==== mm-484 >Subject: HELP: combinatorics straining my limited brain capacity >Desperately need help on this one. Dont even know where to start. Here >goes: >2*n different persons stand in row. The first n persons wear a red hat, the >next n persons wear a blue hat. What color are the rest of the hats? >Hats are distinguishable(even those of the >same color). Also, persons are distinguishable. The hats are thrown up in >the air and land, one hat for each person. Calculate the number of ways the >hats can land so that >a) the n first persons still wear a red hat >a) the n first persons wear a blue hat >c) the color of the hat alternate among the persons, and the first person >wears a red hat >d) formulate an assumption that generalizes the result of a), b) and c) === Subject: Re: HELP: combinatorics straining my limited brain capacity > What color are the rest of the hats? What? n+n=2n. Anyway, the problem is fairly simple. (I say problem because the first three questions are just rewordings of each other and the last is then obvious.) There are n! ways to put n hats on n heads. For each of these states, there are n! ways to put the other n hats on the other n heads. So the number of ways to put a specific set of n out of 2n hats on a specific set of n out of 2n heads is always (n!)^2. === >Subject: Re: HELP: combinatorics straining my limited brain capacity >Message-id: What color are the rest of \ the hats? >What? n+n=2n. Sorry, I misread it. Thought it was 2^n hats. >Anyway, the problem is fairly simple. (I say problem because the >first three questions are just rewordings of each other and the last is >then obvious.) >There are n! ways to put n hats on n heads. For each of these states, >there are n! ways to put the other n hats on the other n heads. So the >number of ways to put a specific set of n out of 2n hats on a specific >set of n out of 2n heads is always (n!)^2. === Subject: 2 problems for you to solve by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1FKF4v06796; CHALLENGE 1: Can anybody prove/disprove: Let a > 0, b > 0 and c > 0 integers, a different from b, and d > 0 with d^2 = a^2 + b^2. If the discriminant of the equation is not 0 and c > 2d, then the cubic equation t^3 + (d^2 - c^2) t^2 + 4 a^2 b^2 c^2 = 0 has no integral roots. Note that if a = b then the cubic has the integer root - 2a^2. Note that if the dsicriminant is 0 then all roots of the cubic are integer roots. CHALLENGE 2: Can anybody prove/disprove: For a > 0 and b > 0 integers, (a^2 + b^2) (b^2 - 8 a^2) is never a square. === Subject: Re: 2 problems for you to solve > CHALLENGE 1: > Can anybody prove/disprove: > Let a > 0, b > 0 and c > 0 integers, a different from b, and d > 0 with d^2 = a^2 + b^2. > If the discriminant of the equation is not 0 and c > 2d, then the cubic equation t^3 + (d^2 - c^2) t^2 + 4 a^2 b^2 c^2 = 0 has no integral roots. Notice about the cubic : - The fact that c > 2d implies three real roots, two positive and one negative (that can be deduced from the stationary points). - Not all roots can be integers, because, if we call them r1, r2 and r3, we have r3 = - r1 r2 / (r1 + r2) since the 1st-deg coefficient of the cubic is zero, and this can only be an integer if r1 = r2 (impossible with a nonzero discriminant). - Let r1 be an integer root and r2 a non-integer one. Then, from the 2nd-deg coefficient of the cubic and the previous equation for r3, we arrive at the equation (r1^3 - r2^3) / (r1^2 - r2^2) = c^2 - d^2. But the LHS cannot be an integer by construction (since r1 is integral and r2 is not). Thus the cubic cannot have any integer roots. Dimitris === Subject: Re: 2 problems for you to solve >CHALLENGE 2: >Can anybody prove/disprove: >For a > 0 and b > 0 integers, (a^2 + b^2) (b^2 - 8 a^2) is never a square. Assume GCD(a, b) = 1. Let c = sqrt(a^2 + b^2). Look at the four values c + 3*a c + a c - a c - 3*a Show that all are squares or all are twice squares. I recall a theorem that there cannot be four squares in arithmetic progression unless all are equal. === Subject: Re: 2 problems for you to solve >CHALLENGE 2: >Can anybody prove/disprove: >For a > 0 and b > 0 integers, (a^2 + b^2) (b^2 - 8 a^2) is never a square. > Assume GCD(a, b) = 1. Let c = sqrt(a^2 + b^2). Look at the four values > c + 3*a > c + a > c - a > c - 3*a > Show that all are squares or all are twice squares. > I recall a theorem that there cannot be four squares > in arithmetic progression unless all are equal. Implicit in the above is that if (a^2 + b^2) (b^2 - 8 a^2) is a square then a^2 + b^2 and b^2 - 8 a^2 are both squares. Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Graph Theory: Number of maximal Matchings > Hi Robert, > you are right, I should more precisely specify my problem. > Let G = (V1, V2, E) be a bipartite graph with > 1) Edges connects vertices from V1 with V2 only > 2) |V1| = |V2| > 3) There exists a perfect matching covering all vertices > My Problem: Is there a tighter bound on the maximal number of perfect > matchings than n!? An upper bound on the number of perfect matchings for your graph with |V1|=|V2|=n and maximum degree d is F(n) where F(n) = F(n-1) + (d-1)^2 F(n-2) (i.e. choose x1 in V1 and a neighbour y1 in V2: either you match x1 with y1 or match x1 with some other member of V2 and y with some other member of V1) F(d) = d!, F(d-1) = (d-1)! The solution of this recurrence is F(n) = (d-1)!(r_1^(n+1-d)(r_2-d) - r_2^(n+1-d)(r_1-d))/(r_2-r_1) where r_1 and r_2 are the roots of z^2 - z - (d-1)^2. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 > For example if |V1| = |V2| = 3 and each vertex has degree one, then I > conclude the number of perfect matchings is one. But is there a general > formula as a function on the number |E| of edges , degree sequence, etc, > which gives a good bound on the number of perfect matchings? > Best wishes > bjj >let G = (V, E) be a bipartite graph with matching number m(G) = n. > Meaning, I guess, that a maximal matching is of size n. >Assume that V is the disjoint union of the subset U1 and U2 both with >cardinality |U1| = |U2| = n. Edges connect each vertex from U1 with one >or more vertices of U2. Is there any result about the number of maximal >matchings in G? > It could be anywhere from 1 (e.g. if every vertex has degree 1) > to n! (if every vertex of U1 is connected to every vertex of U2). > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 === Subject: Re: pentiamond rep-tiles Originator: gls@news.monmouth.com (Col. G. L. Sicherman) > I think I have a dissection of the pentabolo corresponding to the first, > which is composed of copies of itself either of the original orientation > or rotated by 180 degrees. This can be deformed to give a dissection of > the pentiamond. However this is an ir-rep-tiling. Your web site mentions that there are 5 pentiamonds--what is the fifth? -:- POZZO: He used to dance the farandole, the fling, the brawl, the jig, the fandango and even the hornpipe. He capered. For joy. Now that's the best he can do. Do you know what he calls it? ESTRAGON: The Scapegoat's Agony. VLADIMIR: The Hard Stool. POZZO: The Net. He thinks he's entangled in a net. --Samuel Beckett, _Waiting for Godot_ Col. G. L. Sicherman colonel@mail.monmouth.com === Subject: Re: pentiamond rep-tiles I need to solve this equation step by step. >y(x) = y(x)/x + 3(y(x))^2 >Could someone help me? I think that this is Bernoullis equation, But not sure. >Jaakko Others have given solutions, but you are correct that it is a Bernoulli equation. If you wish to solve it that way rewrite it, dividing by y^2: y^(-2) y' -(1/x)y^(-1) = 3 Now let u = y^(-1) (this is the Bernoulli substitution) Then u' = - y^(-2) y'. Plug these in: -u' - (1/x)u = 3 xu' + u = -3x which is a linear equation (already exact). Solve it for u then solve for v. --Lynn === Subject: Re: Please help me >Solve it for u then solve for v. >--Lynn Typo - then solve for y. === Subject: Re: expectation problem === >Subject: Re: expectation problem >Message-id: === >Subject: Re: expectation problem >Message-id: === >Subject: Re: expectation problem >Message-id: To keep total prize money paid to a minimum, how fast do you want >the >voice guessed? > And no one would budget the worse case scenario of 60 quid/day. > They would look at the average number of trials to get the answer, say > 6, and set the budget accordingly. The game planners task is then to > see that the contest remains popular by giving away the amount they > have budgeted for, but not give away too much so as to go over budget > and get their bosses mad at them. > Without altering the prize schedule, the proper way to frame the > original question would be: > Given a budget of x quid/week, how fast do you want the voice guessed? Now thats a better Question >-- >If its Monday then I am a fool but not ignorant. > -- > Mensanator > Ace of Clubs Once every 3 days for MARKETING :) Cost 30 quid + (y-1).10 quid/day Budget x quid/week => a quid/day E[y] == y {not formal E[] expected value} (Budget-Cost)*E[y]>=0 a= 10 quid/day + 20/(E[y]) E[y] is what the producer controls by choice of voice. E[y] >= 1 Carl If its Monday then I am a fool but not ignorant. === Subject: Re: A tricky integral > Does anyone know if there is a function for the following integration > with respect to X please : SQRT[1 + A*(SIN(X))^2 + B*(SIN(X))^4] > where A and B are positive constants and the limits are from zero to Y > (pi/2 < Y < pi). >Mathematica seems to give an answer. It is very lengthy, quite horrific, >in terms of incomplete elliptic integrals of all three kinds. >BTW, I'd almost be willing to bet that it's incorrect for at least some >values of A, B, or Y. terms of cos(2x), the result is the square root of a quadratic in cos(2x). To get a more usual form, using cos(2x) as the new variable gives a ratio of two square roots of quadratics. This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: . Zeno's paradox . X-SessionID: 5sRXb-1326-_4-3037@news.uchicago.edu X-Hash-Info: post-filter,v:1.4 X-Hash: a2e2a8b7 cb2eb6aa 36df0692 d022b290 049ded31 Huh? --Dan Weiner === Subject: . A Zero . Hi Dan Weiner, Taking all of 5 seconds from your life, you pause to ask this very in depth question: Huh ? , On a scale of 0 to 9, I rank your post a Zero . === Subject: Re: . A Zero . X-SessionID: Wo%Xb-3352-_4-5831@news.uchicago.edu X-Hash-Info: post-filter,v:1.4 X-Hash: 206cbaed d439e7bc d67db555 1678d42d 7cddb2ae > Hi Dan Weiner, > Taking all of 5 seconds from your life, > you pause to ask this very in depth question: Huh ? , > On a scale of 0 to 9, I rank your post a Zero . Duly noted. === Subject: Re: . A Zero . X-SessionID: uo%Xb-3351-_4-5658@news.uchicago.edu X-Hash-Info: post-filter,v:1.4 X-Hash: 1e9fc015 dd6b2947 9fa16c07 53d3d8cd d89d3e04 > Hi Dan Weiner, > Taking all of 5 seconds from your life, > you pause to ask this very in depth question: Huh ? , > On a scale of 0 to 9, I rank your post a Zero . ..duly noted... === Subject: A trend ? Hi Dan Weiner, Re: How I feel about your one word posts, You added two words: duly noted , Do I detect a trend here ? === Subject: Re: A trend ? > Hi Dan Weiner, > Re: How I feel about your one word posts, > You added two words: duly noted , > Do I detect a trend here ? But will his next post have three words or four? === Subject: Re: A trend ? X-SessionID: bh5Yb-4311-_4-7062@news.uchicago.edu X-Hash-Info: post-filter,v:1.4 X-Hash: 8bffad27 bd82e25e d4450bc3 f711729a 0b598ed9 An arithmetic series? --Dan Weiner === Subject: Re: . Zeno's paradox . > Huh? > --Dan Weiner Don't worry about it. It's just more grist for his special mill. -Mark Martin === Subject: Re: Tea with Sarfatti > Tea with Sarfatti at Kensington Palace Gardens > So Sarfatti has graduated from having conversations with himself > to having tea with himself, I see. Chatting himself up, as a prelude to going home and having sex with himself. Gib === Subject: Re: Tea with Sarfatti > Tea with Sarfatti at Kensington Palace Gardens > So Sarfatti has graduated from having conversations with himself > to having tea with himself, I see. > Chatting himself up, as a prelude to going home and having sex with himself. Isn't he worried about rejection? John Wilkins Men mark it when they hit, but do not mark it when they miss - Francis Bacon === Subject: Re: Tea with Sarfatti >Tea with Sarfatti at Kensington Palace Gardens >So Sarfatti has graduated from having conversations with himself >to having tea with himself, I see. >Chatting himself up, as a prelude to going home and having sex with himself. > Isn't he worried about rejection? I don't think there's any worry about that. He's his own biggest fan. -E === Subject: test NNTP-Proxy-Relay: library1-aux.airnews.net new account, just testing === Subject: Re: proof of the dimension theorem of linear algebra >Hi all, [text snipped] >But all these countings ... we have just pushed them into the domain >of common sense [my book reads, we must end up with at least one row >of zeroes]. This amounts to, IMHO, the same thing that we are >supposed to prove. Can you guys provide me with an elegant proof of >this fundamental theorem, using rigorous arguments only, extending to >infinite cardinalities? I'm going to give you a proof tailored for the infinite case. First, an easy application of Zorn's lemma gives that any vector space V has maximally linearly independent sets. Lemma: Esubseteq V is a Hamel basis iff it is maximally linearly ordered. proof: This is easy and I'll leave it to you. Arguing by contradiction, it's easy to establish that Lemma: If Esubseteq V is a basis then no proper subset of E generates V. With this lemma, and arguing by contradiction, it's easy to see that Lemma: If Esubseteq V is an infinite basis and S generates V then #E <= #S. With these easy preliminaries out of the way we have Theorem: If E and E' are two infinite basis of V then #E = #E'. proof: For each e in E define the subset B(e) of E' by B(e) = {e'in E': e is a linear combination of e'} This set is finite and it's clear that bigcup_{ein E}B(e) generates V. By the above lemma, #E' <= #(bigcup_{ein E}B(e)) <= #E*#Natural = #E Reverse the roles of E and E' to get the reversed inequality. Q. E. D. G. Rodrigues === Subject: Re: how to calculate the degrees of an angle? > Knowing the sides of the triangle, and knowing that its a right > triangle, how do you find out the degrees of each angles? (One is 90, > I know that.) > For example, if I got a triangle with sides (0,0), (1,0), (0,2) what > are the angles? Use trigonometry. In this case, your legs are both the same length (1) so you know that the angles are 45, 45 and 90. But, in the case of legs of different lengths, use trig functions: A | | o | h +-+ | | +-+--- B a C I'll describe it in pseudo-code below, and assume that the angle at C is theta. Get the length of the legs and the hypoteneuse first using basic algebra: o = Ay - By; a = Cx - Bx; h = sqrt(o^2 + a^2); Now take these values to get the sine for A and C: sinc = o / h; sina = a / h; Now you can use sin^-1 to convert these values into radians and convert the radians into degrees. You're all set! :) > Please don't tell me to find out the tangent and look it up on a > table. If that's the answer, then how do I calculate the values of > that table? How are you going to determine the values? Are you doing this in software or by hand or with a calculator? Darryl L. Pierce Visit the Infobahn Offramp - What do you care what other people think, Mr. Feynman? === Subject: Re: how to calculate the degrees of an angle? ETAtAhUAwOa3EFFl6pEpAU7eOk2+6DClBokCFDHSVqG/ F81FafGgvt4ytH7J2XvD I would find the lengths of the sides from the vertex coordinate, then use the Law of Cosines. --OL === Subject: Re: Rudin's proof of the open mapping theorem >Well, I already knew that, for topological groups, T_0 <=> T_3. > Figured it must be in Rudin, I don't have the book here. I didn't say that it was in Rudin; I haven't seen it there. > This morning I thought that this was the only problem. > Let's see: > We want to show that y_n -> 0. Let O be a neighborhood > of the origin. Choose W, a neighborhood of the origin > such that the closure of W is contained in O. > Now since L is continuous there exists delta > 0 so that > d(x, 0) < delta implies Lx is in W. Hence there exists > N so that L(V_n) is a subset of W for all n > N. So the > closure of L(V_n) is contained in O for all n > N. So > y_n is in O for all n > N. > So for every O there exists N such that y_n is in O for > all n > N. Maybe I'm missing something here... Jose Carlos Santos === Subject: Re: Rudin's proof of the open mapping theorem >Well, I already knew that, for topological groups, T_0 <=> T_3. > Figured it must be in Rudin, I don't have the book here. >I didn't say that it was in Rudin; I haven't seen it there. Great - if it's not in Rudin we can publish it! I imagine it may be in a few other places, but I don't think they count. > This morning I thought that this was the only problem. > Let's see: > We want to show that y_n -> 0. Let O be a neighborhood > of the origin. Choose W, a neighborhood of the origin > such that the closure of W is contained in O. > Now since L is continuous there exists delta > 0 so that > d(x, 0) < delta implies Lx is in W. Hence there exists > N so that L(V_n) is a subset of W for all n > N. So the > closure of L(V_n) is contained in O for all n > N. So > y_n is in O for all n > N. > So for every O there exists N such that y_n is in O for > all n > N. Maybe I'm missing something here... No charge. Let's just be careful about suggesting that results in Rudin's books might be wrong unless we have an actual counterexample, eh?... If graph G is a minimal counter example [mce] to the FCT; then, > G is maximal planar, > G is internally 6-connected and > every vertex in G has minimum degree = 5. > Is this essentially a correct and complete description of an mce? There is no complete description of a hypothetical counterexample to a proposition. The above hypotheses ensure that G is not reducible in any of three ways, but there might be other ways. If you're out to prove FCT by contradiction, you need only prove that the counterexample has _some_ properties -- whichever properties will be needed to reach a contradiction. === Subject: Re: Four Color Theorem > If graph G is a minimal counter example [mce] to the FCT; then, > G is maximal planar, > G is internally 6-connected and > every vertex in G has minimum degree = 5. > Is this essentially a correct and complete description of an mce? The minimum degree is usually defined for a graph... min degree(G) is smallest of degrees over all vertices. What do you mean that *every vertex* has a minimum degree of 5? J === Subject: Re: Four Color Theorem 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: > If graph G is a minimal counter example [mce] to the FCT; then, > G is maximal planar, > G is internally 6-connected and > every vertex in G has minimum degree = 5. > Is this essentially a correct and complete description of an mce? How could it be complete when it omits the more important property of nonexistence? David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science === Subject: Re: Four Color Theorem > If graph G is a minimal counter example [mce] to the FCT; then, > G is maximal planar, > G is internally 6-connected and > every vertex in G has minimum degree = 5. > Is this essentially a correct and complete description of an mce? > How could it be complete when it omits the more important property of > nonexistence? Can you prove it? === Subject: Permutations let S_8 be the set of all possible permutations of N_8. Let pi be any permutation in S_8. Say i want to count the number of permutations pi in S_8 such that , in cycle notation, all the cycle lengths of pi are even. I would proceed like this. Let ^ denote superscript. Let A=[8] be the set of permutations with 1 cycle of length 8 B=[6 2 ] the set of permutations with one cycle of length 6 and one cycle of length 2 C=[4^2 ] the set of permutations with two cycles of length 4 D=[4 2^2 ] the set of permutations with one cycle of length 4 and two cycles of length 2 E=[2^4 ] the set of permutations with four cycles of length 2 summing the cardinality of all the sets would then answer the original question am i right? However i am told that this sum can be written down as an expression with binomial numbers. How can i arrive at such an expression ? === Subject: Re: Stuck on two combinatorics problems En el mensaje:L7vXb.49040$mU6.192397@newsb.telia.net, Valentino Berti escribi.97: > yes i forgot to the condition p =/= . Sloppy of me. > 2) Show that if p and q are odd primes then > 2^(pq+1) [equivalance symbol] 2^(p+q) (mod pq) By Fermat Little Therem, 2^p = 2 (mod p). Then LHS: 2^(pq + 1) = 2*(2^p)^q = 2*2^q (mod p) RHS: 2^(p + q) = 2^p*2^q = 2*2^q (mod p) In the same way, both sides are equals (mod q). Then, as p =/= q and gcd(p, q) = 1, both sides are equals (mod pq). > If p = q odd prime it fails. Did you forgotten the condition p =/= q? Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Stuck on two combinatorics problems > yes i forgot to the condition p =/= . Sloppy of me. > 2) Show that if p and q are odd primes then > 2^(pq+1) [equivalance symbol] 2^(p+q) (mod pq) change the form to 2^(p+q)(2^(pq-p-q+1)-1) =2^(p+q)(2^(p-1)(q-1)-1) Now since p q prime mod p,q -{0} are groups under * and 2^(p-1) is not 0 mod q since q is a odd (prime). Thus by the well known elementary theorem from group theory 2^(p-1)(q-1) is 1 mod q Do the same in reverse and we get 2^(p-1)(q-1) is 1 mod p since p,q relatively prime(since p n.e. q) 2^(p-1)(q-1)-1 is divided by pq Q.E.D. === Subject: Graph theory question Hi all, can somebody help me with this? Let (V,E) be an undirected graph that consists of one component, with n vertices of odd degree. Show that there are n/2 walks, none of all! GB === Subject: Re: Graph theory question >Let (V,E) be an undirected graph that consists of one component, with >n vertices of odd degree. Show that there are n/2 walks, none of >all! Hint: First add n/2 edges to pair up the odd-degree vertices. Now there are no odd-degree vertices any more, so ... Then delete the added edges. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Probability of finding a Prime Number Hello Everybody, I do want to say thank you for your comments -- particularly the more recent ones. I'd like to explain my use of probability. I rely on probability being defined as the ratio of the number of favorable outcomes to the total number construct to search for primes. The table is a variation of the Sieve of Eratosthenes. The rows are numbered from 2 to n and the columns are numbered 1 to m. I put a 1 in a cell of the table if a column can be divided by a row. It is an interesting execise to do this because a variety of patterns emerge. These patterns are particularly evident if you build very large tables and convert them to bitmaps. The most prominent feature is that a diagonal emerges that runs from the top left to the bottom right of the table. This is not surprising because it corresponds to a number being divided by itself. It became apparent to me that the columns that contain all blanks above this central diagonal are prime. So, what is the probability that a column will contain all blanks above the diagonal? There is 1 possible way out of 2 to do this in the row labled 2. In row 3 it is possible to choose a blank two ways out of three. In row 4 the probability becomes 3/4. The probability that all the rows in a given column are blank is the product of these probabilities. This is equivalent to saying that a randomly chosen column corresponds to a prime number. What if the number were simply odd? All of the even numbered rows MUST contain a blank. Therefore, the probability of a blank in an even numbered row is 1. So, the probability of a column containing all blanks becomes 2/3 * 1 * 4/5 * 1 * 6/7 * ... * (r-1)/r. This becomes 1 * Product((r-1)/r) where r is odd or simply Product((r-1)/r) I am going to jump ahead a bit and say that if all the prime rows contain a blank in a randomly chosen column, then the column MUST contain all blanks. In other words, the probability of a non prime row being blank must be 1 if the prime rows are all blank. What is the probability of all the prime rows being blank? Product((p-1)/p) where p is prime from 3 to some number n. This is why I call it a probability. However, many of you (maybe all of you) say this is not a correct use of the term probability. I guess I do not see why not. David > I have placed a paper on my web site at http://www.guffy.net/primes.pdf. > In it I prove the following: > Given the prime numbers 3 to p, the probability of finding the next prime is > exactly: > Product((p-1)/p) > The paper is 3 pages and contains 4 equations. I would appreciate any > comments. > David === Subject: Re: Probability of finding a Prime Number >Therefore, the probability of a blank in an even numbered >row is 1. So, the probability of a column containing all blanks becomes 2/3 >* 1 * 4/5 * 1 * 6/7 * ... * (r-1)/r. But your original assertion did not say this. You said something about the probability of finding the next prime. And that statement *was* nonsense on its face. Furthermore, you did NOT say I think I may have a proof......Will someone please check it? Instead, you asserted that you DID have a proof. Furthermore, your argument is *almost*, but not mine, but part of that seems to have been left out. *Seemingly* the prob that N is prime should be prod(1-1/p) for p < sqrt(N). But there is a subtle problem that I discussed in the prior post, but which seems to have ben omitted. Let N be a large but unspecified integer. By the PNT, the probability that it is prime is ~1/log(N). Now consider the following. *IF* (and as we shall see this is a big if) the probability that a prime p divides N is independent of the probability that other primes divide N, then indeed the probability that N is prime is product of (1-1/p_i) for p_i < SQRT(N). However: Consider P = product(p_i) for p_ i < x. As soon as x ~ log(N), the probabilities are no longer independent. Why? Because by the PNT we have psi(x) ~ x and thus P > N. i.e. once you get enough primes in the probability product, then the product of p will exceed N. Now, the probability of the next prime you add into your product is no longer independent. What you are attempting is a sieve method proof of the PNT and it can't be done. I now have to get technical. Something known as the fundamental lemma of the sieve gets in the way. You can not insert SQRT(N) primes into the product, because you lose independence. (also estimation error starts to creep in, but I don't discuss that here) See Halberstam & Richert Sieve Methods. Be careful; it is a difficult book. I have kicked it across the room more than once in frustration. Also look up Mertens' theorem in Hardy & Wright. The constant one gets in Mertens' theorem (e^gamma) differs from the constant in the PNT (which is 1) precisely because independence is lost as soon as one forces the number of primes in the product beyond log(N). Indeed. If one asks, what is the probability that a large integer N is not divisible by any primes less than x where x is SUFFICIENTLY SMALL WITH RESPECT TO N, then the answer is indeed product(1-1/p) for p < x. But you can't take x beyond log(N). And to determine the probability that N is prime you need to take x up to sqrt(N). Thus you can't really use this argument to estimate the probability that N is prime. You can lead a horse's ass to knowledge, but you can't make him think. === Subject: Re: Probability of finding a Prime Number > I found this in one of your posts (they're easy to find given that you sign > everything with that horse's ass comment): > The probability that a large integer N is prime is the probability that > it is not divisible by 2,3,5,7,11...sqrt(N). This > probability is (1-1/2) * (1-1/3) * (1-1/5) .... > all the primes up to N. My reasons for doing so are in the paper. > Admittedly, stopping at sqrt(N) is the correct approach). The probability that a large number roughly equal to[*] N is prime is 1/log(N). The product (1-1/p) does not equal 1/log(N) until p reaches N^0.5614595 (that exponent is e^(-EulerGamma)), _not_ N^0.5 as Bob appears to have indicated. Amusingly, your figure and Bob's will be remarkably similar. You omit the (1/2) term, but make up for this factor of two by taking p up to (sqrt(N))^2 rather than sqrt(N). This effectively gives you bac a factor of 1/2. I.e. the numeric value for your two approaches agree quite well for any reasonable sized N. Pop to a library and borrow Riesel's /Prime Numbers, and Computer Methods for Factorisation/ (or pick up one second-hand, it's pretty dear brand new). Mertens' Theorem is what you want. Phil [* I like to avoid saying that single numbers have probabilities of anything deterministic; hence a range is considered.] Unpatched IE vulnerability: Security zone transfer Description: Automatically opening IE + Executing attachments Published: March 22nd 2002 Reference: http://security.greymagic.com/adv/gm002-ie/ === Subject: Re: How fast is the Confinued Fraction factorization algorithm? Isn't this usually called the Quadratic Sieve? Where do continued fractions come into it? No, the poster really mean CFRAC and not QS. CFRAC uses trial division, and NOT a sieve to trial factor the potential relations. You can lead a horse's ass to knowledge, but you can't make him think. === Subject: Re: How fast is the Confinued Fraction factorization algorithm? I'm factoring numbers n using the Continued Fraction algorithm, .......I'm able to factor the 26-digit number n_26 in 90 seconds and the 28-digit number n_28 in 9 minutes, These times are slow on modern computers, but not extraordinarily so. I am concerned about the run time ratio for C28/C26. Adding 3 digits should increase run time by a factor of 3-4 for composites in this range. This ratio decreases as the size increases. Around 40 digits or so, adding 3 digits rougly doubles the run time. Note that there will be some variation in run time for numbers of a given size, depending on the quality of the multiplier. Are you using the early abort strategy? 70 digits is too big for CFRAC. MPQS or its variants are uniformly faster for all integers bigger than about 20 digits. (exact crossover will be machine & implementation dependent). Back in the early to mid 80's I was able to do 45 digits with CFRAC in ~10 hours on a VAX780, using a well-tuned CFRAC with early aborts. Quite a few people have experiemented with CFRAC. But we gave it up totally back in '84-'85. ECM is a different *kind* of algorithm. CFRAC succeeds in time that depends only on the size of the composite. ECM succeeeds in random time depending on the size of the *factors*. Now, we can routinely find 25 digit factors in a few 10's of seconds, 30-35 digits in a couple of hours, 40-45 digits in a small number of weeks, and 50+ digits only with a big effort and luck. (these are typical; one can succeed more quickly or much more slowly as ECM is a random algorithm) You should read my two papers: (both in Mathematics of Computation) The Multiple Polynomial Quadratic Sieve (with Sam Wagstaff Jr.) A Practical Analysis of the Elliptic Curve Factoring Algorithm. The first gives run time comparisons (circa 1985) between CFRAC and QS. You can lead a horse's ass to knowledge, but you can't make him think. === Subject: Re: How fast is the Confinued Fraction factorization algorithm? > I'm factoring numbers n using the Continued Fraction algorithm, .......I'm > able to factor the 26-digit number n_26 in 90 seconds and the > 28-digit number n_28 in 9 minutes, Good news! I've found a gross error in my linear algebra programming. I've now fixed that bug and made a few other optimizations, so now I'm factoring C32 in 68 seconds, C34 in 98 seconds, and C36 in 425 seconds. This time I'm using a factor base consisting of the 256 smallest primes. I have another question: In each of the above factorizations, only half (approximately) of the primes in the factor base are used. I read somewhere that N (the composite) must be congruent to a square modulo the primes in the factor base, but I don't see why: If x^2 = y mod N, then x^2 - y = kN, for some k. Now if a prime p (from the factor base) divides y, then we must have p | (x^2 - kN). So I conclude that kN must be congruent to a square modulo p, but that does not say anything about N itself? Please enlighten me. -Michael. === Subject: Re: How fast is the Confinued Fraction factorization algorithm? > I have another question: In each of the above factorizations, only half > (approximately) of the primes in the factor base are used. I read > somewhere that N (the composite) must be congruent to a square modulo > the primes in the factor base, but I don't see why: > If x^2 = y mod N, then x^2 - y = kN, for some k. Now if a prime p (from > the factor base) divides y, then we must have p | (x^2 - kN). So I > conclude that kN must be congruent to a square modulo p, but that does > not say anything about N itself? Please enlighten me. I don't know anything about the continued fraction method (as has been pointed out!) but if it is like the simple quadratic sieve one considers products of numbers of the form Q(x) = x^2 - n (with x close to sqrt{n}) each of which is smooth. So one need only consider small primes p such that n is a quadratic residue mod p. (In the multiple polynomial version one takes Q(x) = ax^2 + 2bx + c, with b^2 - ac = n, so that aQ(x) = (ax + b)^2 - n and much the same is true.) Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: How fast is the Confinued Fraction factorization algorithm? > I'm factoring numbers n using the Continued Fraction algorithm, .......I'm > able to factor the 26-digit number n_26 in 90 seconds and the > 28-digit number n_28 in 9 minutes, > These times are slow on modern computers, > but not extraordinarily so. I am concerned about > the run time ratio for C28/C26. Adding 3 digits > should increase run time by a factor of 3-4 for > composites in this range. This ratio decreases > as the size increases. Around 40 digits or so, > adding 3 digits rougly doubles the run time. I did some crude profiling and it turns out that 75% of the time is spent on housekeeping (manipulating iterators and such) rather than actual multi-precision calculation. Arggh!!! This I expect to be able to fix by organizing the data structures differently. This may also change the ratio above. > Note that there will be some variation in run time > for numbers of a given size, depending on the > quality of the multiplier. Say what? Yes, the CFRAC algorithm does suggest multiplying the original number by a small factor. Currently, I don't do this, because it seems completely arbitrary. Are there any hints to which values to choose as multiplier? How do I know if a given multiplier (incl. 1) is less than optimal? > Are you using the early abort strategy? No. What is that? > 70 digits is too big for CFRAC. Good, that is nice to know. What is the intuitive/fundamental reason for this limitation? Is it because too few y's actually factor over the factor base, even including a single larger prime factor, so we end up with too few useful qudratic relations? > ECM is a different *kind* of algorithm. CFRAC > succeeds in time that depends only on the size > of the composite. ECM succeeeds in random time > depending on the size of the *factors*. I read your words, but I'm not sure what your point is. I mean, if the composite has a small factor, then ECM is ideal, right? But if the composite is constructed as a product of two large primes, then you are saying that ECM is inferior to other modern algorithms? > Now, we > can routinely find 25 digit factors in a few 10's > of seconds, 30-35 digits in a couple of hours, > 40-45 digits in a small number of weeks, and > 50+ digits only with a big effort and luck. (these > are typical; one can succeed more quickly or > much more slowly as ECM is a random algorithm) So a 70-digit composite should take no more than a few hours using ECM, even in worst case? I mean, for such a number, there will be a factor with no more than 35 digits. > You should read my two papers: (both in > Mathematics of Computation) > The Multiple Polynomial Quadratic Sieve > (with Sam Wagstaff Jr.) > A Practical Analysis of the Elliptic Curve Factoring > Algorithm. > The first gives run time comparisons (circa 1985) > between CFRAC and QS. library. -Michael. === Subject: Re: How fast is the Confinued Fraction factorization algorithm? > I'm factoring numbers n using the Continued Fraction algorithm, finding > pairs (x,y) such that x^2==y mod n. The y values are factorized as y = > product(p_i^e_i) * p, where p_i are primes<=307, p<307^2 is another > prime, and e_i>=0. > Isn't this usually called the Quadratic Sieve? > Where do continued fractions come into it? You can generate squares modulo n by looking at continued fraction expansions (of sqrt(kn)). Of course, you have to do bignum operations in order to go from term to term. Yeuch! (The 'x' are the numerators of the CF approximants in the expansion.) Phil Unpatched IE vulnerability: history.back method caching Description: cross-domain scripting, cookie/data/identity theft, command execution Reference: http://safecenter.net/liudieyu/RefBack/RefBack-Content.HTM Exploit: http://www.safecenter.net/liudieyu/RefBack/RefBack-MyPage.HTM === Subject: Re: How fast is the Confinued Fraction factorization algorithm? > Hi! > I'm factoring numbers n using the Continued Fraction algorithm, > finding pairs (x,y) such that x^2==y mod n. The y values are > factorized as y = product(p_i^e_i) * p, where p_i are primes<=307, > p<307^2 is another prime, and e_i>=0. > I'm running a home-made program on a 1 GHz standard desktop computer, > using the GNU MP library for all the basic integer functions. > I'm able to factor the 26-digit number n_26 in 90 seconds and the > 28-digit number n_28 in 9 minutes, where n_26 = > 42240587706227658402774373 = 4646464647359 * 9090909091547 and n_28 = > 4224058769561941597796273857 = 46464646464983 * 90909090909479. > I'm somewhat disappointed at these time measurements. I was expecting > speeds several order of magnitudes faster; considering that it is now > possible to factor 70 digit numbers, albeit with a different algorithm. > So, is it because of inefficiences in my program, or is it simply the > limit of the algorithm? Has anybody else experimented with this > algorithm? That is pretty slow, I'd say. However, I've never implemented CFRAC to be able to compare expected runing times. I believe machines in the 70s were running CFRAC at about that rate. CFRAC is a cranky pre-cursor to QS, and there's absolutely no reason to run it any more since QS was invented. I don't believe it's possible to implement a QS slower than a CFRAC! Definitely have a go at QS implementation. I persnally think that the linear algebra is the hardest part, and you've already got that part written for CFRAC. (But I'm a bit of a sieve-nut.) > I can add that for n_26 a total of 6567 pairs (x,y) are considered, > and for n_28 the count is 11368. > I'm considering putting an effort into optimizing the program, but I > would like to hear other peoples experiences as to whether I can > expect any significant speed-up. I think old Mathematica (4.0 and before, probably) has CFRAC. Maybe you could race your implementation against that? I believe that one of the current Wolfram developers has made comments to the effect that the Mma implementation really wasn't very good, and has been booted, and replaced with a far swifter QS. But I really wouldn't recommend investing too much time in CFRAC, as QS is probably _simpler_, both mathematically and computationally, as well as being much faster. > If not, then I'll switch over to the Elliptic Curve algorithm. I would > like to hear peoples experiences with this algorithm too. ECM's for a different problem. ECM, like Rho, P-1, P+1 and the other mass-GCD algorithms are designed to chip off smallish factors, and the amount of work required to find them is most closely related to the size of the factors. CFRAC, like Fermat's method, QS, and NFS, is designed to split a number into two parts, and the work required is most closely related to the size of the original composite, and independent of the factors. Certainly use ECM on anything over about 50-60 digits, but with a slick QS, if you've got a 50-digit composite, you may as well hit QS straight after you've done some trial-div and a bit of P-1, say. If you don't have them yet, Crandall and Pomerance's /Prime Numbers, a Computational Perspective/ should be your first port of call, and after that I guess Riesel's /Prime Numbers, and Computer Methods for Factorisation/. Phil Unpatched IE vulnerability: protocol control chars Description: Circumventing content filters Reference: http://badwebmasters.net/advisory/012/ Exploit: http://badwebmasters.net/advisory/012/test2.asp === Subject: Re: How fast is the Confinued Fraction factorization algorithm? Definitely have a go at QS implementation. I persnally think that the linear algebra is the hardest part Actually, the hardest part of a QS implementation is not the linear algebra. The size of the matrices one gets is small compared to those in NFS. For factor bases up to (say) 50,000 primies, ordinary Gaussian elim works easily. Above that, even a crude block lanczos implementation will be fine. You don't get matrices more than a few hundred thousand rows, whereas NFS results in matrices with millions of rows. For QS, the tricky part is implementing FAST methods for changing polynomials. Furthermore, for large numbers, you need to do the actual factorizations of the relations by resieving, and not by trial division. You can lead a horse's ass to knowledge, but you can't make him think. === Subject: Re: How fast is the Confinued Fraction factorization algorithm? > Actually, the hardest part of a QS implementation is > not the linear algebra. The size of the matrices one > gets is small compared to those in NFS. For factor bases up to (say) > 50,000 primies, ordinary > Gaussian elim works easily. Above that, even a > crude block lanczos implementation will be fine. > You don't get matrices more than a few hundred > thousand rows, whereas NFS results in matrices > with millions of rows. Are you saying that the Quadratic Sieve is better than the Number Field Sieve for certain sizes of n? Which sizes, roughly speaking? My impression, based on not very scientific (or knowledgeable) experiment is that the Quadratic Sieve (with many polynomials) is unlikely to factorize number with more than 80 digits in a reasonable time on a reasonable computer (say 1 day on my 667MHz laptop). Incidentally (I'm sure this will show my ignorance) I've never understood why the various methods looking for solutions of x^2 = y^2 mod n (to give a factor gcd(x-y,n) of n) always seem to compare a quadratic residue mod n with a perfect square (of a smooth number). Why not compare two quadratic residues, or two perfect squares? Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: How fast is the Confinued Fraction factorization algorithm? > Regarding the speed, there is a standard trick using logarithms > to determine if a number is smooth (has small factors), > using the fact that if it is not smooth > there must be quite a large discrepancy. > Do you use this? > No, and I'm not sure what that trick is. Could you explain it in a few > more words. First of all, are we talking about discrete logarithm or > floating point logarithm? Ordinary floating-point logs. In brief, if one wants to filter out smooth numbers n, instead of dividing n by small primes p, one takes log n and subtracts log p for each prime power p^e dividing n, for primes p <= b, say. Now you should get 0 if n is b-smooth. The point is that if n is not b-smooth then the remainder is at least log b, so the inaccuracy of logs does not matter - one can safely say n is b-smooth if the remainder is < log b/2, say. The use of logs means that one can use ordinary floating-point numbers for this calculation, rather than arbitrary precision numbers. I'm not sure if this is relevant to what you are doing. Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: How fast is the Confinued Fraction factorization algorithm? more words. First of all, are we talking about discrete logarithm or > floating point logarithm? Ordinary floating-point logs. In brief, if one wants to filter out smooth numbers n, Once again, the original poster asked about CFRAC and not QS. Your discussion about a sieve will be confusing to the uninitiated and is not relevant to CFRAC. You can lead a horse's ass to knowledge, but you can't make him think. === Subject: Re: JSH: Splitting field, algebraic integer factors That's a nifty and powerful result. Why am I the one who had to > discover it? > James Harris Because you're so special in God's eyes. The math isn't that complicated. Consider that I'm using x^2 - x + 42 = 0, which has (1 + sqrt(-167))/2 as on of its roots, and y^2 - by - 7 = 0, which has as one of its roots (b + sqrt(b^2 + 28))/2. So I can simply introduce z, where (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2. Now it doesn't take a math genius to see that you can't multiply by an > algebraic integer z, and manage that sqrt(-167) on the left side > without having b^2 + 28 = -167n^2, which is why I brought up the > splitting field. But to see replies in this thread you'd think I brought up the Tooth > Fairy. > You brought up the equation > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0, derived from > (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2, after several false > starts, and claime that that equation > . > But for b = 6, which was an integer last time I looked, > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 > = 7z^4 + 6z^3 + 299z^2 - 36z + 252 > = (z^2 + z + 42)(7z^2 - z + 6). > which certianly fits my definition of being reducible. Which brings up > teh question of what your definition of irreducible versus reducible > must look like. > AARGGHHH!!! and YEAH!!!! > Why did it have to be you Virgil? > Is Virgil your real name or a pseudonym? > Rick Decker and Nora Baron FINALLY concede? > (If you I'm crazy here then you haven't looked carefully at Virgil's > result. Think about it. I'll add more later if necessary.) False alarm. What he pointed out is a trivial result, but it was important for me to include it. My blog has been updated. Well this has been a fruitful weekend. I think I'll take a break. James Harris === Subject: Re: JSH: Splitting field, algebraic integer factors > That's a nifty and powerful result. Why am I the one who had to > discover it? > James Harris Because you're so special in God's eyes. The math isn't that complicated. Consider that I'm using x^2 - x + 42 = 0, which has (1 + sqrt(-167))/2 as on of its roots, and y^2 - by - 7 = 0, which has as one of its roots (b + sqrt(b^2 + 28))/2. So I can simply introduce z, where (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2. Now it doesn't take a math genius to see that you can't multiply by an > algebraic integer z, and manage that sqrt(-167) on the left side > without having b^2 + 28 = -167n^2, which is why I brought up the > splitting field. But to see replies in this thread you'd think I brought up the Tooth > Fairy. You brought up the equation > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0, derived from > (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2, after several false > starts, and claime that that equation . > But for b = 6, which was an integer last time I looked, > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 > = 7z^4 + 6z^3 + 299z^2 - 36z + 252 > = (z^2 + z + 42)(7z^2 - z + 6). which certianly fits my definition of being reducible. Which brings up > teh question of what your definition of irreducible versus reducible > must look like. > AARGGHHH!!! and YEAH!!!! > Why did it have to be you Virgil? > Is Virgil your real name or a pseudonym? > Rick Decker and Nora Baron FINALLY concede? I can't speak for Arturo or Rick, though I greatly doubt they are going to concede, nor will Dik Winter. For myself, absolutely not. Here is why: 1. The argument in your blog spot is incorrect. You assume with no proof that if (b + sqrt(b^ + 28)) is a divisor of (1 + sqrt(-167)), then there must exist an integer n such that b^2 + 28 = -167*n^2. 2. You exclude Virgil's note regarding b = 6 in a totally ad hoc way, not as part of your main argument. Your main argument would not lead to the conclusion that b could not be 6. Therefore your main argument is wrong. Also there may be other algebraic integer values of b which cause your polynomial to be reducible. Your main argument does not exclude this possibilty. 3. I know of 4 really separate proofs that show that your main claim is incorrect. These have been discussed in sci.math at length. You have never refuted any of them. Your basic thinking on these appears to be, These arguments are too hard for me to understand, so I can ignore them. What you are claiming to prove is that one of the roots of x^2 - x + 42 is coprime to 7: namely, the root (1 + sqrt(-167))/2. Since the product of the roots is 7*6, this means that the other root, (1 - sqrt(-167))/2, must be *divisible* by 7. This however as you know is easily shown to be false. Thus if you were right, you would have found a basic contradiction in mathematics. *NOT* a deficiency in the definition of algebraic integers, but an outright contradiction which would imply that mathe- matics is inconsistent. This to me is the most compelling reason that your claims are wrong. You have implied that you are going to write this up for publication. You will be shot down immediately unless you can show what is wrong with elementary Galois theory, elementary algebraic number theory, the class number theorem, and Keith Ramsay's explicit example showing that (1 + sqrt(-167))/2 has a factor in common with the root of an 11th degree polynomial with constant term a power of 7. Nora B. > (If you I'm crazy here then you haven't looked carefully at Virgil's > result. Think about it. I'll add more later if necessary.) > False alarm. What he pointed out is a trivial result, but it was > important for me to include it. My blog has been updated. > Well this has been a fruitful weekend. I think I'll take a break. > James Harris === Subject: Re: JSH: Splitting field, algebraic integer factors > But for b = 6, which was an integer last time I looked, > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 > = 7z^4 + 6z^3 + 299z^2 - 36z + 252 > = (z^2 + z + 42)(7z^2 - z + 6). which certianly fits my definition of being reducible. Which brings up > teh question of what your definition of irreducible versus reducible > must look like. ... > 2. You exclude Virgil's note regarding b = 6 in a totally ad hoc way, > not as part of your main argument. Your main argument would not > lead to the conclusion that b could not be 6. Therefore your > main argument is wrong. Also there may be other algebraic > integer values of b which cause your polynomial to be > reducible. Your main argument does not exclude this possibilty. There certainly are. b = 8i is an example: 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = = 7z^4 + 8.i.z^2 - 301z^2 - 48.i.z + 252 = = (z^2 + i.z - 42)(7z^2 + i.z - 6) where the roots of the first factor are indeed algebraic integers [ (z^2 + i.z - 42)(z^2 - i.z - 42) = z^4 - 83z^2 + 1764 ]. So we now already have 4 values of b for which the quartic in z has algebraic integer roots: b = 6, b = -6, b = 8i and b = -8i. Note that these are all such that the constant term (note, James, we talk about polynomials here) splits in two integer factors. When we allow algebraic integer factors many more will be found. dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Splitting field, algebraic integer factors ... > So we now already have 4 values of b for which the quartic in z has > algebraic integer roots: b = 6, b = -6, b = 8i and b = -8i. Note that > these are all such that the constant term (note, James, we talk about > polynomials here) splits in two integer factors. When we allow > algebraic integer factors many more will be found. O, granted. Only for b = 6 and b = -6 does the quartic split with integer coefficients and one monic polynomial. But that is not sufficient to show that there can not be algebraic integer roots. There is a theorem: The root of a monic polynomial with algebraic integer coefficients is an algebraic integer. So to show that the quartic has no algebraic integer roots James has to show that it does not split over the algebraic integers with one of the factors monic. dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Splitting field, algebraic integer factors > Well this has been a fruitful weekend. I think I'll take a break. Most posters who had launched ten or fifteen threads within a few days, which are riddled with math errors and which consistently arrive at false conclusions, would consider the weekend to be an embarrassing public display of chronic ignorance and would be ashamed of their posts. You, on the other hand, consider it a fruitful weekend. Well, maybe in your case it is. > James Often in error, but never in doubt. Harris 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 === Subject: Re: JSH: Splitting field, algebraic integer factors > That's a nifty and powerful result. Why am I the one who had to > discover it? > James Harris Because you're so special in God's eyes. The math isn't that complicated. Consider that I'm using x^2 - x + 42 = 0, which has (1 + sqrt(-167))/2 as on of its roots, and y^2 - by - 7 = 0, which has as one of its roots (b + sqrt(b^2 + 28))/2. So I can simply introduce z, where (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2. Now it doesn't take a math genius to see that you can't multiply by an > algebraic integer z, and manage that sqrt(-167) on the left side > without having b^2 + 28 = -167n^2, which is why I brought up the > splitting field. But to see replies in this thread you'd think I brought up the Tooth > Fairy. You brought up the equation > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0, derived from > (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2, after several false > starts, and claime that that equation . > But for b = 6, which was an integer last time I looked, > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 > = 7z^4 + 6z^3 + 299z^2 - 36z + 252 > = (z^2 + z + 42)(7z^2 - z + 6). which certianly fits my definition of being reducible. Which brings up > the question of what your definition of irreducible versus reducible > must look like. > AARGGHHH!!! and YEAH!!!! > Why did it have to be you Virgil? > Is Virgil your real name or a pseudonym? > Rick Decker and Nora Baron FINALLY concede? > (If you I'm crazy here then you haven't looked carefully at Virgil's > result. Think about it. I'll add more later if necessary.) > False alarm. What he pointed out is a trivial result, but it was > important for me to include it. My blog has been updated. That trivial result important as a counterexample to JSH's repeated claim that 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 is irreducible for all integer values for b and has no algebraic integer roots for z. > Well this has been a fruitful weekend. I think I'll take a break. The fruit on JSH's side is a bit sour this week. > James Harris === Subject: Re: JSH: Splitting field, algebraic integer factors ... > . But for b = 6, which was an integer last time I looked, > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 > = 7z^4 + 6z^3 + 299z^2 - 36z + 252 > = (z^2 + z + 42)(7z^2 - z + 6). ... > AARGGHHH!!! and YEAH!!!! ... > False alarm. What he pointed out is a trivial result, but it was > important for me to include it. My blog has been updated. Your page does not show *why* it is trivial. Moreover later in your page you never exclude b=6. Especially the part: > which means that (b - sqrt(-167) sqrt(b^2 + 28))/14 would have to be an > algebraic integer. > Notice it can't be an integer with algebraic integer b, as if > b^2 + 28 = -167n^2, with n an integer, b can't be real. What if b = 6? Do you know of any other counterexamples that are trivial results? Moreover, not all algebraic integers are real. dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Splitting field, algebraic integer factors Adjunct Assistant Professor at the University of Montana. > Notice it can't be an integer with algebraic integer b, as if > b^2 + 28 = -167n^2, with n an integer, b can't be real. >What if b = 6? If n is an integer, then -167n^2 is nonpositive; so b^2<= -28, hence b cannot be real. Whether the condition b^2+28=-167n^2 is truly necessary is a different issue, of course... 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 === Subject: Re: JSH: Splitting field, algebraic integer factors >JSH is now taking a new tack: that counterexamplees to his case prove >his case. It is not a tack that many of the mathematically apt will >follow. Surely this isn't a *new* tack? Hasn't he been using Rick Decker's example for a while? Min So where are all the buffaloes? === Subject: Re: JSH: Splitting field, algebraic integer factors Adjunct Assistant Professor at the University of Montana. >Yup, that is easy. Arturo already provided an easy proof. But take the >smallest field that contains all three. It will be the splitting field >of some polynomial. Surely not; to be the splitting field of a polynomial with rational coefficients, the extension must be normal over Q. Simply taking the field Q(a,b,c) does not guarantee that this extension is normal. It would have to contain all conjugates (i.e., all other roots of the minimal polynomials over Q) of a, b, and c. Of course, every field K is a splitting field of a polynomial over K (linear terms), but surely that is not what is being talked about here? Presumably, as in most things algebraicnumbertheory-ites, splitting field without qualifier means splitting field over Q means normal extension of Q. 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 === Subject: Re: JSH: Splitting field, algebraic integer factors >Yup, that is easy. Arturo already provided an easy proof. But take the >smallest field that contains all three. It will be the splitting field >of some polynomial. > Surely not; to be the splitting field of a polynomial with rational > coefficients, the extension must be normal over Q. > Previously I posted that if you can't *see* the factors between > irrational algebraic integers then they're not there, but more > correctly the situation is that two algebraic integers have to be > members of the same splitting field to have a non-unit algebraic > integer in common. This was entirely unclear as written. You provided a proof that it is true with splitting fields over the rationals, I said that it was true when you use any field extension. What I am thinking James is thinking (but you can never be sure of that), is that whenever two algebraic numbers a and b are in a single splitting field, so is their common factor. That is obviously blatantly false. But I can not interprete his statement in any other way. > Of course, every field K is a splitting field of a polynomial over K > (linear terms), but surely that is not what is being talked about > here? Presumably, as in most things algebraicnumbertheory-ites, > splitting field without qualifier means splitting field over Q > means normal extension of Q. What James is talking about is not what you see. dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Splitting field, algebraic integer factors That's a nifty and powerful result. Why am I the one who had to >discover it? >James Harris >Because you're so special in God's eyes. >The math isn't that complicated. Consider that I'm using >x^2 - x + 42 = 0, which has (1 + sqrt(-167))/2 as on of its roots, and >y^2 - by - 7 = 0, which has as one of its roots >(b + sqrt(b^2 + 28))/2. >So I can simply introduce z, where >(1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2. >Now it doesn't take a math genius to see that you can't multiply by an >algebraic integer z, and manage that sqrt(-167) on the left side >without having b^2 + 28 = -167n^2, which is why I brought up the >splitting field. >But to see replies in this thread you'd think I brought up the Tooth >Fairy. >You brought up the equation > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0, derived from > (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2, after several false >starts, and claime that that equation > . >But for b = 6, which was an integer last time I looked, > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 > = 7z^4 + 6z^3 + 299z^2 - 36z + 252 > = (z^2 + z + 42)(7z^2 - z + 6). >which certianly fits my definition of being reducible. Which brings up >the question of what your definition of irreducible versus reducible >must look like. >AARGGHHH!!! and YEAH!!!! >Why did it have to be you Virgil? >Is Virgil your real name or a pseudonym? > How is my name relevant to the correctness of my methematics? >Rick Decker and Nora Baron FINALLY concede? > They have nothing to concede. You do. > Yoou have claimed that there was no raational integer value of b for > which 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 was reducible, and no > altgebraic integer value of b for which the equation in zm > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0, had an algebraic integer > solution. > Both of these claims are now trivially false. >(If you I'm crazy here then you haven't looked carefully at Virgil's >result. Think about it. I'll add more later if necessary.) >James Harris > JSH is now taking a new tack: that counterexamplees to his case prove > his case. It is not a tack that many of the mathematically apt will > follow. Heh. James is going to observe that the four roots of his equation are (-1 +/- sqrt(-167))/2 and (1 +/- sqrt(-167))/14 and with b = 6, (b + sqrt(b^2 + 28))/2 = 7 so his original equation, (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2 becomes (1 + sqrt(-167))/2 = 7z and the root which satisfies this is (1 + sqrt(-167))/14 which of course isn't an algebraic integer, hence, through some sort of magic, (1 + sqrt(-167))/2 is coprime to 7 in the algebraic integers! This in spite of the fact that we have two different ways (at least) of showing that (1 + sqrt(-167))/2 in *not* coprime to 7. Rick === Subject: Re: e is transcendental (was: classes of transcendental numbers ? >So? The real part of exp(i pi) is cos(pi), and its imaginary part >is sin(pi), so all you are saying is that cos(pi) = -1 and >sin(pi) = 0, and we were already aware of these facts. There is >NO reason to conclude that exp(i pi) = 0. > I,have a reason , with my due respect. > Panagiotis Stefanides >Yes, but you DON'T tell us what your reason is. You can't expect us to >accept your claims without giving support for those claims. So what >possible reason could you have for expecting us to agree with your claim >that exp(i pi) = 0? >The reason is simply that exp(ipi0=-1 should be accompanied >by the statement that this is the real part solution. >Is it fair? No, that is not a fair comment. exp(i pi), the complex number, is equal to -1, the complex number. There is no need to appeal to the real part. Also, what does exp(i pi) = -1 is the real part solution mean? You look like you are using terminology in a manner not recognized in mathematics. David McAnally -------------- === Subject: Re: Anybody want to play with a new math of infinities? > Consider the question of what happens when we > divide 1 by 0? A lot of work assumes there is an answer that > correlates to infinity, which is problematic, > because what happens when you multiply this > infinity by 0 again? A simpler solution is to use the symbology: The result of dividing 1 by 0 is 1/0. > The result of dividing 2 by 0 is 2/0. etc. Since the / has a pre-existing meaning, let us say, The result of dividing 1 by 0 is 1~0. > The result of dividing 2 by 0 is 2~0. etc. When you multiply 1~0 by 0, you get 1. > But when you multiple 2~0 by 0, you get 2. Does this lead to any inconsistency? > Yes it does: > 1= 1-0*0 = 1-0*(0+0) = 1-0*0 + 1-0*0 = 1+1 = 2. > ---- David > 1= 1~0*0 = 1~0*(0+0) = 1~0*0 + 1~0*0 = 1+1 = 2. > You are right! > Can we outlaw something here? He used the fact that addition is distributive w/ respect to multiplication. Evidently, that can't be the case in your structure. Unfortunately, that takes away most of the attraction of studying it; if addition is unrelated to multiplication, you might as well study two separate structures, one with addition defined, and one with multiplication defined. If you wish to have both, and wish them to be related to each other, you must outlaw division by zero. Or, perhaps you could come up with something that replaces the distributive property, and relates addition and multiplication without reducing your structure to the trivial one. Then, however, you're straying quite far from your original intention. Nathan === Subject: Re: Anybody want to play with a new math of infinities? > He used the fact that addition is distributive w/ respect to > multiplication. Evidently, that can't be the case in your structure. > Unfortunately, > that takes away most of the attraction of studying it; if addition is > unrelated to multiplication, you might as well study two separate I saw it more like 0 has new properties, not that addition becomes unrelated to multiplication. (Or rather, zero loses its familiar additive identity.) === Subject: Re: Anybody want to play with a new math of infinities? > Consider the question of what happens when we > divide 1 by 0? A lot of work assumes there is an answer that > correlates to infinity, which is problematic, > because what happens when you multiply this > infinity by 0 again? A simpler solution is to use the symbology: The result of dividing 1 by 0 is 1/0. > The result of dividing 2 by 0 is 2/0. etc. Since the / has a pre-existing meaning, let us say, The result of dividing 1 by 0 is 1~0. > The result of dividing 2 by 0 is 2~0. etc. When you multiply 1~0 by 0, you get 1. > But when you multiple 2~0 by 0, you get 2. Does this lead to any inconsistency? Yes it does: > 1= 1-0*0 = 1-0*(0+0) = 1-0*0 + 1-0*0 = 1+1 = 2. > 1= 1~0*0 = 1~0*(0+0) = 1~0*0 + 1~0*0 = 1+1 = 2. > You are right! > Can we outlaw something here? > He used the fact that addition is distributive w/ respect to > multiplication. Evidently, that can't be the case in your structure. > Unfortunately, that takes away most of the attraction of studying it; Multiplication doesn't distribute over addition in interval arithmetic, and yet some people find it attractive to study. > if addition is unrelated to multiplication, you might as well study two > separate structures, one with addition defined, and one with > multiplication defined. If you wish to have both, and wish them to be > related to each other, you must outlaw division by zero. Generally speaking, that's false. There are interesting systems which allow division by zero and in which distributivity holds. But if you were speaking only of bhanwara's proposal, then perhaps you're right. David Cantrell === Subject: Re: Anybody want to play with a new math of infinities? > In sci.math, bhanwara > divide 1 by 0? > A lot of work assumes there is an answer that > correlates to infinity, which is problematic, > because what happens when you multiply this > infinity by 0 again? > A simpler solution is to use the symbology: > The result of dividing 1 by 0 is 1/0. > The result of dividing 2 by 0 is 2/0. etc. > Since the / has a pre-existing meaning, let us say, > The result of dividing 1 by 0 is 1~0. > The result of dividing 2 by 0 is 2~0. etc. > When you multiply 1~0 by 0, you get 1. > But when you multiple 2~0 by 0, you get 2. > Does this lead to any inconsistency? > What about comparisons? > It would seem 1~0 should be greater than 2~0, which > should be greater than 3~0. 0~0 would be greather than 1~0. > Are there theorems than can be proved using > this new symbology? > One has the minor problem of evaluating 0~0, which > is the result of dividing 0/0. This can be gotten by > e.g. subtracting 1~0 - 1~0, and could be any real at all. > Of course, for normal real values r, r - r = 0. > Also, if one multiplies 1~0 * 1~0, what does one get? > Ditto for 1~0 / 1~0 and r / 1~0, which may or may not > be 0. I am not sure what you are saying about subtracting 1~0 - 1~0. Why does 0~0 not work? Re the rest, the points appear valid -- the rules would need to be carefully deduced. === Subject: Re: Anybody want to play with a new math of infinities? In sci.math, bhanwara In sci.math, bhanwara > Consider the question of what happens when we > divide 1 by 0? > A lot of work assumes there is an answer that > correlates to infinity, which is problematic, > because what happens when you multiply this > infinity by 0 again? > A simpler solution is to use the symbology: > The result of dividing 1 by 0 is 1/0. > The result of dividing 2 by 0 is 2/0. etc. > Since the / has a pre-existing meaning, let us say, > The result of dividing 1 by 0 is 1~0. > The result of dividing 2 by 0 is 2~0. etc. > When you multiply 1~0 by 0, you get 1. > But when you multiple 2~0 by 0, you get 2. > Does this lead to any inconsistency? > What about comparisons? > It would seem 1~0 should be greater than 2~0, which > should be greater than 3~0. 0~0 would be greather than 1~0. > Are there theorems than can be proved using > this new symbology? > One has the minor problem of evaluating 0~0, which > is the result of dividing 0/0. This can be gotten by > e.g. subtracting 1~0 - 1~0, and could be any real at all. > Of course, for normal real values r, r - r = 0. > Also, if one multiplies 1~0 * 1~0, what does one get? > Ditto for 1~0 / 1~0 and r / 1~0, which may or may not > be 0. > I am not sure what you are saying about subtracting > 1~0 - 1~0. Why does 0~0 not work? Mostly because a - a = 0 for all real a, but this concept can't apparently be extended to these non-real numbers. > Re the rest, the points appear valid -- the rules would > need to be carefully deduced. Very carefully so. :-) #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Anybody want to play with a new math of infinities? > Mostly because a - a = 0 for all real a, but this concept > can't apparently be extended to these non-real numbers. Ah, yes, because there is a 0~0, distinct from and in addition to 0! === Subject: Re: Anybody want to play with a new math of infinities? > Does this lead to any inconsistency? > Yes it does: > 1= 1-0*0 = 1-0*(0+0) = 1-0*0 + 1-0*0 = 1+1 = 2. > ---- David > 1= 1~0*0 = 1~0*(0+0) = 1~0*0 + 1~0*0 = 1+1 = 2. >You are right! >Can we outlaw something here? Is that we a royal we? Hopefully so, since I don't think that anyone else really cares. adam === Subject: Re: Anybody want to play with a new math of infinities? > Does this lead to any inconsistency? > Yes it does: > 1= 1-0*0 = 1-0*(0+0) = 1-0*0 + 1-0*0 = 1+1 = 2. > ---- David > 1= 1~0*0 = 1~0*(0+0) = 1~0*0 + 1~0*0 = 1+1 = 2. >You are right! >Can we outlaw something here? > Is that we a royal we? > Hopefully so, since I don't think > that anyone else really cares. > adam Would that be a royal nobody? :-) === Subject: Royden, Chapter 10 Section 3 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1G1Yvr30641; T linear subspace of normed space X. y in X. Prove inf{||y-t||:t in T} = sup{f(y): ||f||=1, f(t)=0 for all t in T}. so want to show <= & => & I need help. The one inequality follows basically immediately from Prop7 of the same section. As for the other way, I'm kinda stuck. Help? === Subject: Re: Royden, Chapter 10 Section 3 >T linear subspace of normed space X. y in X. >Prove inf{||y-t||:t in T} = sup{f(y): ||f||=1, f(t)=0 for all t in T}. >so want to show <= & => & I need help. On inequality is immediate from the definitions. The other one follows from the Hahn-Banach theorem. >The one inequality follows basically immediately from Prop7 of the same section. As for the other way, I'm kinda stuck. Help? ************************ David C. Ullrich === Subject: Help binomial/normal distribution Can someone here point me to a straightforward proof that the binomial distribution reduces to the normal distribution as n -> inf? I've seen the statement over and over that it does, and of course it's easy to plot histograms of both dists and show that they get closer and closer as n increases. I'm aware that Stirling's formula is involved, but so far I've been unable to show the equivalence. Would appreciate some help. Jack === Subject: Re: Help binomial/normal distribution Jack > Can someone here point me to a straightforward proof that the binomial > distribution reduces to the normal distribution as n -> inf? > I've seen the statement over and over that it does, and of course it's > easy to plot histograms of both dists and show that they get closer and > closer as n increases. I'm aware that Stirling's formula is involved, > but so far I've been unable to show the equivalence. Would appreciate > some help. > Jack === Subject: Re: Help binomial/normal distribution >Can someone here point me to a straightforward proof that the binomial >distribution reduces to the normal distribution as n -> inf? >I've seen the statement over and over that it does, and of course it's >easy to plot histograms of both dists and show that they get closer and >closer as n increases. I'm aware that Stirling's formula is involved, >but so far I've been unable to show the equivalence. Would appreciate >some help. At , I give the outline of a proof of the Central Limit Theorem that can be completed to work with most probability densities which have a finite variance. This includes the binomial distribution, which is the convolution and contraction of two point masses of weight 1/2 at -1 and 1. Rob Johnson take out the trash before replying === Subject: Re: Help binomial/normal distribution > Can someone here point me to a straightforward proof that the binomial > distribution reduces to the normal distribution as n -> inf? > I've seen the statement over and over that it does, and of course it's > easy to plot histograms of both dists and show that they get closer and > closer as n increases. I'm aware that Stirling's formula is involved, > but so far I've been unable to show the equivalence. Would appreciate > some help. I believe that there is a proof online at mathworld applicable that is specific for the binomial distribution. -michael. === Subject: Re: Help binomial/normal distribution > Can someone here point me to a straightforward proof that the binomial > distribution reduces to the normal distribution as n -> inf? > I've seen the statement over and over that it does, and of course it's > easy to plot histograms of both dists and show that they get closer and > closer as n increases. I'm aware that Stirling's formula is involved, > but so far I've been unable to show the equivalence. Would appreciate > some help. This is a special case of the Central Limit Theorem. The easiest proofs involve transforms (e.g., characteristic funtctions). See Drake AW, Fundamentals of Applied Probability Theory for an elementary outline of most of a proof. Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: branch of log z > What i really like in Your letters, as well in Lynn Kurtz letter to > is the treatment of complex numbers as pairs of two reals, no > differennce between i-axis and y-axis. Is this only for insiders ?! > Why not propagate this? My attempt: > http://i-is-no-longer-imaginary.gmxhome.de > or, the same: http://www.i-z.eu.tt . > Have fun > Hero This is well known. In fact, this is how complex numbers were introduced to me (at the college level, anyways). The i notation is really quite useful, however, since it hides some of the complexity involved in working with vectors (in particular, the notion of vector multiplication defined on R^2 to make it C.) But both approaches are useful. When I took complex analysis, I found myself switching between the two very often in the process of proving theorems (in particular, to apply the theory of real analysis to the complex case) 'cid 'ooh === Subject: Yo mathemeticians i am bad at math. I want to know how to become good at math as quickly as possible. Since you would know better than i would from experience, say whatever silly idea comes to your head. Also, i ing hate math === Subject: Re: Yo mathemeticians > i am bad at math. I want to know how to become good at math as > quickly as possible. Since you would know better than i would from > experience, say whatever silly idea comes to your head. > Also, i ing hate math * But are you good at pumping gas? Flipping burgers? earle * === Subject: Re: Yo mathemeticians > i am bad at math. I want to know how to become good at math as > quickly as possible. Since you would know better than i would from > experience, say whatever silly idea comes to your head. > Also, i ing hate math > But are you good at pumping gas? Flipping burgers? The world needs ditch-diggers, too. Doug === Subject: Re: Yo mathemeticians > i am bad at math. I want to know how to become good at math as > quickly as possible. Since you would know better than i would from > experience, say whatever silly idea comes to your head. > Also, i ing hate math Trolls are notoriously bad at math; it's probably genetic. === Subject: Re: Yo mathemeticians > i am bad at math. I want to know how to become good at math as > quickly as possible. Since you would know better than i would from > experience, say whatever silly idea comes to your head. The best way to get 'good at math' is to do your homework, and when you are done that, find more questions similar to those in either your textbook or in another book related to whatever subject you are interested in and then do all those questions too. If you are talking about real math and not just gaining math skills, then try doing a similar approach but with problem solving books. Try a problem for at least half an hour and once you know you understand the problem and you can't find a solution, then you may allow yourself to look at answers. Then again, if you are only in elementary school, I might recommend learning your times tables. Maybe you want to be a bit more specific. J === Subject: Re: Yo mathemeticians > i am bad at math. I want to know how to become good at math as > quickly as possible. Since you would know better than i would from > experience, say whatever silly idea comes to your head. > Also, i ing hate math Math class is hard. -- Barbie. === Subject: Re: Yo mathemeticians > Also, i ing hate math Doing something about this will be a giant step in solving your other problem. Doug === Subject: lambert w function - fractional derivative hello, is there a known form for the fractional derviative of the lambert W function: cheers moth === Subject: Re: Bound of a sum > I admit I am no math genius, but I need to find some way to bound > k-1 n > sum sum 2/(j-i+1) > i=1 j=k+1 > 1 <= k<= n. > I think (stupid empirical experiment) that the sum is bounded by about > 0.7n for any value of k .. but I really can't figure how to make > something looking like a sound proof. > Any upper bound that is some constant by n is OK. > Any ideas ? > Soren > An overly complicated _start_ to a solution, hopefully. > I assume the double sum is, as it appears, a function of both k and n. > (But k is bounded by n.) > Your double-sum can be written as: > s(k,n) = > 2* sum{j=k+1 to n} (H(j) -H(j-k+1)), > where H(m) = sum{j=1 to m} 1/j. > So, s(k,n) = > 2 * ((n+1)H(n) -n > -(k+1)H(k) +k > -(n-k+2)H(n-k+1) +n-k+1 > -1), > since: sum{k=1 to m} H(k) = > (m+1)*H(m) - m. > So we have: > s(k,n) = 2 * ((n+1)H(n) -(k+1)H(k) -(n-k+2)H(n-k+1)). > But H(n) ~ ln(n) + gamma +1/(2n) - O(1/n^2), > where gamma is Euler's constant. ** > So, s(k,n) = 2(n+1)(ln(n) +gamma) +(1+1/n) > -2(k+1)(ln(k) +gamma) -(1+1/k) > -2(n-k+2)(ln(n-k+1) +gamma) -(1-1/(n-k+1)) > - O(1/n^2) - O(1/k^2)-O(1/(n-k+1)^2) > And the O()'s make sense in terms of n, k, and n-k+1 all increasing to infinity. > Hopefully I did not make an error. But I HAVE made an error, though it does not really matter perhaps. of ln's and O()'s and gamma's (from the ** downward). In any case, hopefully I am still right about: s(k,n) = 2 * ((n+1)H(n) -(k+1)H(k) -(n-k+2)H(n-k+1)). Even if this does not answer your question about the bounds, it still is a somewhat interesting representation of your double sum, I believe. :) === Subject: Re: Bound of a sum >I admit I am no math genius, but I need to find some way to bound >k-1 n >sum sum 2/(j-i+1) >i=1 j=k+1 >1 <= k<= n. >Any upper bound that is some constant by n is OK. >Write the sum as >S = sum_{r=3}^n f(r,k)/r >where f(r,k) = min(k-1,n-r+1) - max(0,k-r+1). Oops, somehow I lost the 2. So my bound would be 2n. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Bound of a sum > k-1 n > sum sum 2/(j-i+1) > i=1 j=k+1 > 1 <= k<= n. > I think (stupid empirical experiment) that the sum is bounded by about > 0.7n for any value of k .. but I really can't figure how to make > something looking like a sound proof. > This will be a little sketchy. Fact: There exists a constant C such that > 1/k + 1(k+1) + ... + 1/n <= C*ln(n/k) for all n, 1 <= k <= n; this is easy > to show. So sum(j=k+1,n) 1/(j-i) <= C*ln[(n-i)/(k+1-i)]. So we just need to > consider C*sum (i=1,k) ln[(n-i)/(k+1-i)]. But that sum of logs is the log > of the product. And that ends up looking a lot like ln(n choose k). Recall > Pascal's triangle shows (n choose k) is maximized when k = n/2 (at least > for even n). So we need a bound on ln(n!/[(n/2)!]^2). But n!/[(n/2)!]^2 <= > 2^n for even n, which is easily verified by induction. So we're getting > ln(2^n) = nln(2), which is looking good. I think the above can be made into > a real proof. Yes, this works out pretty well. We have 1/(m+1) + ... + 1/n < ln(n/m) for 1 <= m < n. So sum(j=k+1,n) 1/(j-i+1) < ln[(n-i+1)/(k+1-i)]. So now we try to estimate sum(i=1,k-1) ln[(n-i+1)/(k+1-i)], which is the ln of the product [n/k]*[(n-1)/(k-1)]* ... *[(n-k+2)/2] <= n!/[k!(n-k)!] <= 2^n. The last inequality is obvious from the expansion of (1+1)^n. So including the factor of 2 you had originally, your sum is < 2*ln(2^n) = 2ln(2)*n. Mmmm ... not quite as good as Robert's answer of n. === Subject: Re: Bound of a sum >(1+1)^n. So including the factor of 2 you had originally, your sum is < >2*ln(2^n) = 2ln(2)*n. Mmmm ... not quite as good as Robert's answer of n. Well, given the quality of Robert's posts one is naturally inclined to assume that they won't ever be wrong, but in this case, as he pointed makes much more sense... Michele > Comments should say _why_ something is being done. Oh? My comments always say what _really_ should have happened. :) - Tore Aursand on comp.lang.perl.misc === Subject: Re: Bound of a sum >This will be a little sketchy. Fact: There exists a constant C such that >1/k + 1(k+1) + ... + 1/n <= C*ln(n/k) for all n, 1 <= k <= n; this is easy >to show. So sum(j=k+1,n) 1/(j-i) <= C*ln[(n-i)/(k+1-i)]. So we just need to >consider C*sum (i=1,k) ln[(n-i)/(k+1-i)]. But that sum of logs is the log >of the product. And that ends up looking a lot like ln(n choose k). Recall >Pascal's triangle shows (n choose k) is maximized when k = n/2 (at least >for even n). So we need a bound on ln(n!/[(n/2)!]^2). But n!/[(n/2)!]^2 <= >2^n for even n, which is easily verified by induction. So we're getting >ln(2^n) = nln(2), which is looking good. I think the above can be made into >a real proof. > Yes, this works out pretty well. We have > 1/(m+1) + ... + 1/n < ln(n/m) for 1 <= m < n. > So sum(j=k+1,n) 1/(j-i+1) < ln[(n-i+1)/(k+1-i)]. So now we try to estimate > sum(i=1,k-1) ln[(n-i+1)/(k+1-i)], > which is the ln of the product [n/k]*[(n-1)/(k-1)]* ... *[(n-k+2)/2] <= > n!/[k!(n-k)!] <= 2^n. The last inequality is obvious from the expansion of > (1+1)^n. So including the factor of 2 you had originally, your sum is < > 2*ln(2^n) = 2ln(2)*n. Mmmm ... not quite as good as Robert's answer of n. But I can't quite see where the constant C from above has gone. I think you have limited the sum to C*ln[(n-i)/(k+1-i)], and them limited a sum of that to 2n ln2 (which I really think can't get lower, I think Robert Israels bound it for HALF my original expression, 1 in the numerator not 2). But again, where did the C go, isn't it really 2 C n ln2 . . ? Meanwhile I have heard that there is some really simple answer to it, somthink about finding all the summands equal to 2/l for any value of l, and then stare at those until one sees that they are bounded by 2 n ln 2 = 2 n (1/1 - 1/2 + 1/3 - 1/4 + 1/5 . . ) I think I will know tomorrow, I'll post the answer. Still I would like to know this one as well, what happent to the C. Fjern de 4 bogstaver i min mailadresse som er indsat for at hindre s... Remove the 4 letter word meaning junk mail in my mail address. === Subject: Re: Bound of a sum >This will be a little sketchy. Fact: There exists a constant C such that >1/k + 1(k+1) + ... + 1/n <= C*ln(n/k) for all n, 1 <= k <= n; this is easy >to show. So sum(j=k+1,n) 1/(j-i) <= C*ln[(n-i)/(k+1-i)]. So we just need to >consider C*sum (i=1,k) ln[(n-i)/(k+1-i)]. But that sum of logs is the log >of the product. And that ends up looking a lot like ln(n choose k). Recall >Pascal's triangle shows (n choose k) is maximized when k = n/2 (at least >for even n). So we need a bound on ln(n!/[(n/2)!]^2). But n!/[(n/2)!]^2 <= >2^n for even n, which is easily verified by induction. So we're getting >ln(2^n) = nln(2), which is looking good. I think the above can be made into >a real proof. > Yes, this works out pretty well. We have > 1/(m+1) + ... + 1/n < ln(n/m) for 1 <= m < n. > So sum(j=k+1,n) 1/(j-i+1) < ln[(n-i+1)/(k+1-i)]. So now we try to estimate > sum(i=1,k-1) ln[(n-i+1)/(k+1-i)], > which is the ln of the product [n/k]*[(n-1)/(k-1)]* ... *[(n-k+2)/2] <= > n!/[k!(n-k)!] <= 2^n. The last inequality is obvious from the expansion of > (1+1)^n. So including the factor of 2 you had originally, your sum is < > 2*ln(2^n) = 2ln(2)*n. Mmmm ... not quite as good as Robert's answer of n. > But I can't quite see where the constant C from above has gone. I think > you have limited the sum to C*ln[(n-i)/(k+1-i)], and them limited a sum > of that to 2n ln2 (which I really think can't get lower, I think Robert > Israels bound it for HALF my original expression, 1 in the numerator not 2). > But again, where did the C go, isn't it really 2 C n ln2 . . ? > Meanwhile I have heard that there is some really simple answer to it, > somthink about finding all the summands equal to 2/l for any value of l, > and then stare at those until one sees that they are bounded by > 2 n ln 2 = 2 n (1/1 - 1/2 + 1/3 - 1/4 + 1/5 . . ) > I think I will know tomorrow, I'll post the answer. > Still I would like to know this one as well, what happent to the C. Well as I said, my first post was sketchy. I knew there was some constant C that would work. In the next post I was more precise: 1/(m+1) + ... + 1/n < ln(n/m). No C is needed here. So I think the estimate 2ln(2)*n is OK. === Subject: Re: Bound of a sum >Well as I said, my first post was sketchy. I knew there was some constant C >that would work. In the next post I was more precise: 1/(m+1) + ... + 1/n < >ln(n/m). No C is needed here. So I think the estimate 2ln(2)*n is OK. It looks like a very nice proof now, and pretty short too. I only know myself that (theta(1) just means some number bounded by -c and c, c is a constant.) H(n) = ln(m) + theta(1) so H(n)-h(m)=ln(n)+theta(1) - (ln(m)+theta(1)) = ln(n/m) + theta(1) So, there must be something I don't know well enough to conclude that, as you say, 1/(m+1) + ... + 1/n < ln(n/m). Ahhh. I get it. d(i) = H(i) - ln(i) is >0 for all i in N, and falls as i grows*** so (talking to myself here) for n>m: H(n) - H(m) = d(n)+ln(n) - d(m) - ln(m) = ln(n/m) - (d(m)-d(n)) --------------- if *** holds then this is >0 I like the proof as is now. I only need to make sure about ***, but I guess that every other textbook has that ? Soren === Subject: Re: Bound of a sum > So, there must be something I don't know well enough to conclude that, > as you say, > 1/(m+1) + ... + 1/n < ln(n/m). The easiest way to see it is to observe that the sum is < the integral from m to n of (1/x) dx = ln(n) - ln(m) = ln(n/m). === Subject: Re: Bound of a sum >So, there must be something I don't know well enough to conclude that, >as you say, >1/(m+1) + ... + 1/n < ln(n/m). > The easiest way to see it is to observe that the sum is < the integral from > m to n of (1/x) dx = ln(n) - ln(m) = ln(n/m). of course .. hmmm I should really have found out that one by myself. But I didn't really do any serious math for several years :) OK, I think that the problem has been solved now, into a nice and short proof. I have seen the problem solved to a 2 n ln2 bound by beginning the same way as Robert Israel, after arguing that the worst case (value of k) is preserved even after assuming that k>=n/2. I don't know if anybody is interested in seeing that. And I think the 2 n ln2 bound really is the lowest one that holds for any k !! - becuase, that's the result I got, empirically, from running the algorithm whose execution time the sum does model. Soren Fjern de 4 bogstaver i min mailadresse som er indsat for at hindre s... Remove the 4 letter word meaning junk mail in my mail address. === Subject: Re: Indexes' Difference Divides Sum Of 2 Terms > Let b(1) = 1; > Let b(m) = lowest positive unpicked integers such that: > (m-k) divides evenly into (b(m) + b(k)) > for EACH k, 1 <= k <= m-1. > I get (again, figured by hand, so not completely believable): > b(m) : 1, 2, 3, 8, 7, 54,... > (That last IS 54, not 5, 4.) > Is this sequence a permutation of the integers >= 1? > (I am not even sure it is infinite, without looking harder at the > sequence's mathematics.) > Leroy Quet Hmmm... I could have sworn this sequence was interesting... :/ === Subject: Re: Filling A Grid By Stepping 1,2,3,4,... Oh, I almost forgot to give the solution to the simple finite maze based on the original puzzle. Solution immediately below copied reply-post: > Start with an infinite (in every direction) rectangular grid/lattice. > Put 1 in a square of the grid (or at a vertex of the lattice). > Place 2, 3, 4, 5, ....infinity, by some algorithm, into the > grid/lattice so that each integer (1+m) is exactly m squares (in only > directions of either up,down,left,or right) from the square/vertex > with the integer m, for all m's. > And only one, no more/ no fewer, integer per vertex/square. I think I found a simple pattern for placing the integers so as to > completely fill the infinite grid with unique positive integers. > It might actually be more difficult to fill the half-plane (grid has > one edge) or the infinite quarter-plane (ie. one quadrant of an > infinite cartesian graph), than to fill an infinite grid. (Do not > place any m's off the grid!) > Is it possible to fill any n-by-n grids for FINITE n, aside from n = > 1?? > Leroy Quet > An EASY puzzle based on the above puzzle, and then a difficult(??) > question: > Easy maze: > Moving as above (by 1 position, then by 2, then by 3, etc., moving > only vertically and horizontally), find a path which lands on each * > below one (and exactly one) time, but misses every x and space. > * x * > * x * * * * x x x x x * * > (view with fixed width font.) >.... 4 x x 3 x 2 5 x 1 10 8 6 x x x x x 7 9 I believe this is a unique answer, but I am not *certain*. === Subject: Re: Dogs, fleas, and hairy balls Finally someone introduced some real math into the thread! > comb the hair in such a way that it is smoothly combed everywhere. You > can't do it, there will be two points (source and sink) where there hair > is not locally all combed in the same direction. A common fallacy. In fact there need only be one point. This can be achieved by taking your source and sink example, (the standard one indeed), and slowly moving them toward each other, while keeping everything smooth. (I guess this is some sort of homotopy.) Now, when they finally converge to the same point, you have a combed surface with just one singularity - which is a dipole singularity. (And amusingly even if the combing is directional, you can no longer tell which was the original source and which the sink.) You can also get more than two singularities, even without this cheating sort of multiple-point effect. There is also a stagnant point singularity; where the directions of flow are (say) south at the top, north at the bottom, right at the east and left at the west. So the flow is sink-like if vertical, and source-like if horizontal, and just off these two, or anywhere in between, it follows hyperbola curves in the appropriate senses. Draw a little arrow-pic for yourself. Each stagnant point can use up one source *or* sink - it is worth drawing flow diagrams to convince yourself of this. And the total number of sources and sinks, minus the number of stagnants, is always the Euler characteristic; two for the sphere, zero for a torus, etc. A dipole counts as plus 2, OC, being a conjoined source+sink. There are also more complex multipole singularities possible. It is fun to look at these possibilities, and discover the idea of a partitioned point! All very kooly-woool stuff! -------------------------------------------------------------- -------------- -- Bill Taylor W.Taylor@math.canterbury.ac.nz -------------------------------------------------------------- -------------- -- May the source be with you. -------------------------------------------------------------- -------------- -- === Subject: DiffEQ - need help to solve a basic linear diffEQ Hi. I'm trying to solve this differential equation: (x + e^y) dy/dx = x * e^(-y) - 1 The techniques that I learned in this section were substitution, exact equations, and Bernoulli equations, but I still do not see a solution. I tried to divide by (x + e^y) but that did not seem to help. A substitution of (x + e^y) did not do much, either. === Subject: Re: DiffEQ - need help to solve a basic linear diffEQ > Hi. I'm trying to solve this differential equation: > (x + e^y) dy/dx = x * e^(-y) - 1 > The techniques that I learned in this section were substitution, exact > equations, and Bernoulli equations, but I still do not see a solution. > I tried to divide by (x + e^y) but that did not seem to help. A > substitution of (x + e^y) did not do much, either. let u = e^y. then y' = u'/e^y = u'/u. (x + u) u'/u = x/u - 1 (x + u)u' = x - u u - x + (x + u)u' = 0 (u - x)dx + (x + u)du = 0 === Subject: Re: DiffEQ - need help to solve a basic linear diffEQ === Subject: DiffEQ Hi. I'm trying to solve this differential equation: >(x + e^y) dy/dx = x * e^(-y) - 1 >I tried to divide by (x + e^y) but that did not seem to help. >A substitution of (x + e^y) did not do much, either. It doesn't? Nay, 'tisn't so! (x + e^y) e^y dy/dx = x - e^y u = e^y (x + u) du/dx = x - u v = x + u; u = v - x v(dv/dx - 1) = 2x - v v dv/dx = 2x 2v dv/dx = 4x w = v^2 dw/dx = 4x w = 2x^2 + c v^2 = 2x^2 + c (x + u)^2 = 2x^2 + c u^2 + 2xu - (x^2 + c) = 0 u = [-2x +- sqr(4x^2 + 4(x^2 + c))]/2 = -x +- sqr(2x^2 + c) y = log(-x + sqr(2x^2 + c)), 0 <= c I leave to you the onus of checking the solution. ---- === Subject: Re: DiffEQ - need help to solve a basic linear diffEQ Following your work I was able to understand up to: > u^2 + 2xu - (x^2 + c) = 0 > u = [-2x +- sqr(4x^2 + 4(x^2 + c))]/2 How did you isolate the u in the first line so you get the result in the second line? === > Subject: DiffEQ > Hi. I'm trying to solve this differential equation: >(x + e^y) dy/dx = x * e^(-y) - 1 >I tried to divide by (x + e^y) but that did not seem to help. >A substitution of (x + e^y) did not do much, either. > It doesn't? Nay, 'tisn't so! > (x + e^y) e^y dy/dx = x - e^y > u = e^y > (x + u) du/dx = x - u > v = x + u; u = v - x > v(dv/dx - 1) = 2x - v > v dv/dx = 2x > 2v dv/dx = 4x > w = v^2 > dw/dx = 4x > w = 2x^2 + c > v^2 = 2x^2 + c > (x + u)^2 = 2x^2 + c > y = log(-x + sqr(2x^2 + c)), 0 <= c > I leave to you the onus of checking the solution. > ---- === Subject: Re: DiffEQ - need help to solve a basic linear diffEQ >Hi. I'm trying to solve this differential equation: >(x + e^y) dy/dx = x * e^(-y) - 1 >The techniques that I learned in this section were substitution, exact >equations, and Bernoulli equations, but I still do not see a solution. >I tried to divide by (x + e^y) but that did not seem to help. A >substitution of (x + e^y) did not do much, either. Did it say anything in this section about integrating factors that are functions of y? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Calculating fraction period lengths in different bases Can anyone prove these statements My interest is recreational. Statement 1 If u=2n^2-n+2 and v=2n^2+n+2 Then u^2 mod (u+v-1) =u-1 and v^2 mod (u+v-1) = u-1 For all n>0 Length of fraction period in base u and v =6 Length of fraction period in base u-1 and v-1 =3 Statement 2 If u=2n^2-3n+3; v=2n^2-n+2; p=u+v Then u^2 and v^2 =(p-1) mod p for all u,v. Length of fraction period in base u and v =4 For all n There is also another case: Where u=n+1; v=n^2+n+1. Then u^2 and v^2 =(p-1) mod p for all u,v. Length of fraction period in base u and v =4 For all n>0 Randall McDonnell === Subject: Re: Calculating fraction period lengths in different bases > Can anyone prove these statements > My interest is recreational. > Statement 1 > If u=2n^2-n+2 and v=2n^2+n+2 > Then u^2 mod (u+v-1) =u-1 and v^2 mod (u+v-1) = u-1 > For all n>0 > Length of fraction period in base u and v =6 I think you have to tell us just what frfaction you have in mind. Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Calculating fraction period lengths in different bases > Can anyone prove these statements > My interest is recreational. > Statement 1 > If u=2n^2-n+2 and v=2n^2+n+2 > Then u^2 mod (u+v) =u+v-1 and v^2 mod (u+v) = u+v-1 > For all n>0 > Length of fraction period in base u and v =6 > I think you have to tell us just what frfaction you have in mind. The fraction in mind is 1/(u+v) Sorry for missing that detail. I'd like to add the next pair If u= 2n+1 and v=2n^2 Then u^2 mod (u + v) =u + v-1 and v^2 mod (u + v) = u + v-1 The fraction period length for 1/(u+v) =4 === Subject: Re: Calculating fraction period lengths in different bases > Can anyone prove these statements > My interest is recreational. > Statement 1 > If u=2n^2-n+2 and v=2n^2+n+2 > Then u^2 mod (u+v-1) =u-1 and v^2 mod (u+v-1) = u-1 > For all n>0 > Length of fraction period in base u and v =6 I think you have to tell us just what fraction you have in mind. Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: The Golden Spiral how to graph the Golden Spiral in parametric mode. I've been looking all over the net for the equation but the ones I have found only confused me. Could someone give me a simple equation that I could input into my calculator to get it to work? Jason O http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: The Golden Spiral X-SessionID: Bp%Xb-3354-_4-5811@news.uchicago.edu X-Hash-Info: post-filter,v:1.4 X-Hash: 87266009 ac0611b0 1918b4b6 8ff5c390 7fb5dc46 > how to graph the Golden Spiral in parametric mode. I've been > looking all over the net for the equation but the ones I have found > only confused me. Could someone give me a simple equation that I > could input into my calculator to get it to work? > Jason O Use polar coordinates. r = e^(theta) --Dan Weiner === Subject: Re: The Golden Spiral X-SessionID: 6n%Xb-3347-_4-5823@news.uchicago.edu X-Hash-Info: post-filter,v:1.4 X-Hash: 3b66a45f eece79f2 5fecb640 dd127b55 5f7bfe81 > how to graph the Golden Spiral in parametric mode. I've been > looking all over the net for the equation but the ones I have found > only confused me. Could someone give me a simple equation that I > could input into my calculator to get it to work? Use polar coordinates, not parametrics: r = e^theta. === Subject: Re: The Golden Spiral X-SessionID: ym%Xb-3346-_4-5650@news.uchicago.edu X-Hash-Info: post-filter,v:1.4 X-Hash: 8ab379ed ac3d3cee d4ed691e 69b27ff5 3d839321 > how to graph the Golden Spiral in parametric mode. I've been > looking all over the net for the equation but the ones I have found > only confused me. Could someone give me a simple equation that I > could input into my calculator to get it to work? I can't imagine why you'd want to do this in parametrics, since it's so simple in polar coordinates: r=e^theta I'm pretty sure that your calculator can do polar graphs. === Subject: Re: JSH: how you can prove to the mathematicians you are right |Thus I was trying to prove Gauss's lemma for | integer polynomials and believed I had succeeded. This means | either I am incredibly smart or incredibly stupid. | I am not sure which. The key thing in this context, I think, is to realize that it's something in need of proof, rather than, say, something that just stands to reason. It's not so terribly difficult to prove. Keith Ramsay === Subject: Re: Checking ring of algebraic integers Yeah, but what about me? I mean, I exist too, you know. > It turns out that there's an incredibly simple way to check the ring > of algebraic integers and see that there's a problem with how it's > currently taught. > You can take two quadratics: > x^2 - x + 42 = 0 > and > y^2 - by - 7 = 0 > where by how algebraic integers are currently taught, you'd think that > there exists an algebraic integer b, such that a root of the second > quadratic is a factor of a root of the first. Intriguingly, there > does not, and it's easy to show. > Now > x^2 - x + 42 = 0 has (1 + sqrt(-167))/2 as on of its roots, and > y^2 - by - 7 = 0, has as one of its roots > (b + sqrt(b^2 + 28))/2. > So I can simply introduce z, where > (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2 > which is > z = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) > where now the question is, can an algebraic b exist such that z is an > algebraic integer? > The square roots don't tell me much, but working to eliminate them > adds solutions, as a first step consider: > 28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0 > where I have *two* solutions, where they are > z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) > and > z_2 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28)) > and working still further I get > 196 z^4 + 28b z^3 + (168b^2 + 2324) z^2 - 168bz + 7056 = 0 > and I can divide both sides by 28 to finally get > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0 > where I have four solutions, which are > z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) > z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28)) > z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28)) > z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28)). > (Those not sure can simply multiply out > (Z - z_1)(Z - z_2)(Z - z_3)(Z - z_4) to verify.) > Now then our question was, could what is now z_1 be an algebraic > integer for an algebraic integer b? > And now an important theorem comes into play, as an algebraic integer > cannot be the root of a non-monic primitive with integer coefficients > that is irreducible over Q. > Going back to x^2 - x + 42, I know that the sum of its roots is 1, so > the roots are coprime to each other. > So looking at the four solutions and assuming that z_1 is an algebraic > integer, that leaves z_4 as a second possibility that is also an > algebraic integer. > But > (Z-z_1)(Z-z_4) = [Z - (1 + sqrt(-167))/(b + sqrt(b^2 + 28))][Z - (1 + > sqrt(-167))/(b + sqrt(b^2 + 28))], which is > Z^2 + (b - sqrt(-167) sqrt(b^2 + 28))Z/14 + 6 = 0, > where the middle coefficient cannot be an integer with algebraic > integer b, as that would require that > b^2 + 28 = -167n^2, with n an integer, which means that b can't be > real. > If you use that irrational b then isolate the irrational term you have > sqrt(-167n^2 - 28) Z/14 = -Z^2 + 167n^2 Z/14 - 6 > which when you square both sides to get read of the square root just > gives you a non-monic quartic again! > Therefore, z cannot be an algebraic integer. > Therefore, in the ring of algebraic integers, the roots of > x^2 - x + 42 = 0 > and > y^2 - by - 7 = 0 > never share non-unit factors in the ring of algebraic integers. > Notice that the key theorem here is that one that prevents an > algebraic integer from being the root of a non-monic primitive with > integer coefficients irreducible over Q. > Take away the condition that you have an algebraic integer factor, and > you're ok, which proves that there must be a ring beyond algebraic > integers, where they *can* share non-unit factors. > I have called the more inclusive ring, the ring of objects, or the > Object Ring. > The Object Ring is a commutative ring that includes all numbers such > that -1 and 1 are the only members that are both a unit and an > integer, where no non-unit member is a factor of any two integers that > are coprime. > One of the most surprising results that follows quickly from > understanding the limitations of the ring of algebraic integers, and > having the fuller definition, is a proof of Fermat's Last Theorem of > around two to three pages. > James Harris === Subject: Re: Checking ring of algebraic integers Adjunct Assistant Professor at the University of Montana. > [.snip.] >Notice that the key theorem here is that one that prevents an >algebraic integer from being the root of a non-monic primitive with >integer coefficients irreducible over Q. >Take away the condition that you have an algebraic integer factor, and >you're ok, which proves that there must be a ring beyond algebraic >integers, where they *can* share non-unit factors. >You're a fascinating guy. I remember a while back when you spent >a few months explaining that this extremely well-known result >was wrong, posting erroneous counterexample after erroneous >counterexample. Finally decided it was correct, eh? >Just curious: You decided it's correct because you knowmathematical establishment? >Took some doing, but he went over the proof and agreed to it. >Well yes, that's more or less what he _said_ happened, but >it doesn't really answer my question - the fact that he says >he agrees a proof is correct doesn't prove that he knows how >to prove the result... To be fair, there are many theorems of which I have, at one time or another, seen a proof, gone over it in detail, am convinced (and was convinced) are true, but would not be able to reproduce on demand without looking them up. Finiteness of the class number is one, to name a relevant item (even though I proved it again from scratch for a seminar talk less than 7 months ago). Granted, this particular result (roots of primitive non-monic irreducibles over Q) is such that most people with only a passing of knowledge about polynomials and basic algebra would be able to come up with a proof from scratch and on the fly, but surely the standard you propose is not a good one. On the other hand, if you brought up, say, the following post: or the final paragraph of So I trust Dedekind's work because I believe that mathematicians were really mathematicians back then--instead of con artists. as an example of just taking the word of the evil mathematical establishment (provided it contains a magic phrase like Dedekind proved or Gauss prove), that would be an excellent place to put to your question... Which I believe you did at sundry points in time... (-: 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 === Subject: Re: Checking ring of algebraic integers > [.snip.] >Notice that the key theorem here is that one that prevents an >algebraic integer from being the root of a non-monic primitive with >integer coefficients irreducible over Q. Take away the condition that you have an algebraic integer factor, and >you're ok, which proves that there must be a ring beyond algebraic >integers, where they *can* share non-unit factors. >You're a fascinating guy. I remember a while back when you spent >a few months explaining that this extremely well-known result >was wrong, posting erroneous counterexample after erroneous >counterexample. Finally decided it was correct, eh? >Just curious: You decided it's correct because you knowhow to prove it, or are you just taking the word of the evil >mathematical establishment? >Took some doing, but he went over the proof and agreed to it. >Well yes, that's more or less what he _said_ happened, but >it doesn't really answer my question - the fact that he says >he agrees a proof is correct doesn't prove that he knows how >to prove the result... >To be fair, Why would one want to be fair here? I mean really, what's wrong with you? Ok, let's be fair, just this once: >there are many theorems of which I have, at one time or >another, seen a proof, gone over it in detail, am convinced (and was >convinced) are true, but would not be able to reproduce on demand >without looking them up. Finiteness of the class number is one, to >name a relevant item (even though I proved it again from scratch for a >seminar talk less than 7 months ago). Granted, this particular result >(roots of primitive non-monic irreducibles over Q) is such that most >people with only a passing of knowledge about polynomials and basic >algebra would be able to come up with a proof from scratch and on the >fly, but surely the standard you propose is not a good one. Well, right. But to be fair, if I asked you whether you knew how to prove something what I'd really be asking is whether you knew how to prove it. Otoh, when I speculate on whether James knows how to prove something I'm just being polite, saying that instead of being explicit about my suspicions that he never did and never will understand enough math to be able to follow the proof. I mean he talks about all this stuff, but he has no idea whethever why the algebraic integers form a ring in the first place. (Hard to see how he could since he's never even got the definition of ring right...) >On the other hand, if you brought up, say, the following post: >or the final paragraph of So I trust Dedekind's work because I believe that mathematicians > were really mathematicians back then--instead of con artists. >as an example of just taking the word of the evil mathematical >establishment (provided it contains a magic phrase like Dedekind >proved or Gauss prove), that would be an excellent place to put to >your question... Which I believe you did at sundry points in time... (-: Yes, I believe I did. ************************ David C. Ullrich === Subject: topology question.... hello....... if topology T is standard topology(=usual topology) on R, (a,b) and [a,b] and [a,b) and (a,b] in T for all a,b in R. thus (a,b) and [a,b] and [a,b) and (a,b] is open set on T. ------------------- it's true??? i think that this is true..... let me advice...please..... thank you. === Subject: Re: topology question.... > hello....... > if topology T is standard topology(=usual topology) on R, > (a,b) and [a,b] and [a,b) and (a,b] in T for all a,b in R. > thus (a,b) and [a,b] and [a,b) and (a,b] is open set on T. > ------------------- Remember that the set of open intervals (a,b), a < b, forms a basis for the euclidean topology on R. Therefore any open set must be equal to the union of an arbitrary number of open intervals. It's easy to see that no union of open sets (in this case, open intervals in R) could be equal to [a,b], [a,b) or (a,b] for any a < b; hence none could define open sets. Also note that [a,b] is a closed set in T. However, people often refer to closed sets as being 'in the topology' even though they are not included in the actual set of subsets of the space that comprises the topology set itself (the set T in this case). Usually, given a rule defining the open sets, it is simple to calculate the closed sets. The closed sets of a topology are equal to the complements of the open sets. More precisely, given any space (X ,T1), and any open set A, X A is a closed set in (X,T1), where X A = {x in X | x not in A }. l8r, Mike N. Christoff === Subject: Re: topology question.... > if topology T is standard topology(=usual topology) on R, > (a,b) and [a,b] and [a,b) and (a,b] in T for all a,b in R. > thus (a,b) and [a,b] and [a,b) and (a,b] is open set on T. > it's true??? No, only (a,b) is in T. If instead of 'and' you mean 'intersection' then yes (a,b) / [a,b] / [a,b) / (a,b] = (a,b) is in T. === Subject: Re: limit of sequence of functions Hmmm....well if a linear functional is continuous, it is bounded, but f_n doesn't seem to be a linear functional of x (although you could define F_n (f(x)) = f_n (x) and that would be a linear functional on the space of functions. Then F_n is continuous at f(0), whence everywhere, whence bounded in the sense that ||F_n f(x)|| < = M ||f(x)|| for f(x) in S^1. But from here I'm not sure how you conclude there is a K such that INT(-pi, pi) p_n <= K, n=1,2,...) How about this? Uniform continuity, etc., gets you a bound for each INT =(-pi, pi) p_n. Now, since lim f_n (0) exists for every f in S^1, consider f = 1. Then lim INT (-pi, pi) p_n exists, call it C. Then given e>0, | INT (-pi, pi) p_n - C | < e. Now INT (-pi, pi) p_n might be just a bit bigger than C (rather than just a bit smaller), but since this limit exists there is certainly some number D such that INT (-pi, pi) p_n < = D for n large enough. Then, take the larger of D and the maximum of the separate bounds obtained by uniform continuity for INT (-pi, pi) p_1, INT (-pi, pi) p_2, INT (-pi, pi) p_3, ..., INT (-pi, pi) p_N (up until n is large enough that we can use the D estimate), and call this number M. Then INT (-pi, pi) p_n < = M, right? === Subject: Re: limit of sequence of functions You know, these conversations would be a lot easier to follow if you didn't delete the entire post you were replying to. Nothing below is anything like a positive answer to the question I asked: >Do you know any results about Banach spaces >where the conclusion is then the norms of ___ are bounded? Do you know _any_ theorems about Banach spaces? If you don't know anything about Banach spaces then you need to _learn_ some basic functional analysis before working on the old exams you say you're working on. If you do know some theorems about Banach spaces, name a few. There really is a very standard basic theorem in the field that gives an immediate answer to the problem, if you look at it right. >Hmmm....well if a linear functional is continuous, it is bounded, but f_n >doesn't seem to be a linear functional of x (although you could define F_n >(f(x)) = f_n (x) and that would be a linear functional on the space of >functions. The notation is confused here. Let's define L_n(f) = int f(x) p_n(-x). Then L_n is a bounded linear functional on C([-pi, pi]) or whatever the interval was (don't delete everything!) Can you say what the norm of L_n is? >Then F_n is continuous at f(0), whence everywhere, whence bounded in >the sense that ||F_n f(x)|| < = M ||f(x)|| for f(x) in S^1. But from here I'm >not sure how you conclude there is a K such that INT(-pi, pi) p_n <= K, >n=1,2,...) >How about this? >Uniform continuity, etc., gets you a bound for each INT =(-pi, pi) p_n. >Now, since lim f_n (0) exists for every f in S^1, consider f = 1. Then >lim INT (-pi, pi) p_n >exists, call it C. Then given e>0, >| INT (-pi, pi) p_n - C | < e. >Now INT (-pi, pi) p_n might be just a bit bigger than C (rather than just a >bit smaller), but since this limit exists there is certainly some number D such >that >INT (-pi, pi) p_n < = D >for n large enough. >Then, take the larger of D and the maximum of the separate bounds obtained by >uniform continuity for INT (-pi, pi) p_1, INT (-pi, pi) p_2, INT (-pi, pi) p_3, >..., INT (-pi, pi) p_N (up until n is large enough that we can use the D >estimate), and call this number M. Then >INT (-pi, pi) p_n < = M, right? ************************ David C. Ullrich === Subject: Re: limit of sequence of functions Sorry! I thought I included the original text but AOL (yes, admittedly inferior) must chop it off! OK, well, using the sup norm, we have | L_n (f) | = | INT f(x) p_n(-x) | < = ||f(x)|| INT |p_n(-x)|, so the norm of L_n is less than or equal to INT |p_n (-x)| ... but you need to know a priori that INT |p_n(-x)| is bounded to conclude that this means L_n is a bounded linear functional, don't you? Certainly, given n, it's bounded, but that's not what we're going for... Some theorems I know...umm...Hahn-Banach Theorem, open graph theorem, closed graph theorem, continuity implies boundedness for linear functionals... Message-id: You know, these conversations would be a lot easier to follow if you didn't delete the entire post you were replying to. Nothing below is anything like a positive answer to the question I asked: >Do you know any results about Banach spaces >where the conclusion is then the norms of ___ are bounded? Do you know _any_ theorems about Banach spaces? If you don't know anything about Banach spaces then you need to _learn_ some basic functional analysis before working on the old exams you say you're working on. If you do know some theorems about Banach spaces, name a few. There really is a very standard basic theorem in the field that gives an immediate answer to the problem, if you look at it right. >Hmmm....well if a linear functional is continuous, it is bounded, but f_n >doesn't seem to be a linear functional of x (although you could define F_n >(f(x)) = f_n (x) and that would be a linear functional on the space of >functions. The notation is confused here. Let's define L_n(f) = int f(x) p_n(-x). Then L_n is a bounded linear functional on C([-pi, pi]) or whatever the interval was (don't delete everything!) Can you say what the norm of L_n is? >Then F_n is continuous at f(0), whence everywhere, whence bounded in >the sense that ||F_n f(x)|| < = M ||f(x)|| for f(x) in S^1. But from here I'm >not sure how you conclude there is a K such that INT(-pi, pi) p_n <= K, >n=1,2,...) >How about this? >Uniform continuity, etc., gets you a bound for each INT =(-pi, pi) p_n. >Now, since lim f_n (0) exists for every f in S^1, consider f = 1. Then >lim INT (-pi, pi) p_n >exists, call it C. Then given e>0, >| INT (-pi, pi) p_n - C | < e. >Now INT (-pi, pi) p_n might be just a bit bigger than C (rather than just a >bit smaller), but since this limit exists there is certainly some number D such >that >INT (-pi, pi) p_n < = D >for n large enough. >Then, take the larger of D and the maximum of the separate bounds obtained by >uniform continuity for INT (-pi, pi) p_1, INT (-pi, pi) p_2, INT (-pi, pi) p_3, >..., INT (-pi, pi) p_N (up until n is large enough that we can use the D >estimate), and call this number M. Then >INT (-pi, pi) p_n < = M, right? ************************ David C. Ullrich === Subject: Re: limit of sequence of functions >Sorry! I thought I included the original text but AOL (yes, admittedly >inferior) must chop it off! >OK, well, using the sup norm, we have >| L_n (f) | = | INT f(x) p_n(-x) | < = ||f(x)|| INT |p_n(-x)|, so the norm of >L_n is less than or equal to INT |p_n (-x)| True. One of the big theorems in measure theory says that the norm is in fact equal to int |p_n|. The Riesz Representation Theorem. >... but you need to know a priori >that INT |p_n(-x)| is bounded to conclude that this means L_n is a bounded >linear functional, don't you? No, we need something else to conclude it's _uniformly_ bounded. >Certainly, given n, it's bounded, but that's not >what we're going for... >Some theorems I know...umm...Hahn-Banach Theorem, open graph theorem, Usually called the open mapping theorem. >closed >graph theorem, continuity implies boundedness for linear functionals... Ok, so you do know some theorems. The one you need is not on this list, but it must be in whatever book you learned the Hahn-Banach theorem and the next two on the list. We need to show those functionals are uniformly bounded. A theorem called the Uniform Boundedness Principle will do this. Also known as the Banach-Steinhaus theorem. >Message-id: you didn't delete the entire post you were replying to. >Nothing below is anything like a positive answer to the question >I asked: >Do you know any results about Banach spaces >where the conclusion is then the norms of ___ are bounded? >Do you know _any_ theorems about Banach spaces? >If you don't know anything about Banach spaces then you >need to _learn_ some basic functional analysis before working >on the old exams you say you're working on. If you do know >some theorems about Banach spaces, name a few. There >really is a very standard basic theorem in the field that gives >an immediate answer to the problem, if you look at it right. >Hmmm....well if a linear functional is continuous, it is bounded, but f_n >doesn't seem to be a linear functional of x (although you could define F_n >(f(x)) = f_n (x) and that would be a linear functional on the space of >functions. >The notation is confused here. >Let's define > L_n(f) = int f(x) p_n(-x). >Then L_n is a bounded linear functional on C([-pi, pi]) or whatever >the interval was (don't delete everything!) Can you say what the >norm of L_n is? >Then F_n is continuous at f(0), whence everywhere, whence bounded in >the sense that ||F_n f(x)|| < = M ||f(x)|| for f(x) in S^1. But from here I'm >not sure how you conclude there is a K such that INT(-pi, pi) p_n <= K, >n=1,2,...) >How about this? >Uniform continuity, etc., gets you a bound for each INT =(-pi, pi) p_n. >Now, since lim f_n (0) exists for every f in S^1, consider f = 1. Then >lim INT (-pi, pi) p_n >exists, call it C. Then given e>0, >| INT (-pi, pi) p_n - C | < e. >Now INT (-pi, pi) p_n might be just a bit bigger than C (rather than just a >bit smaller), but since this limit exists there is certainly some number D >such >that >INT (-pi, pi) p_n < = D >for n large enough. >Then, take the larger of D and the maximum of the separate bounds obtained by >uniform continuity for INT (-pi, pi) p_1, INT (-pi, pi) p_2, INT (-pi, pi) >p_3, >..., INT (-pi, pi) p_N (up until n is large enough that we can use the D >estimate), and call this number M. Then >INT (-pi, pi) p_n < = M, right? >************************ >David C. Ullrich ************************ David C. Ullrich === Subject: Re: limit of sequence of functions Out of curiosity, how does RRT show that ||L_n|| = int p_n ? >Sorry! I thought I included the original text but AOL (yes, admittedly >inferior) must chop it off! >OK, well, using the sup norm, we have >| L_n (f) | = | INT f(x) p_n(-x) | < = ||f(x)|| INT |p_n(-x)|, so the norm of >L_n is less than or equal to INT |p_n (-x)| > True. One of the big theorems in measure theory says that the norm > is in fact equal to int |p_n|. The Riesz Representation Theorem. >... but you need to know a priori >that INT |p_n(-x)| is bounded to conclude that this means L_n is a bounded >linear functional, don't you? > No, we need something else to conclude it's _uniformly_ bounded. >Certainly, given n, it's bounded, but that's not >what we're going for... >Some theorems I know...umm...Hahn-Banach Theorem, open graph theorem, > Usually called the open mapping theorem. >closed >graph theorem, continuity implies boundedness for linear functionals... > Ok, so you do know some theorems. The one you need is not on this > list, but it must be in whatever book you learned the Hahn-Banach > theorem and the next two on the list. > We need to show those functionals are uniformly bounded. A theorem > called the Uniform Boundedness Principle will do this. Also known > as the Banach-Steinhaus theorem. >Message-id: you didn't delete the entire post you were replying to. >Nothing below is anything like a positive answer to the question >I asked: >Do you know any results about Banach spaces >where the conclusion is then the norms of ___ are bounded? >Do you know _any_ theorems about Banach spaces? >If you don't know anything about Banach spaces then you >need to _learn_ some basic functional analysis before working >on the old exams you say you're working on. If you do know >some theorems about Banach spaces, name a few. There >really is a very standard basic theorem in the field that gives >an immediate answer to the problem, if you look at it right. >Hmmm....well if a linear functional is continuous, it is bounded, but f_n >doesn't seem to be a linear functional of x (although you could define F_n >(f(x)) = f_n (x) and that would be a linear functional on the space of >functions. >The notation is confused here. >Let's define > L_n(f) = int f(x) p_n(-x). >Then L_n is a bounded linear functional on C([-pi, pi]) or whatever >the interval was (don't delete everything!) Can you say what the >norm of L_n is? >Then F_n is continuous at f(0), whence everywhere, whence bounded in >the sense that ||F_n f(x)|| < = M ||f(x)|| for f(x) in S^1. But from here I'm >not sure how you conclude there is a K such that INT(-pi, pi) p_n <= K, >n=1,2,...) >How about this? >Uniform continuity, etc., gets you a bound for each INT =(-pi, pi) p_n. >Now, since lim f_n (0) exists for every f in S^1, consider f = 1. Then >lim INT (-pi, pi) p_n >exists, call it C. Then given e>0, >| INT (-pi, pi) p_n - C | < e. >Now INT (-pi, pi) p_n might be just a bit bigger than C (rather than just a >bit smaller), but since this limit exists there is certainly some number D >such >that >INT (-pi, pi) p_n < = D >for n large enough. >Then, take the larger of D and the maximum of the separate bounds obtained by >uniform continuity for INT (-pi, pi) p_1, INT (-pi, pi) p_2, INT (-pi, pi) >p_3, >..., INT (-pi, pi) p_N (up until n is large enough that we can use the D >estimate), and call this number M. Then >INT (-pi, pi) p_n < = M, right? >************************ >David C. Ullrich > ************************ > David C. Ullrich === Subject: Re: limit of sequence of functions Nevermind. I just found the (applicable) statement of the theorem and see how they do it. One question about how it applies here, though: Our linear functional L_n (f) = INT(-pi, pi) p_n (-y) f(y) dy is seen as mapping a subset of L_oo (that's why we used the sup norm) into R. But the RRT (at least the version I'm looking at) says we need to view L_n as mapping a subset of L_p for 1 <= p < oo into R. Does it still work for p = oo ? > Out of curiosity, how does RRT show that > ||L_n|| = int p_n ? >Sorry! I thought I included the original text but AOL (yes, admittedly >inferior) must chop it off! OK, well, using the sup norm, we have | L_n (f) | = | INT f(x) p_n(-x) | < = ||f(x)|| INT |p_n(-x)|, so the > norm of >L_n is less than or equal to INT |p_n (-x)| > True. One of the big theorems in measure theory says that the norm > is in fact equal to int |p_n|. The Riesz Representation Theorem. >... but you need to know a priori >that INT |p_n(-x)| is bounded to conclude that this means L_n is a > bounded >linear functional, don't you? > No, we need something else to conclude it's _uniformly_ bounded. >Certainly, given n, it's bounded, but that's not >what we're going for... Some theorems I know...umm...Hahn-Banach Theorem, open graph theorem, > Usually called the open mapping theorem. >closed >graph theorem, continuity implies boundedness for linear functionals... > Ok, so you do know some theorems. The one you need is not on this > list, but it must be in whatever book you learned the Hahn-Banach > theorem and the next two on the list. > We need to show those functionals are uniformly bounded. A theorem > called the Uniform Boundedness Principle will do this. Also known > as the Banach-Steinhaus theorem. >Message-id: you didn't delete the entire post you were replying to. Nothing below is anything like a positive answer to the question >I asked: Do you know any results about Banach spaces >where the conclusion is then the norms of ___ are bounded? Do you know _any_ theorems about Banach spaces? If you don't know anything about Banach spaces then you >need to _learn_ some basic functional analysis before working >on the old exams you say you're working on. If you do know >some theorems about Banach spaces, name a few. There >really is a very standard basic theorem in the field that gives >an immediate answer to the problem, if you look at it right. >Hmmm....well if a linear functional is continuous, it is bounded, but > f_n >doesn't seem to be a linear functional of x (although you could define > F_n >(f(x)) = f_n (x) and that would be a linear functional on the space of >functions. The notation is confused here. Let's define L_n(f) = int f(x) p_n(-x). Then L_n is a bounded linear functional on C([-pi, pi]) or whatever >the interval was (don't delete everything!) Can you say what the >norm of L_n is? >Then F_n is continuous at f(0), whence everywhere, whence bounded in >the sense that ||F_n f(x)|| < = M ||f(x)|| for f(x) in S^1. But from > here I'm >not sure how you conclude there is a K such that INT(-pi, pi) p_n <= K, >n=1,2,...) >How about this? >Uniform continuity, etc., gets you a bound for each INT =(-pi, pi) p_n. >Now, since lim f_n (0) exists for every f in S^1, consider f = 1. Then >lim INT (-pi, pi) p_n >exists, call it C. Then given e>0, >| INT (-pi, pi) p_n - C | < e. >Now INT (-pi, pi) p_n might be just a bit bigger than C (rather than > just a >bit smaller), but since this limit exists there is certainly some number >such >that >INT (-pi, pi) p_n < = D >for n large enough. >Then, take the larger of D and the maximum of the separate bounds > obtained by >uniform continuity for INT (-pi, pi) p_1, INT (-pi, pi) p_2, INT (-pi, > pi) >p_3, >..., INT (-pi, pi) p_N (up until n is large enough that we can use the D >estimate), and call this number M. Then >INT (-pi, pi) p_n < = M, right? >************************ David C. Ullrich ************************ > David C. Ullrich === Subject: Re: limit of sequence of functions UNIFORM BOUNDEDNESS THEOREM. > Sorry! I thought I included the original text but AOL (yes, admittedly > inferior) must chop it off! > OK, well, using the sup norm, we have > | L_n (f) | = | INT f(x) p_n(-x) | < = ||f(x)|| INT |p_n(-x)|, so the norm of > L_n is less than or equal to INT |p_n (-x)| ... but you need to know a priori > that INT |p_n(-x)| is bounded to conclude that this means L_n is a bounded > linear functional, don't you? Certainly, given n, it's bounded, but that's not > what we're going for... > Some theorems I know...umm...Hahn-Banach Theorem, open graph theorem, closed > graph theorem, continuity implies boundedness for linear functionals... > Message-id: you didn't delete the entire post you were replying to. > Nothing below is anything like a positive answer to the question > I asked: >Do you know any results about Banach spaces >where the conclusion is then the norms of ___ are bounded? > Do you know _any_ theorems about Banach spaces? > If you don't know anything about Banach spaces then you > need to _learn_ some basic functional analysis before working > on the old exams you say you're working on. If you do know > some theorems about Banach spaces, name a few. There > really is a very standard basic theorem in the field that gives > an immediate answer to the problem, if you look at it right. >Hmmm....well if a linear functional is continuous, it is bounded, but f_n >doesn't seem to be a linear functional of x (although you could define F_n >(f(x)) = f_n (x) and that would be a linear functional on the space of >functions. > The notation is confused here. > Let's define > L_n(f) = int f(x) p_n(-x). > Then L_n is a bounded linear functional on C([-pi, pi]) or whatever > the interval was (don't delete everything!) Can you say what the > norm of L_n is? >Then F_n is continuous at f(0), whence everywhere, whence bounded in >the sense that ||F_n f(x)|| < = M ||f(x)|| for f(x) in S^1. But from here I'm >not sure how you conclude there is a K such that INT(-pi, pi) p_n <= K, >n=1,2,...) >How about this? >Uniform continuity, etc., gets you a bound for each INT =(-pi, pi) p_n. >Now, since lim f_n (0) exists for every f in S^1, consider f = 1. Then >lim INT (-pi, pi) p_n >exists, call it C. Then given e>0, >| INT (-pi, pi) p_n - C | < e. >Now INT (-pi, pi) p_n might be just a bit bigger than C (rather than just a >bit smaller), but since this limit exists there is certainly some number D > such >that >INT (-pi, pi) p_n < = D >for n large enough. >Then, take the larger of D and the maximum of the separate bounds obtained by >uniform continuity for INT (-pi, pi) p_1, INT (-pi, pi) p_2, INT (-pi, pi) > p_3, >..., INT (-pi, pi) p_N (up until n is large enough that we can use the D >estimate), and call this number M. Then >INT (-pi, pi) p_n < = M, right? > ************************ > David C. Ullrich === Subject: Re: (p-1)'th cyclotomic polynomial > How come, the sum of the roots of the (p-1)th cyclotomic polynomial(mod p) > where p is a prime, are always equal to 1, -1 or 0? > The sum of the roots (zeroes) of the n-th cyclotomic polynomial > (the primitive n-th roots of unity) is mu(n) where mu is the > Mobius function. Oh, thank you very much :-) === Subject: L^2 Frechet ?? There's a marginal note in my textbook next to the following problem that says L^2 Frechet: Let X,Y be random variables on a probability space such that E(X^2) x) ≤ u} and Y_q = Q_Y(U) = inf{y : P (|Y| > y) ≤ u}, be random variables defined by the quantile transformation. Then given the result P{X>x, Y>y} < = P {X_q > x, Y_q > y}, show E(XY) < = E(X_q Y_q). I get what the L^2 part of the note refers to, but what does the Frechet part refer to? I searched for Frechet inequality and things like that but couldn't find anything that looks like this result === Subject: Re: L^2 Frechet ?? Douglas Muir > There's a marginal note in my textbook next to the following problem > that says L^2 Frechet: ... > I get what the L^2 part of the note refers to, but what does the > Frechet part refer to? I searched for Frechet inequality and things > like that but couldn't find anything that looks like this result. L^2 is a Banach space (look for Fischer-Riesz theorem of Riesz-Fischer theorem) and therefore a Fr.8echet space. A Fr.8echet space is a topological vector space that is complete, metrizable, and various other things; the rather ornate definition can be found in probably any introductory text on functional analysis. === Subject: Difference between polynomial and field arithmetic!! Could someone please explain me the difference between field arithmetic and polynomial arithmetic. Any help would be greatly appreciated. OP. === Subject: Re: Difference between polynomial and field arithmetic!! >Could someone please explain me the difference between field >arithmetic and polynomial arithmetic. >Any help would be greatly appreciated. Well, since fields are not polynomials, it would make more sense to ask whether there is any connection between the two, rather than what is the difference between them. Many fields are of the form F(a), where F is some more basic field. Typically, F is either a finite field F_p of integers mod p, or F is the field of rational numbers, and you assume that you can already do arithmetic in F. The element a could be either transcendental or algebraic over F. In the first case, F(a) is isomorphic to the field of rational functions over F, so arithmetic in F(a) can be carried out using polynomial arithmetic. If a is algebraic over F, then it satisfies a minimal polynomial f(a) = 0, for some polynomial f, and then arithmetic in F(a) can be carried out using polynomial arithmetic modulo f. But for reasonably small finite fields, there are other methods of doing arithmetic, which do not involve polynomial arithmetic, which can be more efficient. One way is to choose a primitive element b of the field, and represent nonzero elements x by the integer n, where x = b^n. Then multiplication is very fast, and addition can be carried out using a stored pre-computed table of Zech logarithms l(n), defined by b^n + 1 = b^{l(n)}. Derek Holt. === Subject: Re: Difference between polynomial and field arithmetic!! >Could someone please explain me the difference between field >arithmetic and polynomial arithmetic. Sure. First explain to us exactly what you mean by field arithmetic and polynomial arithmetic. >Any help would be greatly appreciated. >OP. ************************ David C. Ullrich === Subject: Re: Lattice points, balls, number theory. > What is known about the coupled problem where one looks for pairs of > non-negative integers (k,m) such that there exist a closed ball in R^n > having exactly k lattice points on its boundary and exactly m in its > interior ?? I don't know much, but if the ball has radius r then m = V_n r^n + O(r^{n-1}) and k = O(r^{n-1}) where V_n is the volume of the ball of radius 1. Hence there are no solutions unless k = O(m^{(n-1)/n}). Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Silly question on limits, tensor products |Fine and dandy. But I am stepping back a bit and asking myself the |following: There was one step that was skipped. That is : | |lim [ u(t+h) tensor [v(t+h) - v(t)]/h ] = |h--> 0 | |lim u(t+h) tensor lim ( [v(t+h) - v(t)]/h ) |h ----> 0 h --> 0 | |Why is that true? (I know, it is a stupid stupid question) The way I |justify this is by basically hand-waving and saying it makes sense, but I |am not convinced that you can do this type of thing for every occasion in |which a limit breaks up into two limits. The question amounts to, when is lim(a(t) tensor b(t)) equal to lim(a(t)) tensor lim(b(t))? If the limit of a(t) is A, and the limit of b(t) is B, then a(t) tensor b(t) = (a(t)-A+A) tensor (b(t)-B+B) = (a(t)-A) tensor (b(t)-B) + A tensor (b(t)-B) + (a(t)-A) tensor B + (A tensor B). So it's enough to show that when c(t)=a(t)-A and d(t)=b(t)-B go to 0, then so do c(t) tensor d(t), c(t) tensor B, and A tensor d(t) for any given A, B. How you show that sort of depends on what definitions you've been given. By some definitions, you get a norm on tensor products that satisfies |u tensor v| = |u| |v|, and that's pretty much all you need to know, since if |c(t)| and |d(t)| go to 0, then obviously |c(t) tensor d(t)| goes to 0 too, as well as |c(t) tensor B|=|c(t)| |B| and |A tensor d(t)| = |A| |d(t)|. Keith Ramsay === Subject: Re: The role of infinity in math |Hersh,_The_Mathematical_Experience_: | |The constructivists regard as genuine mathematics only what can be |obtained by a finite construction. The set of real numbers, or any |other infinite set, cannot so be obtained. Davis and Hersh have not tried very hard to understand what is meant by constructivism. Was L.E.J.Brouwer a constructivist? Yes. Did he accept the set of real numbers? Yes. Was Errett Bishop a constructivist? Yes. Did Bishop accept the set of real numbers? Yes. Have a look at any of their work and you'll soon see they aren't the kind of finitist that Davis and Hersh mistakenly thought constructivists are. |Then there is the intuitionist philosophy which is a little more |mainstream but is still not the majority opinion. Their objection is |not based on whether infinity exists or not, but on whether we can |even say anything about infinity if it did exist. That's silly. |What separates them |from the classic mathematicians is that they reject Aristotle's Law of |the Excluded Middle that the proposition (A or not A) is a tautology, Rejecting the law of excluded middle is one of the main differences, although not the only one. |since one can never know for sure if for each x, (A(x) or not A(x)) is |a tautology when x ranges over an infinite domain. Intuitionists accept assumptions that imply that in some cases for each x, A(x) or not A(x) is _false_. Describing it as a matter of not knowing whether it's a tautology is a little odd. In constructive terms, knowing A or B means having a method which in principle will select one of them, with a guarantee that the one selected is true. Certainly one doesn't generally have a procedure for determining whether A(x) is true or false. |While this may seem |crazy and counter-intuitive, intuitionists might respond that the |theorems that result from this assumption are also counter-intuitive - |i.e., Cantor's aleph null and aleph one. Do you mean i.e. or do you mean e.g.? If you really mean i.e., you're indicating that you've just stated all the counterintuitive theorems that result from the assumption. If you only mean for example, that's abbreviated by e.g.. |Why should there be many |types of infinities? Infinity is infinity. This is not the constructivist point of view. Brouwer didn't accept all of Cantor's infinities, for other reasons (for example, not accepting the power set axiom as a general principle), but he did agree that there were different kinds of infinity. For instance, that the continuum was a different kind of infinity than the integers. I would guess Bishop thought there was a somewhat misplaced emphasis in the theory of cardinalities, but had no special problem with the theorem that there isn't an onto mapping from the natural numbers to the real numbers. The counterintuitive consequences of the law of excluded middle have more to do with the claim that mathematical objects can exist allegedly without there being any way of our finding them. | And also, look at all of |the paradoxes of set theory - Is there a set of all sets, etc. When |one eliminates the Law of the Excluded Middle, one avoids a lot of the |crazy paradoxes associated with infinity. Which ones? The existence of a set of all sets still results in a contradiction even without the law of excluded middle. It would imply that the Russell set R={x : x is not a member of x} exists. If R is a member of R, then by the definition of R, we get that R is not a member of R, which is a contradiction. Consequently R is not a member of R. But then by the definition of R, R must be a member of R, a contradiction. I didn't need the law of excluded middle there. I needed that a proof that a proposition P leads to a contradiction constitutes a proof of not P. That's the canonical way to prove a negative statement constructively. I also needed that y is a member of {x: P(x)} if and only if P(y), which is just what {x: P(x)} means. So evidently there's no such thing as the Russell set. |The classic philosophy (Platonism) says that all of this stuff |(infinity) is in fact real. And just because we cannot perceive it |does not mean that it does not exist. This is the majority opinion, |not necessarily because the majority believe it, but more because it |is the most practical way of thinking of mathematics, I would say that in practice it helps you do mathematics if you think of the objects and relationships involves as being just as real as all the stuff of daily life that one deals with the rest of the time. Now, it happens that some of the most philosophically minded constructivists have been people whose philosophy led them to treat all realms the same way. I remember one telling me with a chuckle how it was sometimes claimed that we apply the law of excluded middle in reasoning about real world, but that people in fact hardly ever did. | since the |conclusions have applications in natural sciences. Nonconstructive mathematics is applied in natural science, but I'm not entirely convinced this is helpful overall. |If we were to go to |what my brother-in-law proposes, replacing differential equations with |difference equations, etc., we would get the same theorems, but they |would be much more difficult to write and describe. The derivative |of x^2 would not be 2x but would be 2x+1/M, where M is the largest |number. In practice, this number 1/M would be so small that it |wouldn't even matter, so why write it down on paper? Also, the |Pythagorean theorem would not hold, and it would be much more messy to |describe this relationship. But this way of thinking still avoids |paradoxes. This is a different story entirely. :-) Inquiring minds want to know whether you married his sister, or (perhaps more unsettling?) your sister married him. Perhaps he'd like to hunt down a paper by Jan Mycielski written in about 1980 where he describes a system where you deal with real numbers by pretending that they're ratios of really big numbers. In a model of his axiom system, there are infinite elements, but in any given proof in the system, you only use finitely many axioms, and any finite collection of axioms has a finite model where the elements are merely large. Keith Ramsay === Subject: Re: The role of infinity in math |What is the role of infinity in math: | |How is it defined? It depends on context. Many of the uses of the term infinity are best thought of as idiomatic expressions. For example, the limit of f(x) as x goes to infinity should be thought of as an idiom meaning nothing more or less than what the definition says that it means. The uses of the term which are least metaphorical this way are where infinity means the property of being infinite, in reference to infinite sets. An infinite set S is simply one having the property that for any finite subset {s1,...,sn} of S, there exists an element of S different from s1,...,sn. |Why is it needed? To answer that one has to compare how things work now with how they would work if we didn't consider infinite sets. That's a little slippery, since it's hard to say exactly when one has crossed the line into talking about sets. A set of integers for example corresponds to a property of integers, where two properties are considered equivalent if the same integers satisfy them. So talking about the set of primes is pretty much another way of talking about the property of being prime. One way to analyze the situation is to consider some standard mathematical systems and dump out the infinite sets. Often the resulting system of finite structures turns out to be equivalent to a system of axioms for arithmetic, since finite structures can be encoded as integers. The most stubborn problem with trying to found all of mathematics on such a foundation is that there are properties of finite structures that we ordinarily prove indirectly by reasoning that involves referring to infinite sets along the way. For instance, there's a result called the Paris-Harrington variant of Ramsey's theorem, that's impossible to prove in PA, which is a standard system for axioms for arithmetic, even though PH only talks about finite structures. The usual proof of PH defines a sequence of infinite sets of integers, each of which individually could be defined purely in terms of integers, but which together don't have a definition in the language of elementary arithmetic. |At what precition does math work? This seems like a very strange question. If by precition you mean numerical precision, the answer is absolute precision. If you mean linguistic precision, well, take a look for yourself. The precision of the language of mathematics tends to be rather high, but in practice varies somewhat depending on the context. In an informal talk, one may brush off quite a lot of detail in order to convey the gist of an idea. In publication one needs to be more careful. Imprecision which serves no purpose is considered poor. Imprecision which is due to your not being _able_ to be precise is considered bad. |These are questions which seem to have to accepted answer, what do you |think? People would state things differently. Probably people would give a diversity of answers why we need infinity in mathematics. I don't know that there's much controversy about how actually to use it, though. Keith Ramsay === Subject: Re: The role of infinity in math | |> The classic philosophy (Platonism) says that all of this stuff |> (infinity) is in fact real. And just because we cannot perceive it |> does not mean that it does not exist. This is the majority opinion, |> not necessarily because the majority believe it, but more because it |> is the most practical way of thinking of mathematics | |What?!!!! (read that as I'm stunned in disagreement and intrigue) It's a little hard to accurately characterize the views of typical mathematicians, since they're not usually so interested in these philosophical issues. Realism might be a better term here than Platonism, since the latter implies more of a commitment to the existence of a kind of realm of ideas. Without having any polling data in hand, I would say it seems rather common for mathematicians to have a realist stance (a philosopher might call it naive realism) about the commonplace objects of study in mathematics. For instance, the existence of infinitely many twin primes would be commonly considered either actually true or actually false (almost certainly true), independent of any means we might have or not have of determining which is the case. I think it would take something remarkable to cause a serious change in this situation. To some extent, for people dealing seriously with a subject matter, naive realism is a natural default position, and only really gives way to some form of nonrealism once the nonrealism is able to establish a sufficiently solid basis for itself. Probably formalism is the strongest candidate for such an usurper. If anything makes me cautious about claiming that realism is the usual point of view of mathematicians, it's mainly that I've heard mathematicians make comments somewhat in favor of formalism. For instance, I remember one guy at an afternoon tea remarking that we could just as well choose whichever set of axioms for set theory we wanted, and he thought assuming all sets of reals were measurable (and sticking to just the axiom of dependent choice) would be handy. I think formalism raises enough issues, however, that it hasn't really succeeded in establishing itself as the default understanding by mathematicians of what their subject is about. Saying that there is not just one twin prime conjecture (that there are infinitely many integers n such that n and n+2 are both prime), but an assortment depending on which axiom system you mean to use to prove it, grates against many of our intuitions. I don't see any easy way for a formalist to avoid this kind of pluralism. That leaves the mathematical community with a lot of this kind of squishy realism. Fortunately, not much about the way mathematics is actually done depends on this kind of philosophical issue, so there's not much motivation to settle the issue convincingly one way or the other. Keith Ramsay === Subject: Re: The role of infinity in math |Really? And that part about where intuitionism avers that mathematics |and proofs cannot be communicated? That doesn't conflict with any |mathematicians' experiences? I don't know that anyone has ever followed Brouwer that far. But I think I can see how one might come to some of his pessimistic conclusions about language. On the one hand, we can see people learning to play certain kinds of language games, as a kind of skill. He expressed great misgivings about people adopting such an instrumental attitude toward language, mathematics, or just about anything. On the other hand, if it really is supposed to be a matter of carrying a thought from one mind to another, isn't it somewhat remarkable that we feel such confidence that the thought that arrives at the other end is really the same as the one that set forth? I'm not at all as pessimistic, but I still find language often sliding toward either a successful but robot-like technique that succeeds in conveying basically data, or into poetry that is easily lost in translation. Keith Ramsay === Subject: Re: The role of infinity in math Discussion, linux) > |Really? And that part about where intuitionism avers that mathematics > |and proofs cannot be communicated? That doesn't conflict with any > |mathematicians' experiences? > I don't know that anyone has ever followed Brouwer that far. I'm sure you know a lot more about the state of modern intuitionism than I do, so I'll take your word for it. Frankly, it's good to hear that, since that part of intuitionism is certainly the most dubious to my ears. > But I think I can see how one might come to some of his pessimistic > conclusions about language. On the one hand, we can see people > learning to play certain kinds of language games, as a kind of > skill. He expressed great misgivings about people adopting such an > instrumental attitude toward language, mathematics, or just about > anything. On the other hand, if it really is supposed to be a matter > of carrying a thought from one mind to another, isn't it somewhat > remarkable that we feel such confidence that the thought that > arrives at the other end is really the same as the one that set > forth? I'm not at all as pessimistic, but I still find language > often sliding toward either a successful but robot-like technique > that succeeds in conveying basically data, or into poetry that is > easily lost in translation. Personally, I find that the mathematical constructions that I carry out in my poor addled brain are much more dubious than those that I express in writing. If I understand Brouwer (likely not), it is the former that yields greater certainty and the latter is just an approximation of the former. But I'm not really sure what counts as a mental construction versus an attempt to communicate the same. Jesse F. Hughes Leaving things always seems to fix me, Running seems to ease my worried mind. -- Bad Livers, Honey, I've Found a Brand New Way === Subject: Re: The role of infinity in math by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1G39gB05442; The empty set and the set of all sets Let's start with a set theoretical universe where the empty set is all of the sets. Then, there is the set containing the empty set, and the set of the empty set and the set containing the empty set: any combination is a set. Then there is the set containing those and their subsets in enumeration, there plainly are the sets containing each. A simple rule assigns each ordinal a set representation. The set of all of those sets is a set. The set of each of those sets is a set. Reiterate for each as above as if it were the empty set. Many sets are generated. This doesn't yet have any non-regular sets, as any of these sets. There are infinitely many of the sets. Each subset of each set is a member of a set. None of them is the set of all sets. Which is a mint julep? By simply constructing from the empty set there is no necessary axiomatization of regularity, infinity, or powerset. What other axioms are unnecessary? How low can you go? Why yes, two plus two does equal four. Ross F. === Subject: Re: e is transcendental > Since e^[iPi]=cosPi+isinPi > or , e^[iPi]=-1+i[0] > then there are two solutions here, to the given equatio: > A) e^[ipi]=-1 the real part solution and > B) e^[ipi]=i[0] , or e^[ipi]=0 the imaginary part solutio. >No, the conclusion from e^[iPi]=-1+i[0] is >Re(e^[iPi]) = Re(-1+i[0]) = -1 AND >Im(e^[iPi]) = Im(-1+i[0]) = 0 But he won't accept your word for it. Stefanides firmly believes that he is right. David McAnally At the moment, they (the Time Lords) are far from being all-powerful. That's why it's been left up to me and me and me. quote by: Patrick Troughton in The Three Doctors ------- === Subject: Re: e is transcendental > But he won't accept your word for it. Stefanides firmly believes that he > is right. http://b5.sdvc.uwyo.edu/bab5/snds/foolthng.wav Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W Supersedes: X-Last-Updated: 1999/08/06 === Subject: Invariant Galilean Transformations (FAQ) On All Laws Summary: All laws/equations are Galilean invariant when expressed in the generalized cartesian coordinates demanded by basic analytic geometry, vector algebra, and measurement theory. Originator: faqserv@penguin-lust.MIT.EDU Disclaimer: approval for *.answers is based on form, not content. Opponents of the content should first actually find out what it is, then think, then request/submit-to arbitration by the appropriate neutral mathematics authorities. Flaming the hard- working, selfless, *.answers moderators evidences ignorance and despicable netiquette. Archive-Name: physics-faq/criticism/galilean-invariance Version: 0.04.03 Posting-frequency: 15 days Invariant Galilean Transformations (FAQ) On All Laws (c) Eleaticus/Oren C. Webster Thnktank@concentric.net An obvious typo or two corrected. The Brittanica section revised to less 'pussy-footing' and to more directly anticipate the elementary measurement theory and basic analytic geometry that is applied to the transformation concept. ------------------------------ === Subject: 1. Purpose The purpose of this document is to provide the student of Physics, especially Relativity and Electromagnetism, the most basic princ- iples and logic with which to evaluate the historic justification of Relativity Theory as a necessary alternative to the classical physics of Newton and Galileo. We will prove that all laws are invariant under the Galilean transformation, rather than some being non-invariant, after we show you what that means. We shall also show that another primal requirement that SR exist is nonsense: Michelson-Morley and Kennedy-Thorndike do indeed fit Galilean (c+v) physics. ------------------------------ === Subject: 2. Table of Contents 1. Foreword and Intent 2. Table of Contents 3. The Principle of Relativity 4. The Encyclopedia Brittanica Incompetency. 5. Transformations on Generalized Coordinate Laws 6. The data scale degradation absurdity. 7. The Crackpots' Version of the Transforms. 8. What does sci.math have to say about x0'=x0-vt? 9. But Doesn't x.c'=x.c? 10. But Isn't (x'-x.c')=(x-x.c) Actually Two Transformations? 11. But Doesn't (x'-x.c+vt) Prove The Transformation Time Dependent? 12. But Isn't (x'-x.c')=(x-x.c) a Tautology? 13. But Isn't (x'-x.c')=(x-x.c) Almost the Definition of a Linear Transform? 14. But The Transform Won't Work On Time Dependent Equations? 15. But The Transform Won't Work On Wave Equations? 16. But Maxwell's Equations Aren't Galilean Invariant? 17. First and Second Derivative differential equations. ------------------------------ === Subject: 3. The Principle of Relativity and Transformation If a law is different over there than it is here, it is not one law, but at least two, and leaves us in doubt about any third location. This is the Principle of Relativity: a natural law must be the same relative to any location at which a given event may be perceived or measured, and whether or not the observer is moving. The idea of location translates to a coordinate system, largely because any object in motion could be considered as having a coordinate system origin moving with it. If you perceive me moving relative to you - who have your own coordinate system - will your measurements of my position and velocity fit the same laws my own, different measurements fit? If a law has the same form in both cases it is called covariant. If it is identical in form, var- ables, and output values, it is called invariant. What we're asking is that if the x-coordinate, x, on one coordinate axis works in an equation, does the coordinate, x', on some other, parallel axis work? Speaking in terms of the axis on which x is the coordinate, x' is the 'transformed' coordinate. The situation is complicated because we're talking about coordinates - locations - but in most mean- ingful laws/equations, it is lengths/distances (and time intervals) the equations are about, and x coord- inates that represent good, ratio scale measures of distances are only interval scale measures on the x' axis. [See Table of Contents for discussion of scales.] So, if we have an x-coordinate in one system, then we can call the x' value that corresponds to the same point/location the transform of x. In particular, the Principle of Relativity is embodied in the form of the Galilean transformation, which relates the original x, y, z, t to x', y', z', t' by the transform equations x'=x-vt, y'=y, z'=z, t'=t in the simplified case where attention is focused only on transforming the x-axis, and not y and z. In the case of Special Relativity, the x' transform is the same except that x' is then divided by sqrt(1-(v/c)^2), and t'=(t-xv/cc)/sqrt(1-(v/c)^2). In either case, v is the relative velocity of the coordinate systems; if there is already a v in the equations being trans- formed use u or some other variable name. ------------------------------ === Subject: 4. The Encyclopedia Brittanica Incompetency. One example of the traditional fallacious idea that an equation is not invariant under the galilean transformation comes from the Encyclopedia Brittanica: Before Einstein's special theory of relativity was published in 1905, it was usually assumed that the time coordinates measured in all inertial frames were identical and equal to an 'absolute time'. Thus, t = t'. (97) The position coordinates x and x' were then assumed to be related by x' = x - vt. (98) The two formulas (97) and (98) are called a Galilean transformation. The laws of nonrelativ- istic mechanics take the same form in all frames related by Galilean transformations. This is the restricted, or Galilean, principle of relativity. The position of a light wave front speeding from the origin at time zero should satisfy x^2 - (ct)^2 = 0 (99) in the frame (t,x) and (x')^2 - (ct')^2 = 0 (100) in the frame (t',x'). Formula (100) does not transform into formula (99) using the transform- ations (97) and (98), however. ................................................. Besides the trivially correct statement of what the Galilean 'transform' equations are, there is exactly one thing they got right. I. Eq-100 is indeed the correct basis for discussing the question of invariance, given that eq-99 is the correct 'stationary' (observer S) equation. [Let observer M be the 'moving'system observer.] In particular, eq-100 is of exactly the same form [the square of argument one minus the square of argument two equals zero (argument three).] II. It is nonsense to say eq-99 should be derivable from eq-100; for one thing, the transforms are TO x' and t' from x and t, not the other way around, and the idea that either observer's equation should contain within itself the terms to simplify or rearrange to get to the other is ridiculous. As the transform equations say, the relationship of t', x' to t, x is based on the relative velocity between the two systems, but neither the original (eq-99) equation nor the M observer equation is about a relationship between coordinate systems or observers. One might as well expect the two equations to contain banana export/import data; there is no relevancy. The 'transform' equations are the relationships between x' and x, t' and t and have nothing to do with what one equation or the other ought to 'say'. The equations' content is the rate at which light emitted along the x-axes moves. III. Most remarkable, the True Believer SR crackpots who most despise the consequences of measurement theory (demonstrable fact) contained in this document are those who want to argue against our saying the Britt- anica got eq-100 right; They insist that the correct equation is derived directly from x'=x-vt and t'=t. Solve for x=x'+vt and replace t with t', then substitute the result in eq-99: (x'+vt')^2 - (ct')^2 = 0. Besides the fact that this results in an equation with arguments exactly equal to eq-99, they will insist the transform is not invariant. IV. A major justification they have for their idea of the correct M system equation on which to base the the discussion of invariance, is that the variables are M system variables, never mind the fact that the arguments are S system values. That argument of theirs is arrant nonsense. The velocity v that S sees for the M system relative to herself is the negative of what the M system sees for the S system relative to himself. In other words, x'+vt' is a mixed frame expression and it is x'+(-v)t' that would be strictly M frame notation, and that equation is far off base. [Work it out for yourself, but make sure you try out an S frame negative v so as not to mislead yourself.] V. In I. we said: given that eq-99 is the correct 'stationary' equation. Let's look at it closely: x^2 - (ct)^2 = 0 (99) This whole matter is supposed to be about coordinate transforms. Is that what t is, just a coordinate? No. It isn't, in general. Suppose you and I are both modelling the same light event and you are using EST and I'm using PST. 'Just a time coordinate' is just a clock reading amd your t clock reading says the light has been moving three hours longer than my clock reading says. Well, that's what the idea that t is a coordinate means. Eq-99 works if and only if t is a time interval, and in particular the elapsed time since the light was emitted. Thus, that equation works only if we understand just what t is, an elapsed time, with emissioon at t=0. However, we don't have to 'understand' anything if we use a more intelligent and insightful form of the equation: (x)^2 - [ c(t-t.e) ]^2 = 0, where t.e is anyone's clock reading at the time of light emission, and t is any subsequent time on the same clock. Similarly, x is not just a coordinate, but a distance since emission. (x-x.e)^2 - [ c(t-t.e) ]^2 = 0 (99a) VI. In the spirit of 'there is exactly one thing they got right', the correct M system version of eq-99a is eq-100a: (x'-x.e')^2 - [ c(t'-t.e') ]^2 = 0 (100a) Every observer in the universe can derive their eq-100a from eq-99a and vice versa, not to mention to and from every other observer's eq-99a. Now, THAT's invariance. [You do realize that every eq-100a reduces to eq-99a, when you back substitute from the transforms, right? t.e'=t.e, x.e'=x.e-vt.] ------------------------------ === Subject: 5. Transformations on Generalized Coordinate Laws The traditional Gallilean transform is correct: t' = t x' = x - vt. But remember this: a transform of x doesn't effect just some values of x, but all of them, whether they are in the formula or not. This is important if you want to do things right. The crackpot position is strongly against this sci.math verified position, and the apparently standard coordinate pseudo-transformation they suggest is perhaps the result. {See Table of Contents.] Let's use a simple equation: x^2 + y^2 = r^2, which is the formula for a circle with radius r, centered at a location where x=0. But what if the circle center isn't at x=0? Well, we'd want to use the form analytic geometry, vector algebra, and elementary measurement theory tells us to use, a form where we make explicit just where the circle center is, even if it is at x=x0=0: (x-x0)^2 + (y-y0)^2 = r^2. The circle center coordinate, x0, is an x-axis coordinate, just like all the x-values of points on the circle. So, in proper generalized cartesian coordinate forms of laws/equations we want to transform every occurence of x and x0 - by whatever name we call it: x.c, x_e, whatever. So, what is the transformed version of (x-x0)? Why, (x'-x0'); both x and x0 are x-coordinates, and every x-coordinate has a new value on the new axis. So, what is the value of (x'-x0') in terms of the original x data? is also true for x0'=x0-vt: (x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0). In other words, when we use the generalized coordinate form specified by analytic geometry, we find that the value of (x'-x0') does not depend on either time or velocity in any way, shape, form, or fashion. Similarly for (y-y0). We can treat time the same way if necessary: (t-t0). The above is a proof that any equation in x,y,z,t is invariant under the galilean transforms. Just use the generalized coordinate form, with (x-x0)/etc, in the transformation process, not the incompetently selected privileged form, with just x/etc. [The form is privileged because it assumes the circle center, point of emission, whatever, is at the origin of the axes instead at some less convenient point. After transform the coordinate(s) of the circle center/origin are also changed but the privileged form doesn't make this explicit and screws up the calculations, which should be based on (x'-x0') but are calculated as (x'-0).] The value of (x'-x0') is the same as (x-x0). That makes sense. Draw a circle on a piece of paper, maybe to the right side of the paper. On a transparent sheet, draw x and y coordinate axes, plus x to the right, plus y at the top. Place this axis sheet so the y-axis is at the left side of the circle sheet. Now answer two questions after noting the x-coordinate of the circle center and then moving the axis sheet to the right: (a) did the circle change in any way because you moved the axis sheet (ie because you transformed the coordin- nate axis)? (b) did the coordinate of the circle center change? The circle didn't change [although SR will say it did]; that means that (x'-x0') does indeed equal (x-x0). The coordinate of the circle center did change, and it changed at the same rate (-vt) as did every point on the circle. That means that x0'<>x0, and the fact the circle center didn't change wrt the circle, means that the relationship of x0' with x0 is the same as that of any x' on the circle with the corresponding x: x'=x-vt; x0'=x0-vt. This is to prepare you for the True Believer crackpots that say 'constant' coordinates can't be transformed; some even say they aren't coordinates. These crackpots include some that brag about how they were childhood geniuses, btw. QED: The galilean transformation for any law on generalized Cartesian coordinates is invariant under the Galilean transform. The use of the privileged form explains HOW the transformed equation can be messed up, the next Subject explains what the screwed up effect of the transform is, and how use of the generalized form corrects the screwup. ------------------------------ === Subject: 6. The data scale degradation absurdity. The SR transforms and the Galilean transforms both convert good, ratio scale data to inferior interval scale data. The effect is corrected, allowed for, when the transforms are conducted on the generalized coordinate forms specified by analytic geometry and vector algebra. Both sets of transforms are 'translations' - lateral movements of an axis, increasing over time in these cases - but with the SR transform also involving a rescaling. It is the translation term, -vt in the x transform to x', and -xv/cc in the t transform to t', that degrades the ratio scale data to interval scale data. In general, rescaling does not effect scale quality in the size-of-units sense we have here. SR likes to consider its transforms just rotations, however - in spite of the fact Einstein correctly said they were 'translations' (movements) - and in the case of 'good' rotations, ratio scale data quality is indeed preserved, but SR violates the conditions of good ro- tations; they are not rigid rotations and they don't appropriately rescale all the axes that must be rescaled to preserve compatibility. The proof is in the pudding, and the pudding is the combination of simple tests of the transformations. We can tell if the transformed data are ratio scale or interval. Ratio scale data are like absolute Kelvin. A measure- ment of zero means there is zero quantity of the stuff being measured. Ratio scale data support add- ition, subtraction, multiplication, and division. The test of a ratio scale is that if one measure looks like twice as much as another, the stuff being measured is actually twice as much. With absolute Kelvin, 100 degrees really is twice the heat as 50 degrees. 200 degrees really is twice as much as 100. Interval scale data are like relative Celsius, which is why your science teacher wouldn't let you use it in gas law problems. There is only one mathematical operation interval scales support, and that has to be between two measures on the same scale: subtraction. 100 degrees relative (household) Celsius is not twice as much as 50; we have to convert the data to absolute Kelvin to tell us what the real ratio of temperatures is. However, whether we use absolute Kelvin or relative Celsius, the difference in the two temperature readings is the same: 50 degrees. Thus, if we know the real quantities of the 'stuff' being measured, we can tell if two measures are on a ratio scale by seeing if the ratio of the two measures is the same as the ratio of the known quant- ities. If a scale passes the ratio test, the interval scale test is automatically a pass. If the scale fails the ratio test, the interval scale test becomes the next in line. It isn't just the bare differences on an interval scale that provides the test, however. Differences in two interval scale measures are ratio scale, so it is ratios of two differences that tell the tale. Let's do some testing, and remember as we do that our concern is for whether or not the data are messed up, not with 'reasons', excuses, or avoidance. ------------------------------------------------------ Are we going to take a transformed length (difference) and see whether that length fits ratio or interval scale definitions? Of course, not. Interval scale data are ratio after one measure is subtracted from another. That is the major reason the SR transforms can be used in science. Let there be three rods, A, B, C, of length 10, 20, 40, respectively. These lengths are on a known ratio scale, our original x-axis, with one end of each rod at the origin, where x=0, and the other end at the coordinate that tells us the correct lengths. Note that these x-values are ratio scale only because one end of each rod is at x=0. That may remind you of the correct way to use a ruler or yard/meter-stick: put the zero end at one end of the thing you are measuring. Put the 1.00 mark there instead of the zero, and you have interval scale measures. Let A,B,C, be 10, 20, 40. Let a,b,c be x' at v=.5, t=10. x'=x-vt. A B C a b c ---------------- -------------------- 10 20 40 5 15 35 ---------------- -------------------- B/A = 2 b/a = 3 C/A = 4 c/a = 7 C/B = 2 c/b = 2.333 Obviously, the transformed values are no longer ratio scale. The effect is less on the greater values. C-A = 10 b-a = 10 C-A = 30 c-a = 30 C-B = 20 c-b = 20 Obviously, the transformed values are now interval scale. This will hold true for any value of time or velocity. (C-A)/(B-A) = 3 (c-a)/(b-a) = 3 (C-B)/(B-A) = 2 (c-b)/(b-a) = 2 Obviously, the ratios of the differences are ratio scale, being identical to the ratios of the corresponding original - ratio scale - differences. The main difference between these results and the SR results is that the differences do not correspond so neatly to the original, ratio scale, differences. This is due only to the rescaling by 1/sqrt(1-(v/c)^2). The ratios of the differences on the transformed values do correspond neatly and exactly to the ratio scale results. Using the generalized coordinate form, such as (x-x0), the transform produces an interval scale x' and an interval scale x0'. That gives us a ratio scale (x'-x0'), just like - and equal to - (x-x0). ------------------------------ === Subject: 7. The Crackpots' Version of the Transforms. It has become apparent - whether misleading or not - that the crackpot responses to the obvious derive from a common source, whether it be bandwagoning or their SR instructors. Below, in the sci.math subject, we see that all sci.math respondents agree with the basic controversial position of this faq: every coordinate is transformed, whether a supposed constant or not. Think about it, the generalized coordinate of a circle center, x0, applies to infinities upon infinities of circle locations (given y and z, too); it is a constant only for a given circle, and even then only on a given coordinate axis. And even variables are often held 'constant' during either integration or differentiation. The utility of a variable is that you can discuss all possible particular values without having to single out just one. That utility does not make particular - singled out - values on the variable's axis not values of the variable just because they have become named values. In any case, all that is preamble to the incompetent idea they have proposed for a transform of coordinates. It is based on the idea that the circle center, point of emission, whatever, has coordinates that cannot be transformed. Let there be an equation, say (x)^2 - (ict)^2 = 0. What is the transformed version of that equation? Answer: (x')^2 - (ict')^2 = 0. That's the one thing the Brittanica got right. Note that the leading crackpot just criticized this faq for presuming to correct the Britt- anica, but it then and before poses the incompetent pseudo- transform we analyze here in this section. x to x' and t to t' are obviously coordinate transforms; the x and t coordinates have been replaced by the coord- inates in the primed system. A tranform of an equation from one coordinate system to another is NOT a substitution of the/a definition of x for itself; that is not a coordinate transformation. The most that can said for such a substitution is that it is a change of variable. But the crackpots are calling this a coordinate trans- form of the original equation: (x'+vt)^2 - (ict')^2 = 0. It is not a coordinate transform, of course, except accidentally. (x'+vt) is not the primed system coordinate, it is another form/expression of x. They get that substitution by solving x'=x-vt for x; x=x'+vt. So, by incompetent misnomer, they accomplish what they have been railing against all along. It has been the generalized coordinate form in question all this time: (x-x0)^2 - (ict)^2 = 0. Here they substitute for x instead of transforming to the primed frame: (x'+vt-x0)^2 - (ict')^2. ----- ^ | ^ | It is still x ^ but see what they have accomplished by their mis/malfeasance: [x'+vt-x0]=[x'+(vt-x0)]=[x'-(x0-vt)]. =[x'-x0'] The crackpots have been bragging about how you don't have to transform the circle center's coordinate to transform the circle center's coordinate. Bragging that what they were doing was not what they said they were doing. This does give us insight as to some of the crackpot variations on their x0'<>x0-vt theme, which in all the variations will be discussed in later sections.. They are used to seeing the mixed coordinate form, (x'+vt-x0) without realizing what it respresented, so - accompanied with a lack of understanding of the term 'dependent' - they are used to seeing just the one vt term, and not the one hidden in the defi- nition of x' and are used to imagining it makes the whole expression time dependent and thus not invariant. About which, let x=10, let, x0=20, v=10, and t variously 10 and 23: (x-x0)=-10. Using their (x'+vt-x0): For t=10, we have (x'+vt-x0) = [ (10-10*10) + (10*10) - (20) ] = -90 + 100 - 20 = -10 = (x-x0) For t=23, we have (x'+vt-x0) = [ (10-10*23) + (10*23) - (20) ] = -220 + 230 - 20 = -10 = (x-x0) The result depends in no way on the value of time; we showed the obvious for a couple of instances of t just so you can see that the crackpots not only do not understand the obvious logic of the algebra { (x'-x0')=[ (-vt)-(x0-vt) ]=(x-x0) } - which shows that the transform has no possible time term effect - but they don't understand even a simple arithmetic demonstration of the facts. Oh. Their (x'+vt-x0) or (x'+vt'-x0) reduces the same way since t'=t: (x-vt+vt-x0)=(x-x0). Their process, which says (x'+vt') is the transform of x, says that (x'+vt') is the moving system location of x, but it can't be because x is moving further in the negative direction from the moving viewpoint. That formula will only work out with v<0 which is indeed the velocity the primed system sees the other moving at. However, that formula cannot be derived from x'=x-vt, the formula for transformation of the coordinates from the unprimed to the primed, ------------------------------ === Subject: 8. What does sci.math have to say about x0'=x0-vt? The crackpots' positions/arguments were put to sci.math in such a way that at least two or three who posted re- sponses thought it was your faq-er who was on the idiot's side of the questions. Their responses: ---------------------------------------------------------- I. x0' = x0. In other words: x0' <> x0-vt, or constant values on the x-axis are not subject to the transform. No. x0' = x0 - vt. Well, if you want, you could define constant values on the x-axis, but in the context of the question that is not relevant. The relevant fact is that if the unprimed observer holds an object at point x0, then the primed observer assigns to that object a coordinate x0' which is numerically related to x0 by x0'= x0 -vt. What does this mean? The line x=x0 will give x'=x-v*t=x0-vt', so if x0' is to give the coordinate in the (x',t',)-system, it will be given by x0'=x0-v*t': ie., it is not given by a constant. Thus, being at rest (constant x-coordinate) is a coordinate-dependent concept. Sounds very false. We can say that the representation of the point X0 is the number x0 in the unprimed system, and x0' in the primed system. Clearly x0 and x0' are different, if vt is not zero. However one may say that (though it sounds/is stupid) the point X0 itself is the same throughout the transformation. However that expression sounds meaningless, since a transform (ok, maybe we should call it a change of basis) is only a function that takes the point's representation in one system into the same point's representation in another system. It is preferrable to use three notations: X0 for the point itself and x0 and x0' for the points' representations in some coordinate systems. ------------------------------ === Subject: 9. But Doesn't x.c'=x.c? That idea is one of the most idiotic to come up, and it does so frequently. And in a number of guises. The idea being that x.c' <> x.c-vt, with x.c being what we have called x0 above; the notation makes no difference. Some crackpots have managed to maintain that position even after graphs have illustrated that such an idea means that after a while a circle center represented by x.c' could be outside the circle. The leading crackpot just make that explicit, as far as one can tell from his befuddled post in response to a line about active transforms, which are actually moving body situations, not coordinate transformations: -------------------------------------------------------------- ------ e>An active transform is not a coordinate transform, ... Right, it is a transform of the center (in the opposite direction) done to effect the change of coordinates without a coordinate transform. ... E: Transform of the center? Center of a circle? He really is saying a circle center moves in the opposite direction of the circle! Right? -------------------------------------------------------------- ------ If r=10 and x.c was at x.c=0, then the points on the circle (10,0), (-10,0), (0,10) and (0,-10) could at some time become (-10,0), (-30,0), (-20,10), and (-20,-10), but with x.c'=x.c, the circle center would be at (0,0) still! The circle is here but its center is way, way over there! Indeed, although a change of coordinate systems is not movement of any object described in the coordinates, the x.c'=x.c crackpottery is tantamount to the circle staying put but the center moving away. Or vice versa. ------------------------------ === Subject: 10. But Isn't (x'-x.c')=(x-x.c) Actually Two Transformations? One crackpot puts the (x'-x.c')=(x-vt - x.c+vt) relationship like this: (x-vt+vt - x.c). See, he says, that is transforming x (with x-vt - x.c) and then reversing the transform (x-vt+vt - x.c). That's just another crackpot form of the idiocy that x.c' <> x.c-vt. You'll have noticed the implication is that there is no transform vt term relating to x.c. ------------------------------ === Subject: 11. But Doesn't (x'-x.c+vt) Prove The Transformation Time Dependent? That particular crackpottery is perhaps more corrupt than moronic, since it includes deliberately hiding a vt term from view, and pretending it isn't there. [However, we have seen above that it is a familiar incompetency, and not likely an original.] Look, the crackpots say, there is a time term in the transformed (x' - x.c+vt). The transform isn't invariant! It's time dependent! Just put x' in its original axis form, also, which reveals the other time term, the one they hide: (x'-x.c+vt) = (x-vt - x.c+vt) = (x-x.c). So, at any and all times, the transform reduces to the original expression, with no time term on which to be dependent. Then there is the fact that if you leave the equation in any of the various notation forms - with or without reducing them algebraicly - the arithmetic always comes down to the same as (x-x.c). That means nothing to crack- pots, but may mean something to you. ------------------------------ === Subject: 12. But Isn't (x'-x.c')=(x-x.c) a Tautology? My dictionary relates 'tautology' to needless repetition. That's another form of the x.c' <> x.c-vt idiocy. The repetition involved is the vt transformation term. Apply the -vt term to the x term, and it is needless repetition to apply it anywhere again? The 'again' is to the x.c term. The x.c' = x.c crackpot idiocy. The repetition of the vt terms is required by the presence of two x values to be transformed. Be sure to note the next section. ------------------------------ === Subject: 13. But Isn't (x'-x.c')=(x-x.c) Almost the Definition of a Linear Transform? Now, how on earth can we relate a tautology to a basic definition in math? we get this definition: -------------------------------------------------------------- A linear transformation, A, on the space is a method of corr- esponding to each vector of the space another vector of the space such that for any vectors U and V, and any scalars a and b, A(aU+bV) = aAU + bAV. ------------------------------------------------------------- Let points on the sphere satisfy the vector X={x,y,z,1}, and the circle center satisfy C={x.c,y.c,z.c,1}. Let a=1, and b=-1. Let A= ( 1 0 0 -ut ) ( 0 1 0 -vt ) ( 0 0 1 -wt ) ( 0 0 0 1 ) A(aX+bC) = aAX + bAC. aX+bC = (x-x.c, y-y.c, z-z.c, 0 ). The left hand side: A( x - x.c , y - y.c, z - z.c, 0 ) = ( x-x.c , y-y.c, z-z.c, 0 ). The right hand side: aAX= ( x-ut, y-vt, z-wt, 1 ). bAC= (-x.c+ut, -y.c+vt, -z.c+wt, -1 ). and aAX+bAC = ( x-x.c, y-y.c, z-z.c, 0 ). Need it be said? Sure: QED. On the galilean transform the definition of a linear transform, A(aU+bV)=aAU + bAV, is completely satisfied. The generalized form transforms exactly and non-redundantly - with ONE TRANSFORM, not a transform and reverse transform - and non- tautologically, just as the very definition of a linear transform says it should. And does so with absolute invariance, with this galilean transformation. ------------------------------ === Subject: 14. But The Transform Won't Work On Time Dependent Equations? The main crackpot that has asserted such a thing was referring to equations such as in Subject 4, above. The Light Sphere equation; for which we have shown repeatedly elsewhere that the numerical calculations are identical for any primed values as for the unprimed values. The presence - before transformation - of a velocity term seems to confuse the crackpots. It turns out there is ex- treme historical reason for this, as you will see in the subject on Maxwell's equations. ------------------------------ === Subject: 15. But The Transform Won't Work On Wave Equations? See Subject 17, below, for a discussion of Second Derivative forms and the galilean transforms. ------------------------------ === Subject: 16. But Maxwell's Equations Aren't Galilean Invariant? Oh? Just what is the magical term in them that prevents (x'-x.c')=(x-vt - x.c+vt)=(x-x.c) from holding true? It turns out not to be magic, but reality, that interferes with the application of the galilean transforms to the gen- eralized coordinate form(s) of Maxwell: there are no coordi- nates to transform! When True Believer crackpots are shown the simple demonstration that the galilean transform on generalized cartesian coordinates is invariant, their first defense is usually an incredibly stupid x0'=x0, because the coordinate of a circle center, or point of emission, etc, is a constant and can't be transformed. The last defense is but Maxwell's equations are not invariant under that coordinate transform. When asked just what magic occurs in Maxwell that would prevent the simple algebra (x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0) from working, and when asked them for a demonstration, they will never do so, however many hundreds of times their defense is asserted. The reason may help you understand part of Einstein's 1905 paper in which he gave us his absurd Special Relativity derivation: THERE ARE NO COORDINATES IN THE EQUATIONS TO BE TRANSFORMED. Einstein gave the electric force vector as E=(X,Y,Z) and the magnetic force vector as B=(L,M,N), where the force components in the direction of the x axis are X and L, Y and M are in the y direction, Z and N in the z direction. Those values are not, however, coordinates, but values very much like acceleration values. BTW, the current fad is that E and B are 'fields', having been 'force fields' for a while, after being 'forces'. So, when Einstein says he is applying his coordinate transforms to the Maxwell form he presented, he is either delusive or lying. (a) there are no coordinates in the transform equations he gives us for the Maxwell transforms, where B=beta=1/sqrt(1-(v/c)^2): X'=X. L'=L. Y'=B(Y-(v/c)N). M'=B(M+(v/c)Z). Z'=B(Z+(v/c)M). N'=B(N-(v/c)Y). X is in the same direction as x, but is not a coordinate. Ditto for L. They are not locations, coordinates on the x-axis, but force magnitudes in that direction. Similarly for Y and M and y, Z and N and z. (b) the v of the coordinate transforms is in Maxwell before any transform is imposed; Einstein's transform v is the velocity of a coordinate axis, not the velocity he touched it. (c) if they were honest Einsteinian transforms, they'd be x, which means it is X and L that are supposed to be transformed, not Y and M, and Z and N. And when SR does transform more than one axis, each axis has its own velocity term; using the v along the x-axis as the v for a y-axis and z-axis transform is thus trebly absurd: the axes perpendicular to the motion are not changed according to SR, the v used is not their v, and the v is not a transform velocity anyway. (d) as everyone knows, the effect of E and B are on the direction. Both the speed and direction are changed by E and B, but v - the speed - is a constant in SR. As absurd as are the previously demonstrated Einsteinian blunders, this one transcends error and is an incredible example of True Believer delusion propagating over decades. The components of E and B do differ from point to point, and in the variations that are not coordinate free, they are subject to the usual invariant galilean trans- formation when put in the generalized coordinate form. ------------------------------------------------------------- The SR crackpots don't know what coordinates are. The various things they call coordinates include coordin- nates, but also include a variety of other quantities. ------------------------------------------------------ 1. One may express coordinates in a one-axis-at-a-time manner [like x^2+y^2=r^2] but it is the use of vector notation that shows us what is going on. In vector notation the triplet x,y,z [or x1,x2,x3, whatever] represents the three spatial coordinates, but there are so-called basis vectors that underlie them. Those may be called i,j,k. Thus, what we normally treat as x,y,z is a set of three numbers TIMES a basis vector each. 2. These e*i, f*j, g*k products can have a lot of meanings. If e, f, j are distances from the origin of i,j,k then e*i, f*j, g*k are coordinates: distances in the directions of i,j,k respectively, from their origin. That makes the triplet a coordinate vector that we describe as being an x,y,z triplet; perhaps X=(x,y,z). The e*i, f*j, g*k products could be directions; take any of the other vectors described above or below and divide the e,f,g numbers by the length of the vector [sqrt(e^2+f^2+g^2)]. That gives us a vector of length=1.0, the e,f,g values of which show us the direction of the original vector. That makes the triplet a direction vector that we describe as being an x,y,z triplet; perhaps D=(x,y,z). The e*i, f*j, g*k products could be velocities; take any of the unit direction vectors described above and multiply by a given speed, perhaps v. That gives a vector of length v in the direction specified. That makes the triplet a velocity vector that we describe as being an x,y,z triplet; perhaps V=(x,y,z). Each of the three values, e,f,g, is the velocity in the direction of i,j,k respectively. The e*i, f*j, g*k products could be accelerations; take any of the unit direction vectors described above and multiply by a given acceleration, perhaps a. That gives a vector of length a in the direction specified. That makes the triplet an acceleration vector that we describe as being an x,y,z triplet; perhaps A=(x,y,z). Each of the three values, e,f,g, is the acceleration in the direction of i,j,k respectively. The e*i, f*j, g*k products could be forces (much like accel- erations); take any of the unit direction vectors described above and multiply by a given force, perhaps E or B. That gives a vector of length E or B in the direction specified. That makes the triplet a force vector that we describe as being an x,y,z triplet; perhaps E=(x,y,z) or B=(x,y,z). Each of the three values, e,f,g, is the force in the direction of i,j,k respectively. Einstein's - and Maxwell's - E and B are not coordinate vectors. There is another variety of intellectual befuddlement that misinforms the idea that Maxwell isn't invariant under the galilean transform: confusions about velocities. Velocities With Respect to Coordinate Systems. ----------------------------------------------- Aaron Bergman supplied the background in a post to a sci.physics.* newsgroup: Imagine two wires next to each other with a current I in each. Now, according to simple E&M, each current generates a magnetic field and this causes either a repulsion or attraction between the wires due to the interaction of the magnetic field and the current. Let's just use the case where the currents are parallel. Now, suppose you are running at the speed of the current between the wires. If you simply use a galilean transform, each wire, having an equal number of protons and electrons is neutral. So, in this frame, there is no force between the wires. But this is a contradiction. First of all, the invariance of the galilean transform (x'-x.c') =(x-x.c), insures that it is an error to imagine there is any difference between the data and law in one frame and in another; the usual, convenient rest frame is the best frame and only frame required for universal analysis. [Well, (x'<>x, x,c'<>x.c, but (x'-x.c')=(x-x.c).] Second, given that you decide unnecessarily to adapt a law to a moving frame, don't confuse coordinate systems with meaningful physical objects, like the velocity relative to a coordinate system instead of relative to a physical body or field. In other words, what does current velocity with respect to a coordinate system have to do with physics? Nothing. Certainly not anything in the example Bergman gave. What is relevant is not current velocity with respect to a coordinate system, but current velocity with respect to wires and/or a medium. The velocity of an imaginary coordinate sys- tem has absolutely nothing to do with meaningful physical vel- ocity. You can - if you are insightful enough and don't violate item (e) - identify a coordinate system and a relevant physical object, but where some v term in the pre-transformed law is in use, don't confuse it with the velocity of the coordinate transform. Velocities With Respect to ... What? ----------------------------------------------- Albert Einstein opened his 1905 paper on Special Relativity with this ancient incompetency: The equations of the day had a velocity term that was taken as meaning that moving a magnet near a conductor would create a current in the conductor, but moving a conductor near a wire would not. This was belied by fact, of course. The important velocity quantity is the velocity of the magnet and conductor with respect to each other, not to some absolute coordinate frame (as far as we know) and not to an arbitrary coordinate system. One possible cause was the idea: but the equation says the magnet must be moving wrt the coordinate system or ... the absolute rest frame. There not being anything in the equation(s) to say either of those, it is amazing that folk will still insist the velocity term has nothing to do with velocity of the two bodies wrt each other. ----------------------------------------------------------- ------------------------------ === Subject: 17. First and Second Derivative differential equations. One of the intellectually corrupt ways of denying the very simple demonstration of galilean invariance of all laws expressed in the generalized coordinate form demanded by analytic geometry, vector analysis, and measurement theory [ (x'-x.c')=[ (x-vt)-(x.c-vt) ]=(x-x.c) ] is the assertion that those equations 'over there' (usually Maxwell or wave) are somehow immune to the elementary laws of algebra used to demon- strate the invariance. [Unfortunately, the assertions are never accompanied by reference to the magical math that makes elementary al- gebra invalid. Wonder why that is?] Part of the time it is based on the old lore based on the incompetent transformation of the privileged form of an equation instead of the correct form. [Evidence of this is any reference to an effect due to the velocity of the transform; it falls out algebraicly - as you see above - and cancels out arith- metically - as you can see above.] But usually it is just whistling in the dark, waving the cross (zwastika, I'd say) at the mean old vampire. The most general equation that could be conjured up is a differential with either First or Second Derivatives. Let's examine the plausibility of such magical magical, non-invariance assertions. (a) to get a Second Derivative you must have a First Derivative. (b) to get a First Derivative you must have a function to differentiate. (c) to get a Second Derivative you must have a function in the second degree. So, let us examine the question as to whether any such common Maxwell/wave equation will differ for (a) the common, privileged form, represented as ax^2, with a being an unknown constant function. (b) the generalized cartesian form, represented as a(x-x.c)^2 = ax^2 -2ax(x.c) + ax.c^2, with a being an unknown constant function. (c) the transformed generalized cartesian form, represented as a(x-vt -x.c+vt)^2, same as for (b), = ax^2 -2ax(x.c) + ax.c^2, of course, with a being an unknown constant function. I. for (a), remembering that x.c is a constant, and that this version is only correct because x.c=0, otherwise (b) is the correct form: d/dx ax^2 = 2ax (d/dx)^2 ax^2 = 2a II. for (b), remembering that x.c is a constant. d/dx (ax^2 -2ax(x.c) + ax.c^2) = 2ax - 2ax.c (d/dx)^2 (ax^2 -2ax(x.c) + ax.c^2) = 2a III. for (c); same as for (b). So, what we have seen so far is (1) differential equations in the second degree - the wave equations - must clearly be the same for all forms: the privileged form in x, the generalized cartesian form in x and the centroid, x.c, or the transformed generalized cartesian form. That is, anyone who imagines that correct usage gives different results for galilean transformed frames is at first showing his ignorance, and in the end showing his intellectual corruption. (2) As far as the First Derivatives are concerned, the only cases in which there really is a difference between the two forms is where x.c <> 0, and in that case, the use of the privileged form is obviously incompetent. So, how do you correctly use the differential equations? If you are using rest frame data with the centroid at x=0, etc, you can't go wrong without trying to go wrong. If you are using rest frame data with the centroid not at x=0, you must use (x-x.c) anyplace x appears in the equation. If you are using moving frame data, you must use the moving frame centroid as well as the light front (or whatever) moving frame data itself, perhaps first calculating (x'-x.c'), which equals (x-x.c) which is obviously correct, and which is obviously the plain old correct x of the privileged form. Unless, of course, there really is some magical term or expression that invalidates the obvious and elemen- tary algebra of the invariance demonstration. Or maybe you just whistle when you don't want basic algebra to hold true. Eleaticus !---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?- --!---? ! Eleaticus Oren C. Webster ThnkTank@concentric.net ? ! Anything and everything that requires or encourages systematic ? ! examination of premises, logic, and conclusions ? !---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?- --!---? === Subject: . The hardest of all hard facts . Hi Eleaticus, Re: How you think, Michelson-Morley and Kennedy-Thorndike do indeed fit Galilean ( c + v ) physics , All throughout the annals of history ... No premise has been better tested than this premise: The speed of light is the same for all observers. That makes it: The hardest of all hard facts. As the double-slit experiment so amply illustrates ... Nature is constructed of mass-energy ... Photons, atoms, etc. all have wave-like properties. ( e.g. They all are waves that add and cancel ) And they also have mass-like properties. ( e.g. They all have an exact inertia and position ) And they all are notionally random. ( i.e. Our information about quanta is, by it's very definition, quite incomplete ) Nothing can exceed the speed of light because of this extremely well tested axiom: The laws of physics are the same at each point in space-time, no matter the degree of zooming or precision. Your unarticulated objection is that you are repulsed by this level of zooming. === Subject: Re: . The hardest of all hard facts . > The speed of light is the same for all observers. The speed of light through vacuum is the same for all observers. However, when the speed of light is less than c (eg when passing through air), it would not be the same for all observers. The explains how the Michelson-Morley experiment gave results that were non-null, due to the effect of air in the apparatus. Process Physics http://www.ctr4process.org/publications/PSS/pps_cahill.pdf A related paper is: Gravitation, the 'Dark Matter Effect and the Fine Structure Constant http://xxx.arxiv.cornell.edu/pdf/physics/0401047 Peter === Subject: Re: . The hardest of all hard facts . > The speed of light is the same for all observers. > The speed of light through vacuum is the same for all observers. > However, when the speed of light is less than c (eg when passing > through air), it would not be the same for all observers. > The explains how the Michelson-Morley experiment gave results that > were non-null, due to the effect of air in the apparatus. The MMX has been repeated in a vacuum. No change in the result. David A. Smith === Subject: Re: . The hardest of all hard facts . Hi Peter, You flip, the Michelson-Morley experiment gave results that were non-null, due to the effect of air in the apparatus , You're obviously wrong about that. And you obviously can't explain your reasoning. === Subject: Re: . The hardest of all hard facts . > Hi Eleaticus, Re: How you think, > Michelson-Morley and Kennedy-Thorndike do indeed fit > Galilean ( c + v ) physics , > All throughout the annals of history ... > No premise has been better tested than this premise: > The speed of light is the same for all observers. > That makes it: The hardest of all hard facts. No experiment ever showed that the speed of light is the same for all observers. Indeed any determination of the one way speed of light can be used to demonstrate the speed of light is NOT the same for all observers.... A light----> B <-you < ----------- L --------------> v m/s Use syncronised clocks at A amd B to time how long it takes light to travel a distance of L meters across the laboratory.. Speed of light relative to the laboratory = L/ (tB - tA) = c where 'tA' is the time at which the light left A and 'tB' is the at which the light arrived at B Now repeat the experiment while running towards B at v m/s Note that 'in your frame of reference' the point B is moving , so that the light must travel an extra distance = v * (tB - tA) which is the distance B has moved as the light travels from A to B. Therefore: Speed of light relative to you = (L+ v * (tB - tA)) / (tB - tA) = c + v keith stein === Subject: Re: . The hardest of all hard facts . > Hi Eleaticus, Re: How you think, > Michelson-Morley and Kennedy-Thorndike do indeed fit > Galilean ( c + v ) physics , > All throughout the annals of history ... > No premise has been better tested than this premise: > The speed of light is the same for all observers. > That makes it: The hardest of all hard facts. > No experiment ever showed that the speed of light is > the same for all observers. Indeed any determination > of the one way speed of light can be used to demonstrate > the speed of light is NOT the same for all observers.... > A light----> B <-you > < ----------- L --------------> v m/s > Use syncronised clocks at A amd B to time how long it takes > light to travel a distance of L meters across the laboratory.. > Speed of light relative to the laboratory = L/ (tB - tA) = c > where 'tA' is the time at which the light left A > and 'tB' is the at which the light arrived at B > Now repeat the experiment while running towards B at v m/s > Note that 'in your frame of reference' the point B is moving , > so that the light must travel an extra distance = v * (tB - tA) > which is the distance B has moved as the light travels from > A to B. But this extra distance is compensated by the difference of your evolution through time while runnning. You go slower in time so that light may travel at the same speed for you as irtt does for the laboratory. > Therefore: > Speed of light relative to you = (L+ v * (tB - tA)) / (tB - tA) > = c + v > keith stein === Subject: Re: . The hardest of all hard facts . >Indeed any determination >of the one way speed of light can be used to demonstrate >the speed of light is NOT the same for all observers.... Have such experiments been done? Peter Supersedes: X-Last-Updated: 1999/10/17 === Subject: (SR) Lorentz t', x' = Intervals Summary: The Lorentz transforms themselves are proof t' and x' cannot possibly be just coordinates. Examination of their derivation verifies their identity as intervals. Originator: faqserv@penguin-lust.MIT.EDU Disclaimer: approval for *.answers is based on form, not content. Opponents should first actually find out what the content is, then think, then request/submit-to arbitration by the appropriate neutral mathematics authorities. Flaming the hard- working, selfless, *.answers moderators evidences ignorance and atrocious netiquette. Version: 0.02.1 Archive-name: physics-faq/criticism/lorentz-intervals Posting-frequency: 15 days (SR) Lorentz t', x' = Intervals (c) Eleaticus/Oren C. Webster Thnktank@concentric.net ------------------------------ === Subject: 1. Introduction with the obvious debunking of the use of 'just coordinates' in any scientific formula. Defenders of the Special Relativity faith are especially fond of telling opponents of their space-time fairy tales that they do not know the difference between coordinates and magnitudes. That may often be so, but the fault lies ultimately with SR dogma. The Lorentz-Einstein transformations cannot possibly be 'just coordinates', which is the interpre- tation required to support the many sideshow carnival acts with which the SR faithful bedazzle the public, and establish their moral and intellectual superiority. If I get in my car and drive steadily for a few hours at 50 kilometers per hour, is 50t the distance I travel? Of course not. The last time my hours-counting 'just coord- inates' clock was set to zero was when Zeno first reported one of his paradoxes to Parmenides. That was a long time ago, so my t is not useful for such purposes unless you also use my clock to established the starting time, perhaps t0, and use the formula 50(t-t0) to calculate the distance. In any case, my t is even then not 'just a coordinate' because it always represents particular elapsed times that can be used in the (t-t0) form to calculate perfectly good time intervals (elapsed times). Alternatively, I could (re)set my clock to zero at the start of some meaningful time interval, in which case my t shows a scientifically perfect current and/or end time. In which case, the Lorentz-Einstein t'=(t-vx/cc)/g is a function of an elapsed time interval (not 'just a coordinate') and a time interval (-vx/cc; the interval amount the t' clock is being screwed up at time t) and thus cannot be 'just a coordinate' since neither of the independent variables is such a 'just' thing. {Their meaning is shown below, step-by-step.] If it takes me 50 minutes to cross the Interstate highway, was x/50 my velocity crossing it? Of course not. The origin of all my axes is at the very spot where Zeno first presented his first paradox to Parmenides. That makes my x equal a couple of thousands of miles, plus, and is not useful for such purposes unless you establish the starting x value, perhaps x0, and use the formula (x-x0)/50 to calculate my velocity. In any case, even then my x is not 'just a coordinate' because it always repesents particular distance intervals that can always be used in the (x-x0) form for any and every scientific purose. Alternatively, I could move my x-axis origin to the starting (zero) point of some meaningful distance, in which case my x shows a scientifically perfect current and/or end distance. In which case, the Lorentz-Einstein x'=(x-vt)/g is a function of a current/ending distance interval (not 'just a coordinate') and a distance interval (-vt; the interval amount the x' axis is being screwed up at time t) and thus cannot be 'just a coordinate' since neither of the independent variables is such a 'just' thing. {Their meaning is shown below, step-by-step.] ------------------------------ === Subject: 2. Table of Contents 1. Introduction with the obvious debunking of the use of 'just coordinates' in any scientific formula. 2. Table of Contents. 3. The Lorentz-Einstein transforms. 4. The 'just coordinates' argument. 5. Single-system, little-purpose ambiguity. 6. Relating two coordinate measures/systems. 7. Distances and moving coordinate axes. 8. Time intervals. 9. Einstein's (1905) derivations. 10. A word about intervals. 11. Intervals versus the Twins Paradox. 12. Summary ------------------------------ === Subject: 3. The Lorentz-Einstein transforms Special Relativity's space-time circus is based on the 'transformation' equations by which it is believed one can relate a nominally 'stationary' system's space and time coordinates to those of an inertially (not accelerating) moving other observer. That moving observer's own physical body and coordinate system might have been identical in size to those of the stationary observer before the traveller began moving, but are 'seen' as very different by the stationary observer when the relative velocity of the two is great enough, a high percentage of the velocity of light. Concerning ourselves - as is customary - with just the spatial coordinate axis that lies parallel to the direction of motion, and with time, Einstein arrived at these formulas that relate the moving system measures or coordinates (x' and t') to the stationary system coordinates (x and t): x' = (x - vt)/sqrt(1-vv/cc) (Eq 1x) t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t) The v is for the two systems' relative velocity as seen by the stationary observer, and is positive if the dir- ection is toward higher values of x. By concensus, the moving system x'-axis higher values also lie in that direction, and all axes parallel the other system's corresponding axis. We used vv to mean the square of v but might use v^2 for that purpose below. Similarly for c. Because it is believed that no physical object can reach or exceed c, the square-root term in both denominators is presumed always less than one, which means that the formulas say both x' and t' will tend to be greater than x and t, respectively. However, SRians call the x' result 'contraction' - which means shortening - and the t' result 'dilation' - which means increasing. ------------------------------ === Subject: 4. The 'just coordinates' argument The 'just coordinates' argument is so patently ridiculous that even opponents have a hard time accepting just how simple and obvious its debunking can be, as shown in this section. However, further sections take a more arithmet- ical approach that you'll maybe find more professorial. The 'just coordinates' argument is that t is mot a duration, not a time interval; it's just an arbitrary clock reading. But what if the moving system observer comes speeding by while you make your annual 'spring forward' or 'fall back' change? The formula says that the moving system clock's 'just coordinate' reading can be calculated from yours: t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t) Imagine the moving system oberver's confusion if his clock changes its reading while he's looking at it! If his clock doesn't change when yours does, the formula is wrong; if it is truly a 'just coordinates' formula. And then what happens if you realize you were a day early and put your clock back to what it had said previously? And suppose you are in NYC and your twin in LA and both are watching the moving observer. You'll both be using the same v because you are at rest wrt (with respect to) each other. You're on Eastern Standard Time and your twin is on Pacific Standard Time maybe. You have three hours more on your clock than does your twin. On which 'just coordinate' clock will the Lorentz transforms base the 'just coordinate' time the moving system clock says? The formula applies to both of your t-times: t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t) Sure, the idea that you can change someone else's clock with no connection of any kind is really ridiculous, but Eqs 1x and 1t aren't MY equations. Are they yours? And we aren't the ones to say x, t, x', and t' are just coordinates. If the t' formula is actually either an elapsed time formula, or the basis of a t'/t ratio, then there is no implication that one clock's reading has anything to do with the other's. It can only be rates of clock ticking, or how one time INTERVAL compares to the other that the formula is about. ------------------------------ === Subject: 5. Single-system, little-purpose ambiguity. Since we're going to be comparing measurements on two coordinate systems in the next section, let's go to our supply cabinet and get our yard-stick (which we use to measure things in inches) and our meter-stick (which we use to measure things in centimeters). Here, I'm getting mine. Oh! Oh! There's an ant on mine, and he ... she ... sure is hanging on, right at the 3.5 inch mark of the yard- stick. Let's see if I can wave the stick around enough that she'll let go. Nope. However, before I gave up I waved the stick and the ant 'all over the place. Always, however, the ant was at the 3.5 mark on the yard-stick, and always 3.5 away from the end of the stick, however far and wide I have transported her. Neither of those 3.5 facts means very much. Of the two, the distance aspect meant almost nothing. So the distance was 3.5 from the end. So what? That length, distance, was not in use. And only maybe the ant might have been concerned with just what location, 'just coordinate', on the stick she was at. Just so with x and t. So, is the 3.5 reading just a coordinate? Or a distance/length? It's ambiguous in and of itself, and really makes no difference what you say until you try to make use of the number. Hey, my address is 5047 Newton Street. If you are looking for me and you're at 4120 Newton, it is helpful information, because it tells you which direction to go. Is that 'just coordinate'? Where it really becomes useful, perhaps, is in telling you how far away I am. That's not just a coordinate value, that's a distance, length, interval. However, it is subtracting 4120 from 5047 that tells you which direction and how far. It is only because both 5047 and 4120 are distances from the same point - ANY same point - that the result means anything. My x - my yardstick reading - is always a distance or length; it is impossible to be otherwise with an honest, competently designed yardstick. Whether or not its reading is of good use in some particular scientific formula depends on whether I put the zero end of the yardstick at some useful place. As in the introduction, we should either put it at the starting location/end, or use two readings from it: (x-x0). ------------------------------ === Subject: 6. Relating two coordinate measures/systems. Taking care to not damage our brave little ant, I place my yard-stick onto the table, zero end to the left, 36 end to the right. Now I place the 'just coordinate' meter-stick on the table in the same orientation, in a random location, and find that the ant's coordinate on the meter-stick is 51. The formula relating centimeters to inches is cm=i*2.54 but we want a formula similar to x'=(x-vt)/sqrt(1-vv/cc). That would be c=i/.03937 approximately, but let's use x' for the meter-stick reading, and x for the inch reading: x'=x/.3937. 3.5/.3937 = 8.89 Wait a minute. It's not just science but definition that says c=i/.3937=8.89, so something is wrong. 8.89 is not 51. We already knew that 51 cm was just an arbitrary coordinate. Arbitrary not because that point isn't 51 cm from the zero end of the meter-stick, but because the zero point was in an arbitrary position. Let's put the meter-stick in a position where it's zero point is at the yard-stick zero point. What is the centimeter coordinate now? Hey. 8.89, just like the formula says. The only way for a 'transform' like x'=x/g to work, whatever g might be, is for both coordinate systems to have their zero points aligned, in which case saying the two measures are not intervals is pure idiocy. Noe that with both zero points at the same position both x' and x are great measures for scientific purposes, in any and every case where we were smart enough to put those zero points at a useful location. There is one extension of x'=x/g that will let us use the meter-stick in arbitrary position. When the cm reading was 51, the zero point of the yard-stick read (51-8.89=) 42.11 cm. If we call that point x.z' we get x' = x.z' + x/.3937. = 42.11 + 3.5/.3937 = 42.11 + 8.89 = 51. Obviously, in this formula x/.3937 is the distance from the x' coordinate of the location where x=0. An interval. Just as obviously, the fact that we now have the correct formula for relating an x interval to an arbitrary x' coordinate, does not mean that x' is anything more than nonsense for use in any scientific formula. Unless we were smart enough to put the x zero point in a useful location, and use (x'-x.z') in the scientific formula. (x'-x.z') equals the useful, Ratio Scale value x/.3937. So, we have discovered a basic fact: a transformation formula like x'=x/g works only if the two zero points of the coordinate systems coincide. That makes it non- sense to say the two coodinates are only coordinates and not intervals. Both must be values that represent distances from their respective zero points unless you take the proper steps to adjust for the discrepancy. Make sure you understand that although the inclusion of x.z' made it possible to correctly calculate x', the result is nonsense when it comes to use of x' for general length/distance purposes; it is x'-x.z' that is a useful number in such cases. It could be that we're measuring a sheet of paper with one end at x=0 and the other at x=3.5; x'=51 is nonsense as a centimeter measure of the paper. But, you say, the Lorentz transform contain a -vt term. ------------------------------ === Subject: 7. Distances and moving coordinate axes. We discovered x'=x.z' + x/g as the correct formula for relating one coordinate to another system's. But the Lorentz transform contains another term, -vt/sqrt(1-vv/cc). What is it? Let's start with our x'=51 cm, x=3.5, x.z'=42.11 example. Every minute, let's move the meter-stick one inch to our right. At minute 0, the cm reading was 51 cm. At minute 1, the cm reading is now 50 cm. At minute 2, the cm reading is now 49 cm. In this instance, v=1 inch/minute. And t was 0, 1, 2. What has happened is that we have made our x.z' a lie, and increasingly so. -vt/.3937 is the change in x.z'. x' = (x.z - vt/.3937) + x/.3937. Obviously, vt/.3937 is not a coordinate; even most SRians wouldn't imagine it was. It is an interval, the distance over which the moving system has moved since t=0. And, of course, x/.3937 is the distance of our brave little ant from the point where x=0 and the centimeter reading is x.z'-vt/.3937. Yes, every minute the meter- stick moves to the right and the meter-stick coordinate of the spot where x=0 gets less and less - and eventually negative. Make sure you understand that every minute the x' coordinate, because of -vt/g, becomes a better measure of, say, the 3.5 paper we might be measuring with the yard-stick, given that 51 was too big a number and -vt is negative. That is, until the two origins coincide at x'=x=0, and then it gets worse and worse. With -vt positive (because v<0) the situation is different. With 51 and -vt positive, x' just gets worse and worse over time. Quite obviously, the fact that we now have the correct formula for relating an x interval to an arbitrary x' coordinate even when the x' axis is moving, does not mean that x'is anything more than nonsense for use in any scientific formula. Unless we were smart enough to put the x zero point in a useful location, and use (x'-x.z'+vt/.3937) in the scientific formula. (x'-x.z'+vt/.3937) equals the useful, Ratio Scale value x/.3937. ------------------------------ === Subject: 8. Time intervals. Instead of using our sticks, let's get out two clocks. Mind you, we're not going to deal with different clock rates here, just establish the same basics as for distance. Your clock says 9:00 Eastern Standard Time (EST) and we note that t=540 minutes when we put down the clock. Blindly, let's turn the setting knob of your twin's Pacific Standard Time clock and put it down before us. According to what we see, EST's 540 minutes (9:00) corre- sponds to PST's 14:30; t'=870. We know the formula relating PST to EST is t' (pacific) = t (eastern) - 180 (minutes). Thus, it is not correct that the second clock can have an arbitrary setting, because 870 <> 540-180. We know that the two clocks are related by t' = t/1 since both are using the same second, hour, etc units. But 870 (14:30 in minutes) is not 540/1-180, so once again we know something is wrong. However, t'=t.z' + t/1 works. EST midnight equals PST 0.0 (midnite) - 180, so t.z' = -180, and t' = -180 + 540/1 = 360. Since EST-180=PST, 9:00 EST is 6:00 PST = 360 minutes. We see thus that like distance measures/coordinates, time axis origins (zero points) must either be 'lined up' or adjusted for. So, the Lorentz/Einstein t'=t/sqrt(1-vv/cc) must be the moving system elapsed time interval since the time axes were both at a common zero. There is no t.z' adjustment: t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t) Make sure you understand that in the clock case, if the EST is showing a good number for elapsed time since the travelling observer passed NYC, then the PST clock is silliness. t.z' must be zero or must be taken out of time lapse calculations for the PST clock to be used intelligently, just as was true for x.z'. What is lacking as yet for Lorentz t' is the -vx/cc term that corresponds to the x' formula -vt term. Break it up into two parts: v/c and x/c. v/c is a scaling factor that changes velocity from whatever kind of unit you are using over to fractions of c. x/c is distance divided by velocity, which is time. x/c is thus the time interval since the two time axes had a common zero point - which they have to have in the Lorentz transforms which do not have the t.z' term we learned to use above. Thus, (-vx/cc)/sqrt(1-vv/cc) is the interval amount the moving system clock has been changed - since the common/ adjusted time - over and beyond the elapsed time interval represented by x/sqrt(1-vv/cc). We have discovered that the only way for t' to be t/g is for t' and t to have a common zero point, just as for x' and x. It would be otherwise if the t' formula contained an adjustment t.z' under some name or other, but the necessity to include such a term correlates 100% with t' numbers that aren't directly usable. As for x and x', our knowledge of how to setup a proper formula relating t and t' is of no use unless we use the knowledge in scientific formulas; (t'-t.z'+xv/gcc) gives us the only directly useful value: t/g. ------------------------------ === Subject: 9. Einstein's (1905) derivations. When we return to Einstein's derivations of the transform formulas with a well-focused eye, we find he was a wee bit confused - or at least self-contradictory. When he set up his (at first unknown) tau=moving system time formulas, he created three particular instances of tau. Tau.0 is the time at which light is emitted at the moving origin toward a mirror to the right that is moving at rest wrt that moving origin and at a constant distance from that origin. He lets the stationary time slot have the value t, a constant, the stationary system starting time. Tau.1 is the time at which the light is reflected. He lets the stationary time be t+x'/(c-v); t is still a constant and x'/(c-v) is the time interval since t. Tau.2 is the time at which the light gets (back) to the moving origin. The stationary time value is put as t + x'/(c-v) + x'/(c+v); t is still a constant and x'/(c-v) + x'/(c+v) is the time interval since t. On the thesis that the moving observer sees the time to the mirror as the same as the time back to the origin, he sets .5[ tau.0 + tau.2 ] = tau.1. Tau.0 completely drops out of the analysis and leaves no trace, and has no effect. Further, the t you see in tau.0, tau.1, and tau.2 also completely drops out with no trace and no effect, leaving us with exactly what you'd get if you had explicilty said t' is an interval and so is t. What doesn't drop out in the stationary time values is x'/(c-v) and x'/(c+v), the time interval it takes for light to get to the fleeing mirror, and the time interval it takes for light to get back to the approaching origin. Thus, his resultant t' formula is strictly based on time intervals in the stationary system. Time intervals since some starting time, yes, but time intervals. There is absolutely nothing in the derived formulas that depends on arbitrary coordinates like the constant t in the stationary time arguments. Let's look at the x dimension; it is x'=x-vt [as x increases by vt, the effect over time is x'=(x+vt)-vt)], which Einstein explicitly sets up as a constant stationary distance. He uses that x' not just in the time interval parts of the stationary time arguments, but also in the x (distance) stationary system argument for the tau at the time light is reflected. x' can't be the stationary system coordinate of the mirror at that time. That value is x'+vt. x' is explicitly an interval, distance. Thus, the whole tau derivation of the t' formula is fully and explicitly based on x' - a spatial length/distance/interval - and the two time interals x'/(c-v) and x'/(c+v). While we're at it, if the starting t is not zero, his x'=x-vt formula is complete nonsense also. Given that there was some L that was the mirror x-location and length when the light is emitted, if t was already, say, 500, then x'=L-vt could have been a very negative length. ------------------------------ === Subject: 10. A word about intervals. There are intervals, and there are intervals. If we put our yard stick zero point at one end of a piece of paper and read off the coordinate at the other end of the paper, we have a good measure of the paper's length, a Ratio Scale measure. [Absolute temperature scales are ratio scale.] If instead we put the one end of the paper at the one inch mark (or the zero end of the stick one inch 'into' the length of the paper) we get measures that are one inch off the true, ratio scale length. The two messed up measures are still intervals, but they are Interval Scale measures. [Household temperature scales are interval scale, which is why your physics and chemistry professors won't let you use them without first converting to the ratio scale absolute temperatures.) t'=t/g and x'=x/g represent ratio scale measures, given that t and x were ratio scalae to start with. t'=t.z'+t/g and t'=t/g-vx/gcc are both interval scale measures, even given a good ratio scale t and a good ratio scale x. x'=x.z'+x/g and x'=x/g-vt/g are both interval scale measures, even given a good ratio scale x and a good ratio scale t. Look for the (SR) Lorentz t', x' = degraded measures document soon at a newsgroup near you. ------------------------------ === Subject: 11. Intervals versus the Twins Paradox. t'=(t-vx/cc)/g shows t' being greater than t. The reason Special Relativity will not allow the use of its basic time equation in determining what SR has to say about the twins' ages, is that t' and x' are supposedly just coordinates, and they say you have to take the coordinate pairs (t',x') and (x,t) into consideration in both the time and place the twins' separation started and the time and place the twins reunited. Since t' and x' are actually both intervals, not just coordinates, the 'excuse' is spurious, and is so even without use of the obvious (x_b-x_a) and (t_b-t_a) usages. However, SR is right to be embarrassed by their transformation formulas. Look for the (SR) Lorentz t', x' = degraded measures document at a newsgroup near you. ------------------------------ === Subject: 12. Summary A. t'=t/g and x'=x/g can be almost 'just coordinates' in the sense that the values obtained may not be of much use except in the most primal and useless way: how long and how far since/from the time/ place they were zero. Even here, however, the zero points within each of the two scale pairs (t',t) and (x'.x) must have been lined up. If the zero points have been intelligently selected (such as at the starting point and time of a trip) they can be rationally used 'as is' in any valid sci- entific equation. B. Even the interval scale t'=t.z' - xv/gcc + t/g and x'=x.z' - vt/g + x/g are not 'just coordinates'. They can be used to good effect by establishing the relevant starting times/points and using (t'-t.z'+xv/gcc) and (x'-x.z'+vt/g), as the situation may require. C. When you see vx/gcc or vt/g in use in any guise with non-zero values, you know the resultant t' or x' is a degraded, interval scale value. E-X: Anytime you do not see what amounts to t.z' and xv/gcc in the time case, or x.z' and vt/g in the distance case, you know that the t' and/or x' in use are intervals. Period. Y: Either set your clock to zero at the start of the relevant time interval, or use (t-t0), with both being readings on the same clock. Either move your x-axis origin to the starting end or point, or use (x-x0), with both being readings on the same axis. Z: In _(SR) Lorentz t', x' = Degraded (Interval) Scales_ we see that t' and x' satisfy the mathematical tests for/of interval scales when -vt and -vx/cc are not zero; thus, they must be intervals. When -vt and -vx/cc are zero, t' and x' satisfy the much better mathematical definition of ratio scales, and are thus not just mere intervals, but (rescaled) good ones. Eleaticus !---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?- --!---? ! Eleaticus Oren C. Webster ThnkTank@concentric.net ? ! Anything and everything that requires or encourages systematic ? ! examination of premises, logic, and conclusions ? !---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?- --!---? === Subject: Re: (SR) Lorentz t', x' = Intervals > (SR) Lorentz t', x' = Intervals > (c) Eleaticus/Oren C. Webster > Thnktank@concentric.net [snip 700 lines of trolled garbage] Originally trolled across sci.physics sci.physics.relativity alt.physics sci.math sci.answers alt.answers news.answers http://b5.sdvc.uwyo.edu/bab5/snds/argcstpd.wav Psychotic ineducable boring troll Eleaticus, You see yourself this way, http://www.mazepath.com/uncleal/effete6.jpg The entire remainder of the planet sees you this way, http://www.mazepath.com/uncleal/effete7.jpg http://w0rli.home.att.net/youare.swf http://www.mazepath.com/uncleal/sunshine.jpg http://www.you-moron.com/ http://www.mazepath.com/uncleal/effete0.jpg http://www.mazepath.com/uncleal/effete1.png http://www.mazepath.com/uncleal/effete2.png http://www.mazepath.com/uncleal/effete3.png http://www.mazepath.com/uncleal/effete4.png http://www.mazepath.com/uncleal/effete5.jpg http://www.apa.org/journals/psp/psp7761121.html http://insti.physics.sunysb.edu/~siegel/quack.html [snip 1300 lines of trolled garbage] Originally trolled across sci.physics sci.physics.relativity alt.physics sci.math sci.answers alt.answers news.answers http://b5.sdvc.uwyo.edu/bab5/snds/argcstpd.wav Psychotic ineducable boring troll Eleaticus, You see yourself this way, http://www.mazepath.com/uncleal/effete6.jpg The entire remainder of the planet sees you this way, http://www.mazepath.com/uncleal/effete7.jpg http://w0rli.home.att.net/youare.swf http://www.mazepath.com/uncleal/sunshine.jpg http://www.you-moron.com/ http://www.mazepath.com/uncleal/effete0.jpg http://www.mazepath.com/uncleal/effete1.png http://www.mazepath.com/uncleal/effete2.png http://www.mazepath.com/uncleal/effete3.png http://www.mazepath.com/uncleal/effete4.png http://www.mazepath.com/uncleal/effete5.jpg http://www.apa.org/journals/psp/psp7761121.html http://insti.physics.sunysb.edu/~siegel/quack.html Hey, stooopid troll Eleaticus - Do you want EVIDENCE? Each of the 24 GPS satellites carries either four cesium atomic clocks or three rubidum atomic clocks in orbit, with full relativistic corrections being applied. http://arXiv.org/abs/hep-th/0307140 GR structure, especially Part 4/p. 7 Experimental constraints on General Relativity. http://www.eftaylor.com/pub/projecta.pdf Relativity in the GPS system http://arXiv.org/abs/gr-qc/9909014 falling light http://arXiv.org/abs/astro-ph/0401086 http://arxiv.org/abs/astro-ph/0312071 Deeply relativistic neutron star binaries http://arXiv.org/abs/gr-qc/0301024 Nordtvedt Effect NIM A 355 537 (1995) Physics Letters B 328 103 (1994) Physical Review Letters 64 1697 (1990) Physical Review Letters 39 1051 (1977) Physical Review 135 B1071 (1964) Physics Letters 12 260 (1964) Europhysics Letters 56(2) 170-174 (2001) General Relativity and Gravitation 34(9) 1371 (2002) http://fourmilab.to/etexts/einstein/specrel/specrel.pdf http://users.powernet.co.uk/bearsoft/Paper6.pdf http://users.powernet.co.uk/bearsoft/LPHrel.html Longitudinal and transverse mass http://www.navcen.uscg.gov/pubs/gps/gpsuser/gpsuser.pdf http://www.navcen.uscg.gov/pubs/gps/sigspec/default.htm http://www.navcen.uscg.gov/pubs/gps/icd200/default.htm http://www.trimble.com/gps/index.html http://sirius.chinalake.navy.mil/satpred/ http://www.phys.lsu.edu/mog/mog9/node9.html http://egtphysics.net/GPS/RelGPS.htm http://www.schriever.af.mil/gps/Current/current.oa1 http://edu-observatory.org/gps/gps_books.html Uncle Al http://www.mazepath.com/uncleal/qz.pdf http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Guess who you are fooling ? Hi Eleaticus, Re: Your convoluted, rapidly nymshifting, monologue spewing, spam-bot ... and your complete inability to engage in dialog, You project your faults onto others thusly, The Lorentz-Einstein transformations cannot possibly be ' just coordinates ', which is the interpretation required to support the many sideshow carnival acts with which the SR faithful bedazzle the public, and establish their moral and intellectual superiority , Q. Guess who you are fooling ? A. No one. Like we know the nose on our faces ... Everyone here knows that you are trying to bedazzle the public in a perfectly vain attempt to feign moral and intellectual superiority . Many sideshow carnival acts ? You said it dude. That's exactly what you are doing. Your complete inability to engage in meaningful dialog is what is most telling. All your incessant rambling is saying just this: Because relativity is used to zoom far in or far out, you feel it demeans humans. Why can't you express yourself more clearly ? Why are you rapidly nymshifting and using spam-bots ? === Subject: Re: Guess who you are fooling ? > we know the nose on our faces ... [GH] really...? SCIENTIFIC PROOF OF GOD WEBSITE http://geocities.com/scientific_proof_of_god mirror site: http://proof-of-god.freewebsitehosting.com === Subject: Re: Mahlo Cardinals >I read a definition of Mahlo Cardinals in Rudy Rucker's book (Infinity >and the mind). It's probably not the best place for such definitions, >and I didn't look at more serious books yet. >Rudy gives what he calls a formal definition. He says that a Mahlo >Cardinal is an inaccessibe cardinal which for every set of ordinals >smaller than itself with it's cardinality, there is smaller inaccessible >cardinal kappa, for which the set of all ordinals in that set less than >kappa has a cardinality of kappa. >I'd like to know if that's the way they usually define Mahlo Cardinals. >For the converse, suppose rho is a weakly Mahlo cardinal, and let A be a >set of ordinals less than rho with cardinality rho. Define f : rho -> rho >as follows. f(0) is the smallest element of A. f(a+1) is the smallest >element of A-{f(b) : b < a or b = a} for all ordinals a < rho. If b < rho >is a limit ordinal, then f(b) = lim{f(a) : a < b}. Then f is a normal >function on rho, so that f has a regular fixed point, alpha, since rho is >a weakly Mahlo cardinal. >Thr bit that went here should be replaced by: >For any ordinal c < rho, define g_c : rho -> rho by g_c(a) = f(c+a) for >all a < rho, then g_c is a normal function on rho, and so g_c has a >regular fixed point alpha. Since alpha <= c+alpha <= f(c+alpha) = >g_c(alpha) = alpha, where a <= b denotes that a is less than or equal >to b, then c < alpha, c+alpha = alpha (i.e. alpha >= c omega), and >f(alpha) = alpha, i.e. alpha is a fixed point of f. So for any ordinal >less than rho, there is a regular fixed point of f which exceeds it. >Therefore the set of regular fixed points of f is unbounded, and so the >cardinality of the set B of regular fixed points of f is rho. >Define h : rho -> rho as follows. h(0) is the >smallest element of B. h(a+1) is the smallest element of B-{h(b) : b < a >or b = a} for all ordinals a < rho. If b < rho is a limit ordinal, then >h(b) = lim{h(a) : a < b}. Then h is a normal function on rho, so that h >has a regular fixed point, kappa, since rho is a weakly Mahlo cardinal. >Since kappa is a limit ordinal, then h(kappa) is a limit cardinal, and so >kappa is a limit cardinal since h(kappa) = kappa. Since kappa is a >regular limit cardinal, then kappa is inaccessible, so that kappa is an >inaccessible fixed point of h. Since kappa is a limit of fixed points of >f, then kappa is a fixed point of f, since f is continuous. So kappa is >an inaccessible fixed point of f. Since f(kappa) = kappa, and the set of >successor ordinals less than kappa has cardinality kappa, then the >intersection of A and {f(a) : a < kappa} has cardinality kappa. Since >f(a) < kappa if a < kappa (since f is increasing), then the cardinality of >the intersection of A and kappa is kappa, and so there exists an >inaccessible cardinal kappa < rho such that the set of ordinals in A less >than kappa has cardinality kappa. This demonstrates that Rudy's >condition correctly characterizes the weakly Mahlo cardinals. The above can be rewritten by first noting that any normal function on a weakly Mahlo cardinal has a weakly inacessible fixed point. Let f : delta -> rho be a normal function on a weakly Mahlo cardinal rho. Since rho is regular, then delta = rho. For any ordinal c < rho, define g_c : rho -> rho by g_c(a) = f(c+a) for all a < rho, then g_c is a normal function on rho, and so g_c has a regular fixed point alpha. Since alpha <= c+alpha <= f(c+alpha) = g_c(alpha) = alpha, then c < alpha, c+alpha = alpha, and f(alpha) = alpha, i.e. alpha is a fixed point of f. So for any ordinal less than rho, there is a regular fixed point of f which exceeds it. Therefore the set of regular fixed points of f is unbounded, and so the cardinality of the set B of regular fixed points of f is rho, since rho is regular. Define h : rho -> rho as follows. h(0) is the smallest element of B. h(a+1) is the smallest element of B-{h(b) : b <= a} for all ordinals a < rho. If b < rho is a limit ordinal, then h(b) = lim{h(a) : a < b}. Then h is a normal function on rho, so that h has a regular fixed point, kappa. Since kappa is a limit ordinal, then h(kappa) is a limit cardinal, and so kappa is a limit cardinal since h(kappa) = kappa. Since kappa is a regular limit cardinal, then kappa is weakly inaccessible, so that kappa is a weakly inaccessible fixed point of h. Since kappa is a limit of fixed points of f, then kappa is a fixed point of f, since f is continuous. So kappa is a weakly inaccessible fixed point of f. It follows that any normal function on rho has a weakly inaccessible fixed point. Suppose rho is a weakly Mahlo cardinal, and let A be a set of ordinals less than rho with cardinality rho. Define f : rho -> rho as follows. f(0) is the smallest element of A. f(a+1) is the smallest element of A-{f(b) : b <= a} for all ordinals a < rho. If b < rho is a limit ordinal, then f(b) = lim{f(a) : a < b}. Then f is a normal function on rho, so that f has a weakly inaccessible fixed point, kappa, since rho is a weakly Mahlo cardinal. Since f(kappa) = kappa, and the set of successor ordinals less than kappa has cardinality kappa, then the intersection of A and {f(a) : a < kappa} has cardinality kappa. Since f(a) < kappa if a < kappa (since f is increasing), then the cardinality of the intersection of A and kappa is kappa, and so there exists a weakly inaccessible cardinal kappa < rho such that the set of ordinals in A less than kappa has cardinality kappa. This demonstrates that Rudy Rucker's condition for weakly Mahlo cardinals correctly characterizes the weakly Mahlo cardinals. David McAnally At the moment, they (the Time Lords) are far from being all-powerful. That's why it's been left up to me and me and me. quote by: Patrick Troughton in The Three Doctors ------- === Subject: Re: A question about Kripke semantics and physics (Was: Re: Study groups in science) > if you are interested in scientific online workshops, please visit > site > http://de.geocities.com/scienceworkshops/ > Its goal is to organize study groups on scientific topics like quantum > field theory, > probabilistic inference or neural nets, to name a few topics I am > personally interested in > (of course, arbitray topics may be suggested). > What positive precautions are you taking to prevent the idiots morons and > kooks from taking it over, as has happened in sci.physics? >Hi Franz! >Let's just defuse the reference to morons and kooks by deferring to >Shannon's discussion concerning the statistical character of language. >I want to ask you a question [snip nearly 100 lines...] So what _was_ the question? Richard Herring === Subject: Simple numbers Hallo, I know that every simple number (beside 2 and 3) has its one formula : S.N = 6*n +/- 1 for some n , but not any n={1,2,3,...+} in formula 6*n +/- 1 gives the simple number (for example: n=6 => 6*6-1=35 , 35 is not simple , 5,7|35) so my question is: If n={1,2,3,...} what is the formula which gives the simple numbers (beside 2 and 3) for any n? === Subject: Re: Simple numbers > Hallo, > I know that every simple number (beside 2 and 3) has its one formula : S.N = > 6*n +/- 1 for some n , but not any n={1,2,3,...+} in formula 6*n +/- 1 gives > the simple number (for example: n=6 => 6*6-1=35 , 35 is not simple , 5,7|35) > so my question is: > If n={1,2,3,...} what is the formula which gives the simple numbers (beside > 2 and 3) for any n? **** For those of you that wont understand rest of my message - I will answer to original poster in Serbian. You will not miss anything important - it's all said already in other answers to him. **** Daklem, u engleskom jeziku se prost broj (onaj koji je deljiv BEZ OSTATKA samo sa 1 i samim sobom) zove PRIME number a ne Simple number. Sto se tice formule koja bi davala za n=1,2,..., redom sve proste brojeve - NEMA takve formule barem ne proste formule oblika a*n+b ili tako nesto. Mislim da su u jednom od odgovora vec dali link ka http://www.utm.edu/research/primes/notes/faq/p_n.html gde se opisuju formule za izracunavanje n-tog prostog broja za dato n - medjutim to NISU formule oblika a*n+b ili slicno koje daju direktno n-ti prost broj. Google search po kljucnim recima prime numbers daje gomilu linkova kao sto su http://www.math.utah.edu/~alfeld/math/prime.html http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Prime_ numbers.html Zbog toga sto nema jednostavne formule za pronalazenje prostih brojeva, jako je tesko izracunati (odnosno faktorisati) vrlo velike proste brojeve pa se oni koriste kao osnov za algoritme za sifrovanje kao sto je RSA algoritam. http://world.std.com/~franl/crypto/rsa-guts.html http://renoir.vill.edu/~cbeatty/netclass/portfolio/RSA.html Inace ima ovde na news grupi jedan ludak :-) po imenu James Harris koji tvrdi da ima efikasan alogritam za brzo faktorisanje prostih brojeva - sve njegove price o prostim brojevima mozes slobodno da ignorises. Pozdrav Goran === Subject: Re: Simple numbers Hvala na linkovima ! Puno pozdrava! -------------------------------------------------------------- ------------- > Hallo, > I know that every simple number (beside 2 and 3) has its one formula : S.N = > 6*n +/- 1 for some n , but not any n={1,2,3,...+} in formula 6*n +/- 1 gives > the simple number (for example: n=6 => 6*6-1=35 , 35 is not simple , 5,7|35) > so my question is: > If n={1,2,3,...} what is the formula which gives the simple numbers (beside > 2 and 3) for any n? > **** > For those of you that wont understand rest of my message - I will > answer to original poster in Serbian. You will not miss anything > important - it's all said already in other answers to him. > **** > Daklem, > u engleskom jeziku se prost broj (onaj koji je deljiv BEZ OSTATKA samo > sa 1 i samim sobom) zove PRIME number a ne Simple number. Sto se > tice formule koja bi davala za n=1,2,..., redom sve proste brojeve - > NEMA takve formule barem ne proste formule oblika a*n+b ili tako > nesto. > Mislim da su u jednom od odgovora vec dali link ka > http://www.utm.edu/research/primes/notes/faq/p_n.html > gde se opisuju formule za izracunavanje n-tog prostog broja za > dato n - medjutim to NISU formule oblika a*n+b ili slicno koje daju > direktno n-ti prost broj. > Google search po kljucnim recima prime numbers daje gomilu linkova > kao sto su > http://www.math.utah.edu/~alfeld/math/prime.html > http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Prime_ numbers.html > Zbog toga sto nema jednostavne formule za pronalazenje prostih > brojeva, jako je tesko izracunati (odnosno faktorisati) vrlo velike > proste brojeve pa se oni koriste kao osnov za algoritme za sifrovanje > kao sto je RSA algoritam. > http://world.std.com/~franl/crypto/rsa-guts.html > http://renoir.vill.edu/~cbeatty/netclass/portfolio/RSA.html > Inace ima ovde na news grupi jedan ludak :-) po imenu James Harris > koji tvrdi da ima efikasan alogritam za brzo faktorisanje prostih > brojeva - sve njegove price o prostim brojevima mozes slobodno da > ignorises. > Pozdrav > Goran === Subject: Re: Simple numbers > I know that every simple number (beside 2 and 3) has its one formula : S.N = > 6*n +/- 1 for some n , but not any n={1,2,3,...+} in formula 6*n +/- 1 gives > the simple number (for example: n=6 => 6*6-1=35 , 35 is not simple , 5,7|35) What is a simple number? Do you mean a prime? Lydia === Subject: Re: Simple numbers I don't know what a prime is because I don't know English much , but simpel number (as we call it) is a number which can be only divided by 1 and himself. === Subject: Re: Simple numbers > Hallo, > I know that every simple number (beside 2 and 3) has its one formula : S.N = > 6*n +/- 1 for some n , but not any n={1,2,3,...+} in formula 6*n +/- 1 gives > the simple number (for example: n=6 => 6*6-1=35 , 35 is not simple , 5,7|35) By simple number I think you must mean prime. > so my question is: > If n={1,2,3,...} what is the formula which gives the simple numbers (beside > 2 and 3) for any n? Evidently you don't consider p(n) is the n'th prime to be a formula. Any particular reason? Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: multilinear algebra consider the operator S^k defined on V x...x V k-times with itself V is a finite dimensional vector space over |K , (say |K = C = complex numbers with field structure) S^k takes its values on the k-th symmetric tensor power of V , sometimes denoted by V odot ...odot V (k-times) S^k is defined by S^k : (v_1, ..., v_k )|-->1 / k! sum_{sigma in S_k}v_sigma(1) tensor.. tensor v_sigma(k) I ask you if S^k defined above is or not a surjective map. Tern === Subject: Re: multilinear algebra > consider the operator S^k defined on V x...x V k-times with itself > V is a finite dimensional vector space over |K , (say |K = C = complex > numbers with field structure) > S^k takes its values on the k-th symmetric tensor power of V , sometimes > denoted by V odot ...odot V (k-times) > S^k is defined by > S^k : (v_1, ..., v_k )|-->1 / k! sum_{sigma in S_k}v_sigma(1) tensor.. > tensor v_sigma(k) > I ask you if S^k defined above is or not a surjective map. > Tern evidently there are tow ways to read this - i was taking the view that it was a map from the n'th tensor power to the n'th tensor power, which was probably the wrong one, rereading it. but the other interpretation seems to be trivially true. === Subject: Re: multilinear algebra >consider the operator S^k defined on V x...x V k-times with itself >V is a finite dimensional vector space over |K , (say |K = C = complex >numbers with field structure) >S^k takes its values on the k-th symmetric tensor power of V , sometimes >denoted by V odot ...odot V (k-times) >S^k is defined by >S^k : (v_1, ..., v_k )|-->1 / k! sum_{sigma in S_k}v_sigma(1) tensor.. >tensor v_sigma(k) >I ask you if S^k defined above is or not a surjective map. >Tern > evidently there are tow ways to read this - i was taking the view that it > was a map from the n'th tensor power to the n'th tensor power, which was > probably the wrong one, rereading it. but the other interpretation seems > to be trivially true. I assume you mean by the k^th symmetric tensor power the subspace of the k-fold tensor power of symmetric tensors. To answer your question, consider the case of a finite field and assume k isn't divisible by the charactersitic. Then simply count the number of elements in each. The case k = 2 should suffice. If K has q elements, and V has dimension n over K, V x V has q^(2n) elements. V tensor V has q^(n^2) elements, and the symmetric subspace has q^(n(n+1)/2) elements. === Subject: Re: multilinear algebra > consider the operator S^k defined on V x...x V k-times with itself > V is a finite dimensional vector space over |K , (say |K = C = complex > numbers with field structure) > S^k takes its values on the k-th symmetric tensor power of V , sometimes > denoted by V odot ...odot V (k-times) > S^k is defined by > S^k : (v_1, ..., v_k )|-->1 / k! sum_{sigma in S_k}v_sigma(1) tensor.. > tensor v_sigma(k) > I ask you if S^k defined above is or not a surjective map. > Tern As long as V is at least two dimensional, no this is not a surjection. Its image is the symmetric part of the tensor space. For example, when k=2, the 'V tensor V' is the direct sum of the symmetric and exterior algebra (exterior when you mulitply the permuted component by the sign of the permutation in the sum you have). And this is not a trivial decomposition for dim(V)>1 - the exterior part (also called alternating) has dimension dim(V) choose 2. for its dimension, as I'm to lazy to figure it out for you. Sorry. === Subject: Re: multilinear algebra > consider the operator S^k defined on V x...x V k-times with itself > V is a finite dimensional vector space over |K , (say |K = C = complex > numbers with field structure) > S^k takes its values on the k-th symmetric tensor power of V , sometimes > denoted by V odot ...odot V (k-times) > S^k is defined by > S^k : (v_1, ..., v_k )|-->1 / k! sum_{sigma in S_k}v_sigma(1) tensor.. > tensor v_sigma(k) > I ask you if S^k defined above is or not a surjective map. Assuming that the k-th symmetric tensor power of V is the subspace of the k-th tensor power of V whose elements are invariant under the natural action of the symmetric group, the answer is yes. Simply consider the inclusion i of the k-th symmetric tensor power of V into the k-th tensor power of V and prove that f o i is the identity. Jose Carlos Santos === Subject: Help needed Hallo, I know that every simple number (beside 2 and 3) has its one formula : S.N = 6*n +/- 1 for some n , but not any n={1,2,3,...+} in formula 6*n +/- 1 gives the simple number (for example: n=6 => 6*6-1=35 , 35 is not simple , 5,7|35) so my question is: If n={1,2,3,...} what is the formula which gives the simple numbers (beside 2 and 3) for any n? === Subject: Re: Help needed > Hallo, > I know that every simple number (beside 2 and 3) has its one formula : S.N = > 6*n +/- 1 for some n , but not any n={1,2,3,...+} in formula 6*n +/- 1 gives > the simple number (for example: n=6 => 6*6-1=35 , 35 is not simple , 5,7|35) > so my question is: > If n={1,2,3,...} what is the formula which gives the simple numbers (beside > 2 and 3) for any n? Simple numbers are called Prime numbers in English. There is no formula that gives Prime Numbers for any n. -Michael. === Subject: Re: Help needed > I know that every simple number (beside 2 and 3) has its one formula : S.N > 6*n +/- 1 for some n , but not any n={1,2,3,...+} in formula 6*n +/- 1 > gives > the simple number (for example: n=6 => 6*6-1=35 , 35 is not simple , > 5,7|35) > so my question is: > If n={1,2,3,...} what is the formula which gives the simple numbers > (beside > 2 and 3) for any n? > Simple numbers are called Prime numbers in English. > There is no formula that gives Prime Numbers for any n. If you *really* think that, then I suggest that you *do not* read the page: http://www.utm.edu/research/primes/notes/faq/p_n.html Jose Carlos Santos === Subject: Re: Help needed > I know that every simple number (beside 2 and 3) has its one formula : S.N > = > 6*n +/- 1 for some n , but not any n={1,2,3,...+} in formula 6*n +/- 1 > gives > the simple number (for example: n=6 => 6*6-1=35 , 35 is not simple , > 5,7|35) > so my question is: > If n={1,2,3,...} what is the formula which gives the simple numbers > (beside > 2 and 3) for any n? > Simple numbers are called Prime numbers in English. > There is no formula that gives Prime Numbers for any n. > If you *really* think that, then I suggest that you *do not* read the > page: > http://www.utm.edu/research/primes/notes/faq/p_n.html > Jose Carlos Santos === Subject: Re: Help needed THANK VERY MUCH , YOU JOSE CARLOS SANTOS !!!!!! > I know that every simple number (beside 2 and 3) has its one formula : S.N > = > 6*n +/- 1 for some n , but not any n={1,2,3,...+} in formula 6*n +/- 1 > gives > the simple number (for example: n=6 => 6*6-1=35 , 35 is not simple , > 5,7|35) > so my question is: > If n={1,2,3,...} what is the formula which gives the simple numbers > (beside > 2 and 3) for any n? > Simple numbers are called Prime numbers in English. > There is no formula that gives Prime Numbers for any n. > If you *really* think that, then I suggest that you *do not* read the > page: > http://www.utm.edu/research/primes/notes/faq/p_n.html > Jose Carlos Santos === Subject: Re: A question on infinite integral domains >Has any one seen a proof that showed that an infinite integral domain cannot >be a field? All the books I have prove that a finite intergal domain is a >field but don't give any indication about why infinite domains can't be. For >instance, Z is not a field; I know why it is not; but does the same argument >hold for any infinite domain. >A hint about the proof would be welcome; for I am sure it is trivial. >Navin Kadambi If the infinite integral domain is a field, then it is an infinite field. If no field is infinite, you are done. Otherwise try starting with an infinite field. If your infinite field is an integral domain, you have found a counterexample. If you cannot find a counterexample, explain why an infinite field is never an integral domain. If your algebraic system is K, then you are asking whether the three properties *) K is infinite *) K is an integral domain *) K is a field are compatible. The concepts of integral domain and field play symmetric roles. The two problems Is any infinite integral domain a field? Is any infinite field an integral domain? are equivalent as is If an integral domain is a field, can it be infinite? Changing your perspective may simplify the problem. [Your solution will be much shorter than this posting!] John Adams served two terms as Vice President and one as President, but lost reelection. Later his son became President despite losing the popular vote. That son lost his reelection attempt badly. Now history is repeating itself. pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael, California === Subject: Re: A question on infinite integral domains Yes, that's right: I got my negations wrong. Of course all fields are integral domains and there are infinite fields (e.g. Q). If only I had spent a little more time on this one: man I feel really stupid! Navin > Hi > Has any one seen a proof that showed that an infinite integral domain cannot > be a field? All the books I have prove that a finite intergal domain is a > field but don't give any indication about why infinite domains can't be. For > instance, Z is not a field; I know why it is not; but does the same argument > hold for any infinite domain. > A hint about the proof would be welcome; for I am sure it is trivial. > Navin Kadambi === Subject: Re: A question on infinite integral domains >Yes, that's right: I got my negations wrong. Of course all fields are >integral domains and there are infinite fields (e.g. Q). >If only I had spent a little more time on this one: man I feel really >stupid! You must be new here. You're not supposed to say of course all fields are integral domains so there are infinite integral domains, you're supposed to spend the next few months telling everyone else how stupid they are for not being aware of the fact that there are no infinite integral domains... >Navin > Hi > Has any one seen a proof that showed that an infinite integral domain >cannot > be a field? All the books I have prove that a finite intergal domain is a > field but don't give any indication about why infinite domains can't be. >For > instance, Z is not a field; I know why it is not; but does the same >argument > hold for any infinite domain. > A hint about the proof would be welcome; for I am sure it is trivial. > Navin Kadambi ************************ David C. Ullrich === Subject: Re: A question on infinite integral domains Adjunct Assistant Professor at the University of Montana. >Has any one seen a proof that showed that an infinite integral domain cannot >be a field? All the books I have prove that a finite intergal domain is a >field but don't give any indication about why infinite domains can't >be. You have your negations and quantifiers in the wrong place. Wedderburn's Theorem is: For all D, If D is finite and is an integral domain, then D is a field. An extension to infinite integral domains would say For all D, if D is infinite and an integral domain, then D is a field. This is false; its negation is not For all D, if D is infinite and an integral domain, then D is not a field (what you seem to be claiming). The correct negation is: There exists a D such that D is infinite, and integral domain, and not a field. (The negation of for all x, P(x) is not for all x, not(P(x)); the correct negation is there exists x such that not(P(x)). ) By exhibiting Z, you have already shown that this negation is true. And surely you realize that any field is necessarily also an integral domain, so would not Q show that what you are asking for is impossible? 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 === Subject: Re: A question on infinite integral domains >Has any one seen a proof that showed that an infinite integral domain cannot >be a field? All the books I have prove that a finite intergal domain is a >field but don't give any indication about why infinite domains can't be. For >instance, Z is not a field; I know why it is not; but does the same argument >hold for any infinite domain. >A hint about the proof would be welcome; for I am sure it is trivial. >Navin Kadambi So you think that there are no infinite fields ? Derek Holt. === Subject: topolgy question.... in the usual topology on R^2 let A = {(x,y} | x,y in rational, (x^2) + (y^2) < 1} -------------------- 1. Fr(A) = A , Fr is boundary 2. Int(A) = empty 3. C(A) = {(x,y} | x,y in R, (x^2) + (y^2) <= 1} ,C is closure 4. Fr(A^c) = (Fr(A)) all item(1,2,3,4) is right?? i think that problem 4 is wrong Fr(A^c) = (Fr(A)) <= book.... Fr(A^c) = (Fr(A))^c <= i think...... let me advice.....sir..... thank you... === Subject: Re: topolgy question.... > in the usual topology on R^2 > let A = {(x,y} | x,y in rational, (x^2) + (y^2) < 1} > -------------------- > 1. Fr(A) = A , Fr is boundary The bundary also includes points with x^2+y^2=1, and irrational points. > 2. Int(A) = empty OK > 3. C(A) = {(x,y} | x,y in R, (x^2) + (y^2) <= 1} ,C is closure OK. Boundary = closure minus interior, right? So now you can do the boundary. > 4. Fr(A^c) = (Fr(A)) no > all item(1,2,3,4) is right?? > i think that problem 4 is wrong > Fr(A^c) = (Fr(A)) <= book.... > Fr(A^c) = (Fr(A))^c <= i think...... no > let me advice.....sir..... > thank you... === Subject: Re: topolgy question.... > in the usual topology on R^2 > let A = {(x,y} | x,y in rational, (x^2) + (y^2) < 1} > -------------------- > 1. Fr(A) = A , Fr is boundary > 2. Int(A) = empty > 3. C(A) = {(x,y} | x,y in R, (x^2) + (y^2) <= 1} ,C is closure > 4. Fr(A^c) = (Fr(A)) > all item(1,2,3,4) is right?? No; the answer to 1. is wrong. Fr(A) = C(A) Int(A) = = {(x,y} | x,y in R, (x^2) + (y^2) <= 1}. > i think that problem 4 is wrong No; the answer is right. Jose Carlos Santos === Subject: Re: Hausdorff dimension of graph of Weierstrass function. Hunt, Brian R. The Hausdorff dimension of graphs of Weierstrass functions. Proc. Amer. Math. Soc. 126 (1998), no. 3, 791--800. === Subject: Re: Hausdorff dimension of graph of Weierstrass function. > Hunt, Brian R. > The Hausdorff dimension of graphs of Weierstrass functions. Proc. > Amer. Math. Soc. 126 (1998), no. 3, 791--800. G.C. === Subject: Re: Hausdorff dimension of graph of Weierstrass function. The lower bound of the Hausdorff dimension of the graph of f(x)=sum^{infty}_{i=1}t^{(s-2)i}sin(t^ix) For 11. is s. Is it equal to s? >If anyone is still interested > s >= HausDim(graph f) >= s - c/log(t) > Huh. GIve us a hint/sketch of how you prove that. >It is stated, and a proof method merely hinted at, on page 151 of >Kenneth Falconer Fractal Geometry Mathematical Foundations and >Applications. I'll quote: it - been dusting off more cobwebs: Of course one wouldn't try to get a lower bound on the Hausdorff dimension directly from the definition, one would use Frostman's lemma. Duh. >The known lower bounds come from mass distribution methods Like for example Frostman's lemma... >depending on estimates for L{t: (t, f(t)) in B} where B is >a disc and L is Lebesgue measure. That was the first thing I thought of - then I wondered if simply taking normalized arc length (or rather the limit of normalized arc length on the partial sums) would do just as well. >The rapid small-scale >oscillation of f ensures that the graph is inside B relatively >rarely, so that this measure is small. Precisely. >In this way it is >possible to show that there is a constant c such that > s >= dim_H graph y >= s - c/log(lambda) I was going to go for s. That's because lambda is what it's supposed to be. (Of course t's easier to type. But you can't use i for anything but sqrt(-1) when you're talking about Fourier series, even a sine series - it's just not right...) >so when lambda is large the Hausdorff dimension cannot be >much less than the conjectured value. ************************ David C. Ullrich === Subject: Re: Preparing for lessons, compiling teaching plans, arranging test papers, and writing scientific and technological documents are all handy with SciencePress. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1GDD6817009; I need suggestion. I would like to by Sciencepress software. I visited producer's site but I cannot read chenese language. Please inform me as soon as possible haw I can buy the full last version of this software. Marko Milos Belgrade Serbia&Montenegro (ex Yugoslavia) e-mail: mmarko@eunet.yu === Subject: Re: e is transcendental (was: classes of transcendental numbers ? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1GDD8X17098; >is sin(pi), so all you are saying is that cos(pi) = -1 and >sin(pi) = 0, and we were already aware of these facts. There is >NO reason to conclude that exp(i pi) = 0. > I,have a reason , with my due respect. > Panagiotis Stefanides >Yes, but you DON'T tell us what your reason is. You can't expect us to >accept your claims without giving support for those claims. So what >possible reason could you have for expecting us to agree with your claim >that exp(i pi) = 0? >The reason is simply that exp(ipi0=-1 should be accompanied >by the statement that this is the real part solution. >Is it fair? >No, that is not a fair comment. exp(i pi), the complex number, is >equal to -1, the complex number. There is no need to appeal to the >real part. Also, what does exp(i pi) = -1 is the real part solution >mean? You look like you are using terminology in a manner not >recognized in mathematics. >David McAnally >-------------- e^[i*pi] ,as accepted ,is a phasor. It is only fair to state that its polar representation is : e^[i*pi] = MOD 1 , ARG 180 . Panagiotis Stefanides http://www.stefanides.gr === Subject: Re: e is transcendental (was: classes of transcendental numbers ? >So? The real part of exp(i pi) is cos(pi), and its imaginary part >is sin(pi), so all you are saying is that cos(pi) = -1 and >sin(pi) = 0, and we were already aware of these facts. There is >NO reason to conclude that exp(i pi) = 0. > I,have a reason , with my due respect. > Panagiotis Stefanides >Yes, but you DON'T tell us what your reason is. You can't expect us to >accept your claims without giving support for those claims. So what >possible reason could you have for expecting us to agree with your claim >that exp(i pi) = 0? >The reason is simply that exp(ipi0=-1 should be accompanied >by the statement that this is the real part solution. >Is it fair? >No, that is not a fair comment. exp(i pi), the complex number, is >equal to -1, the complex number. There is no need to appeal to the >real part. Also, what does exp(i pi) = -1 is the real part solution >mean? You look like you are using terminology in a manner not >recognized in mathematics. >David McAnally >-------------- >e^[i*pi] ,as accepted ,is a phasor. In most of the relevant fields of mathematics, e^[i pi] is a complex number. >It is only fair to state that its >polar representation is : >e^[i*pi] = MOD 1 , ARG 180 . That is Arg 180 degrees, not just Arg 180. And so what? That does not lead to your assertion that exp[i pi] = 0, a result for which you have given absolutely no support. Why don't you just give up? David McAnally At the moment, they (the Time Lords) are far from being all-powerful. That's why it's been left up to me and me and me. quote by: Patrick Troughton in The Three Doctors ------- === Subject: Mathematical induction has anyone a good link to some exercises (with solutions !) for mathematical induction ? A PDF file would be great... I have so far only found pages with 1 or 2 exercises or some with execises but no solutions... === Subject: Wiener windowing Hi all. This is a question about signal processing. I have to multiply (in time domain) a signal with a window which should have a Wiener shape. I mean, the shape Wiener filters have in the Power Spectral Density plot. Knowing that the signal is s(t) = w(t)*exp(-t/(2tau))*u(t) (u(t) is the step signal and w(t) is a white gaussian noise), my window should compensate the exponential decay in order to obtain something more similar to w(t). Currently i'm using a x(t) = exp(t/2tau) window; however, because the signal is presented with an additive noise (say r(t) = s(t) + n(t), n(t) noise), I don't want of course to amplify noise samples when the signal has decayed a lot, after something like 3 or 4 tau. I've thought that a solution could be a wiener filtering. However I don't know how to build it in time domain, and how to transform a windowing into a filtering.... Is it possible or should I try another way? If it is possible, can you explain me how to build this filter? (I just need the main directions: i.e. how to set a least squares method, ecc...) If it is not possible, how can I do that in another way? Excuse me for posting this question here... :) capitan harlock === Subject: Re: Wiener windowing > Hi all. This is a question about signal processing. > I have to multiply (in time domain) a signal with a window which should have > a Wiener shape. I mean, the shape Wiener filters have in the Power Spectral > Density plot. > Knowing that the signal is s(t) = w(t)*exp(-t/(2tau))*u(t) (u(t) is the step > signal and > w(t) is a white gaussian noise), my window should compensate the exponential > decay in order to obtain something more similar to w(t). > Currently i'm using a x(t) = exp(t/2tau) window; however, because the signal > is presented with an additive noise (say r(t) = s(t) + n(t), n(t) noise), I > don't > want of course to amplify noise samples when the signal has decayed a lot, > after something like 3 or 4 tau. > I've thought that a solution could be a wiener filtering. However > I don't know how to build it in time domain, and how to transform a > windowing > into a filtering.... > Is it possible or should I try another way? > If it is possible, can you explain me how to build this filter? (I just need > the main directions: i.e. how to set a least squares method, ecc...) > If it is not possible, how can I do that in another way? > Excuse me for posting this question here... :) > capitan harlock You may get better answers in a DSP forum, but if just want suggestions about time-domain filtering (smoothing) take a look at Savitzky-Golay filters. 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 === Subject: Re: Got a speeding ticket and need to fight back X-NewsReader: GRn 3.2n February 9, 1999 > even by your arguement, > take > velocity measure dlamda/ lamda approx v/c ( dopper ) > c = 180 000 * 60*60 mph > 50 mph/ (180 000 * 60 * 60 mph), you know how small that is? > If you were to measure change in the wave length 'lamda', by the > photelectric effect or something, you need extremely precise instruments. > Measuring change in voltage or current of the order 50mph/(180 000 * 60 * > 60) is a painful and complicated process. Right. Dividing one number by another should be avoided at all costs if it taxes your abilities... On the other hand, the speed of light doesn't enter into the calculations in any way since the frequency shift is the quantiity to be measured so the preceding calculation is irrelevant. BTW: Your value for c is in error by more than 3%. If you let me select any random algorithm and choose values by whim, I can compute any value you want to any accuracy you want. [It's no wonder the courts are beginning to look at expert testimony with a rather jaundiced eye!] > In fact, even by this arguement, It can definitely be shown that analog > instruments that measure this level of precision would be extremely flakey, > if not built correctly. I presume you've never heard of DSP (digital signal processing). It's extremely unlikely in this day and age for analog processing to be used for something of the nature of Doppler radar. === Subject: Re: Got a speeding ticket and need to fight back monkey see, monkey do? But a monkey is just a monkey. first of all, if you dont know, one is not measuring the speed of light. Instead one is measuring 'lamda', more exactly the change in 'lamda'. Your argument is like this Suppose 1=0, let me show that 2+2 = 10 But,we are not even dealing with that! Learn math/science first. -suresh > even by your arguement, > take > velocity measure dlamda/ lamda approx v/c ( dopper ) > c = 180 000 * 60*60 mph > 50 mph/ (180 000 * 60 * 60 mph), you know how small that is? > If you were to measure change in the wave length 'lamda', by the > photelectric effect or something, you need extremely precise instruments. > Measuring change in voltage or current of the order 50mph/(180 000 * 60 * > 60) is a painful and complicated process. > Right. Dividing one number by another should be avoided at > all costs if it taxes your abilities... > On the other hand, the speed of light doesn't enter into the > calculations in any way since the frequency shift is the > quantiity to be measured so the preceding calculation is irrelevant. > BTW: Your value for c is in error by more than 3%. > If you let me select any random algorithm and choose > values by whim, I can compute any value you want to any > accuracy you want. [It's no wonder the courts are > beginning to look at expert testimony with a rather > jaundiced eye!] > In fact, even by this arguement, It can definitely be shown that analog > instruments that measure this level of precision would be extremely flakey, > if not built correctly. > I presume you've never heard of DSP (digital signal processing). > It's extremely unlikely in this day and age for analog processing > to be used for something of the nature of Doppler radar. === Subject: Re: Got a speeding ticket and need to fight back >Your argument is like this >Suppose 1=0, let me show that 2+2 = 10 Hey, I disagree. If 1 = 0 then surely 2+2 = 2 anyone can see that! === Subject: Re: What is a number?/What is not a number? 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 X-Sanguinate: themvsguy@email.com X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 01:39 PM, reany@asu.edu (Patrick Reany) said: >I'll offer this definition: A number is an element of a number set. A >number set is either the set of integers Z or any ring/field R that >contains Z. So you're excluding the Cayley Numbers? Elements of finite fields? Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: What is a number?/What is not a number? > at 01:39 PM, reany@asu.edu (Patrick Reany) said: >I'll offer this definition: A number is an element of a number set. A >number set is either the set of integers Z or any ring/field R that >contains Z. > So you're excluding the Cayley Numbers? Elements of finite fields? Yes. It was my intention to eliminate all finite fields or finite rings for that matter. Otherwise, it seems to me that the criteria for defining number would be so weak as to justify calling any element of any group, ring, field, module, or algebra a number. In which case, what does the term distinguish? Perhaps we could consider treating as -- vague and imprecise -- but tolerated just the same. Patrick === Subject: Re: What is a number?/What is not a number? > at 01:39 PM, reany@asu.edu (Patrick Reany) said: >I'll offer this definition: A number is an element of a number set. A >number set is either the set of integers Z or any ring/field R that >contains Z. So you're excluding the Cayley Numbers? Elements of finite fields? > Yes. It was my intention to eliminate all finite fields or finite > rings for that matter. Good, that is an opinion. However, you have to exclude more. Polynomials in one of more variables with integer coefficients for instance. They form a ring and contain Z. Unless you would call them numbers (and that is not common practice). I have no idea how you would exclude them. dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: What is a number?/What is not a number? 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 X-Sanguinate: themvsguy@email.com X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 10:39 PM, Carl Parkes said: >What is a number?/What is not a number? What would you like it to be? Like many other words, people apply it to unrelated things. Mathematics uses distinct names for things that you are subsuming into the vague term number. >As children understanding of numbers starts with the Naturals then >the Intergers, the Rationals, the Reals, the Complex and beyond. Hardly. The public schools are not interested in transmitting such an understanding. >Mathamatics later turns this all on its head No way, Jos.8e. In fact, it was Mathematicians who constructed[1] the progression. >All Computer Data & programs that currently Exist now can be >considered as on very large interger, this Interger is unknowen, >and tecnicaly impossable to calculate Incorrect. [1] Read that as discovered if you're a Platonist. Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Got a speeding ticket and need to fight back The Lord of Chaos (Suresh Devanathan) > The last i checked, I dictate the laws of physics. I am God, to many > people. Your court is beneath me. Your constitution is beneath me. In fact, > you cannot even prosecute me, because the constitution depends on me. I tell > what the laws are. And that's final. Not another case of Narcissistic Personality Disorder... === Subject: Re: Got a speeding ticket and need to fight back > The Lord of Chaos (Suresh Devanathan) > The last i checked, I dictate the laws of physics. I am God, to many > people. Your court is beneath me. Your constitution is beneath me. In fact, > you cannot even prosecute me, because the constitution depends on me. I tell > what the laws are. And that's final. > Not another case of Narcissistic Personality Disorder... I want to know where and when his court date is. If he keeps copping that same attitude in front of the judge, it could be fun to watch. Paul Hovnanian mailto:Paul@Hovnanian.com note to spammers: a Washington State resident -------------------------------------------------------------- ---- Life is like a buffet. Its not very good but there's plenty of it. === Subject: Re: Got a speeding ticket and need to fight back > The Lord of Chaos (Suresh Devanathan) > The last i checked, I dictate the laws of physics. I am God, to many > people. Your court is beneath me. Your constitution is beneath me. In fact, > you cannot even prosecute me, because the constitution depends on me. I tell > what the laws are. And that's final. > Not another case of Narcissistic Personality Disorder... >I want to know where and when his court date is. If he keeps copping >that same attitude in front of the judge, it could be fun to watch. Yes, if you like seeing people get jailtime! === Subject: Re: Got a speeding ticket and need to fight back > The Lord of Chaos (Suresh Devanathan) > The last i checked, I dictate the laws of physics. I am God, to many > people. Your court is beneath me. Your constitution is beneath me. In fact, > you cannot even prosecute me, because the constitution depends on me. I tell > what the laws are. And that's final. > Not another case of Narcissistic Personality Disorder... >I want to know where and when his court date is. If he keeps copping >that same attitude in front of the judge, it could be fun to watch. > Yes, if you like seeing people get jailtime! Heh heh. Would that be before, or after he gets laughed at? Rob === Subject: Re: Got a speeding ticket and need to fight back The Lord of Chaos (Suresh Devanathan) > That's translation of my name. Suresh = lord of the rain, lord of chaos,... > Kumar = Prince, handsome,.... > Devanathan = King of the Gods > rgrds, > Nothing in there about inteligence, huh? > That shows the level of your intelligence. You are definitely not the deep > thinker type. A Prince, is supposed to be intelligent. duh!!! I guess your name is more a sign of high hopes of the parents - which was then deeply disappointed... === Subject: Re: Got a speeding ticket and need to fight back > I guess your name is more a sign of high hopes of the parents - which > was then deeply disappointed... I didnt disppoint my mom and dad. But definitely, i know, it hurts you so much to know that there is atleast someone, who is not you, who didnt. -suresh === Subject: Re: Got a speeding ticket and need to fight back > The Lord of Chaos (Suresh Devanathan) > That's translation of my name. Suresh = lord of the rain, lord of chaos,... > Kumar = Prince, handsome,.... > Devanathan = King of the Gods > rgrds, Nothing in there about inteligence, huh? > That shows the level of your intelligence. You are definitely not the deep > thinker type. A Prince, is supposed to be intelligent. duh!!! > I guess your name is more a sign of high hopes of the parents - which > was then deeply disappointed... My means Lion-hearted Small Ruler of the House. Just right, except for the Small part. === Subject: Re: Got a speeding ticket and need to fight back The Lord of Chaos (Suresh Devanathan) > high school students who take statistics are smarter than you. Besides, > everything i am talking is at the high school level. Any high school can > understand what i am saying. You have got to be stupid. You dont understand > 'posulates'! you think they are signs of weakness. The Radar works based on > posulates. One of them is that speed of light is constant and there is such > a thing as wavelength, such a thing as velocity, etc. and such a thing as Doppler radar. > -suresh === Subject: Re: Got a speeding ticket and need to fight back May be, you know the reason why nobody builds anlog computers? Do i have to explain it to u? -suresh Approximately f lamda = c f lamda = 186 000 mps f -> frequency in hertz h = 60min > = 60 * 60 sec f lamda = 186 000 > = 186 000 mps f -> 1000,000,000 s^-1 (gigahertz) lamda = .000186 miles That's the radar's wavelength approximately. Take the radar's time > window, > it internally uses to compute the speed: 1/10 second ( the human eye > can > see about 30 frames / sec ). Assuming Reaction time of a driver = 1/10 > sec > approx 0.000186/ .1s = .00186 mps = 6.6 mph This has nothing to do with how these radar units actually work. Google > on 'Doppler'. And then understand that most traffic court judges have probably heard > them all. And thrown them all out of court. If you can't identify some > procedural error on the part of the officer, pay the fine. Amid all the noise associated with this posting, I am > wondering why nobody has bothered to question what > computational time and human reaction time have to do > with a Doppler velocity measurement. If your case is weak, try to baffle them with BS. > I will answer your question why reaction time is important > A) even for DSP analysis, you know there are going to be processing > delays. Your MP3 player for example, actually is always a few milli seconds > ahead, processing all the crappy songs you listen to. > Delay in processing does not make the data taken for the reading > any different nor does it change the computed results. > B) it gives a good accurate measure of spontaneous velocity. You point the > radar at some big stupid object. Internally, it(the object) is composed of > billions and billions of molecules. Even air, for example, could causes > interference to measurement. The way, you get rid of the error, is by > averaging, a lot of data, over a period of time. > Which has nothing to do with reaction time. > C) Syncrozning with the Speedometer reading. Speedometer is usally analog. > If you are not that blind, you will notice that needle that does the > reading, is actually bumping up and down, around a certain value. That is > because the car, is actually vibrating at a very high angular velocity. > I don't understand how bumping up and down has anything to do > with angular velocity nor reaction time. > However, by averaging out the measurement and by slowing the speedometer > needle down to your reaction time, you have a steady-state reading. > If things were not measured around your reaction time, you know the speed of > the car can be anywhere from a million 1,000,000,000 mph to 0 mph > Learn some science first, before you spew your crap. > I didn't state anything -- I merely asked a question. > To talk smart, you need to know how smart people talk. > I've been watching for an example from you. > Too bad, catch-22 for u. > According to some two-bit school on the Hill in Boulder, > I knew enough to sneak through their engineering program. > I must have slept through more of those EE classes than I > thought. I guess I should also refund the money I've > been paid me over the years to work on instrumentation > systems. === Subject: Re: Got a speeding ticket and need to fight back even by your arguement, take velocity measure dlamda/ lamda approx v/c ( dopper ) c = 180 000 * 60*60 mph 50 mph/ (180 000 * 60 * 60 mph), you know how small that is? If you were to measure change in the wave length 'lamda', by the photelectric effect or something, you need extremely precise instruments. Measuring change in voltage or current of the order 50mph/(180 000 * 60 * 60) is a painful and complicated process. In fact, even by this arguement, It can definitely be shown that analog instruments that measure this level of precision would be extremely flakey, if not built correctly. -suresh Approximately f lamda = c f lamda = 186 000 mps f -> frequency in hertz h = 60min > = 60 * 60 sec f lamda = 186 000 > = 186 000 mps f -> 1000,000,000 s^-1 (gigahertz) lamda = .000186 miles That's the radar's wavelength approximately. Take the radar's time > window, > it internally uses to compute the speed: 1/10 second ( the human eye > can > see about 30 frames / sec ). Assuming Reaction time of a driver = 1/10 > sec > approx 0.000186/ .1s = .00186 mps = 6.6 mph This has nothing to do with how these radar units actually work. Google > on 'Doppler'. And then understand that most traffic court judges have probably heard > them all. And thrown them all out of court. If you can't identify some > procedural error on the part of the officer, pay the fine. Amid all the noise associated with this posting, I am > wondering why nobody has bothered to question what > computational time and human reaction time have to do > with a Doppler velocity measurement. If your case is weak, try to baffle them with BS. > I will answer your question why reaction time is important > A) even for DSP analysis, you know there are going to be processing > delays. Your MP3 player for example, actually is always a few milli seconds > ahead, processing all the crappy songs you listen to. > Delay in processing does not make the data taken for the reading > any different nor does it change the computed results. > B) it gives a good accurate measure of spontaneous velocity. You point the > radar at some big stupid object. Internally, it(the object) is composed of > billions and billions of molecules. Even air, for example, could causes > interference to measurement. The way, you get rid of the error, is by > averaging, a lot of data, over a period of time. > Which has nothing to do with reaction time. > C) Syncrozning with the Speedometer reading. Speedometer is usally analog. > If you are not that blind, you will notice that needle that does the > reading, is actually bumping up and down, around a certain value. That is > because the car, is actually vibrating at a very high angular velocity. > I don't understand how bumping up and down has anything to do > with angular velocity nor reaction time. > However, by averaging out the measurement and by slowing the speedometer > needle down to your reaction time, you have a steady-state reading. > If things were not measured around your reaction time, you know the speed of > the car can be anywhere from a million 1,000,000,000 mph to 0 mph > Learn some science first, before you spew your crap. > I didn't state anything -- I merely asked a question. > To talk smart, you need to know how smart people talk. > I've been watching for an example from you. > Too bad, catch-22 for u. > According to some two-bit school on the Hill in Boulder, > I knew enough to sneak through their engineering program. > I must have slept through more of those EE classes than I > thought. I guess I should also refund the money I've > been paid me over the years to work on instrumentation > systems. === Subject: Re: Got a speeding ticket and need to fight back > Btw, a wonderful sign of your ignorance. Do you know how the droppler shift Speaking of ignorance, what is a 'droppler shift'? > is analysed? Paul Hovnanian mailto:Paul@Hovnanian.com note to spammers: a Washington State resident -------------------------------------------------------------- ---- Opinions stated herein are the sole property of the author. Standard disclaimers apply. All rights reserved. For external use only. If irritation, rash or swelling occurs, discontinue use immediately and consult a physician. Void where prohibited. === Subject: Re: Got a speeding ticket and need to fight back you are an idiot. Go away. -suresh > Btw, a wonderful sign of your ignorance. Do you know how the droppler shift > Speaking of ignorance, what is a 'droppler shift'? > is analysed? > -- > Paul Hovnanian mailto:Paul@Hovnanian.com > note to spammers: a Washington State resident > -------------------------------------------------------------- ---- > Opinions stated herein are the sole property of the author. Standard > disclaimers apply. All rights reserved. For external use only. If > irritation, rash or swelling occurs, discontinue use immediately > and consult a physician. Void where prohibited. === Subject: Re: Got a speeding ticket and need to fight back Devanathan) So all this was just an excuse to hurl some verbal abuse around. If that's really the case I hope you do try your stunt in the upcoming court appearance. >You should never build a computer. relevance? >Dont ever send your children to college for engineering. relevance? >They are most likely to drive the companies, where they work, >bankrupt. relevance? >If 15 out of every 100 instructions, a computer executes is in >error, you know something, the computer is no more than random number >generator. Exactly what do you know about physics and laboratory measurement techniques? I'm sure Penny will be happy to fill in any gaps in your knowledge when she has a decade or two to spare. >That's the restatement of your idiocy. You do not understand statistics. >Go away. You do not understand pecentages. You know nothing about them. >You do not even know how to use them. If you ever got a Ph.D in psychology, >remember that all you did was bull your way thru college. I never claimed to know volumes about mathematics. What I do know is when and how to apply the right tool to a situation. Statistics is not a tool you can apply in our court system. May you can bamboozle a judge or two in your home country, but you are going to have a difficult time doing that in MY country. >How about if i told u, that there is 15% chance you are going to die >tomorrow? Let's see: 85% chance that you wont die. Let's see the power of >the 15% chance. You know 3/20 people hate you, and think you are ugly. >Out of 20 times you drive, 3 times, the chances are you are going to hit >something. chance are that our of 3 of 20 people, in public, are going to >mug you. >-suresh How about if I tell you that you have a 100% chance of loosing your case, loosing your temper and mouthing off at the judge, getting yourself fined an additional $500 for contempt of court and then spending 24 hours in the drunktank ? Gosh I love these statistics. Can we do some more ? Later, Pepe le Pew aka Pat Sullivan === Subject: re:hyperbolas http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: hyperbolas > I would like to know how to find the gradient of the tangent at the > turning point of a hyperbola that doesn't have asymptotes going > straight up and straight across. Finding the turning point of a nice > simple hyperbola is easy since the gradient is 1 or -1, that's > obvious, but I'm not quite sure how to find them for other Turning point usually means a maximum or a minimum, where the gradient is zero. It looks as if you actually want the gradient at the vertices, i.e. the two points where the hyperbola gets closest to its centre. Is that right? If so, it's equal to the gradient of the conjugate axis. A hyperbola has two axes of symmetry, which are lines bisecting the angles between the asymptotes. The transverse axis is the one which cuts the hyperbola at its two vertices. The conjugate axis is perpendicular to that, and doesn't cut the hyperbola at all. It's parallel to the tangents at the vertices, so its gradient is what I think you're looking for. Any conic in the plane has a second-degree equation a(x^2) + 2hxy + b(y^2) + 2gx + 2fy + c = 0. If this is a hyperbola then ab < h^2. (You can check it in simple cases such as xy = 1, which has h = 1/2, c = -1, and all its other coefficients 0.) I shan't try to give proofs, but the asymptotes are in fact parallel to the line-pair through the origin which has equation a(x^2) + 2hxy + b(y^2) = 0. That factorizes into two first-degree factors giving equations of two lines parallel to the hyperbola's asymptotes. The bisectors of the angles between those two lines form another line-pair h(x^2) - (a - b)xy - h(y^2) = 0. This time the first-degree factors give lines parallel to the hyperbola's axes of symmetry. The gradient of one of those is your answer; so solve the quadratic equation -h(y/x)^2 -(a - b)(y/x) + h = 0 for the gradient y/x = (b - a +/- sqrt((b - a)^2 + 4(h^2))/(2h). Whether to take the plus or minus sign depends on the particular example, but anyway I think that's the formula you asked for. Whether you really want it now that you've got it is another question. :-) Ken Pledger. === Subject: Re: 'erf' function in C > Are there any C compilers which have the erf function (from > probability) as part of their math libraries? Or are there any math > libraries available to download which implement this function? > James Harlacher I have always considered the erf function a damned nuisance---not for its potential use, but for the way that it must be used, and the many complicated methods used to get what can otherwise be expressed by means of simple mathematics and simple programming that lets the CPU do the work for you. erf is often not available on particular systems, (for example on the gcc under DOS that I use), and may rely on obscure methods---fits by piecewise polynomials rational functions, continued fractions, Chebychev or asymptotic expansions for the tail---and have less that hoped-for accuracy. Examples are the widely used C programs based on Algorithm AS66 Applied Statistics v22, p 424 by Hill, from which you are lucky to get 7 digits of accuracy, or the ponderous 300-line code from Sun Microsystems, listed in one of the postings above. Virtually all calls for an erf(z) evaluation are likely to be for the normal probability distribution: Phi(x) = integral exp(-t^2/2), t = -infinity to x, which must be obtained via erf by means of Phi(x) = .5 + .5*erf(x/sqrt(2)), another source of irritation for erf (or rather Phi) users. A procedure that can determine Phi(x) nearly to the limit available in double precision can be based on the following method, which I developed the 1960's: Let phi(t) be the normal density, phi(t) = exp(-t^2/2)/sqrt(2*Pi), so that Phi(x) = integral phi(t), t = -infinity to x. and cPhi(x) = 1-Phi(x) = integral phi(t), t = x to infinity. If you define R(x) = cPhi(x)/phi(x) then the function R is a well behaved function that has a rapidly convergent Taylor expansion, an expansion with terms easily determined by a two-step recursion. (Try to derive the recursion yourself: r[k+1](x)=x*r[k](x)+k*r[k-1](x), where r[k](x) is the kth derivative of R(x).) Furthermore, the convergence is fast ---or at least accurate---enough that the Taylor expansion about x = 0 can be used to determine Phi(x) to some 15-16 decimal places even for x as large as 6 or 7---farther than most applications are likely to require. So, a few lines of code and no tables are all one needs for a very accurate value of Phi(x), the normal probability distribution: ------------------------------------------------------ #include #include double Phi(double x) {long double s,t=0,a=1.2533141373155L,z=0,b=-1,pwr=1; int i; s=a+b*x; for(i=2;s!=t;i+=2) { a=(a+z*b)/i; b=(b+z*a)/(i+1); pwr=pwr*x*x; t=s; s+=pwr*(a+x*b); } return 1.-s*exp(-.5*x*x-.91893853320467274178L); } int main(){ double x; while(1){ printf(Enter your x value:); scanf(%lf,&x); printf(Normal Prob(X<%f)=%20.16fn,x,Phi(x)); } } /* An indication of the accuracy of this Phi function is given by the following examples, with the true value (to 20 places) given under the resulting Phi(x): x Phi(x) 0.123 0.5489464510164369 .54894645101643675909 1.2 0.8849303297782918 .88493032977829173198 2.4 0.9918024640754040 .99180246407540387055 6.1 0.9999999994696578 .99999999946965767370 -6.1 0.0000000005303427 .00000000053034232638 -1.1 0.1356660609463828 .13566606094638267517 7.2 0.9999999999996990 .99999999999969893721 See what your system gives and compare with an erf function if it has one. */ ---------------------------------------------------------- Cut, paste and try. You may like it. George Marsaglia (Note: With long doubles I hope to invoke the 80-bit floating point processor, providing a little more accuracy that the IEEE 64-bit floating point standard. Does your system provide similar accuracy? I use gcc under DOS on Intel machines with W98, Me or XP.) === Subject: Re: 'erf' function in C > Are there any C compilers which have the erf function (from > probability) as part of their math libraries? Or are there any math > libraries available to download which implement this function? > James Harlacher > I have always considered the erf function > a damned nuisance---not for its potential use, but > for the way that it must be used, and the many complicated > methods used to get what can otherwise be expressed > by means of simple mathematics and simple programming > that lets the CPU do the work for you. ... > A procedure that can determine Phi(x) nearly to the limit > available in double precision can be based on the following > method, which I developed the 1960's: ... I once did this (with a small lookup table as you had it in earlier versions) for Excel. Even in that poor setting it worked well (and with some stupid language tricks one could go there for 18 digits). === Subject: Re: 'erf' function in C > I have always considered the erf function > a damned nuisance---not for its potential use, but > for the way that it must be used, and the many complicated > methods used to get what can otherwise be expressed > by means of simple mathematics and simple programming > that lets the CPU do the work for you. > erf is often not available on particular systems, > (for example on the gcc under DOS that I use), and may > rely on obscure methods---fits by piecewise polynomials > rational functions, continued fractions, Obscure? It's hard to find a math function that doesn't use at least one of these techniques. And why should you care, if it gets the right answer with reasonable speed? > Chebychev > or asymptotic expansions for the tail--- Ah, now you're thinking about erfc. erf saturates quickly for values away from zero, so it has no tail problems. > and have > less that hoped-for accuracy. Now *that's* a legitimate gripe. We find accuracies all over the map for even the commonest math functions. For erf and its exotic brethren, the errors are often astonishing. > Examples are the widely used C programs based on > Algorithm AS66 Applied Statistics v22, p 424 by Hill, > from which you are lucky to get 7 digits of accuracy, Indeed. We've found a rich literature of mediocre algorithms for math functions. Sadly, these algorithms are widely used. > or the ponderous 300-line code from Sun Microsystems, > listed in one of the postings above. That ponderousness is mostly about getting the erfc tail right for positive x, which is very difficult to do. The erf part doesn't use it. In fairness to Sun, the function is pretty damned accurate. P.J. Plauger Dinkumware, Ltd. http://www.dinkumware.com === Subject: Re: 'erf' function in C > Are there any C compilers which have the erf function (from > probability) as part of their math libraries? Or are there any math > libraries available to download which implement this function? > James Harlacher [...] >A procedure that can determine Phi(x) nearly to the limit >available in double precision can be based on the following >method, which I developed the 1960's: [...] >/* An indication of the accuracy of this Phi function > is given by the following examples, with the true > value (to 20 places) given under the resulting Phi(x): > x Phi(x) > 0.123 0.5489464510164369 > .54894645101643675909 > 1.2 0.8849303297782918 > .88493032977829173198 > 2.4 0.9918024640754040 > .99180246407540387055 > 6.1 0.9999999994696578 > .99999999946965767370 > -6.1 0.0000000005303427 > .00000000053034232638 > -1.1 0.1356660609463828 > .13566606094638267517 > 7.2 0.9999999999996990 > .99999999999969893721 >See what your system gives and compare with >an erf function if it has one. >---------------------------------------------------------- >Cut, paste and try. You may like it. >George Marsaglia [...] Is there someplace a table available of the true values of the normal distribution up to 20 places? Matthias Kl.8ay www.kcc.ch === Subject: Re: 'erf' function in C > Is there someplace a table available of the true values of the > normal distribution up to 20 places? Let me share one of my favorite hacks with you! Go to integrals.wolfram.com, and submit something like: Erf[0.3`200] (0.3`200 means 0.3, with a precision of 200 digits) You will get back the integral of your input (as a picture, unfortunately), which is just x multiplied by the constant you're after of course; Mathematica is pretty good at getting the precision right. Sidney === Subject: Re: 'erf' function in C Is there someplace a table available of the \ true values of the > normal distribution up to 20 places? > integrals.wolfram.com, and submit something like: > Erf[0.3`200] > (0.3`200 means 0.3, with a precision of 200 digits) > You will get back the integral of your input (as a picture, > unfortunately), Let me share an even better hack with you. Open the picture in a new browser window and change the format=StandardForm part of the URL to format=OutputForm. You'll get the result as text. > which is just x multiplied by the constant you're after of course; > Mathematica is pretty good at getting the precision right. Actually, unless my understanding of the term precision is completely wrong, Mathematica seems to be pretty /bad/ at getting the precision right: It returns a number with only 199 digits after the decimal point! OTOH, the Emacs calculator gets the number /and/ the precision right. :) Martin === Subject: [OT] Re: 'erf' function in C > Let me share an even better hack with you. Open the picture in a new > browser window and change the format=StandardForm part of the URL to > format=OutputForm. You'll get the result as text. Nice one...! To the OP: assuming you're using a Unix-like OS, this means something like curl http://integrals.wolfram.com/ graphics.cgi?format=OutputForm&expr=Erf[0.3`200] ...can be used in a script to generate your table, with a bit of scripting magic. >which is just x multiplied by the constant you're after of course; >Mathematica is pretty good at getting the precision right. > Actually, unless my understanding of the term precision is completely > wrong, Mathematica seems to be pretty /bad/ at getting the precision > right: It returns a number with only 199 digits after the decimal point! This can be because of two things. Mathematica distinguishes precision (total number of significant digits) from accuracy (number of significant digits after the decimal point), so the Erf argument is already given with an accuracy of 199 digits. Alternatively, the result of an unary operation doesn't necessarily have the same precision as the argument, e.g. what is sin(x), with x=1.00000E20 ? > OTOH, the Emacs calculator gets the number /and/ the precision right. :) There's people in the world that use web-browsers but don't use emacs... I kid you not. :-) === Subject: how to create an index I have a dataset of merger cases. I know start date and end date of the merger and the profit that can be made in trading the target company during the merger-period. Now I'd like to create an index which reflects the profit that can be made by investing into companies being acquired. What is the best way to do this given the information mentioned above? any help is appreciated Nicolas === Subject: Need help with proof! Show that for arbitrary natural numbers m and n, such that 1<=m<=n the following holds: the product n(n-1)(n-2) ...(n-m+2)(n-m+1) is dividable by m! === Subject: Re: Need help with proof! > Show that for arbitrary natural numbers m and n, such that 1<=m<=n the > following holds: > the product n(n-1)(n-2) ...(n-m+2)(n-m+1) is dividable by m! Prove that n(n-1)(n-2) ...(n-m+2)(n-m+1)/m! is the number of ways to do something, and conclude that it is an integer. Alternate. Prove it by induction on n. === Subject: Re: Need help with proof! > Show that for arbitrary natural numbers m and n, such that 1<=m<=n the > following holds: > the product n(n-1)(n-2) ...(n-m+2)(n-m+1) is dividable by m! > Prove that n(n-1)(n-2) ...(n-m+2)(n-m+1)/m! > is the number of ways to do something, and conclude > that it is an integer. > Alternate. Prove it by induction on n. See Pascal's Triangle ... (if one row is made up of integers, so is the next) === Subject: Re: Need help with proof! > Show that for arbitrary natural numbers m and n, such that 1<=m<=n the > following holds: > the product n(n-1)(n-2) ...(n-m+2)(n-m+1) is dividable by m! The number n(n - 1)(n - 2)...(n - m + 2)(n - m + 1) is the number of sequences with m distinct elements of a set with n objects. Now, choose a sequence with m elements and consider all the sequences that you get by changing the order of its elements; at the end, you'll get m! sequences. Now take some sequence that you had not obtained before and do the same thing. Keep doing this until there are no more sequences. Then you have proved that the set of all sequences can be devided into subsets with m! elements each, and therefore m! divides the number of elements of the set of all sequences. Of course, if this is homework, as I think it is, this will not be the answer that your teacher wants. :-) Jose Carlos Santos === Subject: Re: plotting n-dimensional data points on a 2d screen > Perhaps the OP should start by listing which properties of the data > set are important to him. (such as distances between images must be > proportional to distances in reality, etc.) The purpose of the n-dimensional plot is for a cellular-automata type system, and so I want to see patterns, groupings, novel behaviors, etc. Therefore, the information I want preserved is local relationships. === Subject: A matrix Hi As far as I　know, we can regard a nx1 vector as a point in n dimensional space, then how about nxn matrix? What is it? === Subject: Re: A matrix > Hi As far as I　know, we can regard a nx1 vector as a point in n > dimensional space, then how about nxn matrix? What is it? n points in an n-dimensional space === Subject: Re: A matrix > Hi As far as I　know, we can regard a nx1 vector as a point in n > dimensional space, then how about nxn matrix? What is it? It is a point in an n^2 dimensional space. But you can do it in more than one way since there is no obvious way of listing the coordinates. === Subject: Re: A matrix > Hi As far as I　know, we can regard a nx1 vector as a point in n > dimensional space, then how about nxn matrix? What is it? a linear transformation of n-dimensional space to itself. === Subject: Re: A matrix > Hi As far as I　know, we can regard a nx1 vector as a point in n > dimensional space, then how about nxn matrix? What is it? > a linear transformation of n-dimensional space to itself To be even more confusing: A basis-specific representation of a linear transformation of one n-dimensional space to another. Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs > Again (as in my other reply), here is the new-and-improved statement > of the problem,...(!) > but this time with an example: > Official (as of now) version: > Let us say we have 2 spherical dice which somehow each represent all > positive integers, each integer represented exactly once per die, on > their countably-infinite number of sides. And each integer is equally > likely to be rolled on both dice. > So, the dice-pair are rolled repeatedly *until* (if ever) > on the m_th roll, > GCD(k,j) = m, > where j and k are the integers of the dice. > What is the probability that we will ever stop rolling the dice > (assuming we are eternal and need to keep rolling until > GCD(j,k) =m on the m_th roll)? > For example: > roll 1: j=3; k = 1026; GCD(3,1026) = 3, not 1. > roll 2: 25 ; 23534545; GCD = 5, not 2. > roll 3: 63 ; 2121; GCD = 21, not 3. > roll 4: 36 ; 20; GCD= 4, which DOES = 4. > So we CAN stop rolling now! > But we will not necessarily be able to ever stop! > By the way, as I calculate, the probability is *between* 0 and 1... > Leroy Quet As pointed out, this has no mathematical meaning. But it can be repaired. Take a large integer M. Use two dice with M (equally likely) sides numbered 1, 2, ..., M. Compute the probability that gcd=m. Take the limit as M -> infinity. You can find the case m=1 in the literature, here LQ is asking about general m. G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >Again (as in my other reply), here is the new-and-improved statement >of the problem,...(!) >but this time with an example: >Official (as of now) version: >Let us say we have 2 spherical dice which somehow each represent all >positive integers, each integer represented exactly once per die, on >their countably-infinite number of sides. And each integer is equally >likely to be rolled on both dice. A uniform distribution on the integers is not possible. Matthew T. Russotto mrussotto@speakeasy.net Extremism in defense of liberty is no vice, and moderation in pursuit of justice is no virtue. But extreme restriction of liberty in pursuit of a modicum of security is a very expensive vice. === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >Again (as in my other reply), here is the new-and-improved statement >of the problem,...(!) >but this time with an example: >Official (as of now) version: >Let us say we have 2 spherical dice which somehow each represent all >positive integers, each integer represented exactly once per die, on >their countably-infinite number of sides. And each integer is equally >likely to be rolled on both dice. >A uniform distribution on the integers is not possible. Just wondering... would this constitute a valid proof of that statement: The probability of picking any one integer from the set of all integers is zero. Therefore you can't pick one at random. Therefore there is no uniform distribution on the integers. Or does it need some boning up with rigorous mathspeak? Patrick Hamlyn posting from Perth, Western Australia Windsurfing capital of the Southern Hemisphere Moderator: polyforms group (polyforms-subscribe@egroups.com) === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs > Again (as in my other reply), here is the new-and-improved statement > of the problem,...(!) > but this time with an example: > Official (as of now) version: > Let us say we have 2 spherical dice which somehow each represent all > positive integers, each integer represented exactly once per die, on > their countably-infinite number of sides. And each integer is equally > likely to be rolled on both dice. > So, the dice-pair are rolled repeatedly *until* (if ever) > on the m_th roll, > GCD(k,j) = m, > where j and k are the integers of the dice. > What is the probability that we will ever stop rolling the dice > (assuming we are eternal and need to keep rolling until > GCD(j,k) =m on the m_th roll)? > For example: > roll 1: j=3; k = 1026; GCD(3,1026) = 3, not 1. > roll 2: 25 ; 23534545; GCD = 5, not 2. > roll 3: 63 ; 2121; GCD = 21, not 3. > roll 4: 36 ; 20; GCD= 4, which DOES = 4. > So we CAN stop rolling now! > But we will not necessarily be able to ever stop! > By the way, as I calculate, the probability is *between* 0 and 1... > Leroy Quet I SHOULD have mentioned, of course, that the dice are *supernatural*, and so allow us to mysteriously pick integers uniformly from the positive integer domain... But, yes, we can consider the positive integers to both be <= n, where n -> infinity. In any case, the probability of (on any particular roll) rolling k and j such that GCD(j,k) equal m is 6/(pi^2 m^2), as n -> oo. So, Rob Johnson has gotten the same solution as I: We found the probability of the dice-rolling can eventually be stopped is: 1 - sin(sqrt(6))/sqrt(6), where the angle is in radians. (I find this result, although simple to find, fascinating, given it involves a trig-function of an angle not a rational multiple of pi {nor even a rational multiple of 1}.) === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs > Again (as in my other reply), here is the new-and-improved statement > of the problem,...(!) > but this time with an example: > Official (as of now) version: > Let us say we have 2 spherical dice which somehow each represent all > positive integers, each integer represented exactly once per die, on > their countably-infinite number of sides. And each integer is equally > likely to be rolled on both dice. > So, the dice-pair are rolled repeatedly *until* (if ever) > on the m_th roll, > GCD(k,j) = m, > where j and k are the integers of the dice. > What is the probability that we will ever stop rolling the dice > (assuming we are eternal and need to keep rolling until > GCD(j,k) =m on the m_th roll)? > For example: > roll 1: j=3; k = 1026; GCD(3,1026) = 3, not 1. > roll 2: 25 ; 23534545; GCD = 5, not 2. > roll 3: 63 ; 2121; GCD = 21, not 3. > roll 4: 36 ; 20; GCD= 4, which DOES = 4. > So we CAN stop rolling now! > But we will not necessarily be able to ever stop! > By the way, as I calculate, the probability is *between* 0 and 1... > Leroy Quet > I SHOULD have mentioned, of course, that the dice are *supernatural*, > and so allow us to mysteriously pick integers uniformly from the > positive integer domain... > But, yes, we can consider the positive integers to both be <= n, > where n -> infinity. > In any case, the probability of (on any particular roll) > rolling k and j such that > GCD(j,k) equal m is > 6/(pi^2 m^2), > as n -> oo. > So, Rob Johnson has gotten the same solution as I: > We found the probability of the dice-rolling can eventually be stopped > is: > 1 - sin(sqrt(6))/sqrt(6), > where the angle is in radians. > (I find this result, although simple to find, fascinating, given it > involves a trig-function of an angle not a rational multiple of pi > {nor even a rational multiple of 1}.) > Leroy Quet My justification for assuming that the dice had infinite number of sides is similar to how some people might justify saying that '1/2 of ALL integers are even'. More rigorously, the number of positive integers <= n which are even is: floor(n/2). And as n -> oo, floor(n/2)/n -> 1/2. (And I am not the first to *reluctantly* {by the way} talk about the probability that '2 uniformly-picked positive integers, chosen from among ALL positive integers, are coprime', or, in this puzzle's case, have a specific GCD.) (You know what I am probably talking about here: the 'probablility is 6/pi^2 that the 2 integers will be coprime'. Of course, in this situation, they{/we} are talking REALLY about the limit as n -> oo, and the 2 integers are both <= n.) But in any case, you can assume the dice are just able to uniformly choose at random from all integers <= n, where n is an arbitrarily huge *finite* integer. :) :/ Anyway, the point of the puzzle was not the infinite dice. I just found the closed-form of the probability (of whatever I was trying to find the probability of, exactly) interesting. PS - At least one of my posts gets some replies, FINALLY!... === Subject: Entering *Real* Math Formulae in Excel Maybe I've missed something obvious but I've now blown a day on this ... How the &*&%$ does one enter a real math formula (simple example: the sum of all N sub x as x goes from x1 to x2) ???! I have the data that I want to operate on. I know some basic university calculus including differentials and integration. I can formualte these simple summation functions on paper ... But then how are they entered into and operated on in Excel??? Am I perhaps using the wrong program? (i.e. Should I be using something like MathCad?) Do I need to learn Excel arrays or other such extra baggage? === Subject: Re: Entering *Real* Math Formulae in Excel > How the &*&%$ does one enter a real math formula (simple example: the > sum of all N sub x as x goes from x1 to x2) ???! I can't imagine doing real math in Excel, but doesn't sum(n23:n97) produce the value n23 + n24 + ... + n97? If it doesn't, something like it does, and it should be possible to find out within Excel because it has some kind of Help thing built in & you can just look for sum and it should tell you how to use it. Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Entering *Real* Math Formulae in Excel Wilhelm wibbled: > Maybe I've missed something obvious but I've now blown a day on this > ... > How the &*&%$ does one enter a real math formula (simple example: the > sum of all N sub x as x goes from x1 to x2) ???! > I have the data that I want to operate on. I know some basic > university calculus including differentials and integration. I can > formualte these simple summation functions on paper ... > But then how are they entered into and operated on in Excel??? Excel is the equivalent of the back of an envelope, with a small calculator attached. It is quite hard to set it up to do what you want. > Am I perhaps using the wrong program? (i.e. Should I be using > something like MathCad?) This would be much better, yes. Have a look at Matlab, too. > Do I need to learn Excel arrays or other such extra baggage? Where oh where is my 32? === Subject: Re: Entering *Real* Math Formulae in Excel If by 'real' you mean algebraic then yes you are using the wrong program. Try Maple, MathCad, or Mathematica. Why you should call arrays 'baggage' is another matter. Have you had a bad day? Bernard Liengme www.stfx.ca/people/bliengme remove CAPS in e-mail address > Maybe I've missed something obvious but I've now blown a day on this > ... > How the &*&%$ does one enter a real math formula (simple example: the > sum of all N sub x as x goes from x1 to x2) ???! > I have the data that I want to operate on. I know some basic > university calculus including differentials and integration. I can > formualte these simple summation functions on paper ... > But then how are they entered into and operated on in Excel??? > Am I perhaps using the wrong program? (i.e. Should I be using > something like MathCad?) > Do I need to learn Excel arrays or other such extra baggage? === Subject: Re: easy...analysis....simple confirm... > let function f : R->R and g : R->R is continuous. > let f(r) = g(r) for all r in rational. > show that f(x) = g(x) > ---------------------------------- > this is my solving process > first......method...... > any r in rational, > if for each e>0 there is a d, > |x-r| |f(x)-f(r)| thus > |{f(x)-g(x)} - {f(r)-g(r)}| > =|{f(x)-f(r)} - {g(x)-g(r)}| > <=|f(x)-f(r)| + |g(x)-g(r)| > thus f(x)-g(x) = f(r)-g(r) = 0 if this were correct, then you could prove by this method that f(x)=f(r)!! > thus f(x) = g(x) not exactly... first of all, you want to prove that f(x)=g(x), for all x in R, so you beginn by: for any x in R then you want to prove it by using the continuity of f and g. so first you have to find rationals that are close to x. so you construct a sequence of rationals that converges to x: let r_n, n=1,2,.. (in Q) be a sequence s.t. for every d>0 there is an N s.t. for every n>N |x-r_n| and....second method..... > let h(x) = f(x) - g(x) > thus h(x) is continuous. > suppose, there is a c in irrational such that f(c) != g(c). > because h(x) is continuous, > for each e>0, |x-c| |h(x)-h(c)| let e = |h(c)|/2 > if x is rational, |h(x)-h(c)| = |h(c)| < |h(c)|/2 > thus contradiction. > thus for x in R, f(x)=g(x) > ----------------------------- > um........my two process is right?? > please....let me point out my errors...... === Subject: Re: easy...analysis....simple confirm... > let function f : R->R and g : R->R is continuous. > let f(r) = g(r) for all r in rational. > show that f(x) = g(x) The easy way is to note that (1) every real is the limit of a sequence of rationals (2) continuous functions commute with limits in the sense that if lim_{n->oo} x_n = L and h is continuous at L then lim_{n->oo} h(x_n) =h(lim_{n->oo} x_n) = h(L) === Subject: Expected mean and VAR with DFT I'm trying to calculate the expected mean and variance of both the real and imaginary part of a DFT with real values in input (I have only 0's and 1's in input). I found this page http://mathworld.wolfram.com/FourierTransform.html (formula 37 and 38) which should be good for me, but unfortunately my math skill is poor. Please, could someone explains the meaning of those formulas? Cristiano === Subject: Re: 3 Squares Covering 1 Circle > Yet another coverings-problem... > (I myself might have already asked this, but I do not believe that I, > at least, have asked this specific question before.) > If we have a unit circle (radius = 1), > and we have 3 unit squares (sides =1), > then what is the maximum area of the circle we can ever cover with the > squares > a) if overlapping of the squares is not allowed? > b) if overlap is allowed? > I may be way off, > but I conjecture that the (a) case is > squares arranged like: > ____ > ! ! > ! ! > --------- > ! ! ! > ! ! ! > --------- I find that the maximal amount of the circle that can covered in the above configuration is pi/6 - 1/2 + sqrt(3)/4 + sqrt(55)/8 + Arccos(3/8), which is about 2.57003584. This is achieved by letting the center of the circle lie on the boundary between the two lower squares, at a distance 1/2 - sqrt(55)/20 (~= 0.12919) below the boundary between the upper and lower squares. -Jim Ferry === Subject: Re: 3 Squares Covering 1 Circle For b, what does a C3v (invariant under 120 deg rotation) symmetric arrangement yield? (Place the outermost edge of the squares tangent to the circle.) Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de als man ankam wollte man werden, die geschichte schreiben, die doofen sollen sterben, der plan als man damals nach hamburg kam (Kettcar) === Subject: Re: 3 Squares Covering 1 Circle > For b, what does a C3v (invariant under 120 deg rotation) > symmetric arrangement yield? (Place the outermost edge of > the squares tangent to the circle.) The amount of the circle this covers is (pi + sqrt(3))/2, which is about 2.43682. Each of the squares covers an area sqrt(3)/4 + pi/6 ~= 0.956611, and the overlap area may be partitioned into 6 right triangles, each of area sqrt(3)/24. This shape, therefore, is not optimal -- compare to the result for part(a) which is slightly more than 2.57. -Jim Ferry === Subject: Re: 3 Squares Covering 1 Circle > This shape, therefore, is not optimal -- compare to the > result for part(a) which is slightly more than 2.57. ... namely about 2.57003584. Rainer Rosenthal r.rosenthal@web.de === Subject: Re: 3 Squares Covering 1 Circle > For b, what does a C3v (invariant under 120 deg rotation) > symmetric arrangement yield? (Place the outermost edge of > the squares tangent to the circle.) Here's a similar situation that does better than the 80.4% with part (a). Use this 120 deg symmetry, but place the diagonals of the square along the radii. The optimum placement of the squares (they can be shifted along the radii) covers 83.0% of the circle's area. This is the best that I've found, but I don't know if it's the best possible. === Subject: Re: 3 Squares Covering 1 Circle > || Yet another coverings-problem... > || (I myself might have already asked this, but I do not believe that I, > || at least, have asked this specific question before.) > || > || If we have a unit circle (radius = 1), > || > || and we have 3 unit squares (sides =1), > || > || then what is the maximum area of the circle we can ever cover with > || the squares > || > || a) if overlapping of the squares is not allowed? > || > || b) if overlap is allowed? > || > || > || I may be way off, > || but I conjecture that the (a) case is > || > || squares arranged like: > || > || ____ > || ! ! > || ! ! > || --------- > || ! ! ! > || ! ! ! > || --------- > || > || But I bet tilting the squares works better. > || > || As for the (b) case, > || > || if we start with the (a) solution I give, then raise the bottom > || squares (or lower the top square by the same amount), then tilt the > || squares somewhat somehow, > || > || PERHAPS we might have an arrangement which maximally covers a unit > || circle placed properly on top of the squares. > || > || Any better ideas or arrangements? (or more precise descriptions of > || possible/actual ideal arrangement(s)?...) > || > || > || Leroy Quet > | Sorry, accidentally posted message prematurely. > | Area of circle is [pi], area of 3 squares is is 3. Maximum possible > | area is 3/[pi]. > Well yes the theoretical maximum is 3/pi assuming we can get all of the > area of the squares within the circle. But we clearly can't, I think the > problem means what is the maximum area of the circle we *can* cover. Yes, of course we cannot get 3/pi of the circle covered, because squares are square... :) === Subject: convergence of measurable functions Hi all, Let X be a measurable space (a set with a sigma-algebra of sets) and B a general complete metric space, that is, not necessarily separable or compact or whatever. Put on B, the measurable structure generated by the open sets of B (the Borel structure). Now, let f_{n}:Xlongrightarrow B be a sequence of measurable functions converging pointwise to some f. Question: Is f measurable? Proof? G. Rodrigues === Subject: Re: convergence of measurable functions > Hi all, > Let X be a measurable space (a set with a sigma-algebra of sets) and > B a general complete metric space, that is, not necessarily separable > or compact or whatever. Put on B, the measurable structure generated > by the open sets of B (the Borel structure). Now, let > f_{n}:Xlongrightarrow B be a sequence of measurable functions > converging pointwise to some f. > Question: Is f measurable? Proof? > G. Rodrigues How about this. Let F be the sigma-algebra on X. To show f is measurable, we must show: if U is open in B, then f^{-1}(U) in F. Fix U. Define phi : X -> R by phi(b) = dist(b,XU). Then phi is continuous. And b in U iff phi(b)>0. All f_n are F-measurable, so phi(f_n(.)) is F-measurable X -> R. phi is continuous, so phi(f_n(.)) -> phi(f(.)) pointwise. Therefore (by the real-valued version of this theorem) phi(f(.)) is F-measurable, so U = inverse image of open set (0,infinity) is in F. === Subject: Re: JSH: What it would take by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1GK1rI16780; > off you stupid dumb. >Yes I'm back to cursing at you Nora Baron. > OFF!!! They are *incapable* ? Do you think that they are *incapable* ? >James Harris Remember James Aaron Ramanujan Harris, you consider them *incapable* ! *Incapable* ! Maurice === Subject: parabola y^2=4px intercepted by the line x=a by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1GK1sR16785; I came across this problem the other day. I actually found it on the math forum. No one replied to this question that someone else posted, and I am stumped. I am wondering how a person would do this problem. Anyone care to try? Find the dimensions of the rectangle of maximum area A that can be inscribed in the portion of the parabola y^2=4px intercepted by the line x=a. -Derrick === Subject: Re: parabola y^2=4px intercepted by the line x=a > I came across this problem the other day. I actually found it on the math forum. No one replied to this question that someone else posted, and I am stumped. I am wondering how a person would do this problem. Anyone care to try? > Find the dimensions of the rectangle of maximum area A that can be > inscribed in the portion of the parabola y^2=4px intercepted by the line x=a. Seems not so hard... Nobody was interested in ? Or thought it was homework ? The sides of the rectangle are 2y and a-x = a-(y^2)/4p. Its area is A = 2y(a-(y^2)/4p)= 2ay-(y^3)/2p The derivative is =0 when y^2 = 4ap/3, that is x = a/3. And the rectangle has dimensions 2a/3 and 2*sqrt(ap/3). philippe (chephip at free dot fr) === Subject: Re: parabola y^2=4px intercepted by the line x=a > I came across this problem the other day. I actually found it on the math forum. No one replied to this question that someone else posted, and I am stumped. I am wondering how a person would do this problem. Anyone care to try? > Find the dimensions of the rectangle of maximum area A that can be > inscribed in the portion of the parabola y^2=4px intercepted by the line x=a. > Seems not so hard... > Nobody was interested in ? Or thought it was homework ? > The sides of the rectangle are 2y and a-x = a-(y^2)/4p. > Its area is A = 2y(a-(y^2)/4p)= 2ay-(y^3)/2p > The derivative is =0 when > y^2 = 4ap/3, that is x = a/3. > And the rectangle has dimensions 2a/3 and 2*sqrt(ap/3). You assume that one side is a segment of x = a. Can you prove that that is optimum? G.C. === Subject: Re: parabola y^2=4px intercepted by the line x=a > I came across this problem the other day. I actually found it on the math > forum. No one replied to this question that someone else posted, and I am > stumped. I am wondering how a person would do this problem. Anyone care to > try? > Find the dimensions of the rectangle of maximum area A that can be > inscribed in the portion of the parabola y^2=4px intercepted by the line > x=a. > Seems not so hard... > Nobody was interested in ? Or thought it was homework ? > .... Or perhaps realized that not all rectangles have horizontal and vertical sides? The maximum area may indeed be achieved by such a rectangle, but how can you first eliminate those oriented in other directions? I suspect a geometrical argument might do it, but otherwise some quite fiddly calculations would be needed. Ken Pledger. === Subject: Re: parabola y^2=4px intercepted by the line x=a Derrick: > Find the dimensions of the rectangle of maximum area A that can be > inscribed in the portion of the parabola y^2=4px intercepted by the line x=a. If we assume that the line x=a is one side of the rectangle, then we can inscribe a rectangle of area 2(a-x)sqrt(4px) for any x in (0,a). You can use calculus to maximize this area or, equally good, maximize the square of the area, which is a cubic function with no sqrt appearing in it. === Subject: Re: parabola y^2=4px intercepted by the line x=a > Derrick: > Find the dimensions of the rectangle of maximum area A that can be > inscribed in the portion of the parabola y^2=4px intercepted by the line > x=a. > If we assume that the line x=a is one side of the rectangle, then we can A big assumption. > inscribe a rectangle of area > 2(a-x)sqrt(4px) > for any x in (0,a). You can use calculus to maximize this area or, equally > good, maximize the square of the area, which is a cubic function with no > sqrt appearing in it. G.C. === Subject: Re: Unknown Functions & Einstein's Incompetence > Unknown Functions & Einstein's Incompetence (FAQ) > (c) Eleaticus/Oren C. Webster > Thnktank@concentric.net [snip trolled garbage] Originally trolled across sci.physics sci.physics.relativity alt.physics sci.math sci.answers alt.answers news.answers http://b5.sdvc.uwyo.edu/bab5/snds/argcstpd.wav Psychotic ineducable boring troll Eleaticus, You see yourself this way, http://www.mazepath.com/uncleal/effete6.jpg The entire remainder of the planet sees you this way, http://www.mazepath.com/uncleal/effete7.jpg http://w0rli.home.att.net/youare.swf http://www.mazepath.com/uncleal/sunshine.jpg http://www.you-moron.com/ http://www.mazepath.com/uncleal/effete0.jpg http://www.mazepath.com/uncleal/effete1.png http://www.mazepath.com/uncleal/effete2.png http://www.mazepath.com/uncleal/effete3.png http://www.mazepath.com/uncleal/effete4.png http://www.mazepath.com/uncleal/effete5.jpg http://www.apa.org/journals/psp/psp7761121.html http://insti.physics.sunysb.edu/~siegel/quack.html Hey, stooopid troll Eleaticus - Do you want EVIDENCE? Each of the 24 GPS satellites carries either four cesium atomic clocks or three rubidum atomic clocks in orbit, with full relativistic corrections being applied. http://arXiv.org/abs/hep-th/0307140 GR structure, especially Part 4/p. 7 Experimental constraints on General Relativity. http://www.eftaylor.com/pub/projecta.pdf Relativity in the GPS system http://arXiv.org/abs/gr-qc/9909014 falling light http://arXiv.org/abs/astro-ph/0401086 http://arxiv.org/abs/astro-ph/0312071 Deeply relativistic neutron star binaries http://arXiv.org/abs/gr-qc/0301024 Nordtvedt Effect NIM A 355 537 (1995) Physics Letters B 328 103 (1994) Physical Review Letters 64 1697 (1990) Physical Review Letters 39 1051 (1977) Physical Review 135 B1071 (1964) Physics Letters 12 260 (1964) Europhysics Letters 56(2) 170-174 (2001) General Relativity and Gravitation 34(9) 1371 (2002) http://fourmilab.to/etexts/einstein/specrel/specrel.pdf http://users.powernet.co.uk/bearsoft/Paper6.pdf http://users.powernet.co.uk/bearsoft/LPHrel.html Longitudinal and transverse mass http://www.navcen.uscg.gov/pubs/gps/gpsuser/gpsuser.pdf http://www.navcen.uscg.gov/pubs/gps/sigspec/default.htm http://www.navcen.uscg.gov/pubs/gps/icd200/default.htm http://www.trimble.com/gps/index.html http://sirius.chinalake.navy.mil/satpred/ http://www.phys.lsu.edu/mog/mog9/node9.html http://egtphysics.net/GPS/RelGPS.htm http://www.schriever.af.mil/gps/Current/current.oa1 http://edu-observatory.org/gps/gps_books.html Uncle Al http://www.mazepath.com/uncleal/qz.pdf http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Re: Einstein (1905) Absurdities > Einstein (1905) Absurdities > (c) Eleaticus/Oren C. Webster > Thnktank@concentric.net [snip 1300 lines of trolled garbage] Originally trolled across sci.physics sci.physics.relativity alt.physics sci.math sci.answers alt.answers news.answers http://b5.sdvc.uwyo.edu/bab5/snds/argcstpd.wav Psychotic ineducable boring troll Eleaticus, You see yourself this way, http://www.mazepath.com/uncleal/effete6.jpg The entire remainder of the planet sees you this way, http://www.mazepath.com/uncleal/effete7.jpg http://w0rli.home.att.net/youare.swf http://www.mazepath.com/uncleal/sunshine.jpg http://www.you-moron.com/ http://www.mazepath.com/uncleal/effete0.jpg http://www.mazepath.com/uncleal/effete1.png http://www.mazepath.com/uncleal/effete2.png http://www.mazepath.com/uncleal/effete3.png http://www.mazepath.com/uncleal/effete4.png http://www.mazepath.com/uncleal/effete5.jpg http://www.apa.org/journals/psp/psp7761121.html http://insti.physics.sunysb.edu/~siegel/quack.html [snip 1300 lines of trolled garbage] Originally trolled across sci.physics sci.physics.relativity alt.physics sci.math sci.answers alt.answers news.answers http://b5.sdvc.uwyo.edu/bab5/snds/argcstpd.wav Psychotic ineducable boring troll Eleaticus, You see yourself this way, http://www.mazepath.com/uncleal/effete6.jpg The entire remainder of the planet sees you this way, http://www.mazepath.com/uncleal/effete7.jpg http://w0rli.home.att.net/youare.swf http://www.mazepath.com/uncleal/sunshine.jpg http://www.you-moron.com/ http://www.mazepath.com/uncleal/effete0.jpg http://www.mazepath.com/uncleal/effete1.png http://www.mazepath.com/uncleal/effete2.png http://www.mazepath.com/uncleal/effete3.png http://www.mazepath.com/uncleal/effete4.png http://www.mazepath.com/uncleal/effete5.jpg http://www.apa.org/journals/psp/psp7761121.html http://insti.physics.sunysb.edu/~siegel/quack.html Hey, stooopid troll Eleaticus - Do you want EVIDENCE? Each of the 24 GPS satellites carries either four cesium atomic clocks or three rubidum atomic clocks in orbit, with full relativistic corrections being applied. http://arXiv.org/abs/hep-th/0307140 GR structure, especially Part 4/p. 7 Experimental constraints on General Relativity. http://www.eftaylor.com/pub/projecta.pdf Relativity in the GPS system http://arXiv.org/abs/gr-qc/9909014 falling light http://arXiv.org/abs/astro-ph/0401086 http://arxiv.org/abs/astro-ph/0312071 Deeply relativistic neutron star binaries http://arXiv.org/abs/gr-qc/0301024 Nordtvedt Effect NIM A 355 537 (1995) Physics Letters B 328 103 (1994) Physical Review Letters 64 1697 (1990) Physical Review Letters 39 1051 (1977) Physical Review 135 B1071 (1964) Physics Letters 12 260 (1964) Europhysics Letters 56(2) 170-174 (2001) General Relativity and Gravitation 34(9) 1371 (2002) http://fourmilab.to/etexts/einstein/specrel/specrel.pdf http://users.powernet.co.uk/bearsoft/Paper6.pdf http://users.powernet.co.uk/bearsoft/LPHrel.html Longitudinal and transverse mass http://www.navcen.uscg.gov/pubs/gps/gpsuser/gpsuser.pdf http://www.navcen.uscg.gov/pubs/gps/sigspec/default.htm http://www.navcen.uscg.gov/pubs/gps/icd200/default.htm http://www.trimble.com/gps/index.html http://sirius.chinalake.navy.mil/satpred/ http://www.phys.lsu.edu/mog/mog9/node9.html http://egtphysics.net/GPS/RelGPS.htm http://www.schriever.af.mil/gps/Current/current.oa1 http://edu-observatory.org/gps/gps_books.html Uncle Al http://www.mazepath.com/uncleal/qz.pdf http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Looking for interesting algebraic numbers (with a major condition) I am implementing various algorithms (especially Gosper's one) for computing exact arithmetic on continued fraction. For the moment, I am interested by defining constants of the following kind: algebraic numbers (degree > 2), single root of a polynomial equation (when there is only one root, there is an exact method for computing the continued fraction). I am interested by important numbers, for which it would be interesting to have them defined as constants. There are important numbers that are algebraic, but many of them are roots of equations having several roots. Do you have any suggestions for building my library of constants ? Cordially, nous devons agir comme si la chose qui peut-.90tre ne sera pas devait .90tre (Kant, M.8etaphysique des moeurs, doctrine du droit, II conclusion) Thomas Baruchel === Subject: Re: Looking for interesting algebraic numbers (with a major condition) > I am implementing various algorithms (especially Gosper's one) > for computing exact arithmetic on continued fraction. For the moment, > I am interested by defining constants of the following kind: > algebraic numbers (degree > 2), single root of a polynomial > equation > (when there is only one root, there is an exact method for > computing the continued fraction). > I am interested by important numbers, for which it would be > interesting to have them defined as constants. > There are important numbers that are algebraic, but many of them > are roots of equations having several roots. > Do you have any suggestions for building my library of constants ? I think your library is empty. A polynomial of degree n > 2 has n roots, guaranteed. The only way it can have a single root is to have that root to multiplicity n, that is, for the polynomial to be of the form (x - a)^n, where a is that single root. That doesn't distinguish any library. The only way I can make sense out of your question is to assume that when you ask for a polynomial to have only one root, you mean for it to have only one *real* root. There are some important algebraic numbers of that type, among the Pisot numbers (formerly known as the Pisot-Vijayaraghavan numbers). These are algebraic integers with a single conjugate outside the unit circle. That conjugate is necessarily real, the others may all be complex. There is plenty of literature about these numbers. Have a look. Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Looking for interesting algebraic numbers (with a major condition) >mean for it to have only one *real* root. There are some sorry for that, but you did well understand my question. >important algebraic numbers of that type, among the Pisot numbers >(formerly known as the Pisot-Vijayaraghavan numbers). These are >algebraic integers with a single conjugate outside the unit circle. >That conjugate is necessarily real, the others may all be complex. >There is plenty of literature about these numbers. Have a look. thank you nous devons agir comme si la chose qui peut-.90tre ne sera pas devait .90tre (Kant, M.8etaphysique des moeurs, doctrine du droit, II conclusion) Thomas Baruchel === Subject: Division and Modulo by 2^n-1 How does one efficiently divide and modulo by $2^n-1$? More specifically, how does one divide and modulo by $2^(2^m)-1$. Also, and most importantly, how does one modulo by $2^(2^m)-1$ with only additions and subtractions modulo 2^(2^m), xors, nots, ands, ors, comparisons, comparisons by and, jumps, and negations? === Subject: Re: Division and Modulo by 2^n-1 > How does one efficiently divide and modulo by $2^n-1$? More specifically, > how does one divide and modulo by $2^(2^m)-1$. > Also, and most importantly, how does one modulo by $2^(2^m)-1$ with only > additions and subtractions modulo 2^(2^m), xors, nots, ands, ors, > comparisons, comparisons by and, jumps, and negations? Without actually answering your question, here are some things to think about. Let's say you had a large number which, when written in base 2**n, has the digits I J K L M where each of I,J,K,L and M is in the range 0 to (2**n)-1. If you multiply it by (2**n)+1, the result is I I+J J+K K+L L+M M Some of these digits might be >=2**n, so use carry to normalize the digits. If you divide I J K L M by (2**n)+1, you get I J-I K-J+I L-K+J-I remainder M-L+K-J+I Some of the digits might be <0, so use borrowing to normalize the digits. You can check that result multiplying the quotient by (2**n)+1. I J-I K-J+I L-K+J-I 0 + I J-I K-J+I L-K+J-I ___________________________ = I J K L L-K+J-I which, when you add in the remainder, gives the original dividend. The extension to base (2**n)-1 is left as an exercise for the reader. When you do, you will probably recognize the form of the remainder because you've used it to find the remainder of a base-10 number divided by 9. --Mike Amling === Subject: Re: Division and Modulo by 2^n-1 > How does one efficiently divide and modulo by $2^n-1$? More specifically, > how does one divide and modulo by $2^(2^m)-1$. > Also, and most importantly, how does one modulo by $2^(2^m)-1$ with only > additions and subtractions modulo 2^(2^m), xors, nots, ands, ors, > comparisons, comparisons by and, jumps, and negations? Do a few by hand. Spot the pattern. Phil Unpatched IE vulnerability: WebFolder data Injection Description: Injecting arbitrary data in the My Computer zone Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0305/13.html === Subject: Re: Division and Modulo by 2^n-1 > How does one efficiently divide and modulo by $2^n-1$? More specifically, > how does one divide and modulo by $2^(2^m)-1$. > Also, and most importantly, how does one modulo by $2^(2^m)-1$ with only > additions and subtractions modulo 2^(2^m), xors, nots, ands, ors, > comparisons, comparisons by and, jumps, and negations? It's always September somewhere on the Internet, isn't it? Suggestion 2: Post your instructor's email address, and I'll send the solutions directly there -- no need for the middleman. -Arthur === Subject: We come from your future Super Cosmos book under construction http://qeedcorp.com/destiny/ The Question is: What is The Question? Wheeler Mitch et-al a = 1/(1 +z) a + az = 1 z = (1 - a)/a so z = 0 for a(now). Note that a < 1 for a retarded signal from the past if space is expanding. a > 1 for advanced signal from future going back in time again if 3D space is expanding. z < 0 is a blue shift from the future, which might be relevant to some phenomena in gamma ray astronomy? It's basically simple. Speed of signals inside expanding 3D space is completely decoupled from expansion rate dR(t)/dt of 3D space itself! Space is assumed isotropic and homogeneous. Motion is inside space. Scale factor R(t) is controlling the space itself. You can think of 3D space imbedded in hyperspace if you like but you do not need to. No speed of light limit in hyperspace beyond 3D space. The Hubble parameter measured in redshifts z is H(t) = R(t)^-1dR(t)/dt Define for FRW metric a(t) = R(t)/R(now) a(past) = [1 + z(now)]-1 For a light signal emitted at a past moment and received now, z(now) is the observed redshift from the expansion of space during the time it took for the light signal to leave the source and arrive at the detector from the POV of the detector of course. You are making a very elementary error. Think about it. Get a balloon. Mark two points A & B on it with ink. Blow it up and see how the points separate. Imagine a light signal starting at A and ending at B moving at finite speed c as the balloon is expanding. [M: Yes, I can imagine all that clearly in my mind. I don't think it will help to actually get a balloon. A & B have some speed of separation, which is much less than c, right?] No not necessarily at all. That's the whole idea of inflation. There is no contradiction with relativity which only works ON THE SURFACE (3D space). There's one error you make at the start. What you have is the FRW metric with a scale factor R(t). t is a cosmic parameter not the t you use below. dR(t)/dt is not limited to c, it is not the speed of anything inside the expanding space. The cosmology solution is for R(t) and that's how it's all computed. [M: If t0 is some time instant not long after the big bang, then at t0 the balloon is very small and A & B are close together, call this distance d0. Let s be the separation rate, light years per year. Then s << c. Let T be the time instant when the light emitted from A at t0 reaches B. Let D be the separation distance at that time. D = d0 + s *(T-t0)] s need not be constant [M: Now if T-t0 is 13 billion years, then D is of the order of 1 billion light years, assuming d0 is very small and s is an order of magnitude smaller than c. But light can travel this distance in only a billion years, which contradicts the assumption that T-t0 is 13 billion years. Call that a paradox or not, what is wrong with the above reasoning?] Everything. Look all these guys like Wheeler, Penrose and Hawking know what they are doing. You need to look at the actual math. Metaphors don't really help. [M: OK, I give up.] Many are stubborn in pursuit of the path they have chosen, few in pursuit of the goal. - Friedrich Nietzsche === Subject: Re: We come from your future Jack Sarfatti > Super Cosmos book under construction > http://qeedcorp.com/destiny/ Spam-in-advance for Sarfatti's next book. === Subject: branch of log z A veil has been woven out of words like imaginary and spread over the real plane, giving the impression of an domain, accessible only for people with special brain-power, sometimes letting the real plane shimmering through. introduced to me (at the college level, anyways). The first time i hear this, except from Caspar Wessel and sincerely Yours for these words. No. It's known to insiders, working under the veil - but it's not made well known. Can You give me a reference, where it's said, that the i-axis and the y-axis are just different names, but with no mathematical difference? The same for R2, the 2D-plane, no difference to the complex plane, argand diagramm, (gauss plane)? Can You give me a reference, where mixed-mode-calculating is used, like : i*(3,4)=(-4,3) (see my calculator)? Can You name someone, who uses the notation 3+i*4 in any 2D-vectorspace in place of (3,4) ? The second is of course an ordered pair of reals and the first is using the linear-combination of two basis-vectors (1,0) and i=(0,1) and, as the real numbers are embedded, you ommit (1,0). And sometimes insiders entangle in their own veil and can't see clear, look at Algebra: What is the relation between (R2,+,r.s.m.) and (R2,+,*)-this is called the vectorspace of complex numbers(and an commutative field)- and (R2,+,r.s.m.,dot)-this is the euclidian vectorspace? branches of log ...This stuff can get messy later. - but may be without imaginary veil a little bit less. Have fun Hero === Subject: Calculating Modulo (very big numbers) Hi. Is there a way to calculate the modulo with numbers like this: 1226^37 mod 4838 ? Or is it impossible to do? === Subject: Re: Calculating Modulo (very big numbers) >Is there a way to calculate the modulo with numbers like this: >1226^37 mod 4838 ? I assume you want a way to do it without needing a 115-digit calculator. Using a 10-digit calculator, I computed this as follows: exponent in base-2 1226^1 = 1226 mod 4838 1 1226^2 = 1503076 mod 4838 10 = 3296 mod 4838 1226^4 = 10863616 mod 4838 100 = 2306 mod 4838 1226^8 = 5317636 mod 4838 1000 = 674 mod 4838 1226^9 = 826324 mod 4838 1001 = 3864 mod 4838 1226^18 = 14930496 mod 4838 10010 = 428 mod 4838 1226^36 = 183184 mod 4838 100100 = 4178 mod 4838 1226^37 = 5122228 mod 4838 100101 = 3624 mod 4838 Where each step is squaring, multiplying by 1226, or modding by 4838. Rob Johnson take out the trash before replying === Subject: Re: Calculating Modulo (very big numbers) > Is there a way to calculate the modulo with numbers like this: > 1226^37 mod 4838 ? > Or is it impossible to do? Exponentiation can be broke down into just a sequence of squarings and multiplies. Google for 'binary exponentiation'. Phil Unpatched IE vulnerability: WMP local file bounce Description: Switching security zone, arbitrary command execution, automatic email-borne command execution Reference: http://www.ntbugtraq.com/default.asp?pid=36&sid=1&A2=ind0307&L =ntbugtraq&F=P& S=&P=6783 Exploit: http://www.malware.com/once.again!.html === Subject: Re: Calculating Modulo (very big numbers) > Is there a way to calculate the modulo with numbers like this: > 1226^37 mod 4838 ? > Or is it impossible to do? > Exponentiation can be broke down into just a sequence of squarings > and multiplies. Google for 'binary exponentiation'. The trick is to reduce modulo 4838 after each multiplication, to keep the operands small. Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Calculating Modulo (very big numbers) >1226^37 mod 4838 ? but 1226^37 is not very big: 187995475307059626169460899238649920814212386397422931520242984 4679255045350 300314081784073749696387332489741336576 and the result of what you are asking is 3624. But maybe it was only an example, or maybe you want a method for computing it by hand ? nous devons agir comme si la chose qui peut-.90tre ne sera pas devait .90tre (Kant, M.8etaphysique des moeurs, doctrine du droit, II conclusion) Thomas Baruchel === Subject: constant Gauss curvature K Which geometrical property should be extremized subject to which constraint to get constant Gauss curvature K as the solution? .. just as we have CMC (constant mean curvature H ) surfaces for maximum volume with given area constraint, Delaunay surfaces belong to the particular axisymmetric class. I tried moment of inertia/perimeter length/area etc. but it does not seem to solve upto a constant K. === Subject: Re: Math Too Advanced For Mainstream Economists with his continual conflating and begging of all questions. Here are some questions he conflates: 1. Are there internally consistent models of wages and employment determined by the intersection of supply and demand curves, as in supposedly neoclassical theory? 2. Do such internally consistent models, if any, have any actually existing labor markets in their scope? 3. What competing theories have actually existing markets within their scope? 4. What empirical evidence, if any, suggests one theory or another better explains some observations in some context for some purpose? Note that empirical evidence is irrelevant for the answer to the first question. But, empirically, production in the U.S.A. usually occurs with the usage of goods previously produced, that is, with capital goods. As I have been pointing out for years, there seems to be no internally consistent neoclassical model of well-behaved labor supply and demand curves that has such an economcy within its scope. This, of course, is not merely my opinion. It is the conclusion of a collection of arguments I have presented. And these arguments follow the literature. (Notice the above paragraph does not imply that there are no internally consistent neoclassical models of well-behaved labor supply and demand curves which, by assumption, exclude time and the existence of capital goods.) If one were serious, unlike Mark Witte, one might address the arguments I have presented. If one wanted to challenge my conclusion, for example, one might outline a coherent neoclassical model of supply and demand incorporating capital goods. On the other hand, some silly person insulting the readership of sci.econ might continue to beg the answers to the first two questions above, while trying to change the subject to some such question as: Do you believe that the supply and demand model is never a reasonable description of what we observe in the world? This is a non sequitur when the topic is the internal consistency and scope of the supply and demand model. It begs the question of the existence of any such model. > This raise an empirical question. Does poor Mark Witte realize > he has no point? > I think it was clear early on that there was never any point to > this thread beyond it being merely a strange polemic. Whatever. > Of course, if > someone would present an interesting model that did a good job of > describing something that we see, I could be persuaded otherwise. > But...that's just not going to happen, is it? lots of empirical data goes along with what I am saying. For example, ... The URL has been under discussion here, inasmuch as other posters can be said to have been discussing anything at all. That URL contains: Consider a firm whose managers know two techniques for producing widgets. Table 1 shows the amount of labor inputs required to produce a widget for each technique. The interesting qualitative properties of this example depend merely on the difference in labor inputs between the techniques each year. Thus, if every input of labor in both techniques was increased by the same amount (e.g., 25 person-years), the results of this example would look much the same. One instance discussed in the literature is a choice between using a plot of land for either grazing or mining; mining requires a high initial expenditure and a costly cleanup phase at the end of the use of the land. Other empirical examples have been discussed in the literature, usually in the context of environmental economics or geography. And, once again from this thread: Here's one example of empirical and applied work in the literature drawing on Sraffa: Raymond Prince and J. Barkley Rosser, Jr., Some Implications of Delayed Environmental Costs for Benefit Cost Analysis: A Study of Reswitching in the Western Coal Lands, _Growth and Change_, V. 16, 18-25, 1985. If one were actually to produce such an argument, one should be aware that both reswitching and capital-reversing can arise in models in which technology is described in other ways. And one might want to address the empirical examples in the literature. Some empirical examples are vaguely alluded to at the above URL, but have been more fully cited in the past on this newsgroup. Here's a list that poor Mark Witte has seen before: Peter Albin, Reswitching: An Empirical Observation, _Kyklos_, 1975, Number 1, 28, pp. 149-54. Geir B. Asheim, The Occurrence of Paradoxical Behavior in a Model where Economic Activity has Environmental Effects, Norwegian School of Economics and Business Administration Discussion Papers, 1980. Trevor Barnes and Eric Sheppard, Technical Choice and Reswitching in Space Economies, _Regional Science and Urban Economics_, V. 14, pp. 345-352, 1984. John Hartwick, Intermediate Goods and the Spatial Integration of Land Use, _Regional Science and Urban Economics_, V. 6, pp. 127-145, 1976. Adam Ozanne, Do Supply Curves Slope Up? The Empirical Relevance of the Sraffian Critique of Neoclassical Production Economics, _Cambridge Journal of Economics_, Volume 20, pp. 749-762, 1996. Raymond Prince and J. Barkley Rosser, Jr., Environment Costs and Reswitching Between Food and Energy Production in the Western United States, mimeo, James Madison University, 1984. Raymond Prince and J. Barkley Rosser, Jr., Some Implications of Delayed Environmental Costs for Benefit Cost Analysis: A Study of Reswitching in the Western Coal Lands, _Growth and Change_, V. 16, 18-25, 1985. U. Schweizer and P. Varaiya, The Spatial Structure of Production with a Leontief Technology-II: Substitute Techniques, _Regional Science and Urban Economics_, V. 7, pp. 293-320, 1977. A. J. Scott, Commodity Production and the Dynamics of Land-Use Differentiation, _Urban Studies_, V. 16, pp. 95-104, 1979. I have still not read every paper listed above. An amusing quote: I spotted no such naive errors in Marglin's estimation techniques. -- Mark Witte, 12 April 1997 > Kind of like the simple question about funding the Sraffa Memorial? Consider: > 1. What tax should the Gov. of NY propose to fund the Sraffa > Memorial? I was amused that Mr. Auld should raise a question of tax incidence for a demonstration of the lack of practical implications of Sraffianism. That reveals that he had very little idea of what he had been arguing about for years. (What is the distinction between basic and non-basic goods? The King of Sweden conferred on Sraffa the Soderstrom gold medal of the Royal Academy of Sciences in March 1961. This award was for a massive work of scholarship. What was the topic of the middle third of the first volume?) -- Robert Vienneau, approximately 18 January 2001 Naturally, the simple questions I asked then have gone unanswered here, as far as I know. Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: wanted: more mathematical false proofs |> ... I would like to know if anyone has more of these false proofs. I'm surprised everybody here missed this unique opportunity of killing two birds with one stone. Starting with a rock solid result from harrissian fundamental tautology space theory and developping at an elementary level, here is a crystal clear proof of a core error in mathematics. (0) 1 = 1 (1) 1 = sqrt( 1 ) (2) 1 = sqrt[ (-1)^2 ] (3) 1 = sqrt[ (-1) * (-1) ] (4) 1 = sqrt( -1 ) * sqrt( -1 ) (5) 1 = i * i (6) 1 = -1 Please note that some nitpickers might argue that sqrt being such an ambiguous and lunatic operator, the odds of sqrt(-1) evaluating to (i) the first time and (-i) the second time (or vice-versa) are not completely negligeable. Hence this proof should be verified by a sufficiently large number of experimented physicists, perhaps in a coordinated net-based effort over a period of at least 8 years. === Subject: number of regions for a regular n-gon with all diagonals drawn... According to the Online Encylopedia of Integer Sequences (A7678), the number of regions in a regular n-gon with all diagonals drawn is (starting with n=3): 1,4,11,24,50,80,154,220,375,444,781,952,1456,1696,2500, 2466,4029,4500,6175,6820,9086,9024,12926,13988,17875, 19180,24129,21480,31900,33856, ... I have three formulas for calculate the number of regions when N is odd: _________________________________________________________ fn=floor(n/2) cn= ceil(n/2) x=(n-5)/2 r=((fn-1)**2-1)*n+fn+(n-2)*fn+cn+x*(x+1)*(2*x+1)/6*n say 'n=' n 'regions=' r _________________________________________________________ r=(n**4-6*n**3+23*n**2-18*n+24)/24-n say 'n=' n 'regions=' r _________________________________________________________ r=(n-1)*(n-2)*(n**2-3*n+12)/24 say 'n=' n 'regions=' r _________________________________________________________ where I believe the last two are just variations of each other. Does anybody have the formula for the number of regions where N is even ? __________________________Gerard S. === Subject: Re: number of regions for a regular n-gon with all diagonals drawn... > According to the Online Encylopedia of Integer Sequences > (A7678), the number of regions in a regular n-gon with > all diagonals drawn is (starting with n=3): > 1,4,11,24,50,80,154,220,375,444,781,952,1456,1696,2500, > 2466,4029,4500,6175,6820,9086,9024,12926,13988,17875, > 19180,24129,21480,31900,33856, ... > I have three formulas for calculate the number of regions > when N is odd: > _________________________________________________________ > fn=floor(n/2) > cn= ceil(n/2) > x=(n-5)/2 > r=((fn-1)**2-1)*n+fn+(n-2)*fn+cn+x*(x+1)*(2*x+1)/6*n > say 'n=' n 'regions=' r > _________________________________________________________ > r=(n**4-6*n**3+23*n**2-18*n+24)/24-n > say 'n=' n 'regions=' r > _________________________________________________________ > r=(n-1)*(n-2)*(n**2-3*n+12)/24 > say 'n=' n 'regions=' r > _________________________________________________________ > where I believe the last two are just variations of each > other. > Does anybody have the formula for the number of regions > where N is even ? __________________________Gerard S. A general formula for the number of regions is given by Bjorn Poonen and Michael Rubinstein: The number of intersection points made by the diagonals of a regular polygon http://math.berkeley.edu/~poonen/papers/ngon.pdf on page 3, Theorem 2. Hugo Pfoertner === Subject: Permutation(?): + Integers Increasingly Arranged Define an integer sequence as follows: a(1) = 1; And place the positive integers, once per integer, into the sequence so that: (m+1) is positioned in the sequence such that there are exactly (m-1) higher-valued terms between it and the term equal to m. And the (m+1) is always less that the m whenever this is possible. As to hopefully eliminate ambiguity, here is the arrangement of the first 16 integers: (figured by-hand) 1, 2, 13, 3, 6, *, 4, 11, *, 9, 5, * 7, *, *, *, *, *, *, 8, *, 10, 16, 12, *, 14, *, *, *, *, *, *, *, *, *, *, *, *, *, 15,... Of course, the *'s are yet to be determined and are each > 16. Because of the higher-valued part of the definition, the sequence will never get 'stuck', ie. unlike if we simply counted the number of terms in-between, where the sequence might get into a region where the nearest unused position is too far away. But I must ask, is this sequence a permutation of the positive integers? In other words, will every position eventually contain an integer? If so, here is the inverse permutation's first few terms: 1, 2, 4, 7, 11, 5, 13, 20, 10,... (Maybe the inverse sequence would be easier to study, actually.) === Subject: Monetary Change Density A fun problem to think about in more than one way. If someone offered you a trunk filled with either pennies, nickels, dimes, or quarters, which would you choose? === Subject: Submodular functions... To whom it may concern, Working on approximation algorithms in combinatorial auctions and am hoping to find some existing implementations with submodular functions. Can anyone point me in the right direction ?? Fred. === Subject: Re: Submodular functions... Fred Mailhot > Working on approximation algorithms in combinatorial auctions and am hoping > to find some existing implementations with submodular functions. Can anyone > point me in the right direction ?? Maybe http://www-fp.mcs.anl.gov/otc/Guide/SoftwareGuide/ Also, www.nist.gov has saved my bacon on occcasion. === Subject: TSP and the Bell curve % Hello everyone % it seems that i have stumbled on something interesting. % Basically, an algorithm that would use the 'Bell curve' assumption to find a very quick O(N^2) sub-optimal solution % Your opinions needed % The source is in MATLAB % n is the number of cities n = 10; % create a random matrix rr = rand(n,n); % city is a symmetric matrix with a_ij = cost of going from city i to city % j city = (rr + rr')/2; % alias il = 1:n; % the max value of the city matrix is at 1. This is equivalent to % normalizing the matrix % Now, the diagonal elements are set to 1. This way they are avoided in % comparisons city( il + (il -1)*n) = 1; % Use sampling to calculate mean and standard deviation % Interestingly, the distribution of 'total costs', along path is % gaussian. % It follows from the fact that % total cost = cost_( a to b) + cost_(b to c) + cost_(c to d) + .... % L is the number of samples L = 20000; p = zeros(1,n); for i = 1:L % the idea behind the following logic is that % randomly sort (1 to n) % after getting a random arragement, look up randrows_i , randrows_(i+1) % in the city matrix and sum up the costs randrows = sortrows([ rand(n,1) (1:n)']); p(i) = sum(city( n*(randrows(1:n-1,2)-1) + randrows( 2:n,2))); end; % Our Traveling Sales Man problem solution starts here % First, find the smallest matrix element fmin = sortrows([city(:) (0:n*n-1)']); pa = []; % pa(1) is the row of the smallest element % pa(2) is the col of the smallest element pa(2) = floor( fmin(1,2)/n) + 1; pa(1) = mod(fmin(1,2),n) +1; % this is the basic alogrithm % after finding the starting point, now basically, what we do is that % we go from link a_(kj) , where 'j' is all cities that the salesman never visited % we travel thru the city j, that yeilds min a_{kj} for k = 2:n-1 indx = setdiff( 1:n, pa); fmin= sortrows([ city(pa(k), indx)' indx']); % and pa, stores the link pa = [pa fmin(1,2)]; end; % this is estimated cost from our suboptimal solution finder solution_cost = sum(city( n*(pa(1:n-1)-1) + pa( 2:n))) % the histogram hist(p,50) mean(p) std(p) % this step gives the estimated accuracy of the solution % based on the gaussian distribution assumption factorial(n) *( 0.5+ 0.5*erf((solution_cost - mean(p))/ (sqrt(2)*std(p)))) === Has anyone heard about this? I found this site, and it really boiled my blood! SaveTheCEO.com. Apparently this CEO wants people to raise $100,000 to hire a lobbyist to ease tax restrictions so he can move his factory to Mexico! As if it's not already easy enough as it is! Someone should find this guy and put him in jail! Where he belongs! === > Has anyone heard about this? I found this site, and it really boiled > my blood! SaveTheCEO.com. Apparently this CEO wants people to raise > $100,000 to hire a lobbyist to ease tax restrictions so he can move > his factory to Mexico! As if it's not already easy enough as it is! > Someone should find this guy and put him in jail! Where he belongs! Just like the e mails I get, but its in a web site format..... === > Has anyone heard about this? I found this site, and it really boiled > my blood! SaveTheCEO.com. Apparently this CEO wants people to raise > $100,000 to hire a lobbyist to ease tax restrictions so he can move > his factory to Mexico! As if it's not already easy enough as it is! > Someone should find this guy and put him in jail! Where he belongs! nice try. i'm not clicking on no ceo porn site, === If you have such issues with it, why post the link to thousands of readers? > Has anyone heard about this? I found this site, and it really boiled > my blood! KillTheCEOs.com. Apparently this CEO wants people to raise > $100,000 to hire a lobbyist to ease tax restrictions so he can move > his factory to Mexico! As if it's not already easy enough as it is! > Someone should find this guy and put him in jail! Where he belongs! And apparently you're helping. Did you think telling us it was a scam would get us to go to this site? Spam! === Subject: Proving the permutations are parities. Hello everyone, I need some help in proving this problem. Your help would greatly be appreciated. The problem is as follows: If a, g E Sn, prove that the permutations a and g'ag have the same parity. I know that the parity of the permutations can either be both even or both odd. Again thank you for your help. === Subject: Re: Proving the permutations are parities. Jordan > Hello everyone, I need some help in proving this problem. Your help > would greatly be appreciated. The problem is as follows: > If a, g E Sn, prove that the permutations a and g'ag have the > same parity. > I know that the parity of the permutations can either be both even or > both odd. No restriction needed. In casual jargon, g and g' involve the same number of transpositions. The sum of their two contributions to g' a g is therefore even. === Subject: Re: Proving the permutations are parities. > .... > If a, g E Sn, prove that the permutations a and g'ag have the > same parity.... You've had an answer to this from Hendrik Maryns on aus.mathematics but perhaps you didn't follow it. If you're familiar with the basic rules for the parity of permutation products (even x odd = odd, etc.) then just remember that each of a and g may be either even or odd, and consider the four possibilities. The only extra thing you need is that g' has the same parity as g. Assuming g' means the inverse, you can get that from the equation g'g = e and the fact that the identity is even. If you get stuck on any details, just explain how far you've reached, preferably cross-posting to both news groups to save people's time. Ken Pledger. === Subject: Re: there is no such thing as infinity >You think you can reach the largest number with a FORTAN program? How >crude! Obviously, you need to write this program in C++. >Anything worth doing can be done in Perl, in a fraction of the memory >space something like C++ would use. > And can be written in exponentially more obfuscated a manner than even C or > assembler... :) Assembler? Bah!! Raw machine language burnt into a ROM is the only way to find the largest number. In any case, I have the world's largest number written on my office blackboard, so you all might as well give up. John Starrett http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- === Subject: Re: there is no such thing as infinity > Assembler? Bah!! Raw machine language burnt into a ROM is the only way > to find the largest number. In any case, I have the world's largest > number written on my office blackboard, so you all might as well give up. > -- > John Starrett * Add one to it and see what the sum is. earle * === Subject: Re: there is no such thing as infinity <4f1e888502f84d511af633bad4d7158d@news.teranews.com> In message , Richard Henry > In message <4f1e888502f84d511af633bad4d7158d@news.teranews.com>, Darryl > You think you can reach the largest number with a FORTAN program? How > crude! Obviously, you need to write this program in C++. Anything worth doing can be done in Perl, in a fraction of the memory > space something like C++ would use. >And can be written in exponentially more obfuscated a manner than even C >assembler... :) > Do you have something against APL? >APL has the benefit of being obvious to anyone who understands the symbols. For a suitable definition of obvious ;-) IMO no language that allows you to do conditional jumps by *multiplying* a GOTO by 0 or 1 is entirely transparent. >Forth, on the other hand... ... allows you to define the symbols yourself. Richard Herring === Subject: Re: there is no such thing as infinity X-SessionID: nt%Xb-3364-_4-5884@news.uchicago.edu X-Hash-Info: post-filter,v:1.4 X-Hash: 139c527e 8241f69b 6a9789f7 0a453e59 4f380a40 Take a joke, people. --Dan Weiner === Subject: Re: there is no such thing as infinity > Take a joke, people. > --Dan Weiner Spu Vomit === Subject: Coprime Grid: Filling Infinite Quadrant In this post, I write of a specific and altered case of the puzzle mentioned in these previous threads: In the puzzle, we try to write, in order, the positive integers into a grid, one integer per grid-square, such that: Each integer n is adjacent (above /left of /right of /below) of the integer (n+1); And each adjacent pair of integers (above /left of /right of /below) are coprime. And the goal is to completely fill the grid. But here I am asking about filling an infinite grid which is bounded along 2 perpendicular sides. ie. the grid is an entire quadrant of the Cartesian plane, bounded by the x-axis and the y-axis. I think I found a simple procedure which *might* ensure a successful filling of the grid with coprime-adjacent integers. (Sorry to those on rec.puzzles, but I will give my procedure below. You can still post your own algorithm, however, or confirm that mine can really work for the entire grid without problems.) I illustrate with the first 99 terms: (figured by hand, so maybe in-error) 99 98 97 96 95 54 55 94 93 92 53 56 57 58 91 90 89 52 51 50 59 60 61 88 87 21 22 49 48 47 62 63 86 20 23 24 25 46 45 64 85 19 18 17 26 27 44 65 84 06 07 16 15 28 43 66 83 82 81 80 79 05 08 09 14 29 42 67 68 69 70 71 78 04 03 10 13 30 41 40 39 38 37 72 77 76 01 02 11 12 31 32 33 34 35 36 73 74 75 Basically, the path swings clockwise and counterclockwise, running along the outside of the already-filled section. When it gets to either the x-axis or y-axis, it forms a 'peninsula', the length of which is the shortest needed to avoid uncoprime integers being placed next to each other in the path-section which runs from that peninsula's axis to the other axis. Now, we do not want a situation where, following the algorith precisely, there is NO peninsula-length which would avoid uncoprime neighbors. I am not certain, but I believe this issue is not a problem. Fun perhaps: Show if my algorithm is foolproof...or just foolish. (I know I do not give the peninsula's length. Also fun perhaps: try to determine the shortest length needed for each peninsula, given that the algorithm never leads to a problem.) === Subject: Re: A series for the inverse sine cardinal function David: In general a rational polynomial approximation [mini-max] determined using say Remez second method will be much more economic than any power series. I have used rational polynomial approximations in digital signal processing (DSP) applications of the sinc function and its' inverse to great practical advantage in terms of demands upon real time storage and execution cycles. Peter Consultant - Signal Processing and Analog Electronics Indialantic By-the-Sea, FL > The sine cardinal function, > ( 1 if x = 0 > sinc(x) = ( > ( sin(x)/x otherwise, > arises in several applications, and so its inverse should also be of > interest. This note gives a series expansion for the inverse of sinc(x), > 0 <= x <= x0, where x0 (approx. 4.4934) denotes the positive value of x > at which sinc(x) reaches its absolute minimum. > [Besides posting this to sci.math, I have also posted it to > sci.math.num-analysis and comp.dsp in the hope that some people might be > able to give references to previous mathematical treatments of the inverse > sinc function. In particular, I'd be interested to know if the series given > here is already known.] > With f(x) = 2*x + 3*x^3/10 + 321*x^5/2800 + 3197*x^7/56000 + > 445617*x^9/13798400 + 1766784699*x^11/89689600000 + > 317184685563*x^13/25113088000000 + > 14328608561991*x^15/1707689984000000 + > 6670995251837391*x^17/1165411287040000000 + > 910588298588385889*x^19/228420612259840000000 + ..., > it can be shown that the desired inverse, abbreviated as Asinc here, is > given by > Asinc(x) = Sqrt(3/2) * f(Sqrt(1 - x)) > -------------------------------------------------------------- --------- > A simple example: > Find the positive root of the equation sin(x) = x/2. > Rewriting the equation as sinc(x) = 1/2, > the solution is x = Asinc(1/2) = Sqrt(3/2) * f(Sqrt(1/2)). > Using just those terms shown above in the series for f, we get > x = 1.8954905... (For comparison, solving by a different method, a much > more accurate approximation is x = 1.8954942668788...) > -------------------------------------------------------------- --------- > The radius of convergence of the above series for f is slightly larger than > 1.1 . As expected, when x is near that value, convergence is very slow. > The series for f was obtained by reversion of series, etc. To be more > specific, for anyone interested, in Mathematica the series may be obtained > using > Simplify[Normal[InverseSeries[1 - Series[Sin[x]/x, {x, 0, n}]]/Sqrt[3/2] > /. x -> x^2], x > 0] > where n (even) should be chosen to be 1 more than the highest power of x > desired in the result. > Unfortunately, I do not know a nice formula for the coefficients in the > series for f. > Thoughtful comments will be appreciated. > David Cantrell === Subject: PDE Max-Min Principle (Help?) A question really about math... Prove that if u(x,t) <= v(x,t) for x=0, for x=l (lower-case L), for t=0, then u<=v for all (x,t) in 0<=x<=l, 0<=t<=oo. I'm not really looking for a solution, just a way to start. My professor is away at a conference and I've been working on this problem set all weekend. After 30 hours or so, if I can't get it, I think I need some help. - Tim Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University === Subject: Re: PDE Max-Min Principle (Help?) > A question really about math... > Prove that if u(x,t) <= v(x,t) for x=0, for x=l (lower-case L), for t=0, > then u<=v for all (x,t) in 0<=x<=l, 0<=t<=oo. Given u(0,0) <= v(0,0) u(L,0) <= v(L,0) Prove that For certian class of PDEs , u(x,t) <= v(x,t) for all (x,t) in 0<=x<=L, 0<=t<=oo Is this correct interpretation of your question? I am missing something. Mohan Pawar ---snip === Subject: Re: PDE Max-Min Principle (Help?) > Prove that if u(x,t) <= v(x,t) for x=0, for x=l (lower-case L), for t=0, > then u<=v for all (x,t) in 0<=x<=l, 0<=t<=oo. Ridiculous as it stands. Please include all hypotheses. (Also, what does t = oo mean here?) === Subject: Re: Mathcad Upgrade Question Don't use mathcad. I installed version 2001i this weekend. After installition I noticed that c-dilla had also been introduced to my registry. Google c-dilla to find out what this program does. Anyhow I have spent 2 days rebuilding my hard drive because of the damage that Mathcad did. > I am thinking of upgrading from an old version of Mathcad (Version 7) > to the most recent. But past experience with upgrades has taught me > to be a bit wary, so 2 questions: > 1. Did Mathsoft do anything to screw up the product, in terms of > usability or features, since I last purchased it (back in 1997)? > 2. Does the latest product come with any kind of registration > feature, like Mathematica has, that locks your software into one > computer, and possibly locks you out of the software if you upgrade > your hardware? (I respect the rights of companies to make money off > their software, but I'm a firm believer that the only fair deal is > one-user/one-app. If I have three computers -- and I do -- I should > not have any worries or complications with loading the same software > package on all three of them.) > Steve O. > Standard Antiflame Disclaimer: Please don't flame me. I may actually *be* an idiot, but even idiots have feelings. === Subject: Sequence: (m-k) divides a(m)*a(k) Here is a sequence similar to the one I describe in the message Indexes' Difference Divides Sum Of 2 Terms, which I posted at: But here I multiply adjacent terms instead of add them. Let a(1) = 1; Let a(m) = lowest positive unpicked integers such that: (m-k) divides evenly into (a(m) * a(k)) for EACH k, 1 <= k <= m-1. I get (again, figured by hand, so not completely believable): a(m) : 1, 2, 4, 3, 12, 30,... Now this looks less like it might be a permutation of the positive integers, less likely a permutation than the sequence in the link above, in any case. Is it a permutation of the positive integers? === Subject: Re calculating fraction period lengths with more info Can anyone prove these statements My interest is recreational. Statement 1 If u=2n^2-n+2 and v=2n^2+n+2 Then u^2 mod (u+v-1) =u-1 and v^2 mod (u+v-1) = u-1 For all n>0 Length of fraction period for 1/(u+v-1) in base u and v =6 Length of fraction period for 1/(u+v-1) in base u-1 and v-1 =3 Statement 2 If u=2n^2-3n+3; v=2n^2-n+2; p=u+v Then u^2 and v^2 =(p-1) mod p for all u,v. Length of fraction period for 1/(u+v) in base u and v =4 For all n There is also another case: Where u=n+1; v=n^2+n+1. Then u^2 and v^2 =(u+v-1) mod (u+v) for all u,v. Length of fraction period for 1/(u+v) in base u and v =4 For all n>0 And another Where u=2n+1; v=n^2. Then u^2 and v^2 =(u+v-1) mod (u+v) for all u,v. Length of fraction period for 1/(u+v) in base u and v =4 For all n>0 Randall McDonnell === Subject: Candy Inspiration (in the news) M & M packing: I guess I am not surprised they pack better than spheres. If M&Ms had been VERY flat, it seems intuitively obvious they would have packed even more efficiently. (as long they were oriented in the barrel so that neighboring oblate spheroids were mostly positioned, on average, in somewhat the same direction.) === Subject: Parity Check Matrix of a Systematic Linear Block Code windows-nt) The generator matrix of a systematic linear block code has the form G = [Ik : P]. How can it be shown that the parity check matrix is of the form H = [-P^T : In-k]? % Randy Yates % So now it's getting late, %% Fuquay-Varina, NC % and those who hesitate %%% 919-577-9882 % got no one... %%%% % 'Waterfall', *Face The Music*, ELO http://home.earthlink.net/~yatescr === Subject: Constraints on Self-Dual Linear Block Codes windows-nt) If (n,k) is a q-ary linear block code of length n and size q^k, then in order for the code to be self-dual it is fairly easy to show that n = 2*k. Are there any other constraints required of a code in order for it to be self-dual? === Subject: Re: One Of Them Sequence-Puzzles Again One more clue below original post. I will post answer in a couple/few days. > Here is the beginning of the sequence: > 1, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, > 38, 40, 42, 44, 3, 52, 54, 56, 58, 60, 62, ... > It, as a whole, has these 2 characteristics: > 1) It is a permutation of the positive integers. > 2) It relates somehow to a post I have made to sci.math/rec.puzzles in > the last few weeks. > (And, yeah, I know, the sequence has an infinite number of > definitions. > But this puzzle is for FUN. And what fun is it reminding us AGAIN of > the fact that there is no specific answer?) > Leroy Quet As with the beginning of the sequence, the even integers are always far more numerous than the odd integers among the first n terms, n >= 3. And the pattern of many strings of consecutive even integers continues indefinitely as well.