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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 04 Dec '06

04 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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Re: [math-fun] Now for something completely different
by Eugene Salamin 04 Dec '06

04 Dec '06

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 ____________________________________________________________________________________ Need a quick answer? Get one in minutes from people who know. Ask your question on www.Answers.yahoo.com

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[math-fun] Now for something completely different
by Daniel Asimov 04 Dec '06

04 Dec '06

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. ------------------------------------------------------------------ (No fair posting an answer if you've seen it before!) --Dan

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Re: [math-fun] Now for something completely different
by Daniel Asimov 04 Dec '06

04 Dec '06

I may be mistaken, but I have the impression that someone is solving a similar but different question (What is the probability that two random integers are relatively prime?) But this question asks for (essentially) the probability of something a bit different: that gcd(i,j) = gcd(k,m), for four random integers i,j,k,m. --Dan

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Re: [math-fun] Now for something completely different
by N. J. A. Sloane 03 Dec '06

03 Dec '06

By the way, the relevant sequence (which does not give the answer away) is A124162 - for those who prefer whole numbers to probabilities. Neil

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Re: [math-fun] Now for something completely different
by Daniel Asimov 03 Dec '06

03 Dec '06

I will add that figuring out how to evaluate the expression one gets for the answer is not intended to be the crux of the problem. (The evaluation part is part of extremely standard and easy-to-find mathematics literature, including on line.) As for posting the answer, I usually like to wait a few days to do that, if only so that those people who haven't checked their e-mail today will get a chance to think about it. (Will post it Tues. or Weds. if no one else has.) --Dan

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[math-fun] [0, 1] to (0, 1)?
by Kerry Mitchell 01 Dec '06

01 Dec '06

More of an intellectual curiosity question than anything useful. Is there a "nice" function that is one-to-one and continuous that maps the closed interval [0, 1] to the open interval (0, 1)? Ideally, such a "nice" function would be not a piecewise function or a power series and something that could be implemented with standard library functions. If such a thing doesn't exist, is there at least a "not nice" function that is one-to-one and maps the intervals? If not, why not? Thanks for any pointers or wisdom, Kerry Mitchell -- lkmitch(a)gmail.com www.fractalus.com/kerry

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RE: [math-fun] Add odd digits, subtract even
by Eric Angelini 01 Dec '06

01 Dec '06

[Joshua Zucker] : > I played around with this a little, and found: (...) ... thanks, Joshua -- very interesting indeed ! I'm trying to find a way to transform what you have found into a genuine sequence for Neil ! Best, É.

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CORRECTION Re: [math-fun] [0, 1] to (0, 1)?
by Daniel Asimov 30 Nov '06

30 Nov '06

I meant to define f(0) = 1/2 (not 1). --Dan

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