mm-61 === =So then the de?ition where we must satisfy = means the same> as(gp)(gv) = p(v). i.e. gp is a map in the dual which is applied to gv, a> vector in V.I just don't really see why this de?ition is made in the ?st place, i.e.> why it is the most sensible de?ition, but maybe I will understand> it better when I work through some examples?> You could de?e the action so that = , but then if the group G is not commutative, then it may not actually be an action, because then g(hp) = (hg)p which is not be what is desired, that is, (gh)p. === >Suppose f is a representation of the group G in the vector space V>over> k.>Then f* is the representation of G in the dual vector space V* de?ed> by>the action (gp)(v) = p((g^-1) v), where g is in G, p is in V*, and v>is> in>V. (Firstly, why is this action de?ed like this?)>We have f:G->L(V), where L(V) denotes the linear operators on V. We> de?e>f*:G->L(V*) so that> = = p(v) de?es the duality between p in V* and v in V. >This> gives> Why exactly do we want to de?e f*:G->L(V*) so that = ?> Why> does this de?ition make sense? Is < > the standard inner product?> What> do you mean by duality between p in V* and v in V?While I don't disagree with anything that was said,it doesn't seem to me essential to answer the original question.A linear map t: V -> V (where V is ?ite-dimensional)is represented by a matrix T _with respect to a given basis e_i of V_;to be precise, t(e_j) = sum_i T_{ij} e_i.The dual space V* has a dual basis p_j given byp_j(e_i) = 1 if i = j, 0 otherwise.If A(g) is the matrix representing the linear map v -> gvwith respect to the basis e_i of V,and the action of G on V* is de?ed as you say,then the matrix representing the linear map f -> gf (g in G, f in V*)with respect to the dual basis p_j is A(g^{-1})', as you say,and as you will ?d if you determine the matrix B carefully fromgf_j = sum_i B_{ij} f_i.Note of course that A(g^{-1}) = A(g)^{-1}since we are talking of a representation.The use of the notation = f(v) (f in V*, v in V)certainly simpli?s computation,but may obscure the basic idea, IMHO.-- Timothy Murphy tel: +353-86-2336090, +353-1-2842366 === > This is a reponse to a comment something like of what use is> abstract math.> Pure math is on the whole distinctly more useful that applied.> For what is useful above all is technique, and math technique> is taught mainly through pure math.Kronecker: God made integers, the rest is the work of man.Number theorists are like lotus-eaters--have once tasted of this food> they can never give it up.Stein: A computer is to a number theorist like a telescope is to> an astronomer. To teach a class without looking at integers through> the lens of a computer is like teaching astronomy without looking> through a telescope.VanQuet: Integers: those numbers possessing edges. Primes: dissonant integers of enlightened madness... ;) <- (obviously) thanks,Leroy Quet === Quet: Integers: those numbers possessing edges. Primes: dissonant integers of enlightened madness... ;) <- (obviously)>Easterly: Integers are an illusion.Russell- The universe is one dimensional === > Easterly: Integers are an illusion.Integers: Easterly is an illusion. === Van Jacques> This is a reponse to a comment something like of what use is> abstract math.> Pure math is on the whole distinctly more useful that applied.> For what is useful above all is technique, and math technique> is taught mainly through pure math.>...>Nowadays many sorts of mathematical and physical things are modelled in>terms of _sets_ of points. But where do you suppose the method was ?st>used? If I mistake not, it was in Dedekind's de?ition of an idealnumber,>as a kind of subset (a submodule) of the ring C. And what was the motivefor>bringing in ideals? Trying to prove FLT, of course! Not really. The reason for bringing in ideals was to present a> concrete counterpart to Kummer's ideal numbers, which was subject to> generalization in arbitrary number ?lds (both Dedekind and Kronecker> had run into trouble in trying to use Kummer's approach in rings of> integers that did not have an integral basis made of powers of the> same element). And while Kummer used his approach to prove FLT for> regular primes, he never considered it particularly important. It was> all an unintentional consequence of his ->true<- interest:> generalising quadratic reciprocity to higher reciprocity laws.I think this is a bit of an overstatement. Kummer was a very complexindividual (aren't we all?). He certainly was *extremely* interested inproving FLT. But he understood at least some of the limitations of hisarguments (clearly not all, or he wouldn't have called regular primesregular). I don't know how strong the publish or perish imperative was atthat time, but I have doubts that his interest in higher reciprocity lawswas stronger than his interest in FLT. However, he could get much morepublishable results about reciprocity laws. I'm certain (noting thedistinction between my certainty and the actual truth) that he did notconsider higher reciprocity laws as his all-encompassing passion and FLT asa mere sideshow, which I think would be a common interpretation of yourassertion by the general public.As an analogy, consider Wiles' proof of FLT -- it's just a corollary tohis true interest, the Taniyama-Shimura conjecture. But it's the reasonhe was studying the conjecture in the ?st place. I haven't the foggiestnotion why Kummer was studying reciprocity laws, but I am certain that hewas excited by the possibility of applying them to FLT.Motivation is such a hard thing to fathom, except in the case of certain?tion writers who bare all.Jon Miller === [.snip.]> Not really. The reason for bringing in ideals was to present a> concrete counterpart to Kummer's ideal numbers, which was subject to> generalization in arbitrary number ?lds (both Dedekind and Kronecker> had run into trouble in trying to use Kummer's approach in rings of> integers that did not have an integral basis made of powers of the> same element). And while Kummer used his approach to prove FLT for> regular primes, he never considered it particularly important. It was> all an unintentional consequence of his ->true<- interest:> generalising quadratic reciprocity to higher reciprocity laws.I think this is a bit of an overstatement. Kummer was a very complex> individual (aren't we all?). He certainly was *extremely* interested in> proving FLT.I was under that impression as well, but when I started reading intoKummer's ideal numbers, I was mostly disabused of this notion. See forexample Edwards book _Fermat's Last Theorem: A Genetic Introduction toNumber Theory_. The proof of quadratic reciprocity of Gauss and cubicreciprocity by Einsenstein were, by his own account, foremoest in hismind.> But he understood at least some of the limitations of his> arguments (clearly not all, or he wouldn't have called regular primes> regular). He was well aware of them, since he noted to the Berlin Academy thathis assumptions probably did not held for p=37 (they did not).>I don't know how strong the publish or perish imperative was at> that time, but I have doubts that his interest in higher reciprocity laws> was stronger than his interest in FLT.The imperative was very different. > However, he could get much more> publishable results about reciprocity laws. I'm certain (noting the> distinction between my certainty and the actual truth) that he did not> consider higher reciprocity laws as his all-encompassing passion and FLT as> a mere sideshow, which I think would be a common interpretation of your> assertion by the general public.As an analogy, consider Wiles' proof of FLT -- it's just a corollary to> his true interest, the Taniyama-Shimura conjecture. But it's the reason> he was studying the conjecture in the ?st place. I haven't the foggiest> notion why Kummer was studying reciprocity laws, but I am certain that he> was excited by the possibility of applying them to FLT.What makes you certain of that? On what do you base this? Wiles isquite forthcoming that he attacked T-S-W ->because<- he was interestedin FLT. But Gauss proved quadratic reciprocity not out of any interestin FLT, and it was seen by him and others as a great achievement.Gauss was particularly interested in extending quadratic reciprocityto higher reciprocity laws, and worked on biquadratic, and gave toEisenstein the job of dealing with cubic; and this is what Kummer'swork did, extend them. In fact, reciprocity laws played no role inKummer's work in FLT. What played a role was the whole scaffolding hehad to build in order to prove the higher reciprocity laws: thecyclotomic number ?lds and unique factorization into ideal primes.But ->not<- the reciprocity laws. So I don't know on what you base theidea that Kummer was excited by the possiblity of applying[reciprocity laws] to FLT.> Motivation is such a hard thing to fathom, except in the case of certain> ?tion writers who bare all.Kummer described his achievement in reciprocity laws as both foremostin his mind, and his most important work. He did not consider hisproof of FLT for regular primes to be in the same league, by his ownadmission, from what I read.Of course if you have some sources, I would be most interested to knowabout them.Arturo Magidin, sans .sig === I believe it is likely the following COULD be true, yet it was ?uredvery unrigorously and could be wrong.For r = integer >= 2,limit{n-> oo} n-1 --- --- 1 j--- > ----------- n / / n^(1/r) - j --- --- k=1 j|k j<=k^(1/r)= 1 + 1/2 + 1/3 +...+ 1/r, the r_th harmonic number.In linear-mode:limit{n-> oo} (1/n) sum{k=1 to n-1} sum{j|k,j<=k^(1/r)} j/(n^(1/r) -j)=1 + 1/2 + 1/3 +...+ 1/r, the r_th harmonic number.(Right?)By the way, the inner-sum of the limit is over the positive divisors,j, of kwhich are <= the r_th root of k.thanks,Leroy Quet === > Then the number x = .(x_1)(x_2)(x_3)... is the required number. It is> not in the list because for each k, x differs from f(k) in the k-th> digit. And note that none of its digits can be 0 or 9, so that it cannot be any> of the numbers having two potential decimal representations, such as> 1.000... = 0.999....As long as the numbers aren't written in base 2 or base 3.Russell- 2 many 2 count === > Then the number x = .(x_1)(x_2)(x_3)... is the required number. It is> not in the list because for each k, x differs from f(k) in the k-th> digit. And note that none of its digits can be 0 or 9, so that it cannot be any> of the numbers having two potential decimal representations, such as> 1.000... = 0.999....As long as the numbers aren't written in base 2 or base 3.> But in base two (or three) one uses pairs of digits instead of single digits, in effect translating into base four (or nine), and in any base greater than 3, one can always use the rule: for the Cantor diagonal number use a 1 to replace all non-1's and use 2 to replace 1. === > Then the number x = .(x_1)(x_2)(x_3)... is the required number. It is> not in the list because for each k, x differs from f(k) in the k-th> digit.> And note that none of its digits can be 0 or 9, so that it cannot be any> of the numbers having two potential decimal representations, such as> 1.000... = 0.999....> As long as the numbers aren't written in base 2 or base 3.> But in base two (or three) one uses pairs of digits instead of single > digits, in effect translating into base four (or nine), and in any base > greater than 3, one can always use the rule: for the Cantor diagonal > number use a 1 to replace all non-1's and use 2 to replace 1.None of that matters, because the theorem merely says, given a mapping f: N -> R, that f is not a surjection. Notice that the theorem does notmention representations of real numbers in any particular base; it onlymentions the reals as the codomain of the mapping. We are free to adoptany representation we choose for those numbers. Why would anyone want tomake things arti?ially dif?ult by specifying base 2 or 3?-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === None of that matters, because the theorem merely says, given a mapping> f: N -> R, that f is not a surjection. Notice that the theorem does not> mention representations of real numbers in any particular base; it only> mentions the reals as the codomain of the mapping. We are free to adopt> any representation we choose for those numbers. Why would anyone want to> make things arti?ially dif?ult by specifying base 2 or 3?Whay does anyone want to make things arti?ially dif?ult byworking with expansions at all?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === None of that matters, because the theorem merely says, given a mapping> f: N -> R, that f is not a surjection. Notice that the theorem does not> mention representations of real numbers in any particular base; it only> mentions the reals as the codomain of the mapping. We are free to adopt> any representation we choose for those numbers. Why would anyone want to> make things arti?ially dif?ult by specifying base 2 or 3?> Whay does anyone want to make things arti?ially dif?ult by> working with expansions at all?That's a good question. My preferred proof of the uncountability of thereals is to show that there is a natural bijection between the Cantor setand the power set of the integers. Of course, the simplest de?ition ofthe Cantor set involves base-3 expansions.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === > None of that matters, because the theorem merely says, given a mapping> f: N -> R, that f is not a surjection. Notice that the theorem does not> mention representations of real numbers in any particular base; it only> mentions the reals as the codomain of the mapping. We are free to adopt> any representation we choose for those numbers. Why would anyone want> to make things arti?ially dif?ult by specifying base 2 or 3?> Whay does anyone want to make things arti?ially dif?ult by> working with expansions at all?That's a good question. My preferred proof of the uncountability of the> reals is to show that there is a natural bijection between the Cantor set> and the power set of the integers. Of course, the simplest de?ition of> the Cantor set involves base-3 expansions.My preferred proof is to invoke the Baire Category Theorem.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === > None of that matters, because the theorem merely says, given a mapping> f: N -> R, that f is not a surjection. Notice that the theorem does not> mention representations of real numbers in any particular base; it only> mentions the reals as the codomain of the mapping. We are free to adopt> any representation we choose for those numbers. Why would anyone want> to make things arti?ially dif?ult by specifying base 2 or 3?>Whay does anyone want to make things arti?ially dif?ult by> working with expansions at all?That's a good question. My preferred proof of the uncountability of the> reals is to show that there is a natural bijection between the Cantor set> and the power set of the integers. Of course, the simplest de?ition of> the Cantor set involves base-3 expansions.> My preferred proof is to invoke the Baire Category Theorem.There's also the proof from measure theory, but the Cantor set argument issimpler than those.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === > My preferred proof is to invoke the Baire Category Theorem.There's also the proof from measure theory, but the Cantor set argument is> simpler than those.Examining the proof of BCT in this context leads to the following argument--- which could hardly be simpler and clearly exhibits the role of completeness.Let (x_n) be a sequence of reals.Let a_0 < b_0 be any real numbers, and recursively de?e a_n andb_n to satisfy a_{n-1} <= a_n < b_n <= b_{n-1} and x_n notin [a_n, b_n].Let A = lim a_n (this is a bounded increasing sequence). Then A is in[a_n, b_n] for all n, and so A =/= x_n for all n.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === >My preferred proof is to invoke the Baire Category Theorem.There's also the proof from measure theory, but the Cantor set argument is> simpler than those.> Examining the proof of BCT in this context leads to the following argument> --- which could hardly be simpler and clearly exhibits the role of > completeness.> Let (x_n) be a sequence of reals.> Let a_0 < b_0 be any real numbers, and recursively de?e a_n and> b_n to satisfy a_{n-1} <= a_n < b_n <= b_{n-1} and x_n notin [a_n, b_n].> Let A = lim a_n (this is a bounded increasing sequence). Then A is in> [a_n, b_n] for all n, and so A =/= x_n for all n.For each f: X -> P(X), S = { x in X : not(x in f(x)) } is not in ran(f).But x |-> {x} is an injection, hence |X| < |P(X)|. But f: P(N) -> Cgiven by S |-> sum_{k in S} 2/3^k is a bijection between P(N) and C,hence |N| < |C| <= |R|.I can omit the second sentence if we assume AC.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === In <200401022149.i02LnMq13132@proapp.mathforum.org>, on 01/03/2004 at 02:23 AM, nico80@jazzfree.com (Nicolas de la Foz) said:>As the number of digits of the natural numbers increases as its value> grows, we will add enough zeroes on the left of each natural in the >list,That has no meaning. Don't confuse a representation of a number withthe number itself.>This is a neutral transformation, and it will always be possible. FSVO possible. It's possible to add a ?ite number of zeros to theleft of a decimal representation of a natural number. However, that isnot what you are asking for. -- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === [reposting with correct (shorter) title. sorry.]I believe it is likely the following COULD be true, yet it was ?uredvery unrigorously and could be wrong.For r = integer >= 2,limit{n-> oo} n-1 --- --- 1 j--- > ----------- n / / n^(1/r) - j --- --- k=1 j|k j<=k^(1/r)= 1 + 1/2 + 1/3 +...+ 1/r, the r_th harmonic number.In linear-mode:limit{n-> oo} (1/n) sum{k=1 to n-1} sum{j|k,j<=k^(1/r)} j/(n^(1/r) -j)=1 + 1/2 + 1/3 +...+ 1/r, the r_th harmonic number.(Right?)By the way, the inner-sum of the limit is over the positive divisors,j, of kwhich are <= the r_th root of k.thanks,Leroy Quet === Defending myself:By the way, I am very aware that base-1 is not a base in the samesense that we typically refer to our commonly-used number-system asbase-10and to binary as base-2.(Although some repliers seem to believe I am unaware of myless-than-literal use of the word base.)My particular de?ition of the term base-1 is not my own, yet Icannot recall where else I have seen it used in the sense I use it inmy original post, but I have seen the term used this way in severaldifferent (and reputable) places, I am sure.thanks,Leroy Quet> I am posting this as more a fun challenge rather than a serious> question.> {So, that is why I have cross-posted this to rec.puzzles AND> sci.math.}We almost all are aware that, for n = integer >= 2, we can write a> non-integer real with base-n digits (0 through {n-1}), some digits> following after a decimal-point if necessary.But what about in base-1?Integers are easy (though base-one representations are not exactly> analogous to higher bases, since we do not write base-1 integers using> only zeros).Example: 7 (base 10) => 1111111 (base 1)But what about non-integers?Have you any clever schemes for writing, say, 1/2 or pi in base-1??[The best I can come up with right now is to write the continued> fraction of the real, with each term consisting of a base-1 positive> integer. But this is really a list of base-1 integers. Still,> anything better??]> thanks,> Leroy Quet === > Can we use a fraction as a radix, such as r = 3/2?and the Mensanator replied:> I don't know. How many digits are in Radix 3/2? One and a half?I'd say two-- zero and one. I'm making this up as I go, but for radix r,digits could range from 0 to ceiling(r)-1.So for example in base 3/2 11 represents 3/2 + 1 = 5/2 101 represents 9/4 + 1 = 13/4 .1 represents 2/3I think we can represent any number. For example, how do we represent 2? 2 = 3/2 + 1/2 = 3/2 + 4/9 + 1/18 = 3/2 + 4/9 + 1/18 = ... = 3/2 + 4/9 + 256/6561 + 2048/177147 + ...Working out more digits, you can get two can be represented by 1.00100 00010 01001 01000 ...I also think we probably have more than one representation for most numbers.For example, 1 represents 1, but we also have 1 = 2/3 + 1/3 = 2/3 + 8/27 + 1/27 = 2/3 + 8/27 + 512/19683 + ...which leads to one can be represented by .10100 00010 01001 01000 ...In fact, given any representation of a number with a 1 in some position, wecould subtract the 1 from that position, and add the string 1010000010... atthe next position to get another (probably) representation of the samenumber.When adding two numbers, carries are funky. If you add 1 and 1 at someposition, you get a result the has to be equivalent to a two (at thatposition). But two can't be represented by a ?ite string. So the carrypropogates one digit to the left *and* in?itely to the right.Bob H === In Base-0 the integers exist, but you can't tell two integers apart. ObPuzzle: Do non-integers exist?You might be able to use some geometric method, or the fact that the integersare countable. === > In Base-0 the integers exist, but you can't tell two integers apart. ObPuzzle: Do non-integers exist?Yes. Either of these newsgroups constitutes a constructive proof.-- Aatu Koskensilta (aatu.koskensilta@xortec.?Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === > Example: 7 (base 10) => 1111111 (base 1)... and just for fun, given that usually numbers in base N are written>using alpha-numeric digits 0 to (N-1), whereby symbols representing>digits N >= 10 are A, B, C etc or any other representation. My>question is, why in base N=1, we do not write:7 (base 10) = 0000000 (base 1) ?Hey... are we kidding on this Base 1 stuff or what?The whole point of, say, a base 2 number like 1 is that it reallyrepresents 00000001 or 000001 or any number of turned offswitches followed by some number of turned on switches.--dgates@spamfreelinkline.com === I'm currently a 2nd semester (junior) math/physics double majorstudent at a .. less-than-exemplary state school. Physics is the?ld I intend to work in, despite math being my ?st love. The planI came up with at the beginning of my undergrad work was as follows:double major math/phys (physics primary, math secondary), masters inmath, masters in physics, PhD physics.First, would a graduate degree in math be helpful for my occupation inphysics (theoretical work as a professor is my hope)?Second, what schools would be good for getting a masters in math, butnot necessarily a doctorate? Note: Please answer this one even if the?st answer is negative.Third, what schools would be good for my physics graduate work? Iintend to, though it's not imperative, attend seperate universitiesfor my masters and doctoral work. === > Second, what schools would be good for getting a masters in math, but not> necessarily a doctorate? Note: Please answer this one even if the ?st> answer is negative.One thing you might ?d interesting is that some schools that normally onlygive doctorates will allow for a masters if you are getting a doctoratethere in another ?ld.Caltech used to be like that, for example (I don't know if they are now).That would probably be your best bet: get into a good doctoral program inphysics, at a school with a good doctoral math program that will let you geta masters in math there, unless for some reason you want the math masters tocome from someplace other than the physics doctorate.-- --Tim Smith === > Second, what schools would be good for getting a masters in math, but not> necessarily a doctorate? Note: Please answer this one even if the ?st> answer is negative.unless for some reason you want the math masters to> come from someplace other than the physics doctorate.well, I was hoping to get my math masters _?st_.. to ensure Iunderstand the material I cover in physics.. I de?itely don't wantall three graduate degrees from the same school. but I guess it wouldbe a possibility to get my masters in physics, then my MS in mathafterwards.. to take advantage of such a thing === hi,i cannot answer your questions about the universities, because ipersonally feel that amount of learning and knowledge one aquiresdoesnot directly come from the university in which he is studying, butfrom the amount of efforts he puts into learning. There wouldcertainly be exceptions if you are in a university where theprofessors are all noble laurites and have contributed major pathbreaking inventions to their ?ld. In that case, you would get a lotof motivation, a lot of insight and a lot of knowledge from theuniversity, bacause of the professors. Otherwise, all universities aresame, with minor differences which does not matter. This is howeveronly my view.The question: Will masters in math be helpful for theoritical work asa professor, my answer would be, you would never make a good professorif you dont know enough math. The best way to know enough math is todo your masters in math, atleast, taht is the disciplined way. Butanyway, this is the conclusion: Good enough math is a must foreverything more than a bachelor's degree.Good l.Happy time.Prasanna.> Second, what schools would be good for getting a masters in math, but not> necessarily a doctorate? Note: Please answer this one even if the ?st> answer is negative.One thing you might ?d interesting is that some schools that normally only> give doctorates will allow for a masters if you are getting a doctorate> there in another ?ld.Caltech used to be like that, for example (I don't know if they are now).> That would probably be your best bet: get into a good doctoral program in> physics, at a school with a good doctoral math program that will let you get> a masters in math there, unless for some reason you want the math masters to> come from someplace other than the physics doctorate. === I just checked out the Wolfram research Web pages recommended by otherpeople in this thread. To my mind, they are too dense and concise forsomeone who has no idea what Div, Gradient, and Curl really are.The Feynman discussions may be helpful -- I haven't checked -- but thebest treatment of these topics I've ever seen (in words, pictures, andthe essential math) is in the physics book I used 20+ years ago incollege: Electricity and Magnetismby Edward M. PurcellVolume 2 of the Berkeley Physics CourseISBN: 0070049084 About $100I don't know if the most recent version retains the treatment used 20year ago, but the version I used 20 years back had an excellent,carefully drawn out, step-by-step approach to div, grad, and curl (aswell as the foundations of E&M). Maybe someone else can comment on whether the current version of thetext is still as strong.Steve O. === There is a formula to ?d distance between two points on plane. Iwant to learn if there is a formula to ?d distance between twopoints on cube's surface.General distance formula for space gives the distance with usinginside of the cube, I want the distance with using surface of cube.This is similar with walking ant problem on the cube. === >There is a formula to ?d distance between two points on plane. I>want to learn if there is a formula to ?d distance between two>points on cube's surface.>General distance formula for space gives the distance with using>inside of the cube, I want the distance with using surface of cube.>This is similar with walking ant problem on the cube.Maybe this is a naive solution but you could unwrap the cube so thefaces all lie in the same plane, then use the usual distance formula.This unwrap would be done using a combination of rotations andtranslations. === >There is a formula to ?d distance between two points on plane. I>want to learn if there is a formula to ?d distance between two>points on cube's surface.>Maybe this is a naive solution but you could unwrap the cube so the>faces all lie in the same plane, then use the usual distance formula.>This unwrap would be done using a combination of rotations and>translations.Yes, this is the best answer for the OP. (Any attempt to write it outas a formula is probably more trouble than it will prove to be worth.)But note that it is not always clear which way to unwrap the cube tocompute the distance. Exercise: Consider the 1 x 1 x 2 box stretching from the originin R^3 to the point (1,1,2). Which pair of points is furthest apart?dave === Typo:Expressions should read 9/10 not 1/10. (I was thinking of the binary seriesbinary)> 1/10 + 1/10^2 + ...+1/10^n = 1 - 1/10^n This is true for all (?ite) values of n. (1 - 1/10^n) > 1 This is also true for all ?ite values of n (and trans?ite values too)> Only when n achieves absolute in?ity does 1/10^n become zero. Mathematicians avoid this awkwardnwess by the weasel words: the limit of 1/10^n tends to zero as n tends to in?ity. No reaching of in?ity is intended by this pristine phrase, so we are> justi?d in maintainig that no limit of any kind could ever be realisedby> its use. Furthermore, if the ?tending' is towards something in?te short ofabsolute> in?ity the the limit of zero cannot be claimed either. Tony Thomas> Sorry, the ridiculous assertion is 0.999.... does NOT equal 1.> It certainly does!> Just try to subtract 0.999... from 1:> 1 - 0.999... = 0> Reason:> There is no real number between 0.999... and 1, and, therefore, they> must be one and the same number!> PH Neat proof, but you are playing foot loose with the de?ition of =.> 0.999... is an in?ite series, a shorthand notation for .9, .99,> .999, ... 1 is an integer. The relationship is that 1 is a (the)> value for which, for every D>0 there is an N such that for all M>N the> value of the Mth number in the series is between 1-D and 1+D. The problem occurs when people start saying that .999... equals 1. Charlie Volkstorf === I thought there was a proof for this... long ago I had seen one forrepeated decimals, how to change them into the accurate fraction formah.. and a simple search provides it..de?e .999... as X10X = 9.999... = 9 + .999... = 9 + X10X = 9 + x10X-X = 99X = 9X = 1.999... = 1I notice now that it looks strange for .999... but for any otherrepeating decimal, it is more lucid... === Dear all,My problem is to construct some integer matrices of certain pattern tosatisfy a matrix equation... There are so many unknowns, and there are alsoso many equations, (number of equations > number of unknowns)...Since the required matrices are integer, I cannot use Newton, or Conjugatedescent methods, etc... I have to do some search algorithm... but sincethere are so many unknowns, the search space is too large...So I am thinking of having an initial matrix, then do some change, see ifthe matrix equation(left hand side - right hand side) approximately morenear zero and ?d the minimization point...But even this, still make a large search space when the size of matrix getslarger... moreover, another constraint is that the number of non-zeroelements of this matrix should be as small as possible...Please tell me what kind of problem is this? I even don't know how tocategorize my problem and where to ?d answers on google...Please give me some pointers!-Walala === > I wanted to ask if anyone knows the following regarding the period> doubling quadratic maps, generated by these equations:> ax(1-x)> a-x^2> 1-ax^2How does one determine the parameter value of ?a' (along the x axis)> where period 2^14 resides ? How do I then also determine ?a' (where> the period is 2^14) within a particular subdomain I zoom in to, take> for example the Period 3 window top bifurcation diagram ?I think the Feigenbaum delta will help you. We have:delta_n := (a_{n} - a_{n-1}) / (a_{n+1} - a_{n})where a_{n} is the value of a for which period-n behavior gives birthto period-(2n) behavior. Feigenbaum showed that:lim (n->infty) delta_n = delta = 4.66920161...Rearranging this result, and taking n=2, we obtain (approximatlely,since we are using the limiting value of delta_n):a_3 = ((a_2 - a_1) / delta) + a_2for the next a, we obtain:a_4 = ((a_3 - a_2) / delta) + a_3 = (a_2 - a_1)(1/delta + 1/delta/delta) + a_2continuing on, we ?d (summing over the geometric series in delta):a_infty = (a_2 - a_1)*(1/(delta - 1)) + a_2It is then possible to show that:(a_infty - a_n)*delta^n = (a_2 - a_1)*(delta^2/(delta-1))Therefore, one should be able to ?d ?st the a_infty value for theparameter, and then the value for a speci? value of a_n. This is0198507232.So I think you could ?d the value of a for which period 2^14 resultsif you do enough calculations. I'm not sure about the subdomain,however.-John === Greetings.Happy... new year.Explain.I am quite sincere in my public development of logical systems. Perhaps the Great Pumpkin is Schultz' work, via extant sincerety.Explain.Basically I expect the computer to ?ure it out later, among thosewho ?ure it out, to help clarify our mutual understandings.Etcetera.We each have read this. Legitimate mathematical discord is a sign of progress.Have a nice day, if you would, Ross === I came across a Paper of W.Noll based on Truesdell's lectures writtenin German-titled Die Herleitung der Grundgleichungen derThermomechanik der Kontinua aus der statistischen Mechanik, 1955.Can somebody help me in ?ding out an English Translation of thispaper. === a loose translation :-)A derivation of the fundamental equations of thermodynamics of mechanics of continua based on statistical mechanics.> I came across a Paper of W.Noll based on Truesdell's lectures written> in German-titled Die Herleitung der Grundgleichungen der> Thermomechanik der Kontinua aus der statistischen Mechanik, 1955.> Can somebody help me in ?ding out an English Translation of this> paper.> === I heard this yesterday and I still won't understand it...we got this:next_x = S * prv_x * (1-prv_x)where S = 3.998 I thinkand starting x = .4and that equation is repeated MANY times...I heard it produces pseudo-random numbers, and I said it's probablysome computer bug but I was told that it is a math bug...How can that equation give pseudo-random numbers!?!?!?!? It's reallyweird isn't it?cmad === On 5 Jan 2004 01:03:35 -0800, cmad_x@yahoo.com (Chris Mantoulidis)>I heard this yesterday and I still won't understand it...we got this:next_x = S * prv_x * (1-prv_x)AKA the logistic equation. A well known example of chaotic behaviourin a nonlinear difference equation.Google gives this link which explains the behaviour of the equationquite well:>I heard it produces pseudo-random numbers, and I said it's probably>some computer bug but I was told that it is a math bug...It is of course neither. === Does exist a sort of set theory in wich we have to deal with a set of allsets i.e. a set that does not satisfy the Cantor's theorem (instead of|P(S)|>|S| we have |P(S)|=|S|) ? === >Does exist a sort of set theory in wich we have to deal with a set of all>sets i.e. a set that does not satisfy the Cantor's theorem (instead of>|P(S)|>|S| we have |P(S)|=|S|) ?In Quine's _New Foundations_, one cannot even discuss thefunctions mapping a set X into its subsets; the functionf de?ed by f(x) = {x} does not exist automatically inthis system. Otherwise, the paradox cannot be avoided.In NBG, there are classes with this property, as a properclass cannot be an element. A class S with the propertyyou pose must be such.-- 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 === > Does exist a sort of set theory in wich we have to deal with a set of all> sets i.e. a set that does not satisfy the Cantor's theorem (instead of> |P(S)|>|S| we have |P(S)|=|S|) ?Quine's New Foundations (NF) and its slightly less eccentric variantNFU (New Foundations with Ur-elements) both prove the existence of auniversal set. This set is not, however, de?ed in either theory as a set, s.t. |P(s)|=|s| but as a set of which everything is a member. Others might be able to enlighten you as to whether this set satis?s |P(s)|=|s|.-- Aatu Koskensilta (aatu.koskensilta@xortec.?Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === |This is called Graph Covering in general, and what you're looking for > |is a solution for the Minimum dominating set. Unfortunately, it's an> |NP problem.Being in NP is not a bad sign. All ef?iently solvable problems are in> NP. What suggests it's dif?ult is its being NP-complete, one of the> NP problems to which all the others can be reduced in polynomial time.Keith Ramsay> Well, I don't know whether it's in P. Do you?It doesn't really matter though - what really matters is the size of the problem, rather than its class. This is a speci? problem, and all asymptotic classi?ations are meaningless in that respect. We have 50 states, not n states, and we're looking for a 12-vertex solution, not a k-vertex solution. It doesn't matter whether you can ?d a solution which is O(n) but takes seven years to run, if you can ?d an O(2^n) solution which takes an hour. === >Yes, a posted list would be good, to clarify for example whether you>count corner adjacencies (like UT and NM, or AZ and CO), and which>underwater boundaries (as between HI and AK, MN and MI, or RI and NY)>you treat as adjacencies. Without specifying a list you probably>won't get useful answers.HI and AK??? There's a heck of a lot of international water between> them. Or are you claiming American sovereignty over the whole North> Paci??I'm not making that claim -- Russia probably would object -- but merely wanted the OP to give his or her adjacency list. On the three cases above, I would say no, yes, and maybe.The maps I looked at show a Minnesota - Michigan boundary within Lake Superior, and don't indicate if Rhode Island and New York have a boundary in Long Island Sound.-jiw === >Yes, a posted list would be good, to clarify for example whether you>count corner adjacencies (like UT and NM, or AZ and CO), and which>underwater boundaries (as between HI and AK, MN and MI, or RI and NY)>you treat as adjacencies. Without specifying a list you probably>won't get useful answers.HI and AK??? There's a heck of a lot of international water between> them. Or are you claiming American sovereignty over the whole North> Paci??I'm not making that claim -- Russia probably would object -- > but merely wanted the OP to give his or her adjacency list. > On the three cases above, I would say no, yes, and maybe.> The maps I looked at show a Minnesota - Michigan boundary > within Lake Superior, and don't indicate if Rhode Island > and New York have a boundary in Long Island Sound.> -jiwI did a quick searh for Rhode Island maps, and came up with a map thatshows Massachusetts, Connecticut and Rhode Island, with a boundarysaying New York in between Long Island and the rest of the water.The image is about 300 kilobytes, and the link is:http://www.hognews.com/states/ri/connect_mass_ rhode.jpgAnthony === >On Sat, 3 Jan 2004 00:22:48 +0000 (UTC), rob@trash.whim.org (RobOn 1 Jan 2004 10:27:56 -0800, denoncou@euclid.colorado.edu (Hugh>I have been having some dif?ulties with the following problem:>It is from Royden (p127 #17). The space throughout is assumed to be>[0,1], though I think any ?ite measure space will work. >Suppose p > 1. Suppose f_n -> f a.e. , f is in L_p and f_n is in>L_p for all n. Suppose there exists M such that >|| f_n || (p-norm) <= M for all n. Suppose g is in L_q. >Show that g*f_n -> g*f in the L_1 norm. If epsilon > 0 then there exists delta > 0 such that if m(A) < delta>then the L^q norm of g*chi_A is less than epsilon, because ___.>Now there exists a set A of measure less than delta such that>f_n -> f ___ly on the complement of A, by ___'s theorem...>[...]>It is not mentioned here, but in the problem, is 1/p + 1/q <= 1 assumed?>If not, then g*f might not even be in L^1.Well of course. In careful writing one would certainly want to state>1/p + 1/q = 1 explicitly, but in the present context, given that the >OP clearly has some idea what he's talking about, that's very clearly>an implicit assumption.>[...]>One must have p > 1. That's correct. Lily p > 1 _was_ given explicitly in the problem.>[...]I was responding to the OP while replying to your post. In his post, hementions having some problem with using the assumption p > 1. I wastrying to point out that is was necessary by showing that p = 1 fails.I wanted to quote your post to show where your argument broke down forp = 1.I had intended only to add to your post, not to detract from it.Rob Johnson take out the trash before replying === On Mon, 5 Jan 2004 11:57:37 +0000 (UTC), rob@trash.whim.org (Rob>On Sat, 3 Jan 2004 00:22:48 +0000 (UTC), rob@trash.whim.org (Rob>[...]It is not mentioned here, but in the problem, is 1/p + 1/q <= 1 assumed?>If not, then g*f might not even be in L^1.>Well of course. In careful writing one would certainly want to state>1/p + 1/q = 1 explicitly, but in the present context, given that the >OP clearly has some idea what he's talking about, that's very clearly>an implicit assumption.>[...]One must have p > 1. >That's correct. Lily p > 1 _was_ given explicitly in the problem.>[...]I was responding to the OP while replying to your post. In his post, he>mentions having some problem with using the assumption p > 1. I was>trying to point out that is was necessary by showing that p = 1 fails.>I wanted to quote your post to show where your argument broke down for>p = 1.Oh. Never mind then...>I had intended only to add to your post, not to detract from it.Rob Johnson was discovered at Los Alamos laboratories on News Years 2004. This>right in the middle of an.., experiment, has Los Alamos scientists>greatly perplex over chronom tunneling since it was detected a few>days before it's discovery.>Its 01/01/04, not 04/01/04!>Huh?> The Fourth of January xxxx.That would have been 04.01.xxxx:I.v9 === >was discovered at Los Alamos laboratories on News Years 2004. This>right in the middle of an.., experiment, has Los Alamos scientists>greatly perplex over chronom tunneling since it was detected a few>days before it's discovery.>Its 01/01/04, not 04/01/04!>Huh?> The Fourth of January xxxx.That would have been 04.01.xxxxThe same as 04/01/xxxx then.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === > If G is a ?ite p-group such that Aut(G) is Abelian must G be cyclic ?> A non-cyclic example is the semi-direct product of the cyclic group of order8 with the dihedral group of order 8. A presentation is < a,b,c | a^8 = b^4 = c^2 = 1, b^c=b^-1, a^b=a^5, a^c=a >= < a,b,c | a^8 = b^4 = c^2 = 1, bcb=c, ab=ba^5, ac=ca >where x^y denotes y^-1*x*y. is normal, is dihedral of order 8,and intersect is trivial. This is a nonabelian p-group of order 64,yet its group of automorphisms is isomorphic to the elementary abelian groupof order 128 (the direct product of 7 cyclic groups of order 2).There are two other examples of order 64, which are also nonabelian semidirectproducts.http://www-gap.dcs.st-and.ac.uk/~gap/ is a wonderful package for ?dingexamples. === Is there a concept called harmonic dynamics and if so where might Ilearn about it?Stig Holmquist === On Mon, 05 Jan 2004 08:24:05 -0500, Stig Holmquist>Is there a concept called harmonic dynamics and if so where might I>learn about it?Google gives a few hundred hits on harmonic dynamics(don't forget the quote marks).>Stig Holmquist************************David C. Ullrich === says...> * James Harris> Now math history is full of people like David Ullrich, a guy with a> title, ?hting against some new idea. In the past it was sqrt(-1),> as mathematicians fought against an idea they thought of as silly.are dead and buried, (the ideas).Most people show worth through effort. For instance, if you value>your job, one would assume that you put a lot of effort into it. >Similarly, if you value a relationship, you're willing to work to keep>it going.People like Ullrich value trying to attack the new, as evidenced by>his *efforts* in that regard.>Or it could evidence the fact that he values correct mathematics and he believesyour work here to be incorrect. Note for this to be the case does not requireyour work to actually be incorrect, merely for David Ullrich to believe it tobe.If you could try to embrace this kind of attitude - allowing people to disagreewith your mathematics (and you theirs) without ascribing negative motives to thedisagreement - you would get a much better reception. Which in turn would meanthat you could develop your ideas in collaboration with others, rather thanconstantly ?hting them.Wouldn't that be much less frustrating? === >My research can be dif?ult to understand, so I thought I'd try out>yet another way of explaining it. Some of you may have ?ured out>that I test out explanations on Usenet for use elsewhere, to re?emy>own understanding, or just in case someone out there might ?allyget>it.>Now then, again here's my discovery:>(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)>where b_3(x) = a_3(x) - 3 and the a's are roots of>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)>and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.> So far, no discovery. We agree on this part and it has> no particular signi?ance.>In that form it's hard to understand what follows next unless you pay>attention to what you have, speci?ally that cubic de?ing the a's.>I can get it because of the symmetry of>(5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)>where I've gone ahead and substituted a_3(x) back in to replace>b_3(x), and it's important that you focus on that symmetry.>It's that symmetry which allows the cubic>(*) a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)>to de?e ALL the a's, but something happens when I divide by 49.> Only if you divide by 49 in a certain way: factoring it as> 7, 7, 1. There is *another* factorization which works as> desired when x <> 0.>Then the symmetry is broken.> Physics jargon, used super?ially here to give the> impression that the writer sees and understands a pattern.> I doubt anyone is either fooled or impressed by it.>Without that symmetry it's impossible to>?d a SINGLE cubic to handle what results when you divide both sides>by 49.> False. There is a cubic. But it does not correspond> to the 7, 7, 1 factorization.> But ironically symmetry IS the key to all this: symmetry in the> form of Galois permutations of the roots of irreducible polynomials.> That is what tells you that if one of a_i(x) is non-coprime to 7,> then they all are. And that, of course, tells you that your> factorization of 49 as 7, 7, 1 is wrong, wrong, wrong whenever> your polynomial in the a's is irreducible - which it is for> almost all x.>That's important because it's why the functions are NOT algebraic>integer functions!!!> I think you have accepted this fact - that a_1(x)/7 is> not an algebraic integer - which of course we pointed> out months and months ago, and you fought tooth and nail for> a very long time.> But I bet you don't really understand the proof of it. As a> test, why don't you explain to the folks here in your own> words why it is true?>Now then, I'll recap. Symmetry allows the a's to be de?ed by a>cubic, which shows them to be algebraic integer functions, but>dividing by 49 *breaks* that symmetry, taking away the ability to?d>some cubic to de?e the results, which proves that the resulting>functions are not algebraic integer functions.>(5 b_1(x) + 1)(5 b_2(x) + 1)(5 b_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22>where the b's are roots of>b^3 + ? b^2 + ? b - (2401 x^3 - 147 x^2 + 3x)>and when x=0, b_1(0) = b_2(0) = b_3(0) = 0.>My point is that the second and third coef?ients are impossible to>de?e in general.> If by impossible to de?e in general you mean that they cannot be> algebraic integers, I agree. That is because 7, 7, 1 is the> wrong factorization.>You may ?d them for some particular x, but in general, they are>forever hidden from you.>Notice that doing that substitution with a_3(x) for b_3(x) gives me>(5 b_1(x) + 1)(5 b_2(x) + 1)(5 a_3(x) + 7) => 300125 x^3 - 18375 x^2 - 360 x + 22>but you have broken symmetry since the other constant terms are 1 and>1, so you're still st.> Right. b_1(x) and b_2(x) cannot be algebraic integers.> We all agree on this. It comes back to your having made the> wrong choice in factoring 49: 7, 7, 1 doesn't work. Something> else does.>Now by emphasizing what happens *after* 49 is divided from both sides>I'm trying to get at least some of you to face the mathematical>realities here, and I've made other posts pointing it out as well.> Only if you divide by 49 in the wrong way, as you keep insisting> on doing.> OK, here is another way to think about this. Consider yourpolynomial> in a,> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).> Notice that the constant term always has a factor of 49.[SNIP] James, now that the little diversion has been beaten to death concerningwhat the constant term of the polynomial: a^3 + A*a^2 + B, where the coef?ients are not a function of a,is (hint: consider setting ?a' to zero) I apologize to Nora Baron forhijacking her thread. Please go back and read her post starting just abovemy [SNIP]. I think you'll ?d it instructive and I'd be interested in yourcomments (if you con?e them to the math).KeithK> James Harris === >My research can be dif?ult to understand, so I thought I'd try> out>yet another way of explaining it. Some of you may have ?ured> out>that I test out explanations on Usenet for use elsewhere, to> re?e> my>own understanding, or just in case someone out there might> ?ally> get>it.>Now then, again here's my discovery:>(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)>where b_3(x) = a_3(x) - 3 and the a's are roots of>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)>and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.> So far, no discovery. We agree on this part and it has> no particular signi?ance.>In that form it's hard to understand what follows next unless you> pay>attention to what you have, speci?ally that cubic de?ing the> a's.>I can get it because of the symmetry of>(5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)>where I've gone ahead and substituted a_3(x) back in to replace>b_3(x), and it's important that you focus on that symmetry.>It's that symmetry which allows the cubic>(*) a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)>to de?e ALL the a's, but something happens when I divide by 49.> Only if you divide by 49 in a certain way: factoring it as> 7, 7, 1. There is *another* factorization which works as> desired when x <> 0.>Then the symmetry is broken.> Physics jargon, used super?ially here to give the> impression that the writer sees and understands a pattern.> I doubt anyone is either fooled or impressed by it.>Without that symmetry it's impossible to>?d a SINGLE cubic to handle what results when you divide both> sides>by 49.> False. There is a cubic. But it does not correspond> to the 7, 7, 1 factorization.> But ironically symmetry IS the key to all this: symmetry in the> form of Galois permutations of the roots of irreducible> polynomials.> That is what tells you that if one of a_i(x) is non-coprime to 7,> then they all are. And that, of course, tells you that your> factorization of 49 as 7, 7, 1 is wrong, wrong, wrong whenever> your polynomial in the a's is irreducible - which it is for> almost all x.>That's important because it's why the functions are NOT algebraic>integer functions!!!> I think you have accepted this fact - that a_1(x)/7 is> not an algebraic integer - which of course we pointed> out months and months ago, and you fought tooth and nail for> a very long time.> But I bet you don't really understand the proof of it. As a> test, why don't you explain to the folks here in your own> words why it is true?>Now then, I'll recap. Symmetry allows the a's to be de?ed by a>cubic, which shows them to be algebraic integer functions, but>dividing by 49 *breaks* that symmetry, taking away the ability to> ?d>some cubic to de?e the results, which proves that the resulting>functions are not algebraic integer functions.>(5 b_1(x) + 1)(5 b_2(x) + 1)(5 b_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22>where the b's are roots of>b^3 + ? b^2 + ? b - (2401 x^3 - 147 x^2 + 3x)>and when x=0, b_1(0) = b_2(0) = b_3(0) = 0.>My point is that the second and third coef?ients are impossible> to>de?e in general.> If by impossible to de?e in general you mean that they> cannot be> algebraic integers, I agree. That is because 7, 7, 1 is the> wrong factorization.>You may ?d them for some particular x, but in general, they are>forever hidden from you.>Notice that doing that substitution with a_3(x) for b_3(x) gives> me>(5 b_1(x) + 1)(5 b_2(x) + 1)(5 a_3(x) + 7) => 300125 x^3 - 18375 x^2 - 360 x + 22>but you have broken symmetry since the other constant terms are 1> and>1, so you're still st.> Right. b_1(x) and b_2(x) cannot be algebraic integers.> We all agree on this. It comes back to your having made the> wrong choice in factoring 49: 7, 7, 1 doesn't work. Something> else does.>Now by emphasizing what happens *after* 49 is divided from both> sides>I'm trying to get at least some of you to face the mathematical>realities here, and I've made other posts pointing it out as> well.> Only if you divide by 49 in the wrong way, as you keep insisting> on doing.> OK, here is another way to think about this. Consider your> polynomial> in a,> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).> Notice that the constant term always has a factor of 49.> Oh Nora, Nora dude, how can you be so mean? This is blowing James'> mind!> He's going crazy here setting x's to zero trying to ?d the> constant> term.> No James it's true! The constant term of your polynomial in a> is:> - 49(2401 x^3 - 147 x^2 + 3x) !!> Here is a _constant_ term of a polynomial that is a function of> x!!!> It> changes when x changes!!!> Your silent admirer,> KeithK>Well then it's not then constant now is it? That's why I used to talk> about being polynomial-like with another more complicated expression> where coef?ients also varied.> Mathematicians haven't done much work in this area, eh?> So I guess you can get confused enough from precedent to think it> sounds like a good idea to call that the constant term, but then you> might notice that it is variable dependent!>It's the constant term of the given polynomial in a. That polynomial> was:> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)> which is a cubic polynomial in the form:> a^3 + A*a^2 + B> where since ?B' = B(x) is independent of ?a' it is the constant term> of> the polynomial.> What you fail to understand is that in your polynomial, the coef?ients> are> simply _functions_ of ?x', where ?x' is independent of ?a', which means> they> are not treated as polynomials but rather are to be evaluated to a> numeric> value for a given choice of ?x'.> KeithK It seems to me that possibly the complexity has you confused,What complexity? You have a monic single-variate polynomial in ?a' given> by:> a^3 + A*a^2 + BActually x is still a variable, so it's just wishing to claim a singlevariable ?a'.Now then, can you understand that just *saying* something isn't avariable doesn't mathematically make it so?Givena^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)there's just no way to claim that - 49(2401 x^3 - 147 x^2 + 3x)is a constant term, as x is a variable. Understand?> where the coef?ients are independent of a, for which you solved for the> roots as a function of A and B and then plugged into that solution the> values> A(x) = 3(-1 + 49x),> B(x) = - 49(2401 x^3 - 147 x^2 + 3x)>so consider x^2 + xy + y^2. Now then, what is the constant term? > We're discussing single-variate polynomials.Maybe that's what *you* are discussing, but I see two variables,including x itself, so why would you try to call a variable expressiona constant term?Now then, back to x^2 + xy + y^2, what is the constant term to you?James Harris === >My research can be dif?ult to understand, so I thought I'dtry> out>yet another way of explaining it. Some of you may have?ured> out>that I test out explanations on Usenet for use elsewhere, to> re?e> my>own understanding, or just in case someone out there might> ?ally> get>it.>Now then, again here's my discovery:>(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)>where b_3(x) = a_3(x) - 3 and the a's are roots of>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)>and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.> So far, no discovery. We agree on this part and it has> no particular signi?ance.>In that form it's hard to understand what follows next unlessyou> pay>attention to what you have, speci?ally that cubic de?ingthe> a's.>I can get it because of the symmetry of>(5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)>where I've gone ahead and substituted a_3(x) back in toreplace>b_3(x), and it's important that you focus on that symmetry.>It's that symmetry which allows the cubic>(*) a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)>to de?e ALL the a's, but something happens when I divide by49.> Only if you divide by 49 in a certain way: factoring it as> 7, 7, 1. There is *another* factorization which works as> desired when x <> 0.>Then the symmetry is broken.> Physics jargon, used super?ially here to give the> impression that the writer sees and understands a pattern.> I doubt anyone is either fooled or impressed by it.>Without that symmetry it's impossible to>?d a SINGLE cubic to handle what results when you divideboth> sides>by 49.> False. There is a cubic. But it does not correspond> to the 7, 7, 1 factorization.> But ironically symmetry IS the key to all this: symmetry inthe> form of Galois permutations of the roots of irreducible> polynomials.> That is what tells you that if one of a_i(x) is non-coprime to7,> then they all are. And that, of course, tells you that your> factorization of 49 as 7, 7, 1 is wrong, wrong, wrong whenever> your polynomial in the a's is irreducible - which it is for> almost all x.>That's important because it's why the functions are NOTalgebraic>integer functions!!!> I think you have accepted this fact - that a_1(x)/7 is> not an algebraic integer - which of course we pointed> out months and months ago, and you fought tooth and nail for> a very long time.> But I bet you don't really understand the proof of it. As a> test, why don't you explain to the folks here in your own> words why it is true?>Now then, I'll recap. Symmetry allows the a's to be de?edby a>cubic, which shows them to be algebraic integer functions,but>dividing by 49 *breaks* that symmetry, taking away theability to> ?d>some cubic to de?e the results, which proves that theresulting>functions are not algebraic integer functions.>(5 b_1(x) + 1)(5 b_2(x) + 1)(5 b_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22>where the b's are roots of>b^3 + ? b^2 + ? b - (2401 x^3 - 147 x^2 + 3x)>and when x=0, b_1(0) = b_2(0) = b_3(0) = 0.>My point is that the second and third coef?ients areimpossible> to>de?e in general.> If by impossible to de?e in general you mean that they> cannot be> algebraic integers, I agree. That is because 7, 7, 1 is the> wrong factorization.>You may ?d them for some particular x, but in general, theyare>forever hidden from you.>Notice that doing that substitution with a_3(x) for b_3(x)gives> me>(5 b_1(x) + 1)(5 b_2(x) + 1)(5 a_3(x) + 7) => 300125 x^3 - 18375 x^2 - 360 x + 22>but you have broken symmetry since the other constant termsare 1> and>1, so you're still st.> Right. b_1(x) and b_2(x) cannot be algebraic integers.> We all agree on this. It comes back to your having made the> wrong choice in factoring 49: 7, 7, 1 doesn't work. Something> else does.>Now by emphasizing what happens *after* 49 is divided fromboth> sides>I'm trying to get at least some of you to face themathematical>realities here, and I've made other posts pointing it out as> well.> Only if you divide by 49 in the wrong way, as you keepinsisting> on doing.> OK, here is another way to think about this. Consider your> polynomial> in a,> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).> Notice that the constant term always has a factor of 49.> Oh Nora, Nora dude, how can you be so mean? This is blowingJames'> mind!> He's going crazy here setting x's to zero trying to ?d the> constant> term.> No James it's true! The constant term of your polynomial ina> is:> - 49(2401 x^3 - 147 x^2 + 3x) !!> Here is a _constant_ term of a polynomial that is a function of> x!!!> It> changes when x changes!!!> Your silent admirer,> KeithK> Well then it's not then constant now is it? That's why I used totalk> about being polynomial-like with another more complicatedexpression> where coef?ients also varied.> Mathematicians haven't done much work in this area, eh?> So I guess you can get confused enough from precedent to think it> sounds like a good idea to call that the constant term, but thenyou> might notice that it is variable dependent!>It's the constant term of the given polynomial in a. Thatpolynomial> was:> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)> which is a cubic polynomial in the form:> a^3 + A*a^2 + B> where since ?B' = B(x) is independent of ?a' it is the constantterm> of> the polynomial.> What you fail to understand is that in your polynomial, thecoef?ients> are> simply _functions_ of ?x', where ?x' is independent of ?a', whichmeans> they> are not treated as polynomials but rather are to be evaluated to a> numeric> value for a given choice of ?x'.> KeithK> It seems to me that possibly the complexity has you confused, What complexity? You have a monic single-variate polynomial in ?a'given> by:> a^3 + A*a^2 + B Actually x is still a variable, so it's just wishing to claim a single> variable ?a'. ..x is .. just wishing to claim a single variable ?a' ???Hogwash. You are perfectly aware that in your development: (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where b_3(x) = a_3(x) - 3 and the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) that ?x' is not a function of ?a', it is an independent variable into whichone plugs a number in order to evaluate the functions a_1(x), a_2(x),a_3(x). Now then, can you understand that just *saying* something isn't a> variable doesn't mathematically make it so? Given a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) there's just no way to claim that - 49(2401 x^3 - 147 x^2 + 3x) is a constant term, as x is a variable. Understand?>That expression varies with ?x' as any fool can plainly see. Ah cahn see, asLil Abner used to say. It does _not_ vary with ?a' and so is treated as the(and I put this all by itself so you can absorb this phrase): constant term of the polynomial in ?a'when calculating the roots of that polynomial to ?d a_1(x), a_2(x), a_3(x)in your development. where the coef?ients are independent of a, for which you solved forthe> roots as a function of A and B and then plugged into that solution the> values> A(x) = 3(-1 + 49x),> B(x) = - 49(2401 x^3 - 147 x^2 + 3x) >so consider> x^2 + xy + y^2.> Now then, what is the constant term?We're discussing single-variate polynomials. Maybe that's what *you* are discussing, but I see two variables,> including x itself, so why would you try to call a variable expression> a constant term? Now then, back to x^2 + xy + y^2, what is the constant term to you?>Now stop procrastinating and go back to Nora Baron's example in this threadthat begins: OK, here is another way to think about this. Consider yourpolynomial in a, a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). Notice that the constant term always has a factor of 49.and try to follow what she's saying.Keith James Harris === What complexity? You have a monic single-variate polynomial in ?a'> given> by:> a^3 + A*a^2 + B Actually x is still a variable, so it's just wishing to claim a single> variable ?a'. ..x is .. just wishing to claim a single variable ?a' ???> Hogwash. You are perfectly aware that in your development:No, it's quite clear that there is more than one variable ina^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).Now then, here's a reality check, can you or can you not admit thatthere are TWO variables in that expression?I will actually give them speci?ally to make sure you don't havewiggle room: the two variables are ?a', and x.Understand?I'm serious here, can you admit that there are TWO variables?> that ?x' is not a function of ?a', it is an independent variable into which> one plugs a number in order to evaluate the functions a_1(x), a_2(x),> a_3(x).So then it's a variable!!! However, your claims include the wordsmonic single-variate polynomial. Now then, can you understand that just *saying* something isn't a> variable doesn't mathematically make it so?Readers note, the poster didn't answer. Given a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) there's just no way to claim that - 49(2401 x^3 - 147 x^2 + 3x) is a constant term, as x is a variable. Understand? > That expression varies with ?x' as any fool can plainly see. Ah cahn see, as> Lil Abner used to say. It does _not_ vary with ?a' and so is treated as theWell, it's clear this poster isn't interested in adult communcation,but probably is yet ANOTHER poster playing on some stage in their ownminds.The reality of the world stage here is that it's about mathematics,not smart-ass comments.> (and I put this all by itself so you can absorb this phrase): constant term of the polynomial in ?a'when calculating the roots of that polynomial to ?d a_1(x), a_2(x), a_3(x)> in your development.Regardlessa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)is NOT single variable!It has *two* variables, and those are ?a' and x. > where the coef?ients are independent of a, for which you solved for> the> roots as a function of A and B and then plugged into that solution the> values> A(x) = 3(-1 + 49x),> B(x) = - 49(2401 x^3 - 147 x^2 + 3x)>so consider> x^2 + xy + y^2.> Now then, what is the constant term?>We're discussing single-variate polynomials.Now is where you lose all credibilty, as at a minimum you might haveclaimed that y^2 is the constant term in x. Maybe that's what *you* are discussing, but I see two variables,> including x itself, so why would you try to call a variable expression> a constant term? Now then, back to x^2 + xy + y^2, what is the constant term to you? > Now stop procrastinating and go back to Nora Baron's example in this thread> that begins:It occurs to me that you might *be* Nora Baron posting under anotherpseudonym!!!After all, like that poster you're showing an odd recalcitrance atadmitting even basic points like thata^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)has two variables, while stumbling over a simple check with theexpressionx^2 + xy + y^2.I'm currently considering odd posters like you which is why I'mreplying here, and part of my point to other readers is a bizarreirrationality where you will reply, and reply, and reply without everconceding even minor points!!!James Harris === > [snip]> Now stop procrastinating and go back to Nora Baron's example in this thread> that begins:It occurs to me that you might *be* Nora Baron posting under another> pseudonym!!!> Nope, it's not me. You will have to invent another excuse tostop replying to Keith K. Nora B. > After all, like that poster you're showing an odd recalcitrance at> admitting even basic points like thata^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)has two variables, while stumbling over a simple check with the> expressionx^2 + xy + y^2.I'm currently considering odd posters like you which is why I'm> replying here, and part of my point to other readers is a bizarre> irrationality where you will reply, and reply, and reply without ever> conceding even minor points!!!> James Harris === Now math history is full of people like David Ullrich, a guy with a> title, ?hting against some new idea.David Ullrich has a title??Good grief. Why didn't somebody tell us this before? I hate making> these social errors through ignorance. (See .sig quote.)And what is his title, anyway? Sir David? Lord Ullrich? Baron> Okstate?David Ullrich is a math professor at Oklahoma State University, so ata minimum he has the title: mathematician.I notice that you ran away from further discussions on the math.Have you yet found anyone in math literature you can cite who usesnon-uniqueness of polynomial factorization? Or have you now ?uredout how not allowing members that are both integers and units notequal to -1 or 1, can be a property used to de?e a ring?The point I want to make to you is that mathematics, no matter howmuch modern mathematicians might be trying to make it such, is notjust some social activity with gangs ?hting for turf!So now your ego might be bruised. You might have had a lot ofcon?ence before only to now need to re-think what you thought youknew. Shifting to a *social* position where you look to other peopleto make you feel con?ent again, to tell you that it's ok, is not thebest way.Possibly you should trust mathematics as more than just somethingother people tell you is correct.James Harris === > Now math history is full of people like David Ullrich, a guy with a> title, ?hting against some new idea. David Ullrich has a title?? Good grief. Why didn't somebody tell us this before? I hate making> these social errors through ignorance. (See .sig quote.) And what is his title, anyway? Sir David? Lord Ullrich? Baron> Okstate? David Ullrich is a math professor at Oklahoma State University, so at> a minimum he has the title: mathematician. I notice that you ran away from further discussions on the math. Have you yet found anyone in math literature you can cite who uses> non-uniqueness of polynomial factorization? Or have you now ?ured> out how not allowing members that are both integers and units not> equal to -1 or 1, can be a property used to de?e a ring? The point I want to make to you is that mathematics, no matter how> much modern mathematicians might be trying to make it such, is not> just some social activity with gangs ?hting for turf! So now your ego might be bruised. You might have had a lot of> con?ence before only to now need to re-think what you thought you> knew. Shifting to a *social* position where you look to other people> to make you feel con?ent again, to tell you that it's ok, is not the> best way. Possibly you should trust mathematics as more than just something> other people tell you is correct.> James HarrisJames, Hasn't it occurred to you that people have other things to do thanmeet your challenges? People have lives other than mathematics, you know.David Moran === | Have you yet found anyone in math literature you can cite who uses| non-uniqueness of polynomial factorization?A lot of the problem with these questions of yours is that people arefairly unlikely to use the terminology you're using, so even if we ?dexamples, they are unlikely to be saying exactly that they are studyingnon-uniqueness of polynomial factorization, for instance. This makesit hard to prove to you that my citations are genuine. But if you likeI can mention some.When mathematicians nowadays talk about non-uniqueness of factorization,they usually talk in terms of what is called the class group or moreformally the divisor class group. If you look in the index ofHartshorne's _Algebraic Geometry_ for divisor class group, you can seethat one of the subentries is for =0 <-> UFD, where UFD stands forunique factorization domain. The theorem referenced says that if A isa Noetherian ring, A is a unique factorization domain if and only if theclass group of Spec A is 0. That means the divisor class