>I read somewhere that a countable set is finite iff there does not >exist a bijection from it to any proper subset of it. How to express this in mathematical notation? I'll cut some corners by >using exists f to mean there is a function f. > > [.snip.] > >> Unless you want it to say exactly a bijection. If so, you could always >> write > >> not (exists B subset A: B!=A & exists f:A->B, exists g:B->A >> fg=id_B, gf=id_A). You mean that fg and gf are the combined functions of f and g, and >respectively of g and f, and id_B and id_A are the identity functions of >B and A? > > Yes. It is an easy exercise for students when they first learn about > functions that a (set theoretic) function is injective if and only if > it has a left inverse if and only if it is left cancellabe; a function > is surjective if and only if it has a right inverse if and only if it > is right cancellabe; and a function is bijective if and only if it has > two-sided inverse if and only if it has both a left and a right > inverse if and only if it is cancellabel on either side. It is also interesting to point out that the proof of the characterization of surjectivity requires the Axiom of Choice while that of the chzracterization of the injectivity doesn't. Felix. ==== |I read somewhere that a countable set is finite iff there does not |exist a bijection from it to any proper subset of it. You don't have to assume it's countable. An uncountable set is not finite, and there exists a bijection from it to a proper subset. Keith Ramsay ==== >|I read somewhere that a countable set is finite iff there does not >|exist a bijection from it to any proper subset of it. > You don't have to assume it's countable. An uncountable set > is not finite, and there exists a bijection from it to a proper > subset. Assuming the axiom of choice. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== AyAx(y=x <-> y in {x}). > Find a copy of David Lewis' Parts of Classes for a clearer discussion (not necessarily clear) of this relation to singletons. It was published in 1991 There is a good formal discussion of the mathematics behind this approach. Run a google search on A necessary relation algebra for mereotopology The relationship of mereologies to quantum mechanics can be seen by the fact that mereologies relate to one another as an orthocomplemented lattice. This is the same as for the Boolean subblocks of von Neumann's quantum logic. The relationship of classical mereologies to set theory can be seen by virtue of Tarski's result that a classical mereology can be generated from a Boolean algebra without 0. This is the same basic structure into which separative partial orders are embedded in order to generate Cohen generic models for Zermelo-Fraenkel set theory. :-) mitch ==== : The relationship of classical : mereologies to set theory can be seen by virtue of Tarski's : result that a classical mereology can be generated from a : Boolean algebra without 0. There is NO SUCH THING as a Boolean algebra without 0. Seriously, if you google Boolean algebra without 0, you will get hits, but on NONE of them will the prepositional phrase without 0 show up as modifying Boolean algebra. Of course, any less stupid than Mitch could've just looked at the axioms defining a boolean algebra and noticed that 0 occurs in them. Obviously mitch knew that Boolean algebras normally have an 0. The question I am trying to ask now is, just how much damage is done when the 0 is removed? Does what is left really DESERVE to be called ANY kind of Boolean algebra? -- --- It's difficult ... you need to be united to have any strength, but internal issues have to be addressed. --- E. Ray Lewis, on liberalism in America ==== > f:R+---> R+*, C^1, f strictely increasing > > An = int_[0;n] dt/(f(t) + f '(t)) dt <= 1 > > Prove that there exist a real M >0 such that > > Bn = int_[0;n] dt/f(t) <= M for any n in N > Actually we can simplify this argument somewhat, and remove the use of measure theory. Simply note: 1/f(t) <= 2/(f(t)+f'(t)) + f'(t)/f(t)^2 (Proof: if f'(t) <= f(t) then the first term on the RHS is >= the LHS, otherwise the second term on the RHS is > the LHS.) Integrating from 0 to n gives: Bn <= 2An + (1/f(0)-1/f(n)) <= 2 + 1/f(0) So you can take M = 2 + 1/f(0). Michael ==== In today's NYT (page C6 in my edition) there's an ad by DHL that lists 400 zip codes. Are they in random order or is there a pattern? ==== Several people on sci.math dual with James Harris on various mathematical (and other) topics. Currently, much energy is being expended on Lemma 1 and Lemma 2. Here is a suggestion for a plan of activity. I am well aware that this suggestion is likely to be ignored. Do the following steps in numerical order. No jumping ahead is ever allowed. Step 1: Produce a wording for Lemma 1 which is acceptable to both sides. Only standard math terminology should be used. (But further terms could be defined, of course.) Step 2: Produce an example of the application of Lemma 1 which is acceptable to both sides. Step 3: Produce a proof for Lemma 1 which is acceptable to both sides. Only standard math terminology should be used. (But further terms could be defined, of course.) Step 4: Produce a wording for Lemma 2 which is acceptable to both sides. Only standard math terminology should be used. (But further terms could be defined, of course.) Step 5: Produce an example of the application of Lemma 2 which is acceptable to both sides. Step 6: Produce a proof for Lemma 2 which is acceptable to both sides. Only standard math terminology should be used. (But further terms could be defined, of course.) If any step, beyond step 1, proves impossible: Go back to step 1. If step 1 proves impossible abandon further discussion of Lemma 1 and Lemma 2. All definitions, examples and proofs should be too verbose, rather than too concise. -- Clive Tooth http://www.clivetooth.dk ==== * The Last Danish Pastry > Several people on sci.math dual with James Harris on various mathematical > (and other) topics. Currently, much energy is being expended on Lemma 1 and > Lemma 2. Here is a suggestion for a plan of activity. I am well aware that > this suggestion is likely to be ignored. The problem with your suggestion is that it is likely to settle the case. And who would want that? However, I guess that it would strand on the very first paragraph in step 1. -- Jon Haugsand ==== * The Last Danish Pastry > Several people on sci.math dual with James Harris on various mathematical > (and other) topics. Currently, much energy is being expended on Lemma 1 and > Lemma 2. Here is a suggestion for a plan of activity. I am well aware that > this suggestion is likely to be ignored. The problem with your suggestion is that it is likely to settle the case. And who would want that? However, I guess that it would strand on the very first paragraph in step 1. -- Jon Haugsand ==== >>Is there established mathematical notation for this set is finite, >>this set is countable and this set is uncountable? >>I think one way of writing A is finite would be #A in N (where in >>is the is an element of symbol and N is the set of natural numbers), >>but how to write the others? Would A is countable be #A = #N and A >>is uncountable be #A > #N? (The N is again the set of natural >>numbers.) >In the context of set theory, one could write |A| < w and |A| >= w, >where I use w in place of lower-case omega. Outside of set theory, I >don't recall ever seeing a case where this wasn't just written out in words. Without the Axiom of Choice, the standard terminology is to use uncountable to mean |A| ~<= aleph_0, and to use transfinite for |A| >= aleph_0. Dedekind-finite infinite cardinals are uncountable but not transfinite. -- 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 ==== Wow. Even when people try to explain the difference and why >you might want to take note of it you don't see the distinction >between sounding like a lunatic and sounding like a >_dangerous_ lunatic. Wow. > > If you math guys were in high school, JH would be in jail right > now for making those kinds of threats. This has become completely > out of hand. > > /BAH >>What threats? Not to speak for /BAH, but: While threat may not be accurate, >strictly speaking, the distinction between an actual threat and >your statement that entities that _you_ can hear but we cannot >hear are going to kill us is a little subtle. /BAH ==== >> >>Wow. Even when people try to explain the difference and why >>you might want to take note of it you don't see the distinction >>between sounding like a lunatic and sounding like a >>_dangerous_ lunatic. >>Wow. >> >> If you math guys were in high school, JH would be in jail right >> now for making those kinds of threats. This has become completely >> out of hand. >> >> /BAH What threats? The ones you posted. Don't do that. /BAH ==== > It's like from the movie Ghostbusters where the evil thing tells the > Ghostbusters to make a choice so they all try to blank their minds. As if angry mobs, the FBI, CIA, NSA, the US Army, etc weren't enough, he's now going to bring forth the Stay-Puft Marshmallow Man to punish the evil liars! Well I'm convinced. -- Dave Taylor 'Ray, when someone asks you if you're a god, you say, Yes!' ==== > If I have AxB=C and I know A and C, how do I find B? Besides brute-force > computation. Firstly we hope A.C=0. Otherwise there cannot be any solutions for B. The question assumes B is unique. It is not. This is clearly seen because given any solution x for B, (x+A),(x+2A), .... are also solutions. We can however identify a line of solutions. The easiest member of the solution set is the vector perpendicular to C and A (lets call it x0). We can see it is in the direction of C x A. It's length must be |C|/|A|. Therefore x0 = C x A / (|A|^2) We pointed out before that we can add any muliple of A to x0 to get a solution, so the complete set of solutions is: {x | x = x0 + kA, for any k real} ==== Greetings. I just finished some FEM work which required numerical integration of a triangle. I got the technique for performing this integration from Jianming Jin's book on FEM. Using his formalism, you take the (x,y,z) coordinate of each of the three vertices and plug them into a formula which multiplies these coordinates by special factors (abscissae?) to obtain the actual points of integration within the triangle. Each point is given an associated weight. I need to obtain a similar table for the arbitrary three-dimensional tetrahedron. I have seen some tables out there (given by Ronald Cools) on the net, but I do not know how to use them, and I've seen nothing like Jin's table for the triangle applied to solid elements anywhere. Could someone please help me with the problem of obtaining a robust, understandable way to numerically integrate over an arbitrary tetrahedron? Michael ==== > Typing Monkeys Don't Write Shakespeare A rebuttal: Typing Monkeys Don't Write Shakespeare = Keen networked simian sets typography -- Morgan Lewis mrl@efn.org mlewis@cs.uoregon.edu ==== > Typing Monkeys Don't Write Shakespeare A rebuttal: Typing Monkeys Don't Write Shakespeare >= >Keen networked simian sets typography Stinky Morgan ponders the kewpie, Yeats! ========================= Endeavor to persevere ========================= ==== Over a hundred years ago the ring of algebraic integers became a part of the mathematical lexicon, waiting until now for an intriguing problem with the ring to be revealed by the use of advanced polynomial factorization techniques, which work by using non-polynomial factors of a polynomial. In what follows variables, unless otherwise noted, are in the ring of algebraic integers. Lemma 1: Given a factor G(X) of a polynomial P(X), R(X) and C exists such that G(X)=R(X) + C, where C=G(0), and R(X)=G(X)-C. Proof: Consider that C is a factor of the constant term P(0), which follows as G(X) is a factor of P(X), so G(0) is a factor of P(0), but C=G(0); therefore, C exists and is a factor of the constant term P(0). R(X) = G(X)-C, so it exists as well. Proof Complete. Lemma 2: Now consider P(X) such that it has b^2 as a factor. Further consider that P(0)/b^2 is coprime to b; then if P(X) has the factor G(X) then C, from lemma 1, must either be coprime to b or have a factor in common with b^2. Proof: Since C is a factor of P(0), and P(0)/b^2 is coprime to b, if C is not already coprime to b, C must have a factor in common with b^2 such that when b^2 is divided off of P(0), that factor divides off of C leaving a result coprime to b. Proof Complete. Now consider P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - 3(-1 + b^2 X )t u^2 + b u^3) where the odd grouping is so that I can factor P(X) into non-polynomial factors. Doing so I have the factorization P(X) = (r_1(X) t + bu)(r_2(X) t + bu)(r_3(X) t + bu) where again the purpose above with the special grouping was to get that factorization. Now letting G_1(X) = r_1(X) t + bu, consider that X=0, gives P(0) = b^2 (3t u^2 + b u^3) = b^2 u^2 (3t + bu), so at X=0, (r_1(0) t + bu)(r_2(0) t + bu)(r_3(0) t + bu) = b^2 u^2 (3t + bu) which shows that at least two of the r's go to 0, when X=0, and choosing r_1 to be one of them, as the indices are arbitrary, I have from lemma 1 that G_1(0) = bu, so C_1 = bu. Then R_1(X) = G_1(X) - C_1 = r_1(X) t + bu - bu = r_1(X) t. Given that one other of the r's goes to 0, I have a similar result for G_2(X) = r_2(X) t + bu. But one of the r's does not go to 0, and letting that one be r_3, I have for G_3(X) = r_3(X) t + bu, that G_3(0) = 3t + bu, so C_3 = 3t + bu and that gives G_1(0) G_2(0) G_3(0) = b^2 u^2 (3t + bu) = P(0) which is correct. Now consider that C_1 = bu, which shows it has a factor that is b. Notice also that the constant term P(0) = b^2 u^2 (3t + bu), so it has a factor that is b^2, but dividing that factor off gives P(0)/b^2 = u^2 (3t + bu) so if b is coprime to 3 and t, then P(0) is coprime to b. Assume that b is coprime to 3 and t. Now C_1=bu, and it is a factor of the constant term P(0), so when b^2 is divided off of P(X), and as shown divides off of P(0), it MUST divide off of C_1, from lemma 2. Therefore, r_1(X) t should have a factor that is b, and given that b is coprime to t, then r_1(X) should have a factor that is b. I say should because oddly enough, though you can find a case where it is, like using b=sqrt(2), and t=1, there are also cases where you are pushed out of the ring of algebraic integers, which proves a problem with the ring. James Harris ==== Revision 1. Over a hundred years ago the ring of algebraic integers became a part of the mathematical lexicon, waiting until now for an intriguing problem with the ring to be revealed by the use of advanced polynomial factorization techniques, which work by using non-polynomial factors of a polynomial. In what follows variables, unless otherwise noted, are in the ring of algebraic integers. Lemma 1: Given a factor G(X) of a polynomial P(X), R(X) and C exist such that G(X)=R(X) + C, where C=G(0), and R(X)=G(X)-C. Proof: Consider that C is a factor of the constant term P(0), which follows as G(X) is a factor of P(X), so G(0) is a factor of P(0), but C=G(0); therefore, C exists and is a factor of the constant term P(0). R(X) = G(X)-C, so it exists as well. Proof Complete. Lemma 2: Now consider P(X) such that it has b^2 as a factor. Further consider that P(0)/b^2 is coprime to b; then if P(X) has the factor G(X) then C, from lemma 1, must either be coprime to b or have a factor in common with b^2 which divides off when b^2 is divided off of P(X). Proof: Since C is a factor of P(0), and P(0)/b^2 is coprime to b, if C is not already coprime to b, C must have a factor in common with b^2 such that when b^2 is divided off of P(X), that factor divides off of C leaving a result coprime to b. Proof Complete. Now consider P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - 3(-1 + b^2 X )t u^2 + b u^3) where the odd grouping is so that I can factor P(X) into non-polynomial factors. Doing so I have the factorization P(X) = (r_1(X) t + bu)(r_2(X) t + bu)(r_3(X) t + bu) where again the purpose above with the special grouping was to get that factorization. Now letting G_1(X) = r_1(X) t + bu, consider that X=0, gives P(0) = b^2 (3t u^2 + b u^3) = b^2 u^2 (3t + bu), so at X=0, (r_1(0) t + bu)(r_2(0) t + bu)(r_3(0) t + bu) = b^2 u^2 (3t + bu) which shows that at least two of the r's go to 0, when X=0, and choosing r_1 to be one of them, as the indices are arbitrary, I have from lemma 1 that G_1(0) = bu, so C_1 = bu. Then R_1(X) = G_1(X) - C_1 = r_1(X) t + bu - bu = r_1(X) t. Given that one other of the r's goes to 0, I have a similar result for G_2(X) = r_2(X) t + bu. But one of the r's does not go to 0, and letting that one be r_3, I have for G_3(X) = r_3(X) t + bu, that G_3(0) = 3t + bu, so C_3 = 3t + bu and that gives G_1(0) G_2(0) G_3(0) = b^2 u^2 (3t + bu) = P(0) which is correct. Now consider that C_1 = bu, which shows it has a factor that is b. Notice also that the constant term P(0) = b^2 u^2 (3t + bu), so it has a factor that is b^2, but dividing that factor off gives P(0)/b^2 = u^2 (3t + bu) so if b is coprime to 3, u and t, then P(0) is coprime to b. Assume that b is coprime to 3, u and t. Now C_1=bu, and it is a factor of the constant term P(0), so when b^2 is divided off of P(X), and as shown divides off of P(0), it MUST divide off of C_1, from lemma 2. Therefore, r_1(X) t should have a factor that is b, and given that b is coprime to t, then r_1(X) should have a factor that is b. I say should because oddly enough, though you can find a case where it is, like using b=sqrt(2), and t=1, there are also cases where you are pushed out of the ring of algebraic integers, which proves a problem with the ring. James Harris ==== >Revision 1. [.snip social commentary reflecting James ignorance of history.] >In what follows variables, unless otherwise noted, are in the ring of >algebraic integers. Correction (for the N-th time): In what follows, the variables, unless otherwise noted, are assumed to take algebraic integer values. >Lemma 1: Given a factor G(X) of a polynomial P(X), R(X) and C exist such that >G(X)=R(X) + C, where C=G(0), and R(X)=G(X)-C. Proof: >Consider that C is a factor of the constant term P(0), which follows >as G(X) is a factor of P(X), so G(0) is a factor of P(0), but C=G(0); >therefore, C exists and is a factor of the constant term P(0). R(X) = >G(X)-C, so it exists as well. Proof Complete. You do not specify the ring in which G(X) is a factor of P(X). I've already offered you a corrected version of this which is complete and accurate. Is your pride preventing you from adopting it? You do not have to attribute the wording to me if you do not want to. You shoudl state that G(X) is a factor of P(X) in A^A (lest people think that G(X) is also a polynomial and that it is a factor in A[x], the natural reading of your words). You should also add that C divides P(0) in A since you talk about it in your proof and you use it later. >Lemma 2: Now consider P(X) such that it has b^2 as a factor. IN A[x]. Further consider that P(0)/b^2 is coprime to b; then if P(X) has the >factor G(X) then C, from lemma 1, must either be coprime to b or have >a factor in common with b^2 which divides off when b^2 is divided off of P(X). This lemma still has an empty first clause. It says that C is either coprime to b, or else it is not coprime to b. The second mysterious clause, which divdes off when b^2 is divided off of P(X) seems to be trying to say the following: Say G(X) is a factor of P(X) in A^A; write P(X) = G(X)*H(X). Then write G(X)=R(X)+G(0); (R(X) is just G(X)-G(0)). Since we are assuming that every coefficient of P(X) is a multiple of b^2 in A, we can write P(X) = b^2*Q(X) for some Q(X) in A[X]. Then we have b^2*Q(X) = (R(X)+G(0))*H(X). You seem to be trying to claim that: (1) If G(0) is coprime to b, then Q(X) = (R(X)+G(0))*(H(X)/b^2), so that H(X)/b^2 is still a function of algebraic integer values; that is, if G(0) is coprime to b, then each value of H(X) is a multiple of b^2; and (2) If G(0) is not coprime to b, and s is a greatest common divisor of G(0) and b^2 in A, then Q(X) = ( (R(X)+G(0))/s) * (H(X)/t) where t is the algebraic integer such that s*t=b^2. That is, R(X)+G(0) is always a multiple of s, and H(X) is always a multiple of t, in A. In particular, R(X) is always a multiple of s in A. Is this what you had in mind? I ask because the divides off of C seems to suggest you have such a division in mind, as do your attempts at applying, but your lemma is at best unclear here. >Proof: >Since C is a factor of P(0), and P(0)/b^2 is coprime to b, if C is not >already coprime to b, C must have a factor in common with b^2 such >that when b^2 is divided off of P(X), that factor divides off of C >leaving a result coprime to b. See, here? Since P(X) = (R(X)+C)H(X), dividing off C must mean dividing off the entire linear term R(X)+C; so you must be thinking that R(X) is also a multiple of the common factor of C and b^2. Have you proven this? I do not see it. > Proof Complete. Now consider P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - > > 3(-1 + b^2 X )t u^2 + b u^3) where the odd grouping is so that I can factor P(X) into >non-polynomial factors. Doing so I have the factorization P(X) = (r_1(X) t + bu)(r_2(X) t + bu)(r_3(X) t + bu) where again the purpose above with the special grouping was to get >that factorization. Now letting G_1(X) = r_1(X) t + bu, That is, G_1 is obtained by thinking of X, u, and b as fixed values, and factoring the resulting polynomial in t into linear terms over the algebraic integers. That a factorization always exists follows from results that have been discussed in the newsgroup and that appear in the literature; but the results do not guarantee that the constant terms can be forced to be equal to bu in each case. You need to prove that such a factorization always exists; it is not immediate that this is so. Assuming you can prove such a factorization exists with those given properties, still keeping b and u as fixed, now we think of t also as a parameter and think of the coefficients r_1, r_2, r_3 as functions of X; they are really functions of X, b, and u, but since b and u are being kept fixed for this, they will only depend on how X varies. Yes? > consider that X=0, gives P(0) = b^2 (3t u^2 + b u^3) = b^2 u^2 (3t + bu), so at X=0, (r_1(0) t + bu)(r_2(0) t + bu)(r_3(0) t + bu) = b^2 u^2 (3t + bu) which shows that at least two of the r's go to 0, when X=0, and >choosing r_1 to be one of them, as the indices are arbitrary, I have >from lemma 1 that G_1(0) = bu, so C_1 = bu. Then R_1(X) = G_1(X) - C_1 = r_1(X) t + bu - bu = r_1(X) t. Given that one other of the r's goes to 0, I have a similar result for G_2(X) = r_2(X) t + bu. But one of the r's does not go to 0, and letting that one be r_3, I >have for G_3(X) = r_3(X) t + bu, that G_3(0) = 3t + bu, so C_3 = 3t + bu and that gives G_1(0) G_2(0) G_3(0) = b^2 u^2 (3t + bu) = P(0) which is correct. In other words, when X=0, the polynomial becomes 3tb^2u^2 + b^3u^3 = b^2u^2 (3t+bu) = (0+bu)(0+bu)(3t+bu). >Now consider that C_1 = bu, which shows it has a factor that is b. Notice also that the constant term P(0) = b^2 u^2 (3t + bu), so it has >a factor that is b^2, but dividing that factor off gives P(0)/b^2 = u^2 (3t + bu) so if b is coprime to 3, u and t, then P(0) is coprime to b. Assume that b is coprime to 3, u and t. >Now C_1=bu, and it is a factor of the constant term P(0), so when b^2 >is divided off of P(X), and as shown divides off of P(0), it MUST >divide off of C_1, from lemma 2. Lemma 2 says that EITHER C_1 is coprime to b^2, or else it has a common factor with b^2. You have that C_1 is not coprime to b, so then you conclude that it has a common factor with b^2; yes, the factor is b. >Therefore, r_1(X) t should have a factor that is b, and given that b >is coprime to t, then r_1(X) should have a factor that is b. And here it is: you are trying to use that mysterious second clause of Lemma 2 which is not actually proven as far as I can see. That is, you are trying to claim that the equality P(X) = G_1(X)*G_2(X)*G_3(X), G_1(X) = r_1(X)*t+ub implies that r_1(X) must be a multiple of b. That does not follow from your work. This is where everything really goes to hell. You are assuming that r_3(X) will be coprime to to b because r_3(0) is coprime to b; this is not necessarily the case. You are also assuming that r_1(X) is a multiple of b always because r_1(0) is a multiple of b; and that r_2(X) is a multiple of b always because r_2(0) is a multiple of b. None of these three things follow, none of these three things are justified. The extension of Lemma 2 that you are trying to apply, namely Say G(X) is a factor of P(X) in A^A; write P(X) = G(X)*H(X). Then write G(X)=R(X)+G(0); (R(X) is just G(X)-G(0)). Since we are assuming that every coefficient of P(X) is a multiple of b^2 in A, we can write P(X) = b^2*Q(X) for some Q(X) in A[X]. Then we have b^2*Q(X) = (R(X)+G(0))*H(X). You seem to be trying to claim that: (1) If G(0) is coprime to b, then Q(X) = (R(X)+G(0))*(H(X)/b^2), so that H(X)/b^2 is still a function of algebraic integer values; that is, if G(0) is coprime to b, then each value of H(X) is a multiple of b^2; and (2) If G(0) is not coprime to b, and s is a greatest common divisor of G(0) and b^2 in A, then Q(X) = ( (R(X)+G(0))/s) * (H(X)/t) where t is the algebraic integer such that s*t=b^2. That is, R(X)+G(0) is always a multiple of s, and H(X) is always a multiple of t, in A. In particular, R(X) is always a multiple of s in A. has not bee shown to be true. In fact, as we have shown over and over and over again, it can be shown to be FALSE, since the conclusion you derive from it can explicitly be shown to be false, all your confusion notwithstanding. >I say should because oddly enough, though you can find a case where >it is, like using b=sqrt(2), and t=1, there are also cases where you >are pushed out of the ring of algebraic integers, which proves a >problem with the ring. Or, (1) with your argument; (2) with your understanding; (3) with your lemma. In short, you are claiming, again, that you can have an equality of algebraic integers a*b = r*s in which r is not a unit, and a and b are both coprime to r in the ring of all algebraic integers. You have failed to exhibit any such examples; all your attempts have been explicitly disproven, and your claim contradicts well known results about the algebraic integers, established by Dedekind among others (you know, the guy you've said you would trust blindly, contradicting your claims about what should and should not be done). ====================================================================== Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== >Over a hundred years ago the ring of algebraic integers became a part >of the mathematical lexicon, waiting until now for an intriguing >problem with the ring to be revealed by the use of advanced polynomial >factorization techniques, which work by using non-polynomial factors >of a polynomial. In what follows variables, unless otherwise noted, are in the ring of >algebraic integers. > (snip proof) It's interesting that you do not ever use the fact that any quantity in your proof is an algebraic integer (or a root of a monic polynomial) >I say should because oddly enough, though you can find a case where >it is, like using b=sqrt(2), and t=1, there are also cases where you >are pushed out of the ring of algebraic integers, which proves a >problem with the ring. > I have no idea wat you mean by you are pushed out of the ring means, or what your conclusion here is at all. Do you think that? 1. The definition of what is and what isn't an algebraic integer is unclear 2. The algebraic integers do not form a ring 3. The ring of the algebraic integers lacks some property that you need for your proof of FLT. 4. The ring of algebraic integers lacks some property that is used in one or more of the proofs that som your claims are false? 5. There is a number of wich you can both prove that it is not an algebraic integer. 6. anything else? -- Wim Benthem ==== [.philosophical and historical claptrap removed.] >In what follows variables, unless otherwise noted, are in the ring of >algebraic integers. Nonsense as written. You mean In what follows, the variables RANGE OVER the ring of algebraic integers, unless otherwise stated. Again, variables are not ELEMENTS of the ring of algebraic integers. To say that a variable is in the ring of algebraic integers is to speak nonsense. >Lemma 1: Given a factor G(X) of a polynomial P(X), R(X) and C exists such that >G(X)=R(X) + C, where C=G(0), and R(X)=G(X)-C. Proof: >Consider that C is a factor of the constant term P(0), which follows >as G(X) is a factor of P(X), so G(0) is a factor of P(0), but C=G(0); >therefore, C exists and is a factor of the constant term P(0). R(X) = >G(X)-C, so it exists as well. Proof Complete. Will you EVER learn to write correctly and what you (apparently) mean? I already offered a correct and complete version of this that you apparently accepted. Why do you go back to your nonsense-as-written? What you mean is: LEMMA 1. Let A be the ring of all algebraic integers. Let P(X) be a polynomial in A[x], and let G(X) be a factor of P(X) in A^A, the ring of all functions of algebraic integer valued functions of an algebraic integer variable. Then G(X) can be written as G(X)=R(X)+G(0), with R(X) in A^A, and G(0) is a factor of P(0) in A. Proof. Letting R(X):A->A be given by R(X)=G(X)-G(0) gives the first clause. For the second, note that P(X) = G(X)*S(X) for some S(X) in A^A; this equality holds in A^A, hence holds for each value of X. In particular, at X=0 we have P(0)=G(0)*S(0); since G(0),S(0), and P(0) are in A, this proves that G(0) divides P(0) in A. QED And, since in fact what you want is for multiple variable functions, you should probably write something like: LEMMA 1.5 Let A be the ring of all algebraic integers, and let P(X_1,...,X_n) be a polynomial with coefficients in A and in n variables. Let G(X_1,...,X_n) be an element of A^{A^n}, the ring of all algebraic integer valued functions of n algebraic integer variables, such that G(X_1,...,X_n) divides P(X_1,...,X_n) in A^{A^n}. Then G(0,a_2,...,a_n) is a factor of P(0,a_2,...,a_n) in A for each (n-1)-tuple of algebraic integers (a_2,...,a_n). A[X_1,...,X_n] to A[X_1] and from A^{A^n} to A^A. Apply Lemma 1 to P(X_1,a_2,...,a_n) and G(X_1,a_2,....,a_n). QED >Lemma 2: Now consider P(X) such that it has b^2 as a factor. Further consider that P(0)/b^2 is coprime to b; then if P(X) has the >factor G(X) then C, from lemma 1, must either be coprime to b or have >a factor in common with b^2. This lemma is empty. It says that given two algebraic integers C and b, either C is coprime to b or else it is not coprime to b. This is just the excluded middle, a logical tautology, and everything else is nothing but chaff and distraction. >Proof: >Since C is a factor of P(0), and P(0)/b^2 is coprime to b, if C is not >already coprime to b, C must have a factor in common with b^2 such >that when b^2 is divided off of P(0), that factor divides off of C >leaving a result coprime to b. Proof Complete. Now consider P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - > > 3(-1 + b^2 X )t u^2 + b u^3) where the odd grouping is so that I can factor P(X) into >non-polynomial factors. P(X) is a function on 4 variables: X, y, u, and b. It should be written as such: P(X,t,u,b) = b^2 ((b^4 X^3 - 3b^2X^2 + 3X)t^3 - 3(-1 + b^2X)tu^2 + bu^3) A specific choice of t, u, and b gives a natural map from A[X,t,u,b] to A[X]. >Doing so I have the factorization P(X) = (r_1(X) t + bu)(r_2(X) t + bu)(r_3(X) t + bu) where again the purpose above with the special grouping was to get >that factorization. This is incomplete and unclear. I have a somewhat lengthy discussion on this, but since the mistake in this post is so glaring, I will remove it to the end so it does not confuse and interrupt. >Now letting G_1(X) = r_1(X) t + bu, consider that X=0, gives G_1 is clearly a function of 4 variables, X, t, b, and u; having fixed values for b and u, and leaving t as a polynomial variable, we can think of G_1 as a function of 1 algebraic integer valued variable, with values in A[t], the ring of polynomials with algebraic integer coefficients in the polynomial variable t. This is ALREADY different from the setting on Lemma 1, and even on the extension Lemma 1.5. This equality is apparently taking place in (A[x])^{A} (after fixing values of b and u). > P(0) = b^2 (3t u^2 + b u^3) = b^2 u^2 (3t + bu), so at X=0, (r_1(0) t + bu)(r_2(0) t + bu)(r_3(0) t + bu) = b^2 u^2 (3t + bu) which shows that at least two of the r's go to 0, when X=0, and >choosing r_1 to be one of them, as the indices are arbitrary, I have >from lemma 1 that G_1(0) = bu, so C_1 = bu. In this particular instance (when X=0), the value of r_1 and r_2 are constant polynomials, so they can be considered elements of A. The difficulty in the wrong domain above gets magically solved for this instance, through a transference. >Then R_1(X) = G_1(X) - C_1 = r_1(X) t + bu - bu = r_1(X) t. Given that one other of the r's goes to 0, I have a similar result for G_2(X) = r_2(X) t + bu. But one of the r's does not go to 0, and letting that one be r_3, I >have for G_3(X) = r_3(X) t + bu, that G_3(0) = 3t + bu, so C_3 = 3t + bu and that gives G_1(0) G_2(0) G_3(0) = b^2 u^2 (3t + bu) = P(0) which is correct. Now consider that C_1 = bu, which shows it has a factor that is b. In standard mathematical terminology (as opposed to Harrispeak), C_1 is a multiple of b in A. >Notice also that the constant term P(0) = b^2 u^2 (3t + bu), Constant term with respect to WHAT? Not with respect to t; so suddenly, we go from t being a polynomial variable to it being a parameter/given value. This is already a source of much confusion, so let's try to clear it up. The definitions of r_1, r_2, r_3 (assuming James could prove that they actually exist, see below) are made by taking X, b, and u as parameters, and t as a polynomial variable. Then G_1, which is a function of 4 variables (the parameters X, b, and u; and the polynomial variable t) is re-interpreted as a family of functions on the variable X, with parameters t, u, and b. It is in this interpretation that Lemma 1 is applied, fixing values of t, u, and b for all subsequent applications. > so it has >a factor that is b^2, but dividing that factor off gives P(0)/b^2 = u^2 (3t + bu) so if b is coprime to 3 and t, then P(0) is coprime to b. At this point it is clear that t is no longer being considered to be a variable, but a parameter. Otherwise, talking about an algebraic integer being coprime to t makes no sense. Which is why it must be that it is being considered a parameter now, as mentioned above. >Assume that b is coprime to 3 and t. Now C_1=bu, and it is a factor of the constant term P(0), so when b^2 >is divided off of P(X), and as shown divides off of P(0), it MUST >divide off of C_1, from lemma 2. Alas, no. Lemma 2's conclusions are that EITHER C_1 is coprime Lemma 2 was: >Lemma 2: Now consider P(X) such that it has b^2 as a factor. Further consider that P(0)/b^2 is coprime to b; then if P(X) has the >factor G(X) then C, from lemma 1, must either be coprime to b or have >a factor in common with b^2. Now you are claiming that one of the two clauses does not apply. You must explain why. The conclusion from Lemma 2 is that EITHER C_1 must be coprime to b, or else it must have a factor in common with b^2. It is NOT that it must be divisible by b. Why do you claim that Lemma 2 implies that C_1 is a MULTIPLE of b^2? >Therefore, r_1(X) t should have a factor that is b, and given that b >is coprime to t, then r_1(X) should have a factor that is b. False. As to the unclear stuff from above: What you mean is that you pick functions r_1(X), r_2(X), and r_3(X) (which apparently depend only on X, and not on u, b, and t) with the properties that the above is an equality in the ring of function A[X,t,u,b]. It's existence would have to be first of all established, which you have not done. It will follow from work of Cohn, the theorem that David McKinnon and I reproved independently (and which YOU DENY is true; how do you get it)? To find the functions, note that for each specific choice of X, u, and b, the resulting polynomial P(t) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - 3(-1 + b^2 X )t u^2 + b u^3) is a polynomial in t with algebraic integer coefficients. Therefore, there exists a factorization into linear terms with algebraic integer coefficients P(t) = (r_1*t + s_1)(r_2*t + s_2)(r_3*t + s_3); This factorization is unique up associates of b^2, the content. IF you can prove that the choice s_1=s_2=s_3=b*u can always be made, then one can choose ANY assignments of r_1(X), r_2(X), r_3(X) into the coefficients r_1, r_2, r_3. Invoking the axiom of choice, you obtain functions r_1(X,b,u), r_2(X,b,u), r_3(X,b,u) of three algebraic integer valued functions with algebraic integer values which will yield the equality. exists (this factorization is in A[t]) with r_1, r_2, and r_3 algebraic integers. James is really working with a modification of his old polynomial Q(x,v) = (v^3+1)x^3 - 3vx + 1, He is taking the polynomial (v^3+1)t^3 - 3v*u^2*f^{2j}*t + u^3*f^{3j}, with v=-1+mf^{2j} (this can be obtained by replacing X by t/uf^j and then multiplying through by (uf^j)^3 to clear denominators). Expanding we get (m^3f^{6j} - 3m^2f^{4j} + 3mf^{2j})t^3 - 3(-1+mf^{2j})u^2f^{2j}t +u^3f^{3j}; Settting b = f^j, and X=m, we get the expression above: b^2(b^4X^3 - 3b^2X^2 + 3X)t^3 - 3(-1+b^2X)u^2t + u^3b. exactly equal to his expression above. He is assuming that he can find algebraic integers r_1, r_2, r_3 for each value of b_0, u_0, and X_0 that give an equality P(t)|(X=X_0,u=u_0,b=b_0) = = (r_1*t + b_0*u_0)(r_2*t + b_0*u_0)(r_3*t + b_0*u_0); and these values will be assigned to the functions r_1, r_2, r_3, and thus yield factors in A^A. This is slight modification of James's early attempts, at which he claimed that a factorization of the form P(t)/b^2 = (r_1*t + u_0)(r_2*t+u_0)(r_3*t+b_0*u_0) could be found with r_1, r_2, r_3 algebraic integers. This has been disproven many times, recently at the end of Nora's post: http://groups.google.com/groups?selm=36024859.0309071429.4e09bc20%40posting. google.com so I won't go into the details here. The new factorization proffered: P(t) = (r_1*t + u*b)(r_2*t+u*b)(r_3*t+u*b) is luckily valid in several instances. In particular, from previous observations, since we are dealing with P(t) = (v^3+1)t^3 - 3vu^2*t + u^3*b^3, we would have that -ub/r is a root of P(t), so that -(v^3+1)(u^3b^3/r^3) + 3vu^2(ub/r) + u^3b^3 = 0 Multiplying through by r^3 we have -(v^3+1)(u^3b^3) + 3vu^3b*r^2 + u^3b^3*r^3 = 0 Factorign out u^3b we have -(v^3+1)b^2 + 3vr^2 + b^2r^3=0. So if 3v is divisible by b^2, we can factor b^2 and thus we conclude that r is indeed an algebraic integer. However, if b^2 does NOT divide 3v, then letting s=gcd(b^2,3vr) we would have that r^3 is the root of f(X)=(b^2/s)X^3 + (3v/s)X^2 - (v^3+1)(b^2/s) which is a primitive, non monic polynomial. If this polynomial turns out to be irreducible, then the factorization as above does not exists. I ->believe<- that in all applications James has in mind, the values of v, b, and u are so contrived that it is possible this does not occur, but there is certainly no guarantee ->right now<- that the factorization is possible. ====================================================================== Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== > Lemma 1: Given a factor G(X) of a polynomial P(X), R(X) and C exists such that > G(X)=R(X) + C, where C=G(0), and R(X)=G(X)-C. Proof: > Consider that C is a factor of the constant term P(0), which follows > as G(X) is a factor of P(X), so G(0) is a factor of P(0), but C=G(0); > therefore, C exists and is a factor of the constant term P(0). R(X) = > G(X)-C, so it exists as well. Proof Complete. Shorter version: Lemma 1: Given a G(X), G(0) exist and G(X)-G(0) exist. Proof: duh! ==== > Over a hundred years ago the ring of algebraic integers became a part > of the mathematical lexicon, waiting until now for an intriguing > problem with the ring to be revealed by the use of advanced polynomial > factorization techniques, which work by using non-polynomial factors > of a polynomial. > > In what follows variables, unless otherwise noted, are in the ring of > algebraic integers. > > Lemma 1: > > Given a factor G(X) of a polynomial P(X), R(X) and C exists such that > G(X)=R(X) + C, where C=G(0), and R(X)=G(X)-C. > > Proof: > Consider that C is a factor of the constant term P(0), which follows > as G(X) is a factor of P(X), so G(0) is a factor of P(0), but C=G(0); > therefore, C exists and is a factor of the constant term P(0). R(X) = > G(X)-C, so it exists as well. Proof Complete. > > Lemma 2: > > Now consider P(X) such that it has b^2 as a factor. > > Further consider that P(0)/b^2 is coprime to b; then if P(X) has the > factor G(X) then C, from lemma 1, must either be coprime to b or have > a factor in common with b^2. > > Stop right there. The conclusion of Lemma 2 is: *** C must either be coprime to b or have a factor *** in common with b^2. Of course, clearly if C has a factor in common with b^2 it must have a factor in common with b. Thus the conclusion can be restated as: *** C must either be coprime to b or have a factor *** in common with b. Or even more simply, *** C must either be coprime to b or not be coprime *** to b. Therefore this lemma is completely vacuous. It cannot be used to prove anything. Better revise and try again, eh? Nora B. [remainder deleted pending correction] ==== > Over a hundred years ago the ring of algebraic integers became a part > of the mathematical lexicon, waiting until now for an intriguing > problem with the ring to be revealed by the use of advanced polynomial > factorization techniques, which work by using non-polynomial factors > of a polynomial. > > In what follows variables, unless otherwise noted, are in the ring of > algebraic integers. > > Lemma 1: > > Given a factor G(X) of a polynomial P(X), R(X) and C exists such that > G(X)=R(X) + C, where C=G(0), and R(X)=G(X)-C. > > Proof: > Consider that C is a factor of the constant term P(0), which follows > as G(X) is a factor of P(X), so G(0) is a factor of P(0), but C=G(0); > therefore, C exists and is a factor of the constant term P(0). R(X) = > G(X)-C, so it exists as well. Proof Complete. > > Lemma 2: > > Now consider P(X) such that it has b^2 as a factor. > > Further consider that P(0)/b^2 is coprime to b; then if P(X) has the > factor G(X) then C, from lemma 1, must either be coprime to b or have > a factor in common with b^2. > > > > Stop right there. The conclusion of Lemma 2 is: > > *** C must either be coprime to b or have a factor > *** in common with b^2. Hmmm...that is correct. > Of course, clearly if C has a factor in common with > b^2 it must have a factor in common with b. > > Thus the conclusion can be restated as: > > *** C must either be coprime to b or have a factor > *** in common with b. > > Or even more simply, > > *** C must either be coprime to b or not be coprime > *** to b. > > > Therefore this lemma is completely vacuous. It cannot > be used to prove anything. > > Better revise and try again, eh? Yeah, you're right, so, um, thanks. > > Nora B. > > > [remainder deleted pending correction] Well that looks like enough for me to do an updated version. James Harris ==== >> Over a hundred years ago the ring of algebraic integers became a part >> of the mathematical lexicon, waiting until now for an intriguing >> problem with the ring to be revealed by the use of advanced polynomial >> factorization techniques, which work by using non-polynomial factors >> of a polynomial. >> >> In what follows variables, unless otherwise noted, are in the ring of >> algebraic integers. >> >> Lemma 1: >> >> Given a factor G(X) of a polynomial P(X), R(X) and C exists such that >> G(X)=R(X) + C, where C=G(0), and R(X)=G(X)-C. >> >> Proof: >> Consider that C is a factor of the constant term P(0), which follows >> as G(X) is a factor of P(X), so G(0) is a factor of P(0), but C=G(0); >> therefore, C exists and is a factor of the constant term P(0). R(X) = >> G(X)-C, so it exists as well. Proof Complete. >> >> Lemma 2: >> >> Now consider P(X) such that it has b^2 as a factor. >> >> Further consider that P(0)/b^2 is coprime to b; then if P(X) has the >> factor G(X) then C, from lemma 1, must either be coprime to b or have >> a factor in common with b^2. >> >> >> >> Stop right there. The conclusion of Lemma 2 is: >> >> *** C must either be coprime to b or have a factor >> *** in common with b^2. Hmmm...that is correct. > >> Of course, clearly if C has a factor in common with >> b^2 it must have a factor in common with b. >> >> Thus the conclusion can be restated as: >> >> *** C must either be coprime to b or have a factor >> *** in common with b. >> >> Or even more simply, >> >> *** C must either be coprime to b or not be coprime >> *** to b. >> >> >> Therefore this lemma is completely vacuous. It cannot >> be used to prove anything. >> >> Better revise and try again, eh? Yeah, you're right, so, um, thanks. Pfeh. same thing: -- Begin Insert -- >Now then, as P(X) has b^2 as a factor, P(0) has b^2 as a factor, and >G(0) is a factor of P(0) as well. So dividing off b^2 will divide >some factor off of G(0), if G(0) isn't already coprime to b, since >P(0)/b^2 IS coprime to b. Your Lemma is empty, as Nora noted. You are saying that either G(0) is > not coprime to b, or else it is coprime to b. Big surprise. Well it's not empty and it's not complicated. It's like how with P(x) = 2x^2 + 4x + 2, for G(x) = 2x+2, you know that dividing the 2 off of P(x) divides it off of G(x). Or if you have G(x) = x+1, nothing divides off, but you know that G(0) is coprime to 2. I've broken things up into *very* simple pieces to take away room for specious objections. -- End Insert -- Doesn't that mean mean I should expect a retraction on claiming my objection was specious? Or will, you as usual, simply dismiss this error of yours as a simple mistake, and ignore all the insult you have heaped and attacks you have made based on your belief that this lemma was not content-free? ====================================================================== Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== ... > In what follows variables, unless otherwise noted, are in the ring of > algebraic integers. > > Lemma 1: > > Given a factor G(X) of a polynomial P(X), R(X) and C exists such that > G(X)=R(X) + C, where C=G(0), and R(X)=G(X)-C. > > Proof: > Consider that C is a factor of the constant term P(0), which follows > as G(X) is a factor of P(X), so G(0) is a factor of P(0), but C=G(0); > therefore, C exists and is a factor of the constant term P(0). R(X) = > G(X)-C, so it exists as well. Proof Complete. > > Lemma 2: > > Now consider P(X) such that it has b^2 as a factor. > > Further consider that P(0)/b^2 is coprime to b; then if P(X) has the > factor G(X) then C, from lemma 1, must either be coprime to b or have > a factor in common with b^2. > > Proof: > Since C is a factor of P(0), and P(0)/b^2 is coprime to b, if C is not > already coprime to b, C must have a factor in common with b^2 such > that when b^2 is divided off of P(0), that factor divides off of C > leaving a result coprime to b. Proof Complete. ... You can greatly simplify the above because it's irrelevant for Lemma 2 what C is a factor of, what P is, etc. That is, Lemma: C must either be coprime to b or have a factor in common with b^2 can be proved as follows: Case i. If C is coprime to b, ok. Case ii: Suppose C not coprime to b, and non-unit f is a common factor of b and C. Then f is a non-unit common factor of b^2 and C. Cases i and ii complete the proof. -jiw ==== > Now consider > P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - > 3(-1 + b^2 X )t u^2 + b u^3) > where the odd grouping is so that I can factor P(X) into > non-polynomial factors. > > Doing so I have the factorization > P(X) = (r_1(X) t + bu)(r_2(X) t + bu)(r_3(X) t + bu) > where again the purpose above with the special grouping was to get > that factorization. Doing this we get that the r_i(X) are roots of the polynomial: R^3 - 3(1 - b^2.X).R - (b^6.X^3 - 3.b^4.X^2 + 3.b^2.X) if I did not screw up somewhere... So we find that when X -> 0, two of the roots go to 0 and one goes to 3. > and that gives > G_1(0) G_2(0) G_3(0) = b^2 u^2 (3t + bu) = P(0) > which is correct. Indeed. G1(0) = G2(0) = bu, G3(0) = 3t + bu. Note however that indeed the G_i(X) are some strange functions of X, including cube roots etc.. > so if b is coprime to 3 and t, then P(0) is coprime to b. Requires also coprime to u... > Now C_1=bu, and it is a factor of the constant term P(0), so when b^2 > is divided off of P(X), and as shown divides off of P(0), it MUST > divide off of C_1, from lemma 2. > > Therefore, r_1(X) t should have a factor that is b, and given that b > is coprime to t, then r_1(X) should have a factor that is b. *Why* should the factor in r_1(X) be exactly b? We have: P(X) = (R0(X)t + C0)(R1(X)t + C1)(R2(X)t + C2) we have further that b^2 divides P(X) and b divides C0 and C1 so that the quotients are coprime to b, and C2 coprime to b. That is all trivial. But nothing states that (R2(X)t + C2) is coprime to b for all X. It is true for X = 0, but not necessarily so when X != 0. Indeed, *when* R2(X)t + C2 is coprime to b, R1(X) and R2(X) must be divisible by b. But as the condition is not yet met... -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== ... > > Therefore, r_1(X) t should have a factor that is b, and given that b > > is coprime to t, then r_1(X) should have a factor that is b. > > *Why* should the factor in r_1(X) be exactly b? We have: > P(X) = (R0(X)t + C0)(R1(X)t + C1)(R2(X)t + C2) > we have further that b^2 divides P(X) and b divides C0 and C1 so that > the quotients are coprime to b, and C2 coprime to b. That is all > trivial. But nothing states that (R2(X)t + C2) is coprime to b for > all X. It is true for X = 0, but not necessarily so when X != 0. > Indeed, *when* R2(X)t + C2 is coprime to b, R1(X) and R2(X) must be > divisible by b. But as the condition is not yet met... Argh. Not even that. R0(X)t + C0 can be divisible by b^2 while R1(X)t + C1 is coprime to b. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== > > Now consider > > P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - > > 3(-1 + b^2 X )t u^2 + b u^3) > > where the odd grouping is so that I can factor P(X) into > > non-polynomial factors. > > > > Doing so I have the factorization > > P(X) = (r_1(X) t + bu)(r_2(X) t + bu)(r_3(X) t + bu) > > where again the purpose above with the special grouping was to get > > that factorization. > > Doing this we get that the r_i(X) are roots of the polynomial: > R^3 - 3(1 - b^2.X).R - (b^6.X^3 - 3.b^4.X^2 + 3.b^2.X) > if I did not screw up somewhere... So we find that when X -> 0, > two of the roots go to 0 and one goes to 3. Yes, two of the r's are 0, and one equals 3, when X=0. > > and that gives > > G_1(0) G_2(0) G_3(0) = b^2 u^2 (3t + bu) = P(0) > > which is correct. > > Indeed. G1(0) = G2(0) = bu, G3(0) = 3t + bu. Note however that indeed > the G_i(X) are some strange functions of X, including cube roots etc.. That is of side interest. > > Now consider P(X) such that it has b^2 as a factor. > > Further consider that P(0)/b^2 is coprime to b; then if P(X) has the > > factor G(X) then C, from lemma 1, must either be coprime to b or have > > a factor in common with b^2. > > *What* must be coprime to b? P(X)? G(X)? What does it *mean* that the > strange function G(X) is coprime to b? I didn't say that G(X) is coprime to b, and I didn't say that P(X) is coprime to b either. What is coprime to b is P(0)/b^2, and since the variables are algebraic integers, it is an algebraic integer, so the usual definition applies. Similarly, C is an algebraic integer, so as it is a factor of P(0), if it has a factor in common with b^2, then that factor gets divided off when b^2 is divided off of P(0). > > Notice also that the constant term P(0) = b^2 u^2 (3t + bu), so it has > > a factor that is b^2, but dividing that factor off gives > > P(0)/b^2 = u^2 (3t + bu) > > so if b is coprime to 3 and t, then P(0) is coprime to b. > > Requires also coprime to u... Hmmm...yup, you're right. I'll add that to the list of fixes. > > Now C_1=bu, and it is a factor of the constant term P(0), so when b^2 > > is divided off of P(X), and as shown divides off of P(0), it MUST > > divide off of C_1, from lemma 2. > > > > Therefore, r_1(X) t should have a factor that is b, and given that b > > is coprime to t, then r_1(X) should have a factor that is b. > > *Why* should the factor in r_1(X) be exactly b? We have: > P(X) = (R0(X)t + C0)(R1(X)t + C1)(R2(X)t + C2) > we have further that b^2 divides P(X) and b divides C0 and C1 so that > the quotients are coprime to b, and C2 coprime to b. That is all > trivial. But nothing states that (R2(X)t + C2) is coprime to b for > all X. It is true for X = 0, but not necessarily so when X != 0. > Indeed, *when* R2(X)t + C2 is coprime to b, R1(X) and R2(X) must be > divisible by b. But as the condition is not yet met... That's why I have lemma 2 to handle objections where it is applied. If you feel that there is a problem with lemma 2, then present it. James Harris ==== > Now consider > P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - > 3(-1 + b^2 X )t u^2 + b u^3) > where the odd grouping is so that I can factor P(X) into > non-polynomial factors. > > Doing so I have the factorization > P(X) = (r_1(X) t + bu)(r_2(X) t + bu)(r_3(X) t + bu) > where again the purpose above with the special grouping was to get > that factorization. Doing this we get that the r_i(X) are roots of the polynomial: > R^3 - 3(1 - b^2.X).R - (b^6.X^3 - 3.b^4.X^2 + 3.b^2.X) >if I did not screw up somewhere... So we find that when X -> 0, >two of the roots go to 0 and one goes to 3. and that gives > G_1(0) G_2(0) G_3(0) = b^2 u^2 (3t + bu) = P(0) > which is correct. Indeed. G1(0) = G2(0) = bu, G3(0) = 3t + bu. Note however that indeed >the G_i(X) are some strange functions of X, including cube roots etc.. Now consider P(X) such that it has b^2 as a factor. > Further consider that P(0)/b^2 is coprime to b; then if P(X) has the > factor G(X) then C, from lemma 1, must either be coprime to b or have > a factor in common with b^2. *What* must be coprime to b? P(X)? G(X)? What does it *mean* that the >strange function G(X) is coprime to b? [The] C, from Lemma 1, [associated to G(X)] must be either coprime to b or have a factor in common with b^2. The content of this lemma, as observed by Nora, seems to be: Let G(X) is a function from A to A, and let b be an arbitrary algebraic integer. Then G(0) is either coprime to b or else it is not coprime to b. [.snip.] ====================================================================== [Gabriele Rossetti] has left a vast body of writings... in which he has attempted to prove the truth of his unorthodox interpre- tation of medieval literature. They present a formidable record of unsystematic research in which we see an enthusiast plunging farther and farther and farther from the logic of facts and good sense until truth is lost in the dreadful nightmare of an idee fixe. There is no real evolution of the Theory although it grows and expands until it embraces ever wider horizons. The numerous inaccuracies of deduction, mis-statements of historical fact, and self-contradictions...have caused critics to turn away from them in disgust... [...] It is impossible to read far... without realizing that we have to deal with a work of faith and imagination rather than of reasoning. There is an appearance of reason, for the author is set on proving by logic the truth of what he already believes by intuition. The truth is plain to him and he cannot comprehend why others do not immediately accept it, but as they desire demonstration he has multiplied his proofs. It is the redundancy and confusion of a prophet expounding by a familiar method the truth revealed to his own simple soul in a flash of inspiration... In such work as this... it is idle to look for the calm reasoning of a scholar; we do not find it, and there is little or no advantage in attacking the obvious inconsistencies and absurdities that abound. -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in _The Shakespearan Ciphers Examined_, by William F. Friedman and Elizebeth S. Friedman ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== > Lemma 1: Given a factor G(X) of a polynomial P(X), R(X) and C exists such that > G(X)=R(X) + C, where C=G(0), and R(X)=G(X)-C. I think exists should be exist. ==== > > Lemma 1: Given a factor G(X) of a polynomial P(X), R(X) and C exists such that > G(X)=R(X) + C, where C=G(0), and R(X)=G(X)-C. > > I think exists should be exist. mistake. Other readers should note that because my arguments *are* mathematically correct I'm quite interested in comments, or any findings of minor errors, including grammatical ones or typo's. It IS fun having a correct math argument, especially one that proves something wacky like the problem with the ring of algebraic integers. I've been working rather hard in answering a LOT of posts to show you that some people you might have trusted have betrayed your trust, and are attempting to teach you bogus math. They should not be working to feed you all false information, but it's their choice. I'll return in a bit to see if anyone else found any other errors, and post an update in this thread, underneath the original. If it's just chock full of errors, I'll make a REVISED thread. James Harris ==== > For some time several posters have gotten away with pushing bogus > mathematics in their efforts to argue with me. Possibly they got > sucked in as I worked out the mathematical ideas, and when faced with > finally correct arguments, after my many failures, decided to just > keep arguing with me, and in doing so taught those of you who trusted > them bogus math. > > Here finally I've chased down their objection and can explain it to > you simply enough. > > Given an expression like > > P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - > > 3(-1 + b^2 X )t u^2 + b u^3) > > where the odd grouping is so that I can factor P(X) into > non-polynomial factors, for instance, > > P(X) = (r_1(X) t + bu)(r_2(X) t + bu)(r_3(X) t + bu) > > they would claim that my factorization meant that *really* the > polynomial is P(t) or P(X,t). Just as below, this factorization is a consequence of treating P(X) as a polynomial in another variable. In this case, t. > The gist of their complaint being that the *factorization* changed the > polynomial. (Mathematical version of tail wagging the dog?) No, that when you treat a quantity other than x as a free variable, that that other quantity is a free variable. > > Luckily for me, as it is mathematics and a general principle, I could > switch to something less complicated and used > > P(x) = 11^2 + 11x + 2 Which you factor, in effect, by creating a new polynomial in a new free variable y. You know this, you just state it strangely, to whit: > so I've just used the 11's before as placemarkers, so I > can *act* like it's the polynomial y^2 + xy + 2, to get the > factorization It acts like this polynomial in the sense that you use a factorization theorem which relies on the presence of this new variable. > > P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2) You can then substitute the particular value of y back in. But clearly if instead of fixed values like 11, 11^2, and 2, you have free parameters which depend on things like t and m, then all the terms here will also depend on t and m. Are you claiming that if you had 13 instead of 11, i.e., 13^2 + 13x + 2, you'd still have (x+sqrt(x^2-8)/2) in your non-polynomial factorization? - Randy ==== > For some time several posters have gotten away with pushing bogus > mathematics in their efforts to argue with me. Possibly they got > sucked in as I worked out the mathematical ideas, and when faced with > finally correct arguments, after my many failures, decided to just > keep arguing with me, and in doing so taught those of you who trusted > them bogus math. > > Here finally I've chased down their objection and can explain it to > you simply enough. > > Given an expression like > > P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - > > 3(-1 + b^2 X )t u^2 + b u^3) > > where the odd grouping is so that I can factor P(X) into > non-polynomial factors, for instance, > > P(X) = (r_1(X) t + bu)(r_2(X) t + bu)(r_3(X) t + bu) > > they would claim that my factorization meant that *really* the > polynomial is P(t) or P(X,t). > > Just as below, this factorization is a consequence of treating > P(X) as a polynomial in another variable. In this case, t. Actually, no, though you can look at it that way. It goes back to the fact that you can have 3^2 + 2(3) + 1 where the expression isn't a polynomial. But it still factors as (3+1)(3+1). Similarly, 6=2(3) doesn't require polynomials, though you can use P(x) = (x+1)(x+2), with x=1. The *factorization* is independent of whether or not it's a polynomial. > The gist of their complaint being that the *factorization* changed the > polynomial. (Mathematical version of tail wagging the dog?) > > No, that when you treat a quantity other than x as a free > variable, that that other quantity is a free variable. If the polynomial is P(x) then it's P(x) *regardless* of how it's factored. Just like 6 is still 6, no matter how you factor it. > > Luckily for me, as it is mathematics and a general principle, I could > switch to something less complicated and used > > P(x) = 11^2 + 11x + 2 > > Which you factor, in effect, by creating a new polynomial > in a new free variable y. You know this, you just state > it strangely, to whit: The polynomial is P(x). Now readers have been lead astray by posters on this issue for months, so I'll emphasize that what Randy Poe is trying to do is distract you from the fact that the polynomial is P(x), where P(x) = 11x + 123. The factorization does NOT change the polynomial!!! > so I've just used the 11's before as placemarkers, so I > can *act* like it's the polynomial y^2 + xy + 2, to get the > factorization > > It acts like this polynomial in the sense that you use > a factorization theorem which relies on the presence of > this new variable. What variable? It's 11, and 11 is NOT a variable. The equivalent would be saying that 6=2(3) forces a new polynomial, like, maybe P(x) = (x+1)(x+2) with x=1. It's just a lot of unnecessary extra, which Randy Poe is trying to push on readers. The gist of his position has to be that the factorization changes the polynomial, which is nonsense. The polynomial is P(x) = 11x + 123. > > P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2) > > You can then substitute the particular value of y back in. > > But clearly if instead of fixed values like 11, 11^2, and 2, > you have free parameters which depend on things like t and > m, then all the terms here will also depend on t and m. The factorization does NOT change the polynomial. That simply does not change. > Are you claiming that if you had 13 instead of 11, i.e., > 13^2 + 13x + 2, you'd still have (x+sqrt(x^2-8)/2) in your > non-polynomial factorization? > > - Randy That *does* change the polynomial. The polynomial I'm using is P(x) = 11x + 123. It can be factored into non-polynomial factors as (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2) and similarly P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - 3(-1 + b^2 X )t u^2 + b u^3) can be factored into non-polynomial factors *without* changing it from being P(X), though several posters have managed to get away with claiming my work is false based on the notion that how the polynomial is factored changes the polynomial. I've switched variable names to make it harder for them, as for a while I've used P(m) = f^2(( m^3 f^4 - 3 m^2 f^2 + 3m) x^3 - 3(-1 + m f^2)x u^2 + f u^3) and they'd try to push the idea that it was really P(x) because of the grouping which is for the non-polynomial factorization. My guess is that they relied on people being used to P(x), so I changed variable names. James Harris ==== > For some time several posters have gotten away with pushing bogus > mathematics in their efforts to argue with me. Possibly they got > sucked in as I worked out the mathematical ideas, and when faced with > finally correct arguments, after my many failures, decided to just > keep arguing with me, and in doing so taught those of you who trusted > them bogus math. > > > You wish! Taunts are one thing, but math is another. The mathematics backs me up. > Here finally I've chased down their objection and can explain it to > you simply enough. > > Given an expression like > > P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - > > 3(-1 + b^2 X )t u^2 + b u^3) > > where the odd grouping is so that I can factor P(X) into > non-polynomial factors, for instance, > > P(X) = (r_1(X) t + bu)(r_2(X) t + bu)(r_3(X) t + bu) > > they would claim that my factorization meant that *really* the > polynomial is P(t) or P(X,t). > > The gist of their complaint being that the *factorization* changed the > polynomial. (Mathematical version of tail wagging the dog?) > > > Not at all. You have been doing factorizations in this > form for months: before, they looked like > > a1*x + u*f, > > where a1 was a function of m (the variable m has now been > transformed into X for some reason). We didn't object > to that. When I had a_1 x + uf it was possible for posters to claim the polynomial was P(x), and rely on people being used to seeing P(x), so I switched variable names to take that way. The poster Nora Baron keeps trying to go back though, which confirms my suspicions that I needed the new variable names. After all it IS algebra, so switching the variable names shouldn't be significant. That is, I'm emphasizing to readers that this poster is putting emphasis on something that *should* be immaterial. The poster and others would claim that my P(m) was really P(x) because of how I factored P(m), which is bogus. > What you have been doing here with all this noise about > substituting 11 for t is trying to refute something > that I have not said. You are pretending that I said > you cannot factor polynomials with non-polynomial factors, > but I have never said that. We have been accepting your > concept of doing that for many months. That's disingenuous. When faced with non-polynomial factors the poster kept trying to claim they really were polynomial factors, claiming I had a different polynomial from what I said was the polynomial. For instance, with my current P(X) this poster would try to claim it was really P(X,t), based on the *factorization*, so I've started saying things like 6 is still 6 despite how it's factored. The poster has refused to back down and follow the math, and is still posting apparently in an attempt to convince at least some of you that bogus math is correct. That's just wrong. > You have made up a straw man. Now you are frantically > trying to beat it to death to gain some credibility with > somebody. Who, I don't know. I've proven my case mathematically. However this poster has made a career out of making false claims about my work, so I'm in the process of shutting the poster down. > Luckily for me, as it is mathematics and a general principle, I could > switch to something less complicated and used > > P(x) = 11^2 + 11x + 2 > > where again you have a special grouping simply to allow the > non-polynomial factorization, as you can see that P(x) is also > > P(x) = 11x + 123 > > so I've just used the 11's before as placemarkers, so I > can *act* like it's the polynomial y^2 + xy + 2, to get the > factorization > > P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2) > > but the polynomial is STILL P(x)= 11x + 123. These posters worked to > convince that the factorization, with more complicated expressions, > changed the polynomial in some way. > > > What? How can a factorization change a polynomial ??? We never > said any such thing! Yes Nora Baron, it is true that a factorization can't change a polynomial. And by we I guess you mean yourself and other posters like Arturo Magidin, who once actually *said* my factorization was making a new polynomial which I referenced in the post that started this thread. > Apparently plenty of sci.math > readers and especially alt.math.undergrad readers, who may be more > susceptible, were convinced by them. > > > What is the evidence for that ? Have people been telling > you that privately ? Or is that really just your OWN > reaction that you are suppressing ? I've noticed replies on the newsgroups where posters would claim that I'd not refuted various claims by posters like yourself Nora Baron. There were also posts praising you in particular. Therefore, I concluded that you were getting your message across, and that enough people believed in you that some would make the effort to come out and post their support. > They were very successful which can be seen by the date on the > following and the fact that they've been arguing with me up until now. > Think of all the months pumping false information out to readers. > > Consider with my early use of non-polynomial factorization Arturo > Magidin's reply claiming that it is a new polynomial. > > > > the original polynomial; the coefficients a1,a2,a3,b1,b2,b3 have > nothing to do with this new polynomial. They correspond to the > original one. > > ====================================================================== > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > ====================================================================== > > Arturo Magidin > magidin@math.berkeley.edu > And oddly enough for the complicated expression people seemed to > believe him. > > Now Arturo Magidin has a PhD from Berkeley in mathematics, and is > talking about a factorization changing the polynomial, where also he > has a telling statement at the end of his posts? What would you > think? > > I think that some of you would agree with posters like Arturo Magidin > and Nora Baron on anything or sit idly by without caring about some > people thinking they're learning correct mathematics from them, as > long as it makes me miserable. > > > > Let's cut to the chase. > > But first let me say I am gratified to see that you have > abandoned the argument you were pursuing so tenaciously and > obtusely for the last several months: claiming with no hint of > proof that properties of the factorization for m = 0 had to > be the same for m <> 0. Evidently we finally got through > to you on that, though you kicked and screamed every inch > of the way. And again the poster is referring back to an old variable name!!! That's not algebra, as in fact, the variable is now X so it IS in the argument. Bizarre. > Now for the heart of the difficulties with your new > *present* argument. First, you have two lemmas. In both, > you assume that a polynomial P(X) has a factor G(X). You > let C be the constant term in G(X), i.e., C = G(0). You > note that C must divide the constant term P(0) of P(X). > You define another function R(X) as > > R(X) = G(X) - C = G(X) - G(0). > > Of course, > > G(X) = R(X) + C. > > It is important to note here that > > [0] R(0) = G(0) - G(0) = 0. > > That is the kind of function R(X) that you need in your > lemmas, right ? And the G(X) is the kind of function that > you intend to consider in your factorization of your > polynomial, right? Everything OK so far? Any bogus rules > invoked so far anywhere? Any lying or confusing going on? Well Nora Baron, you're talking too much. Otherwise it looks ok so far. > Then you consider your beloved polynomial, > > [1] P(X, t) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 > > - 3(-1 + b^2 X)*t*u^2 + b*u^3) > > You let X = 0 to get the constant term. It is > > [2] P(0, t) = b^2*u^2*(3*t + b*u). > > Save that for reference below. > > Now you assume a factorization of the form > > P(X, t) = (r1(X)*t + b*u)*(r2(X)*t + b*u)*(r3(X)*t + b*u). > > Choose one of your factors: say > > G1(X) = r1(X)*t + b*u. > > This is, as you intend, of the form G1(X) = R1(X) + C1. As > in your lemmas, C1 = G1(0) [because, by [0] above, R1(0) = 0], > and C1 must divide P(0, t). Also, obviously, C1 = b*u. > > Similarly for C2 and C3: C2 = b*u and C3 = b*u. That is false. > Notice here that I am not saying anything about G1(X), G2(X), > or G3(X) being a *polynomial* factor. That is not important at > the moment. I don't care whether they are or not. They aren't polynomials as it's a non-polynomial factorization. > Now of course the product of the constant terms of the > three factors, G1(X), G2(X), and G3(X) must equal the constant > term of the original polynomial, P(X, t). That is, > > C1*C2*C3 = (b*u)*(b*u)*(b*u) = P(0, t), > > or, from [2] above, > > [3] (b*u)^3 = b^2*u^2*(3*t + b*u). > > Whoa! Not so good. The right side of [3] is not > constant with respect to t. See that 3*t in there? Yeah, it comes from one of the C's, which refutes your previous claim about their values. > But the left side IS constant with respect to t. > Do you see any factors of t on the left side? Any t's > in there anywhere? Is b or u a function of t? Gee, I > don't think so. I think b and u are constants. > > The left side cannot equal the right side. > > Why don't you straighten out this little problem, > Mr. Harris, and then maybe we can talk further sometime. > > > Nora B. There's no problem as one of the C's equals 3t + bu. James Harris ==== >And notice that David Ullrich deleted out the gist of Nora Baron and >Arturo Magidin's complaints. I've also noticed an outpouring of >hostile posts recently, as apparently, mathematics isn't good enough >for many of you. Here's an excerpt that David > > being necessary. > > G C entirely different thread!!! You do realize that's not exactly rational, right? Remember the big picture which is that mathematicians pulled in an idea without fully working it out over a *hundred* years ago, so you have this wacky problem with algebraic integers. I've simply pointed that out, and if mathematicians aren't themselves wacky, they can just work to figure out the full ramifications of the situation, which could be an incredible opportunity for some of you. Instead, I see posters working to destroy math society by fighting mathematics as if you were fighting me, a person, as if you could actually win against the math. But then what would be the point of being mathematicians? Here's what was deleted out, yet again. Luckily for me, as it is mathematics and a general principle, I could switch to something less complicated and used P(x) = 11^2 + 11x + 2 where again you have a special grouping simply to allow the non-polynomial factorization, as you can see that P(x) is also P(x) = 11x + 123 so I've just used the 11's before as placemarkers, so I can *act* like it's the polynomial y^2 + xy + 2, to get the factorization P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2) but the polynomial is STILL P(x)= 11x + 123. These posters worked to convince that the factorization, with more complicated expressions, changed the polynomial in some way. Apparently plenty of sci.math readers and especially alt.math.undergrad readers, who may be more susceptible, were convinced by them. It's actually rather amazing that a few posters could convince so many people to follow along with something so odd, as to believe that the *factorization* changed the polynomial. But then, maybe for many of you an expression like P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - 3(-1 + b^2 X )t u^2 + b u^3) where the odd grouping is so that I can factor P(X) into non-polynomial factors, for instance, P(X) = (r_1(X) t + bu)(r_2(X) t + bu)(r_3(X) t + bu) is just so weird and outside of your mathematical knowledge that you'd figure, hey, if those people say it's no longer P(X) because of the factorization, you'd just go along and trust the majority. But you see, mathematics isn't a democracy. James Harris ==== >>And notice that David Ullrich deleted out the gist of Nora Baron and >>Arturo Magidin's complaints. I've also noticed an outpouring of >>hostile posts recently, as apparently, mathematics isn't good enough >>for many of you. >>Here's an excerpt that David >> >> being necessary. >> >> G C entirely different thread!!! You do realize that's not exactly rational, right? If _you_ thought my behavior was rational I'd be worried. David C. Ullrich ************************** As far as I'm concerend you're trying to wait until I die, so I figure maybe you should die instead. How about that, eh? Wouldn't that be a better twist? You refuse to follow the math, so the great Powers that control reality and *speak* in mathematics decide to kill you instead of me. So what do you think about that, eh? Oh, can't hear Them talking? Well, I guess that's because you don't really understand Mathematics, the true language, which is THE language. They're talking about you now, and They agree with my assessment, and will not penalize me as They allowed the others like Galois and Abel to be penalized. They will kill you instead. James Harris speaking on Weird factorization, genius X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h81FbOc24006; ==== > Bruce Percy > If I have a function, the derivative f' of which is say, > f'(x) = x^2 > Is f' differentiable as x->infty? >> Do you mean f instead of f' here? I mean df/dx = x^2. Sorry about the confusion.. f in this case would be >x^3/3. > The reason I ask is because as I understand the definition of > differentiability at a point, it is that the limit that defines the > derivative must both exist and be finite at this point. Does f' meet the > criteria of the definition as x->infty. >> f' doesn't have a limit at infinity, but even if it did, there is no meaning >> to the statement that f is differentiable at infinity or differentiable >> in the limit. If we _adjoin_ a point at infinity, and f behaves suitably in >> a neighbourhood of the new point (this f doesn't) then the notion of >> differentiable at infinity can be validated; but that expression is not in >> conventional use, as far as I know. >> LH Okay, since f'= x^2 is defined for all x, f is differentiable for all x. The question is what do you mean by differentiable as x-> infinity? I would interpret that to mean differentiable for x very large. Since it is obviously differentiable for all x, it is certainly differentiable for x large. However, you seem to be asking if it is differentiable AT infinity which doesn't make since unless you are talking about extending the real number system in some way. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h81GMre27196; ==== it's a big problem for me to find the way to solve any of these two >given differential equations (second-order and ordinary, with variable >coefficients). x^2*y'' - x*(x+2)*y' + (x+2)*y=x^3 and (1-x^2)*y'' - 4*x*y' - (1+x^2)*y(x) = x Solutions are not given, but solutions with mathematics software are >possible. Please give me a hint. A. Kratzer Use series solutions: y= Sum(n=0 to infinity) a_n x^n. If you want solutions about singularities (x=0 for the first equation, x= 1 or x= -1 for the second) look up Frobenius' method,which uses y= Sun(n=0 to infinty)a_n x^(n+c) for appropriate c. ==== >>x^2*y'' - x*(x+2)*y' + (x+2)*y=x^3 Reduction of order will give you the complete solution. >>and >>(1-x^2)*y'' - 4*x*y' - (1+x^2)*y(x) = x Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h81GsWU29109; ==== I've been trying to solve the following problem with no success. Find all functions f: N>=0 -> N>=0 such that f(n^2+m^2)=f(n)^2+f(m)^2 for all n,m in N>=0. (where N>=0 means the natural numbers plus the zero). Any ideas? nojb. How about f(n) = n ? X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h81NHBM21772; ==== >don««t know it. Can anyone help me ? Well, Atom Penis, it equals 2. Or 10. Or 100. Or, if you wanna stretch it, 0 and -1. Go watch some Nicktoons. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h81NH9a21759; ==== >you're saying: >We have a 2-dimensional observer stationed at the origin of a cartesian >plane, with all lattice points (i.e. points whose coordinates are both >integers) marked. The observer can see the point (3,4), but no other >points (in the first quadrant) of the form (3p,4p), because (3,4) is in >the way. >You want to find the ratio of visible lattice points to all lattice >points, and hope to find pi somewhere along the way. >Do I understand you so far? >If so ... >What you are doing is analogous (and probably isomorphic) to finding the >ratio of >reduced fractions to all fractions whose numerator n and denominator d are >both integers. >Note that if you find an answer to the case for n, d both positive, you >can quit; the other three quadrants will have the same arrangement of >points, thus the same ratio. Note also that if you find an answer to the >case for n < d (which can correspond to that part of the first quadrant >above the line y = x), the arrangement for n > d (which can correspond to >the part of the first quadrant below the line y=x) is a reflection of it, >and thus has the same ratio. Happy hunting. Please let me know what you turn up. Ted Shoemaker >shoematr@uwec.edu >> Please forgive the vague phrasing of this question, but I recall a TV >> show on geometery demontrating that pi shows up in what was refered to >> as the ratio of intesrections visible from the origin >> (over total intersections?) on a cartesian plane- I assume this was a >> limit as the plane became infinitly large. >> I tried to demonstrate this with a small program, but without a decent >> reference, I was unable. >> Any ideas on this? >> This was shown in the same context >> of the Buffon needle problem. But >> I don't think it is the same. > X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h81NH9F21755; ==== What is the time difference from Athens, Greece to California ==== >What is the time difference from Athens, Greece to California If you have Windows XP [similar things probably exist in other versions but this is what I'm using at the moment], right click on the time in your a drop-down list of time zones where it gives your current time zone. standard time in Athens is 10 hours ahead of California. I'm not sure whether the dates for start and end of Daylight Saving Time are the same in both places though. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >>What is the time difference from Athens, Greece to California If you have Windows XP [similar things probably exist in other versions >but this is what I'm using at the moment], right click on the time in your >a drop-down list of time zones where it gives your current time zone. >standard time in Athens is 10 hours ahead of California. I'm not sure >whether the dates for start and end of Daylight Saving Time are the same >in both places though. Not always, in fact I nearly missed a morning flight from Thessaloniki to Athens (with a connection to New York) on Sunday 3/30/03 ... precisely for this reason; I understood when they called my name :-) baloglouAToswego.edu X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h81NHAW21763; ==== >Does anyone have a good reference for mathematical explanation of the >degrees of freedom for the sum of squares equation ? thanks If the number of squares in the White House is >/= 1, Freedom is I want know about of amount children are in age group the 0-2 years in >the country: Canada, Brasil, China, Venezuela, Guatemala, El Salvador, >Nicaragua, Panama, Mexico, Ecuador, Honduras and Peru I don't know about the other countries, but for Canada you could probably find something at http://www.statcan.ca. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h81FbNj23995; ==== >> don««t know it. >> >> Can anyone help me ? Yes, it's 0 (sometimes). :-) Nijmegen, Netherlands > I could be 10 also! ==== > don««t know it. > > Can anyone help me ? >>Yes, it's 0 (sometimes). :-) >>Nijmegen, Netherlands > I could be 10 also! There are 10 kinds of people in this world... those who understand binary, and those who don't. ==== >I could be 10 also! There are 10 kinds of people in this world... those who understand binary, and > those who don't. Sure, but what would have answered the buggy pentium processor ? 0, 1, 11, 10.000001 ? ;-) J-L. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h82E22t25851; ==== Let sigma(n) be the sum of the positive divisors of n. There is a term(Hardy ?; Ramanujan?) for sigma(n): sigma(n)=Pi^2n/6( 1+(-1)^n/4 + 2cos((2/3)nPi)/9 + 2cos((1/2)nPi)/16 + 2(cos((2/5)nPi)+cos((4/5)nPi))/25 + 2cos((1/3)nPi)/36 + ...) for which I have lost the reference. Is there a general expression for this term or how can I calculated the next elements? GG X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h82JpWe22463; ==== What does F(a,b;c;d) mean, where a,b,c,d are numbers? Is this function in Mathematica? ==== > What does F(a,b;c;d) mean, where a,b,c,d are numbers? Is this function in Mathematica? > Hypergeometric function. -- Julien Santini > X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h82Ft8c03390; ==== El 12 de oct 1997 Terry de Moore escribi.97: >
  > en el art.92culo < 
61bfbn$70k$1@hp.fciencias.unam.mx >, Unam < fcueto@usa.net > escribi.97: 
 > Joe escribi.97 en el mensaje < 
343924å.E388B5å@psido.exp.univie.ac.at >...  > Yampolskiy 
escribi.97:  > hace cualquier persona saben que un algoritmo 
eficaz para encontrar n.9cmeros primeros > yo ha hecho dos algoritmos: 
 À> primer, aplicaciones el tamiz de Erathosthenes (es 
deletre.97 correctamente?  en > Eratosthenes > espa.96ol es Criva de 
Erat.97stenes), y lo utilizo conjuntamente con dividirse > solamente por 
n.9cmeros impares > el otro algoritmo (que sucede ser m.87s lento), 
tambi.8en utilizo concepto principal de > el primer, pero en vez de 
dividirse por n.9cmeros impares, se divide por ya > encontrado prepara > 
ni unos ni otros es _ _ el tamiz verdadero de Eratosthenes.  Usted no 
necesita dividirse > por cualquier cosa.  Usted salto justo con el arsenal, 
con una distancia del salto > igual a la prim!
a lo m.87s recientemente posible encontrada, limpiando fuera de cada 
n.9cmero > usted satisface.  Esto requiere solamente la adici.97n, no 
dividi.8endose > asume un arsenal, P:  ARSENAL [ 2..CN ] DE BOLEANO;  > 
llenado ya de VERDAD;  > en Modula 2 esto estar.92a > i: =  2;  j: =  2;  > 
MIENTRAS QUE LO HACE i < = TRUNC(SQRT(N))) > (* ninguna necesidad de mirar 
m.87s lejos que sqrt(N) - pruebe esto *) > MIENTRAS QUE P[i ] y (i < 
TRUNC(SQRT(N))) HACEN INC[i, 1 ] EXTREMO;  > (* busque para el n.9cmero 
siguiente demostrado no previamente para ser compuesto, > el m.87s 
peque.96o que tales deben ser primeros *) > MIENTRAS QUE j < N HACEN INC[j, 
i ];  P[j ]: =  EXTREMO FALSO > (* limpie fuera de todos los m.9cltiplos, 
k*i, k > 1 *) > EXTREMO;  Á> (* ahora P[i ] = verdad si y 
solamente si i es primero *) > el intento esto contra su algoritmo y 
considera cu.87l es el mejor > - - > Terry Moore, departamento de la 
estad.92stica, universidad de Massey, zealand. nuevo > los teoremas!  
Necesito teoremas!
.  D.8eme los teoremas y encontrar.8e > impermeabilizo f.87cilmente 
bastante.  Bernard Riemann >
X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h83CuHB16215; ==== > Is the name Godel pronounced Go - Dell or is it, Go - dull? >>Perhaps the best advice to an American English speaker is to say >>girdle but without quite getting to the r. English and Australian >>speakers of English don't ever get to the r anyway, so they can >>pronounce it as if it were girdle. >>-- > >Oh Great. This is going to be just like my Goethe fiasco. I >couldn't read anything by Goethe for so long because I was >embaressed to say his name to librarians. adam In what part of Germany? My understanding is that in Berlin, they are likely to pronounce 0 umlaut like an English long a so it would be Gay- del. In Bavaria it might sound like Grrrr-del and in most of the country Goo- del. Of course, in Swabia, all bets are off! X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h83EbuT24912; ==== >can anyone help me with these pleeeeease!!! : 1. y=4x|X|-3 >2. y=|2X-1| >3. y=3x|X+4|-2 >4. y=-0.5|X+4|+3 i really need your help on this, thanks to anyone who can reply. And you would have us do what with them? Graph them, Find Domain and Range , ... ? X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h83IpAV13290; ==== Why are we so concerned about the arrow of time? The basic laws of physics,as proposed by Newton and extended by Einstein and Schrodinger, work equally well in the forward and reverse time directions. So as far as these laws are concerned, there is no arrow of time, just as there is no arrow of space. For the purpose of measurement, we humans impose the artificial construct of scale and zero-point for time and space, but the physics we observe is invariant with respect to fixed reference points. The only time we observe an arrow of time is when we analyze thermodynamic properties (i.e., statistical properties). But in this case, we have prepared a physical state which is highly ordered, so we expect statistically that a more disordered state will evolve. If we imagine a state of maximal entropy, then in fact we will see order emerging from chaos. For the biologists out there, the appreciation of order out of chaos is understood as the wonderment of life. Life itself would seem to violate the principle of a time's arrow. A time's arrow is a mere convenience to reflect our prejudice of our memories and cognitions. We imagine a future, and recall a past. If we take a simpler model of a memory that is processed, say, the modern computer, then we see, in fact, that we can recall the future and imagine the past, for in this case, we understand how to everse the processes, as long as we do not discard any information (i.e., memory configurations of the computer). > One might, however, reasonably surmise that no state ever occurs twice in >> actuality so the uniqueness of the state that actually follows a given >state >> is automatically true. Or maybe not? See Process Physics: Modelling Reality as Self-Organising >Information - Reginald T. Cahill, Christopher M. Klinger and Kirsty Kitto >(http://www.scieng.flinders.edu.au/cpes/people/cahill_r/processphysics /00090non-linear and stochastic; these QSD terms are ultimately responsible for >the emergence of classicality via an objectification process, but in >particular they produce wave-function(al) collapses during quantum >measurements; a mechanism that eluded quantum theory since its discovery and >which is finally seen to have its explanation with Geodel's incompleteness >theorem and its associated Self Referential Noise within a process-system. >The random click of the detector is then a manifestation of Geodel's >profound insight that truth has no finite description in self-referential >systems; the click is simply a random contingent truth. The SRN is thus seen >to be a major missing component of the modelling of reality. In the above we >have a deterministic and unitary evolution, tracking and preserving >topologically encoded information, together with the stochastic QSD terms, >whose form protects that information during localisation events, and which >also ensures the full matching in Quantum Homotopic Field Theory of process >time to real time: an ordering of events, an intrinsic direction or `arrow' >of time and a modelling of the contingent present moment effect. > X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h83CskZ16044; ==== can anyone help me with these pleeeeease!!! : 1. y=4x|X|-3 2. y=|2X-1| 3. y=3x|X+4|-2 4. y=-0.5|X+4|+3 i really need your help on this, thanks to anyone who can reply. X-Received: from home.mathforum.org (home-1.mathforum.org [144.118.94.17]) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id h842SVt15356 X-Received: (from approve@localhost) by home.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.4 nullclient) id h842SUb02637; ==== >-- What is the smallest square (x^2) with nine 9's? .. The smallest square (x^2) with four 9's? >The smallest square (x^2) with three 9's? >(Answer: 1999) The smallest square (x^2) with two 9's? >(Answer: 199) The smallest square (x^2) with one 9? >(Answer: 19) The smallest square (x^2) with zero 9's? >(Answer: 2) John D. >-- Plug in the number on the right as x^2 and you will have the number of 9's that I listed(and yes, these are the first time that such a pattern works) First with 1 = 3 First with 2 = 63 first with 3 = 173 first with 4 = 1414 first with 5 = 17313 first with 6 = 53937 first with 7 = 138923 first with 8 = 953937 first with 9 = 3082207 first with 10 = 31622764 X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h83JBr014861; ==== Given the two polynomials q(x)=x^{17}+9 and r(x)=(x+1)^{17}+9, there are polynomials s(x) and t(x) with integer coefficients such that s(x)*q(x)+t(x)*r(x)=R is an integer (independent of x). This integer is the resultant of q and r. There are efficient ways of computing R, using linear algebra. Now suppose we have an integer n and a prime p such that p divides gcd(n^{17}+9,(n+1)^{17}+9), that is, p divides n^{17}+9=q(n) and p also divides (n+1)^{17}+9=r(n). Then p also divides s(n)*q(n)+t(n)*r(n), so that p divides R. So by computing R and finding its factors, we know the possible values for p. Assuming R is a prime (I did not check this), so that p=R, we can calculate those n for which p|n^{17}+9, and there will be only a handful to check. All of this could be done with a computer algebra package. Don Coppersmith >recently someone on sci.math pointed to the law of small numbers page >(http://prime s.utm.edu/glossary/page.php?sort=LawOfSmall). >It says a.o. the first value for n | gcd(n^17+9,(n+1)^17+) > 1 is >8424432925592889329288197322308900672459420460792433. >I can imagine there are tricks to find certain values for n (like you do to >find Mersenne primes), but how do you know it's the _first_ value for n to >satisfy the given condition? >Obviously you can't try them all starting at n = 1. I'm not a mathematician, so please type slowly :-) -- >Steven > X-Received: from home.mathforum.org (home-1.mathforum.org [144.118.94.17]) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id h84Bnst20132 X-Received: (from approve@localhost) by home.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.4 nullclient) id h84Bnr103250; ==== >Any general algorithm to solve such exponential equation. First, this is not an exponential equation. That term is applied to equations in which the unknown, x, is an exponent. This is a polynomial equation. There is no algorithm for general solution of any polynomial equation of degree greater than 4. If you know specific values for a, you could use Newton's method to get a numerical solution. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h84JPsS22500; ==== [ ... ] >But even in A[x], the two statements are not equivalent: 2 and x have >no common divisors other than units, but there do not exist >polynomials f(x) and g(x) in A[x] such that f(x)*x + g(x)*2 = 1. How about f(x)=0 and g(x)=1/2? Cron ==== > > [ ... ] > >But even in A[x], the two statements are not equivalent: 2 and x have >no common divisors other than units, but there do not exist >polynomials f(x) and g(x) in A[x] such that f(x)*x + g(x)*2 = 1. > > How about f(x)=0 and g(x)=1/2? Last I checked, 1/2 was not an element of A[x], the ring of polynomials with algebraic integer coefficients. Arturo Magidin, sans .sig X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h84IYjd17832; ==== >Im attempting to prove that 2^n + 1 is a prime for any non-negative int n. >My reasoning is that no matter what int you use you always get some prime >number. However I was going to use a 'case' proof but obviously the list of Before trying to prove something you better make sure it's actually true... 2^3+1=9=3*3 2^5+1=33=3*11 2^6+1=65=5*13 ... Gal X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h84JQ3622518; ==== >I started reading Ireland and Rosen's A Classical Introduction to >Modern Number Theory and noticed a remark that in an arbitrary ring >the greatest common divisor not necessarily exists. I couldn't exhibit >an example. Could you give an example of integral domain such that there exist two >elements a,b, but there is no their greatest common divisor? Not that it matters, since there are examples with integral domains, but you did notice, assuming you quoted accurately, that Ireland-Rosen states that gcds need not exist in an arbitrary ring, not arbitrary integral domain? >An element d in R is called the greatest common divisor (gcd) of >elements a and b from R, if >(a) d|a and d|b; >(b) d'|a and d'|b implies d'|d. If R is integral domain, then any two gcd's of a and b are associated >(differ by a unit). Consider the ring Z[sqrt(-5)]; all elements are of the form x+y*sqrt(-5), with x and y integers, and the obvious addition and multiplication. Let a = 6, b = 2+2*sqrt(-5). Note that 2 divides both a and b; note also that 1+sqrt(-5) divides both 6 and 2+2*sqrt(-5); the former, since (1+sqrt(-5))(1-sqrt(-5))=6. Let N:Z[sqrt(-5)]->Z be the norm function, N(x+y*sqrt(-5)) = x^2 + 5y^2. Note that: (i) N is multiplicative. (ii) If x+y*sqrt(-5) divides z+w*sqrt(-5) in Z[sqrt(-5)], then N(x+y*sqrt(-5)) divides N(z+w*sqrt(-5)) in Z. (iii) x+y*sqrt(-5) is a unit in Z[sqrt(-5)] if and only if N(x+y*sqrt(-5)) = 1, if and only if x+y*sqrt(-5)=1 or x+y*sqrt(-5)=-1. N(6) = 36; N(2+2*sqrt(-5)=4+20 = 24. So any common divisor of a and b must have norm dividing both 36 and 24; that is, we must have that the norm divides 12. Since N(2)=4 and N(1+sqrt(-5))=6, any common multiple of the two must have norm a multiple of 12. So if 6 and 2+2*sqrt(-5) have a greatest common divisor, it must be a number of norm 12 which is a multiple of 2 and of 1+sqrt(-5). Call it x+y*sqrt(-5). Its norm is x^2+5y^2=12. Since x+y*sqrt(-5) must be a multiple of 2, both x and y are even, so we actually have x = 2m, y=2n with m and n integers, so the norm is 12 = x^2 + 5y^2 = 4m^2 + 20n^2. This is impossible: n must equal zero, and then m must have square equal to 3. Therefore, 6 and 2+2*sqrt(-5) have no greatest common divisor in Z[sqrt(-5)]. Arturo Magidin, sans .sig X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h83GY8m01926; ==== all my sums are hard in class and i thinck u can help me they are to hard every sum is hard i get every one wrong can you help pleaseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e thanck f r o m-----b e n X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h83Kpeu24034; ==== Does somebody could consider following input: X=m+v ; Y=m ; Z=m+w When beginning with n=3, primitive set of X;Y;Z also numbers of gdc=1 will provide to us also m;v and w of gdc=1 . once we'll have: w^3 - v^3 = m [ m^2 - 3m(w-v) - 3(w^2 - v^2)] so consistency of this expression needs to factorise m by w - v ; eventually we'll have then more than single factor at Right side, what it is to avoid once taking w - v = 1 . But again taking superposition of fixed numbers: m+v = c ; then m = c-v and m+w = c+1 for to achieve previously taken w - v = 1 So using similar developments we'll come to condition 1+v = 1 Once m was some of smaller numbers so w is some natural number and from condition w - v = 1 v should be also some natural number . ( previously it can be considered as some integer) The last statement 1 + v = 1 can not be true for v as natural number and so on FLT is true for n = 3. Very similar and general developments can be extended to every prime number bigger than 3 so is there some fault or this is just this very hidden proof of FLT ? Ro X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h84Mamq04164; ==== Why isn't the number one prime? X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h8542Li25882; ==== > If I throw 8 dice, what are my odds of getting at least 3 sixes? >> And if I throw 6 dice? The binomial distribution solves this problem. There are lots of >explanations of the binomial dostribution on the net. If p is the probability of one success in one trial (0 < p < 1) then the probability of exactly k successes in n trials is given by n!/[k!*(n-k)!]*p^k*(1-p)^(n-k) found this site tryin to remember how to calculate odds of a dice roll......English Mother F*&^ER, DO YOU SPEAK IT In the great words of Samuel L. Jackson in Pulp Fiction. chicky's tryin to get answers and you throw her an alphebet soup...try to simplify things for her..although im sure she appreciates the answer....as would i if i knew what the hell it meant.this is why i hate math. no one tells you what n,k,p represent. maybe thats why im looking for the answer to a math question on the net but who knows! Sorry if i bother ya'll but maybe soeone will find this and find it agreeable X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h8414VP10129; ==== Any general algorithm to solve such exponential equation. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h832q7328501; ==== >
>[Deleted]
>>Fuck off.
>>Get the fuck out of nanae and sci.math, dickhead.
You're just jealous of my greatness.-JismMonkeY
--
>      Lord Gilbert T. Sullivan
>      Ruler of: alt.evil.
>                alt.flame.
>                alt.A.V.F.F.F.
>                alt.F.K.M.N.
>
Your greatness is highly dubious.-MeatSmokeR
Archie Leach
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> In a 2-dimensional lattice random walk with balanced probabilities,
> i.e. four of 1/4 each, starting at <0,0>; what is the probability
> that it gets to <0,1> before returning to <0,0> ?
 The answer was given as exactly 1/2, and my crude numerical estimations
> seem to bear this out.  At the time, I think a full proof was given,
> but it seemed a rather long one, and it struck me that for such a simple
> result, there ought to be a simple proof.      But I never found one.
 So, can anyone find a quick proof of the result?
 Or failing that, can anyone recall the earlier longish proof?
Here's a quick justification that relies on physical intuition.
Let f(x,y) be a point response: 
f(0,0)=1, 
[limit (x or y -> infinity or -infinity) f(x,y)]=0,
f(x,y)=(1/4)[f(x+1,y)+f(x-1,y)+f(x,y+1)+f(x,y-1)] if (x,y) is not (0,0).
f enjoys symmetry: f(x,y)=f(y,x)=f(-x,y)=...
Define g(x,y)=(1/2)+b*[f(x,y)-f(x,y-1)]
where b is chosen to make g(0,0)=1.
By symmetry we will have g(0,1)=0.
g satisfies the difference equation except at (0,0) and (0,1).
g(x,y) is the probability that starting from (x,y), the 
first visit to (0,0) precedes the first visit to (0,1), 
counting the starting point as a first visit.  (g has the 
right boundary conditions and right difference equation.)
But we are not counting the starting point as a first visit.
So the probability that, starting from (0,0) and not 
counting the starting point, we revisit (0,0) before (0,1)
is  P=(1/4)(g(0,1)+g(0,-1)+g(1,0)+g(-1,0)).
P=(1/2)+(b/4)*[f(0,1)-f(0,0)+f(0,-1)-f(0,-2)+
   +f(1,0)-f(1,-1)+f(-1,0)-f(-1,-1)]
Substituting f(0,0)+f(0,-2)+f(1,-1)+f(-1,-1)=4f(0,-1),
(from the difference equation), we get
P=(1/2)+(b/4)*[f(0,1)-3f(0,-1)+f(1,0)+f(-1,0)]
But symmetry of f gives f(0,1)=f(0,-1)=f(1,0)=f(-1,0)
so the bracketed expression is 0 and p=1/2.
Don Coppersmith
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I am willing to test the programs also.  I will compile them on a MAC.  This 
was we can have multiple runs on different machines.
j_r_o_d_g_e_r_s@a_u_g.e_d_u
just get rid of all the underscores.
I will compile them with as little change as possible for my machine and 
post the times.
Jason Rodgers
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On my machine.  A 1GHz Mac the time to run the 1,000,000 problem is .06 
seconds
to run the 20,000,000 problem was 2.55 seconds.
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The few so far that I have  are as follows
All time is in seconds.
James Harris original           
size,   time    
1000000,          0.64  
10000000        ,  7.12 
20000000,  14.72
                
James Harris improved by Stan Gula              
size    time    
1000000,        0.11    
10000000,  1.19
20000000,  2.52 
                
Jason Rodgers           
size    time    
1000000,  0.04  
10000000        ,  1.11
20000000,  2.42
        
                
Modification of Christian Bau's by C Bond               
size    time    
1000000,   0.44 
10000000        ,   11.68
20000000,  31.21
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     I'm testing a procedure I'm using to solve solvable sextics.
It works so far, but I need to test it more. So, if you know
any other solvable sextic, pls post it. And not those that factor
over sqrt(D) for some D. Those are too easy.
--Tito
====
>     I'm testing a procedure I'm using to solve solvable sextics.
>It works so far, but I need to test it more. So, if you know
>any other solvable sextic, pls post it. And not those that factor
>over sqrt(D) for some D. Those are too easy.
Try these.  Each has a different, solvable, Galois group, according
to Maple.
                                     6
                          3 + 3 x + x 
                                 2    3    6
                    2 - 2 x - 3 x  + x  + x 
                               2      3    6
                    2 - 2 x + x  + 2 x  + x 
                             2      3    6
                     -3 + 3 x  + 2 x  + x 
                           2    3    4    6
                     -1 - x  + x  + x  + x 
                             2    3    4    6
                  1 - x - 2 x  + x  + x  + x 
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2
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Here are the times I have
James Harris original           
size    time    
1000000,  0.64  
10000000,  7.12 
20000000        ,  14.72        
50000000,  38.31        
                
James Harris improved by Stan Gula              
size    time    
1000000,  0.11  
10000000,  1.19 
20000000        ,  2.52 
50000000        ,  6.46 
                
Jason Rodgers           
size    time    
1000000,  0.04  
10000000        ,  1.11 
20000000        ,  2.42 
50000000        ,  6.6  
                
Modification of Christian Bau's by C Bond               
size    time    
1000000,  0.44  
10000000        ,  11.68        
20000000        ,  31.21        
50000000        ,  115.35       
                
Daniel Jimenez  (WOW!!!)        
size    time    
1000000,  0.01  
10000000,  0.09 
20000000        ,  0.19 
50000000        ,  0.54
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>#4: A straight line is a line which lies evenly with the points on itself.
#7: A plane surface is a surface which lies evenly with the straight lines 

>on itself.
[Definitions listed as in Heath, p. 153; of course line = curve in 
#4.]
My interpretation(s): lies evenly with the points on itself = lies 
evenly
>with the [straight] tangent line drawn at each of the points on itself 
(#4);
>lies evenly with the straight lines on itself = lies evenly with any
>straight tangent line (drawn at each of the points) on itself (#7).
Justification: definitions #8 (A plane angle is the inclination to one 
>another of two lines in a plane which meet one another and do not lie in a 

>straight line) and #9 (And when the lines containing the angle are 
straight,
>the angle is called rectilineal) make it clear that Euclid had no qualms
>about discussing the angle between two curves (lines), a strong 
indication
>(if not proof) that tangentiality did cast its shadow on his definitions 
>*and* that vicious circularity -- certainly a problem in the interpretation 

>of #4 above -- was not absent from them.  
I do not see these possibilities mentioned in Heath's 1925 edition, perhaps 

>they have been raised elsewhere since then? Any such information or other 
>comments would be appreciated!
                                               baloglouAToswego.edu
P.S. Speaking of angles, I suspect that the term right angle instead 
of
>upright angle in definition #10 is due to the double meaning of Greek
>orthos/orthi as both vertical and correct :-)
  How does adding a NEW undefined term (tangent line rather than 
line) make things clearer??  Do you have any reason to believe that 
tangent line is a SIMPLER notion than line itself?
====
>>#4: A straight line is a line which lies evenly with the points on 
itself.
>>#7: A plane surface is a surface which lies evenly with the straight lines 

>>on itself.
>>[Definitions listed as in Heath, p. 153; of course line = curve in 
#4.]
>>My interpretation(s): lies evenly with the points on itself = lies 
evenly
>>with the [straight] tangent line drawn at each of the points on itself 
(#4);
>>lies evenly with the straight lines on itself = lies evenly with 
any
>>straight tangent line (drawn at each of the points) on itself (#7).
[snip]
>  How does adding a NEW undefined term (tangent line rather than 
line) 
> make things clearer??  Do you have any reason to believe that tangent 
> line is a SIMPLER notion than line itself?
Not quite, but, as I explained in my first posting in this thread, the
concept of a tangent line appears to have been simple enough for Euclid so
that it is present in definition #8 (angle between two curves) without
being explicitly mentioned; and then I explained in my second posting
(replying to Nat Silver) how my tangential interpretation simplifies the
matter by using one point instead of two (in both #4 and #7) *and* by 
explaining the appeal to surface's straight lines (rather than merely
points) in #7.
Of course all this is just a possibility, I may well have cut myself with 
Occam's razor :-)
                                                baloglouAToswego.edu
====
>#4: A straight line is a line which lies evenly with the points on
itself.
>>#7: A plane surface is a surface which lies evenly with the straight
lines
>>on itself.
>>[Definitions listed as in Heath, p. 153; of course line = curve 
in
#4.]
>>My interpretation(s): lies evenly with the points on itself = 
lies
evenly
>>with the [straight] tangent line drawn at each of the points on 
itself
(#4);
>>lies evenly with the straight lines on itself = lies evenly with 
any
>>straight tangent line (drawn at each of the points) on itself (#7).
 [snip]
  How does adding a NEW undefined term (tangent line rather than
line)
> make things clearer??  Do you have any reason to believe that tangent
> line is a SIMPLER notion than line itself?
 Not quite, but, as I explained in my first posting in this thread, the
> concept of a tangent line appears to have been simple enough for Euclid 
so
> that it is present in definition #8 (angle between two curves) without
> being explicitly mentioned; and then I explained in my second posting
> (replying to Nat Silver) how my tangential interpretation simplifies the
> matter by using one point instead of two (in both #4 and #7) *and* by
> explaining the appeal to surface's straight lines (rather than merely
> points) in #7.
 Of course all this is just a possibility, I may well have cut myself with
> Occam's razor :-)
>
I see this all of the time when Euclid's Elements is being discussed.  
I
think it is important to remember that Euclid was writing a textbook and
never intended this to be the seminal work on Geometry for close to 3000
years.  Does Euclid have gaps and ommissions and tacit assumptions.  You
bet!!!  BUT, I challenge anyone to pick up ANY high-school geometry 
textbook
and not find gaps and ommissions and tacit assumptions.
-Tralfaz
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>I've followed sci.math.num-analysis for the better part of 5 years as
>a grad student.  While I rarely post, I have learned a great deal from
>a few of you who have answered my questions.  Sadly, the signal to
>noise level on this newsgroup has simply become unbearable.  Cross
>posting to sci.physics, sci.math, sci.math.num-analysis, and
>alt.writing a few posters have effectively saturated the mathematical
>science usenet groups with nonsense and destroyed this valuable
>resource.
do you know how to use kill files? if not, could you try learning how to?
>Are there any moderated forms where one can post numerical analysis
>questions?
====
>>I've followed sci.math.num-analysis for the better part of 5 years as
>>a grad student.  While I rarely post, I have learned a great deal from
>>a few of you who have answered my questions.  Sadly, the signal to
>>noise level on this newsgroup has simply become unbearable.  Cross
>>posting to sci.physics, sci.math, sci.math.num-analysis, and
>>alt.writing a few posters have effectively saturated the mathematical
>>science usenet groups with nonsense and destroyed this valuable
>>resource.
You could try sci.math.research.
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2
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It is very easy to see this result algebrically.
Let x, y and C be some constants. Let a_n be defined as:
a_0 = x
a_n = C a_(n-1) + y
Then:
a_n - a_(n-1) = C a_(n-1) + y - (C a_(n-2) + y) = C(a_(n-1) - a_(n-2))
So the difference between two consecutive elements is multiplied by C in 
each iteration. This is a simpler way of saying what you said in your post.
>I should have thought of it sooner but I must correct myself.
 I had stated that the 3x+y of the Collatz conjecture was a series
>that has a structure where a value fits in some series where the
>difference between two members of that series is a common value times
>some power of three.
 Example 47:
Enter a value for x:47
>Arithmetic series report for 3(47) + 1
> The X = 0 : The result > 47   < Calling value
> The X = 1 : The result > 142
> The X = 2 : The result > 427
> The X = 3 : The result > 1282
> The X = 4 : The result > 3847
> The X = 5 : The result > 11542
> The X = 6 : The result > 34627
> The X = 7 : The result > 103882
> The X = 8 : The result > 311647
> The X = 9 : The result > 934942
>Primary Diff is 95
 This means that 47 is the base or root, as it might be called, of a
>sequence who has a relationship of 142 - 47 = 95 and is 3^0 * 95.  The
>second and third value are 427 - 142 =  3^1 * 95. The third and fourth
>1282 - 427 = 855 or 3^2 *95. And so on... Each iteration of this
>formula creates a higher power of 3 times the primary difference 95.
 As I was eating lunch tonight at work I thought that perhaps 3 wasn't
>the only value that might create a sequence and I was correct.
 I will assume for this post that the form of A(x)+y holds this
>structure. I will also add the quick hack of the old program to the
>end of this message.
Example of 7(x)+1 x = 47
Arithmetic series report for 7(47) + 1
> The X = 0 : The result > 47   < Calling value
> The X = 1 : The result > 330
> The X = 2 : The result > 2311
> The X = 3 : The result > 16178
> The X = 4 : The result > 113247
> The X = 5 : The result > 792730
> The X = 6 : The result > 5549111
> The X = 7 : The result > 38843778
> The X = 8 : The result > 271906447
> The X = 9 : The result > 1903345130
>Primary Diff is 283
>So if it holds then the second minus the first is 7^0 * 283.  The
>difference between the third and the second should be 7^1 * 283 and it
>is 2311 - 330 = 1981  ans is 7 * 283.
 So I think this is true then. That for all A of A(x)+y a sequence is
>formed where the difference between the second and the first terms is
>raised to powers of A in succesive iterations.
 Very cool.
 
>Ernst
// This program displays the series a given value is in or creates.
>//
>#include  int main(void)
>{
>unsigned int x,y,z,a,b,c,k,e;
>printf(Arithmetic sequence display for A(x) + ynBy Ernst Berg
>printf( Enter a value for A of A*x+yn);
>scanf(%u,&e);
>printf(Enter the value for Y of %u(x)+y:,e);
>k=0;
>scanf(%u,&k);
>if( k == 0 ) k = 1;
>printf(Formula is %u(x) + %unn,e,k);
do{
>printf(Enter a value for x:);
>scanf(%u,&z);
>printf(Arithmetic series report for %u(%u) + %un,e,z,k);
>c = z;
for(;;)
>{
>if((c>k)&&!((c-k)%e)) { c = ((c-k)/e); }
>else break;
>}
>b = c; // get a copy of the base value
>  printf( The X = 0 : The result > %u,c);
>for( x = 1; x < 10; x++ )
>{
>  if( c == z ) printf(   < Calling value n);
>  else printf(n);
>  y = (e*c)+k;
>  printf( The X = %u : The result > %u,x,y);
  if(x==1) a = y;
>  c = y;
>}
printf(nPrimary Diff is %un,a-b);
>}while(1);
}
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> I am interested in Theoretical computer science and I am considering
>> going back to school to pursue an MS degree. Should I work towards an
>> MS in mathematics taking many (if not all) electives in Comp Theory or
>> should I pursue an MS in Comp Sci taking relevant electives in math?
>> I've noticed that most people involved in Theoretical comp science
>> hold degrees in mathematics. As an example, most of the individuals
>> listed in the book People and Ideas in Theoretical Computer Science
>> have PhDs in math. Maybe one reason is that there was no Computer
>> Science department back when they were students. Now that many
>> colleges and universities offer MS and PhDs in Comp Sci does it make
>> more sense to pursue a graduate degree in Comp Sci instead of
>> Mathematics for someone interested in Theory? Comments please!
piggyback
i have some bad news for you.
unless you really love math and or comp.sci for their own sake, i advice you 
keep away from them. advanced degrees in either of those subjects, especially 
mathematics, are bound to lead you nowhere careerwise.
>Depends on the place.  At some universities, the CS dept is not
>theoretical at all.  At others it is quite theoretical.
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>This is a rather simple proof.  It can easily be seen that the chance
>of an integer lower than 10^n having a 6 in it is 1-(9/10)^n.  As
>n->oo, 1-(9/10)^10 approaches 1.  Basically, the longer a number is,
>the lower it's chances of not having a 6 are.  Of course, this applies
>to all numbers.  It can even be extended to show that most integers
>contain the string 1348945238742, and applies to all bases.  In
>another discussion group, there has been debate about this recently. 
>Several people beleive it is untrue, and that this proof make errors
>in working with infinites.  Here's one persons counterproof:

>This list is the 1st integer containing a 6, then one that doesn't,
>followed by another with a 6, and two without one.  The spacing
>between the numbers with 6's continues to increase:
6,1,16,2,3,26,4,5,7,36,8,9,10,11,46....
Because the frequency of numbers with a 6 decreases, most numbers
>don't have a 6 in them.
I agree that most numbers do contain a 6.  Apparently, it is proof 143
>in Hardy and Wright.  I don't have the book, and it's not in the
>library here.  I would appreciate any ideas you have on this.
Your proposition may not be meaningful. You are tring to take the
ratio of two countably infinite numbers. Any result is possible.
phil
====
> 
>>This is a rather simple proof.  It can easily be seen that the chance
>>of an integer lower than 10^n having a 6 in it is 1-(9/10)^n.  As
>>n->oo, 1-(9/10)^10 approaches 1.  Basically, the longer a number is,
>>the lower it's chances of not having a 6 are.  Of course, this applies
>>to all numbers.  It can even be extended to show that most integers
>>contain the string 1348945238742, and applies to all bases.  In
>>another discussion group, there has been debate about this recently. 
>>Several people beleive it is untrue, and that this proof make errors
>>in working with infinites.  Here's one persons counterproof:
>>This list is the 1st integer containing a 6, then one that doesn't,
>>followed by another with a 6, and two without one.  The spacing
>>between the numbers with 6's continues to increase:
>>6,1,16,2,3,26,4,5,7,36,8,9,10,11,46....
>>Because the frequency of numbers with a 6 decreases, most numbers
>>don't have a 6 in them.
>>I agree that most numbers do contain a 6.  Apparently, it is proof 143
>>in Hardy and Wright.  I don't have the book, and it's not in the
>>library here.  I would appreciate any ideas you have on this.
> 
> 
> Your proposition may not be meaningful. You are tring to take the
> ratio of two countably infinite numbers. Any result is possible.
> 
> phil
> 
An even stronger statement is this:
The sum of the reciprocals of all the positive integers
that do not contain a 6 (or, mor generally, any specified digit
in any base) converges.
This contrasts with the harmonic series, whose sum diverges.
I have some computations at home, which I did over 30 years ago,
computing these sums to 6 or more decimal places.
A relatively simple upper bound can be gotten by considering the
integers from B^n to B^(n+1)-1, where B is the base.
For base 10, I think the sums for any digit are less than 23.
Martin Cohen
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>I am not familiar with the terms arithmetic and quadratic means  and
>their relative sizes .
Any help appreciated.

The previous poster provided his answer in technical language that you may 
not understand, since you are a beginner.  I'll try to keep it really simple
Arithmetic mean:  This is what most people call the average of two 
numbers.
Quadratic mean:  This is the formula your teacher gave you.
Relative size:  This means to check which is bigger.  Just try a few 
calculations, and try to see which type of mean is usually bigger.  Your 
educated guess is called a conjecture.  In a more advanced course, you may 
learn ways to prove your conjecture.
-Jonathan
====
I am not familiar with the terms arithmetic and quadratic means  and
>their relative sizes .
Any help appreciated.

 The previous poster provided his answer in technical language that you 
may
not understand, since you are a beginner.  I'll try to keep it really 
simple
 Arithmetic mean:  This is what most people call the average of two
numbers.
 Quadratic mean:  This is the formula your teacher gave you.
 Relative size:  This means to check which is bigger.  Just try a few
calculations, and try to see which type of mean is usually bigger.  Your
educated guess is called a conjecture.  In a more advanced course, you may
learn ways to prove your conjecture.
 -Jonathan
>
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It is fractal spiral, a spiral made of spirals made of spirals, etc. It is 
one continuous line, and a closed shape. The form is asymmetric and the lines 
can overlap / cross each other many times. It is produced by an iterative 
process. The self similarity is not of the strictest sense of the term, but 
more conceptual. Spirals are made of spirals, but they are not truly 
identical self similar pieces that occur at different scales. The only 
continuous line fractals I'm aware of are L-systems. It looks nothing like 
the IFS dust fractals, or Smale's Horseshoe or the Lorenz Attractor. I am 
going to wait for comments before I post the formula and an explanation at my 
website.
So here are my questions:
What other types of fractals are one continuous, closed line? 
Does anyone know of any non IFS fractal spirals?
If I did create this, how can I inform people and get credit for it?
-Kevin Doughty
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>
> 
>> 
>> Well, as I write (from Waukon, Iowa), I'm listening to Nimette
>> 
>> And I was listening on Saturday to WQED, Chicago, in Stereo,
>> whilst sitting at home 75km south of Paris! Beat that.
Hate to tell you, but WQED Radio is in Pittsburgh, PA.  89.3 FM, as well
>as WQED-TV, Channel 13.  But, hey, from either Chicago or Pittsburgh to
>Paris is decent.
Unless the call WQED is shared by every public radio station in America.
And what's after a trillion?  A trillion plus one.  Unless you have an
>older Pentium... 1,000,000,000,000.999999999999999999999999999609127348 :)
> 
>G. Gollinger
>http://keystone.westminste
r.edu/~gollingj (Now with JavaScript!)
>gollingj {at} keystone {dot} westminster {dot} edu
>---------------------------------------------------------------------
>A UNIX saleslady, Lenore
>Likes work, but likes the beach more.
>She found a clever way
>To mix work with play...
>She sells C shells by the seashore.

>

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> The spacing
>between the numbers with 6's continues to increase:
6,1,16,2,3,26,4,5,7,36,8,9,10,11,46....
>
Could you explain what you mean here?  Why is 1 between 6 and 16?  Why are 2 
and 3 between 16 and 26?
====
>> The spacing
>>between the numbers with 6's continues to increase:
>>6,1,16,2,3,26,4,5,7,36,8,9,10,11,46....
>Could you explain what you mean here?  Why is 1 between 6 and 16?  Why
>are 2 and 3 between 16 and 26?
He seems to be making a list like this: 
(first number with a 6)
(first without a 6)
(next one with a 6)
(next two without a 6)
(next one with a 6)
(next three without a 6)
...
In the initial part of this sequence, of course, the
numbers with a 6 are greater than those without a 6.
But this is quite misleading, as after a while that
relation will be reversed.  If my calculations are correct,
the first integer x such that as many positive integers
<= x have a 6 as don't is 6377290.
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2
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>|graph of n nodes that their exists either a clique or an anti-clique
>|(independent set) that contains at least (log_2 n) / 2 nodes.
>|
>|I have done a little bit of research to try to find a solution to this
>|(probably simple) problem.  I know that this question relates to
>|Ramsey's
>|Theorem, and hence to Ramsey Numbers.  (Though I must admit to only
>|rudimentary understanding of thse concepts at this point.)
Well, this *is* a result of Ramsey theory. Let R(m,n) be the smallest
>N such that every graph with N verticies has either a clique of m
>verticies or an anti-clique of n verticies. These are called Ramsey
>numbers. Famously R(3,3)=6. The claim is equivalent to
>R(k,k)<=2^{2(k-1)}+1. Evidently they think you might be able to figure
>out a little Ramsey theory for yourself.
There's a story about Erdos. He said that if a demon were to come
>threatening the human race if we didn't find the value of R(5,5), we
>should get to work finding it, whereas if the demon demanded R(6,6),
>we'd better try to find some way of getting rid of the demon, because
>if we were so smart we could figure out R(6,6) just like that, we'd be
>smart enough that a demon wouldn't be a problem.
Every so often, somebody moves one of the upper and lower bounds a
>little, though.
The first little bit of _Ramsey Theory_ by Graham, Rothschild, and
>Spencer gives an argument which is good enough for your exercise.
>Sketch: show by induction that there is a sequence of verticies
>v1,...,v_{2k-1} such that for each v_i, either v_i has an edge with
>v_j for all j>i, or v_i does not have an edge with v_j for any j>i.
>sequence is more numerous.
One defines a more general Ramsey number R(n1,...,nk) by coloring a
>complete graph with k colors, and asking how many verticies are
>required to force the existence of a clique having n_i verticies
>connected by color i, for some 1<=i<=k. The R(m,n) are a special case
>where we can color an edge red if it's in the original graph and blue
>if it isn't.
Then there's an even more general Ramsey number R(s; n1,...,nk), where
>instead of an ordinary graph (where the edges correspond to subsets
>of order 2 of the verticies) we take a multigraph-- where the
>multiedges correspond to subsets of order s.
The finite Ramsey theorem is that these numbers exist. Ramsey's
>original theorem was the infinite theorem, that if there are
>infinitely many vertices, then there is an infinite monochromatic
>clique. He was proving these results for an application to logic, a
>decision procedure for statements in predicate calculus having some
>restricted form (only universal quantifiers or something like that).
>The algorithm isn't very effective; it relies upon the fact that
>searching all the possibilities which aren't large enough to force a
>monochromatic clique is finite.
Transfinite versions of Ramsey's theorem appear in set theory; some of
>them are independent of ZF, naturally.
Frank Plimpton Ramsey was sort of an interesting guy, although he
>didn't live very long (1903-1930). Economists remember him for some
>of mathematics. He had something to say about probability theory, I
>think. He once said that he didn't regard human beings as small and
>insignificant; he saw us as being in the foreground, prominently, much
>more important than all those large structures in the background.
My father and I thought it would be amusing to prove some result in
>Ramsey theory as a kind of joke. :-) Kind of difficult joke, though.
Keith Ramsay
>Boulder, CO      But not related to JonBenet Ramsey either!

>
X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h87CkpG15912; ==== Steve: Consider (-3,-3) Gary Tupper Terrace BC >>I came across the follwing statement in a book, and I still can't seem to prove it. >>|a - b| < |b|/2 => (implies) |b|/2 < |a|. >>i have tried all kinds of things using the triangle inequality... >>thanks for any help you can give, >>this is not homework, by the way. > >|b| - |a| <= | |b| - |a| | <= | b - a | -- >Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== > Steve: > > Consider (-3,-3) > > Gary Tupper > Terrace BC > > >|b| - |a| <= | |b| - |a| | <= | b - a | -- >Stephen J. Herschkorn herschko@rutcor.rutgers.edu a=-3, b=-3 then |a| = 3 |b| = 3 b - a = -3 - (-3) = -3 + 3 = 0 |b| - |a| <= ||b| - |a|| <= | b - a | so all of the above are = 0 in this case and 0<=0<=0 is a true statement... X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h83Ec4924945; ==== >An amusing commentary on the state of physics knowledge among >journalists and suicidal husbands . . . An English teacher tied an exercise machine to himself and jumped >160ft to his death after his wife left him, an inquest heard yesterday. >Christopher Peers leapt from the 21st floor of a hotel with the device >attached to his chest after he realised his Thai bride had left him for >good. The 49-year-old had used a bed sheet to tie the 3ft steel walking >machine to himself, which acted as a weight to quicken his fall. The rest of the story's at http://www.thisisdevon.co.uk/displayNode.jsp?nodeId=103354&command=disp la Perhaps quicken his fall in the sense of preventing him from changing his mind once he had the machine out the window! X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h87CkqG15927; ==== I'm need for mythesis about problemes : 1-winner take all 2-taravel sales man 3-knap sack 4-assingment problem and their history. thanks; X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h88C0Vc04931; ==== >fault. It gave me some control again, and I made the mistake of saving my >latest edits in the file I had been working on. Now, whenever I load that >file I get the invalid page fault listed at the bottom. The file seems to still have the information somewhat intact. After I get >rid of the error window, the file appears to be blank. But, if I use the >mouse to select regions, all the dashed selection boxes show up. If I then >click in one of those regions, the correct equation pops up. But, if I >scroll it off the screen and back, it's gone again. After the error occurs >upon loading the file, Mathcad is basically fried - I cannot print, or open >a new doc for cut/paste etc. I loaded the file into a text editor and it appears fine - compared to >others I've looked at. I'm not sure what all the commands are in a mcd file >though, so if something were wrong I would not recognize it. I have a backup of the file, but it's several edits old. I could save >several hours of work if the bad data within the file could be removed or >edited. Someone who knows MCAD-speak in the mcd files could probably spot >the problem. Anyone else come across this problem and find a solution? Any hints Tom ----- >MCAD caused an invalid page fault in >module MCAD.EXE at 03f7:00477d06. >Registers: >EAX=00000000 CS=03f7 EIP=00477d06 EFLGS=00010216 >EBX=00793370 SS=03ff ESP=0092f238 EBP=0096ea94 >ECX=00792f10 DS=03ff ESI=00793a50 FS=1097 >EDX=00793b40 ES=03ff EDI=0096ea8c GS=0000 >Bytes at CS:EIP: >8b 78 10 8b 44 24 14 89 38 8b 44 24 18 8b 5b 10 >Stack dump: >0096ea8c 0096e99c 0096ea88 00478e81 >00793640 0096ea80 0096ea84 0096ea88 >0096ea8c 0096ea90 0096ea94 00793b60 >00792f30 0096e99c 0000c119 bff718ce > X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h83CsOT15996; ==== >A cyclic digraph is a digraph in which every vertex is in a cycle. In >other words, cyclic digraps are non-acyclic digraphs. >In a digraph, a source is a vertex with in-set of size zero; a sink is >a vertex with out-set of size zero. I would like to know the number of labeled cyclic digraphs on n >vertices without sources nor sinks. Many thanks Here is a way to count labelled digraphs in which every node belongs to a directed cycle (clearly they have neither sources nor sinks). Let K be a class of strongly connected digraphs (closed wrt relabelling the nodes) and s_n(K) denote the number of digraphs with n labelled nodes in it. a_n(K) denotes the number of all digraphs with n nodes whose strong components belong to K. Introduce the following formal generating functions: s(z,K):= sum_{n>0}s_n(K)z^n/n!, a(z,K):= sum_{n>0}a_n(K)z^n/(n!2^{n(n-1)/2}), u(z,K):= sum_{n>0}u_n(K)z^n/n! := 1-1/(1+a(z,K)) and v(z,K):= sum_{n>0}v_n(K)z^n/n!, where v_n(K):= 2^{n(n-1)/2}u_n(K). Then the following general equation takes place: s(z,K) = -log(1-v(z,K)) (*) (see V.A.Liskovets, On a general enumerative scheme for labelled graphs, Dokl. AN BSSR, 21:6 (1977), 496-499 (in Russian); MR58#21797). In particular for the class K = D of ALL strongly connected digraphs (without loops and multiple edges), _including_ the 1-node digraph [.], we have s(z,D) = -log(1-v(z,D)) and a_n(D) = 2^{n(n-1)} = #(all digraphs). Thus, a(z,D) = sum_{n>0}2^{n(n-1)/2}z^n/n!, u(z,D) = sum_{n>0}u(n)z^n/n! = 1-1/(1+a(z,D)) and v(z,D) = sum_{n>0}2^{n(n-1)/2}u_n(D)z^n/n!. s_n(D) for n=1,2,3,... is the OEIS sequence A003030 =_1_,1,18,1606,... The digraph [.] is the ONLY type of strong components whose nodes do not belong to any cycle. In order to exclude such nodes, let us eliminate [.] from D and denote C:= D-{[.]}. Then we have s(z,C) = s(z,D)-z. By (*), s(z,C) = -log(1-v(z,C)), whence v(z,C) = 1-exp(z-s(z,D)) = 1-exp(z+log(1-v(z,D))) = 1-exp(z)(1-v(z,D)). Now knowing v(z,C), we obtain u(z,C) = sum_{n>0}v_n(C)z^n/(n!2^{n(n-1)/2}) and a(z,C) = 1/(1-u(z,C))-1 = sum_{n>0}a_n(C)z^n/(n!2^{n(n-1)/2}). The coefficients a_n(C) are the desired numbers of cyclic digraphs: the ones in which all n labelled nodes belong to (directed) cycles. Numerically a_n(C) for n=1,2,3,... are 0,1,18,1699,587940, 750744901,... The difference a_4(C)-s_4(D) = 93 is verifiable easily by hand, as well as A086193(5)-a_5(C) = 5*6*9*16 = 4320 (it's evident that a_4(C) = A086193(4)). Btw, the acyclic digraphs can be counted in the same way with 1-node components only: K:= {[.]}. Besides (curiously), u_n(D) is the number of strong tournaments (A054946). Valery Liskovets X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h83Ebv624923; ==== >I'm looking for R< (R<)-1. For a set (0,1,2,3,4,5) R< gives me {<0,2>, <0,3>, <0,4>, <0,5>, <1,3>, <1,4>, <1,5>, <2,4>, <2,5>, ><3,5>} (R<)-1 gives me the reverse {<2,0>, <3,0>, <4,0>, <5,0>, <3,1>, <4,1>, ><5,1>, <4,2>, <5,2>, <5,3>} Now here I am stuck. I'm tyring to use set comprehension to show the above >memberships. I've done R< R< and got { | m , n ? Nat, n >= (m + >2) }. This wasnt too bad. But R< (R<)-1 seems to be the empty set but I'm >not sure I'm right. Any help or direction would be appreciated. Dermot. > It's not clear to me what you are asking. I take it that R< is the less than relation: is in the set if and only if a< b for a,b in {1,2,3,4,5}. R<-1 (perhaps better written (R<)^-1) is the inverse relation and simply reverses the pairs. If that's correct then I don't know what you mean by use set comprehension to show the above memberships- they follow from the definition. I'm also not sure what you mean by R< (R<)-1. Applying the relationships in order? As in gives ? In that case, from the definition of inverse, R< (R<)^-1 is the identity relation: {<1,1>,<2,2>,<3,3>,<4,4>,<5,5>}. ==== >>I'm looking for R< (R<)-1. >>For a set (0,1,2,3,4,5) >>R< gives me {<0,2>, <0,3>, <0,4>, <0,5>, <1,3>, <1,4>, <1,5>, <2,4>, <2,5>, >><3,5>} >>(R<)-1 gives me the reverse {<2,0>, <3,0>, <4,0>, <5,0>, <3,1>, <4,1>, >><5,1>, <4,2>, <5,2>, <5,3>} >[...] > I'm also not sure what you mean by R< (R<)-1. > Applying the relationships in order? As in gives ? > > In that case, from the definition of inverse, > R< (R<)^-1 is the identity relation: {<1,1>,<2,2>,<3,3>,<4,4>,<5,5>}. Not quite. The identity relation is always a subset of the above composite. In the above example you have <0,3> in R< and <3,1> in (R<)^-1, hence <0,1> in (R<) (R<)^-1. Marc P.S.: I do find the OP notation ugly. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h88C0PR04921; ==== I have a problem: when sending a function (such as sin(x)) to a proc, when I try to use it, I get sin(x)(0) instead of my desired sin(0). Can anybody help me? ==== > I have a problem: when sending a function (such as sin(x)) to a proc, when I try to use it, I get sin(x)(0) instead of my desired > sin(0). There is a Maple newsgroup. I added it to the header. The answer to your question depends on whether you need to know the variable 'x' within the procedure. If you don't need to know it, then simply pass `sin` rather than sin(x). ==== > I have a problem: when sending a function (such as sin(x)) to a proc, when I try to use it, I get sin(x)(0) instead of my desired > sin(0). Can anybody help me? Can you give an example of what you are trying to do? Note that sin(x) is not a function; sin is a function, but sin(x) is an expression representing the value of the sin function at a point x. Perhaps what you need to do is specify the function sin as an argument to your proc instead of the expression sin(x). Without seeing actual code, it's difficult to say more than that. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h83Crlt15916; ==== Just to show the values on another machine Using a similar program a. The time was about .9664 per loop. Using a similar program b. The time was about 1.0236 per loop. Here is the program I used compiled on Gcc on the MAC OS X platform, with a 1 Ghz processor. The code is very similar to what you had, I just had to change a few thigs to make it compile. For example, I changed the declarations of the arrays to using pointers, and the #defines, I changed to const ints just because I wanted to. And the adding of the arrays I used the += method. Here is the start of the program. #include #include inline double seconds(); int main(void) { const int N = 16000000; const int M = 50; int i,j,k,l; double *a = new double[N]; double *b = new double[N]; double *c = new double[N]; for (i=0; iI would be grateful if, in vector space theory, anyone could say, what, if >any, difference there is between a scalar product and an inner product. >Ron Jones > Actually, there is a slight technical difference. The dot product is originally defined in R^n as the sum of the products of corresponding components. Once can then extend it to an arbitrary (finite dimensional) vector space by: First choose a specific basis. To find the dot product of two vectors, u and v, write u and v in terms of the given basis. Now form the dot product of the coefficients. That, of course, depends upon the specific basis chosen. An inner product is defined more abstractly. An inner product on a vector space V is any function from VxV to R (C if V is over the complex numbers) such that: i) It is linear: (au+bv,w)= a(u,w)+ b(v,w) ii) It is symmetric: (u,v)= (v,u) (If V is over the complex numbers then (u,v)= complex conjugate of (v,u).) iii) (u,u) is greater than or equal to 0 iv) (u,u)= 0 if and only if u= 0. It is easy to prove that the dot product on R^n is an inner product and that the dot product defined (for a specific basis) on vector space V is an inner product. It is the theoretical meat of the Gram-Schmidt Orthonalization Process that, given any inner product over a vector space V, there exist a basis such that the inner product is the dot product defined using that basis. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h83KULH22210; ==== >> |I look in this newsgroup since quite >> |often I find things that are interesting enough that I want to spend a >> |few hours of my spare time looking at them. However nothing that you >> |have posted has inspired me in this way though occasionally the >> |rebuttals of others have. >> >> You might be entertained to figure out which algebraic numbers >> u have the property that u is contained in a subring R of the complex >> numbers, with the property that 1 and -1 are the only units in R, >> and that any integers m and n having no common factor greater >> than 1 in the integers also have no common factors in R other >> than units of R. (This is an exercise suggested by James' definition >> of object ring. I don't see that he necessarily means to restrict >> only to algebraic numbers, though, and it's looking like he might >> want number to be broader than complex number.) You might >> also enjoy determining whether the set of all such u's is a ring. >> >> Keith Ramsay That one is beyond me. I had read it before but I could not make >sense of it. I vaguely recall reading the phrase: The integers of a field and >was wondering what it might mean. The ring of integers of the field is more common, but it is usually reserved for number fields (finite extensions of Q) and function fields (fields of transcendence degree 1 over an algebraically closed field). Let F be a field and let R be a subring of F. We say that R is integrally closed in F if and only if any element of F which satisfies a monic polynomial with coefficients in R lies in R. Given an integral domain D, we say that D is integrally closed if and only if it is integrally closed in its field of fractions. Let D be a domain, and F a field containing D. Then we can talk about the D-integral elements of F, namely, all elements which satisfy monic polynomials with coefficients in D. The ring of integers of a number field F is simply the collection of all Z-integral elements of F: all elements of F that satisfy a monic polynomial with integer coefficients. That is, the collection of all algebraic integers in F. A similar definition is made in function fields, although in that case there is no canonical choice for the distinguished subring, and depends on the choice of transcendence base. > I tried to consider how I could define >the integers of an arbitrary field. One possible definition would be >the smallest subring containing 0 and 1. I think that it is clear >that this does exist and is unique. Unfortunately it does not seem >very interesting since depending on the characteristic it would be >either Z or Zp. Also the integers of the algebraic numbers would >not be the algebraic integers. To make the integers more interesting and to address the algebraic >integer / number case, I thought to add the condition that the >quotients of the integers must be the full field. But it is no longer >clear whether this is well defined. It is not. In any quadratic extension F of Q, you could take any ring of the form Z[r] with r in F-Q. > Is there a unique subring >containing 0 and 1 such that the quotients of its member are the whole >field? The obvious place to start is the intersection of all subrings >with these properties. This is clearly a subring containing 0 and 1. >But does it satisfy my last requirement? I have not yet figured that >one out for the general case yet. I do not think this will work except in some cases with positive characteristic. Let K be any field of characteristic 0, different from Q. Let {a_1,...,a_n,...} be a basis for K over Q, and take Z[a_1,...,a_n,...]. This satisfies the condition that its field of fractions equals K. Two bases will in general yield different rings; I think they only yield the same ring if the change of basis matrix has integer coefficients. So by judiciously choosing your bases, you should be able to get intersections which are too small (if not equal to Z itself) to give all of K. >A last point on this subject. Is there a common name for Q[i]? e.g. >Gaussian rationals? I've heard them called gaussian numbers, but not often. Arturo Magidin, sans .sig ==== was wondering what it might mean. > > The ring of integers of the field is more common, but it is usually reserved for number fields (finite extensions of Q) and function fields (fields of transcendence degree 1 over an algebraically closed field). That may have been the term that I was thinking of. > Let F be a field and let R be a subring of F. We say that R is integrally closed in F if and only if any element of F which satisfies a monic polynomial with coefficients in R lies in R. > > Given an integral domain D, we say that D is integrally closed if and only if it is integrally closed in its field of fractions. > > Let D be a domain, and F a field containing D. Then we can talk about the D-integral elements of F, namely, all elements which satisfy monic polynomials with coefficients in D. > > The ring of integers of a number field F is simply the collection of all Z-integral elements of F: all elements of F that satisfy a monic polynomial with integer coefficients. That is, the collection of all algebraic integers in F. I was not very close to the track then. > A similar definition is made in function fields, although in that case there is no canonical choice for the distinguished subring, and depends on the choice of transcendence base. > > I tried to consider how I could define >the integers of an arbitrary field. One possible definition would be >the smallest subring containing 0 and 1. I think that it is clear >that this does exist and is unique. Unfortunately it does not seem >very interesting since depending on the characteristic it would be >either Z or Zp. Also the integers of the algebraic numbers would >not be the algebraic integers. To make the integers more interesting and to address the algebraic >integer / number case, I thought to add the condition that the >quotients of the integers must be the full field. But it is no longer >clear whether this is well defined. > > It is not. In any quadratic extension F of Q, you could take any ring of the form Z[r] with r in F-Q. I suspected that but I had not found a counter example yet. > Is there a unique subring >containing 0 and 1 such that the quotients of its member are the whole >field? The obvious place to start is the intersection of all subrings >with these properties. This is clearly a subring containing 0 and 1. >But does it satisfy my last requirement? I have not yet figured that >one out for the general case yet. > > I do not think this will work except in some cases with positive characteristic. Let K be any field of characteristic 0, different from Q. Let {a_1,...,a_n,...} be a basis for K over Q, and take Z[a_1,...,a_n,...]. This satisfies the condition that its field of fractions equals K. Two bases will in general yield different rings; I think they only yield the same ring if the change of basis matrix has integer coefficients. So by judiciously choosing your bases, you should be able to get intersections which are too small (if not equal to Z itself) to give all of K. I will try to find a concrete example. >A last point on this subject. Is there a common name for Q[i]? e.g. >Gaussian rationals? > > I've heard them called gaussian numbers, but not often. I did a web search and found a few people using Gaussion Rationals for Q[i]. I will look for Gaussian numbers and see if it is more popular. Possibly not many people care about Q[i] at all. > Arturo Magidin, sans .sig J also senza .sig X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h83LgCR28005; ==== [snip] >HERE'S THE SETUP TO THE QUESTIONS<<<<<<<<<<<<<<<<<<<< >> The problem in the ring of algebraic integers is >> that you can have algebraic integers 'a', 'b', 'c', >> and a prime p, such that >> abc = p >> but neither 'a', 'b', nor 'c' shares any non-unit >> factors with p. >>If you claim to have proven this, then clearly your methodology >>is flawed. After all, we have >> 1. If 'a' is not a unit, then 'a' itself is >> a non-unit factor of 'a' and 'p'. >> 2. If 'b' is not a unit, then 'b' itself is >> a non-unit factor of 'b' and 'p'. >> 3. If 'c' is not a unit, then 'c' itself is >> a non-unit factor of 'c' and 'p'. Not that I really think that James has the faintest clue what he >is talking about, but isn't it reasonable to assume that he actually >means that neither 'a', 'b', nor 'c' share any non-unit, non-trivial >factors with p? Not really. The situation that shows up in his arguments is slightly different, and slightly more complicated: there, he has a*b*c=r, with a,b,c algebraic integers, r an integer, and r divisible by a prime p. He claims that there situations where none of a, b, and c have nonunit common factors with p. This is much harder to disprove, and it eventually depends on a very deep result of Dedekind, which he calls not at all easy to prove. It is a consequence of the finiteness of the class number, that the ring of all algebraic integers is a Bezout Domain, and therefore, the claim as I just outlined is necessarily false: one can show that gcd(a*b*c,p) divides gcd(a,p)*gcd(b,p)*gcd(c,p) (just as in the integers), and therefore, the latter cannot be a unit; so at least one of gcd(a,p), gcd(b,p), and gcd(c,p) is not a unit. (Note, however, that gcd(x,y) is only defined up to units in the algebraic integers; it is better to think of it as an ideal). >I have no idea if this revised statement is true or not but I'm equally >sure that (a) James doesn't either (b) he can't prove it either way >and (c) if true, it won't help his argument. The revised argument is also trivially false: if y is an algebraic integer, then so is sqrt(y); for if y is a root of x^n + a_{n-1}*x^{n-1} + ... + a_1*x + a_0, then sqrt(y) is a root of x^{2n} + a_{n-1}x^{2n-2} + ... + a_1*x^2 + a_0, hence also an algebraic integer. If sqrt(y) is a unit, then so is y; and if y is a unit, then so is sqrt(y) (factors of units are units; products of units are units). So, if a*b*c=p, and p is not a unit, then sqrt(a) is a common factor of a and p; sqrt(b) is a common factor of b and p; and sqrt(c) is a common factor of c and p. If all of a, b, c are units, then p would be a unit; so at least one of sqrt(a), sqrt(b), sqrt(c) is a proper factor of a, b, c, respectively, giving the falsity of the claim. Arturo Magidin, sans .sig X-Received: from home.mathforum.org (home-1.mathforum.org [144.118.94.17]) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id h852Act19165 X-Received: (from approve@localhost) by home.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.4 nullclient) id h852AbO05846; ==== >|graph of n nodes that their exists either a clique or an anti-clique >|(independent set) that contains at least (log_2 n) / 2 nodes. >| >|I have done a little bit of research to try to find a solution to this >|(probably simple) problem.  I know that this question relates to >|Ramsey's >|Theorem, and hence to Ramsey Numbers.  (Though I must admit to only >|rudimentary understanding of thse concepts at this point.) Well, this *is* a result of Ramsey theory. Let R(m,n) be the smallest >N such that every graph with N verticies has either a clique of m >verticies or an anti-clique of n verticies. These are called Ramsey >numbers. Famously R(3,3)=6. The claim is equivalent to >R(k,k)<=2^{2(k-1)}+1. Evidently they think you might be able to figure >out a little Ramsey theory for yourself. There's a story about Erdos. He said that if a demon were to come >threatening the human race if we didn't find the value of R(5,5), we >should get to work finding it, whereas if the demon demanded R(6,6), >we'd better try to find some way of getting rid of the demon, because >if we were so smart we could figure out R(6,6) just like that, we'd be >smart enough that a demon wouldn't be a problem. Every so often, somebody moves one of the upper and lower bounds a >little, though. The first little bit of _Ramsey Theory_ by Graham, Rothschild, and >Spencer gives an argument which is good enough for your exercise. >Sketch: show by induction that there is a sequence of verticies >v1,...,v_{2k-1} such that for each v_i, either v_i has an edge with >v_j for all j>i, or v_i does not have an edge with v_j for any j>i. >sequence is more numerous. One defines a more general Ramsey number R(n1,...,nk) by coloring a >complete graph with k colors, and asking how many verticies are >required to force the existence of a clique having n_i verticies >connected by color i, for some 1<=i<=k. The R(m,n) are a special case >where we can color an edge red if it's in the original graph and blue >if it isn't. Then there's an even more general Ramsey number R(s; n1,...,nk), where >instead of an ordinary graph (where the edges correspond to subsets >of order 2 of the verticies) we take a multigraph-- where the >multiedges correspond to subsets of order s. The finite Ramsey theorem is that these numbers exist. Ramsey's >original theorem was the infinite theorem, that if there are >infinitely many vertices, then there is an infinite monochromatic >clique. He was proving these results for an application to logic, a >decision procedure for statements in predicate calculus having some >restricted form (only universal quantifiers or something like that). >The algorithm isn't very effective; it relies upon the fact that >searching all the possibilities which aren't large enough to force a >monochromatic clique is finite. Transfinite versions of Ramsey's theorem appear in set theory; some of >them are independent of ZF, naturally. Frank Plimpton Ramsey was sort of an interesting guy, although he >didn't live very long (1903-1930). Economists remember him for some >of mathematics. He had something to say about probability theory, I >think. He once said that he didn't regard human beings as small and >insignificant; he saw us as being in the foreground, prominently, much >more important than all those large structures in the background. My father and I thought it would be amusing to prove some result in >Ramsey theory as a kind of joke. :-) Kind of difficult joke, though. Keith Ramsay >Boulder, CO But not related to JonBenet Ramsey either! > X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h86MPD127195; ==== > Does somebody could consider following input: > X=m+v ; Y=m ; Z=m+w > When beginning with n=3, primitive set of X;Y;Z > also numbers of gdc=1 will provide to us also > m;v and w of gdc=1 . > once we'll have: > w^3 - v^3 = m [ m^2 - 3m(w-v) - 3(w^2 - v^2)] > so consistency of this expression needs to factorise > m by w - v ; eventually we'll have then more than > single factor at Right side, *** the very small correction need to be done in this place: we can consider w - v as number 3 and once m will be divided by 3 so at Rs we'll have 3^3 but Ls 3^2 ; using w-v divided by 3^2 and m by 3 we achieve Rs divided by 3^3 and Ls also by 3^3 and so on some combinations of w-v divided by 3^(3k-1) but m divided by 3^k . Than with similar developments like down we'll come to condition 3^(3k-1) + v = 1 Because we can not complete the deal using the secound of smaller number to be divided by 3. With such alterations we are coming to deal of FLT also for all bigger primes. Ro > what it is to avoid once taking w - v = 1 . > But again taking superposition of fixed numbers: > m+v = c ; then m = c-v and m+w = c+1 for to > achieve previously taken w - v = 1 > So using similar developments we'll come to > condition 1+v = 1 > Once m was some of smaller numbers so w is some > natural number and from condition w - v = 1 > v should be also some natural number . > ( previously it can be considered as some integer) > The last statement 1 + v = 1 can not be true for > v as natural number and so on FLT is true > for n = 3. > Very similar and general developments > can be extended to every prime number bigger > than 3 so is there some fault or this > is just this very hidden proof of FLT ? > Ro > > X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h87IdLf05320; ==== >> Does somebody could consider following input: X=m-v ; Y=m ; Z=m+w For m;v;w as natural numbers also once taking X < Y < Z When beginning with n=3, primitive set of X;Y;Z also numbers of gcd=1 will provide to us also m;v and w of gcd=1 . once we'll have: w^3 + v^3 = m [ m^2 - 3m(w+v) - 3(w^2 - v^2) so consistency of this expression needs to factorise m by w + v ; eventually we'll have then more than single such factor at Right side. For to avoid such inconsistency we can take v + w = 1 . But this is also inconsistency once v;w are natural numbers. Also we need consider w + v as number 3 and once m will be divided by 3 so at Rs we'll have 3^3 but Ls 3^2 ; using w+v divided by 3^2 and m by 3 we achieve Rs divided by 3^3 and Ls also by 3^3 and so on some combinations of w+v divided by 3^(3k-1) but m divided by 3^k . Very similar procedures can be taken for every prime exponent bigger than 3 and so on we've completed I fall of FLT. Some difference, that is shown with this method, it is direct pointing to the middle valued number as to be divided by exponent. Some more procedures,once basing on this first result have guided me to the full elementary proof of FLT. Anybody like to find some more faults? Sorry for my not so controlled enthusiasm in the first topic and first corrections. Ro X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h87HSRP00711; ==== can u please tell me which country does not use the metric system other than the US. is it liberia or burma thats all i could find on the net. it would be soo much apreciated if u can let me know today ASAP thanx ==== > can u please tell me which country does not use the metric system other than the US. is it liberia or burma thats all i could find on the net. > it would be soo much apreciated if u can let me know today ASAP > thanx The UK. -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ B-but Angus! You're a dragon! - Mickey Mouse X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h880RBU27575; ==== Nathaniel, you seem to be missing the point of the intention of my post, and have missed reading the last few paragraphs. >the cells in Conway's multicellular Life creatures? >> That is the reason why matter and space cannot have a smallest >> unit because they would have to move as is described above. Not necessarily. Conway's Glider is stationary at every instant, but its >speed (1/4 dist/time) and direction can be determined. Again, my intention in my post was to suggest that the way out of the paradox is to reject matter as a substance. What I am saying is that time and space DO have a smallest unit and that movement and speed should be redefined as change in position through subsequent instants of time. >> Logically >> instant in time. No. Gliders follows predictalbe paths. >> Nothing links the two's identities together. The passage of time, and the rules of Conway's game, link lifeforms to >cells. Likewise, the passage of time, and the rules of the universe, link Again, the rules of Conway's game have already been established for you. In the physical world, we don't know the real rules, only the one's that we make up based on our reason that we project onto the world. Think of an object moving around on your computer screen. Is there something that constantly exists transfered to the adjacent pixel in the subsequent moment? No, and there is no reason why the physical world can't have the same form of change. >> Secondly, in our present definition of matter, matter is a >> substance, and a substance is not annihilated and then created at >> every consecutive instant in time. There is no matter, there is no substance. There is only form. This was the conclusion of my argument. >> There are many other odd >> conclusions from this. >> So what then is the solution? It has certainly not been solved >> by mathematics. Huh? Your questions are easy to answer. You must have mathematicians >confused with Cantorians. >> The problem of assuming the infinite divisibility of >> space is demonstrated logically impossible in the dichotomy. The >> alternative on finite points constituting space, a smallest >> been shown to be no less full of contradictions. Only *your interpretation* of the alternative has been shown to be full of >contradictions. The last line of what I said here is stated in the context that matter is the substance of the physical world. Regarding what you said here, if this is so why did you only state this instead of presenting even one contradiction that I made? Not that I am saying I made any but do you know the difference between a contradiction and a fallacy? It appears that you were arguing with my presentation of the problem as if that is what I asserting while my final conclusion was entirely consistent with the analogy you presented. The only exception is that you don't appear to be familiar with the traditional defintion of substance. If you would like to see this in detail, in better detail visit my webpage at http://www.geocities.com/Roddys2rad/matter.html Roddy X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h881hGF32349; ==== this is a test. JSH X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h83GY6C01914; ==== >Last-modified: 1996 October 02 q10. FAQ: Numerical Analysis & Associated Fields Resource Guide Copyright 1995-6 S. J. Sullivan Welcome! My intent here is to provide reviews of software, texts, >and other resources instead of simply a listing. My experience is >that for someone looking for a package or system, reviews by previous >users can be a lifesaver. If you have any suggestions, comments, or contributions please send >them to me at: sullivan@mathcom.com Other reviews would be most welcome! If you use a mathematical >package or set of programs and would care to write one to twenty >sentences on it, please let me know. If you have a favorite text >or two you'd like to recommend, please let me know. Sigh, and now the legalities ... The information contained in this document is believed to be true, >but no guarantees of accuracy are made, and there is no liability >of any sort for any consequences of its use. Copyright 1995-6 S. Sullivan: This document may be copied and/or >reproduced providing that: > * the use is for non-commercial purposes only, and > * all copies contain this copyright notice >See: > * q20, NA FAQ: Introduction > * q30, NA FAQ: Overview of Recent Additions > * q50, NA FAQ: Acknowledgements >Steve Sullivan >sullivan@mathcom.com Mathcom, Inc. >8555 Hollyhock St., Lafayette, CO 80026 USA >========================================================================= q20. NA FAQ: Introduction Where to find this FAQ: On the web: >http://www.indra.com/~sulliv an/q10.html Mathcom This FAQ is usually available from MIT's rtfm and its mirrors: >ft p://rtfm.mit.edu/pub/usenet/news.answers/num-analysis/faq/part1 MIT's rtfm If not, it is at: >ftp://ftp.mathcom.com/mathcom/na faq Mathcom ftp >A compressed (with gzip) version is at: >ftp://ftp.mathcom.com/mathcom /nafaq.gz Mathcom ftp >Abbreviations used in this document: NA Numerical Analysis ### Denotes to be filled in later. This is work in progress, >and probably always will be. [] Reviews are associated with the name of the reviewer in brackets. >Those reviews marked [SJS] are by myself. [author] indicates text taken from a package documentation. 10^12 10 to the power 12. In Fortran, that's 10**12; >in C that's pow(10,12). x(i) the i'th element of vector x. In C, that would be x[i]. >Instead of the normal question/answer form, this FAQ is organized >as an outline ... hopefully, you'll find your questions answered here. >========================================================================= q30. NA FAQ: Overview of Recent Additions >q150, Electronic Journals for NA >The Southwest Journal of Pure and Applied Mathematics q165, Books, With and Without Software >Mathews, John H. 1992: >NUMERICAL METHODS: for Mathematics, Science & Engineering. >Source code is available in several languages. q160.3, Modula-3 NA Library q160.4, Forth Numerical/Scientific Library q210.7, Tests for Randomness >The Diehard suite of RNG tests, by George Marsaglia, is available >with a rotating 3-D graph showing the collinearities of >tuples. q210.1, Web Sites for Random Number Generators >Additional pLab info. q260.1, PDE and FEM Web Sites >MGNet at Yale has codes, preprints, virtual proceedings, >a large bibliography, and more dealing with multigrid and/or >domain decomposition methods for solving PDE's. q260.2.10, Madpack5 >abstract solver. It is PDE, domain, and discretization independent. q265.1, Optimization, Linear and Non-Linear Programming >Xu's list of public optimization codes. q505, Probability and Statistics >Software: U. Texas Dept of Biomathematics software. >This is mostly Fortran, but some is also in C. DCDFLIB is a >good set of functions for computing cumulative distribution >function values. q520.2.6, GRTensorII >GRTensorII is a computer algebra package for performing calculations >in the general area of differential geometry. >=================================== Regarding random number generation, please note ... Call For Papers >ACM Transactions on Modeling and Computer Simulation Special Issue on Uniform Random Number Generation Raymond Couture and Pierre L'Ecuyer >University of Montreal, Guest Editors >For details, please consult the ACM TOMACS Web page: >http://www.acm.org/pubs/tomacs/ ACM TOMACS >or write: >Pierre L'Ecuyer, Departement d'Informatique et de Recherche Operationnelle >Universite de Montreal ========================================================================= * q10, FAQ: Numerical Analysis & Associated Fields Resource Guide > * q20, NA FAQ: Introduction > * q30, NA FAQ: Overview of Recent Additions > * q50, NA FAQ: Acknowledgements > * > * q105, What is Numerical Analysis? > * q110, Indices of NA Software on the Net > * q112, Indices of Commercial NA Software > * q115, Libraries of NA Software on the Net > * q120, NA Packages on the Net > * q125, Commercial NA Libraries and Packages > * q140, Professional Societies for NA > * q145, Electronic Newsletters for NA > * q150, Electronic Journals for NA > * q155, Online Preprints for NA > * q160, Miscellaneous Web Sites for NA > * q165, Books, With and Without Software, for NA Specialized Subfields Within Numerical Analysis > * q205, Dense (Non-Sparse) Linear Algebra Systems > * q207, Sparse Linear Algebra Systems > * q210, Random Number Generators (RNGs) > * q215, Function Evaluation > * q220, Finding Roots > * q230, Curve Fitting, Data Modelling, Interpolation, Extrapolation > * q240, Transforms (FFT, etc) and digital signal processing (DSP) > * q245, Wavelets > * q250, Integration and Ordinary Differential Equations (ODEs) > * q260, Partial Differential Equations (PDEs) and Finite Element Modeling (FEM) > * q265, Operations Research: Minimization, Optimization > * q270, Computational Geometry > * q285, Graphics and Scientific Visualization > * q290, Miscellaneous NA Software Associated Fields > * q505, Probability and Statistics > * q510, Chaos Theory (Nonlinear Dynamics) > * q520, Symbolic Algebra > * q530, Cryptography (Cryptology) > * q540, Fractals > * q550, Neural Networks > * q560, Discrete algorithms > * q570, Constraints > * q580, Genetic Algorithms > * q590, Simulated Annealing Teaching and Academic Software > * q800, Teaching and Academic Software =========================================================================== >q50. NA FAQ: Acknowledgements Many thanks to all those who've given their time and advice >in creating this FAQ, including: Bob Berman berman@FERMAT.macsyma.com > Ronald F Boisvert boisvert@cam.nist.gov > Ted Brown tbrown@tekotago.ac.nz > John Chandler jpc@a.cs.okstate.edu > Luiz Henrique de Figueiredo lhf@csgrs6k1.uwaterloo.ca > Bill Frensley frensley@utdallas.edu > Pawel Gora gora@if.uj.edu.pl > Amara Graps agraps@netcom.com > Vijay Gupta gupta@acsu.Buffalo.edu > Doug Hart hart@de01.denver.waii.com > Dave Linder dwl@apmaths.uwo.ca > George Marsaglia geo@stat.fsu.edu > Pierre Maxted pflm@star.maps.susx.ac.uk > Allen Mcintosh mcintosh@bellcore.com > Sean O riordain sor@inrets.fr > Daniel Pfenniger pfennige@scsun.unige.ch > Daniel Pick pick@lune.math.tau.ac.il > Brian Ripley ripley@stats.ox.ac.uk > Ramin Samadani ramin@leland.Stanford.EDU > Robert Schneiders robert@Informatik.RWTH-Aachen.DE > Peter Somlo somlo@zeta.org.au > Tim Strotman tim.strotman@sdrc.com > N. Sukumar n-sukumar@nwu.edu > Stephen Vavasis vavasis@CS.Cornell.EDU > Dave Watson watson@maths.uwa.edu.au Many thanks also to the organizers of the many services >listed herein - Netlib, the NIST guide, NA-Net, CAIN, the NASA >Graphics site, and numerous other indices and informative web pages. >=========================================================================== q105. What is Numerical Analysis? NA is the union of theoretical and computational investigation into >the computer solution of mathematical problems. NA generally includes >those problems involving continuous functions of real or complex >variables, as opposed to solely discrete variables and functions. >The mixing of theoretical and computational concerns >gives NA its particular character. The compuational aspects of NA usually take place within >the scope of floating-point arithmetic, and are implemented on >machines ranging from super-computers through PCs to hand-calculators. >The theoretical aspects extend into fields such as Calculus, >Differential Equations, and Analysis. The field of Linear Algebra >is so often used to model physical systems that the theoretical >study of Linear Algebra is in itself often considered to be >NA at work. Primary areas of theoretical concern in NA are: > * global/local error bounding > * stability of algorithms > * rates of convergence of algorithms Primary areas of computational concern in NA are: > * roundoff error > * global/local error and its tolerance > * time and memory requirements of computation > * parallel computing > * architechture/platform specific details. >=========================================================================== >q110. Indices of NA Software on the Net For indices of packages oriented towards symbolic algebra, >see q520, Symbolic Algebra. >The NIST Guide to Available Mathematical Software (Formerly called GAMS) http://gams.nist.gov/ NIST Guide to Mathematical Software >or telnet to: gams.nist.gov [SJS]: >Maintained by National Institute of Standards and Technology (NIST) >An index and server for a wide variety of mathematical >software, including most of netlib (see q115.1, Netlib). >Much of the software is in Fortran. If you prefer to speak C++ or C, >see q160.1, C++ Resources, and q115.2, Fortran, C, and f2c. [Ronald Boisvert]: >The main focus is on fine-grained software components, e.g. >subroutines, although information about some larger packages are >included. As of November 1995, nearly 10,000 components from more >than 90 packages have been cross-indexed using a detailed >tree-structured problem classification system. Both freely available >software (from netlib or developed at NIST) and commercial packages >(used by NIST) are indexed, although source code is available only for >non-commercial software. =================================== q112. Indices of Commercial NA Software A large list of commercial NA products may be found at: >http://www.cray.com/PUBLIC/APPS /DAS/ Cray The Directory of commercial software, by International Computer >Programs, Inc., is at: >http://www.icp.com/softinfo/ ICP Finally, for packages oriented towards symbolic algebra, >see q520, Symbolic Algebra. =========================================================================== q115. Libraries of NA Software on the Net Libraries are collections of source code, and source code packages. >Much of the code is in Fortran. If you prefer to speak C++ or C, >see q160.1, C++ Resources, and q115.2, Fortran, C, and f2c. The main library by far is q115.1, Netlib. >For statistical software, the best resource is q115.3, Statlib. >Other libraries are q115.4, NCAR's Mathematical and Statistical Libraries >and q115.5, Hensa Unix Parallel Archive. * q115.1, Netlib > * q115.2, Fortran, C, and the f2c Translator > * q115.3, Statlib > * q115.4, NCAR's Mathematical and Statistical Libraries > * q115.5, Hensa Unix Parallel Archive =========================================================================== q115.1. Netlib >NetLib is probably the world's largest repository of numerical >methods programs. It is located at Oak Ridge National Laboratory, > netlib@ornl.gov > netlib@research.att.com http://www.netlib.org Netlib main >http://www.netlib.org/n etlib/netlib_faq.html Netlib FAQ >http://www.netlib .org/master/expanded_liblist.html Netlib index >ftp://netlib.att.com/netlib Netlib via ftp Netlib mirrors: http://www.netlib.no/ Netlib in Norway http://www.hensa.a c.uk/ftp/mirrors/netlib/master/ Netlib in England http://elib.zib- berlin.de/netlib/master/readme.html Netlib in Germany ftp://draci.cs.uow.edu.au/netlib/< /a> Netlib in Australia Some gems of netlib: Machine/architecture dependant Basic Linear Algebra Subroutines >(BLAS) are the keystone of Netlib. LAPACK, in Fortran 77, is the modern replacement >of EISPACK, LINPACK, etc. CLAPACK is a C version of LAPACK. See the Caution on >Using Arrays in q115.2, Fortran, C, and f2c. LAPACK++ is a C++ version of, sadly, only a subset of LAPACK. >LAPACK++ is work in progress, and hopefully the full >functionality of LAPACK will be supported soon. ScaLAPACK is for distributed memory machines. >=================================== >q115.2. Fortran, C, and the f2c Translator For C++ and C resources, see q160.1, C++ Resources. Most of the programs in netlib are in Fortran. However, netlib >contains an excellent Fortran-to-C conversion utility, f2c. >While f2c produces working C code, it is visually complex >and ugly. Using f2c on a large package like LAPACK can require >a good deal of time to get all the options correct. >Fortunately, LAPACK has already be converted to C: see CLAPACK. to netlib@research.att.com, with the subject execute f2c, >and body containing the non-confidential Fortran program to be converted. >since a resulting C program of any size must be linked with the >f2c libraries. Usually one will have to download the f2c package >anyway to generate the libraries. Generally it's easier >to download the f2c package, build the libraries and the >f2c conversion program, and do the conversion locally. CAUTION: Programs created by f2c conversion use parameter passing >conventions different from most C or C++ programs. Their >callers must create the appropriate parameters before using them. >See the file f2c.ps in the f2c distribution. >A good description of this issue may also be found in >the readme file for clapack in netlib. >=================================== q115.3. Statlib Statlib is a huge repository of statistics related software and info. >Probability, statistics, random variables, distribution functions. http://lib.stat.cmu.edu/ Statlib at CMU >ftp://lib.stat.cmu.edu Statlib via ftp >=================================== >q115.4. NCAR's Mathematical and Statistical Libraries NCAR's libraries contain some overlap with netlib. http://http.ucar.edu/SOFTL IB/mathlib.html NCAR >=================================== q115.5. Hensa Unix Parallel Archive >http://www.hensa.a c.uk/parallel/environments/pcn/ Hensa >Note: this web server can be very slow! >=========================================================================== >q120. NA Packages on the Net Packages generally include an NA library and an interpretive >language for a front end. Also see q520, Symbolic Algebra, for free symbolic algebra packages. * q120.1, Octave > * q120.2, RLaB > * q120.3, Scilab > * q120.6, Medal > * q120.7, Euler > * q120.8, Prophet > * q120.9, Yorick =================================== q120.1. Octave http://bevo.che.wisc.edu/octave .html Octave >ftp://www.che.wisc.edu/pub/octave< /a> Octave via ftp [Dave Lindner]: Octave is considered the closest-to-Matlab >of the Matlab clones. [author]: >Octave is a high-level language, primarily intended for >numerical computations. It provides a convenient command line >interface for solving linear and nonlinear problems >numerically. Octave can do arithmetic for real and complex scalars and matrices, >solve sets of nonlinear algebraic equations, integrate functions over >finite and infinite intervals, and integrate systems of ordinary >differential and differential-algebraic equations. The Octave distribution includes a 200+ page Texinfo manual. >Two and three dimensional plotting is fully supported using gnuplot. The underlying numerical solvers are currently standard >Fortran ones like Lapack, Linpack, Odepack, the Blas, >etc., packaged in a library of C++ classes. =================================== q120.2. RLaB http://www.eskimo.com/~ians/rl ab.html RLaB >ftp://csi.jpl.nasa.gov/pub/ma tlab/RLaB RLaB via ftp >ftp://evans.ee.adfa.oz.au/pub/RLa B RLaB via ftp in Australia [author]: >Rlab is an interactive, interpreted scientific programming >environment. Rlab is a very high level language intended to provide >fast prototyping and program development, as well as easy >data-visualization, and processing. Rlab is not a clone of languages such as those used by tools like >Matlab or Matrix_X/Xmath. However, as Rlab focuses on creating a good >experimental environment (or laboratory) in which to do matrix math, >it can be called MATLAB-like since the programming language >possesses similar operators and concepts. =================================== q120.3. Scilab http://zenon.inria.f r/Logiciels/SCILAB-eng.html Scilab >ftp://ftp.inria.fr/INRIA/Scilab Scilab via ftp [Dave Lindner]: Scilab is another good Matlab clone. [author]: >Scilab is a high-level language for numerical computations >in a user-friendly environment. It features: >Elaborate data structures (polynomial, rational and string >matrices, lists, multivariable linear systems,...). >Sophisticated interpreter and programming language with >Matlab-like syntax. >Hundreds of built-in math functions (new primitives can easily be >added). >Stunning graphics (2d, 3d, animation). >Open structure (easy interfacing with Fortran and C via online >dynamic link). Many built-in libraries : > * Linear Algebra (including sparse matrices, Kronecker > form, ordered Schur,...). > * Control (Classical, LQG, H-infinity, ...). > * Signal processing. > * Simulation (various ode's, dassl,...). > * Optimization (differentiable and non-differentiable, LQ solver). > * Metanet (network analysis and optimization). >Symbolic capabilities through Maple interface. =================================== General NA package with graphics, linear algebra, FFT, etc. >Is this another Matlab clone? [author]: >It is mainly targeted for prototyping large-scale >numerical simulations and doing pre- and postprocessing for them, and >it replaces a compiled language like C++ or Fortran in this respect. >The feature set is therefore biased to operations needed in partial >differential equation solvers. >=================================== q120.6. Medal ftp://excel2.uwaterloo.ca/pub Medal Apparently there is also available is a commercial version of Medal: [author]: >MEDAL is a novel expert system development environment which is integrated >within a control system design environment, and which supports a tight >coupling of symbolic and numeric processing. MEDAL supports the development >of coupled systems in engineering and science. MEDAL (Matrix and Expert system Development Aid Language) is an interactive >program. The language syntax of MEDAL is similar to the popular MATLAB >(Matrix Laboratory) language. MEDAL retains all of the main features of >MATLAB, including the MATLAB syntax and M-files. >In addition, MEDAL includes an integrated expert system shell for the >development of knowledge-based systems which can perform >sophisticated numeric calculations. Hence, the additional >expert system predicates extends the MATLAB command language syntax. >Also, MEDAL supports a rich set of data structure for representing >objects in the programming environment. Knowledge can be >represented using facts, rules and frames. Main features of MEDAL : >------------------------ > * interactive computing environment ( command-drive ) > * language syntax and user-interface similar to MATLAB > * all basic MATLAB-type of matrix functions are provided > * flexible 2-D graphics > * design of linear control systems > * packed matrix representation, as well as regular matrices > * automatic loading of M-files ( open philosophy ) > * build-in knowledge base development facilities (expert shell ) > * knowledge repesentation : rules, facts, objects ( frames ) > * simple knowledge base of the Systematic Design Approach is included > * runs on Sun Sparc workstations (X-window), PC (DOS), DEC (Ultrix) >(1) Pang, G.K.H.,``Knowledge-based Control System Design'', in >Recent Advances in Computer-Aided Control Systems Engineering, >Jamshidi, M and Herget, C.J. (ed.), Elsevier Science Publishers, 1992. (2) Pang, G.K.H., ``A Knowledge Environment for an Interactive Control >System Design Package'', Automatica, Vol. 28. No. 3, pp. 473-491, May 1992. =================================== q120.7. Euler http://www.ku-eichstaett. de/MGF/euler.html Euler >ftp://ftp.k.-eichstaett.de/pub/m ath Euler via ftp >ftp://ftp.k.-eichstaett.de/ pub/unix/math Euler for unix via ftp [author]: >Free MatLab like program, with real and complex >scalars and vectors, 2D/3D grafics, programming language. >The idea of EULER is a system with the following features > * Interactive evaluation of numerical expressions with real or > complex values, vectors and matrices, including use of variables. > * Builtin functions that can take vectors as input and are then > evaluated for each element of the vector or matrix. > * Matrix functions. > * Interval arithmetic for result verification. > * Exact scalar product. > * Optimization, statistical functions and random numbers. > * 2D- and 3D-plots. > * A builtin programming language with parameters and local > variables. > * An online help. > * A tracing feature for the programming language. > * Possibility to read and write raw numerical data or even binary > data from and to files. These features make EULER an ideal tool for the tasks such as > * Inspecting and discussing functions of one real or complex > variable. > * Viewing surfaces in parameter representation. > * Linear algebra and eigenvalue computation. > * Testing numerical algorithms. > * Solving differential equations numerically. > * Computing polynomials. =================================== q120.8. Prophet http://www-prophet.bbn.com/ Prophet >ftp://www-prophet.bbn.com Prophet via ftp [author]: >Prophet is an NIH-sponsored Unix workstation software package for life >science computing. Prophet includes tools for data management, >statistical analysis, curve fitting, data graphing, mathematical >modeling, and genetic sequence analysis. One of PROPHET's greatest assets is its new graphical >user interface . Employing the latest advances in software >technology, PROPHET lets you store, >analyze and present Data Tables, Graphs, Statistical Analyses and >Mathematical Modeling, and Sequence Analyses with high-resolution >graphics and multiple windows. Anyone, from the computer-naive to the >computer-sophisticate, can learn to use it quickly and effectively. >PROPHET is a National Computing Resource for Life Science Research >sponsored by the National Center for Research Resources of the >National Institutes of Health. Unfortunately, prophet is distributed in binary form only. >It is large: it takes something like 65 MB disk space. =================================== q120.9. Yorick > wuarchive.wustl.edu: /languages/yorick/yorick-1.2.tar.gz > sunsite.unc.edu: /pub/languages/yorick/yorick-1.2.tar.gz > netlib.att.com: /netlib/env/yorick-1.2.tar.gz > netlib2.cs.utk.edu: /env/yorick-1.2.tar.gz [author] >Yorick is an interpreted language. It has: > * A C-like language, but without declarative statements. Operations > between arrays require no explicit loops, which accounts for > Yorick's high speed. Scientific computing and numerical analysis > are the goals of most Yorick sessions. > * An X window system interactive graphics package. > * A library of functions written in the Yorick language. >Because Yorick can read either text or binary files, it can be used >out of the box as a pre- and post-processor for most existing >physics simulation programs. As a pre-processor, you can write a Yorick program that produces >complicated input files for a simulation. These might be based on >output from other programs, or might require evaluation of complicated >functions or involve a lot of repetition. As a post-processor, Yorick allows you to compare the results of >several simulations or to analyze results of a single simulation in >ways you did not forsee when you ran it. =========================================================================== q125. Commercial NA Libraries and Packages Commercial libraries and packages tend to merge, so I've combined >them in one category. Typically a commercial product contains: > * a library of numerical routines > * graphics routines > * an interactive interpreted language Many symbolic algebra packages also contain NA packages. >For info on these packages, see q520, Symbolic Algebra. Braham, Robert. Math & Visualization: new tools, new frontiers, >IEEE Spectrum 32, 11 (November 1995), p. 19-36. >There is no mention of the many excellent free products though. * q125.1, NAG > * q125.2, IMSL and PVWAVE > * q125.3, Matlab and Simulink > * q125.4, WavBox > * q125.5, CraySoft Libraries > * q125.6, IDL > * q125.7, Comparison of IDL and Matlab > * q125.8, Mlab > * q125.9, Gauss > * q125.10, MathViews > * q125.11, Matcom: Matlab to C++ Compiler =================================== >q125.1. NAG http://www.nag.co.uk:70/ NAG in England >http://www.nag.com/ NAG in USA [SJS]: Numerical, symbolic, statistical, and visualization libraries in >Fortran 77, Fortran 90, C, Pascal, Ada, and parallel machine versions. NAG Ltd (The Numerical Algorithms Group) Wilkinson House > OXFORD > OX2 8DR > UK NAG Inc > 1400 Opus Place > Suite 200 > Downers Grove > IL 60515-5702 > USA >=================================== q125.2. IMSL and PVWAVE http://www.vni.com/indexall.html Visual Numerics, Inc. [SJS]: IMSL is a set of routines in C, C++, and Fortran >libraries for general NA, statistics and graphics. >PVWAVE is a visual programming environment built on top of IMSL. Visual Numerics, Inc. > IMSL and Stanford Graphics Products > 9990 Richmond Avenue, suite 400 > Houston, Texas 77042-4548 > USA > FAX: 713-781-9260 Visual Numerics, Inc > PV-WAVE Products Division > 6230 Lookout Road > Boulder, Colorado 80301 > USA > FAX: 303-530-9329 > info@boulder.vni.com [author]: > * Comprehensive Mathematical Functionality > * integration and differentiation > * transforms > * differential equations > * linear systems > * interpolation and approximation > * eigensystem analysis > * optimization > * special functions > * basic matrix/vector operations > * nonlinear equations > * utilities * Extensive Statistical Functionality > * basic statistics > * tests of goodness-of-fit > * time series analysis and forecasting > * analysis of variance > * regression > * nonparametric statistics > * correlation > * random number generation > * cluster analysis > * categorical and discrete data analysis > * probability distribution functions and inverses > * factor analysis > * utilities * Exponent Graphics includes: > * Presentation quality graphs for application development > * Application program interface provides easy access to either > FORTRAN or C > * Two function calls can automatically produce one of over 30 > different plot types. > * Maximum flexibility for modifying plot chacteristics > * Powerful interactive editing and customization tools > * CGM, PostScript, HPGL and other device drivers > * Support for popular graphics accelerators and output systems > * Full Windows-based online documentation with hypertext links PV-WAVE is a software environment for solving problems requiring the >application of graphics, mathematics, numerics and statistics to data >and equations. >PV-WAVE uses an intuitive fourth generation language (4GL) that >analyzes and displays data as you enter commands. With it you can >perform complex analysis, visualization, and application >development quickly and interactively. Robust integrated graphics, numerics, data I/O, and data management >has made PV-WAVE the number one selling Visual Data Analysis software >family. PV-WAVE and the IMSL numerical and statistical routines, which are >seamlessly integrated in PV-WAVE Advantage, are being used by more >than 300,000 technical professionals on workstations worldwide. >=================================== q125.3. Matlab and Simulink http://www.mathworks.com/ Mathworks The MathWorks, Inc. > 24 Prime Park Way > Natick, MA 01760-1500 > (508) 653-1415 For a comparison of Matlab and IDL, see >q125.7, Comparison of IDL and Matlab. [SJS]: Matlab is an interactive general NA package, including graphics. >A huge variety of toolboxes are available, both from the >vendor and on the net, for various specialized NA areas: >control systems, neural nets, optimization, symbolic math, >and on and on. >Simulink is modeling, simulation, and system analysis tool. [author]: >MATLAB is a technical computing environment for high-performance >numeric computation and visualization. MATLAB integrates numerical >analysis, matrix computation, signal processing, and graphics in an >easy-to-use environment where problems and solutions are expressed >just as they are written mathematically - without traditional >programming. MATLAB has evolved over a period of years with input from many users. >In university environments, it has become the standard instructional >tool for introductory co ==== > > It's NOT the same story if instead of being Tom Sawyer the main > character is called Tim Silversmith. > > Mathematics works one way, but literature works another. > > Character names are *part* of the story and aren't arbitrary like > variable names. > > It makes a difference what name a writer picks for a character. > > I do find it fascinating that some of you don't get it. > > Since you, presumably, get it, then please explain to us what > the difference would be. How would calling the main character Tim > Silversmith change the story? Give specific examples from specific > passages, please. I notice James walked away from this one. Gee, what a surprise. - Randy