >> of one set and provably non-empty. the choiceless formula would seem to actually need to be something like U aleph > aleph aleph aleph 0 > 0 0 0 > ( 2 ) = 2 if we aren't going to assume a choice is made I can't understand this. Presumably the second exponent means the >> cardinality of some countable union of countable sets, but we know >> this depends on the particular sets. The only countably infinite set >> we've got to play with is N, but in that case its aleph_0 x aleph_0 >> anyway. > > i must be looking at things from the wrong direction > > i think it may be a fundamental problem > > i've been trying hard not to assume choices > and i think i've been trying to making unclear distinctions along the way > > ignore cardinalities for a second > this is one way i learned currying: > > Y > Z > > is the space of functions from Y to Z > where Y and Z are some given set > > by the graph theorem > for every element y of Y > a function f: Y -> Z produces f(y) e Z > so the space of functions is bijective with > > (x) Z > Y > > so > for instance > > 3 > Z = Z x Z x Z > > now > when one has nested exponents > > Y S > ( Z ) > > one can prove currying holds by noting > > Y S Y > ( Z ) = (x) ( Z ) (by the graph theorem) > S > > (+) Y > S > = Z (by the product rule) > > S x Y > = Z (disjoint sum of identical sets as cartesian product) > > writing it out this way > uses only elementary relations of exponents in the proof > and directly proves the currying relationship > > when writing out the currying operation as successive applications > e.g. f(a)(b) = g(a, b) > i saw a shorthand that explicitly follows elements > and so was a place that i thought could be hiding a choice > > by looking at the proof that goes through explicitly > i saw steps that introduced the forms which are questionably reducible > when choice is not available > in particular the infinite product and sum forms > > so > in my admittedly confused mind > i thought that this proof of curryability > either through the demonstration of repeated application > which required member placeholders > or the path of direct derivation > which went through these questionable forms > could be indicating hiding the use of choice > > or > maybe to make my thinking clearer > i'll just ask some questions - isn't the existence of a function that chooses an element of c > for each of the sets of cardinality (2^(aleph_0))^(aleph_0) > and each element of the set of cardinality aleph_0 > and produces the appropriate function > equivalent to a choice function on (2^(aleph_0))^(aleph_0)? > What sets of cardinality (2^(aleph_0))^(aleph_0)? Again we've only got >> one. I think you mean something like ... No. I just can't work out >> what you mean. I hope I answered it above anyway. >> (Incidentally I don't think c is necessarily a set of cardinality c >> (if we're using Scott's trick), so you need to use 2^N.) > > now let me describe > again in my own confused manner here > the way i have been viewing cardinals and choice > and where things may be going wrong in my understanding > > cardinality has meant to me the existence of a bijection > e.g. iff card(X) = card(Y) then thereExists bijection f: X <-> Y > > choice has meant > if a collection C has card(C) infinite > choice shows thereExists function f > suchThat forAll sets S e C > f(S) e S > > in my unclear state > i've been thinking something along: > > let T = some collection suchThat card(T) = card((2^N)^N) > > look at functions N -> 2^N > > the existence of the cardinality bijection > implies that forEach t e T thereExists a function f_t: N -> 2^N > > the reason i have become suspicious is that > once you can talk about applying elements n of N > you are talking about a function apply_n > suchThat forEach t e T > apply_n(f_t) is an element of 2^N > > and if you view each function N -> 2^N as it's graph in N x 2^N > apply_n(f_t) is producing an element of f_t > > ( even more explicit example: > if every set S in some collection C is of the form > {(1, a1), (2, a2), ...} > then apply_1 could select out the element (1, a1) e S > forEach S in C ) > > and i thought > isn't that choice? > > avoiding the element-wise solution > seemed to bring me back to countable unions and such > again choicy areas to be in > > maybe if someone showed me > using one of these models where every cardinal is > a countable union of strictly smaller cardinals > how even then c^(aleph_0) = c > maybe i'd finally get it and stop being such a nuisance? I don't have access to their papers, or any knowledge of forcing and >> model theory, so I can't help you there. > > it doesn't matter > > in my hunting expeditions trying to understand this > i've just found > > http://aps.arxiv.org/abs/0704.3998 > > which explicitly states that this whole thing with c^(aleph_0) = c > is true in ZF for the reasons given > (see second part to theorem 13) > > it's clear my concerns were ill-founded > > i'm sorry to have wasted anyone's time set theory without the axiom of choice, I'd say one of the clearest is probably David Libert (I don't here discount in any way contributions in sci.math or sci.logic by Fred Galvin and others). For example, there's this post from 2000: I haven't looked hard at that particular post, but I think he's good at explaining things. David Bernier === Subject: Re: -- closure of a subset of N under some operations > > Here's an alternate version of one of my previous > conjectures ... > > Conjecture: > > If S is a nonempty subset of {n in N | n > 2} such > that > > x,y in S => xy, (x-1)(y-1), xy-x+1 in S > > then S contains at least one pair of consecutive > positive integers. > > quasi > > x = 7 y = 11 > > xy = 77 > > (x-1)(y-1) = 60 > > xy - x + 1 = 71 > > 77*60 = 4620 > > 71*71 = 5041 > > 70*70 = 4900 > > 76*76 = 5776 > > 77*71 = 5467 > > note that xy - x + 1 is not symmetric so it has no > well-defined inverse iteration > > thus i believe the conjecture is false ... > > ( unless i misunderstood ) > > > tommy1729 Actually none of the three rules has a well defined inverse iteration. For example, from the value of xy one cannot backtrack to the particular x,y which produced it. Think of the inverses as number-to-set-of pairs functions. For example the inverse of 6 under the map (x,y)->xy consists of the pairs (1,6), (2,3), (3,2), (6,1). The fact that these are not number-to number inverses has nothing to do with whether the conjecture is true. === Subject: Run your own forums. posting-account=HZNw1AoAAACj5O59E7bG9-eTCyZA24mT .NET CLR 1.1.4322; .NET CLR 2.0.50727; Zango 10.0.370.0),gzip(gfe),gzip(gfe) spider-ntc-tc05.proxy.aol.com[CFC87045] (Prism/1.2.1), HTTP/1.1 cache-ntc-ac02.proxy.aol.com[CFC87483] (Traffic-Server/6.1.5 [uScM]) Run your own forums. ForumPro 1.0 Professional Run your own forums. Easiest software to setup in existence. All Written in One exe Unlimited Groups, Boards, Messages, And Users Very Easy To Use, Fully Configurable Lowest Cost Software Available http://www.sharewaresoftware.org/forumpro/forumpro.htm === Subject: Q on Riemann Hypothesis posting-account=yXs7PwoAAABxalAQoV3duo5HJUDgYk1_ MathPlayer 2.10d; .NET CLR 2.0.50727; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) Why is the following not a correct proof of the Riemann Hypothesis? Definitions: z(s)=Riemann zeta function=sum_n=1_to_inf{(-1)^(n+1)*(1/n^s)}, valid for Re(s)>0 eps=the variable epsilon with 0<=|eps|<1/2 non-trivial root s_r = 1/2-eps+i*t, with t, real exp - the exponential function ln - the inverse exponential function s** - the complex conjugate of s s0=s_r+eps=1/2+i*t sinh(x) - hyperbolic sine, evaluated at x Proof: The non-trivial zeros of z(s) are known to reside in the critical strip 0 posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) On Oct 27, 10:42am, stevendaryl3...@yahoo.com (Daryl McCullough) > MoeBlee says... Whatever that is supposed to mean, it doesn't prove that every model >of ZF has a universe that has every natural number as a member. And >it's trivial to prove that (if ZF has a model then) ZF has a model in >which it is not the case that every natural number is a member of the >universe. for every n, there is an element e n of the model that > has exactly n elements. Why isn't e n identified with the > natural number n? Of course, it's just a *representation* > of the natural number n. I think we can agree that what you've argued is that in any model of ZFC there is a system ISOMORPHIC to the system of natural numbers. > It doesn't really make sense to > ask whether e n is *really* the natural number n. I didn't use the word 'really'. Rather, we have a definition of 'is a natural number' (whatever definition that may be, whether they be the finite ordinals or some other definition that provides a system isomorphic to the system of finite ordinals). Then it is trivial to prove that there are models of ZFC such that not every natural number is a member of the universe of that model. That every model has a system ISOMORPHIC to the system of natural numbers is, of course, not contested. MoeBlee === Subject: Re: Why meta diagonals are irrelevant posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > So that WM claims that an inability to name some numbers means that > those unnamed numbers are all equal? There is no chance to decide that question, because there is no chance > to choose one of such unnamed numbers. If one knows that unnamed numbers belong to disjoint sets, one can konw > that they are not equal. If one knows that unnamed numbers belong to disjoint sets, and if one knows that unnamed numbers are not equal, then one can know that they are not equal. But one does not know anything concerning unnamed numbers. For example a positive number and a negative, even when both are > unnamed, can be distinguished. Every immaterial ghost is an immaterial ghost. Does this knowledge help you or the ghosts? So any number known to be a member of some proper subset of the reals, > such as an interval, can be distinguished from any real non-member. But you cannot take one or two individuals of such numbers. Of course > you can take all the real numbers between 1 and 2, but that is not a > non-baptized number. It certainly contains baptized numbers, and they can be clearly > distinguished from all the numbers, baptized or not, which lie outside > that interval. It is an interval of rational numbers with a few > irrational numbers among them. A FEW? There are 'more' irrationals than rationals in any interval of > postive length. Not those which you can distinguish by names or other means. There is no chance to use any non- > baptized number. You can also call the baptized numbers selectable > numbers or individualizable numbers or briefly numbers in > contrast to non-numbers. The more standard term is inaccessible numbers. A number is inaccessible if there is some positive epsilon such that the > number's value cannot be found to within less than that epsilon. A number is accessible if for every positive epsilon, its value can be > determined with error less than that epsilon. Most reals are inaccessible. with epsilon = oo. Therefore most reals are not real. Regars, WM === Subject: Re: Why meta diagonals are irrelevant > So that WM claims that an inability to name some numbers means that > those unnamed numbers are all equal? There is no chance to decide that question, because there is no chance > to choose one of such unnamed numbers. If one knows that unnamed numbers belong to disjoint sets, one can konw > that they are not equal. > > If one knows that unnamed numbers belong to disjoint sets, and if one > knows that unnamed numbers are not equal, then one can know that they > are not equal. But one does not know anything concerning unnamed > numbers. One knows enough about unnamed numbers to know that the ones in the set of all positive reals are distinct from the ones in the set of all negative reals. And similarly for any non-empty real interval having a non-empty compliment in the reals. So one can know something about the unnamed numbers which belong to some known interval. For example a positive number and a negative, even when both are > unnamed, can be distinguished. > > Every immaterial ghost is an immaterial ghost. Does this knowledge > help you or the ghosts? Knowing that a number is, or is not, a member of some given interval, certainly helps. At least it helps everyone other than WM. So any number known to be a member of some proper subset of the reals, > such as an interval, can be distinguished from any real non-member. But you cannot take one or two individuals of such numbers. Of course > you can take all the real numbers between 1 and 2, but that is not a > non-baptized number. It certainly contains unbaptized numbers, and they can be clearly > distinguished from all the numbers, baptized or not, which lie outside > that interval. It is an interval of rational numbers with a few > irrational numbers among them. A FEW? There are 'more' irrationals than rationals in any interval of > postive length. > > Not those which you can distinguish by names or other means. There are at least as many nameable irrationals, since the countable set of algebraic numbers contains as many irrationals as rationals, and there are at least countably many transcendentals that can be named (the set of rational multiples of pi, for example). There is no chance to use any non- > baptized number. You can also call the baptized numbers selectable > numbers or individualizable numbers or briefly numbers in > contrast to non-numbers. The more standard term is inaccessible numbers. A number is inaccessible if there is some positive epsilon such that the > number's value cannot be found to within less than that epsilon. A number is accessible if for every positive epsilon, its value can be > determined with error less than that epsilon. Most reals are inaccessible. > > with epsilon = oo. If WM insists on taking epsilon to be only oo, then every one of the uncountably many reals is accessible by his own standard. It is only if one allows epsilon to take arbitrarily small positive real (or rational) values that the definitions make any sense mathematically. But we already know that the overlap between the world of mathematics and the world of WM contains very little of the world of mathematics. === Subject: Re: Why meta diagonals are irrelevant posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > To claim that this sequence has limit 0 is of same impertinence as the > claim > > LIM{n --> oo} (1 + 2 + 3 + ... + n) = 0 > > In fact, Dik T. Winter did so. But that is not what I call mathematics > (notwithstanding Euler's erroneous results). It was a *definition*. With > standard definitions that limit does not exist. So in mathematics it > is possible to *provide* a definition. You gave this definition. It is unimportant whether this limit does exist or does not exist in standard definitions. Your claim that this definition is in agreement with mathematics is contradicting mathematics. === Subject: Re: Why meta diagonals are irrelevant > To claim that this sequence has limit 0 is of same impertinence as the > claim > > LIM{n --> oo} (1 + 2 + 3 + ... + n) = 0 > > In fact, Dik T. Winter did so. But that is not what I call mathematics > (notwithstanding Euler's erroneous results). It was a *definition*. With > standard definitions that limit does not exist. So in mathematics it > is possible to *provide* a definition. > > You gave this definition. It is unimportant whether this limit does > exist or does not exist in standard definitions. Your claim that this > definition is in agreement with mathematics is contradicting > mathematics. To claim such reveals WM's ignorance of mathematics. While such a definition might be unusual and of little practical use, it is in no way in contradiction with any standard mathematics. === Subject: Re: Why meta diagonals are irrelevant Nntp-Posting-Host: hera.cwi.nl > > To claim that this sequence has limit 0 is of same impertinence as the > > claim > > > > LIM{n --> oo} (1 + 2 + 3 + ... + n) =3D 0 > > > > In fact, Dik T. Winter did so. But that is not what I call mathematics > > (notwithstanding Euler's erroneous results). > > It was a *definition*. With > standard definitions that limit does not exist. So in mathematics it > is possible to *provide* a definition. > > You gave this definition. It is unimportant whether this limit does > exist or does not exist in standard definitions. Your claim that this > definition is in agreement with mathematics is contradicting > mathematics. What part of mathematics does it contradict? BTW, I just remember that I sum{n = 1...oo} (1 + 2 + 3 + ... + n) = 0 As that sum is not defined within standard mathematics, that definition is not in disagreement with standard mathematics. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Why meta diagonals are irrelevant posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) >Therefore we can construct a sequence of initial segments that for >*every* natural n is a_n = omega - |{1,2,3,...,n}| = omega To claim that this sequence has limit 0 is of same impertinence as the >claim LIM{n --> oo} (1 + 2 + 3 + ... + n) = 0 In fact, Dik T. Winter did so. But that is not what I call mathematics >(notwithstanding Euler's erroneous results). Therefore the union of all finite initial segments does never yield a >set of aleph_0 elements That's the *definition* of aleph_0. aleph_0 is the set containing > all natural numbers (which is the same as the union of all initial > segments of the naturals). So of course the union of all finite > initial segments yields a set with aleph_0 elements. The union > of all initial segments of the naturals includes *every* natural. > So the union is the set of all naturals, which is aleph_0. The definition is aleph_0 > n for evey n in N. That is a contradiction. {1,2,3,...,n+1} U {1,2,3,...,n+2} U {1,2,3,...,n+3} U ... = >{1,2,3,...} = N (U) You can drop every finite initial segment that ends with a natural >number, because we can prove that it is not required to form the set >N. Obviously, that's incorrect. No, obviously that is correct for every natural number. There is no natural number left over that is required on the left hand side. This shows that set theory is incorrect, and for some reason, you are unable to understand that. If we prove that every natural number is divisible by itself without remainder. Does that leave over natural numbers that are not proven to be divisible without remainder? Once again, if we let U_n = the set of all initial segments of N > with at least n elements, then we have forall n, union(U_n) = N And, if existing, then precisely this would be the union of numbers that can be dropped on the left hand side of (U). For every n the segment {1,2,3,...,n} is not required in the union (U) >given above. Yes, that's right. We know that because union(U_{n+1}) = N. Would you say that a proof for every n leaves some n over? No, it's true for all n that union(U_n) = N. So you recognize the contradiction. Why don't you admit it? === Subject: Re: Why meta diagonals are irrelevant That's the *definition* of aleph_0. aleph_0 is the set containing > all natural numbers (which is the same as the union of all initial > segments of the naturals). So of course the union of all finite > initial segments yields a set with aleph_0 elements. The union > of all initial segments of the naturals includes *every* natural. > So the union is the set of all naturals, which is aleph_0. > > The definition is aleph_0 > n for evey n in N. That is a > contradiction. That is NOT the definition of aleph_0, at least anywhere outside of WM's weird weird world, since that would allow aleph_0 to be uncountable. {1,2,3,...,n+1} U {s,2,3,...,n+2} U {1,2,3,...,n+3} U ... = >{1,2,3,...} = N (U) You can drop every finite initial segment that ends with a natural >number, because we can prove that it is not required to form the set >N. Obviously, that's incorrect. > > No, obviously that is correct for every natural number. WM conflates every with any. While one can drop ANY finite initial segment of N and have something left, dropping EVERY finite initial segment leaves nothing left. At least in English. If German is ambiguous on this issue then it is not a good language to use for mathematics. let S = {1,2}. If one removes ANY element of S, what remains will not be empty, but if one removes EVERY element of S, only an empty set remains. If the same is not true in the German language , then German is of dubious use in mathematics. === Subject: Re: Why meta diagonals are irrelevant WM says... >>Therefore the union of all finite initial segments does never yield a >>set of aleph_0 elements That's the *definition* of aleph_0. aleph_0 is the set containing >> all natural numbers (which is the same as the union of all initial >> segments of the naturals). So of course the union of all finite >> initial segments yields a set with aleph_0 elements. The union >> of all initial segments of the naturals includes *every* natural. >> So the union is the set of all naturals, which is aleph_0. The definition is aleph_0 > n for evey n in N. There are alternative definitions. One is that it is the cardinality of the set of natural numbers. Another is that it is the smallest infinite cardinality. They are equivalent. >That is a contradiction. After all this time, you still don't know what a contradiction is? A contradiction is a statement of the form Phi & ~Phi (or a collection of statements that imply such a conclusion). What statement Phi are you talking about such that Phi & ~Phi is provable? >>{1,2,3,...,n+1} U {1,2,3,...,n+2} U {1,2,3,...,n+3} U ... = >>{1,2,3,...} = N (U) You can drop every finite initial segment that ends with a natural >>number, because we can prove that it is not required to form the set >>N. Obviously, that's incorrect. No, obviously that is correct for every natural number. Look, the unambiguous, correct statement is this: If we let U_n = the set of all finite initial segments S of the naturals such that size(S) >= n, then Claim 1: forall n, union(U_n) = N That's all there is to say about it. It isn't contradictory, it is just true. What you always like to do is to start with a precise mathematical statement such as Claim 1, and then make it ambiguous by rendering it into natural language, then reinterpret the ambiguous statement so that it means something different from the original statement, and something provably false. That's the way you go about proving contradictions. Claim 1 is true. You paraphrase Claim 1 as You can drop every finite initial segment that ends with a natural number, and the union is still N. That is ambiguous. It either means Interpretation 1: Forall natural numbers n, you can drop the first n initial segments, and the union of the remaining sets is N. That's true, and is equivalent to Claim 1. Alternatively, it means Interpretation 2: You can remove *all* initial segments and the union of the remaining sets is N. That's clearly false, and is not equivalent to Claim 1. You would *know* that if you actually used mathematical notation, instead of sloppy natural language. Look at a simple, finite example. I have a box that has three objects in it: An apple, a pear, and an orange. Then: Claim 1 (True claim): For all f such that f is a fruit in the box, removing f results in a nonempty box. Claim 2 (False claim): Removing all fruits from the box results in a nonempty box. Surely, you understand that Claim 1 is true, but Claim 2 is false? Then look at the following analogy I have a set, U that contains the finite initial segments {}, {0}, {0,1}, {0,1,2}, ... Claim 1 (True claim): Forall finite initial segments S, removing S (and all smaller initial segements) from U results in a nonempty set. Claim 2 (False claim): Removing all finite initial segments from U results in a nonempty set. Claim 1 is true, and claim 2 is false. You understand the difference when we are talking about a box of fruit. Why can't you understand the difference when we are talking about sets of initial segments of the naturals? -- Daryl McCullough Ithaca, NY === Subject: Re: Two More New Math Ideas posting-account=SWr-zgoAAABXP_I1KRp8SBR28UDaox2f Gecko/2008092510 Ubuntu/8.04 (hardy) Firefox/3.0.3,gzip(gfe),gzip(gfe) Isn't it interesting that both (pi) and (-1) don't seem to have a square root in regular numbers. But when we consider the natural united of c, it can be considered the square root of -1, and pi itself, or at least 2pi, might also be considered to be equal to c, in this way. considered square root of -1 in way I described. Furthermore if (h/ (hx2pi) which is = to (cx2pi) and a measure of its wavelength, than (c x c or c^2) = (c x 2 pi) which makes (c = 2pi) in this special case. It is very interesting that the constants of the universe, including pi seem to be somehow unified around the speed of light. Just as (h = c = 2pi) in this case, (c^2 = h/2pi = G) and the formula used to calculate the Planck scale which involves h/2pi, G and c instead of c^2 may be incorrect. The point at which the constants of E (energy), M (mass) T (time) G (gravity) Q (charge), are unified is at (c^2) not the so called Planck scale, which according to :http:// www.phys.unsw.edu.au/einsteinlight/jw/module6_Planck.htm is way out of our reach at The Planck length is 1.6 x 10-35 metres, and that is a good thing. Conrad Countess === Subject: eigenvalue monotonicity posting-account=z7MvkwoAAABTr6FuOIl2OaQFQ_irgZzA Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) I was hoping that somebody could provide me with some insights into the following problem. Given a real matrix A and its non-negative counterpart, |A| (wherein all the matrix entries have been replaced by their absolute values), it is known that rho(A) <= rho(|A|) (e.g., theorem 8.1.18 in Horn and Johnson, 1985), where given a nxn matrix X, rho(X) is its spectral radius (absolute value of the eigenvalue with largest magnitude). However, can it be shown possibly under certain restrictions on A, that rho(A) increases (possibly monotonically) with rho(|A|)? Mhasoba