So then the definition 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 definition is made in the first place, i.e. > why it is the most sensible definition, but maybe I will understand > it better when I work through some examples? > You could define 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* defined >> 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 defined like this?) We have f:G->L(V), where L(V) denotes the linear operators on V. We >> define >f*:G->L(V*) so that > = = p(v) defines the duality between p in V* and v in V. >This >> gives Why exactly do we want to define f*:G->L(V*) so that = ? > Why >> does this definition 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 finite-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 by p_j(e_i) = 1 if i = j, 0 otherwise. If A(g) is the matrix representing the linear map v -> gv with respect to the basis e_i of V, and the action of G on V* is defined 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 find if you determine the matrix B carefully from gf_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 simplifies 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. Van Quet: 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 first >used? If I mistake not, it was in Dedekind's definition of an ideal number, >as a kind of subset (a submodule) of the ring C. And what was the motive for >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 fields (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. But he understood at least some of the limitations of his arguments (clearly not all, or he wouldn't have called regular primes regular). 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. 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 first 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. Motivation is such a hard thing to fathom, except in the case of certain fiction 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 fields (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 into Kummer's ideal numbers, I was mostly disabused of this notion. See for example Edwards book _Fermat's Last Theorem: A Genetic Introduction to Number Theory_. The proof of quadratic reciprocity of Gauss and cubic reciprocity by Einsenstein were, by his own account, foremoest in his mind. > 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 that his 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 first 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 is quite forthcoming that he attacked T-S-W ->because<- he was interested in FLT. But Gauss proved quadratic reciprocity not out of any interest in FLT, and it was seen by him and others as a great achievement. Gauss was particularly interested in extending quadratic reciprocity to higher reciprocity laws, and worked on biquadratic, and gave to Eisenstein the job of dealing with cubic; and this is what Kummer's work did, extend them. In fact, reciprocity laws played no role in Kummer's work in FLT. What played a role was the whole scaffolding he had to build in order to prove the higher reciprocity laws: the cyclotomic number fields and unique factorization into ideal primes. But ->not<- the reciprocity laws. So I don't know on what you base the idea 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 > fiction writers who bare all. Kummer described his achievement in reciprocity laws as both foremost in his mind, and his most important work. He did not consider his proof of FLT for regular primes to be in the same league, by his own admission, from what I read. Of course if you have some sources, I would be most interested to know about them. Arturo Magidin, sans .sig ==== I believe it is likely the following COULD be true, yet it was figured very 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 k which 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 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 artificially difficult by specifying base 2 or 3? -- Dave Seaman Judge 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 artificially difficult by specifying base 2 or 3? Whay does anyone want to make things artificially difficult by working with expansions at all? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless 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 artificially difficult by specifying base 2 or 3? > Whay does anyone want to make things artificially difficult 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 definition of the Cantor set involves base-3 expansions. -- Dave Seaman Judge 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 artificially difficult by specifying base 2 or 3? > >> Whay does anyone want to make things artificially difficult 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 definition 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.html Needless 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 artificially difficult by specifying base 2 or 3? >Whay does anyone want to make things artificially difficult 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 definition 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 is simpler than those. -- Dave Seaman Judge 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 define 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. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless 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 define 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) -> C given 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 Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-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 with the number itself. >This is a neutral transformation, and it will always be possible. FSVO possible. It's possible to add a finite number of zeros to the left of a decimal representation of a natural number. However, that is not what you are asking for. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not 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 figured very 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 k which 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 same sense that we typically refer to our commonly-used number-system as base-10 and to binary as base-2. (Although some repliers seem to believe I am unaware of my less-than-literal use of the word base.) My particular definition of the term base-1 is not my own, yet I cannot recall where else I have seen it used in the sense I use it in my original post, but I have seen the term used this way in several different (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 2/3 I 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, we could subtract the 1 from that position, and add the string 1010000010... at the next position to get another (probably) representation of the same number. When adding two numbers, carries are funky. If you add 1 and 1 at some position, you get a result the has to be equivalent to a two (at that position). But two can't be represented by a finite string. So the carry propogates one digit to the left *and* infinitely 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 integers are 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.fi) 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 really represents 00000001 or 000001 or any number of turned off switches followed by some number of turned on switches. -- dgates@spamfreelinkline.com ==== I'm currently a 2nd semester (junior) math/physics double major student at a .. less-than-exemplary state school. Physics is the field I intend to work in, despite math being my first love. The plan I came up with at the beginning of my undergrad work was as follows: double major math/phys (physics primary, math secondary), masters in math, masters in physics, PhD physics. First, would a graduate degree in math be helpful for my occupation in physics (theoretical work as a professor is my hope)? 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 first answer is negative. Third, what schools would be good for my physics graduate work? I intend to, though it's not imperative, attend seperate universities for 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 first > answer is negative. One thing you might find interesting is that some schools that normally only give doctorates will allow for a masters if you are getting a doctorate there in another field. 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. -- --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 first > 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 _first_.. to ensure I understand the material I cover in physics.. I definitely don't want all three graduate degrees from the same school. but I guess it would be a possibility to get my masters in physics, then my MS in math afterwards.. to take advantage of such a thing ==== hi, i cannot answer your questions about the universities, because i personally feel that amount of learning and knowledge one aquires doesnot directly come from the university in which he is studying, but from the amount of efforts he puts into learning. There would certainly be exceptions if you are in a university where the professors are all noble laurites and have contributed major path breaking inventions to their field. In that case, you would get a lot of motivation, a lot of insight and a lot of knowledge from the university, bacause of the professors. Otherwise, all universities are same, with minor differences which does not matter. This is however only my view. The question: Will masters in math be helpful for theoritical work as a professor, my answer would be, you would never make a good professor if you dont know enough math. The best way to know enough math is to do your masters in math, atleast, taht is the disciplined way. But anyway, this is the conclusion: Good enough math is a must for everything more than a bachelor's degree. Good luck. 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 first > answer is negative. One thing you might find interesting is that some schools that normally only > give doctorates will allow for a masters if you are getting a doctorate > there in another field. 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 other people in this thread. To my mind, they are too dense and concise for someone who has no idea what Div, Gradient, and Curl really are. The Feynman discussions may be helpful -- I haven't checked -- but the best treatment of these topics I've ever seen (in words, pictures, and the essential math) is in the physics book I used 20+ years ago in college: Electricity and Magnetism by Edward M. Purcell Volume 2 of the Berkeley Physics Course ISBN: 0070049084 About $100 I don't know if the most recent version retains the treatment used 20 year ago, but the version I used 20 years back had an excellent, carefully drawn out, step-by-step approach to div, grad, and curl (as well as the foundations of E&M). Maybe someone else can comment on whether the current version of the text is still as strong. Steve O. ==== There is a formula to find distance between two points on plane. I want to learn if there is a formula to find 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. ==== >There is a formula to find distance between two points on plane. I >want to learn if there is a formula to find 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 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. ==== >There is a formula to find distance between two points on plane. I >want to learn if there is a formula to find 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 out as 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 to compute the distance. Exercise: Consider the 1 x 1 x 2 box stretching from the origin in 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 series binary) > 1/10 + 1/10^2 + ...+1/10^n = 1 - 1/10^n This is true for all (finite) values of n. (1 - 1/10^n) > 1 This is also true for all finite values of n (and transfinite values too) > Only when n achieves absolute infinity 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 infinity. No reaching of infinity is intended by this pristine phrase, so we are > justified in maintainig that no limit of any kind could ever be realised by > its use. Furthermore, if the 'tending' is towards something infinte short of absolute > infinity 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 definition of =. > 0.999... is an infinite 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 for repeated decimals, how to change them into the accurate fraction form ah.. and a simple search provides it.. define .999... as X 10X = 9.999... = 9 + .999... = 9 + X 10X = 9 + x 10X-X = 9 9X = 9 X = 1 .999... = 1 I notice now that it looks strange for .999... but for any other repeating decimal, it is more lucid... ==== Dear all, My problem is to construct some integer matrices of certain pattern to satisfy a matrix equation... There are so many unknowns, and there are also so many equations, (number of equations >> number of unknowns)... Since the required matrices are integer, I cannot use Newton, or Conjugate descent methods, etc... I have to do some search algorithm... but since there are so many unknowns, the search space is too large... So I am thinking of having an initial matrix, then do some change, see if the matrix equation(left hand side - right hand side) approximately more near zero and find the minimization point... But even this, still make a large search space when the size of matrix gets larger... moreover, another constraint is that the number of non-zero elements of this matrix should be as small as possible... Please tell me what kind of problem is this? I even don't know how to categorize my problem and where to find 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^2 How 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 birth to 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_2 for the next a, we obtain: a_4 = ((a_3 - a_2) / delta) + a_3 = (a_2 - a_1)(1/delta + 1/delta/delta) + a_2 continuing on, we find (summing over the geometric series in delta): a_infty = (a_2 - a_1)*(1/(delta - 1)) + a_2 It 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 find first the a_infty value for the parameter, and then the value for a specific value of a_n. This is 0198507232. So I think you could find the value of a for which period 2^14 results if 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 figure it out later, among those who figure 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 written in German-titled Die Herleitung der Grundgleichungen der Thermomechanik der Kontinua aus der statistischen Mechanik, 1955. Can somebody help me in finding out an English Translation of this paper. ==== 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 finding 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 think and starting x = .4 and that equation is repeated MANY times... 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... How can that equation give pseudo-random numbers!?!?!?!? It's really weird 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 behaviour in a nonlinear difference equation. Google gives this link which explains the behaviour of the equation quite 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 all sets 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 the functions mapping a set X into its subsets; the function f defined by f(x) = {x} does not exist automatically in this system. Otherwise, the paradox cannot be avoided. In NBG, there are classes with this property, as a proper class cannot be an element. A class S with the property you pose must be such. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University ==== > 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 variant NFU (New Foundations with Ur-elements) both prove the existence of a universal set. This set is not, however, defined 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 satisfies |P(s)|=|s|. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) 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 efficiently solvable problems are in > NP. What suggests it's difficult 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 specific problem, and all asymptotic classifications 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 find a solution which is O(n) but takes seven years to run, if you can find 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 > Pacific? 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 > Pacific? 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 I did a quick searh for Rhode Island maps, and came up with a map that shows Massachusetts, Connecticut and Rhode Island, with a boundary saying 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.jpg Anthony ==== >On Sat, 3 Jan 2004 00:22:48 +0000 (UTC), rob@trash.whim.org (Rob On 1 Jan 2004 10:27:56 -0800, denoncou@euclid.colorado.edu (Hugh >I have been having some difficulties with the following problem: >>It is from Royden (p127 #17). The space throughout is assumed to be >>[0,1], though I think any finite 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. Luckily 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. 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. Luckily 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.xxxx The same as 04/01/xxxx then. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== > If G is a finite 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 order 8 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 group of order 128 (the direct product of 7 cyclic groups of order 2). There are two other examples of order 64, which are also nonabelian semidirect products. http://www-gap.dcs.st-and.ac.uk/~gap/ is a wonderful package for finding examples. ==== Is there a concept called harmonic dynamics and if so where might I learn 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, fighting 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 believes your work here to be incorrect. Note for this to be the case does not require your work to actually be incorrect, merely for David Ullrich to believe it to be. If you could try to embrace this kind of attitude - allowing people to disagree with your mathematics (and you theirs) without ascribing negative motives to the disagreement - you would get a much better reception. Which in turn would mean that you could develop your ideas in collaboration with others, rather than constantly fighting them. Wouldn't that be much less frustrating? ==== >My research can be difficult to understand, so I thought I'd try out >yet another way of explaining it. Some of you may have figured out >that I test out explanations on Usenet for use elsewhere, to refine my >own understanding, or just in case someone out there might finally 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 significance. > >In that form it's hard to understand what follows next unless you pay >attention to what you have, specifically that cubic defining 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 define 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 superficially 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 >find 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 defined by a >cubic, which shows them to be algebraic integer functions, but >dividing by 49 *breaks* that symmetry, taking away the ability to find >some cubic to define 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 coefficients are impossible to >define in general. > If by impossible to define in general you mean that they cannot be > algebraic integers, I agree. That is because 7, 7, 1 is the > wrong factorization. > >You may find 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 stuck. > 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. [SNIP] James, now that the little diversion has been beaten to death concerning what the constant term of the polynomial: a^3 + A*a^2 + B, where the coefficients are not a function of a, is (hint: consider setting 'a' to zero) I apologize to Nora Baron for hijacking her thread. Please go back and read her post starting just above my [SNIP]. I think you'll find it instructive and I'd be interested in your comments (if you confine them to the math). KeithK > James Harris ==== >My research can be difficult to understand, so I thought I'd try > out >yet another way of explaining it. Some of you may have figured > out >that I test out explanations on Usenet for use elsewhere, to > refine > my >own understanding, or just in case someone out there might > finally > 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 significance. >In that form it's hard to understand what follows next unless you > pay >attention to what you have, specifically that cubic defining 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 define 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 superficially 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 >find 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 defined by a >cubic, which shows them to be algebraic integer functions, but >dividing by 49 *breaks* that symmetry, taking away the ability to > find >some cubic to define 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 coefficients are impossible > to >define in general. > If by impossible to define in general you mean that they > cannot be > algebraic integers, I agree. That is because 7, 7, 1 is the > wrong factorization. >You may find 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 stuck. > 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 find 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 coefficients 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 coefficients > 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 + B Actually x is still a variable, so it's just wishing to claim a single variable 'a'. 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? > where the coefficients 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 expression a constant term? Now then, back to x^2 + xy + y^2, what is the constant term to you? James Harris ==== >My research can be difficult to understand, so I thought I'd try > out >yet another way of explaining it. Some of you may have figured > out >that I test out explanations on Usenet for use elsewhere, to > refine > my >own understanding, or just in case someone out there might > finally > 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 significance. >In that form it's hard to understand what follows next unless you > pay >attention to what you have, specifically that cubic defining 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 define 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 superficially 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 >find 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 defined by a >cubic, which shows them to be algebraic integer functions, but >dividing by 49 *breaks* that symmetry, taking away the ability to > find >some cubic to define 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 coefficients are impossible > to >define in general. > If by impossible to define in general you mean that they > cannot be > algebraic integers, I agree. That is because 7, 7, 1 is the > wrong factorization. >You may find 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 stuck. > 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 find 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 coefficients 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 coefficients > 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 + 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 which one 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, as Lil 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 find a_1(x), a_2(x), a_3(x) in your development. where the coefficients 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 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: 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. 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 in a^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 that there are TWO variables in that expression? I will actually give them specifically to make sure you don't have wiggle 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 words monic 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 the Well, it's clear this poster isn't interested in adult communcation, but probably is yet ANOTHER poster playing on some stage in their own minds. 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 find a_1(x), a_2(x), a_3(x) > in your development. Regardless a^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 coefficients 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 have claimed 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 another pseudonym!!! After all, like that poster you're showing an odd recalcitrance at admitting even basic points like that a^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 expression x^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 ==== > [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 to stop replying to Keith K. Nora B. > After all, like that poster you're showing an odd recalcitrance at > admitting even basic points like that a^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 > expression x^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, fighting 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 figured out how not allowing members that are both integers and units not equal to -1 or 1, can be a property used to define 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 fighting for turf! So now your ego might be bruised. You might have had a lot of confidence 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 confident 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 Harris ==== > Now math history is full of people like David Ullrich, a guy with a > title, fighting 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 figured > out how not allowing members that are both integers and units not > equal to -1 or 1, can be a property used to define 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 fighting for turf! So now your ego might be bruised. You might have had a lot of > confidence 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 confident 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 Harris James, Hasn't it occurred to you that people have other things to do than meet 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 are fairly unlikely to use the terminology you're using, so even if we find examples, they are unlikely to be saying exactly that they are studying non-uniqueness of polynomial factorization, for instance. This makes it hard to prove to you that my citations are genuine. But if you like I 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 more formally the divisor class group. If you look in the index of Hartshorne's _Algebraic Geometry_ for divisor class group, you can see that one of the subentries is for =0 <-> UFD, where UFD stands for unique factorization domain. The theorem referenced says that if A is a Noetherian ring, A is a unique factorization domain if and only if the class group of Spec A is 0. That means the divisor class group of a space of the form Spec A for A Noetherian is studied only because of the possibility of A having non-unique factorization in it. Noetherianness is a property usually assumed in algebraic geometry (named after Emmy Noether). The next little twist is that mathematicians are much more interested in divisor class groups in general, rather than just the special case of a ring with polynomials in it. I just did a Google[tm] search for class group of an affine, and one of the hits is to an item titled New Public-Key Cryptosystem Using Divisor Class Groups. It refers to doing arithmetic in the divisor class group of an affine subring of K[X,Y]. Well, K[X,Y] is the ring of polynomials in variables X and Y with coefficients in K. So there you go. A cryptographic system based on non-unique factorization of polynomials of a particular kind. | Or have you now figured | out how not allowing members that are both integers and units not | equal to -1 or 1, can be a property used to define a ring? Again this tends to be referred to in other terms. If some integer n other than +-1 is a unit in a ring R, then some prime p is a unit in R. That p is a unit in R is equivalent to pR=R, where pR refers to the set of elements of the form p*r where r is an element of R. Write Z for the ring of integers. Then one uses some results from the theory of ideal divisors (as developed by Kummer and applied to Fermat's Last Theorem). pR is an ideal in R. If pR=R, then obviously there doesn't exist a prime ideal of R containing p, since the definition of prime ideal excludes the whole ring R. On the other hand, if pR is properly contained in R, then there exists a prime ideal containing pR, call it q. If q contained any integer m which was not a multiple of p, then q would also contain 1, because for relatively prime integers p and m there always exist u and v such that pu+mv=1. So the integers in q are just the multiples of p, which we can write as pZ. All of this is translated into the language of algebraic geometry. The set of prime ideals of a ring R is denoted by Spec R. The function which takes a prime ideal of R and intersects it with the integers Z to get a prime ideal in the integers is written Spec R -> Spec Z. (It's the only mapping from Spec R to Spec Z.) The condition you describe is equivalent, then, to the map Spec R->Spec Z being an *onto* map, i.e., each prime p in Spec Z is the intersection of some prime ideal q in R with the integers Z. So the phrase to search for is something like onto Spec Z. That gets at least some hits. The problem is that I don't see any of them as maps from Spec of something to Spec Z. There's an example in a book called _Arithmetic Geometry_ by Cornell and Silverman in the last chapter. There's a diagram on page 350 which is like this: Spec Z[X1,...,Xn]/(f_i) ---> Spec Z ^ | | Spec Z with a big = sign between the two Spec Z. It means the result we get at the end is supposed to be the same as what we started with. So given a prime ideal pZ in Spec Z, we get a prime ideal in Z[X1,...,Zn]/(f_i), which when intersected with the integers gives us pZ again. The ring Z[X1,...,Xn]/(f_i) consists of polynomials in variables X1,...,Xn with integer coefficients, subject to a set of polynomial conditions f1(X1,...,Xn)=f2(X1,...,Xn)=...=f_m(X1,...,Xn)=0. |James, Hasn't it occurred to you that people have other things to do than |meet your challenges? People have lives other than mathematics, you know. Probably the person who came closest to not having any life outside of doing mathematics was Erdos. I find it a little hard to imagine Erdos would have spent a lot of time answering questions on sci.math either! Keith Ramsay ==== > Now math history is full of people like David Ullrich, a guy with a >> title, fighting 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? I was formerly the Supreme Ruler of sci.math (I lost that title when I posted a wrong, wrong, wrong solution to a Putnam problem.) ************************ David C. Ullrich ==== > Now math history is full of people like David Ullrich, a guy >> with a title, fighting 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? I was formerly the Supreme Ruler of sci.math (I lost that title > when I posted a wrong, wrong, wrong solution to a Putnam problem.) Would that make you Supreme Ruler Emeritus, then, Your Consistency[?]? (I'm sorry about the form of address. Google wasn't any help at all. I had to wing it.) Jim Burns ==== Does anybody know a good optimization application which can handle with some milions of variables, using linear programming. It can be commercial. Please answer on lciesiel@poczta.onet.pl Haladir ==== can talk about the problem with conventional thinking on algebraic integers using someone else's example. In his post Decker claimed to mirror my argument using a quadratic instead of a cubic, where he has (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). Decker wanted to cast doubt on my reliance on using the constant terms to see how 7 divides both sides, by using an example where you can check at x=1, to see what the factors are, as then you have both a_1(1) and a_2(1) equal to sqrt(14). However, Decker may have naively thought he was refuting my argument, when he didn't follow through to the logical conclusion from his own analysis, which supports it. You see, the conventional thinking is that you can divide 7 from both sides and still be in the ring of algebraic integers, because algebraic integer are infinitely decomposable, so from a common sense perspective you might think that you can always find two algebraic integer factors of 7 to divide the two algebraic integer factors (5a_1(x) + 7) and (5a_2(x) + 7) on the left hand side. So assume there exists algebraic integer functions w_1(x) and w_2(x), such that w_1(x) w_2(x) = 7, and (5a_1(x)+ 7)/w_1(x) (5a_2(x) + 7) /w_2(x) = 25x^2 + 30 x + 2. The assumption is that you're still in the ring of algebraic integers with an algebraic integer x, so consider algebraic integer functions f_1(x), and f_2(x), such that f_1(x) f_2(x) = 25x^2 + 30x + 2, and f_1(x) = (5a_1(x) + 7)/w_1(x) and f_2(x) = (5a_2(x) + 7)/w_2(x). Checking at x=0, gives that f_1(0) = 1, and f_2(0) = 2. Now then, replacing them with f_1(0) = g_1(x) + 1, and f_2(0) = g_2(x) + 2, I have (g_1(x) + 1)(g_2(x) + 2) = 25x^2 + 30x + 2 and letting g_1(x) = 5 b_1(x), and g_2(x) = 5b_2(x), I have (5 b_1(x) + 1)(5 b_2(x) + 2) = 25x^2 + 30x + 2. Pushed to reply further on his original post by me, Decker actually went about calculating b_1(x), and his result is 2b_1(x)^2 - x b_1(x) + x^2 + x = 0. See: where A, B, C, and of course, 7 are algebraic integers. But divide both sides by 7, and because of that non-monic, you must have cases where DE = C, where D and E cannot be algebraic integers!!! Dividing 7 from both sides results in factors that are NOT in general algebraic integers as shown by Decker's own result, proving that there does NOT always exist a decomposition of 7 in the ring of algebraic integers that will do the job, since the quadratic he found isn't always reducible over Q, for an algebraic integer x. It is my hope that my use of Decker's own example, and his own analysis to get that non-monic quadratic might in some small way break through the logjam created by various posters who never back down no matter how often their positions are proven wrong. Want more? See my blog archives: If you wish to interpret Decker's result some different way, then you are free to try, but the gist of it is that you're pushed out of the ring of algebraic integers by dividing both sides by 7, defying conventional thinking that algebraic integer factors of 7 would always exist that can be divided from both of the algebraic integer factors on the left hand side. James Harris ==== > can talk about the problem with conventional thinking on algebraic > integers using someone else's example. In his post Decker claimed to mirror my argument using a quadratic > instead of a cubic, where he has (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). Decker wanted to cast doubt on my reliance on using the constant terms > to see how 7 divides both sides, by using an example where you can > check at x=1, to see what the factors are, as then you have both > a_1(1) and a_2(1) equal to sqrt(14). > Almost. The set {a_1(1), a_2(1)} is {sqrt(-14), -sqrt(-14)}. You can choose their values arbitrarily subject to this constraint. > However, Decker may have naively thought he was refuting my argument, > when he didn't follow through to the logical conclusion from his own > analysis, which supports it. You see, the conventional thinking is that you can divide 7 from both > sides and > still be in the ring of algebraic integers, because algebraic integer > are infinitely decomposable, so from a common sense perspective you > might think that you can always find two algebraic integer factors of > 7 to divide the two algebraic integer factors (5a_1(x) + 7) and > (5a_2(x) + 7) on the left hand side. > I never claimed that. > So assume there exists algebraic integer functions w_1(x) and w_2(x), > such that w_1(x) w_2(x) = 7, and (5a_1(x)+ 7)/w_1(x) (5a_2(x) + 7) /w_2(x) = 25x^2 + 30 x + 2. The assumption is that you're still in the ring of algebraic integers > with an algebraic integer x, so consider algebraic integer functions > f_1(x), and f_2(x), such that f_1(x) f_2(x) = 25x^2 + 30x + 2, and f_1(x) = (5a_1(x) + 7)/w_1(x) and f_2(x) = (5a_2(x) + 7)/w_2(x). Checking at x=0, gives that f_1(0) = 1, and f_2(0) = 2. Now then, replacing them with f_1(0) = g_1(x) + 1, and f_2(0) = g_2(x) + 2, I have (g_1(x) + 1)(g_2(x) + 2) = 25x^2 + 30x + 2 and letting g_1(x) = 5 b_1(x), and g_2(x) = 5b_2(x), I have (5 b_1(x) + 1)(5 b_2(x) + 2) = 25x^2 + 30x + 2. Pushed to reply further on his original post by me, Decker actually > went about calculating b_1(x), and his result is 2b_1(x)^2 - x b_1(x) + x^2 + x = 0. > Careful here. The functions I called b_1(x) and b_2(x) are not the ones you derived above. Mine are roots of an entirely different polynomial, in general, than yours are. In fact, there's no hope of knowing what polynomial your b's satisfy until you specify what the functions w_1 and w_2 are. > See: But 2b_1(x)^2 - x b_1(x) + x^2 + x = 0, is a non-monic, and not generally reducible over Q, with an > integer x, proving that b_1(x) is not in general an algebraic integer. > Exactly! You've gotten the point I was making. Good for you. > Now then, what conventional wisdom has clearly fallen? Well consider that the factorization I have looks something like AB = 7C where A, B, C, and of course, 7 are algebraic integers. But divide both sides by 7, and because of that non-monic, you must > have cases where DE = C, where D and E cannot be algebraic integers!!! > Actually, all I showed was that that you can't simultaneously satisfy the conditions you wanted, that there are two functions b_1(x), b_2(x) with: 1. (5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2 2. Both are roots of b^2 + C(x)b + (x^2 + x) 3. Both take algebraic integer values for each rational integer x. > Dividing 7 from both sides results in factors that are NOT in general > algebraic integers as shown by Decker's own result, proving that there > does NOT always exist a decomposition of 7 in the ring of algebraic > integers that will do the job, since the quadratic he found isn't > always reducible over Q, for an algebraic integer x. That hardly proves what you want. It shows that my two b functions can't be used to provide a factorization. My point, though, was that splitting 7 between the two factors as 7, 1 won't always work. You can't, for example, always divide one of the factors by 7 in P(x) = (5a_1(x) + 7)(5a_2(x) + 7) to get P(x)/7 = (5a_1(x)/7 + 1)(5a_2(x) + 7) although in the x = 1 case you can split 7 as sqrt(7), sqrt(7) since P(1)/7 = 57 = (5sqrt(-2) + sqrt(7))(-5sqrt(-2) + sqrt(7)). May I take it that we're in agreement about this? Rick ==== >can talk about the problem with conventional thinking on algebraic >integers using someone else's example. In his post Decker claimed to mirror my argument using a quadratic >instead of a cubic, where he has (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). Decker wanted to cast doubt on my reliance on using the constant terms >to see how 7 divides both sides, by using an example where you can >check at x=1, to see what the factors are, as then you have both >a_1(1) and a_2(1) equal to sqrt(14). However, Decker may have naively thought he was refuting my argument, >when he didn't follow through to the logical conclusion from his own >analysis, which supports it. You see, the conventional thinking is that you can divide 7 from both >sides and >still be in the ring of algebraic integers, because algebraic integer >are infinitely decomposable, What??? Who thinks that? The only person I know who thinks this thing you call the conventional thinking is _you_. >[...] If you wish to interpret Decker's result some different way, then you >are free to try, but the gist of it is that you're pushed out of the >ring of algebraic integers by dividing both sides by 7, That's right. >defying >conventional thinking that algebraic integer factors of 7 would always >exist that can be divided from both of the algebraic integer factors >on the left hand side. Again, what reason do you have for thinking that this thinking is conventional? As far as I can see from following the discussion the whole point is that you simply _can't_ do what you say defies conventional thinking - this would be a problem if it were in fact conventional, but it's not. >James Harris ************************ David C. Ullrich ==== snakes and snails and puppy dog tails ------------ ==== I am interested in know all the 14 groups of order 16, by mean fo it generators relation or by its multiplication tables. Reinaldo Giudici Universidad Sim.97n Bol.92var Venezuela ==== Reinaldo, Check out the Finite group page at Mathworld (Wolfram.com) http://mathworld.wolfram.com/FiniteGroup.html which lists small groups up to order 31. There is also the Small Groups Library, http://www-public.tu-bs.de:8080/~hubesche/small.html which is available with the GAP package. Will Orrick Indiana University > I am interested in know all the 14 groups of order 16, by mean fo it > generators relation or by its multiplication tables. > Reinaldo Giudici > Universidad Sim.97n Bol.92var > Venezuela ==== Well I just made a post talking about the reality of decomposition in algebraic integers, and then as usual I started thinking about my own post. After a while I started worrying about x=1, as for *that* value, you can clearly find a b_1(1) that's an algebraic integer, which is when I realized just what that meant. Basically Decker has given me a way to prove that a non-monic irreducible over Q can in fact have an algebraic integer root, and he has done a lot of the analysis himself! The outline is simple enough: Decker gave a quadratic which has the nice property of having a simple solution at x=1, which is probably why he chose it. I pressured him about what happens when 7 is divided off from a factorization that he also gave, and he came back with an analysis, where at first he'd dropped the lead coefficient of 2 when talking about the quadratic defining b_1(x), though he'd finished out solving it correctly. Well when I realized that the correct quadratic defining b_1(x) was non-monic, I immediately realized that Decker had offered a way to use *his* work to prove *my* point. And just now I realized that the full point is that at x=1, his quadratic is a non-monic irreducible over Q, but because his a's have an easy solution at x=1, it should be possible to actually find an algebraic integer solution for b_1(1). It just so happens it has another non-algebraic integer solution. Oh yeah, you can find the posts easily enough, to see all the actual equations. I will at least though give the quadratic that *Decker* found! 2b_1(x)^2 - x b_1(x) + x^2 + x = 0 And at x=1, it's 2b_1(1)^2 - b_1(1) + 2 = 0 which is irreducible over Q. James Harris ==== [snip] > Basically Decker has given me a way to prove that a non-monic > irreducible over Q can in fact have an algebraic integer root, and he > has done a lot of the analysis himself! Since you seem to believe that you can now prove that a non-monic polynomial, irreducible over Q, can have an algebraic integer root, you will be expected to provide that proof or an example. To date, your so-called 'proof' of FLT has been thoroughly discredited and you have repudiated your own prime counting function because of the alleged *inherent* ambiguity of the square root function (function, not operator), so this breakthrough may be your only remaining claim to fame (other than being an arrogant, ineducable blowhard).. OK, James, give us a non-monic polynomial, irreducible over Q, which has an algebraic integer root. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== And just now I realized that the full point is that at x=1, his > quadratic is a non-monic irreducible over Q, but because his a's have > an easy solution at x=1, it should be possible to actually find an > algebraic integer solution for b_1(1). It depends on what you want. If you want a factorization of the form (5b_1 + 1)(5b_2 + 2) = P(1)/7 = 57 with the b's being algebraic integers, then you could use b_1 = -4, b_2 = -1. On the other hand, if you want a factorization like (5b_1 + sqrt(7))(5b_2 + sqrt(7)) = 57 then you can use b_1 = sqrt(-2), b_2 = -sqrt(2) Rick ==== >Well I just made a post talking about the reality of decomposition in >algebraic integers, and then as usual I started thinking about my own >post. After a while I started worrying about x=1, as for *that* >value, you can clearly find a b_1(1) that's an algebraic integer, >which is when I realized just what that meant. Basically Decker has given me a way to prove that a non-monic >irreducible over Q can in fact have an algebraic integer root, Oh for heaven's sake, we're back to that? Why not just prove that there exists an even number which is not divisible by 2? That's a lot easier, and just as earthshattering. irreducible polynomial with rational coefficients has an algebraic integer root the _rational_ thing to do is to look for the _error_ in one's argument. >and he >has done a lot of the analysis himself! In this case you don't even have to look - people have found the error _for_ you! >The outline is simple enough: Decker gave a quadratic which has the >nice property of having a simple solution at x=1, which is probably >why he chose it. I pressured him about what happens when 7 is divided off from a >factorization that he also gave, and he came back with an analysis, >where at first he'd dropped the lead coefficient of 2 when talking >about the quadratic defining b_1(x), though he'd finished out solving >it correctly. Well when I realized that the correct quadratic defining b_1(x) was >non-monic, I immediately realized that Decker had offered a way to use >*his* work to prove *my* point. And just now I realized that the full point is that at x=1, his >quadratic is a non-monic irreducible over Q, but because his a's have >an easy solution at x=1, it should be possible to actually find an >algebraic integer solution for b_1(1). It just so happens it has another non-algebraic integer solution. Oh yeah, you can find the posts easily enough, to see all the actual >equations. I will at least though give the quadratic that *Decker* found! 2b_1(x)^2 - x b_1(x) + x^2 + x = 0 And at x=1, it's 2b_1(1)^2 - b_1(1) + 2 = 0 which is irreducible over Q. >James Harris ************************ David C. Ullrich ==== > Well I just made a post talking about the reality of decomposition in > algebraic integers, and then as usual I started thinking about my own > post. After a while I started worrying about x=1, as for *that* > value, you can clearly find a b_1(1) that's an algebraic integer, > which is when I realized just what that meant. Basically Decker has given me a way to prove that a non-monic > irreducible over Q can in fact have an algebraic integer root, and he > has done a lot of the analysis himself! The outline is simple enough: Decker gave a quadratic which has the > nice property of having a simple solution at x=1, which is probably > why he chose it. I pressured him about what happens when 7 is divided off from a > factorization that he also gave, and he came back with an analysis, > where at first he'd dropped the lead coefficient of 2 when talking > about the quadratic defining b_1(x), though he'd finished out solving > it correctly. He probably didn't give into your pressure. You can't pressure anyone into doing anything for you, WHO DO YOU THINK YOU ARE? Lately, you've been acting like you're above everyone else even moreso. Get a clue, James. Well when I realized that the correct quadratic defining b_1(x) was > non-monic, I immediately realized that Decker had offered a way to use > *his* work to prove *my* point. And just now I realized that the full point is that at x=1, his > quadratic is a non-monic irreducible over Q, but because his a's have > an easy solution at x=1, it should be possible to actually find an > algebraic integer solution for b_1(1). It just so happens it has another non-algebraic integer solution. Oh yeah, you can find the posts easily enough, to see all the actual > equations. I will at least though give the quadratic that *Decker* found! 2b_1(x)^2 - x b_1(x) + x^2 + x = 0 And at x=1, it's 2b_1(1)^2 - b_1(1) + 2 = 0 which is irreducible over Q. > James Harris David Moran ==== > Dear all, How to describe the following matrix operation formally? A big matrix, A, if we divide it into blocks > A=[A11 A12 A13; > A21 A22 A23] There A11, A12, A13, A21, A22, A23 are blocks... now I want to multiply each > block with a scalar C11, C12, C13, C21, C22, C23,(these C's are scalar > numbers) how to describe this operation mathematically and formally? Result=[A11*C11 A12*C12 A13*C13; > A21*C21 A22*C22 A23*C23] > Let 1 be the matrix of all ones having the same dimension as the blocks in A. Let C be the matrix of scalar values. First form the Kronecker product K = C * 1 Next take the Hadamard (aka Schur aka elementwise) product to get your Result R = K @ A Since the Hadamard product can be expresses in terms of the Kronecker product and the conventional matrix product as: K @ A = E . K * A . E^T Where E is a fixed rectangular matrix of dimension nxn^2 whose columns are the those of the identity matrix with n columns of 0s inserted between adjacent columns Putting it all together: R = E . C * 1 * A . E^T -Walala You're welcome, ~Glynne ==== Dear all, How to describe the following matrix operation formally? A big matrix, A, if we divide it into blocks > A=[A11 A12 A13; > A21 A22 A23] There A11, A12, A13, A21, A22, A23 are blocks... now I want to multiply > each > block with a scalar C11, C12, C13, C21, C22, C23,(these C's are scalar > numbers) how to describe this operation mathematically and formally? Result=[A11*C11 A12*C12 A13*C13; > A21*C21 A22*C22 A23*C23] > Let 1 be the matrix of all ones having the same dimension as the blocks in > A. > Let C be the matrix of scalar values. First form the Kronecker product > K = C * 1 Next take the Hadamard (aka Schur aka elementwise) product to get your > Result > R = K @ A > Since the Hadamard product can be expresses in terms of the Kronecker > product and the conventional matrix product as: > K @ A = E . K * A . E^T Where E is a fixed rectangular matrix of dimension nxn^2 > whose columns are the those of the identity matrix with n columns of 0s > inserted between adjacent columns Putting it all together: > R = E . C * 1 * A . E^T -Walala You're welcome, > ~Glynne Dear Glynne, man, You come at good time, thank you so much for your help. I previously just had a side-problem about element wise matrix, now it is by chance I know from your posting that it is called Hadamard product as formal name... would you mind give me some pointers about this Hadamard product thing? I need to digest it with your posting to see how your complex mathematical formations work... -Walala ==== > This is a special case of a Kronecker Product. Ask Google. Markus Dear all, How to describe the following matrix operation formally? A big matrix, A, if we divide it into blocks > A=[A11 A12 A13; > A21 A22 A23] There A11, A12, A13, A21, A22, A23 are blocks... now I want to multiply each > block with a scalar C11, C12, C13, C21, C22, C23,(these C's are scalar > numbers) how to describe this operation mathematically and formally? Result=[A11*C11 A12*C12 A13*C13; > A21*C21 A22*C22 A23*C23] > -Walala Dear Markus, Kronecker product... Can you explain more? The result is the same size as A, (since the C's are scalars...) If it is a Kronecker product, the size of the result should be size(A)*size(C)... So this is not a Kronecker product... Best, -Walala ==== ok, it isn't a special case of the Kronecker product, nor is it a generalization of it - I just was too hasty. Unfortunately I don't know another interpretation. Markus > >>This is a special case of a Kronecker Product. Ask Google. >>Markus >Dear all, How to describe the following matrix operation formally? A big matrix, A, if we divide it into blocks >A=[A11 A12 A13; > A21 A22 A23] There A11, A12, A13, A21, A22, A23 are blocks... now I want to multiply each > >block with a scalar C11, C12, C13, C21, C22, C23,(these C's are scalar >numbers) how to describe this operation mathematically and formally? Result=[A11*C11 A12*C12 A13*C13; > A21*C21 A22*C22 A23*C23] >-Walala > Dear Markus, Kronecker product... Can you explain more? The result is the same size as A, (since the C's are scalars...) If it is a Kronecker product, the size of the result should be > size(A)*size(C)... So this is not a Kronecker product... Best, -Walala ==== you are right - I was too hasty. It is a Kronecker product only if all Aij are equal, or if all Cij are equal (and possibly are matrices instead of scalars). BTW: I interpreted scalars as 1x1 matrices. Let's call it the Walala-product. Markus > >>This is a special case of a Kronecker Product. Ask Google. >>Markus >Dear all, How to describe the following matrix operation formally? A big matrix, A, if we divide it into blocks >A=[A11 A12 A13; > A21 A22 A23] There A11, A12, A13, A21, A22, A23 are blocks... now I want to multiply each > >block with a scalar C11, C12, C13, C21, C22, C23,(these C's are scalar >numbers) how to describe this operation mathematically and formally? Result=[A11*C11 A12*C12 A13*C13; > A21*C21 A22*C22 A23*C23] >-Walala > Dear Markus, Kronecker product... Can you explain more? The result is the same size as A, (since the C's are scalars...) If it is a Kronecker product, the size of the result should be > size(A)*size(C)... So this is not a Kronecker product... Best, -Walala ==== Oops, wrong once more. Better I don't make any more statements about the Walala product. you are right - I was too hasty. It is a Kronecker product only > if all Aij are equal, or if all Cij are equal (and possibly are > matrices instead of scalars). BTW: I interpreted scalars as 1x1 > matrices. Let's call it the Walala-product. Markus > This is a special case of a Kronecker Product. Ask Google. Markus >> Dear all, >> How to describe the following matrix operation formally? >> A big matrix, A, if we divide it into blocks >> A=[A11 A12 A13; >> A21 A22 A23] >> There A11, A12, A13, A21, A22, A23 are blocks... now I want to multiply >> each >> block with a scalar C11, C12, C13, C21, C22, C23,(these C's are scalar >> numbers) how to describe this operation mathematically and formally? >> Result=[A11*C11 A12*C12 A13*C13; >> A21*C21 A22*C22 A23*C23] >> -Walala >> Dear Markus, >> Kronecker product... Can you explain more? >> The result is the same size as A, (since the C's are scalars...) >> If it is a Kronecker product, the size of the result should be >> size(A)*size(C)... >> So this is not a Kronecker product... >> Best, >> -Walala >> ==== I am trying to solve a problem (of personal interest), part of which involves converting a series expansion back to its analytic function form. In case I am not being clear/precise enough I present an example: The infinite series: S = 1 + x + x^2 + x^3 + ... (*1) can also be expressed in the form S = 1/(1-x) (*2) by noticing that S is the infinite geometric series. I want to be able to express an infinite series expansion such as (*1) as a finite analytic function (involving no integrals either) such as (*2) . I know very little in this field (if it is a field?), and I realise immeadiately that it cannot always be possible to do what I am asking. However, what I would like to know is if there is a general rule which tells you whether a given function defined as an infinte series can be expressed as a finite analytic function? For example, if the given series converges for all real x then it can be expressed as a finite analytic function? (A false example I think, but I give it to indicate a general idea of what I am looking for) Further to this, is there any literature (preferably electronic ie. accessible on the internet) on this subject which anyone can direct me towards? I havent been able to locate anything concerning this online, possibly because I am using the wrong key words in google. Is there a name for the process of converting an infinite series expansion into a finite analytic function? Finally, is there any software available which attempts to convert a defined infinite series expansion back to a finite analytic function form? Any answers, responses or contributions would be greatly, greatly appreciated. Dave ==== >I am trying to solve a problem (of personal interest), part of which >involves converting a series expansion back to its analytic function >form. In case I am not being clear/precise enough I present an >example: The infinite series: S = 1 + x + x^2 + x^3 + ... (*1) can also be expressed in the form S = 1/(1-x) (*2) by noticing that S is the infinite geometric series. I want to be able to express an infinite series expansion such as (*1) >as a finite analytic function (involving no integrals either) such as >(*2) . I know very little in this field (if it is a field?), and I realise >immeadiately that it cannot always be possible to do what I am asking. However, what I would like to know is if there is a general rule which >tells you whether a given function defined as an infinte series can be >expressed as a finite analytic function? For example, if the given series converges for all real x then it can >be expressed as a finite analytic function? (A false example I think, >but I give it to indicate a general idea of what I am looking for) Further to this, is there any literature (preferably electronic ie. >accessible on the internet) on this subject which anyone can direct me >towards? I havent been able to locate anything concerning this online, >possibly because I am using the wrong key words in google. Is there a >name for the process of converting an infinite series expansion into a >finite analytic function? Finally, is there any software available which attempts to convert a >defined infinite series expansion back to a finite analytic function >form? Any answers, responses or contributions would be greatly, greatly >appreciated. > Try googling: The Book A=B. Not exactly what you want, but it may help a bit. rich ==== I am trying to solve a problem (of personal interest), part of which >involves converting a series expansion back to its analytic function >form. In case I am not being clear/precise enough I present an >example: The infinite series: S = 1 + x + x^2 + x^3 + ... (*1) can also be expressed in the form S = 1/(1-x) (*2) by noticing that S is the infinite geometric series. I want to be able to express an infinite series expansion such as (*1) >as a finite analytic function (involving no integrals either) such as >(*2) . I know very little in this field (if it is a field?), and I realise >immeadiately that it cannot always be possible to do what I am asking. It's _not_ possible in general. I doubt that anyone's going to be able to help you with your problem unless you say what the power series _is_. (Well, in a mathematical sense if you know the power series then you _have_ your analytic function. A better way to state what you're trying to do is this: you have an analytic function given by a power series and you want to convert it to closed form.) >However, what I would like to know is if there is a general rule which >tells you whether a given function defined as an infinte series can be >expressed as a finite analytic function? For example, if the given series converges for all real x then it can >be expressed as a finite analytic function? No. >(A false example I think, >but I give it to indicate a general idea of what I am looking for) Further to this, is there any literature (preferably electronic ie. >accessible on the internet) on this subject which anyone can direct me >towards? I havent been able to locate anything concerning this online, >possibly because I am using the wrong key words in google. Is there a >name for the process of converting an infinite series expansion into a >finite analytic function? Finally, is there any software available which attempts to convert a >defined infinite series expansion back to a finite analytic function >form? I'm not certain, and I don't know how well they accomplish this in general, but I believe packages like Maple and Mathematica will attempt this. >Any answers, responses or contributions would be greatly, greatly >appreciated. >Dave ************************ David C. Ullrich ==== I am trying to solve a problem (of personal interest), part of which > involves converting a series expansion back to its analytic function > form. In case I am not being clear/precise enough I present an > example: The infinite series: S = 1 + x + x^2 + x^3 + ... (*1) can also be expressed in the form S = 1/(1-x) (*2) by noticing that S is the infinite geometric series. I want to be able to express an infinite series expansion such as (*1) > as a finite analytic function (involving no integrals either) such as > (*2) . I know very little in this field (if it is a field?), and I realise > immeadiately that it cannot always be possible to do what I am asking. However, what I would like to know is if there is a general rule which > tells you whether a given function defined as an infinte series can be > expressed as a finite analytic function? For example, if the given series converges for all real x then it can > be expressed as a finite analytic function? (A false example I think, > but I give it to indicate a general idea of what I am looking for) Further to this, is there any literature (preferably electronic ie. > accessible on the internet) on this subject which anyone can direct me > towards? I havent been able to locate anything concerning this online, > possibly because I am using the wrong key words in google. Is there a > name for the process of converting an infinite series expansion into a > finite analytic function? Finally, is there any software available which attempts to convert a > defined infinite series expansion back to a finite analytic function > form? Any answers, responses or contributions would be greatly, greatly > appreciated. > ****************************************** The answer to your question depends on what you mean by finite analytic function. Certainly, if an infinite power series has a positive radius of convergence, then it represents an analytic function. However, this analytic function may not be a finite combination of nice functions such as polynomials, trig. functions, exponentials, and logarithmic functions. _________________________________________________________ Eric J. Wingler (wingler@math.ysu.edu) Dept. of Mathematics and Statistics Youngstown State University One University Plaza Youngstown, OH 44555-0001 330-941-1817 ==== > 1) In my textbook there is an unproved proposition (the proof is said > to be obvious) which states that function f: X subset R -> R is > differentiable at point a in X if and only if there exist A = lim_{x > -> a+} ((f(x) ? f(a)) /(x-a)), B = lim_{x -> a-} ((f(x) ? f(a)) > /(x-a)) and A = B. I do not understand this, given that the textbook > definition of the derivative of function f at point a does not require > a to be interior. Does the statement of the theorem imply that the > condition of point a being interior is implied in the definition of > the derivative or I miss something? In short, what must the derivative > definition look like for this theorem to hold true? 2) Which conditions are needed so that differentiability at a point > would imply differentiability on some interval embracing the point? 3) What are the most common and accepted precise definitions of the > conditions of a function to be continuous at a point and to be > differentiable at a point? Consider identity function f: [0,1] - [0,1]. Is it continuous at 0? Is it differentiable at 0? (according to > the definitions you provide). > given I've only been through the standard calc courses and not touched topology, the derivative was introduced as not existing on the boundaries of a function's domain. for the identity function, it is continuous, but not differentiable at that point I think the statement your book made makes perfect sense, the derivative of a point is the limit of the slope from both sides, by the definition I've used in several courses (CalcI,II,III,DE).. therefore it must match, or the derivative is non-existant. the limit as f(x) approaches zero for the function f(x) given by x from (0,arbitrary number larger than zero] is non-existant from the left, and 0 from the right, and is therefore non-existant. derivatives are simply limit definitions, so the same applies > 2) Which conditions are needed so that differentiability at a point > would imply differentiability on some interval embracing the point? smooth (and thus also continuous) in that interval.. open interval. closed cannot be verified without checking if the function is still smooth infinitessimally (sp) on either side of the interval, so open works more easily for quick verification ==== >> [...] > 2) Which conditions are needed so that differentiability at a point >> would imply differentiability on some interval embracing the point? smooth (and thus also continuous) in that interval.. Huh? What do you mean by smooth? The word means different things in different contexts - one of the usual meanings is infinitely differentiable. If that's what you mean by smooth then you're saying that if we know the function is infinitely differentiable in an interval we can use that to verify it's differentiable - true, but it seems like it can't be what you meant. But I can't figure out what you _do_ mean here. >open interval. >closed cannot be verified without checking if the function is still >smooth infinitessimally (sp) on either side of the interval, so open >works more easily for quick verification Huh? ************************ David C. Ullrich ==== > 1) In my textbook there is an unproved proposition (the proof is said >> to be obvious) which states that function f: X subset R -> R is >> differentiable at point a in X if and only if there exist A = lim_{x >> -> a+} ((f(x) ö f(a)) /(x-a)), B = lim_{x -> a-} ((f(x) ö f(a)) >> /(x-a)) and A = B. I do not understand this, given that the textbook >> definition of the derivative of function f at point a does not require >> a to be interior. Does the statement of the theorem imply that the >> condition of point a being interior is implied in the definition of >> the derivative or I miss something? In short, what must the derivative >> definition look like for this theorem to hold true? 2) Which conditions are needed so that differentiability at a point >> would imply differentiability on some interval embracing the point? 3) What are the most common and accepted precise definitions of the >> conditions of a function to be continuous at a point and to be >> differentiable at a point? Consider identity function f: [0,1] -> [0,1]. Is it continuous at 0? Is it differentiable at 0? (according to >> the definitions you provide). >> Consider X as the set of rationals, then X does not have any interior >points (as a subset of R), but functions from X to R can still have >derivatives. According to whom? Where I come from functions have derivatives at interior points of their domains. >On the other hand, The function f(x) = sqrt(x^3) from the non-negative >reals to the reals clearly has a derivative at x = 0. This is not clear to me. >So that the if-and-only-if fails, at least for many defintions of >derivative. The precise wording of the definition of derivative in your text may be >critical to the issue. ************************ David C. Ullrich windows-nt) Cancel-Lock: sha1:dezzbmX+PEpIP42M9BapG5IBIsg= ==== >> Consider X as the set of rationals, then X does not have any >> interior points (as a subset of R), but functions from X to R can >> still have derivatives. According to whom? Where I come from functions have derivatives > at interior points of their domains. The rationals are a perfectly good metric space in their own right. Nothing stops one from trying to define continuity and differentiability on that space. (Virgil appears not to have noticed that the rationals can be regarded as a space by itself; there are indeed interior points.) OTOH, I've never seen anybody bother with it--and the loss of least upper bounds means that lots of interesting theorems will fail. Len. ==== Consider X as the set of rationals, then X does not have any > interior points (as a subset of R), but functions from X to R can > still have derivatives. >> According to whom? Where I come from functions have derivatives >> at interior points of their domains. The rationals are a perfectly good metric space in their own >right. Nothing stops one from trying to define continuity and >differentiability on that space. The original question was not about what definitions _can_ be made, the question was about what the standard definitions _are_. >(Virgil appears not to have noticed >that the rationals can be regarded as a space by itself; there are >indeed interior points.) OTOH, I've never seen anybody bother with it So evidently you don't disagree with what I said about what the definitions _are_. So I'm missing your point again. >--and the loss of least >upper bounds means that lots of interesting theorems will fail. Len. ************************ David C. Ullrich windows-nt) Cancel-Lock: sha1:axDucMXVb5vp/iUAMUZ5ymM1H2w= ==== >> The rationals are a perfectly good metric space in their own >> right. Nothing stops one from trying to define continuity and >> differentiability on that space. The original question was not about what definitions _can_ be > made, the question was about what the standard definitions _are_. The previous poster quibbled that differentiability over Q can't be defined on interior points, because Q has no interior points. He was confused. >> OTOH, I've never seen anybody bother with it So evidently you don't disagree with what I said about what the > definitions _are_. So I'm missing your point again. You seem convinced that I'm arguing with you. The statement means neither more nor less than itself. Here it is again, in greater generality: One can define difference quotients for real-valued functions over any metric space one wishes. Many metric spaces, such as manifolds, are handled in this way. In particular, the rationals could also be handled in this way, but nobody bothers: the result looks exactly like real analysis, except that lots of existence proofs fail. In general, completeness is a prerequisite to developing an interesting calculus. This observation might relieve the previous poster's confusion about Q having no interior points. Len. ==== The rationals are a perfectly good metric space in their own > right. Nothing stops one from trying to define continuity and > differentiability on that space. >> The original question was not about what definitions _can_ be >> made, the question was about what the standard definitions _are_. The previous poster quibbled that differentiability over Q can't be >defined on interior points, because Q has no interior points. He was >confused. OTOH, I've never seen anybody bother with it >> So evidently you don't disagree with what I said about what the >> definitions _are_. So I'm missing your point again. You seem convinced that I'm arguing with you. The statement means >neither more nor less than itself. Oh. Ok, you weren't arguing. I do have some arguments with your comments below. >Here it is again, in greater >generality: One can define difference quotients for real-valued functions over > any metric space one wishes. I don't see how, unless when X is a metric space and f : X -> R you propose to define the derivative as the limit of (f(x) - f(y)) / d(x,y) (y -> x). One could define that to be the derivative, but then for example if we define f : R -> R by f(x) = x then f no longer has a derivative... If you had some other definition in mind what was it? > Many metric spaces, such as manifolds, > are handled in this way. ??? Differentiation of functions defined on manifolds uses a lot more than just the metric-space structure of the manifold. >In particular, the rationals could also be > handled in this way, but nobody bothers: the result looks exactly > like real analysis, except that lots of existence proofs fail. In > general, completeness is a prerequisite to developing an interesting > calculus. The rationals have the advantage of being a subset of R, so we could talk about the limit of (f(x) - f(y)) / (x-y) as usual. >This observation might relieve the previous poster's confusion about >Q having no interior points. Len. ************************ David C. Ullrich windows-nt) Cancel-Lock: sha1:Wl4kNBqbDxkceOkseLg4Ad30vp8= ==== >> One can define difference quotients for real-valued functions over >> any metric space one wishes. I don't see how, unless when X is a metric space and f : X -> R you > propose to define the derivative as the limit of (f(x) - f(y)) / d(x,y) (y -> x). Got it in one. But dagnabbit--you're correct: lack of a sense of direction means that the derivative will not be defined. Only the absolute value of the derivative can be defined in this way. > The rationals have the advantage of being a subset of R, so we > could talk about the limit of (f(x) - f(y)) / (x-y) as usual. More generally, for my statement to make sense, we also need a sense of direction. I.e., a vector space or decently-behaved manifold. You are quite right. Len. windows-nt) Cancel-Lock: sha1:HEW+njh2nIB8AyUnPU6B3RcGyyQ= ==== More generally, for my statement to make sense, we also need a sense > of direction. I.e., a vector space or decently-behaved manifold. You > are quite right. Or, of course, a metric space that happens also to be an ordered space... Len. ==== 1) In my textbook there is an unproved proposition (the proof is said >> to be obvious) which states that function f: X subset R -> R is >> differentiable at point a in X if and only if there exist A = lim_{x >> -> a+} ((f(x) ö f(a)) /(x-a)), B = lim_{x -> a-} ((f(x) ö f(a)) >> /(x-a)) and A = B. I do not understand this, given that the textbook >> definition of the derivative of function f at point a does not require >> a to be interior. Does the statement of the theorem imply that the >> condition of point a being interior is implied in the definition of >> the derivative or I miss something? In short, what must the derivative >> definition look like for this theorem to hold true? 2) Which conditions are needed so that differentiability at a point >> would imply differentiability on some interval embracing the point? 3) What are the most common and accepted precise definitions of the >> conditions of a function to be continuous at a point and to be >> differentiable at a point? Consider identity function f: [0,1] -> [0,1]. Is it continuous at 0? Is it differentiable at 0? (according to >> the definitions you provide). >> Consider X as the set of rationals, then X does not have any interior >points (as a subset of R), but functions from X to R can still have >derivatives. According to whom? Where I come from functions have derivatives > at interior points of their domains. But using the topology on Q induced by R, all points of Q are interior points. Or am I missing something? > >On the other hand, The function f(x) = sqrt(x^3) from the non-negative >reals to the reals clearly has a derivative at x = 0. This is not clear to me. D = Dom(f) is the set of non-negative reals, [0, +oo). In the topology on D induced by R is 0 an interior point of D or not? My topology is a bit rusty. In any case, f(x) = sqrt(x^3) has a one-sided derivative at x = 0. > >So that the if-and-only-if fails, at least for many defintions of >derivative. The precise wording of the definition of derivative in your text may be >critical to the issue. > ************************ David C. Ullrich ==== 1) In my textbook there is an unproved proposition (the proof is said > to be obvious) which states that function f: X subset R -> R is > differentiable at point a in X if and only if there exist A = lim_{x > -> a+} ((f(x) ö f(a)) /(x-a)), B = lim_{x -> a-} ((f(x) ö f(a)) > /(x-a)) and A = B. I do not understand this, given that the textbook > definition of the derivative of function f at point a does not require > a to be interior. Does the statement of the theorem imply that the > condition of point a being interior is implied in the definition of > the derivative or I miss something? In short, what must the derivative > definition look like for this theorem to hold true? 2) Which conditions are needed so that differentiability at a point > would imply differentiability on some interval embracing the point? 3) What are the most common and accepted precise definitions of the > conditions of a function to be continuous at a point and to be > differentiable at a point? Consider identity function f: [0,1] - [0,1]. Is it continuous at 0? Is it differentiable at 0? (according to > the definitions you provide). > >>Consider X as the set of rationals, then X does not have any interior >>points (as a subset of R), but functions from X to R can still have >>derivatives. According to whom? Where I come from functions have derivatives >> at interior points of their domains. But using the topology on Q induced by R, all points of Q are interior >points. Or am I missing something? An interior point is something that a subset of a topological space has - if you change the space the same point can be or not be an interior point of the same set. I was talking about interior points of X _as_ a subset of R. And no, actually I don't believe that you didn't understand that that was what I meant - I don't think you were missing anything. More to the point: can you give us a _reference_ where the author talks about derivatives of functions defined on the rationals? >>On the other hand, The function f(x) = sqrt(x^3) from the non-negative >>reals to the reals clearly has a derivative at x = 0. This is not clear to me. D = Dom(f) is the set of non-negative reals, [0, +oo). In the topology >on D induced by R is 0 an interior point of D or not? My topology is a >bit rusty. In any case, f(x) = sqrt(x^3) has a one-sided derivative at x = 0. Yes it does. What it doesn't have is a derivative. >>So that the if-and-only-if fails, at least for many defintions of >>derivative. >>The precise wording of the definition of derivative in your text may be >>critical to the issue. >> ************************ David C. Ullrich ************************ David C. Ullrich ==== On 3 Jan 2004 15:23:00 -0800, korovyev@rambler.ru (Alexander Korovyev) >1) In my textbook there is an unproved proposition (the proof is said >to be obvious) which states that function f: X subset R -> R is >differentiable at point a in X if and only if there exist A = lim_{x >-> a+} ((f(x) ö f(a)) /(x-a)), B = lim_{x -> a-} ((f(x) ö f(a)) >/(x-a)) and A = B. I do not understand this, given that the textbook >definition of the derivative of function f at point a does not require >a to be interior. Does the statement of the theorem imply that the >condition of point a being interior is implied in the definition of >the derivative or I miss something? In short, what must the derivative >definition look like for this theorem to hold true? I dn't know what your textbook says - possibly the author disagrees with me, possibly there's a word left out that he meant to include, possilby you misunderstood some detail. But _I_ would certainly say that for f to be differentiable at a point that point must be an interior point of the domain of f. (Or to put the same thing differently: I would only speak of functions defined in _open sets_ as being differentiable or not.) >2) Which conditions are needed so that differentiability at a point >would imply differentiability on some interval embracing the point? I really don't think there are any. A function can be very strange in a heighborhood of a point and still be differentiable at that point; to make it nice in a neighborhood you're going to have to assume it's nice in a neighborhood. >3) What are the most common and accepted precise definitions of the >conditions of a function to be continuous at a point and to be >differentiable at a point? As has come up recently in another thread, calculus books and mathematicians seem to have different definitions here. If you're trying to understand what it says in the book you need to use the definitions in the book, not the most common definitions. But regarding the definitions that I think are standard among mathematicians: >Consider identity function f: [0,1] -[0,1]. Is it continuous at 0? Yes. >Is it differentiable at 0? (according to >the definitions you provide). No. ************************ David C. Ullrich Cancel-Lock: sha1:3KYH79l0QWqfV2kSzM3WqhNAIyg= ==== >> I do not understand this, given that the textbook definition of the >> derivative of function f at point a does not require a to be >> interior... I dn't know what your textbook says - possibly the author disagrees > with me, possibly there's a word left out that he meant to include, > possilby you misunderstood some detail. But _I_ would certainly say > that for f to be differentiable at a point that point must be an > interior point of the domain of f. Same here. But some textbooks apply the terms left differentiable and right differentiable to the cases that only the left- or right- handed limits exist. > (Or to put the same thing differently: I would only speak of > functions defined in _open sets_ as being differentiable or not.) In boundary value problems, one is also interested in whether functions and their derivatives can be continuously extended to the boundary. One oddity introduced by this is that the geometry of the boundary plays a part as well. Some PDE researchers ignore boudary behavior almost entirely, while others specialize in studying it. Len. ==== I do not understand this, given that the textbook definition of the > derivative of function f at point a does not require a to be > interior... >> I dn't know what your textbook says - possibly the author disagrees >> with me, possibly there's a word left out that he meant to include, >> possilby you misunderstood some detail. But _I_ would certainly say >> that for f to be differentiable at a point that point must be an >> interior point of the domain of f. Same here. But some textbooks apply the terms left differentiable >and right differentiable to the cases that only the left- or right- >handed limits exist. > (Or to put the same thing differently: I would only speak of >> functions defined in _open sets_ as being differentiable or not.) In boundary value problems, one is also interested in whether >functions and their derivatives can be continuously extended to the >boundary. Yes. And one certainly does not say that f is differentiable at a boundary point just because the derivative has a limit at that point, so I don't see your point. >One oddity introduced by this is that the geometry of the >boundary plays a part as well. Some PDE researchers ignore boudary >behavior almost entirely, while others specialize in studying it. Len. ************************ David C. Ullrich windows-nt) Cancel-Lock: sha1:mICMwPTcMKUA5LxYIDyxKRNbwmk= ==== >> In boundary value problems, one is also interested in whether >> functions and their derivatives can be continuously extended to the >> boundary. Yes. And one certainly does not say that f is differentiable at a > boundary point just because the derivative has a limit at that > point, so I don't see your point. The statement above is its own point. Do you believe I'm arguing with you in some way? If so, why? In casual speech, Sobolev spaces on closed regions are still referred to as differentiable functions on their regions. In practice, I haven't seen anyone fussing over the distinction that these functions are technically differentiable only on interior points. In particular, some proofs involve convergent difference quotients at boundary points, and the prof didn't stop to say, These limits of difference quotients are technically not derivatives... Len. ==== In boundary value problems, one is also interested in whether > functions and their derivatives can be continuously extended to the > boundary. >> Yes. And one certainly does not say that f is differentiable at a >> boundary point just because the derivative has a limit at that >> point, so I don't see your point. The statement above is its own point. Do you believe I'm arguing with >you in some way? If so, why? In casual speech, Sobolev spaces on closed regions are still referred >to as differentiable functions on their regions. I guess. >In practice, I >haven't seen anyone fussing over the distinction that these functions >are technically differentiable only on interior points. In particular, >some proofs involve convergent difference quotients at boundary >points, and the prof didn't stop to say, These limits of difference >quotients are technically not derivatives... Hmm. Where I come from this sort of quibbling is a large part of the game: If the boundary is smooth (for example if the boundary is the unit circle; I'm still stuck on that case) then one can regard the function on the boundary as having a derivative _as_ a function defined on a manifold (eg if the region is the unit disk one can consider whether f(exp(it)) is a differentiable function of t); then an interesting question is when is it true that if the derivative has a limit at a boundary point (or some analogous condition holds) it follows that the boundary function is actually differentiable at that point, and conversely. >Len. ************************ David C. Ullrich windows-nt) Cancel-Lock: sha1:6EInajcEC9C92N8D/k1BF3Pm3yA= ==== >> ...some proofs involve convergent difference quotients at boundary >> points, and the prof didn't stop to say, These limits of >> difference quotients are technically not derivatives... Hmm. Where I come from this sort of quibbling is a large part of > the game: If the boundary is smooth (for example if the boundary > is the unit circle; I'm still stuck on that case) then one can > regard the function on the boundary as having a derivative _as_ > a function defined on a manifold (eg if the region is the unit > disk one can consider whether f(exp(it)) is a differentiable > function of t); then an interesting question is when is it true > that if the derivative has a limit at a boundary point (or some > analogous condition holds) it follows that the boundary function > is actually differentiable at that point, and conversely. That is an interesting question. I see no a priori reason that the restriction of the extension to the boundary manifold should necessarily be differentiable in that context. Boundary geometry would certainly play a part. In particular, in my remarks above it is actually assumed that the difference quotient is taken only over interior points, as they approach a particular boundary point. This is even worse than left- and right-differentiability (so called), because we disqualify honest to goodness points in the domain. Professor Verchota at Syracuse University specializes in scale-invariant boundary-value estimates for PDEs. For him, smooth boundaries are bad: invoking smoothness amounts to assuming that the boundary is a straight line. He is constrained to use nothing better than Lipschitz continuity of the boundary. Len. ==== >I'm not sure about math not qualifying as science, however. I think that it >may. What is the definition of a science ? Ask any American dean: a field is a science if the major funding comes from the National Science Foundation. Or: who's going to vote your way in the university-wide decisions? Math obviously differs from biology or geology, but it's a science by these criteria! dave ==== >>I'm not sure about math not qualifying as science, however. I think that it >>may. What is the definition of a science ? Ask any American dean: a field is a science if the major funding > comes from the National Science Foundation. Dutch deans evidently have a different rule. My philosophy department receives most of their funding from the NWO. Sadly, not the New World Order, nor the National Wrestling Nonetheless, they stuck this department under Technical Management. Oh, the agony. I'm practically part of a business school. -- Jesse F. Hughes [Mathematical] society has evolved far enough away from mainstream society that it has become rogue, and now is willing to push its needs against that of the majority. -- James S. Harris ==== >>I'm not sure about math not qualifying as science, however. I think that it >>may. What is the definition of a science ? Results must be verifiable, >>falsifiable, and reproducible. I think that math satisfies these three >>better than any physical science, and it does so with complete precision and >>an exactness which is unique to math. If you know something that I dont >>please post. Where are the experiments? They are called conjectures and proofs of special cases of conjectures. > Is the scientific method taught in math classes? Why is this relevant to whether or not mathematics, as practiced, employ the scientific method or not? >Ever try to verify a number is not computable experimentally? Why are there >colleges of science *and* math? There are? I usually see math in the College of Arts and Sciences. -- ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== >>I'm not sure about math not qualifying as science, however. I think that it >>may. What is the definition of a science ? Results must be verifiable, >>falsifiable, and reproducible. I think that math satisfies these three >>better than any physical science, and it does so with complete precision and >>an exactness which is unique to math. If you know something that I dont >>please post. Where are the experiments? They are called conjectures and proofs of special cases of conjectures. Yes - essentially experiments in abstract. You experiment with conjectures, etc, to determine if they are valid within the framework of a given abstract model. So what. I call that an experiment. > Is the scientific method taught in math classes? Why is this relevant to whether or not mathematics, as practiced, > employ the scientific method or not? If it does not employ the scientific method, then JSH's conjectures might as well be taught at Harvard or Yale. Would you want that ? No. Math would disintegrate without the scientific method. >Ever try to verify a number is not computable experimentally? Why are there >colleges of science *and* math? There are? I usually see math in the College of Arts and Sciences. The distinction is made because there is a big difference between someone like Stephen Hawking and someone like Thomas Edison. Universities have always tried to accomidate both types people to some degree. Also, making math a distinct dept. removes it from the various social and political influences which can affect the findings of researchers and others. > ====================================================================== > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > ====================================================================== Arturo Magidin > magidin@math.berkeley.edu > ==== [.snip.] >> Is the scientific method taught in math classes? >> Why is this relevant to whether or not mathematics, as practiced, >> employ the scientific method or not? If it does not employ the scientific method, then JSH's conjectures might as >well be taught at Harvard or Yale. Would you want that ? No. Math would >disintegrate without the scientific method. I think you have me confused. I happen to think that mathematics is science and uses the scientific method, though the way in which math is usually presented hides this. -- ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== [.snip.] > Is the scientific method taught in math classes? >> Why is this relevant to whether or not mathematics, as practiced, >> employ the scientific method or not? If it does not employ the scientific method, then JSH's conjectures might as >well be taught at Harvard or Yale. Would you want that ? No. Math would >disintegrate without the scientific method. I think you have me confused. I happen to think that mathematics is > science and uses the scientific method, though the way in which math > is usually presented hides this. I agree. I think that math is a science as well. Definately a science. > ====================================================================== > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > ====================================================================== Arturo Magidin > magidin@math.berkeley.edu > ==== I'm not sure about math not qualifying as science, however. I think that it >may. What is the definition of a science ? Results must be verifiable, >falsifiable, and reproducible. I think that math satisfies these three >better than any physical science, and it does so with complete precision >and >an exactness which is unique to math. If you know something that I dont >please post. >Where are the experiments? They are called conjectures and proofs of special cases of conjectures. > True enough. I suppose experiments can also be used to formulate conjectures. > Is the scientific method taught in math classes? Why is this relevant to whether or not mathematics, as practiced, >employ the scientific method or not? > It isn't, really. I just assumed that if the scientific method was a vital part of mathematics it would be a vital part of math courses. Bad assumption. Mathematics, as practiced, is clearly different from mathematics, as taught. Why? >>Ever try to verify a number is not computable experimentally? Why are there >>colleges of science *and* math? There are? I usually see math in the College of Arts and Sciences. > There are. Although not as many a I first thought. rich ==== >>I'm not sure about math not qualifying as science, however. I think that it >>may. What is the definition of a science ? Results must be verifiable, >>falsifiable, and reproducible. I think that math satisfies these three >>better than any physical science, and it does so with complete precision >>and >>an exactness which is unique to math. If you know something that I dont >>please post. Where are the experiments? >>They are called conjectures and proofs of special cases of conjectures. True enough. I suppose experiments can also be used to formulate > conjectures. Conjectures don't seem to play the same role as experiments in empirical science[1]. Conjectures in mathematics are surely analogous to conjectures in science. So, if there are analogues to experiments in mathematics, then they must provide (partial) evidence for or refutation of conjectures. Proofs of special cases can't be the analogues of experiments either. At best, they would have to be the analogues of positive outcomes of particular experiments, since such proofs provide some support (but never any negative support) for the conjecture. If there is anything that corresponds to experiments in mathematics, it would have to be the activity of *searching* for proofs or counterexamples. But in the empirical sciences, an important feature of experimentation is its repeatability. In order to ensure that what you've told me is correct, I may repeat your experiment. In mathematics, one does not really repeat the search, but only evaluates the outcome of the search (proof or counterexample) for correctness[2]. So, even here, the analogy is a bit strained -- at least if one takes repeatability as an essential feature of experimentation in science. As a matter of personal taste, I've never thought that the similarities between mathematics and empirical sciences were sufficient to support the claim that mathematics is a kind of science. The fundamental feature of empirical science is its use of inductive, not deductive, reasoning. Empirical science is always open to later correction due to new data, whereas mathematics is only later corrected if there was an error on the part of mathematicians. Mathematics deals with (hypothetical) certainty and science with theory and conjecture as their basic bits. At least, if one stresses these aspects of the two activities, then there is reason to reject the claim that mathematics is a science. I recognize that one may think that other aspects of each of these fields are more essential, and so one may find the claim that mathematics is a science more plausible. Hence my explicit admission that these comments are really a matter of my personal taste. Footnotes: [1] Throughout, I'll take a traditional view of the scientific method --- something like Hempel. One may complain that these explanations about the role of experiment and theory in empirical science are naive and maybe just plain wrong, but I think they'll serve well enough for now. Besides, naivete was always my forte (ooh, I sound so French). [2] Of course, repeating the search even when one has the proof or counterexample in front of him may be useful, just to give one additional understanding of *why* the proof/counterexample works. But this additional understanding is not necessary to confirm that the proof/counterexample is correct. -- [R]eality has a fascinating ability to check us when we get a little too big for our britches... Make no mistake. There isn't a mathematician alive today that I can't now touch, and not a mathematical career on the planet that I can't now affect. --James Harris, render of worlds ==== >>I'm not sure about math not qualifying as science, however. I think that it >>may. What is the definition of a science ? Results must be verifiable, >>falsifiable, and reproducible. I think that math satisfies these three >>better than any physical science, and it does so with complete precision >>and >>an exactness which is unique to math. If you know something that I dont >>please post. Where are the experiments? >>They are called conjectures and proofs of special cases of conjectures. True enough. I suppose experiments can also be used to formulate conjectures. Exactly. You do special cases, you do a bunch of examples, you test a number of specific instances, and you make the conjecture. Sometimes, the conjecture is immediately followed by proof. Sometimes there is a large gap between one and the other in time. Often, parts of the conjecture are proven, others not. The main difference between mathematics and other sciences is that the standard of proof required in mathematics is far more stringent than that required in, say, physics. > Is the scientific method taught in math classes? >>Why is this relevant to whether or not mathematics, as practiced, >>employ the scientific method or not? It isn't, really. I just assumed that if the scientific method was a vital >part of mathematics it would be a vital part of math courses. Bad >>assumption. Mathematics, as practiced, is clearly different from mathematics, as taught. >Why? Because mathematics places such a premium on proof. Just like in physics you don't describe all the experiments that you attempted to design but was unable to, and you only describe the experiments that were actually performed, if you find a proof for a theorem you do not usually go through explaining all the conjectures you made which proved to be false, or all the specific examples you worked out (especially if you have a proof for the general case). Nonetheless, you will sometimes see a textbook (or more frequently, a or discussing possible guesses followed by counterexamples, or guesses followed by proofs that need only to fix certain details on the guesses, and so on. Because mathematics does have an ultimate arbiter (proof), it is not necessary in most instances to discuss the prior steps in the scientific method as you do in empirical sciences, where no amount of verification is enough to establish truth. There is also a tradition to polish proofs, which usually tends to draw them away from the experimentation or the thought process that led to the proof. Gauss used to say that the thought process that led to a proof was like the scaffolding in construction: once you have the building (the proof/theorem), you take away the scaffolding. (This tendency seems to be eroding some, in that more and more papers are taking the time to describe the intuition behind certain particularly technical proofs). But if you read papers that discuss evidence for conjectures, or that propose conjectures, you will see much of the same sort of things that you see in physics and other empirical sciences. -- ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== However, based on > my sensate human experiences in this world, I find it absolutely and > patently absurd in the extreme to say the least, that someone could claim to > have collected data about things like bigfoot, UFO's, alien abductions, > Wolfman (for e's g), when it is rather obvious that these items cannot be > studied due to their failure to even exist. Well, that argument was once made about rocks falling from the sky. > Suppose, for example, that I published information regarding the average > height, weight and IQ of a sample population of ghosts, and I made the claim > that the average IQ was higher and standard deviation was exactly 1/2 of a > similar sized sample from a population of angels. You dont have a problem with that ? If this turned up in the MAA Math > Magazine or somewhere, and it was presented as factual data - you are > telling me that you would not question how I took the samples ? Where I > found such populations in the first place ? Such things show up in word problems all the time. It's a cute way to inject some levity into a problem that is really simply about the relationships between things. just as true if you replace point, line and plane with table, chair, and beer stein. > Without rewriting the question, or rewording the statement, I give you my > claim. > I claim that if you do statistics on ivalid data, then your results are > invalid by definition, regardless of whether the derived solutions come > out correct or not. Yabbut. In real life, data are not really valid or invalid, simply more or less correct. If your data pretty much reflect reality, then your conclusions will again pretty much reflect reality as well. To find out how well, you need to do research on robustness of statistical techniques. Some things are much like walking a tightrope -- one little slip and you're history. Other things are more like walking -- a little trip and nothing much happens (usually). The concept of robustness has also found its way into artificial intelligence and lots of other engineering applications. Again, it's the idea that you can recover from incomplete data or errors introduced somewhere in the process. Also, we don't really care if you are talking about manufacturing widgets or sighting ghosts. What matters mathematically is the process. It's well known that applying the right process to the wrong data generates nonsense, as A N Niel said at the beginning of this subthread. Jon Miller ==== > However, based on > my sensate human experiences in this world, I find it absolutely and > patently absurd in the extreme to say the least, that someone could claim > to > have collected data about things like bigfoot, UFO's, alien abductions, > Wolfman (for e's g), when it is rather obvious that these items cannot be > studied due to their failure to even exist. Well, that argument was once made about rocks falling from the sky. Suppose, for example, that I published information regarding the average > height, weight and IQ of a sample population of ghosts, and I made the > claim > that the average IQ was higher and standard deviation was exactly 1/2 of a > similar sized sample from a population of angels. You dont have a problem with that ? If this turned up in the MAA Math > Magazine or somewhere, and it was presented as factual data - you are > telling me that you would not question how I took the samples ? Where I > found such populations in the first place ? Such things show up in word problems all the time. It's a cute way to > inject some levity into a problem that is really simply about the > relationships between things. > it's > just as true if you replace point, line and plane with table, chair, and > beer stein. Without rewriting the question, or rewording the statement, I give you my > claim. > I claim that if you do statistics on ivalid data, then your results > are > invalid by definition, regardless of whether the derived solutions come > out correct or not. Yabbut. In real life, data are not really valid or invalid, simply more or > less correct. If your data pretty much reflect reality, then your > conclusions will again pretty much reflect reality as well. To find out > how well, you need to do research on robustness of statistical techniques. > Some things are much like walking a tightrope -- one little slip and you're > history. Other things are more like walking -- a little trip and nothing > much happens (usually). The concept of robustness has also found its way into artificial > intelligence and lots of other engineering applications. Again, it's the > idea that you can recover from incomplete data or errors introduced > somewhere in the process. Also, we don't really care if you are talking about manufacturing widgets or > sighting ghosts. What matters mathematically is the process. It's well > known that applying the right process to the wrong data generates nonsense, > as A N Niel said at the beginning of this subthread. Jon Miller recall an issue relating to dangling chads, where a certain set of data was drawn into serious question, and sealt a serious blow to the very credibility of elections in the US as far as I'm concerned. The point being, however, that trials in which discrete decisions are being recorded such as in an election - how does it come out so screwed up ?? And if we are having thes issues with discrete data from a poll booth, then how much more questionable is psychological data where EVERYTHING is subjective ?? My concern here is not to attack statistics - which is sound theoretically, but rather the complete misapplication of statistics by psychology. The only thing in life that I compare this to is that there is nothing wrong with a sportscar. The car is theoretically sound. But the operator is a madman, and he slams the car into a brick wall. This is what psychology does with statistics. They crash it into a brick wall with their stupidity and their erroneous subjective idiocy. I will demonstrate elsewhere, that their usage of statistics is false. My main concern here, is to raise the issue of data and conclusions. It may sound pedantic, btu I am trying to get an AMEN to the follwoing claim: Performing statistics upon bad data is not a sound statistical practice. If you have incorrect data to work with, you cannot produce conclusions which are correct. It is impossible to derive real world truths based uopn data which is corrupted. Now - confidence levels notwithstanding, I am not talking about errors due to approximation or levels of confidence. Here's an example of what I am after. I am hired by Ford Corp to do a statistical analysis of a sample of Ford automobiles. Unfortunately, all I have is a parking lot full of Chevy's and BMW's. Can I perform the task which is required of me based on the population that I have been given ?? Can I turn in my results to Ford, an analysis of Ford vehicles, if I do not have any Ford cars to study ?? Is it valid to study a BMW or a Chevy and turn in the results as if having studied a Ford ? ==== I'd say a pure scientist would observe this is not a 100 % repeatable observation A psychologist would say given enough people the numbers will repeat about the same time every year. In reality, some will write books about the space invasion and others about ghosts and still others will buy them and talk about it in all sorts of ways. > Example: I am a researcher. I study UFO's and ghosts. I have determined that %77 of all ghosts are spotted within a 5 mile radius > of UFO sightings, and that %82 of all UFO sightings are accompanied by a > general increase in the intensity and overall magnitude of ghost related > hauntings. Therefore, I conclude, via statistics, that ghosts and UFO's are somehow > related, and that ghosts are in all probability using UFO's for > transportation purposes. ----------------------------------------------- Will someone in the sci.math or sci.math.stat please stand up and tell me > why this is flawed, and no I am not kidding. I need an independent opinion. People are telling me I'm crazy - where did I go wrong ????? Can I call this > science ?? Is my reasoning flawed somehow ?? Is there a problem with my > populations - or is this valid usage of stats ?? ==== Can anyone help me with the following problems coming from the above-mentioned (most-interesting) handbook?: 1.Page 117,problem 5: If E is the set of points in [0,1] which have no 4 in their decimal representation, show that E is a set of measure zero. (This problem really puzzles me). 2.Page 127,problem 4: If f belongs to L,show that there is a sequence(fk) of continuous functions such that limfk(x)= f(x) for a.a.x. As a hint, problem 4d,p.124 is suggested, however on that page 4.d is not given (deleted?).Has anyone an idea what this hint and/or the correct steps towards the solution of the problem might be? 3.Page 212,problem 1: If f is a function from R into a normed linear space Y such that f is defined in a neighborhood of a point a and f'(a) exists, show that f is differentiable at a and dfa(h)= hf'(a). In my opinion, if a function is defined for every point in R (so R is the domain of f), then it seems superfluous to me to specify that f is defined in a neighborhood of a point, because any neigborhood is always a subset of R.Or is my reasoning wrong? I'd be most grateful if anyone could help with one of these problems and I thank You on beforehand, Rik.Verhoestraete@village.uunet.be ==== Dear all, My problem is to construct some integer matrices of certain pattern to > satisfy a matrix equation... There are so many unknowns, and there are also > so many equations, (number of equations >> number of unknowns)... Since the required matrices are integer, I cannot use Newton, or Conjugate > descent methods, etc... I have to do some search algorithm... but since > there are so many unknowns, the search space is too large... So I am thinking of having an initial matrix, then do some change, see if > the matrix equation(left hand side - right hand side) approximately more > near zero and find the minimization point... But even this, still make a large search space when the size of matrix gets > larger... moreover, another constraint is that the number of non-zero > elements of this matrix should be as small as possible... Please tell me what kind of problem is this? I even don't know how to > categorize my problem and where to find answers on google... > Possible keywords: (linear or lonlinear) mixed integer programming constraint programming NEOS Writing your own algorithm is likely to be inefficient unless > you are very experienced. Thus take one of the packages that > already exist! > Arnold Neumaier Dear Arnold, I will dive out for library and books... and go on GOOGLE for those keywords... Unfornately my problem is non-linear. The equation that I want my integer matrices to satisfy is: (( A * A )V1 + ( B * B) V2) V - (D * D) = 0 where D is known, all other matrices are remain to solve. A, B, V1, V2 should be 2's power; V1 and V2 should be diagonal; V should be diagonal too, but can be any real/integer number... The * symbol represents Kronecker product... With so many unknown and with such a non-linear equation... I guess there is no existing codes that can be plugged-and-played... would you mind giving me more inputs? -Walala problems? ==== > Dear all, > My problem is to construct some integer matrices of certain pattern to > satisfy a matrix equation... There are so many unknowns, and there are > also > so many equations, (number of equations >> number of unknowns)... > Since the required matrices are integer, I cannot use Newton, or > Conjugate > descent methods, etc... I have to do some search algorithm... but since > there are so many unknowns, the search space is too large... > So I am thinking of having an initial matrix, then do some change, see > if > the matrix equation(left hand side - right hand side) approximately more > near zero and find the minimization point... > But even this, still make a large search space when the size of matrix > gets > larger... moreover, another constraint is that the number of non-zero > elements of this matrix should be as small as possible... > Please tell me what kind of problem is this? I even don't know how to > categorize my problem and where to find answers on google... > Possible keywords: (linear or lonlinear) mixed integer programming constraint programming NEOS Writing your own algorithm is likely to be inefficient unless > you are very experienced. Thus take one of the packages that > already exist! > Arnold Neumaier Dear Arnold, > I will dive out for library and books... and go on GOOGLE for those > keywords... Unfornately my problem is non-linear. The equation that I want my integer > matrices to satisfy is: (( A * A )V1 + ( B * B) V2) V - (D * D) = 0 where D is known, all other matrices are remain to solve. A, B, V1, V2 > should be 2's power; V1 and V2 should be diagonal; V should be diagonal too, > but can be any real/integer number... The * symbol represents Kronecker product... With so many unknown You didn't say how many integer variables and how many real variables you have. and with such a non-linear equation... I guess there is > no existing codes that can be plugged-and-played... would you mind giving me > more inputs? I gave you enough inputs - try them first before asking for more! You have a mixed-integer nonlinear programming problem without an objective (i.e., a constraint satisfaction problem). So this is what you need to search for; two codes are in NEOS and several other codes exist; they probably make a convexity assumption and hence do not necessarily solve your problem, but you'd try them first. There are probably several ways to formulate this in a modeling language like GAMS or AMPL, and you'd have to try and find which formulation is most easily solvable. Arnold Neumaier ==== | |> My problem is to construct some integer matrices of certain pattern to > |> satisfy a matrix equation... There are so many unknowns, and there are also > |> so many equations, (number of equations >> number of unknowns)... > | |> Since the required matrices are integer, I cannot use Newton, or Conjugate > |> descent methods, etc... I have to do some search algorithm... but since > |> there are so many unknowns, the search space is too large... > | |> So I am thinking of having an initial matrix, then do some change, see if > |> the matrix equation(left hand side - right hand side) approximately more > |> near zero and find the minimization point... > | |> But even this, still make a large search space when the size of matrix gets > |> larger... moreover, another constraint is that the number of non-zero > |> elements of this matrix should be as small as possible... > | |> Please tell me what kind of problem is this? I even don't know how to > |> categorize my problem and where to find answers on google... > | |> Possible keywords: > | |> (linear or lonlinear) mixed integer programming > | |> constraint programming See Knuth, volume II, for the Spectral test. It does precisely > this. There are a zillion other potential keywords, too, because > it is a common problem. For example, much of seriation and even > multi-dimensional scaling uses these techniques and has done since > the 1950s. Another relevant concept is linear programming. |> NEOS > | |> Writing your own algorithm is likely to be inefficient unless > |> you are very experienced. Thus take one of the packages that > |> already exist! Grrk. Well, yes, but .... This is a classically foul problem, because generic solutions just > don't work. As a result, virtually EVERY useful approach is a hack, > and the difficulty is getting one that will work on YOUR problem. > You often need to roll your own, whether or not you have the skills > to do it - that is, after all, how most people get into this area! But I agree that trying the standard ones BEFORE starting to hack > is a good idea, whether you are experienced or not. I may be a mad > hacker from way back, but even I will take a look around for existing > code first :-) > Nick Maclaren. Dear Nick, I will dive out for library and books... But before that, I guess I'd get some keywords more precise from you: I will search for spectral test but what are the precise statement of seriation and even multi-dimensional scaling as you've mentioned in your posting? Unfornately my problem is non-linear. The equation that I want my integer matrices to satisfy is: (( A * A )V1 + ( B * B) V2) V - (D * D) = 0 where D is known, all other matrices are remain to solve. A, B, V1, V2 should be 2's power; V1 and V2 should be diagonal; V should be diagonal too, but can be any real/integer number... The * symbol represents Kronecker product... With so many unknown and with such a non-linear equation... You think spectral test does precisely the same thing? Would you mind giving me more inputs? -Walala ==== Can a line intersect a circle in more than two points over a _ring_? Ditto for a line and a general conic? If one point of intersection is given is the other point uniquely determined? Note that a quadratic equation may have more than two roots over a ring, e.g. x^2 = 1 for x = +-1,+-3 (mod 8). ==== I have two questions in rings theorie : 1- If we denote by M(n,K) the K-vector space of n x n matrices with coefficients in K, we know that (for dimensions question) M(n,K) and M(n+1,K) are not isomorphic as K-vector spaces, but how to show that they are not isomorphic as rings ? 2- It's not difficult to show that for a unitary ring A for which (x+y)^2 = x^2 + y^2 for all x and y in A, we have A commutative. Is this result is true for non-unitary rings ? counterexample ? -- Ce message a ete poste via la plateforme Web club-Internet.fr This message has been posted by the Web platform club-Internet.fr http://forums.club-internet.fr/ ==== >I have two questions in rings theorie : ... , one of which has been answered. >2- It's not difficult to show that for a unitary ring A for which (x+y)^2 = x^2 >+ y^2 for all x and y in >A, we have A commutative. >Is this result is true for non-unitary rings ? counterexample ? Let R = Z a + Z b + Z c where c = a b = - b a and a^2=b^2=0. Then the square of every element is zero and so your premise is satisfied, but R is not commutative. dave ==== I have two questions in rings theorie : 1- If we denote by M(n,K) the K-vector space of n x n matrices with coefficients >in K, we know that >(for dimensions question) M(n,K) and M(n+1,K) are not isomorphic as K-vector >spaces, but how to >show that they are not isomorphic as rings ? I replied to this question once before - in February 2002. Here it is again: >If M_n(R) and M_m(R) are isomorphic as rings must we have m=n ? Yes, I think this is true for any commutative ring R with 1. The centre of M_n(R) is easily seen to consist of the scalar matrices only, and so is a subring isomorphic to R. A ring isomorphism from M_n(R) to M_m(R) would have to map the centre of the first onto the centre of the second, and so M_n(R) and M_m(R) would be isomorphic as modules over their centres - i.e. as R-modules. But they are free R-modules of dimensions n^2 and m^2. You can prove that the free R-modules R^a and R^b are isomorphic if and only a=b by factoring out a maximal ideal of R, and reducing to the vector space case. Somebody later showed me another proof of this, involving polynomial identities. Any two 1x1 matrices satisfy AB - BA = 0, but this is not true for n>1. Then any 4 2x2 matrices A,B,C,D satisfy the identity ABCD - ABDC - ACBD + ACDB + ... + DCBA = 0, where there are 24 sums in the product, one for each permutation of {A,B,C,D}, and the sign is the sign of the permutation. More generally, any 2n nxn matrices satisfy the corresponding identity formed as a sum of (2n)! products, one for each permutation. These idenitites are not hard to prove, but unfortunately I cannot remember how you do it right now! Derek Holt. ==== > In Grothendieck's EGA(Elements de Geometrie Algebrique) I (1960) Proposition (9.5.1) p.176 states; > Let f:X --> Y be a morphism of schemes such that > f_*(O_X) is quasi-coherent sheaf of O_Y module. > Then there exists a closed subscheme Y' of Y with > the following property. > f splits into X --> Y' --> Y and Y' is the smallest > closed subscheme of Y with this property. However, I think this proposition holds without the condition > on O_X. Am I missing here? My proof(sketch) of the proposition (9.5.1) If Y is an affine scheme Spec(A), then f: X --> Y is > determined by homomorphism h: A --> O_X(X). > Let I = Ker(h). Then Y' = Spec(A/I) satisfies the property > of the proposition. If Y is not affine, then glue affine shemes > obtained in the above method. N. Saito The last bit where you say glue affine schemes obtained in the above method will work only if the kernel of O_Y ---> f_*(O_X) is quasicoherent, which would follow from f_*(O_X) being quasicoherent. Closed subschemes are defined by quasicoherent sheaves of ideals. If the kernel is not quasicoherent, you might still succeed by taking the sum, J, of all the quasicoherent ideals contained in the kernel and using that instead. If J is quasicoherent it defines a subscheme Y' with the desired property. To prove J is quasicoherent, observe that it's the image of the map from the direct sum of all those ideals to O_Y, that a direct sum of quasicoherent sheaves is again quasicoherent, and so is the image of a map between quasicoherent sheaves. Even if I'm correct, the stronger result may not be interesting, because the formation of the subscheme Y' may not be local on Y, i.e., if you consider replacing Y by an open subset U and X by its preimage, you may have to replace Y' by something bigger than its intersection with U. That might be the reason Grothendieck didn't state it that way. For an example where f_*(O_X) is not quasicoherent, take dim Y > 0 and let X be an infinite disjoint union of copies of Y. In this case, Y' = Y. ==== I see that in Hartshorne's _Algebraic Geometry_ exercise II.3.11d reads as follows: |Let f:Z->X be a morphism. Then there is a unique closed subscheme |Y of X with the following property: the morphism f factors through Y, |and if Y' is any other closed subscheme of X through which f factors, |then Y->X factors through Y' also. We call Y the scheme-theoretic |image of f. If Z is a reduced scheme, then Y is just the reduced induced |structure on the closure of the image f(Z). Better still, my past self alleges to have done that exercise once upon a time! Keith Ramsay ==== > taking it in high school. I enjoy math as a hobby and was looking for a good > book on calculus for this summer so that I can learn a little before > college. A few books I saw at the bookstore were: Calculus Made Easy - Silvanus Philips Thompson > Calculus for Dummies > Complete Idiots Guide to Calculus Another interesting book is Calculus: The Elements, which seems more like a > textbook with problems. What aspects of calculus should a beginning book include and does anyone > know if any of these, or any others, are any good? I am not looking for an If you are looking to learn calculus _techniques_, then I cannot help much except to say that almost all calculus textbooks teach techniques. If you are interested in the theory. I would like to recommend Yet Another Introduction to Analysis by Victor Bryant. It is an easy read, and covers many of the highlights of calculus. It also has solutions to all exercises. The only downside is that it focuses on sequences. If you intend to study analysis more deeply, then I would suggest looking at some more advanced books (Rudin, etc...). ==== > taking it in high school. I enjoy math as a hobby and was looking for a good > book on calculus for this summer so that I can learn a little before > college. A few books I saw at the bookstore were: Calculus Made Easy - Silvanus Philips Thompson > Calculus for Dummies > Complete Idiots Guide to Calculus Another interesting book is Calculus: The Elements, which seems more like a > textbook with problems. What aspects of calculus should a beginning book include and does anyone > know if any of these, or any others, are any good? I am not looking for an You have received recommendations for books by Spivak and Apostol. These are high-class books but they may not be suitable for a beginner's self-study. The preface to Spivak's second edition (reprinted in the third edition) states that his book is most used as a follow-up for those who are already skilled in techniques of calculus. While you could use his book, it is likely that you would have an easier time with a conventional college textbook. Although I have not had time to look up the Apostol book, if I recall correctly, it begins with integration instead of differentiation, and this is the opposite of conventional textbooks such as you would be assigned in college. It would be difficult to find a conventional textbook that it faulty or bad. David Ames ==== >> taking it in high school. I enjoy math as a hobby and was looking for a good >> book on calculus for this summer so that I can learn a little before >> college. A few books I saw at the bookstore were: >> Calculus Made Easy - Silvanus Philips Thompson >> Calculus for Dummies >> Complete Idiots Guide to Calculus >> Another interesting book is Calculus: The Elements, which seems more like a >> textbook with problems. >> What aspects of calculus should a beginning book include and does anyone >> know if any of these, or any others, are any good? I am not looking for an >You have received recommendations for books by Spivak and Apostol. >These are high-class books but they may not be suitable for a >beginner's self-study. >The preface to Spivak's second edition (reprinted in the third >edition) states that his book is most used as a follow-up for those >who are already skilled in techniques of calculus. While you could >use his book, it is likely that you would have an easier time with a >conventional college textbook. >Although I have not had time to look up the Apostol book, if I recall >correctly, it begins with integration instead of differentiation, and >this is the opposite of conventional textbooks such as you would be >assigned in college. It does, and this is good. Unfortunately, it does not begin with abstract measure and integration, which can, and should, be taught in high school. Integration with discrete measures is about 5000 years old, and students who have had differentiation first get confused. Fermat computed the integral of x^k before calculus. >It would be difficult to find a conventional textbook that it faulty >or bad. Most of them are both faulty and bad. As long as students want to be told HOW without understanding WHY, and can downrate professors who insist on understanding, this will remain. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University ==== Let U be an open set in |R. Let I_p denote the ideal in C^{infty}(U) of real valued functions of class C^infty which vanish at p. (Precisely, p in Usubseteq |R , f : U --> |R is in I_p if and only if f(p) = 0 and f is C^infty) . Now consider ideal I^k_p in C^infty (U) defined in this manner: I^k_p : = I_p times .... times I_p (k-times), in other words, I^k_p is the the product of I_p k-times with itself. By definition, f in C^infty (U) has a k-order zero in p if exists lim_{ xto p} frac { f(x) } { || x-p ||^(k-1) } = 0. It is clear that f in I^k_p implies f in C^infty (U) has a k-order zero in p . Is the converse true? In other words, is it true the follwing ? : let f in C^infty (U) have a k-order zero in p , i.e. exists lim_{ xto p} frac { f(x) } { || x-p ||^(k-1) } = 0. Is it true that f in I^k_p ? Tern ==== > Let U be an open set in |R. > Let I_p denote the ideal in C^{infty}(U) of real valued functions of > class C^infty which vanish at p. (Precisely, p in Usubseteq |R , f : > U --> |R is in I_p if and only if f(p) = 0 and f is C^infty) . Now consider ideal I^k_p in C^infty (U) defined in this manner: > I^k_p : = I_p times .... times I_p (k-times), in other words, > I^k_p is the the product of I_p k-times with itself. By definition, f in C^infty (U) has a k-order zero in p if > exists lim_{ xto p} frac { f(x) } { || x-p ||^(k-1) } = 0. > It is clear that f in I^k_p implies f in C^infty (U) has a k-order zero in > p . Is the converse true? In other words, is it true the follwing ? : > let f in C^infty (U) have a k-order zero in p , > i.e. exists lim_{ xto p} frac { f(x) } { || x-p ||^(k-1) } = 0. > Is it true that f in I^k_p ? Take p = 0 for convenience. Show the following: If f in C^infinity(U) and f(0) = 0, then f(x) = x*g(x), where g is in C^infinity(U). So by induction, if f in C^infinity(U) has a kth order zero at 0, then f(x) = x^k*g(x), for g in C^infinity(U). That should answer your question. ==== Do you think this book is a valuable choice just for geometry? I mean: is it possible to study one-variable calculus and multivariable calculus from different books, but learn geometry on Apostol's 2 volume? ==== Earlier I posted about a sequence of integers defined as: a_n = min(m) s.t. n^2 + m^2 = k^2 where (n, m, k) is a Pythagorean triplet and m > 0 for n > 2. Let a_1 = a_2 = 0. which starts like this: n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 a_n 0 0 4 3 12 8 24 6 12 24 60 5 84 48 8 12 144 24 180 15 20 Question: Are there infinitely many pairs (m,n) such that a_m = n and a_n = m? It seems intuitively true but the proof isn't immediately obvious to me. For example, one pair is (415, 996) since a_415 = 996 and a_996 = 415 (415^2 + 996^2 = 1079^2). ==== On Mon, 05 Jan 2004 20:44:31 +0200, Toni Lassila Pythagorean triplet and m > 0 for n > 2. Let a_1 = a_2 = 0. which starts like this: n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 >a_n 0 0 4 3 12 8 24 6 12 24 60 5 84 48 8 12 144 24 180 15 20 Another feature. Plotting only the non-prime values gives a plot which is maximized by f(n) = (1/4) n^2 - 1. This is obvious since the maximum a_n for any non-prime n is exactly (1/4) n^2 - 1 since the largest factors giving this term are (2, (1/2)n^2) and ((1/2)n^2 - 2) / 2 = (1/4) n^2 - 1. But notice how almost no non-prime terms fall in between the maximum curve and, say 10n? It seems like all numbers in the sequence that are greater than a certain linear term are either primes or composites of the form (1/4) n^2 - 1. For example the term a_14 = 48 which is the first composite to fulfill the condition a_n > a_(n+1) belongs to this group as 14^2 - 1 = 48. ==== Can someone please help me to determine an integral which i can not find in Gradshteyn & Rhyzik (3.541) ? It is to integrate cosh(a*x+b)^(-d)*exp(-c*x) over the positive axis (x=the variable, the others are hopefully appropriate real constants). ==== The Power of Imagination By Acharya Keshav Dev The Times of India Hursday, November 22, 2001 This term conjures an image of a mind that is active all the time. Power of imagination can be increased with determination. Nothing in the world - from literature to music, scientific progress to artistic achievements, new inventions, discoveries and attainment of high targets can be done without imagination. The domain of kingdoms, business or industry cannot be increased without imagination. Power of imagination has a Midas touch. Once the power of imagination starts working on any person or thing, the whole structure undergoes a transformation. By a simple touch of imagination, thoughts can be given a shape and form like a base metal iron converting into gold at the touch of the philosopher's stone. What is so mysterious in power of imagination? It might not be intelligible to the ordinary man with lowly thinking that all supernatural and mysterious mental powers originate from imagination. Freedom of the country was made possible due to very high imaginative power of Mahatma Gandhi. Consider imagination to be 'fire' and thoughts as 'air' - when the two come together, fire flares up and spreads within no time. Similarly, fire of imagination ignited by fuel of the air of thoughts, assumes a gigantic shape strongly effecting the human mind, body and spirit simultaneous. The largest revolutions of the world are a testimony to this. Strong and determined thoughts encouraged by the power of imagination and far-sight expand and spread at lightening speed without hindrance. As soon as thoughts acquire the wings of determination, discipline, power of imagination, they spread like wild fire and expand in geometric progression. You might not be able to comprehend fully the intensity of your own imagination. In fact it is nothing that has a shape or form which could be seen, touched or measured. It is dimensionless but is capable of manifesting itself in any imaginable form and can assume any shape. No one has witnessed its origin or its end. It is just like zero power. The power of zero depends on its placement. The value of figure or digit becomes ten time more powerful of zero is added after it. Its value increases to millions, trillions or zillions and so on when zero is added after the number. Its value can only be evaluated on assigning a place. The same analogy can be applied to supreme power of the Universe. No one can assess the size, shape and form of the Shoonya Brahm but it is omnipotent, omnipresent and omniscient. Imagination power, like the power of zero, is super power, which in fact is nothing in itself but assumes paramount importance once it manifests itself. What is there that cannot be made possible with power of imagination? It penetrates with ease where no one can dare to reach. What cannot be seen with the naked eyes can very well be imagined and visualised. That which cannot be known by any other power can be understood with the power of imagination. There are some things regarding which cannot be commented upon, cannot be understood, felt, seen or touched but can be comprehended and seen with the power of imagination. Imagining while meditating deeply, one can reach the inner state of consciousness and original form and in due course of time it becomes a reality. Similarly, words too assume a physical form. In the musical scriptures notes arranged in a particular manner constitute a swara lahri or a raga (symphony). Indian musicians presented these raags in a manifested form before the world. In the 16th century, history witnessed ample incidents of great music maestros endowed with devotion, determination and imagination performing miracles with their music. Baiju Bawra, Swami Hari Das and Tansen are a few examples of those who gave shape and form to their imagination. They presented their imagination in manifested form and amazed the world with their performance. Any seeker, who treads on their footsteps even today with devotion, determination and perseverance can develop power of imagination and become master of his branch to accomplish seemingly impossible feats. Read the complete news at: http://www.timesofindia.com Jai Maharaj Creator of newsgroups alt.jyotish, alt.language.hindi, alt.religion.hindu http://www.mantra.com/jyotish http://www.mantra.com/jai Om Shanti Panchaang for 14 Paush 5104, Monday, January 5, 2004: Shubhanu Nama Samvatsare Uttarayane Moksha Ritau Dhanush Mase Shukl Pakshe Indu Vasara Yuktayam Mrgashir Nakshatr Shukl-Brahm Yog Gar-Vanij Karan Chaturdashi Yam Tithau http://www.mantra.com/holocaust http://www.hindu.org http://www.hindunet.org The truth about Islam and Muslims http://www.flex.com/~jai/satyamevajayate o Not for commercial use. Solely to be fairly used for the educational purposes of research and open discussion. The contents of this post may not have been authored by, and do not necessarily represent the opinion of the poster. The contents are protected by copyright law and the exemption for fair use of copyrighted works. considered or answered if it does not contain your full legal name, are not necessarily those of the poster. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i03Df0R16413; ==== >>the naturals to this* you are supposing that S is *the set of ALL >>reals*, then it is a false premise because breaks the rules of >>naturals What? f:N->R, f(n) = n is an injection from the naturals to reals. No, >it does not get all real values since it's not a surjection - that >would be impossible. What do mean breaks the rules of naturals? It means the same you are saying, ăthat would be impossibleä. >but if S represents an INCOPLETE set of reals, then the >>premise is worse than false, it is stupid, since in this case you >>donât need the diagonal argument to find out that at least one real >>number is not in S. Of course it's not stupid. It shows that any mapping between naturals >and reals can only be to a countable subset of reals. We can not start from an incomplete mapping and complete it by induction (adding reals one-by-one) because there's always some reals that we can add. Well, let me see. If we know beforehand that f:N->R cannot be a surjection, as you have pointed out above, Why must we use the diagonal argument to get the same conclusion again?. Moreover, the result that Cantor gets applying the diagonal argument is false, since the premise is useless. f:N->R is not a surjection, but no because there are more reals (aleph 1) than naturals (aleph 0), but because R and N have a different grade of infinity, due to their respective mathematical definitions. >To get a good conclusion, the proof by contradiction needs coherent >>premises, and this is not the case. Define coherent. > Coherent premise: Not false, not useless, and not meaningless before starting any mathematical reasoning. >>To assume that sqrt(2) is rational is not a false and stupid premise >>because it doesnât break any rule and moreover, the probability of >>being rational before the proof is one half. >First you must prove that sqrt(2) is real. Then the probability of any real number being rational is 0. > For me the rationals and the irrationals belong to the set of reals (in some books even the naturals and integers are reals). So, before carrying out any mathematical reasoning, every number with decimal expansion has a probability of 1/2 of being rational. >Sqrt(2) is of the form n/m where n and m are relatively prime and m >is not 0 is a false premise, which leads to a direct contradiction >with itself. There exists a bijection between N and R is a similarly >false premise, which also leads to a direct contradiction with itself. I fail to see the difference between your suggested logical values of >false and false and stupid. Obviously weâve got a different point of view about the concept of ăfalse premiseä. For me Sqrt(2) is of the form n/m where n and m are relatively prime and m is not 0 is a premise which becomes false after a bit of good mathematical reasoning, so there is a proof by contradiction. Therefore, the premise is useful and the proof by contradiction works. However, the premise There exists a bijection between N and R is false before the beginning of the proof, and so that there is not proof by contradiction, but ? (replace ? yourself with a suitable qualifier). In this case, the premise is useless and the proof by contradiction doesnât work, giving a meaningless conclusion. Nicolas de la Foz ==== On Sat, 3 Jan 2004 13:41:02 +0000 (UTC), nico80@jazzfree.com (Nicolas >[...] Obviously weâve got a different point of view about the concept >of ăfalse premiseä. For me Sqrt(2) is of the form n/m where n and m >are relatively prime and m is not 0 is a premise which becomes false >after a bit of good mathematical reasoning, so there is a proof by >contradiction. Therefore, the premise is useful and the proof by >contradiction works. However, the premise There exists a bijection >between N and R is false before the beginning of the proof, and so >that there is not proof by contradiction, but ? (replace ? yourself >with a suitable qualifier). In this case, the premise is useless and >the proof by contradiction doesnât work, giving a meaningless >conclusion. This is gibbering nonsense. Sqrt(2) is rational is not false at the start of the proof but becomes false after the proof, while there is a bijection between N and R is already false at the start of the proof? There's no difference between the two cases except in your imagination - both premises have _always_ been false and always will be. >Nicolas de la Foz ************************ David C. Ullrich X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i03Dex816381; ==== Is it possible to solve the same problem but with 10 balls instead of 9? I am tryting to find the solution to the problem for a while now and I don't think that there is a solution so I wanted to check with you guys. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i03Dexe16377; ==== >Your table is more like what I am looking for. Any functions can be >used in the formula, even conditional operations and conditions. The >formula will be put into a Excel sheet to generate graph from >different Ammo capacity, rate of fire and reload time. I need to find the number of bullets shot for a specified amount of >time, and not the time for a specified number of bullets. > Sniped to shorten overall message(...) Maybe this is what you are looking for? After 4 bullets are fired in each cycle, have a 1 second time interval after the 4th shot and before a reload. Then a 2 second interval between end of reload and first shot fired in next cycle. The following table shows these changes. Other values could be used for each of the 5 parameters used. It makes the calculations simple and also creates a realistic situation, time wise, for firing and reloading a weapon. I used a long list to prove my calculations, so see my calculations after long list below. I am sure the mathematicians can give a much simpler explanation of my time lines fitting into the calculations then I can, so please excuse the numerical presentation. Number of parameters (5) = 1)Ammo, 2)RateofFire, 3)ReloadTime, 4)BeforeReload, 5)ToNextShot Parameters /# of Seconds 1)Ammo : 4 2)RateOfFire : 2 3)ReloadTime : 5 4)Time after last shot BeforeReload : 1 5)Time after end of reload (Next cycle) ToNextShot : 2 Long list below SecondsElapsed (*)= Bullet fired 0 1 * First bullet fired @ 0 seconds 1 1 2 2 * 3 2 4 3 * 5 3 6 4 * 7 4 Reload begins @ 7 seconds 8 4 Reloading 9 4 Reloading 10 4 Reloading 11 4 Reloading 12 4 Reload ends 1st cycle ends 13 4 Starts 2nd cycle. 14 5 * 5th bullet fired @ 14 seconds 15 5 16 6 * 17 6 18 7 * 19 7 20 8 * 21 8 Reload begins @ 21 seconds. 22 8 Reloading 23 8 Reloading 24 8 Reloading 25 8 Reloading 26 8 Reload ends 2nd cycle ends. 27 8 Starts 3rd cycle. 28 9 * 9th bullet fired @ 28 seconds. 29 9 30 10 * 31 10 32 11 * 33 11 34 12 * 35 12 Reload begins at 35 seconds. 36 12 Reloading 37 12 Reloading 38 12 Reloading 39 12 Reloading 40 12 Reload ends 3rd cycle ends. 41 12 Starts 4th cycle. 42 13 * 13th bullet fired @ 42 seconds. 43 13 44 14 * 45 14 46 15 * 47 15 48 16 * 49 16 Reload begins @ 49 seconds. 50 16 Reloading 51 16 Reloading 52 16 Reloading 53 16 Reloading 54 16 Reload ends 4th cycle ends. 55 16 Starts 5th. cycle. 56 17 * 17th bullet fired @ 56 seconds. 57 17 58 18 * 59 18 60 19 * 61 19 62 20 * 63 20 Reload begins @ 63 seconds. 64 20 Reloading 65 20 Reloading 66 20 Reloading 67 20 Reloading 68 20 Reload ends 5th cycle ends. 69 20 Starts 6th. cycle. 70 21 * 21st bullet fired @ 70 seconds. 71 21 72 22 * 73 22 74 23 * 75 23 76 24 * 77 24 Reload begins @ 77 seconds. 78 24 Reloading 79 24 Reloading 80 24 Reloading 81 24 Reloading 82 24 Reload ends 6th cycle ends. 83 24 Starts 7th. cycle. 84 25 * 25th bullet fired @ 84 seconds. 85 25 86 26 * 87 26 88 27 * 89 27 90 28 * 91 28 Reload begins @ 91 seconds. 92 28 Reloading 93 28 Reloading 94 28 Reloading 95 28 Reloading 96 28 Reload ends @ 96 seconds. 97 28 Starts 8th cycle. Etc. --- Note: These calculations below are good for input values =>12 seconds that will give the correct # of bullets fired from 12 sec.--->oo. ;-) The first 11 seconds of input should be easy to calculate and find the number of bullets fired for any particular input from 0 to 11 seconds. The first complete cycle takes 12 seconds; all following cycles take 14 seconds. The reason for this is, at zero seconds in the first cycle the first shot is fired eliminating the first 2 seconds required between shots. Good for finding number of bullets fired where input in seconds =>12. Given a number as an example, picking seconds at random find total number of bullets fired for value (s)? Where s is the total number of seconds required for input, then given s = 93. Using modulus arithmetic, given 93 seconds from time of first shot fired my calculations would be --- floor((s+2)/14) = 6 (adding 2 to 93 because of the two second lost on the first cycle). 6*14 = 84 seconds is the point where the first bullet is fired in the 7th cycle. Also it is the 7th cycle because (6+1) = 7th cycle. s+2)- 84 = 11 Seconds (residual) into 7th cycle. 4*7 = 28 maximum bullets fired. Where (4) is the maximum bullets fired for each cycle and (7) is the cycle therefore maximum number of bullets fired @ end of 7 cycle is (4*7)= 28. If residual < 8 you would use this table below otherwise if residual =>8 no table is required and use maximum bullets fired for that cycle and for this cycle the residual is 11 so the maximum of 28 bullets fired for this cycle would be the answer. Residual If Residual <8 and is a [6 or 7] then = (28-1) = 27 bullets fired. ă ă <8 ă ă [4 or 5] ă = (28-2) = 26 bullets fired. ă ă <8 ă ă [2 or 3] ă = (28-3) = 25 bullets fired. ă ă <8 ă ă [0 or 1] ă = (28-4) = 24 bullets fired. The answer is, @ 93 seconds 28 bullets would have been fired. For this example the above table is not used because residual 11 > 8. Proof: Using the 7th cycle as an example. Seconds 7th Cycle into the Seconds 7th.cycle elapsed (total #) (value of) (residual) 1 83 24 Starts 7th. cycle @ 83 seconds. 2 84 25 * 25th bullet fired @ 84 seconds. 3 85 25 4 86 26 * 5 87 26 6 88 27 * 7 89 27 8 90 28 * 9 91 28 Reload begins @ 91 seconds. 10 92 28 Reloading 11 93 28 Reloading ----------------------------------------- 94 28 Reloading 95 28 Reloading 96 28 Reload ends and 7th cycle ends. This means it is at the 11th second into the 7th cycle. Bullets fired, 7*4 = 28 total bullets fired at end of 7th cycle. If seconds into the cycle <8 then each set in table below where you have to calculate beforehand the maximum # of bullets fired to that point of the end of a cycle which in this case = 28 bullets fired. Residual(r) if r<8 then [6,7] = (28-1) = 27 bullets fired. r<8 then [4,5] = (28-2) = 26 bullets fired. r<8 then [2,3] = (28-3) = 25 bullets fired. r<8 then [0,1] = (28-4) = 24 bullets fired. What this means, if you requested for the # of bullets fired @ (85) seconds instead of (93) seconds then your residual would be (3) second residual instead of an (11) second residual. So (85+2)- (6*14) = 3 seconds would fall on residuals [2,3] thus 25 bullets fired. In this case it is >8 so 11 seconds into the cycle ends up with the full 7th. cycle value of 28 bullets fired and thus no table needed. To double check these calculations I will randomly pick an input number of (57) seconds. floor(57+2)/14 = 4 4*14 = 56 seconds the first bullet is fired in the (4+1) cycle. (57+2)-56 = 3 Seconds (residual) into 5th cycle. 4*5 = 20 maximum bullets fired for the fifth cycle. We Need the above table because residual seconds is < 8. Residual(r) if r<8 then [6,7] = (20-1) = 19 bullets fired. r<8 then [4,5] = (20-2) = 18 bullets fired. r<8 then [2,3] = (20-3) = 17 bullets fired. r<8 then [0,1] = (20-4) = 16 bullets fired. Residual is 3 seconds so in the above table for residual [2,3] = (20-3) = 17. Answer is, in 57 seconds 17 bullets were fired. Proof below. Seconds into 5th cycle the 5th Seconds cycle elapsed 1 55 16 Starts 5th. cycle. 2 56 17 * First bullet fired @ 56 seconds. 3 57 17 ------------------------------------ 58 18 * 59 18 60 19 * 61 19 62 20 * fired at 62 seconds. 63 20 Reload begins at 63 seconds. 64 20 Reloading 65 20 Reloading 66 20 Reloading 67 20 Reloading 68 20 Reload ends at 68 seconds. So @ 57 seconds there was 17 bullets fired. Another example, say @ (1,039) seconds how many bullets fired? floor((1,039+2)/14) = 74 74*14 = 1036 seconds where first bullet is fired in the (74+1) cycle. (1,039+2)- 1036 = 5 seconds (residual) into 75th cycle. 4*75 = 300 maximum bullets fired for all of the 75 cycles. Needing the table again because residual is < 8 seconds. Residual(r) if r<8 then [6,7] = (300-1) = 299 bullets fired. r<8 then [4,5] = (300-2) = 298 bullets fired. r<8 then [2,3] = (300-3) = 297 bullets fired. r<8 then [0,1] = (300-4) = 296 bullets fired. Using residual [4,5] because residual (5) < 8 therefore use (300-2) = 298 bullets fired @ 1039 sec. Proof below! Sec. 75th cycle into 75th Seconds cycle elapsed (*)= Bullet fired. 1 1035 296 75th. Cycle starts @ 1035 sec. 2 1036 297 * First bullet fired @ 1036 sec. 3 1037 297 4 1038 298 * 5 1039 298 ------------------------------------- 1040 299 * 1041 299 1042 300 * 1043 300 Reload begins @ 1043 seconds. 1044 300 Reloading 1045 300 Reloading 1046 300 Reloading 1047 300 Reloading 1048 300 Ends reload @ 1048 sec. Simple enough to work out an algorithm for an Exsel or spreadsheet or c. program. I am sure other values can be calculated in the same manner given different values for the 5 parameters! This could change cycle length, but that should be no problem because you have to determine the first cycle length by what values are given each parameter and thus make minor adjustments to your algorithm. I believe this to be correct but please point out any errors I overlooked. Any input welcome! Dan X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i041SJZ28806; ==== > Let R be a ring and [,]:R*R -> R, [x,y] = xy - yx the commutor function. Are those rings for which for all x,y,z in R: > [x,y] = 0 and [y,z] = 0 -> [x,z] = 0 somehow simpler than other rings? If so, >do any other of any such ring's properties follow >automatically from this? Substituting 1 for y in the above formula, one has [x,z]=0 >for all x,z in R. >Hence, such a ring must be commutative. > The following question did not need the additional complication >> of R ring. Let G be a group. What is known about >> groups with the property that there are elements >> x,y,z in G such that xy=yx, xz=zx, but yz != zy ? I admit, >> I«m asking this question somewhat rhetorically- >> because I have found such a group. Btw., this property >> has very little to do with the reasons for me investigating the >> group found as of yet. Every noncommutative group satisfies the above with x=1. Marc I love it! I've missed something trivial again... however, none of the 3 elements I was thinking of from my group were 1. C.Dement X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i041SLN28881; ==== >On Sat, 3 Jan 2004 02:23:43 +0000 (UTC), nico80@jazzfree.com (Nicolas >If you don't mind, Iâm going to use your own proof, with some >>variations, in order to prove the same, but only with naturals. It's been done before, you know. > I didnât know it, but it's not a surprise. >>Proposition: Let f: N -> Nâ be given. Then f is not a surjection. Counterexample: Let f be the identity mapping. Then f is a bijection. > This confirms that the direct proof is as useless (or false) as the proof by contradiction. Since your counterexample is undoubtedly true, and my proof with naturals is also ăcorrectä (itâs exactly to the one you admit as valid working with reals) then, from the contradiction we come to the conclusion that, as your counterexample is true, my proof proves nothing, and the proof with reals neither. >>Proof. We are to show that there exists n in Nâ such that n is not >>in the range of f. That is, n != f(k) for any k in N. >>We do this by defining, for each k, the k-th digit in the natural >>representation of n. Given k > 0, we first look at d_k, the k-th >>digit following the first digit in the representation of f(k) from >>our list. We next define the k-th digit of n, n_k, as follows: >> If d_k is a 1, set n_k = 2. >> If d_k is not a 1, set n_k = 1. >>Then the number n = (n_1)(n_2)(n_3)... is the required number. It is >>not in the list because for each k, n differs from f(k) in the k-th >>digit. It is not in the list because it's not a natural number. Natural >numbers are finite, yours is infinite. I don't know why you think that the naturals in the list are infinite. Anyway, remember: ăARBITRARY LISTä. In order to state that, you are supposing something about the list and these are not the rules of the game. My proof just prove that the mapping f: N->N' can't be a surjection. Nicolas de la foz X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i041SNk28989; ==== What if I could fly like Superman. >Then I can jump out this window. >(jumps out window). >Ow, that hurt, I guess I cannot fly like superman. With what you were saying about cantor's proof vs the sqrt(2) proof, >how can one tell between something that is obviously false (like >cantor) and something that is only false after some reasoning ( >sqrt(2)) ? The basis of these arguments is that either a statement is true or >false (goedel's not here right now). Sometimes you cannot directly >prove that something is true, so you see what happens if it were >false. If you run into problems, then it must have been true to begin >with. The fact is that the human brain is a computer that, like any other computer, it needs to be fed with trusty information to get reasonable outcomes. It is not easy to know in advance whether that information is good or not. Thus, when the brain runs a program (the proof)- supposing that the program is reliable -only we can decide whether the input information is good or not, comparing the aftermath with our referential knowledge. With Cantor's proof the problem is that the result is on the limit between the credible and the incredible, i.e., it seems possible, but no intuitive. In this case has failed our knowledge of reference, probably because, as you have said, G.9adel was not around. Nicolas de la Foz X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i041SKx28869; ==== The subtlety is ... how do we know when our manipulations are introducing spurious results? Are there rules to follow that will always prevent this? phil X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i041SLN28898; ==== > If you don't mind, Iâm going to use your own proof, with some >> variations, in order to prove the same, but only with naturals. That's a good exercise. It's important to understand why the argument >succeeds in one case and fails in the other. > Well I think the argument fails in both cases, and this is precisely what we have to elucidate. >> Proposition: Let f: N -> Nâ be given. Then f is not a surjection. >> Proof. We are to show that there exists n in Nâ such that n is not >> in the range of f. That is, n != f(k) for any k in N. >> We do this by defining, for each k, the k-th digit in the natural >> representation of n. Given k > 0, we first look at d_k, the k-th >> digit following the first digit in the representation of f(k) from >> our list. We next define the k-th digit of n, n_k, as follows: >> If d_k is a 1, set n_k = 2. >> If d_k is not a 1, set n_k = 1. > Then the number n = (n_1)(n_2)(n_3)... is the required number. It is >> not in the list because for each k, n differs from f(k) in the k-th >> digit. The problem is that the string (n_1)(n_2)(n_3)... has all digits nonzero, >and therefore does not represent a natural number. The fact that x = >.(x_1)(x_2)(x_3)... has all digits nonzero is not a problem in the >corresponding real-number case. For example, 1/9 = 0.111111.... has all >digits nonzero after the decimal point. > Firstly, I would like to know why a sequence of nonzero digits cannot be or cannot represent to a natural number. Are you thinking perhaps that the sequences of N' have infinite figures? Remember: ăarbitrary listä. Do not assume something in advance, but the mapping f: N -> N'. In any case, the diagonal argument does not set limits to the process or manner to generate the number. So that, if necessary, I'll do a process or program to ensure that the synthesized number fulfils all requisites. For instance, the following code could serve for that purpose for(i = 0 ; d[i] == 0; i++) k[i] = 0; while(i < END_OF_LIST) //here begin the significant digits { if(d[i] == 1) k[i] = 2; else k[i] = 1; } Nicolas de la Foz ==== > If you don't mind, Iâm going to use your own proof, with some > variations, in order to prove the same, but only with naturals. >>That's a good exercise. It's important to understand why the argument >>succeeds in one case and fails in the other. [ ... ] > If d_k is a 1, set n_k = 2. > If d_k is not a 1, set n_k = 1. > Then the number n = (n_1)(n_2)(n_3)... is the required number. It is > not in the list because for each k, n differs from f(k) in the k-th > digit. >>The problem is that the string (n_1)(n_2)(n_3)... has all digits nonzero, >>and therefore does not represent a natural number. The fact that x = >>.(x_1)(x_2)(x_3)... has all digits nonzero is not a problem in the >>corresponding real-number case. For example, 1/9 = 0.111111.... has all >>digits nonzero after the decimal point. > Firstly, I would like to know why a sequence of nonzero digits cannot > be or cannot represent to a natural number. That's a simple proof by induction. See below. >Are you thinking perhaps > that the sequences of N' have infinite figures? Why would you think that I thought that? >Remember: ăarbitrary > listä. Do not assume something in advance, but the mapping f: N -> N'. The mapping is f: N -> N. I assume nothing else. > In any case, the diagonal argument does not set limits to the process > or manner to generate the number. So that, if necessary, I'll do a > process or program to ensure that the synthesized number fulfils all > requisites. For instance, the following code could serve for that > purpose As you said above (and I quote): > If d_k is a 1, set n_k = 2. > If d_k is not a 1, set n_k = 1. This guarantees that every n_k is nonzero. Thus, the sequence of digits n_k does not represent a natural number. Proposition. Let n be a natural number. Then n has only finitely many nonzero digits in decimal notation. Proof. By induction on n. The proposition obviously holds for n = 0. If we suppose (induction hypothesis) that a given natural number n has only finitely many nonzero digits (k, say), then we can conclude that n+1 has at most k+1 nonzero digits, still a finite number. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i041SK128858; ==== To correct myself, I noticed that there are 3 distinct elements x, y, z of the group such that xy = yx, xz = zx but yz != zy. (What I forgot to include:) none of these elements are from the group center. C. Dement X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i042Cgc32381; ==== >
  How many solutions has the equation sin(z)=z (in complex
>>numbers)?
Warning: Complex analysis is not my forte. Piano would be
>more like it. Nonetheless...
How about beginning with a related equation which is easy to
>solve precisely (in closed form)?
Specifically, solve sin(z)=Re(z).
{I'll be happy to show my simple solution later if requested,
>but perhaps some readers would enjoy solving the equation
>themselves.)
Then show that the solutions of sin(z)=Re(z), except for those
>with smallest nonzero modulus, approximate the solutions of sin
>(z)=z. I presume that this may be hard to do; I have no proof,
>alas. The approximations appear to improve as the modulus of z
>increases.
Out on a limb here,
>  David Cantrell


The fastest and easiest way to search and participate in Usenet - Free!

>
Ok. Here is a nice proof of this result. Consider the function: g(z)= z/sinz This function g is analytic in C(pi)Z and never vanishes. You can check that the singularity at 0 is removeble since the limit as z approaches 0 is 1. Thus, this function is meromorphic and it has a sequence of poles that go to infinity. Therefore, you can also check that when this happens, it follows that the function has a non-isolated essential singularity at infinity. No, in a neighborhood of an essential singularity like the one at infinity, the function assumes all the values of the complex plane infinitely many times with a possible exception of one. This possible exception is 0 in this case. Thu, the function assumes the value 1 infinitely many times. Also, this proof shows that sinz = cz has an infinite number of solutions for all nonzero complex numbers c. Ziad Adwan X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i041SNv28978; ==== It is usually best not to change the subject line unless the topic of >the post has changed substantially from the original. > [.snip.] > This is not my question, however. It is: > (Idea- since {1,-1} is a nontrivial group, ...), >>What does -1 mean in the context of an arbitrary group? Of course, you are correct here. I suspect one could say >> the following: Given an arbitrary group (G, *), let (R, *, +) >> be any ring such that G is a subgroup with respect to * and >> in a way that every additive inverse of G is in G, There are too many such rings. For any ring R, the natural choice is >R[G], the group ring over R, which consists of all polynomials of >the form sum_{g in G} a_g*g where a_g is in R, only finitely many nonzero, with multiplication >given by multiplying monomials as (a_g*g)(a_h*h) = (a_g*a_h)(gh) > ^ ^ > | |____ multiplication in G > multiplication > in R. Then G embedds as the collection of all elements of the form 1*g, 1 >the multiplicative identity in R; its additive inverse is (-1)*g, >where -1 is the additive inverse of 1 in R. If you choose R to be >any ring of characteristic 2, then -1 = 1 in R, so the additive >inverse of the element 1*g is again 1*g, which is in the image of the >canonical embedding of G. In fact, this is the only situation in which >this will happen in the group ring. I guess there could be other rings >in which this can be done, quotients of group rings over >characteristic different from 2. > then -1(R) is the additive inverse of 1 with respect to + What does -1(R) mean? R is a ring. Are you multiplying every element >of R by -1? Or do you mean, the -1 from R? There is no reason to assume this is >different from 1, either, as noted above. > and is in G. If no such ring exists then G does not have >> a -1. And what makes you think that (1) -1 will be different from 1, the identity of G? > (2) In any two such rings, -1 will be the same element of G? Please forgive me for writing back immediately without first having gone over your interesting example point by point (which I definitely plan to do...). What made me think (2)? That is exactly what I meant by the statement My Guess is as Good as Yours Theorem 1.1 and as the theorem goes My Guess is as Good as Yours. What made me think (1) is that I was unaware of such a ring- so why not just exclude 'em? My Guess is as Good as Yours Theorem 1.2b: Let (G,*) be a (finite) non-abelian group and assume it possible to find a ring (R, *, +) such that G is a subgroup of R with respect to *. In addition, assume that for all x in G: -x(R) in G (where -x(R) is defined as the additive inverse of x) and 1 != -1(R). Then G is of even order (by the first posting, this would also follow from another property: {1, -1(R)} is a subgroup) and has a nontrivial center. My Guess is as Good as Yours Theorem 1.1b: -x(R) is independent of R iff 1 != -1(R), thus we may write -x for it. >Unless both questions have an affirmative answer (and clearly, (1) >certainly does not), it still makes no sense to talk about -1 in an >arbitrary group G. > My Guess is as Good as Yours Theorem 1.1: >> the element -1 designated as above is independent >> of the encompassing ring (R, *, +). In a group ring over a ring of characteristic 2, it will be equal to >1. In other rings, it need not be. >And what makes you think that in your group, even if -1 makes sense, >>-1 is not equal to 1? Using the technique above, it would follow that 1 is no longer >> a unit, am I right? I do not see how that follows. In (F_2)[G], 1, by which you presumably >mean 1*e, e the identity of G, is a unit. Arturo Magidin, sans .sig X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i041SMm28955; ==== It is usually best not to change the subject line unless the topic of >the post has changed substantially from the original. > [.snip.] > This is not my question, however. It is: > (Idea- since {1,-1} is a nontrivial group, ...), >>What does -1 mean in the context of an arbitrary group? Of course, you are correct here. I suspect one could say >> the following: Given an arbitrary group (G, *), let (R, *, +) >> be any ring such that G is a subgroup with respect to * and >> in a way that every additive inverse of G is in G, There are too many such rings. For any ring R, the natural choice is >R[G], the group ring over R, which consists of all polynomials of >the form sum_{g in G} a_g*g where a_g is in R, only finitely many nonzero, with multiplication >given by multiplying monomials as (a_g*g)(a_h*h) = (a_g*a_h)(gh) > ^ ^ > | |____ multiplication in G > multiplication > in R. Then G embedds as the collection of all elements of the form 1*g, 1 >the multiplicative identity in R; its additive inverse is (-1)*g, >where -1 is the additive inverse of 1 in R. If you choose R to be >any ring of characteristic 2, then -1 = 1 in R, so the additive >inverse of the element 1*g is again 1*g, which is in the image of the >canonical embedding of G. In fact, this is the only situation in which >this will happen in the group ring. I guess there could be other rings >in which this can be done, quotients of group rings over >characteristic different from 2. > then -1(R) is the additive inverse of 1 with respect to + What does -1(R) mean? R is a ring. Are you multiplying every element >of R by -1? Or do you mean, the -1 from R? There is no reason to assume this is >different from 1, either, as noted above. > and is in G. If no such ring exists then G does not have >> a -1. And what makes you think that (1) -1 will be different from 1, the identity of G? > (2) In any two such rings, -1 will be the same element of G? Please forgive me for writing back immediately without first having gone over your interesting example point by point (which I definitely plan to do...). What made me think (2)? That is exactly what I meant by the statement My Guess is as Good as Yours Theorem 1.1 and as the theorem goes My Guess is as Good as Yours. What made me think (1) is that I was unaware of such a ring- so why not just exclude 'em? My Guess is as Good as Yours Theorem 1.2b: Let (G,*) be a (finite) non-abelian group and assume it possible to find a ring (R, *, +) such that G is a subgroup of R with respect to *. In addition, assume that for all x in G: -x(R) in G (where -x(R) is defined as the additive inverse of x) and 1 != -1. Then G is of even order (by the first posting, this would also follow from another property: {1, -1} is a subgroup) and has a nontrivial center. My Guess is as Good as Yours Theorem 1.1b: -x(R) is independent of R iff 1 != -1, thus we may write -x for it. >Unless both questions have an affirmative answer (and clearly, (1) >certainly does not), it still makes no sense to talk about -1 in an >arbitrary group G. > My Guess is as Good as Yours Theorem 1.1: >> the element -1 designated as above is independent >> of the encompassing ring (R, *, +). In a group ring over a ring of characteristic 2, it will be equal to >1. In other rings, it need not be. >And what makes you think that in your group, even if -1 makes sense, >>-1 is not equal to 1? Using the technique above, it would follow that 1 is no longer >> a unit, am I right? I do not see how that follows. In (F_2)[G], 1, by which you presumably >mean 1*e, e the identity of G, is a unit. Arturo Magidin, sans .sig X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i04Dl9t15584; ==== Hai srini, I just happen to find your address. I know pretti clearly that you are my pal. I am srini reddi who? sterling engines at jntu. er. srini reddi >> Can somebody point me to a fast, sequential implementation >>of the 'mimimum vertex cover' problem in C/C++ ? >>Murthy Andukuri >> X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i04Dl9Y15603; ==== HAS ANYONE RESEARCHED THE RESONANCE OF CERTAIN (BACTIORIAL) DNA FOR COMUNICATION PURPOSES----UNCANNY ABILITY OF IDENTICAL TWINS TO COMUNICATE---EINSTEINS PROBLEM WITH ACTION AT A DISTANCE,ALPHA,BETA SEEMS DISTANCE IS NOT A PROBLEM,NO TIME LAG,LIGHT IS SIMPLY WHAT IT LOOKS LIKE(SIGNATURE) WHEN MATTER MOVES FROM 3D TO HIGHER D . HYPERGEOMETRIC RESONANCE (ACTION AT A DISTANCE) REIMANN TENSOR ALL INFORMATION IS ALREADY HERE , TO CAPTURE IT ,A RESONATE GEOMETRIC ARANGEMENT IS NECESSARY SIMPLE HU.STRING THEORY IS DAM CLOSE! LOOK CLOSE AT FIBANCCI SERIES.TRANSLATE TO D 4,D6,--D22.ALL CURLED UP. WHAT IS THE RESONATE FREQUENCY OF 3D GEOMETRY? COMPOSITE?WE ARE IN A RESONATE CAVITY.IMAGINE A BUBBLE IN JELLO.PLEASE RESPOND HERE. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i04Dl9v15617; ==== This transformation requires that one preepend infinitely many zeroes to >each natural (in decimal notation), since there is nno finite upper >bound on the number of digits in naturals. Whatever finite number of >zeroes you prepend, there are naturals requiring more than that number >of digits for their expression. Quite possible, but not of great use. And a diagonal construction does not work here, because an infinite >string of digits containing more than finitely many non-zero digits does >not represent a natural. I think you have not completely understood how the neutral transformation works. Let's suppose that in the list a natural [f(j)] has *s* significant digits. Although we don't know whether the list is ordered or not, (since it is an arbitrary one) we'll assume that it is. Then, going down in the list we'll met a natural [f(k), k > j] with (s + 1) significant digits. At that point we will fill the gap of f(j) appending one zero on the left. Thereafter we do the same with f(j- 1), f(j-2),..., f(1). When this process has finished all the natural above f(k) in N' they will have the same amount of digits than f(k). Going down in the list again we will find another natural [f(h)] with (s + 2) digits, therefore we should add another zero on the left, from f(1) to f(h-1), and so on. As all the naturals that we will met down in the list they will always have a finite number of digits (by definition of natural number) it is not possible to find in N' an infinite string of digits. Certainly, there is not a hypothetic finite upper bound on the number of digits in naturals, but naturals run asymptotically towards the infinite, never reaching it. So, they always have a finite number of digits in N', and therefore all the elements in N' are naturals. Nicolas de la Foz X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i04DlAH15626; ==== >In <200401010931.i019VuS17850@proapp.mathforum.org>, on 01/01/2004 > at 01:53 PM, nico80@jazzfree.com (Nicolas de la Foz) said: >The diagonal argument needs a premise GBN or ZF will do nicely. > Sorry, but I donât understand why. >>(a list, otherwise it can®t work). The list is not a premise. > Call it X if you want to, but f: N -> R is the raw material in the demonstration by contradiction, and it is a list. >>If you know a *neutral* premise, then I believe that the proof by >>contradiction would be valid. You don't even need a proof by contradiction. You simply prove >(Ax){x:N- (Er)~(En)x(n)=r} > Thatâs right, we are working on it. >-- > Shmuel (Seymour J.) Metz, SysProg and JOAT >not reply to spamtrap@library.lspace.org X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i04I7kt00391; ==== >Still being kind of a newcomer to (finite) group theory, >I read over the proof of Lagrange's theorem: >H subgroup G -> |G| = |H||G:H|. And of the many >subsequent examples (and problems). The author >of the introductory text book I was reading missed >a very easy example, in my opinion: If G has a subgroup H of even order, then G is even. Proof: If G has a subgroup H of even order, then > |G|= |H||G:H|. The index of H in G, |G:H|, is > either even or odd. It doesn't matter; > since odd*even = even and even*even = even, q.e.d. > My Guess is as Good as Yours Theorem 1.2: Let (G,*) be a (finite) non-abelian group and assume it possible to find a ring (R, *, +) such that G is a subgroup of R with respect to *. In addition, assume that for all x in G: -x(R) in G (where -x(R) is defined as the additive inverse of x) and 1 != -1(R). Then G is of even order (by the above, this would also follow from another property: {1, -1(R)} is a subgroup). Also, -1(R) commutes with all x in G; from which it follows that G has a nontrivial center. My Guess is as Good as Yours Theorem 1.1: -x(R) is independent of R iff 1 != -1(R), thus we may write -x for it. Note: Assuming that x != -x(R) also follows for all x in G (after 1 != -1(R) has been confirmed as a prerequisite in Theorem 1.2), then G must trivially be even- since G could then be written as the union of two equal sets: {1,x_1, x_2, ..., x_n } and {-1,-x_1, -x_2, ..., -x_n }. But consider that we also have the contrapositive of Theorem 1.2: If G is odd, then no such ring exists. C. Dement ==== >>Still being kind of a newcomer to (finite) group theory, >>I read over the proof of Lagrange's theorem: >>H subgroup G -> |G| = |H||G:H|. And of the many >>subsequent examples (and problems). The author >>of the introductory text book I was reading missed >>a very easy example, in my opinion: >>If G has a subgroup H of even order, then G is even. >>Proof: If G has a subgroup H of even order, then >> |G|= |H||G:H|. The index of H in G, |G:H|, is >> either even or odd. It doesn't matter; >> since odd*even = even and even*even = even, q.e.d. My Guess is as Good as Yours Theorem 1.2: >Let (G,*) be a (finite) non-abelian group >and assume it possible to find a ring (R, *, +) >such that G is a subgroup of R with respect to *. In >addition, assume that for all x in G: -x(R) in G (where >-x(R) is defined as the additive inverse of x) and >1 != -1(R). Then G is of even order What your entire thing amounts to is: Assume that there is a rather complicated condition involving embeddings of G into multiplicative subgroups of rings, which implies that G has a central element of order two. Then G is of even order. Which of course is an immediate consequence of Lagrange's Theorem, since the order of any element in a finite group divides the order of the group. Your Theorem reminds me of attempting to scratch your left ear with your right hand by crossing it over your head. >(by the above, >this would also follow from another property: {1, -1(R)} is a >subgroup). Also, -1(R) commutes with all x in G; from >which it follows that G has a nontrivial center. My Guess is as Good as Yours Theorem 1.1: -x(R) is independent >of R iff 1 != -1(R), thus we may write -x for it. This statement does not even make sense. The first statement is -x(R) is indenpendent of R. The only way I can interpret this is as follows: For all x in G there exists y_x in G such that, for all rings R, for all embeddings i:G->U(R) of G into the group of units of R with the property that for all g in G, -i(g) in i(G), -i(x)=i(y_x). The second statement is 1 is not equal to -1 in R. But R was quantified universally in the first statement, so there is no way to make an if and only if connection between this statement and the first statement, unless you mean it to mean for all rings R, 1 is not equal to -1 in R, which is of course a ->false<- statement. -- ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i04I7lR00431; ==== >Find all integral solutions {x,y,z,u} (if any) of the equations >z^2 - 9 x^2 = u^2 & x^2 + y^2 = z^2. Kent Holing I have finally succeeded to show that the system has no non-trivial solutions in integers. If so, we have the Pythagorean triangles (z,x,y) and (z,3x,u) which are primitive P-triples and we can show that x must be even. We also then easily see that z-3x, z-x, z+x and z+3x are all squares. But this cannot happen due to a wellknown theorem of Fermat: No arithmetical progression can contain 4 squares in a row. Kent Holing X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i052QrB01221; ==== >>James Harris Dear Sir, >> >> I am sorry, that You are working so hard I'm sorry but I'm focused on my own work. I say that if you have your >own to present: present it. I know that that involves putting yourself out there, but I hope >you'll respect my efforts to promote my *own* work in threads I've >created, and I will not go to any threads you create promoting my work >to the potential detriment of yours. >James Harris X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i052Qs001259; ==== >>James Harris Dear Sir, >> >> I am sorry, that You are working so hard I'm sorry but I'm focused on my own work. I say that if you have your >own to present: present it. I know that that involves putting yourself out there, but I hope >you'll respect my efforts to promote my *own* work in threads I've >created, and I will not go to any threads you create promoting my work >to the potential detriment of yours. >James Harris Anyhow it is nice to hear Your opinion: I'am doing it in my way My way is, that I can start always from one kind of 1-st plotted equation f(t): t^n = 2 n^u abtp + a^n + b^n once Z;X;Y not divided by n ....(*) or n^(nu-1) t^n = 2 n^u abtp + a^n + b^n once Z divided by n ..(**) or t^n = 2 n^u abtp + n^(nu-1) a^n + b^n once Y divided by n ..(***) or t^n = 2 n^u abtp + a^n + n^(nu-1) b^n once X divided by n (the last one is just inverse of the parameters a to b or our unknown numbers X to Y ) Using t as a changing value we have for example for n=5 T^n = 5AB(2T + A + B)[2 T^2 + 2(A + B)T + A^2 + AB + B^2] what for T = 5^u abtp we can express as: 5^5u a^5 b^5 t^5 p^5 = 5AB F1(t) F2(t) and once simultanously should be completed: t^5 = F1(t) ; p^5 = F2(t) so for to find neccessary obstacles for one possible solution for t>0 it will be sufficient to check only all kinds of t^5 = F1(t) completations: So far we'll have to check 3 kinds of equations of f(t): ( for n=3 it will be obviously only 2 kins of equations f(t) once always Z or X or Y should be divided by 3 . For next bigger prime numbers it is not so easy for to find such option: You can see works of Kummer etc. .Also taking such option, we doesn't need to judge the 1-st fall of FLT with traditional methods...) The focus of the next step is to do some input in the place of a^n + b^n = tC or n^nu-1 a^n + b^n = tC and again to replace some power of a^n = tC - b^n or b^n = tC - n^(nu-1) a^n so, that we'll achieve new equation of (n-1)n degree using only a;C or b:C with apprioprate composition of n and p values also F(t) Such equation F(t) or polynomian of (n-1)n degre with changing value f should be able to be factorised to (n-1)equations f(t) also polynomians of n degree. Now it will be suffiucient to judge only the first derivative: F'(t) = 0 will mean, that it will be some possibility for to complete at least double root of t for such purposes. ( It could be also 2k roots where 2k = n-1 but first diverative can't judge more exactly ) Now with certain F'(t)= 0 equation we can close the deal: It is like we've given some more bigger chance for the tiny ideal and correct factorisation and anyhow the result is 0. What could be wrong with such developments ??? Roman B. Binder X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i052Qtd01296; ==== >>Find all integral solutions {x,y,z,u} (if any) of the equations >>z^2 - 9 x^2 = u^2 & x^2 + y^2 = z^2. You must mean simultaneous solutions of the pair of equations. Solutions of the second equation are well known to be integer scalings of > x = r^2 - s^2 y = 2 r s z = r^2 + s^2 >where r and s are integers restricted only by the other equation: > (r^2 + s^2)^2 - 9 (r^2 - s^2)^2 = u^2 >that is, you need integers r,s so that > - 8 r^4 + 20 r^2 s^2 - 8 s^4 >is a perfect square. Well, there may be slick ways to find all the >solutions to this particular equation, but in general the solutions >of (homogeneous quartic in r,s) = (perfect square) correspond to >rational points on an elliptic curve, which can be sought systematically. >Like with a machine. Which in this case tells me that there are no >nontrivial (nontorsion) solutions, more precisely in this case >meaning |r|=|s|, which then forces x = 0. You can figure out the rest... dave In your last eqn let m = 2r^2 and n = 2s^2 . Then we need - 2m^2 + 5mn - 2n^2 to be a perfect square, or - 2(m - n)^2 + mn = q^2 phil X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i052QtT01307; ==== Try http://www.mathforum.org/discuss/sci.math/ for a laugh phil X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i052Qta01284; ==== need help with figuring out how to make a tree diagram for this problem. If blood types can be A, B, AB, and O and Rh+ and Rh-, draw a tree diagram for the possibilities X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i052Qt601315; ==== See some f(x): x^n - 2 n^u a b x - a^v - b^v = 0 also such polynomial of x where coefficients are expressed stationary with some parameters a;b and fixed values 2;n;u;v using input a^v + b^v = x C .................(*) primary we'll have x^(n-1) - 2 n^u ab - C = 0 { 2 n^u ab = x^(n-1) - C }^v 2^nv a^v b^v = [x^(n-1) - C]^v now inputting from (*) a^v = x C - b^v 2^nv (x C - b^v) b^v = [x^(n-1) - C]^v ................(**) and so on we'll have F(x) as a function of (n-1)v degree of x as changing value. Now coefficients of F(x) are expressed with a;C parameters and stationary fixed values of 2;n;u;v Could we use 1-st derivative for to judge correct completation of such F(x) with some real x value ? Also once n is some odd number, so at least first derivative should equal 0. 2^nv C = v [(n-1) x^(n-2) - C]^(v-1) v[(n-1) x^(n-2) - C]^(v-1) - 2^nv C = 0 Could we use now Eisenstein criterion for to express C as at least C = product of (ci)^2 ?? Ro X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i052QvP01389; ==== Now, I dare take out the My Guess is as Good as Yours in My Guess is as Good as Yours Theorem 1.2. However, at the same time I have demoted it from theorem all the way down to property since I have now seen that there is no need to use a sledgehammer (Lagrange«s theorem) to kill a butterfly (G even order); nevertheless, thanks to Robin Chapman and Arturo Magidin who both showed me that property (ii), below, is not always true for an arbitrary ring. Let (G,*) be a (finite) group (abelian or not) and assume it possible to find a ring (R, *, +) such that (i), (ii) hold below: (i) G is a subgroup of R with respect to * (ii) 1 != -1(R) in G. Then (iv) G is of even order. (v) -1(R) commutes with all x in G; from which it follows that G has a nontrivial center. Note: two proofs of property (iv) are given, with and without the sledgehammer Proof of (iv): (using Lagrange«s theorem) In order to prove G even, (as stated before, because of Lagrange«s theorem) we only have to prove {1,-1(R)} is a subgroup of G since |{1,-1(R)}| = 2 by the prerequisite 1 != -1(R) For short, write -1 for -1(R): We already know that 1*(-1) = -1 in {1,-1} and 1*1 = 1 in {1,-1}, for, otherwise, 1 would not be a unit. The only thing left to show is that (-1)*(-1) = 1. However, we know 1+ (-1) = 0 -> 0 = -1*(1 + (-1)) = -1*1 + (-1)*(-1) = -1 + (-1)*(-1) -> (-1)*(-1) = 1, q.e.d. Proof of (iv): (without Lagrange«s theorem) Again, for short, write -x for -x(R). -1 in G by (ii) and x in G, -1*x and x*(-1) in G. We first need to show -1*x = -x = x*(-1) for all x in G. This is easy: assume x*(-1) != -x then x*(-1) + x != 0 -> x*(-1 +1) != 0 -> x*0 != 0, a contradiction. -1*x = -x is shown similarly. Thus, for all x in G: -x in G. Now, G can be written as G = {1, -1, x_1, x_2, ..., x_n } and since the additive inverse of an element is unique, we should get pairings {1, -1}, {x_1, -x_1}, {x_2, -x_2}, exhausting all the elements of G. Each of these pairings has two elements -> G of even order, q.e.d. Proof of (v): excersise. (q.e.d.) My Guess is as Good as Yours Theorem 1.1: -x(R) is independent of R iff 1 != -1(R), thus we may write -x for it. Notive also the contrapositive of Theorem 1.2: If G is odd, then no such ring exists. C. Dement X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i052QsZ01233; ==== Now, I dare take out the My Guess is as Good as Yours in My Guess is as Good as Yours Theorem 1.2. However, at the same time I demote it to proposition or even property; unless someone gives me reason to see it otherwise. Proposition 1.2: Let (G,*) be a (finite) group (abelian or not) and assume it possible to find a ring (R, *, +) such that (i), (ii), (iii) hold below: (i) G is a subgroup of R with respect to * (ii) for all x in G: -x(R) in G (where -x(R) is defined as the additive inverse of x) (iii) 1 != -1(R). Then (iv) G is of even order. (v) -1(R) commutes with all x in G; from which it follows that G has a nontrivial center. Proof of (iv): In order to prove G even, (as stated before, because of Lagrange«s theorem) we only have to prove {1,-1(R)} is a subgroup of G since |{1,-1(R)}| = 2 by the prerequisite 1 != -1(R) For short, write -1 for -1(R): We already know that 1*(-1) = -1 in {1,-1} and 1*1 = 1 in {1,-1}, for, otherwise, 1 would not be a unit. The only thing left to show is that (-1)*(-1) = 1. However, we know 1+ (-1) = 0 -> 0 = -1*(1 + (-1)) = -1*1 + (-1)*(-1) = -1 + (-1)*(-1) -> (-1)*(-1) = 1, q.e.d. Showing that -1 commutes should be similar... although at the moment I have not completed a proof. My Guess is as Good as Yours Theorem 1.1: -x(R) is independent of R iff 1 != -1(R), thus we may write -x for it. Note: Assuming that x != -x(R) also follows for all x in G (after 1 != -1(R) has been confirmed as a prerequisite in Theorem 1.2), then G must trivially be even- since G could then be written as the union of two sets of equal order: {1,x_1, x_2, ..., x_n } and {-1,-x_1, -x_2, ..., -x_n }. But consider that we also have the contrapositive of Theorem 1.2: If G is odd, then no such ring exists. C. Dement X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i052QwE01412; ==== > at 08:07 PM, nico80@jazzfree.com (Nicolas de la Foz) said: >You can freely suppose that sqrt (2) is rational because this >>supposition does not run against any mathematical concept. No you can't, because it *DOES* run against Mathematical concepts. >However, in the case of Cantor®s proof, if we initially suppose that >> we have a one-to-one correspondence between N and R, then we are >>breaking off the mathematical rules. And, in fact, we don't suppose that. What we do is to establish that >supposing it would lead to a contradiction. >This means that our initial premise is false, Google for reductio ad absurdum. I think that you cannot or you don't want to understand my reasons because, in this case, you do not admit the concept of TIME in mathematics. I imagine that for you f: N -> R is instantaneous. It doesn't matter whether the mapping involves two elements or infinitely many. Therefore, if you don't want to see that a mapping is a process, where time has something to say, and that a proof is also a process, then we will never understand one another. The words before and after are essential to understand my reasoning. Please, do not ask me for its mathematical definition. And please don't top post. > I just would like to know how to avoid it. I always reply to the ăreply to this messageä option, but many times my post appears placed at the root of the tree. Nicolas de la Foz >-- > Shmuel (Seymour J.) Metz, SysProg and JOAT >not reply to spamtrap@library.lspace.org X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i05Bc8A05603; ==== I know the dirac delta is a distribution but my question is essentially is it locally integrable when alone over the distribution space? I don't think so because it either forces its indicating functions to vanish at a bounded rate or else it has to be non supported. Given that, how can the the dirac delta still be a member of the integrables when it clearly disobeys lebesgues integral theory in the limit of no indicative set (or some would say the dominated convergence theorem - but i dunno how that works)??? :( I am confused please point me to something that makes sense.....someone ..... ==== On Mon, 5 Jan 2004 11:38:08 +0000 (UTC), jackbusiness31@tiscali.it You need to find a keyboard with an Enter key - your post was one long line, which is not easy to read. Inserting line breaks: >I know the dirac delta is a distribution but my question is essentially is it locally >integrable when alone over the distribution space? It's not locally integrable - it can't be, since it's not even a function. I have no idea what locally integrable when alone over the distribution space means. >I don't think so because it either >forces its indicating functions to vanish at a bounded rate or else it has to be non >supported. Given that, how can the the dirac delta still be a member of the integrables >when it clearly disobeys lebesgues integral theory in the limit of no indicative set >(or some would say the dominated convergence theorem - but i dunno how that works)??? >:( I am confused please point me to something that makes sense.....someone ..... If you want something that makes sense don't look at that last paragraph. What you should look at depends on what your question is. If all you wanted to know was whether the delta function was locally integrable then now you know, no it's not. ************************ David C. Ullrich X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i05BcBN05726; ==== >that the rep-tile is made up of k copies of itself differing only in >position, the positions of these copies corresponding to the positions >of the k hexes of the corresponding k-hex. I introduce the term hextal >to cover the members of this set of fractals. The set of hextals is uncountably infinite, and includes some well known >fractals such as the fudgeflake and flowsnake (Peano-Gosper curve). For each hextal there is a characteristic angle, which is the angle, >modulo pi/3, by which each copy is rotated relative to the overall >figure[1]. Additional hextals can be produced by rotating the copies by >an additional n.pi/3 radians, or by reflecting them in a suitably chosen >axis. This leads to 12 potential hextals for each polyhex, but not all >potential hextals are necessarily present[2], distinct, and connected. On the basis of examination of a considerable number of hextals, I >conjecture that the characteristic angle depends only on k, and not on >the arrangement of elements. Any ideas how this may be formally proven? >or is it an already known result? I believe that similar rules also apply to the fractal rep-tiles >corresponding to polyominoes and polyiamonds. I have observed that the >characteristic angle for polyominoes and polyiamonds with k^2 elements >is the same (0) as for polyhexes with k^2 elements. However the only >additional value (arctan(0.5)) that I know of for polyominoes >corresponds to rep-tiles with 5k^2 elements, and I believe that there >are no hextals with these numbers of elements. Hence there is little >evidence to conjecture that the same characteristic angles transfer over >to the rep-tiles corresponding to polyiamonds and polyominoes. It is >known that the characteristic angle for the two grids (corresponding to >particular rectangles and parallelograms) is pi/2 irrespective of k (but >in this case the vectors defining the grid depend on k, whereas for >those corresponding to polyhexes, polyiamonds and polyominoes the >vectors are independent of k). [1] With the exception of bar-hextals, in which the k-copies are laid >out in a straight line. In this case rotation by 0 or pi radians does >not generate a 2 dimensional figure. This halves the number of distinct >and present bar-hextals with a specific number of elements from 4 to 2. [2] In those known hextals with only translational symmetry, reflection >of copies in a suitably chosen axis in not productive of additional >hextals, and there are 6 potential and distinct hextals. Otherwise there >are 2, 4, or 6 hextals for each polyhex. >-- >http://www.meden. demon.co.uk/Fractals/fractal.html X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i05BcAh05702; ==== >i have to prove if sum(an) diverges then sum(sqrt(an)) diverges too. >this is what i think is a correct proof: since sum(an) diverges, there exists a real p, 0=1/n^(p/2). Since sum(1/n^(p/2)) diverges it >follows that sum(sqrt(an)) diverges. im making the obvious assumtion that an>=0 for all n, but i dont need >to say that in the proof do i? as i am writting this i also see i need >to exclude sequences that diverge themselves, since obviously if >an=(-1)^n+1 then my argument will not hold. is there anything else >wrong with this proof? thank you very much X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i05EYQC18043; ==== I am working on an engineering problem that has got stuck at a point where I need to find a closed form expression for the INVERSE of the following n-by-n matrix: -- -- | 1 1 . . . 1 | | x1 x1^2 . . . x1^n | | x2 x2^2 . . . x2^n | | . . . . . . | | xn-1 xn-1^2 . . . xn-1^n| -- -- Is this some kind of standard matrix(like a variant of the vandermonte matrix or something)? If it is, does a closed form expression exist for it's inverse. Please Help!! Ankit Seedher ==== I am working on an engineering problem that has got stuck at a point where I need to find a closed form expression for the INVERSE of the following n-by-n matrix: > -- -- > | 1 1 . . . 1 | > | x1 x1^2 . . . x1^n | > | x2 x2^2 . . . x2^n | > | . . . . . . | > | xn-1 xn-1^2 . . . xn-1^n| > -- -- Could you describe what problem you are trying to solve? It looks like you are trying to interpolate a polynomial, perhaps finding coefficients of a polynomial through given points. Well, this can easily be accomplished using Lagrange interpolation. However, often you don't need the explicit values of the coefficients, but merely an efficient means of evaluating the polynonial at arbitrary points. In that case you might consider Chebyshev polynomials. I'll happily expand on these matters if you are interested. If you really want an analytic expression for the inverse of the matrix, I'd guess a simple formula could be given, but I don't have it here. I suggest considering a 3x3 matrix and see if you can determine a pattern by visual inspection. -Michael. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i05HcvF32056; ==== When I showed the property and two proofs of G even ordered to my math professor today (who I hadn«t talked to on that subjet before) she reminded me at once that there are rings of characterstic 2 such that x = -x (i.e., thanks to Magidin and Chapman, it was a reminder to me). Then she advised me to take out the statement (i) G is a subgroup of R with respect to * -which is technically disturbing since R needn«t be a group- and replace it with something like (i)« there exists a homomorphismus taking G into R, etc. In addition, she said that both proofs aren`t really that different from each other, since I am basically using Lagrange«s theorem -albeit discretely- in the second proof, too. C.Dement ==== k in CQ such that A = {a + bk : a,b in Z} is a ring > with +, * operations of C. > Suppose A has exactly 4 elements invertible. > Prove that A = Z[i] But 1/u is also a unit so it must be equal to one of -1,1,u,-u. Only the last choice is possible, else u would equal -1 or 1. Therefore 1/u = -u, i.e. u^2 = -1 -Bill Dubuque ==== > .999... is an infinite series. > The limit of .999... is 1. > .999... is not 1, because it is an infinite series and 1 is not an > infinite series. > 1, as a decimal is actually 1.0000.... > But the value of .999... *is* the limit of the partial sums. > .9999.... is a different series of digits from 1.0000.... > But they are the same number. .9999... and 1.0000... are two different infinite series, with the > same limit (the possibility of which should be of no surprise.) Charlie Volkstorf Says he who swallows camels but srains at gnats. You can swallow whatever you please, man (except the truth, apparently.) C-B ==== The square of .999... is not equal to 1, therefore .999... is not equal to 1. 9^2 = 81 .9^2 = .81 99^2 = 9801 .99^2 = .9801 999^2 = 998001 .999^2 = .998001 9999^2 = 99980001 .9999^2 = .99980001 99999^2 = 9999800001 .99999^2 = .9999800001 999999^2 = 999998000001 .999999^2 = .999998000001 9999999^2 = 99999980000001 .9999999^2 = .99999980000001 99999999^2 = 9999999800000001 .99999999^2 = .9999999800000001 999999999^2 = 999999998000000001 .999999999^2 = .999999998000000001 9999999999^2 = 99999999980000000001 .9999999999^2 = .99999999980000000001 99999999999^2 = 9999999999800000000001 .99999999999^2 = .9999999999800000000001 999999999999^2 = 999999999998000000000001 .999999999999^2 = .999999999998000000000001 9999999999999^2 = 99999999999980000000000001 .9999999999999^2 = .99999999999980000000000001 99999999999999^2 = 9999999999999800000000000001 .99999999999999^2 = .9999999999999800000000000001 999999999999999^2 = 999999999999998000000000000001 .999999999999999^2 = .999999999999998000000000000001 9999999999999999^2 = 99999999999999980000000000000001 .9999999999999999^2 = .99999999999999980000000000000001 99999999999999999^2 = 9999999999999999800000000000000001 .99999999999999999^2 = .9999999999999999800000000000000001 999999999999999999^2 = 999999999999999998000000000000000001 .999999999999999999^2 = .999999999999999998000000000000000001 9999999999999999999^2 = 99999999999999999980000000000000000001 .9999999999999999999^2 = .99999999999999999980000000000000000001 99999999999999999999^2 = 9999999999999999999800000000000000000001 .99999999999999999999^2 = .9999999999999999999800000000000000000001 etc., etc., etc. The limited minds of the limited mathematicians saith not. Garry Denke, Geologist Denoco Inc. of Texas ==== >The limited minds of the limited mathematicians saith not. Namecalling will get you nowhere. Doug ==== > >The limited minds of the limited mathematicians saith not. Namecalling will get you nowhere. Doug There's nothing wrong with having a limited mind. That's why there is education. Garry Denke, Geologist Denoco Inc. of Texas ==== >The limited minds of the limited mathematicians saith not. Namecalling will get you nowhere. But he's getting there _fast_! ==== In sci.logic, Garry Denke on 5 Jan 2004 00:10:21 -0800 : > The square of .999... is not equal to 1, > therefore .999... is not equal to 1. 9^2 = 81 > .9^2 = .81 99^2 = 9801 > .99^2 = .9801 999^2 = 998001 > .999^2 = .998001 [rest snipped for brevity] Interesting logic, but doesn't quite work... :-) The square of .999... is .999... , as can readily be proven. If one solves the equation x^2 = x, one gets two values: 1 and 0. -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== >The square of .999... is not equal to 1, >therefore .999... is not equal to 1. .9^2 = .81 >.99^2 = .9801 [snip] >.99999999999999999999^2 = .9999999999999999999800000000000000000001 etc., etc., etc. The limited minds of the limited mathematicians saith not. Saith not what? Note that your argument has EXACTLY the same structure as: .9 does not equal 1 .99 does not equal 1 ... .99999999999999999999 does not equal 1 etc., etc., etc. You then infer that since none of the individual terms in the sequence equals 1, the *limit* of the entire sequence can not equal 1 either. This is mistaken, since the limit of a sequence of real numbers can be another real number that is NOT one of the terms of the sequence. And note that (.999...)^2 = lim(n->oo) (1 - 1/n)^2 = lim(n->oo) (1 - 2/n - 1/(n^2)) = 1 - 0 - 0 = 1. >Garry Denke, Geologist >Denoco Inc. of Texas -- --------------------------- | B B aa rrr b | | BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ----------------------------- ==== > >The square of .999... is not equal to 1, >therefore .999... is not equal to 1. .9^2 = .81 >.99^2 = .9801 > [snip] >.99999999999999999999^2 = .9999999999999999999800000000000000000001 etc., etc., etc. The limited minds of the limited mathematicians saith not. Saith not what? The square of .999... is not equal to 1, therefore .999... is not equal to 1. [snip your inference] > And note that (.999...)^2 > = lim(n->oo) (1 - 1/n)^2 > = lim(n->oo) (1 - 2/n - 1/(n^2)) > = 1 - 0 - 0 > = 1. The limited mind of the limited mathematician saith 1. Garry Denke, Geologist Denoco Inc. of Texas ==== >> >>The square of .999... is not equal to 1, >>therefore .999... is not equal to 1. >>.9^2 = .81 >>.99^2 = .9801 >> [snip] >>.99999999999999999999^2 = .9999999999999999999800000000000000000001 >>etc., etc., etc. >>The limited minds of the limited mathematicians saith not. Saith not what? > The square of .999... is not equal to 1, > therefore .999... is not equal to 1. Can you name a number between (0.999...)^2 and 1? -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== The limited mind of the limited mathematician saith 1. And what saith the limited mind of the limited usenet crank? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== > The limited mind of the limited mathematician saith 1. And what saith the limited mind of the limited usenet crank? Garry Denke, Geologist Denoco Inc. of Texas ==== What do you mean refer to? Its limit is 1 but the series itself is > not equal to 1. Nor does it ever reach 1. .999999... is a name; it's a symbol. Just as 1 is a symbol. The > symbol refers to a thing: the limit of a series of partial sums. As > it happens, the following are two different names, but names for the > same thing: 0.999999.... > 1.000000.... We use quotation marks when we speak of the name, and drop the marks > when we refer to the thing that the name names. So we have: 0.999999... is not equal to 1.000000... > but > 0.999999... = 1.000000... That is, these are two names for the same thing. We even have a rule > for which number a decimal names: it names the limit of a series of > partial sums. You are mistaken if you think that 1 behaves in some special way > that 0.999.... does not: they both work the same way: they are both > names for a number. Thomas I would argue that .999... is the term resulting from substituting 9 for N in the term .NNN... which represents the infinite series .N, .NN, .NNN, ... which has a limit of N/9. But then, I just like to argue. Charlie Volkstorf ==== > One of the common ways of defining the real numbers is to consider the > equivalence classes of Cauchy sequences of rationals. According to this > definition, the sequences > > 9/10, 99/100, 999/1000, ... > and > 1, 1, 1, 1, ... > > are members of the same equivalence class, and therefore are > representatives of the same real number. > >> Abe Lincoln: If you call a tail a leg, how many legs does a dog >> have? >> Spectator: 5. >> Abe Lincoln: No, 4. Just because you call a tail a leg doesn't make >> it a leg. > >> Not the least bit relevant. > > define = call > 9/10, 99/100, 999/1000, ... = tail > 1, 1, 1, 1, ... = leg > >> There are two common ways of defining the >> real numbers. One is using equivalence classes of Cauchy sequences of >> rationals, as I indicated. The other is using Dedekind cuts. By either >> method, it can be rigorously shown that 0.999... and 1 represent the same >> real number. > > That is what is not relevant. (It is a red herring.) Surely the definition of what the real numbers are is of some relevance > in deciding whether two real numbers are equal or not. You are misusing the word definition. Real numbers existed long before Cauchy and Dedekind. If we were talking about a particular term that someone coined to define a new construction (e.g. triangle), then we could refer to its definition to make sure we were using the term correctly. But the term real number does not refer to something originated by either of these two definitions. Real number is a term coined to represent the intuitive concept of (e.g.) measurement of a physical quantity to unlimited precision. It has its meaning independent of Cauchy sequences and Dedekind cuts. The above uses of Cauchy sequences and Dedekind cuts are attempts to construct mathematical objects that correspond to the real numbers. They define the construction of these objects and the correspondance with the real numbers. They do not define the real numbers themselves. > I have explained to you the following facts: 1. The real numbers are equivalence classes of Cauchy sequences > of rationals. 2. The real number 0.999... is the equivalence class containing > the Cauchy sequence 9/10, 99/100, 999/1000, .... 3. The real number 1.000... is the equivalence class containing > the Cauchy sequence 1, 1, 1, 1, .... 4. The equivalence classes mentioned in statements 2 and 3 are > in fact the same equivalence class. .999... is an infinite series, not an integer, with a limit of 1, an Nobody said anything about integers. The statement that .999... = 1 refers to the integer 1. > The fact that you have not seen this definition before How can you substantiate the above assertion? > does not change > the fact that it is indeed a standard way of defining the real numbers. No, it is a way of modeling the real numbers. Real number was defined long before that. Charlie Volkstorf ==== > There are two common ways of defining the > real numbers. One is using equivalence classes of Cauchy sequences of > rationals, as I indicated. The other is using Dedekind cuts. By either > method, it can be rigorously shown that 0.999... and 1 represent the same > real number. >> That is what is not relevant. (It is a red herring.) >> Surely the definition of what the real numbers are is of some relevance >> in deciding whether two real numbers are equal or not. > You are misusing the word definition. Real numbers existed long > before Cauchy and Dedekind. You don't understand how the term definition is used in mathematics. The real numbers were only vaguely understood before Cauchy and Dedekind. There were no definitions to make the concepts precise. > If we were talking about a particular term that someone coined to > define a new construction (e.g. triangle), then we could refer to > its definition to make sure we were using the term correctly. But the > term real number does not refer to something originated by either of > these two definitions. Real number is a term coined to represent > the intuitive concept of (e.g.) measurement of a physical quantity to > unlimited precision. It has its meaning independent of Cauchy > sequences and Dedekind cuts. If you don't have a definition, then how are you going to tell whether you are using the term correctly or not? Consider the title of Dedekind's paper: Was sind und was sollen die Zahlen?. Dedekind was concerned with the question of how the real numbers *ought* to be defined, precisely because they hadn't been defined yet. We now know that there are other number systems that could also be used to represent the intuitive concept of measurement of a physical quantity to unlimited precision., but these other number systems (such as the hyperreals or the surreals) do not have the property of being isomorphic to the real numbers. > The above uses of Cauchy sequences and Dedekind cuts are attempts to > construct mathematical objects that correspond to the real numbers. > They define the construction of these objects and the correspondance > with the real numbers. They do not define the real numbers > themselves. Wrong. By definition, the real numbers are the objects constructed by these methods. >> I have explained to you the following facts: >> 1. The real numbers are equivalence classes of Cauchy sequences >> of rationals. >> 2. The real number 0.999... is the equivalence class containing >> the Cauchy sequence 9/10, 99/100, 999/1000, .... >> 3. The real number 1.000... is the equivalence class containing >> the Cauchy sequence 1, 1, 1, 1, .... >> 4. The equivalence classes mentioned in statements 2 and 3 are >> in fact the same equivalence class. >> .999... is an infinite series, not an integer, with a limit of 1, an >> Nobody said anything about integers. > The statement that .999... = 1 refers to the integer 1. Only in the sense that the integers are a subring of the reals. The question of whether .999... = 1 makes no sense unless both sides are understood to be real numbers. >> The fact that you have not seen this definition before > How can you substantiate the above assertion? It's obvious that you have completely misunderstood the definition, if you have seen it. >> does not change >> the fact that it is indeed a standard way of defining the real numbers. > No, it is a way of modeling the real numbers. Real number was > defined long before that. By whom? All this is by the way, since you have already conceded the argument in another branch of this thread by admitting that there is no problem in defining 0.999... to mean the limit of the sequence 0.9, 0.99, 0.999, .... -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== >> .999... is an infinite series, not an integer, with a limit of 1, an > >> Nobody said anything about integers. > > The statement that .999... = 1 refers to the integer 1. Only in the sense that the integers are a subring of the reals. Concession accepted. >> The fact that you have not seen this definition before > > How can you substantiate the above assertion? It's obvious that you have completely misunderstood the definition, if > you have seen it. So you don't know that I haven't seen it after all. Now how have I demonstrated a misunderstanding? >> does not change >> the fact that it is indeed a standard way of defining the real numbers. > > No, it is a way of modeling the real numbers. Real number was > defined long before that. By whom? By its use in texts since antiquity. > All this is by the way, since you have already conceded the argument in > another branch of this thread by admitting that there is no problem in > defining 0.999... to mean the limit of the sequence 0.9, 0.99, 0.999, > .... You can define a symbol (e.g. 0.999...) to mean anything you want, as long as it has no meaning already. You can't define an intuitive concept (e.g. real number or integer) because it already has an intuitive meaning. You can only derive properties of it. That is the distinction that you are missing. C-B ==== > .999... is an infinite series, not an integer, with a limit of 1, an > Nobody said anything about integers. >> The statement that .999... = 1 refers to the integer 1. >> Only in the sense that the integers are a subring of the reals. > Concession accepted. An integer may be an equivalence class of ordered pairs of natural numbers, where (a,b) ~ (c,d) if a+d = b+c. Or an integer may be an equivalence class of ordered pairs of Cauchy sequences of rationals, with the obvious equivalence relation, in which one of the members of the equivalence class has the form of a constant sequence with all terms equal to n/1 for some integer n. I wanted to make it clear that I was talking about the second kind of integer, not the first. Both 0.999... and 1 = 1.000... are of this type. > The fact that you have not seen this definition before >> How can you substantiate the above assertion? >> It's obvious that you have completely misunderstood the definition, if >> you have seen it. > So you don't know that I haven't seen it after all. Now how have I > demonstrated a misunderstanding? You demonstrated that with your tail = leg nonsense, in which you clearly did not realize that when a real number is defined to be an equivalence class of Cauchy sequences of rationals, that means precisely that a real number is an equivalence class of Cauchy sequences of rationals. > does not change > the fact that it is indeed a standard way of defining the real numbers. >> No, it is a way of modeling the real numbers. Real number was >> defined long before that. >> By whom? > By its use in texts since antiquity. As I have said, the ancients had only an intuitive idea of what a real number was. >> All this is by the way, since you have already conceded the argument in >> another branch of this thread by admitting that there is no problem in >> defining 0.999... to mean the limit of the sequence 0.9, 0.99, 0.999, >> .... > You can define a symbol (e.g. 0.999...) to mean anything you want, > as long as it has no meaning already. You can't define an intuitive > concept (e.g. real number or integer) because it already has an > intuitive meaning. You can only derive properties of it. > That is the distinction that you are missing. There is no such distinction in mathematics. You can't derive properties of a mathematical object unless it either (1) has a definition, or (2) has properties that are derived from the basic axioms. The real numbers are an example of case (1). Sets and set membership are examples of case (2). Before the reals had a definition, people were merely assuming without proof that the reals had certain properties. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== > .999... is an infinite series. > The limit of .999... is 1. > .999... is not 1, because it is an infinite series and 1 is not an > infinite series. > Get over it. Talk about something significant (e.g. Program Synthesis > or Quine Atoms.) > Charlie Volkstorf > In that case, 0.333... is not 1/3 and 0.666... is not 2/3, and it must > be wrong to write 0.333... = 1/3 or 0.666... = 2/3, or any of the other > non-terminating repeating decimals as fractions!. No. The limit of .3, .33, .333, ... is in fact 1/3 just as the limit > of .6, .66, .666, ... is 2/3, of course. But when you leave out the > limit of (either in an English description or the definition of your > mathematical notation), then you are making a mistake and you > introduce the problems we are now seeing. You've contradicted yourself there. How 'zat? > Answer yes or no > 0.333... = 1/3 If you mean the infinitely long string 0.333... then no. If you mean the limit of the infinite series .3, .33, .333, ... then yes. Charlie Volkstorf > Although you have an interesting point is 0.99... an integer? Fuck no. > Herc ==== 1/10 + 1/10^2 + ...+1/10^n = 1 - 1/10^n This is true for all (finite) values of n. (1 - 1/10^n) > 1 This is also true for all finite values of n (and transfinite values too) Only when n achieves absolute infinity 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 infinity. No reaching of infinity is intended by this pristine phrase, so we are justified in maintainig that no limit of any kind could ever be realised by its use. Furthermore, if the 'tending' is towards something infinte short of absolute infinity 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 definition of =. > 0.999... is an infinite 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 ==== > 1/10 + 1/10^2 + ...+1/10^n = 1 - 1/10^n This is true for all (finite) values of n. (1 - 1/10^n) > 1 This is also true for all finite values of n (and transfinite values too) > Only when n achieves absolute infinity does 1/10^n become zero. Wow. TWO mistakes in three lines. (The sum is 1/9, not 1-1/10^n. There are no values of n with 1-1/10^n > 1.) That's an 0.666... batting average :-) Or 0.333..., depending on what you count as a success. And of course, 0.666... + 0.333... =========== = 0.999... --Ron Bruck ==== > 1/10 + 1/10^2 + ...+1/10^n = 1 - 1/10^n This is true for all (finite) values of n. (1 - 1/10^n) > 1 This is also true for all finite values of n (and transfinite values too) > Only when n achieves absolute infinity does 1/10^n become zero. I find that (1 - 1/10^n) > 1 is false for all real n. Did you mean (1 - 1/10^n) < 1 or did you mean (1 - 1/10^n) > 0 ? For natural number values of n, the values of 1/10^n form a sequence of rationals decreasing towards 0 as n increases without bound. But n is never infinite since natural numbers are all finite, so 1/10^n never becomes zero. The sequence is a Cauchy sequence and therefore has a limiting value, but that limit need not be one of the numbers in the sequence itself, just as least upper bounds and greatest lower bounds, where they exist, of sets of real numbers need not be members of the sets of which they are bounds. windows-nt) Cancel-Lock: sha1:K19z/4ttNyaXSuTtHlswcPFU1QY= ==== Mathematicians avoid this awkwardnwess by the weasel words: > the limit of 1/10^n tends to zero as n tends to infinity. Nit: the limit *IS* zero. 1/10^n *TENDS TO* zero. The terms are going someplace; the limit is not. Len. ==== Yes, the limit is zero. The phrase should be: the expression tends to a limit as n tends to infinity. However, the point is that such limits must remain unrealised 'at' infinity. Tony Thomas Mathematicians avoid this awkwardnwess by the weasel words: > the limit of 1/10^n tends to zero as n tends to infinity. Nit: the limit *IS* zero. 1/10^n *TENDS TO* zero. The terms are going > someplace; the limit is not. Len. > ==== > Yes, the limit is zero. The phrase should be: > the expression tends to a limit as n tends to infinity. However, the point is that such limits must remain unrealised 'at' > infinity. Perhaps I don't understand what you mean by must remain unrealised. But I suppose that I disagree with you. Of course, one may _choose_ to speak of only what happens as n increases without bound, thereby avoiding any reference to anything happening 'at' infinity. There's nothing wrong with doing that. But if one works in, say, the extended reals, then 1/10^x is continuous (from the left) at x = +oo, having the value 0 there. I suppose that is what you would call realisation of the limit at infinity. David Cantrell > Mathematicians avoid this awkwardnwess by the weasel words: > the limit of 1/10^n tends to zero as n tends to infinity. Nit: the limit *IS* zero. 1/10^n *TENDS TO* zero. The terms are going > someplace; the limit is not. ==== > .999... is an infinite series. > The limit of .999... is 1. > .999... is not 1, because it is an infinite series and 1 is not an > infinite series. > Get over it. Talk about something significant (e.g. Program Synthesis > or Quine Atoms.) > Charlie Volkstorf In that case, 0.333... is not 1/3 and 0.666... is not 2/3, and it must > be wrong to write 0.333... = 1/3 or 0.666... = 2/3, or any of the other > non-terminating repeating decimals as fractions!. No. The limit of .3, .33, .333, ... is in fact 1/3 just as the limit > of .6, .66, .666, ... is 2/3, of course. But when you leave out the > limit of (either in an English description or the definition of your > mathematical notation), then you are making a mistake and you > introduce the problems we are now seeing. Duh, that's what the ellipsis in .999... are for. Nobody claims that .9 =1, or .99 =1, etc. But .99..., which is to say, the limit of a series, is 1. Same with .3..., .6..., and so on. Ever heard of context sensitivity? 'cid ==== > .999... is an infinite series. > The limit of .999... is 1. > .999... is not 1, because it is an infinite series and 1 is not an > infinite series. > Get over it. Talk about something significant (e.g. Program Synthesis > or Quine Atoms.) > Charlie Volkstorf > In that case, 0.333... is not 1/3 and 0.666... is not 2/3, and it must > be wrong to write 0.333... = 1/3 or 0.666... = 2/3, or any of the other > non-terminating repeating decimals as fractions!. No. The limit of .3, .33, .333, ... is in fact 1/3 just as the limit > of .6, .66, .666, ... is 2/3, of course. But when you leave out the > limit of (either in an English description or the definition of your > mathematical notation), then you are making a mistake and you > introduce the problems we are now seeing. Duh, that's what the ellipsis in .999... are for. Nobody claims that > .9 =1, or .99 =1, etc. But .99..., which is to say, the limit of a > series, is 1. Same with .3..., .6..., and so on. Ever heard of > context sensitivity? I am referring to the statement that 'There is no real number between 0.999... and 1, and, therefore, they must be one and the same number!' treating .999... as a number in the same vein as 1, leading to two different (literal) numbers being equal. This problem also pops us in when we diagonalize and produce .999... and mathematicians say Oh no - that's 1 and 1 IS in our list! and needlessly jump through hoops to fix it. Charlie Volkstorf > 'cid ==== In sci.logic, |-|erc on Sun, 4 Jan 2004 15:16:36 +1000 > >> .999... is an infinite series. >> The limit of .999... is 1. >> .999... is not 1, because it is an infinite series and 1 is not an >> infinite series. >> Get over it. Talk about something significant (e.g. Program Synthesis >> or Quine Atoms.) >> Charlie Volkstorf >> In that case, 0.333... is not 1/3 and 0.666... is not 2/3, and it must >> be wrong to write 0.333... = 1/3 or 0.666... = 2/3, or any of the other >> non-terminating repeating decimals as fractions!. >> No. The limit of .3, .33, .333, ... is in fact 1/3 just as the limit >> of .6, .66, .666, ... is 2/3, of course. But when you leave out the >> limit of (either in an English description or the definition of your >> mathematical notation), then you are making a mistake and you >> introduce the problems we are now seeing. You've contradicted yourself there. > Answer yes or no > 0.333... = 1/3 Although you have an interesting point is 0.99... an integer? 0.333... = 1/3, and 0.999... is in fact an integer, although extremely technically speaking the ellipsis can be interpreted as ...444... or ...12385382... or any sequence. Most people assume unending 3's and 9's in the above cases. Herc > -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > 2 - 1 is a function, 1 is a number they're still equal. >> >> A number is a special case of a function (zero arguments - actually, >> everything is a special case of a function.) A number is not a >> special case of an infinite series (which is not what we're talking >> about anyway - we're talking about the limit of .999...) One of the common ways of defining the real numbers is to consider the >> equivalence classes of Cauchy sequences of rationals. According to this >> definition, the sequences 9/10, 99/100, 999/1000, ... >> and >> 1, 1, 1, 1, ... are members of the same equivalence class, and therefore are >> representatives of the same real number. Abe Lincoln: If you call a tail a leg, how many legs does a dog >have? >Spectator: 5. >Abe Lincoln: No, 4. Just because you call a tail a leg doesn't make >it a leg. But in mathematics if we say Definition: a _tail_ is a leg then a tail _is_ a leg. > What is the number between 0.99... and 1? >> >> There is none, because .999... isn't a number, remember? Wrong. Are you one of those fame-worshipers? .999... is an infinite series. >The limit of .999... is 1. >.999... is not 1, because it is an infinite series and 1 is not an >infinite series. Uh, no. By definition 0.999... is the _sum_ _of_ a certain infinite series. That infinite series has sum 1. So 0.999... = 1. >Get over it. Talk about something significant (e.g. Program Synthesis >or Quine Atoms.) Charlie Volkstorf ************************ David C. Ullrich ==== > >Abe Lincoln: If you call a tail a leg, how many legs does a dog >have? >Spectator: 5. >Abe Lincoln: No, 4. Just because you call a tail a leg doesn't make >it a leg. But in mathematics if we say Definition: a _tail_ is a leg then a tail _is_ a leg. Any mathematician who redefines an existing term ought to lose his job (IMHO.) >Get over it. Talk about something significant (e.g. Program Synthesis >or Quine Atoms.) Charlie Volkstorf ************************ David C. Ullrich ==== >> >>Abe Lincoln: If you call a tail a leg, how many legs does a dog >>have? >>Spectator: 5. >>Abe Lincoln: No, 4. Just because you call a tail a leg doesn't make >>it a leg. But in mathematics if we say Definition: a _tail_ is a leg then a tail _is_ a leg. Any mathematician who redefines an existing term ought to lose his job (IMHO.) Well, that's a bit more evidence regarding how much you know about how math works - mathematicians redefine existing terms all the time. >>Uh, no. By definition 0.999... is the _sum_ _of_ a certain infinite >>series. That infinite series has sum 1. So 0.999... = 1. That's not a _re_definition, by the way, it's simply the definition. Anyone pretending to mathematical competence who insists that 0.999... <> 1 has no idea what he's talking about (in the literal sense - he doesn't know the _meaning_ of the symbols he's using.) >>Get over it. Talk about something significant (e.g. Program Synthesis >>or Quine Atoms.) >>Charlie Volkstorf ************************ David C. Ullrich ************************ David C. Ullrich ==== > >> >>Abe Lincoln: If you call a tail a leg, how many legs does a dog >>have? >>Spectator: 5. >>Abe Lincoln: No, 4. Just because you call a tail a leg doesn't make >>it a leg. But in mathematics if we say Definition: a _tail_ is a leg then a tail _is_ a leg. Any mathematician who redefines an existing term ought to lose his job (IMHO.) Well, that's a bit more evidence regarding how much you know about how > math works - mathematicians redefine existing terms all the time. So what does integer mean these days? Do we have to reprint all our old math books now? Maybe if you programmed computers for awhile you'd learn to pay more attention to details. Or have you? > >>Uh, no. By definition 0.999... is the _sum_ _of_ a certain infinite >>series. That infinite series has sum 1. So 0.999... = 1. That's not a _re_definition, by the way, it's simply the definition. > Anyone pretending to mathematical competence who insists > that 0.999... <> 1 has no idea what he's talking about (in > the literal sense - he doesn't know the _meaning_ of the > symbols he's using.) You're just being sloppy with your statements. If you treat 0.999... as a literal equal to 1.000... then you have two different literals being equal, which is not a good idea. But if you are clear that you mean the limits of two series, then there is no problem. >>Get over it. Talk about something significant (e.g. Program Synthesis >>or Quine Atoms.) >>Charlie Volkstorf > ************************ David C. Ullrich ==== > .999... is an infinite series. > The limit of .999... is 1. > .999... is not 1, because it is an infinite series and 1 is not an > infinite series. > Get over it. Talk about something significant (e.g. Program Synthesis > or Quine Atoms.) > Charlie Volkstorf In that case, 0.333... is not 1/3 and 0.666... is not 2/3, and it must > be wrong to write 0.333... = 1/3 or 0.666... = 2/3, or any of the other > non-terminating repeating decimals as fractions!. No. The limit of .3, .33, .333, ... is in fact 1/3 just as the limit > of .6, .66, .666, ... is 2/3, of course. But when you leave out the > limit of (either in an English description or the definition of your > mathematical notation), then you are making a mistake and you > introduce the problems we are now seeing. Charlie Volkstorf There is no problem in logic or mathematics _defining_ such symbols as 0.999... or 0.333... as the limit of partial sums rather than as the sequence of partial sums. And there is no obligation on the rest of the world to reject such a useful and generally accepted definition simply because some dingbat gets anal retentive about it. Live with it. ==== > The limit of .3, .33, .333, ... is in fact 1/3 just as the limit > of .6, .66, .666, ... is 2/3, of course. But when you leave out the > limit of (either in an English description or the definition of your > mathematical notation), then you are making a mistake and you > introduce the problems we are now seeing. Charlie Volkstorf There is no problem in logic or mathematics _defining_ such symbols as > 0.999... or 0.333... as the limit of partial sums rather than as the > sequence of partial sums. Read what I said (above): when you leave out 'the limit of' in the definition you introduce problems. You are not leaving out the limit of and thus my statement does not apply to what you are saying. > And there is no obligation on the rest of the world to reject such a > useful and generally accepted definition want, as long as it is defined only once. Again my statement does not apply to what you are saying. > simply because some dingbat gets anal retentive about it. What's worse, being anal or being illiterate? > Live with it. Charlie Volkstorf ==== >> The limit of .3, .33, .333, ... is in fact 1/3 just as the limit >> of .6, .66, .666, ... is 2/3, of course. But when you leave out the >> limit of (either in an English description or the definition of your >> mathematical notation), then you are making a mistake and you >> introduce the problems we are now seeing. Charlie Volkstorf There is no problem in logic or mathematics _defining_ such symbols as >> 0.999... or 0.333... as the limit of partial sums rather than as the >> sequence of partial sums. > Read what I said (above): when you leave out 'the limit of' in the > definition you introduce problems. You are not leaving out the > limit of and thus my statement does not apply to what you are saying. Then you agree that 0.999... = 1. Because: (1) 0.999... means the limit of the sequence 0.9, 0.99, 0.999, ..., (2) The limit of that sequence is 1, and (3) Two things equal to the same thing are equal to each other. Simple, no? -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== > Then you agree that 0.999... = 1. Because: (1) 0.999... means the limit of the sequence 0.9, 0.99, 0.999, ..., > (2) The limit of that sequence is 1, and > (3) Two things equal to the same thing are equal to each other. Simple, no? If 0.999... means the limit of .9, .99, .999... (and is consistently used to mean that) then certainly 0.999...=1 C-B ==== > An interesting deduction chain but I do have to ask the obvious > question as to whether induction works on uncountable ordinals. In general, no. Induction is based on the well ordering principle > (every subset of natural numbers has a lowest term within it), which > does not apply for real subsets. It would also impose an ordering on > the reals, which is not kosher. There is lots more you can say about > it. Huh? An ordinal is by definition well-ordered. Thomas But simple induction does not work beyond the first infinite ordinal because there will be more than one member without predecessor. ==== > But this is exactly the definition of an infinite decimal. What do > you thing .999... means, if not the limit of the partial sums? > .999 means whatever mathematicians decide it means - but only once. > An infinite decimal is an infinite series of digits. An infinite series of digits is typographical. Which number does it refer to? Mathematicians have decided (once): > the limit of the partial sums. Thomas What do you mean refer to? Its limit is 1 but the series itself is > not equal to 1. Nor does it ever reach 1. Charlie Volkstorf If 0.999..., as shorthand for an infinite sequence of 9's following the decimal point, is a number then it equals 1. If 0.999...is NaN (not a number) then it is meaningless to ask whether 0.999... equals, or does not equal, some number, since such comparisons can only hold between numbers. So is it a number or not? Note that if 0.999... is NaN then there are a lot of rationals, and all irrationals, that can have no decimal representation. If 0.999... is NaN, why do you ask silly questions that would only make sense if 0.999... were a n umber? ==== > 3df1e59f.0401030355.35b0f9dc@posting.google.com>... > There is none, because .999... isn't a number, remember? 0.999... doesn«t refer to the sequence of the partial sums {0.9, 0.99, 0.999, ...}, which, in fact, is not a single number but a set of numbers. It refers to the limit of the sequence of the partial sums {0.9, 0.99, 0.999, ...}, which is identical with the sum of the series 0.9 + 0.09 + 0.009 + ... [= n=1|Sigma|inf:9/(10^n)]. And it is this very sum to which 0.999... refers! Sums, if existent, are numbers, and therefore 0.999... is a graphic representation of a number: the number 1. Basically, (convergent) series are sums, and sums are numbers. PH ==== > William Elliot >I can't prove that this equation : y^2=x^3 + 7 >has no integer solutions. Can you help me? Assume some integral solution for y^2 = x^3 + 7 > not 7|x. Otherwise: 7|x, 7|y > 0 = y^2 = x^3 + 7 = 7 (mod 7^2) which cannot be > x odd. Otherwise: x even, y odd, let y = 2n+1, x = 2m > 4n^2 + 4n + 1 = 8m^3 + 7; 1 = 3 (mod 4) which cannot be *** Is Z[sqr 7] a UFD, ie a unique factorization domain? *** > Yes, and a Euclidean domain. > LH So (1+sqrt(7))(-1+sqrt(7))=6=2*3. Thus 3 is not prime. What are its prime factors? Jon Miller ==== I've just bought a combination lock and was wondering how many combinations > it has - and more importantly how secure stuff secured using this lock > really is. I had a 3 number combnation lock and someone flicked through > them all and guessed it (only 999 combinations) which I experimented with > and it takes a meer 8 minutes. if the lock has digits from 0-9 then it would be 10^3 = 1000 combinations if the lock has digits from 1-9 then it would be 9^3 = 729 combinations I guess that this is a mathematics problem so would like to appeal to your > good nature in finding the answer. The new lock is as follows: It has 10 push buttons on the front, numbered 0 up to and including 9. Each > button has two positions, in or out. All numbers can be pushed in or out, in any order or combination. It has a preset to unlock it - 5 numbers must be pushed in, the remaining > five left out - these have to be pushed in, but not in any particular order. So essentially to unlock it, 5 out of the 10 numbers must be pushed in and 5 > must not. Given that the order the numbers are pushed in does not matter, and that we > know there are 5 numbers that must be pushed in to unlock it. What's the > maximum possible number of combinations of buttons one would have to try in > order to open the lock. anybody else? Also If the number of buttons that must be pushed in can be any number other > than 5 (this may be the case with these locks) how many combinations are > there then? then it would be 2^10 = 1024 combinations > Mike. > > ==== So essentially to unlock it, 5 out of the 10 numbers must be pushed in and 5 > must not. Given that the order the numbers are pushed in does not matter, and that we > know there are 5 numbers that must be pushed in to unlock it. What's the > maximum possible number of combinations of buttons one would have to try in > order to open the lock. Also If the number of buttons that must be pushed in can be any number other > than 5 (this may be the case with these locks) how many combinations are > there then? Are you asking anything different from the standard problem of how many ways there are of selecting k things from a set of n things, with your first example having k=5, n=10 ? ==== | | > | > So essentially to unlock it, 5 out of the 10 numbers must be pushed in and 5 | > must not. | > | > Given that the order the numbers are pushed in does not matter, and that we | > know there are 5 numbers that must be pushed in to unlock it. What's the | > maximum possible number of combinations of buttons one would have to try in | > order to open the lock. | > | > Also If the number of buttons that must be pushed in can be any number other | > than 5 (this may be the case with these locks) how many combinations are | > there then? | | Are you asking anything different from the standard problem of how many | ways there are of selecting k things from a set of n things, with your | first example having k=5, n=10 ? I don't think it is, but the poster doesn't seem to have any knowledge about combinatorics, so: choose 5 out of 10 (usually written (10 5)) is 10!/5!^2 = 252 for another number k of digits: n!/k!(n-k)!, where n is always 10 and n! = 1.2.3....(n-1).n You can now find the other answers yourselve I think. Greeting, Hendrik boundary=----=_NextPart_000_000A_01C3D7AF.D0A79C80 ==== --------------------------------------------------------------------- Rounding a number to the nearest 10,100,1000! If a number is exactly halfway to the next multiple of 10 say, the number is rounded up or down to the nearest next even multiple of 10, e.g. 35 is rounded up to 40 but 65 is rounded down to 60. Is this statement true? For instance can 14 525 be rounded to 14000 or 15000? ==== > Rounding a number to the nearest 10,100,1000! > If a number is exactly halfway to the next multiple of 10 say, the number > is rounded up or down to the nearest next even multiple of 10, e.g. 35 is > rounded up to 40 but 65 is rounded down to 60. Is this statement true? For > instance can 14 525 be rounded to 14000 or 15000? What you describe is a variation of bankers rounding. http://www.teradataforum.com/teradata/20021106_141731.htm You can define your own rounding algorithm to be anything you want. gtoomey ==== > Rounding a number to the nearest 10,100,1000! > If a number is exactly halfway to the next multiple of 10 say, the number is > rounded up or down to the nearest next even multiple of 10, e.g. 35 is > rounded up to 40 but 65 is rounded down to 60. Is this statement true? For > instance can 14 525 be rounded to 14000 or 15000? > The rule that is in place needs to be applied consistently, so if the last digit is 5 then you will always round in the same direction. The statement that 35 is rounded up to 40 but 65 is rounded down to 60 is untrue if there is a consistent rule in place. The convention is to round up if the value is exactly half way. Rounding to the nearest multiple of 1000, 14525 is rounded up to 15000. It is not good accounting to round a number multiple times, which seems to be what you are suggesting in this instance. ==== Rounding a number to the nearest 10,100,1000! > If a number is exactly halfway to the next multiple of 10 say, the number is > rounded up or down to the nearest next even multiple of 10, e.g. 35 is > rounded up to 40 but 65 is rounded down to 60. Is this statement true? For > instance can 14 525 be rounded to 14000 or 15000? > The rule that is in place needs to be applied consistently, so if the > last digit is 5 then you will always round in the same direction. The statement that 35 is rounded up to 40 but 65 is rounded down to 60 > is untrue if there is a consistent rule in place. Rounding halfway values to make the previos digit even is a consistent and widely used rule. That you may not be familiar with it does not invalidate its consistency. The convention is to round up if the value is exactly half way. Rounding to the nearest multiple of 1000, 14525 is rounded up to 15000. It is not good accounting to round a number multiple times, which seems > to be what you are suggesting in this instance. If you use the nearest even rule for half-way points, multiple rounding is no problem, as the final result is not affected by the number of roundoffs it takes to get there, whereas the round up for half way points can introduce significant errors in multiple roundoffs. Consider rounding off 4.4444445 one digit at a time until you have an integer. By the round even rule you get 4 but by the round up rule you get 5. Which is better? ==== > Rounding a number to the nearest 10,100,1000! > If a number is exactly halfway to the next multiple of 10 say, the number is > rounded up or down to the nearest next even multiple of 10, e.g. 35 is > rounded up to 40 but 65 is rounded down to 60. Is this statement true? For > instance can 14 525 be rounded to 14000 or 15000? There are different rules for rounding decimals to a particular place value. As far as I kno,w all of them round to the nearest number having all zeros to the right of the given place value, but differ on what to do with values falling exactly half way between two such numbers. One such rule says always to round the exact half way values up, though what up means for negatives is a bit ambiguous. This is easiest to program, so it what many computers do. Your rule seems to be to round such midway values so that the digit in the given place value is even. This is the preferred method in statistical work for a variety of reasons, e.g., (1) Rounding in this fashion does not tend to introduce a systematic bias as round up does. to the data as the above round up rule does. (2) Successive roundings, a bit at a time, always give the same value as a single large roundoff. For example, rounding off 0.4444445 one digit at a time until you have an integer gives 1.0 for round ups, but 0.0 for round even, which is a better result.