mm-31 === Subject 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 thecharacterization of surjectivity requires the Axiom of Choice whilethat of the chzracterization of the injectivity doesn't.Felix. === Subject | 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 setis not finite, and there exists a bijection from it to a propersubset.Keith Ramsay === Subject >|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.-- === Subject >> AyAx(y=x <-> y in {x}).>Find a copy of David Lewis' Parts of Classes for a clearer discussion (not necessarily clear) of thisrelation to singletons. It was published in 1991There is a good formal discussion of the mathematics behind this approach. Run a google search onA necessary relation algebra for mereotopologyThe relationship of mereologies to quantum mechanics can be seen by the fact that mereologies relate toone another as an orthocomplemented lattice. This is the same as for the Boolean subblocks of vonNeumann's quantum logic.The relationship of classical mereologies to set theory can be seen by virtue of Tarski's result that aclassical mereology can be generated from a Boolean algebra without 0. This is the same basic structureinto which separative partial orders are embedded in order to generate Cohen generic models forZermelo-Fraenkel set theory.:-)mitch === Subject : 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 googleBoolean algebra without 0, you will get hits,but on NONE of them will the prepositional phrasewithout 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 algebrasnormally have an 0. The question I am tryingto ask now is, just how much damage is done whenthe 0 is removed? Does what is left really DESERVEto 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 === Subject > 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 ofmeasure 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 === Subject 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? === Subject Several people on sci.math dual with James Harris on various mathematical(and other) topics. Currently, much energy is being expended on Lemma 1 andLemma 2. Here is a suggestion for a plan of activity. I am well aware thatthis 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. Onlystandard math terminology should be used. (But further terms could bedefined, of course.)Step 2:Produce an example of the application of Lemma 1 which is acceptable to bothsides.Step 3:Produce a proof for Lemma 1 which is acceptable to both sides. Only standardmath terminology should be used. (But further terms could be defined, ofcourse.)Step 4:Produce a wording for Lemma 2 which is acceptable to both sides. Onlystandard math terminology should be used. (But further terms could bedefined, of course.)Step 5:Produce an example of the application of Lemma 2 which is acceptable to bothsides.Step 6:Produce a proof for Lemma 2 which is acceptable to both sides. Only standardmath terminology should be used. (But further terms could be defined, ofcourse.)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 Lemma2.All definitions, examples and proofs should be too verbose, rather than tooconcise.-- Clive Toothhttp://www.clivetooth.dk === Subject * 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 thecase. And who would want that?However, I guess that it would strand on the very first paragraph instep 1.-- Jon Haugsand === Subject * 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 thecase. And who would want that?However, I guess that it would strand on the very first paragraph instep 1.-- Jon Haugsand === Subject >>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 isto use uncountable to mean |A| ~<= aleph_0, and to usetransfinite for |A| >= aleph_0. Dedekind-finite infinitecardinals are uncountable but not transfinite.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === Subject >> >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 === Subject >> >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 === Subject > 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 punishthe evil liars! Well I'm convinced.--Dave Taylor'Ray, when someone asks you if you're a god, you say, Yes!' === Subject > 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 seenbecause given any solution x for B, (x+A),(x+2A), .... are alsosolutions.We can however identify a line of solutions.The easiest member of the solution set is the vector perpendicular toC 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 asolution, so the complete set of solutions is: {x | x = x0 + kA, for any k real} === Subject Greetings.I just finished some FEM work which required numerical integration of atriangle. I got the technique for performing this integration from JianmingJin's book on FEM. Using his formalism, you take the (x,y,z) coordinate ofeach of the three vertices and plug them into a formula which multipliesthese coordinates by special factors (abscissae?) to obtain the actualpoints of integration within the triangle. Each point is given an associatedweight.I need to obtain a similar table for the arbitrary three-dimensionaltetrahedron. I have seen some tables out there (given by Ronald Cools) onthe net, but I do not know how to use them, and I've seen nothing like Jin'stable for the triangle applied to solid elements anywhere. Could someoneplease help me with the problem of obtaining a robust, understandable way tonumerically integrate over an arbitrary tetrahedron?Michael === Subject > Typing Monkeys Don't Write Shakespeare A rebuttal:Typing Monkeys Don't Write Shakespeare=Keen networked simian sets typography-- Morgan Lewismrl@efn.orgmlewis@cs.uoregon.edu === Subject > Typing Monkeys Don't Write Shakespeare >>A rebuttal:>>Typing Monkeys Don't Write Shakespeare>=>Keen networked simian sets typographyStinky Morgan ponders the kewpie, Yeats! === Subject === Subject=== Subject=== Subject=== Subject=== Subject= Endeavor to persevere === Subject === Subject=== Subject=== Subject=== Subject=== Subject=== Subject=Over a hundred years ago the ring of algebraic integers became a partof the mathematical lexicon, waiting until now for an intriguingproblem with the ring to be revealed by the use of advanced polynomialfactorization techniques, which work by using non-polynomial factorsof a polynomial.In what follows variables, unless otherwise noted, are in the ring ofalgebraic integers.Lemma 1:Given a factor G(X) of a polynomial P(X), R(X) and C exists such thatG(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 followsas 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 thefactor G(X) then C, from lemma 1, must either be coprime to b or havea 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 notalready coprime to b, C must have a factor in common with b^2 suchthat when b^2 is divided off of P(0), that factor divides off of Cleaving 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) intonon-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 getthat 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, andchoosing r_1 to be one of them, as the indices are arbitrary, I havefrom 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, Ihave for G_3(X) = r_3(X) t + bu, that G_3(0) = 3t + bu, so C_3 = 3t + buand 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 hasa 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^2is divided off of P(X), and as shown divides off of P(0), it MUSTdivide off of C_1, from lemma 2.Therefore, r_1(X) t should have a factor that is b, and given that bis 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 whereit is, like using b=sqrt(2), and t=1, there are also cases where youare pushed out of the ring of algebraic integers, which proves aproblem with the ring.James Harris === Subject Revision 1.Over a hundred years ago the ring of algebraic integers became a partof the mathematical lexicon, waiting until now for an intriguingproblem with the ring to be revealed by the use of advanced polynomialfactorization techniques, which work by using non-polynomial factorsof a polynomial.In what follows variables, unless otherwise noted, are in the ring ofalgebraic integers.Lemma 1:Given a factor G(X) of a polynomial P(X), R(X) and C exist such thatG(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 followsas 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 thefactor G(X) then C, from lemma 1, must either be coprime to b or havea 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 notalready coprime to b, C must have a factor in common with b^2 suchthat when b^2 is divided off of P(X), that factor divides off of Cleaving 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) intonon-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 getthat 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, andchoosing r_1 to be one of them, as the indices are arbitrary, I havefrom 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, Ihave for G_3(X) = r_3(X) t + bu, that G_3(0) = 3t + bu, so C_3 = 3t + buand 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 hasa 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^2is divided off of P(X), and as shown divides off of P(0), it MUSTdivide off of C_1, from lemma 2.Therefore, r_1(X) t should have a factor that is b, and given that bis 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 whereit is, like using b=sqrt(2), and t=1, there are also cases where youare pushed out of the ring of algebraic integers, which proves aproblem with the ring.James Harris === Subject >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'vealready offered you a corrected version of this which is complete andaccurate. Is your pride preventing you from adopting it? You do nothave 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 peoplethink 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 Cdivides P(0) in A since you talk about it in your proof and you useit 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 eithercoprime to b, or else it is not coprime to b. The second mysterious clause, which divdes off when b^2 is dividedoff 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). Thenwrite G(X)=R(X)+G(0); (R(X) is just G(X)-G(0)). Since we are assumingthat every coefficient of P(X) is a multiple of b^2 in A, we can writeP(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 Cseems to suggest you have such a division in mind, as do yourattempts 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 meandividing off the entire linear term R(X)+C; so you must be thinkingthat R(X) is also a multiple of the common factor of C and b^2. Haveyou 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 overthe algebraic integers. That a factorization always exists followsfrom results that have been discussed in the newsgroup and that appearin the literature; but the results do not guarantee that the constantterms can be forced to be equal to bu in each case. You need to provethat such a factorization always exists; it is not immediate that thisis so. Assuming you can prove such a factorization exists with those givenproperties, still keeping b and u as fixed, now we think of talso as a parameter and think of the coefficients r_1, r_2, r_3 asfunctions of X; they are really functions of X, b, and u, but sinceb and u are being kept fixed for this, they will only depend on how Xvaries. 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 becomes3tb^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 acommon factor with b^2. You have that C_1 is not coprime to b, so then you conclude that ithas 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 ofLemma 2 which is not actually proven as far as I can see. That is, youare trying to claim that the equalityP(X) = G_1(X)*G_2(X)*G_3(X),G_1(X) = r_1(X)*t+ubimplies that r_1(X) must be a multiple of b. That does not follow fromyour work. This is where everything really goes to hell. You areassuming that r_3(X) will be coprime to to b because r_3(0) is coprimeto b; this is not necessarily the case. You are also assuming thatr_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 multipleof b. None of these three things follow, none of these three thingsare justified. The extension of Lemma 2 that you are trying toapply, namely Say G(X) is a factor of P(X) in A^A; write P(X) = G(X)*H(X). Thenwrite G(X)=R(X)+G(0); (R(X) is just G(X)-G(0)). Since we are assumingthat every coefficient of P(X) is a multiple of b^2 in A, we can writeP(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 overand over again, it can be shown to be FALSE, since the conclusion youderive from it can explicitly be shown to be false, all your confusionnotwithstanding.>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 yourlemma.In short, you are claiming, again, that you can have an equality ofalgebraic integersa*b = r*sin which r is not a unit, and a and b are both coprime to r in thering of all algebraic integers. You have failed to exhibit any suchexamples; all your attempts have been explicitly disproven, and yourclaim contradicts well known results about the algebraic integers,established by Dedekind among others (you know, the guy you've saidyou would trust blindly, contradicting your claims about what shouldand should not be done). === Subject === Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject === Subject === Subject=== Subject==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 === Subject === Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject === Subject === Subject=== Subject==Arturo Magidinmagidin@math.berkeley.edu=== Subject>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 inyour 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 ringmeans, 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 unclear2. The algebraic integers do not form a ring3. The ring of the algebraic integers lacks some property that youneed for your proof of FLT.4. The ring of algebraic integers lacks some property that is usedin 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 notan algebraic integer.6. anything else?-- Wim Benthem === Subject [.philosophical and historical claptrap removed.]>In what follows variables, unless otherwise noted, are in the ring of>algebraic integers.Nonsense as written. You meanIn what follows, the variables RANGE OVER the ring of algebraicintegers, unless otherwise stated.Again, variables are not ELEMENTS of the ring of algebraicintegers. To say that a variable is in the ring of algebraicintegers 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 youapparently 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 apolynomial in A[x], and let G(X) be a factor of P(X) in A^A, the ringof all functions of algebraic integer valued functions of an algebraicinteger variable. Then G(X) can be written as G(X)=R(X)+G(0), withR(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 firstclause. For the second, note that P(X) = G(X)*S(X) for some S(X) inA^A; this equality holds in A^A, hence holds for each value of X. Inparticular, 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. QEDAnd, 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 letP(X_1,...,X_n) be a polynomial with coefficients in A and in nvariables. Let G(X_1,...,X_n) be an element of A^{A^n}, the ring ofall algebraic integer valued functions of n algebraic integervariables, such that G(X_1,...,X_n) divides P(X_1,...,X_n) inA^{A^n}. Then G(0,a_2,...,a_n) is a factor of P(0,a_2,...,a_n) in Afor 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 toP(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 andb, either C is coprime to b or else it is not coprime to b. This isjust the excluded middle, a logical tautology, and everything else isnothing 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 bewritten 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 discussionon 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, givesG_1 is clearly a function of 4 variables, X, t, b, and u; having fixedvalues for b and u, and leaving t as a polynomial variable, we canthink of G_1 as a function of 1 algebraic integer valued variable,with values in A[t], the ring of polynomials with algebraic integercoefficients in the polynomial variable t. This is ALREADY differentfrom the setting on Lemma 1, and even on the extension Lemma1.5. This equality is apparently taking place in (A[x])^{A} (afterfixing 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 areconstant polynomials, so they can be considered elements of A. Thedifficulty in the wrong domain above gets magically solved for thisinstance, 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_1is 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; sosuddenly, we go from t being a polynomial variable to it being aparameter/given value. This is already a source of much confusion, solet's try to clear it up.The definitions of r_1, r_2, r_3 (assuming James could prove that theyactually exist, see below) are made by taking X, b, and u asparameters, and t as a polynomial variable. Then G_1, which is afunction of 4 variables (the parameters X, b, and u; and thepolynomial variable t) is re-interpreted as a family of functions onthe variable X, with parameters t, u, and b. It is in thisinterpretation that Lemma 1 is applied, fixing values of t, u, and bfor 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 avariable, but a parameter. Otherwise, talking about an algebraicinteger being coprime to t makes no sense. Which is why it must bethat 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 coprimeLemma 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. Youmust explain why. The conclusion from Lemma 2 is that EITHERC_1 must be coprime to b, or else it must have a factor in common withb^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 pickfunctions r_1(X), r_2(X), and r_3(X) (which apparently depend only onX, and not on u, b, and t) with the properties that the above is anequality in the ring of function A[X,t,u,b]. It's existence would haveto be first of all established, which you have not done. It willfollow from work of Cohn, the theorem that David McKinnon and Ireproved independently (and which YOU DENY is true; how do you getit)? To find the functions, note that for each specific choice of X,u, and b, the resulting polynomialP(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 integercoefficientsP(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 youcan prove that the choice s_1=s_2=s_3=b*u can always be made, then onecan choose ANY assignments of r_1(X), r_2(X), r_3(X) into thecoefficients r_1, r_2, r_3. Invoking the axiom of choice, you obtainfunctions r_1(X,b,u), r_2(X,b,u), r_3(X,b,u) of three algebraicinteger valued functions with algebraic integer values which willyield the equality.exists (this factorization is in A[t]) with r_1, r_2, and r_3algebraic integers.James is really working with a modification of his old polynomialQ(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 multiplyingthrough 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 foreach value of b_0, u_0, and X_0 that give an equalityP(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, andthus yield factors in A^A.This is slight modification of James's early attempts, at which heclaimed 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 beendisproven many times, recently at the end of Nora's post:http://groups.google.com/groups?selm= 36024859.0309071429.4e09bc20%40posting.google.comso 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 severalinstances. In particular, from previous observations, since we aredealing withP(t) = (v^3+1)t^3 - 3vu^2*t + u^3*b^3, we would have that -ub/r is aroot of P(t), so that-(v^3+1)(u^3b^3/r^3) + 3vu^2(ub/r) + u^3b^3 = 0Multiplying through by r^3 we have-(v^3+1)(u^3b^3) + 3vu^3b*r^2 + u^3b^3*r^3 = 0Factorign 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 concludethat r is indeed an algebraic integer. However, if b^2 does NOT divide3v, then letting s=gcd(b^2,3vr) we would have that r^3 is the root off(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 turnsout to be irreducible, then the factorization as above does not exists.I ->believe<- that in all applications James has in mind, the valuesof v, b, and u are so contrived that it is possible this does notoccur, but there is certainly no guarantee ->right now<- that thefactorization is possible. === Subject === Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject== === Subject === SubjectWhy 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 === Subject === Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject === Subject === Subject=== Subject==Arturo Magidinmagidin@math.berkeley.edu=== Subject> 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! === Subject > 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 withb^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 cannotbe used to prove anything. Better revise and try again, eh? Nora B.[remainder deleted pending correction] === Subject > > 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 === Subject >> 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 forspecious objections. -- End Insert --Doesn't that mean mean I should expect a retraction on claiming myobjection was specious? Or will, you as usual, simply dismiss thiserror of yours as a simple mistake, and ignore all the insult you haveheaped and attacks you have made based on your belief that this lemmawas not content-free? === Subject === Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject= === Subject === Subject=== Subject=== Subject=== Subject=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 === Subject === Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject === Subject === Subject=== Subject==Arturo Magidinmagidin@math.berkeley.edu=== Subject...> 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 === Subject > 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 indeedthe 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 thatthe quotients are coprime to b, and C2 coprime to b. That is alltrivial. But nothing states that (R2(X)t + C2) is coprime to b forall 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 bedivisible by b. But as the condition is not yet met...-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject ... > 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 whileR1(X)t + C1 is coprime to b.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject > 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) iscoprime to b either.What is coprime to b is P(0)/b^2, and since the variables arealgebraic integers, it is an algebraic integer, so the usualdefinition applies.Similarly, C is an algebraic integer, so as it is a factor of P(0), ifit has a factor in common with b^2, then that factor gets divided offwhen 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 === Subject > 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 tob 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 arbitraryalgebraic integer. Then G(0) is either coprime to b or else it is notcoprime to b. [.snip.] === Subject === Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject== === Subject === Subject=== Subject=== Subject [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 === Subject === Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject== === Subject === Subject=== Subject=== SubjectArturo Magidinmagidin@math.berkeley.edu=== Subject> 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. === Subject > 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 anyfindings of minor errors, including grammatical ones or typo's.It IS fun having a correct math argument, especially one that provessomething wacky like the problem with the ring of algebraic integers.I've been working rather hard in answering a LOT of posts to show youthat some people you might have trusted have betrayed your trust, andare attempting to teach you bogus math. They should not be working tofeed you all false information, but it's their choice.I'll return in a bit to see if anyone else found any other errors, andpost an update in this thread, underneath the original.If it's just chock full of errors, I'll make a REVISED thread.James Harris === Subject > 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 treatingP(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 freevariable, 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 + 2Which you factor, in effect, by creating a new polynomialin a new free variable y. You know this, you just stateit 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> factorizationIt acts like this polynomial in the sense that you usea factorization theorem which relies on the presence ofthis 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 andm, 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 yournon-polynomial factorization? - Randy === Subject > 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) + 1where 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 apolynomial.> 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'sfactored.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 youfrom 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 topush on readers.The gist of his position has to be that the factorization changes thepolynomial, 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?> - RandyThat *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 frombeing P(X), though several posters have managed to get away withclaiming my work is false based on the notion that how the polynomialis factored changes the polynomial.I've switched variable names to make it harder for them, as for awhile 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 thegrouping which is for the non-polynomial factorization.My guess is that they relied on people being used to P(x), so Ichanged variable names.James Harris === Subject > 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 meup. > 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 thepolynomial was P(x), and rely on people being used to seeing P(x), soI switched variable names to take that way.The poster Nora Baron keeps trying to go back though, which confirmsmy suspicions that I needed the new variable names.After all it IS algebra, so switching the variable names shouldn't besignificant.That is, I'm emphasizing to readers that this poster is puttingemphasis on something that *should* be immaterial.The poster and others would claim that my P(m) was really P(x) becauseof 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 claimthey really were polynomial factors, claiming I had a differentpolynomial from what I said was the polynomial.For instance, with my current P(X) this poster would try to claim itwas really P(X,t), based on the *factorization*, so I've startedsaying things like 6 is still 6 despite how it's factored.The poster has refused to back down and follow the math, and is stillposting apparently in an attempt to convince at least some of you thatbogus 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 aboutmy 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 apolynomial. And by we I guess you mean yourself and other posterslike Arturo Magidin, who once actually *said* my factorization wasmaking a new polynomial which I referenced in the post that startedthis 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 thatI'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, andthat enough people believed in you that some would make the effort tocome 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 constant term here does NOT correspond to the constant term of> 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.> > === Subject === Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject== === Subject === Subject> It's not denial. I'm just very selective about> what I accept as reality.> --- Calvin (Calvin and Hobbes)> > === Subject === Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject=== Subject== === Subject === Subject> 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 theargument.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 sofar.> 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 claimabout 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 === Subject >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 Centirely different thread!!!You do realize that's not exactly rational, right?Remember the big picture which is that mathematicians pulled in anidea without fully working it out over a *hundred* years ago, so youhave this wacky problem with algebraic integers. I've simply pointedthat out, and if mathematicians aren't themselves wacky, they can justwork to figure out the full ramifications of the situation, whichcould be an incredible opportunity for some of you.Instead, I see posters working to destroy math society by fightingmathematics as if you were fighting me, a person, as if you couldactually 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 couldswitch to something less complicated and used P(x) = 11^2 + 11x + 2where again you have a special grouping simply to allow thenon-polynomial factorization, as you can see that P(x) is also P(x) = 11x + 123so I've just used the 11's before as placemarkers, so Ican *act* like it's the polynomial y^2 + xy + 2, to get thefactorization 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 toconvince that the factorization, with more complicated expressions,changed the polynomial in some way. Apparently plenty of sci.mathreaders and especially alt.math.undergrad readers, who may be moresusceptible, were convinced by them.It's actually rather amazing that a few posters could convince so manypeople 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) intonon-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'dfigure, hey, if those people say it's no longer P(X) because of thefactorization, you'd just go along and trust the majority.But you see, mathematics isn't a democracy.James Harris === Subject >>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 figuremaybe you should die instead. How about that, eh? Wouldn't that be abetter twist?You refuse to follow the math, so the great Powers that controlreality 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, andwill not penalize me as They allowed the others like Galois and Abelto be penalized.They will kill you instead.James Harris speaking on Weird factorization, geniusX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h81FbOc24006; === Subject > 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; === Subject >>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. === Subject >>x^2*y'' - x*(x+2)*y' + (x+2)*y=x^3Reduction of order will give you the complete solution.>>and>>(1-x^2)*y'' - 4*x*y' - (1+x^2)*y(x) = xRobert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h81GsWU29109; === Subject >>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; === Subject >dont 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; === Subject >
>It's
possible that I misunderstood your question. If I get it
right,>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; === Subject What is the time difference from Athens, Greece to California === Subject >What is the time difference from Athens, Greece to CaliforniaIf 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 youra 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 samein both places though.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject >>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 toAthens (with a connection to New York) on Sunday 3/30/03 ... precisely for this reason; I understood when they called my name :-) baloglouAToswego.eduX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h81NHAW21763; === Subject >Does anyone have a good reference for mathematical explanation of the>degrees of freedom for the sum of squares equation ?>>thanksIf 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 probablyfind something at http://www.statcan.ca.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h81FbNj23995; === Subject >> dont know it.>>> Can anyone help me ?>>Yes, it's 0 (sometimes). :-)>>Nijmegen, Netherlands>I could be 10 also! === Subject > dont 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, andthose who don't. === Subject >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; === Subject 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; === Subject What does F(a,b;c;d) mean, where a,b,c,d are numbers?Is this function in Mathematica? === Subject > 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; === Subject 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 <
343924.E388B5@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:  > 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; > (* 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; === Subject > 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; === Subject >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; === Subject 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/00090>23.pdf) There are additional Quantum State Diffusion (QSD) terms which are>non-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; === Subject can anyone help me with these pleeeeease!!! :1. y=4x|X|-32. y=|2X-1|3. y=3x|X+4|-24. y=-0.5|X+4|+3i 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 h842SVt15356X-Received: (from approve@localhost) by home.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.4 nullclient) id h842SUb02637; === Subject >-->>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 = 3First with 2 = 63first with 3 = 173first with 4 = 1414first with 5 = 17313first with 6 = 53937first with 7 = 138923first with 8 = 953937first with 9 = 3082207first with 10 = 31622764X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h83JBr014861; === Subject 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 coefficientssuch 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 dividesgcd(n^{17}+9,(n+1)^{17}+9), that is, p divides n^{17}+9=q(n) andp also divides (n+1)^{17}+9=r(n). Then p also dividess(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://primes.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 h84Bnst20132X-Received: (from approve@localhost) by home.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.4 nullclient) id h84Bnr103250; === Subject >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; === Subject [ ... ]>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 === Subject > [ ... ]> >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 ofpolynomials with algebraic integer coefficients.Arturo Magidin, sans .sigX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h84IYjd17832; === Subject >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 ofBefore trying to prove something you better make sure it's actually true...2^3+1=9=3*32^5+1=33=3*112^6+1=65=5*13...GalX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h84JQ3622518; === Subject >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 havex = 2m, y=2n with m and n integers, so the norm is12 = 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 .sigX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h83GY8m01926; === Subject 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 pleaseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee eeeeeeeeeeeeee eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee eeeeeeeeeeeeee eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeethanck f r o m-----b e nX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h83Kpeu24034; === Subject 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; === Subject 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; === Subject > 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 ITIn 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 agreeableX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h8414VP10129; === Subject 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; === Subject >
>[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.-MeatSmokeRArchie LeachX-Received: (from
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Subject
> 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
getP=(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 CoppersmithX-Received: (from
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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_ujust get
rid of all the underscores.I will compile them with as little
change as possible for my machine and post the times.Jason
RodgersX-Received: (from approve@localhost) by
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Subject
On my machine. A
1GHz Mac the time to run the 1,000,000 problem is .06
secondsto run the 20,000,000 problem was 2.55
seconds.X-Received: (from approve@localhost) by
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Subject
The few so far
that I have are as followsAll 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, 1.1920000000, 2.52 Jason Rodgers size time
1000000, 0.04 10000000 , 1.1120000000, 2.42 Modification of
Christian Bau's by C Bond size time 1000000, 0.44 10000000 ,
11.6820000000, 31.21X-Received: (from approve@localhost) by
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Subject
 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 knowany other solvable
sextic, pls post it. And not those that factorover sqrt(D) for
some D. Those are too easy.--Tito
===
Subject
> 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, accordingto 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.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2X-Received: (from
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Subject
Here are the times I haveJames 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 ,
0.54X-Received: (from approve@localhost) by
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Subject
>#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?
===
Subject
>>#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,
theconcept of a tangent line appears to have been simple enough
for Euclid sothat it is present in definition #8 (angle between
two curves) withoutbeing explicitly mentioned; and then I
explained in my second posting(replying to Nat Silver) how my
tangential interpretation simplifies thematter by using one
point instead of two (in both #4 and #7) *and* by explaining
the appeal to surface's straight lines (rather than
merelypoints) in #7.Of course all this is just a possibility,
I may well have cut myself with Occam's razor :-)
baloglouAToswego.edu
===
Subject
>>>#4: A straight line is a line
which lies evenly with the points onitself.>>#7: A plane
surface is a surface which lies evenly with the straightlines>
>>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 =
liesevenly>>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 thanline)> 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.
Ithink it is important to remember that Euclid was writing a
textbook andnever intended this to be the seminal work on
Geometry for close to 3000years. Does Euclid have gaps and
ommissions and tacit assumptions. Youbet!!! BUT, I challenge
anyone to pick up ANY high-school geometry textbookand not
find gaps and ommissions and tacit
assumptions.-TralfazX-Received: (from approve@localhost) by
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Subject
>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?
===
Subject
>>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.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2X-Received: (from
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Subject
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 = xa_n = C a_(n-1) + yThen: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);>>}X-Received: (from approve@localhost)
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Subject
> 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!piggybacki 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
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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 theratio of two
countably infinite numbers. Any result is possible.phil
===
Subject
>
>>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 integersthat do not
contain a 6 (or, mor generally, any specified digitin 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
theintegers 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 CohenX-Received: (from approve@localhost) by
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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 simpleArithmetic 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
===
Subject
>>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 maynot 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
twonumbers.>> Quadratic mean: This is the formula your teacher
gave you.>> Relative size: This means to check which is bigger.
Just try a fewcalculations, and try to see which type of mean
is usually bigger. Youreducated guess is called a conjecture.
In a more advanced course, you maylearn ways to prove your
conjecture.>> -Jonathan>X-Received: (from approve@localhost)
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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
DoughtyX-Received: (from approve@localhost) by
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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.westminster.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?
===
Subject
>> 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, thenumbers with a 6 are greater than those without a
6.But this is quite misleading, as after a while thatrelation
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.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2X-Received:
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>
>|Essentially, the question ask for a
proof that for any (undirected)>|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!>
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Steve:Consider
(-3,-3)Gary TupperTerrace 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
===
Subject
>
Steve:> Consider (-3,-3)> Gary Tupper> Terrace BC>
>|b| - |a| <= | |b| - |a| | <= | b - a |>>-- >Stephen J.
Herschkorn herschko@rutcor.rutgers.edua=-3, b=-3 then|a| = 3|b|
= 3b - a = -3 - (-3) = -3 + 3 = 0|b| - |a| <= ||b| - |a|| <= |
b - a | so all of the above are = 0 in this caseand 0<=0<=0 is
a true statement...X-Received: (from approve@localhost) by
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>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=displa Perhaps
quicken his fall in the sense of preventing him from
changing his mind once he had the machine out the
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I'm need for
mythesis about problemes :1-winner take all2-taravel sales
man3-knap sack4-assingment problemand their
history.thanks;X-Received: (from approve@localhost) by
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>
>I was
using Mathcad 6+ the other day, and it crashed with an invalid
page>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 >
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>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
thanksHere is a way to count labelled digraphs in which every
node belongs to a directed cycle (clearly they have neither
sourcesnor 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 multipleedges), _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 verifiableeasily 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 waywith 1-node
components only: K:= {[.]}. Besides (curiously),u_n(D) is the
number of strong tournaments (A054946).Valery
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Subject
>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>}.
===
Subject
>>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.MarcP.S.: I do find the OP notation
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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?
===
Subject
> 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 thevariable 'x'
within the procedure. If you don't need to know it, thensimply
pass `sin` rather than sin(x).
===
Subject
> 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 argumentto your proc instead
of the expression sin(x). Without seeing actualcode, it's
difficult to say more than that.-- Dave SeamanJudge Yohn's
mistakes revealed in Mumia Abu-Jamal
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Just to show the
values on another machineUsing 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 ofcorresponding 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
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>> |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
===
Subject
>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).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 positivecharacteristic. Let K be any field of
characteristic 0, different fromQ. Let {a_1,...,a_n,...} be a
basis for K over Q, and takeZ[a_1,...,a_n,...]. This satisfies
the condition that its field offractions equals K. Two bases
will in general yield different rings; Ithink they only yield
the same ring if the change of basis matrix hasinteger
coefficients. So by judiciously choosing your bases, youshould
be able to get intersections which are too small (if not
equalto 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 Rationalsfor
Q[i]. I will look for Gaussian numbers and see if it is
morepopular. Possibly not many people care about Q[i] at all.
> Arturo Magidin, sans .sigJ also senza .sigX-Received: (from
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>>[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
ofx^n + a_{n-1}*x^{n-1} + ... + a_1*x + a_0,then sqrt(y) is a
root ofx^{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 .sigX-Received: from
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>
>|Essentially, the question ask for a
proof that for any (undirected)>|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!>
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> 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
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>> 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
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Subject
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
ASAPthanx
===
Subject
> 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> thanxThe 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
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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, linkAgain, 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.htmlRoddyX-Received: (from approve@localhost) by
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Subject
this is a test.
JSHX-Received: (from approve@localhost) by
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>
>Archive-name:
num-analysis/faq/part1>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>
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Subject=>>q20. NA FAQ: Introduction>>Where to find
this FAQ:>>On the web:>http://
www.indra.com/~sullivan/q10.html Mathcom>>This FAQ is
usually available from MIT's rtfm and its mirrors:>ftp://rtfm.mit.edu/pub/usenet/news.answers/
num-analysis/faq/part1 MIT's rtfm>>If not, it is at:>ftp://
ftp.mathcom.com/mathcom/nafaq 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.>
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Subject>>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.>
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Subject>>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>>
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Subject>> * 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>>
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Subject==>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.>
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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.>
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Subject>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.>>
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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.>>
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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>>
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Subject>>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/netlib/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.ac.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/ 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.>
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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.>
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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>
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Mathematical and Statistical Libraries>>NCAR's libraries
contain some overlap with netlib.>>http://
http.ucar.edu/SOFTLIB/mathlib.html
NCAR>
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Parallel Archive>http
://www.hensa.ac.uk/parallel/environments/pcn/ Hensa>Note:
this web server can be very
slow!>
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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>>
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Octave>>http://
bevo.che.wisc.edu/octave.html Octave>ftp://
www.che.wisc.edu/pub/octave 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.>>
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RLaB>>http://
www.eskimo.com/~ians/rlab.html RLaB>ftp://
csi.jpl.nasa.gov/pub/matlab/RLaB RLaB via ftp>ftp://
evans.ee.adfa.oz.au/pub/RLaB 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.>>
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Scilab>>http:/
/zenon.inria.fr/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.>>
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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
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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.>>
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Subject>>q120.7. Euler>>http://
www.ku-eichstaett.de/MGF/euler.html Euler>ftp://ftp.k.-
eichstaett.de/pub/math 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.>>
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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.>>
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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.>>
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Subject>>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>>
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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>
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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.>
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Subject>>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
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Subject
> 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
asurprise. - Randy
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 thecharacterization
of surjectivity requires the Axiom of Choice whilethat of the
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 setis not
finite, and there exists a bijection from it to 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 SeamanJudge Yohn's mistakes revealed
in Mumia Abu-Jamal
ruling. 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 ofmeasure 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
edition) there's an ad by DHL that lists 400 zip codes. Are
on sci.math dual with James Harris on various mathematical(and
other) topics. Currently, much energy is being expended on
Lemma 1 andLemma 2. Here is a suggestion for a plan of
activity. I am well aware thatthis 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. Onlystandard math
terminology should be used. (But further terms could
bedefined, of course.)Step 2:Produce an example of the
application of Lemma 1 which is acceptable to bothsides.Step
3:Produce a proof for Lemma 1 which is acceptable to both
sides. Only standardmath terminology should be used. (But
further terms could be defined, ofcourse.)Step 4:Produce a
wording for Lemma 2 which is acceptable to both sides.
Onlystandard math terminology should be used. (But further
terms could bedefined, of course.)Step 5:Produce an example of
the application of Lemma 2 which is acceptable to
bothsides.Step 6:Produce a proof for Lemma 2 which is
acceptable to both sides. Only standardmath terminology
should be used. (But further terms could be defined,
ofcourse.)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 Lemma2.All definitions, examples and
proofs should be too verbose, rather than tooconcise.-- Clive
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 thecase. And who would want
that?However, I guess that it would strand on the very first
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 thecase. And who
would want that?However, I guess that it would strand on the
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
isto use uncountable to mean |A| ~<= aleph_0, and to
usetransfinite for |A| >= aleph_0. Dedekind-finite
infinitecardinals are uncountable but not transfinite.-- This
address is for information only. I do not claim that these
viewsare those of the Statistics Department or of Purdue
University.Herman Rubin, Department of Statistics, Purdue
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
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
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 punishthe evil liars! Well I'm
convinced.--Dave Taylor'Ray, when someone asks you if you're a
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
seenbecause given any solution x for B, (x+A),(x+2A), .... are
alsosolutions.We can however identify a line of solutions.The
easiest member of the solution set is the vector perpendicular
toC 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 asolution, so the complete set of solutions is:
some FEM work which required numerical integration of
atriangle. I got the technique for performing this integration
from JianmingJin's book on FEM. Using his formalism, you take
the (x,y,z) coordinate ofeach of the three vertices and plug
them into a formula which multipliesthese coordinates by
special factors (abscissae?) to obtain the actualpoints of
integration within the triangle. Each point is given an
associatedweight.I need to obtain a similar table for the
arbitrary three-dimensionaltetrahedron. I have seen some
tables out there (given by Ronald Cools) onthe net, but I do
not know how to use them, and I've seen nothing like
Jin'stable for the triangle applied to solid elements
anywhere. Could someoneplease help me with the problem of
obtaining a robust, understandable way tonumerically
Monkeys Don't Write Shakespeare A rebuttal:Typing Monkeys
Don't Write Shakespeare=Keen networked simian sets
typography-- Morgan
Don't Write Shakespeare >>A rebuttal:>>Typing Monkeys Don't
Write Shakespeare>=>Keen networked simian sets
typographyStinky Morgan ponders the kewpie,
ago the ring of algebraic integers became a partof the
mathematical lexicon, waiting until now for an
intriguingproblem with the ring to be revealed by the use of
advanced polynomialfactorization techniques, which work by
using non-polynomial factorsof a polynomial.In what follows
variables, unless otherwise noted, are in the ring ofalgebraic
integers.Lemma 1:Given a factor G(X) of a polynomial P(X), R(X)
and C exists such thatG(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 followsas 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 thefactor G(X) then C, from lemma 1,
must either be coprime to b or havea 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 notalready coprime to b, C must have a factor in
common with b^2 suchthat when b^2 is divided off of P(0), that
factor divides off of Cleaving 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) intonon-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 getthat 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, andchoosing r_1 to be one of them, as
the indices are arbitrary, I havefrom 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, Ihave
for G_3(X) = r_3(X) t + bu, that G_3(0) = 3t + bu, so C_3 = 3t
+ buand 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 hasa 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^2is divided off of P(X), and as
shown divides off of P(0), it MUSTdivide off of C_1, from
lemma 2.Therefore, r_1(X) t should have a factor that is b,
and given that bis 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 whereit is, like using b=sqrt(2), and t=1,
there are also cases where youare pushed out of the ring of
algebraic integers, which proves aproblem with the ring.James
algebraic integers became a partof the mathematical lexicon,
waiting until now for an intriguingproblem with the ring to be
revealed by the use of advanced polynomialfactorization
techniques, which work by using non-polynomial factorsof a
polynomial.In what follows variables, unless otherwise noted,
are in the ring ofalgebraic integers.Lemma 1:Given a factor
G(X) of a polynomial P(X), R(X) and C exist such thatG(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 followsas 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 thefactor G(X) then C, from lemma 1, must either be
coprime to b or havea 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 notalready
coprime to b, C must have a factor in common with b^2 suchthat
when b^2 is divided off of P(X), that factor divides off of
Cleaving 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)
intonon-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
getthat 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, andchoosing r_1 to be one of them, as the
indices are arbitrary, I havefrom 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, Ihave for
G_3(X) = r_3(X) t + bu, that G_3(0) = 3t + bu, so C_3 = 3t +
buand 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 hasa 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^2is divided off of P(X), and
as shown divides off of P(0), it MUSTdivide off of C_1, from
lemma 2.Therefore, r_1(X) t should have a factor that is b,
and given that bis 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 whereit is, like using b=sqrt(2), and t=1,
there are also cases where youare pushed out of the ring of
algebraic integers, which proves aproblem with the ring.James
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'vealready offered you a
corrected version of this which is complete andaccurate. Is
your pride preventing you from adopting it? You do nothave 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 peoplethink
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 Cdivides P(0) in A since you talk about it in your
proof and you useit 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 eithercoprime to b, or else it is
not coprime to b. The second mysterious clause, which divdes
off when b^2 is dividedoff 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). Thenwrite G(X)=R(X)+G(0); (R(X) is just
G(X)-G(0)). Since we are assumingthat every coefficient of
P(X) is a multiple of b^2 in A, we can writeP(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 Cseems to suggest you
have such a division in mind, as do yourattempts 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 meandividing
off the entire linear term R(X)+C; so you must be thinkingthat
R(X) is also a multiple of the common factor of C and b^2.
Haveyou 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 overthe algebraic
integers. That a factorization always exists followsfrom
results that have been discussed in the newsgroup and that
appearin the literature; but the results do not guarantee that
the constantterms can be forced to be equal to bu in each case.
You need to provethat such a factorization always exists; it is
not immediate that thisis so. Assuming you can prove such a
factorization exists with those givenproperties, still keeping
b and u as fixed, now we think of talso as a parameter and
think of the coefficients r_1, r_2, r_3 asfunctions of X; they
are really functions of X, b, and u, but sinceb and u are
being kept fixed for this, they will only depend on how
Xvaries. 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
becomes3tb^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 acommon factor with b^2. You
have that C_1 is not coprime to b, so then you conclude that
ithas 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 ofLemma 2 which is not actually proven as far as
I can see. That is, youare trying to claim that the
equalityP(X) = G_1(X)*G_2(X)*G_3(X),G_1(X) =
r_1(X)*t+ubimplies that r_1(X) must be a multiple of b. That
does not follow fromyour work. This is where everything really
goes to hell. You areassuming that r_3(X) will be coprime to to
b because r_3(0) is coprimeto b; this is not necessarily the
case. You are also assuming thatr_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 multipleof b. None of
these three things follow, none of these three thingsare
justified. The extension of Lemma 2 that you are trying
toapply, namely Say G(X) is a factor of P(X) in A^A; write
P(X) = G(X)*H(X). Thenwrite G(X)=R(X)+G(0); (R(X) is just
G(X)-G(0)). Since we are assumingthat every coefficient of
P(X) is a multiple of b^2 in A, we can writeP(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 overand over again,
it can be shown to be FALSE, since the conclusion youderive
from it can explicitly be shown to be false, all your
confusionnotwithstanding.>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 yourlemma.In short, you are claiming, again, that you
can have an equality ofalgebraic integersa*b = r*sin which r is
not a unit, and a and b are both coprime to r in thering of all
algebraic integers. You have failed to exhibit any
suchexamples; all your attempts have been explicitly
disproven, and yourclaim contradicts well known results about
the algebraic integers,established by Dedekind among others
(you know, the guy you've saidyou would trust blindly,
contradicting your claims about what shouldand should not be
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
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 inyour 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 ringmeans, 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 unclear2. The algebraic integers
do not form a ring3. The ring of the algebraic integers lacks
some property that youneed for your proof of FLT.4. The ring
of algebraic integers lacks some property that is usedin 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 notan
[.philosophical and historical claptrap removed.]>In what
follows variables, unless otherwise noted, are in the ring
of>algebraic integers.Nonsense as written. You meanIn what
follows, the variables RANGE OVER the ring of
algebraicintegers, unless otherwise stated.Again, variables
are not ELEMENTS of the ring of algebraicintegers. To say that
a variable is in the ring of algebraicintegers 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 youapparently
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 apolynomial in A[x], and let G(X) be a
factor of P(X) in A^A, the ringof all functions of algebraic
integer valued functions of an algebraicinteger variable. Then
G(X) can be written as G(X)=R(X)+G(0), withR(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 firstclause. For the second,
note that P(X) = G(X)*S(X) for some S(X) inA^A; this equality
holds in A^A, hence holds for each value of X. Inparticular,
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. QEDAnd, 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 letP(X_1,...,X_n) be a
polynomial with coefficients in A and in nvariables. Let
G(X_1,...,X_n) be an element of A^{A^n}, the ring ofall
algebraic integer valued functions of n algebraic
integervariables, such that G(X_1,...,X_n) divides
P(X_1,...,X_n) inA^{A^n}. Then G(0,a_2,...,a_n) is a factor of
P(0,a_2,...,a_n) in Afor 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 toP(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 andb,
either C is coprime to b or else it is not coprime to b. This
isjust the excluded middle, a logical tautology, and
everything else isnothing 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 bewritten 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 discussionon 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, givesG_1 is clearly a
function of 4 variables, X, t, b, and u; having fixedvalues
for b and u, and leaving t as a polynomial variable, we
canthink of G_1 as a function of 1 algebraic integer valued
variable,with values in A[t], the ring of polynomials with
algebraic integercoefficients in the polynomial variable t.
This is ALREADY differentfrom the setting on Lemma 1, and
even on the extension Lemma1.5. This equality is apparently
taking place in (A[x])^{A} (afterfixing 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 areconstant polynomials, so they can be considered
elements of A. Thedifficulty in the wrong domain above gets
magically solved for thisinstance, 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_1is 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; sosuddenly, we go from t being a polynomial
variable to it being aparameter/given value. This is already a
source of much confusion, solet's try to clear it up.The
definitions of r_1, r_2, r_3 (assuming James could prove that
theyactually exist, see below) are made by taking X, b, and u
asparameters, and t as a polynomial variable. Then G_1, which
is afunction of 4 variables (the parameters X, b, and u; and
thepolynomial variable t) is re-interpreted as a family of
functions onthe variable X, with parameters t, u, and b. It is
in thisinterpretation that Lemma 1 is applied, fixing values of
t, u, and bfor 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 avariable, but a parameter.
Otherwise, talking about an algebraicinteger being coprime to
t makes no sense. Which is why it must bethat 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
coprimeLemma 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. Youmust explain why. The conclusion
from Lemma 2 is that EITHERC_1 must be coprime to b, or else
it must have a factor in common withb^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
pickfunctions r_1(X), r_2(X), and r_3(X) (which apparently
depend only onX, and not on u, b, and t) with the properties
that the above is anequality in the ring of function
A[X,t,u,b]. It's existence would haveto be first of all
established, which you have not done. It willfollow from work
of Cohn, the theorem that David McKinnon and Ireproved
independently (and which YOU DENY is true; how do you getit)?
To find the functions, note that for each specific choice of
X,u, and b, the resulting polynomialP(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
integercoefficientsP(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 youcan prove that the choice s_1=s_2=s_3=b*u can
always be made, then onecan choose ANY assignments of r_1(X),
r_2(X), r_3(X) into thecoefficients r_1, r_2, r_3. Invoking
the axiom of choice, you obtainfunctions r_1(X,b,u),
r_2(X,b,u), r_3(X,b,u) of three algebraicinteger valued
functions with algebraic integer values which willyield the
equality.exists (this factorization is in A[t]) with r_1, r_2,
and r_3algebraic integers.James is really working with a
modification of his old polynomialQ(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 multiplyingthrough 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 foreach value of b_0, u_0, and X_0 that
give an equalityP(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, andthus yield
factors in A^A.This is slight modification of James's early
attempts, at which heclaimed 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
beendisproven many times, recently at the end of Nora's
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
severalinstances. In particular, from previous observations,
since we aredealing withP(t) = (v^3+1)t^3 - 3vu^2*t + u^3*b^3,
we would have that -ub/r is aroot of P(t), so
that-(v^3+1)(u^3b^3/r^3) + 3vu^2(ub/r) + u^3b^3 = 0Multiplying
through by r^3 we have-(v^3+1)(u^3b^3) + 3vu^3b*r^2 +
u^3b^3*r^3 = 0Factorign 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 concludethat r is indeed an algebraic integer.
However, if b^2 does NOT divide3v, then letting s=gcd(b^2,3vr)
we would have that r^3 is the root off(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 turnsout to be irreducible,
then the factorization as above does not exists.I ->believe<-
that in all applications James has in mind, the valuesof v, b,
and u are so contrived that it is possible this does notoccur,
but there is certainly no guarantee ->right now<- that
thefactorization is possible.
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
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)
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 withb^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 cannotbe used to prove anything. Better revise and try
> 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
pending correction]Well that looks like enough for me to do an
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
>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 forspecious objections. -- End Insert
--Doesn't that mean mean I should expect a retraction on
claiming myobjection was specious? Or will, you as usual,
simply dismiss thiserror of yours as a simple mistake, and
ignore all the insult you haveheaped and attacks you have made
based on your belief that this lemmawas not
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
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
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
indeedthe 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 thatthe quotients are coprime to b, and
C2 coprime to b. That is alltrivial. But nothing states that
(R2(X)t + C2) is coprime to b forall 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 bedivisible by b. But as
the condition is not yet met...-- dik t. winter, cwi, kruislaan
413, 1098 sj amsterdam, nederland, +31205924131home: bovenover
215, 1025 jn amsterdam, nederland;
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 whileR1(X)t + C1 is coprime to b.-- dik t. winter, cwi,
kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131home: bovenover 215, 1025 jn amsterdam,
= 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) iscoprime to b either.What is
coprime to b is P(0)/b^2, and since the variables arealgebraic
integers, it is an algebraic integer, so the usualdefinition
applies.Similarly, C is an algebraic integer, so as it is a
factor of P(0), ifit has a factor in common with b^2, then
that factor gets divided offwhen 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
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 tob 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 arbitraryalgebraic integer. Then G(0) is
either coprime to b or else it is notcoprime to b.
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.
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
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 anyfindings of minor errors, including
grammatical ones or typo's.It IS fun having a correct math
argument, especially one that provessomething wacky like the
problem with the ring of algebraic integers.I've been working
rather hard in answering a LOT of posts to show youthat some
people you might have trusted have betrayed your trust, andare
attempting to teach you bogus math. They should not be working
tofeed you all false information, but it's their choice.I'll
return in a bit to see if anyone else found any other errors,
andpost an update in this thread, underneath the original.If
it's just chock full of errors, I'll make a REVISED
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 treatingP(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 freevariable,
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 +
2Which you factor, in effect, by creating a new polynomialin a
new free variable y. You know this, you just stateit 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>
factorizationIt acts like this polynomial in the sense that
you usea factorization theorem which relies on the presence
ofthis 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 andm, 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
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) + 1where 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 apolynomial.> 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'sfactored.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 youfrom 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 topush on readers.The gist of his position has to be
that the factorization changes thepolynomial, 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?> - RandyThat *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 frombeing P(X), though several posters have
managed to get away withclaiming my work is false based on the
notion that how the polynomialis factored changes the
polynomial.I've switched variable names to make it harder for
them, as for awhile 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 thegrouping which
is for the non-polynomial factorization.My guess is that they
relied on people being used to P(x), so Ichanged variable
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 meup. >
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 thepolynomial was P(x), and rely on people being used to
seeing P(x), soI switched variable names to take that way.The
poster Nora Baron keeps trying to go back though, which
confirmsmy suspicions that I needed the new variable
names.After all it IS algebra, so switching the variable names
shouldn't besignificant.That is, I'm emphasizing to readers
that this poster is puttingemphasis on something that *should*
be immaterial.The poster and others would claim that my P(m)
was really P(x) becauseof 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 claimthey really were polynomial factors,
claiming I had a differentpolynomial from what I said was the
polynomial.For instance, with my current P(X) this poster
would try to claim itwas really P(X,t), based on the
*factorization*, so I've startedsaying things like 6 is still
6 despite how it's factored.The poster has refused to back
down and follow the math, and is stillposting apparently in an
attempt to convince at least some of you thatbogus 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 aboutmy 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 apolynomial. And by we I
guess you mean yourself and other posterslike Arturo Magidin,
who once actually *said* my factorization wasmaking a new
polynomial which I referenced in the post that startedthis
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 thatI'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, andthat enough people
believed in you that some would make the effort tocome 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 constant term here does NOT correspond
to the constant term of> 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.>
>
what I accept as reality.> --- Calvin (Calvin and Hobbes)>
>
> 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 theargument.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 sofar.> 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
claimabout 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
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 Centirely different
thread!!!You do realize that's not exactly rational,
right?Remember the big picture which is that mathematicians
pulled in anidea without fully working it out over a *hundred*
years ago, so youhave this wacky problem with algebraic
integers. I've simply pointedthat out, and if mathematicians
aren't themselves wacky, they can justwork to figure out the
full ramifications of the situation, whichcould be an
incredible opportunity for some of you.Instead, I see posters
working to destroy math society by fightingmathematics as if
you were fighting me, a person, as if you couldactually 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
couldswitch to something less complicated and used P(x) = 11^2
+ 11x + 2where again you have a special grouping simply to
allow thenon-polynomial factorization, as you can see that
P(x) is also P(x) = 11x + 123so I've just used the 11's before
as placemarkers, so Ican *act* like it's the polynomial y^2 +
xy + 2, to get thefactorization 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 toconvince that
the factorization, with more complicated expressions,changed
the polynomial in some way. Apparently plenty of
sci.mathreaders and especially alt.math.undergrad readers, who
may be moresusceptible, were convinced by them.It's actually
rather amazing that a few posters could convince so manypeople
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) intonon-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'dfigure, hey, if those people say it's no longer P(X)
because of thefactorization, you'd just go along and trust the
majority.But you see, mathematics isn't a democracy.James
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 figuremaybe you should
die instead. How about that, eh? Wouldn't that be abetter
twist?You refuse to follow the math, so the great Powers that
controlreality 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, andwill not penalize me as They allowed the others
like Galois and Abelto be penalized.They will kill you
instead.James Harris speaking on Weird factorization,
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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
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
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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
will give you the complete solution.>>and>>(1-x^2)*y'' -
4*x*y' - (1+x^2)*y(x) = xRobert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
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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
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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
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
possible that I misunderstood your question. If I get it
right,>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
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
time difference from Athens, Greece to CaliforniaIf 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 youra 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 samein both places
though.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
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 toAthens (with a connection to New
York) on Sunday 3/30/03 ... precisely for this reason; I
understood when they called my name :-)
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
a good reference for mathematical explanation of the>degrees of
of squares in the White House is >/= 1, Freedom is 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 probablyfind something at
http://www.statcan.ca.Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
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?>>Yes, it's 0 (sometimes). :-)>>Nijmegen, Netherlands>I could
?>>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, andthose
people in this world... those who understand binary,and>
those who don't.Sure, but what would have answered the buggy
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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
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F(a,b;c;d) mean, where a,b,c,d are numbers?>> Is this function
in Mathematica?>Hypergeometric function.--Julien
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
Terry de Moore escribi.97: >
 en el art.92culo <
61bfbn$70k$1@hp.fciencias.unam.mx >, Unam < fcueto@usa.net
343924.E388B5@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:  > 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; > (* 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 >
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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,
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:>>1. y=4x|X|-3>2. y=|2X-1|>3. y=3x|X+4|-2>4. y=-0.5|X+4|+3>>i
reply.And you would have us do what with them?Graph them, Find
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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/00090>23.pdf) There are
additional Quantum State Diffusion (QSD) terms which
are>non-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
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
me with these pleeeeease!!! :1. y=4x|X|-32. y=|2X-1|3.
y=3x|X+4|-24. y=-0.5|X+4|+3i really need your help on this,
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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 = 3First with 2 = 63first with 3 =
173first with 4 = 1414first with 5 = 17313first with 6 =
53937first with 7 = 138923first with 8 = 953937first with 9 =
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r(x)=(x+1)^{17}+9,there are polynomials s(x) and t(x) with
integer coefficientssuch 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 dividesgcd(n^{17}+9,(n+1)^{17}+9), that is, p divides
n^{17}+9=q(n) andp also divides (n+1)^{17}+9=r(n). Then p also
dividess(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://primes.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
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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
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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 +
>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 ofpolynomials with
algebraic integer coefficients.Arturo Magidin, sans
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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 ofBefore trying to prove something you
better make sure it's actually
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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
havex = 2m, y=2n with m and n integers, so the norm is12 = 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
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can help me they are to hard every sum is hard i get every
one wrong can you help
pleaseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
eeeeeeeeeeeeee
eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
eeeeeeeeeeeeee
eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeethanck f r o
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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 ITIn 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
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support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id
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
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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
getP=(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
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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_ujust get
rid of all the underscores.I will compile them with as little
change as possible for my machine and post the times.Jason
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
1GHz Mac the time to run the 1,000,000 problem is .06
secondsto run the 20,000,000 problem was 2.55
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
that I have are as followsAll 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, 1.1920000000, 2.52 Jason Rodgers size time
1000000, 0.04 10000000 , 1.1120000000, 2.42 Modification of
Christian Bau's by C Bond size time 1000000, 0.44 10000000 ,
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
procedure I'm using to solve solvable sextics.It works so far,
but I need to test it more. So, if you knowany other solvable
sextic, pls post it. And not those that factorover sqrt(D) for
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, accordingto 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.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
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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 ,
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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
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
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,
theconcept of a tangent line appears to have been simple enough
for Euclid sothat it is present in definition #8 (angle between
two curves) withoutbeing explicitly mentioned; and then I
explained in my second posting(replying to Nat Silver) how my
tangential interpretation simplifies thematter by using one
point instead of two (in both #4 and #7) *and* by explaining
the appeal to surface's straight lines (rather than
merelypoints) in #7.Of course all this is just a possibility,
I may well have cut myself with Occam's razor :-)
which lies evenly with the points onitself.>>#7: A plane
surface is a surface which lies evenly with the straightlines>
>>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 =
liesevenly>>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).>>
line rather thanline)> 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.
Ithink it is important to remember that Euclid was writing a
textbook andnever intended this to be the seminal work on
Geometry for close to 3000years. Does Euclid have gaps and
ommissions and tacit assumptions. Youbet!!! BUT, I challenge
anyone to pick up ANY high-school geometry textbookand not
find gaps and ommissions and tacit
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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
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.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
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(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
algebrically.Let x, y and C be some constants. Let a_n be
defined as:a_0 = xa_n = C a_(n-1) + yThen: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
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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!piggybacki 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
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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 theratio of two
>>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 integersthat do not
contain a 6 (or, mor generally, any specified digitin 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
theintegers 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
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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 simpleArithmetic 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
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 maynot 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
twonumbers.>> Quadratic mean: This is the formula your teacher
gave you.>> Relative size: This means to check which is bigger.
Just try a fewcalculations, and try to see which type of mean
is usually bigger. Youreducated guess is called a conjecture.
In a more advanced course, you maylearn ways to prove your
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
>> 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.westminster.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
approve@localhost) by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
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
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, thenumbers with a 6 are greater than those without a
6.But this is quite misleading, as after a while thatrelation
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.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
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proof that for any (undirected)>|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
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
(-3,-3)Gary TupperTerrace 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
homework, by the way.>>|b| - |a| <= | |b| - |a| | <= | b - a
Steve:> Consider (-3,-3)> Gary Tupper> Terrace BC>
>|b| - |a| <= | |b| - |a| | <= | b - a |>>-- >Stephen J.
Herschkorn herschko@rutcor.rutgers.edua=-3, b=-3 then|a| = 3|b|
= 3b - a = -3 - (-3) = -3 + 3 = 0|b| - |a| <= ||b| - |a|| <= |
b - a | so all of the above are = 0 in this caseand 0<=0<=0 is
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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=displa Perhaps
quicken his fall in the sense of preventing him from
changing his mind once he had the machine out the
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
mythesis about problemes :1-winner take all2-taravel sales
man3-knap sack4-assingment problemand their
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
using Mathcad 6+ the other day, and it crashed with an invalid
page>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
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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
node belongs to a directed cycle (clearly they have neither
sourcesnor 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 multipleedges), _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 verifiableeasily 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 waywith 1-node
components only: K:= {[.]}. Besides (curiously),u_n(D) is the
number of strong tournaments (A054946).Valery
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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:
(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.MarcP.S.: I do find the OP notation
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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
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 thevariable 'x'
within the procedure. If you don't need to know it, thensimply
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 argumentto your proc instead
of the expression sin(x). Without seeing actualcode, it's
difficult to say more than that.-- Dave SeamanJudge Yohn's
mistakes revealed in Mumia Abu-Jamal
ruling.#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; iany, 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 ofcorresponding 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
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|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
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).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 positivecharacteristic. Let K be any field of
characteristic 0, different fromQ. Let {a_1,...,a_n,...} be a
basis for K over Q, and takeZ[a_1,...,a_n,...]. This satisfies
the condition that its field offractions equals K. Two bases
will in general yield different rings; Ithink they only yield
the same ring if the change of basis matrix hasinteger
coefficients. So by judiciously choosing your bases, youshould
be able to get intersections which are too small (if not
equalto 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 Rationalsfor
Q[i]. I will look for Gaussian numbers and see if it is
morepopular. Possibly not many people care about Q[i] at all.
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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
ofx^n + a_{n-1}*x^{n-1} + ... + a_1*x + a_0,then sqrt(y) is a
root ofx^{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
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proof that for any (undirected)>|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
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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
approve@localhost) by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
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
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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
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> thanxThe 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
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
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, linkAgain, 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/
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id
num-analysis/faq/part1>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
this FAQ:>>On the web:>http://
www.indra.com/~sullivan/q10.html Mathcom>>This FAQ is
usually available from MIT's rtfm and its mirrors:>ftp://rtfm.mit.edu/pub/usenet/news.answers/
num-analysis/faq/part1 MIT's rtfm>>If not, it is at:>ftp://
ftp.mathcom.com/mathcom/nafaq 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
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
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
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
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
the many services>listed herein - Netlib, the NIST guide,
NA-Net, CAIN, the NASA>Graphics site, and numerous other
indices and informative web
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
the Net>>For indices of packages oriented towards symbolic
algebra,>see q520, Symbolic Algebra.>The NIST Guide to
href=http://gams.nist.gov/>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
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
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
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/netlib/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.ac.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/ 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
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
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
Mathematical and Statistical Libraries>>NCAR's libraries
contain some overlap with netlib.>>http://
http.ucar.edu/SOFTLIB/mathlib.html
Parallel Archive>http
://www.hensa.ac.uk/parallel/environments/pcn/ Hensa>Note:
this web server can be very
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> *
Octave>>http://
bevo.che.wisc.edu/octave.html Octave>ftp://
www.che.wisc.edu/pub/octave 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++
RLaB>>http://
www.eskimo.com/~ians/rlab.html RLaB>ftp://
csi.jpl.nasa.gov/pub/matlab/RLaB RLaB via ftp>ftp://
evans.ee.adfa.oz.au/pub/RLaB 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
Scilab>>http:/
/zenon.inria.fr/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
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
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
href=http://www.ku-eichstaett.de/MGF/euler.html>http://
www.ku-eichstaett.de/MGF/euler.html Euler>ftp://ftp.k.-
eichstaett.de/pub/math 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
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
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
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++
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>
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
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
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
asurprise. - Randy