December 2006 - mathfun@mailman.xmission.com

[math-fun] Is this familiar? For any f,
by R. William Gosper 10 Dec '06

10 Dec '06

j j ==== /===\ \ 2 j n | | n n 2 j > ( ) (- 1) | | (f(k) - -) (1 + -) = ( ). / j - n | | k k j ==== k = 1 n = 0 E.g., for f:=1-ap^(k-1) j j ==== /===\ \ 2 j n | | n n k-1 > ( ) (- 1) | | (1 + -) (1 - - - a p ) / j - n | | k k ==== k = 1 n = 1 (a; p) = 1 - ----------------------------------------------------- . j 2 j ( ) j --rwg

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RE: [math-fun] Ambiguously placed cities
by Andy Latto 09 Dec '06

09 Dec '06

> Q2. Are there 10 road lengths that lead to distinct town > configurations? > >> > -------------------------------------------------------------- > ---------------------------- > > Q1. Draw a non-isosceles triangle ABC with rational sides. Let M be > the midpoint of AB, and let DD' be a segment drawn through M that's > perpendicular to CM, such that the distances d(D,M) and > d(D',M) are equal > and rational. > > This implies distinct town configurations A,B,C,D and A,B,C,D' having > the same sest of 6 distances. The fact that DMD' is perpendicular to CM shows that the distances CD and CD' are equal. But I don't see why AD and AD' should be equal in this contstruction. Andy Latto andy.latto(a)pobox.com

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[math-fun] phone bills, dollars and cents
by Russ Cox 08 Dec '06

08 Dec '06

http://verizonmath.blogspot.com/ This is amazing. This guy has an unlimited data plan in the US and called to check on the roaming rates before going to Canada. The rep told him it was point zero zero two cents/KB, he asked for clarification that it was really 0.002 cents/KB and not 0.002 dollars/KB and the rep said yes, it is 0.002 cents/KB. Then he got the rep to write in the file that he was being quoted 0.002 cents/KB. He went to Canada, used 35000KB of traffic, and went back home to a bill of $70, not 70 cents. So he called Verizon customer service, tried to explain to the rep that he had been misquoted, and the rep couldn't get it. So he got bumped to a supervisor, and started recording the call. Both the supervisor, and then the floor manager simply cannot understand that 0.002 cents is not the same as 0.002 dollars. The floor manager even says "what do you mean .002 dollars?" the first time he tries to explain that the rate is actually $.002/KB. A sample (around 14:00 in the MP3): Guy: Do you recognize that there is a difference between one dollar and one cent? Verizon Floor Manager: Definitely. G. Do you recognize that there is a difference between half a dollar and half a cent? V. Definitely. G. Then do you therefore recognize that there is a difference between point zero zero two dollars and point zero zero two cents? V. No G. No? V. I mean there's no point zero zero two dollars. There's point zero zero two cents is what you're quoted and that I do show that you're paying for the kilobyte usage. I put a re-encoded copy of the MP3 at http://pdos.lcs.mit.edu/~rsc/verizon-math.mp3 since the one linked on his page is complicated to get at (10 clicks) and unnecessarily large. Russ

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[math-fun] Volume puzzle
by Daniel Asimov 08 Dec '06

08 Dec '06

The regular n-simplex delta_n(S_n) of side S_n is the convex span of n+1 points in n-space any two of which are a distance of S_n apart. PUZZLE: Assume S_n = K*n for some constant K > 0. Define f(n) = the n-volume of delta_n(S_n) Given that f(n) approaches a finite limit L > 0 as n -> oo, find L. --Dan

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[math-fun] another triumph in mathematical journalism
by Thane Plambeck 08 Dec '06

08 Dec '06

http://www.bbc.co.uk/berkshire/content/articles/2006/12/06/divide_zero_feat… -- Thane Plambeck tplambeck(a)gmail.com [*note new address*] http://www.plambeck.org/ehome.htm

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Re: [math-fun] phone bills, dollars and cents
by Daniel Asimov 08 Dec '06

08 Dec '06

For about 20 years, stores -- especially supermarkets -- have labeled especiallly produce bins with prices like .89 ¢ -- and despite my best efforts, no language columnist has chosen to write about this. Time to write an op-ed article. One more thing. I have the mistfortune of having Verizon as my telephone service provider (they virtually have the market cornered where I live) and no other company has ever struck me as nearly as disorganized with so many untrained employees. --Dan Russ wrote: << This is amazing. This guy has an unlimited [Verizon] data plan in the US and called to check on the roaming rates before going to Canada. The rep told him it was point zero zero two cents/KB, he asked for clarification that it was really 0.002 cents/KB and not 0.002 dollars/KB and the rep said yes, it is 0.002 cents/KB. Then he got the rep to write in the file that he was being quoted 0.002 cents/KB. . . . >>

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[math-fun] Integers j where zeta(j) is irrational
by Daniel Asimov 05 Dec '06

05 Dec '06

<< The last time I looked, essentially nothing was known about the odd zeta's. In fact, zeta(3) was shown irrational in 1978 by Roger Apéry. In 2001, Tanguy Rivoal showed that there are infinitely many odd (positive) integers at which zeta is irrational, including at least one value j in the range 5 <= j <= 21 (refined the same year by Zudilin to 5 <= j <= 11), at which zeta(j) is irrational. See < http://algo.inria.fr/seminars/sem01-02/rivoal.ps > for proofs of Rivoal's results, and citations for the others. (A classical result is that zeta(N) is rational for N <= 0, with zeta(2N) = 0 for 2N < 0.) --Dan

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Re: [math-fun] Now for something completely different
by Eugene Salamin 05 Dec '06

05 Dec '06

----- Original Message ---- From: Eugene Salamin <gene_salamin(a)yahoo.com> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Monday, December 4, 2006 7:57:46 AM Subject: Re: [math-fun] Now for something completely different The same reasoning used for the original problem applies more generally. The probability that gcd(i[1],...,i[n]) = gcd(j[1],...,j[n]) is zeta(2n)/zeta(n)^2. It seems strange that for n even the answer is rational, while for n odd the result is of unknown character, presumably transcendental. Gene Further results: The probability that a quantity k of r-tuples of random integers all have the same gcd is zeta(kr)/zeta(r)^k. The probability the gcd of an r-tuple of random integers divides the gcd of an n-tuple of random integers is zeta(n+r)/zeta(n). I wonder if it's possible to find relations between different patterns of this kind, that lead to a formula connecting zeta functions of odd argument to each other or to pi. Is it known if odd zeta's are algebraic or transcendental in Q(pi)? Gene ____________________________________________________________________________________ Need a quick answer? Get one in minutes from people who know. Ask your question on www.Answers.yahoo.com _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ____________________________________________________________________________________ Do you Yahoo!? Everyone is raving about the all-new Yahoo! Mail beta. http://new.mail.yahoo.com

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Re: [math-fun] Ambiguously placed cities
by Daniel Asimov 05 Dec '06

05 Dec '06

Ed writes: << Refering to http://www.mathpuzzle.com/AmbiguousTowns.gif . . . Between four towns are roads of length 3, 5, 6, 7, ~3.10, ~5.44 . There are two distinct town configurations. (See picture in link) Q1. Are there 6 integer road lengths that lead to distinct town configurations? Q2. Are there 10 road lengths that lead to distinct town configurations? >> ------------------------------------------------------------------------------------------ Q1. Draw a non-isosceles triangle ABC with rational sides. Let M be the midpoint of AB, and let DD' be a segment drawn through M that's perpendicular to CM, such that the distances d(D,M) and d(D',M) are equal and rational. This implies distinct town configurations A,B,C,D and A,B,C,D' having the same sest of 6 distances. By scaling everything up by the product of all 6 denominators, the distances also become integral. Q2. Given an answer to Q1, add a 5th town on the perpendicular bisector of the segment DD', and far away from the other 4. This guarantees 10 road lengths with two distinct town configurations. It's unclear to me if this can be done with all distances integral. (Apropos this last comment, it's an unsolved problem whether there is a point in the plane at integral distances from all 4 vertices of an NxN square, for any integer N.) --Dan

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Re: [math-fun] Now for something completely different
by Daniel Asimov 05 Dec '06

05 Dec '06

I asked: << I found this problem someone had posed, and after having fun solving it, I thought it might amuse some folks on math-fun: ------------------------------------------------------------------ Let f(N) be the probability that 4 random integers i,j,k,m in the range 1 <= i,j,k,m <= N satisfy gcd(i,j) = gcd(k,m) . Find the limit of f(N) as N -> oo. >> It looks like a number of folks have solved this just fine, but here is my solution: ------------------------------------------------------------------------------------ Suppose given integer N >= 1, then for each prime p, and k = 0,1,2,.... the probability that two random integers i,j in the range 1 <= i,j <= N have p^k as the power-of-p factor of gcd(i,j) is Prob(p^k | i && p^k | j && !(p^(k+1) | i && p^(k+1) | j) ) = (1/p^k) * (1/p^k) *(1 - 1/p^2), (where ! denotes falsity), at least in the limit as N -> oo. (All probabilities are calculated in the limit as N -> oo.) Thus Prob(gcd(i,j) and gcd(k,m) have the same highest power-of-p factor = p^k) = (1/p^4k) * (1 - 1/p^2)^2, and summing over k we get Prob(gcd(i,j) and gcd(k,m) have the same highest power-of-p factor) = (1 - 1/p^2)^2 * Sum{k = 0,...,oo} of 1/p^4k) = (1 - 1/p^2)^2 * 1/(1 - 1/p^4). Taking the product over all p gives via the Euler product Prob( gcd(i,j) = gcd(k,m) ) = zeta(4) / zeta(2)^2 = 36/90 = 2/5. ----------------------------------------------------------------------------------- The problem, by the way, comes from Noam Elkies's interesting Mathematical Miscellany website, at < http://www.math.harvard.edu/~elkies/Misc/index.html >, where it is mentioned (as someone asked about) that it's a long-standing unsolved problem to show Prod{p prime} of (p^2 - 1)/(p^2 + 1) = 2/5 by elementary means. --Dan

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