December 2006 - mathfun@mailman.xmission.com

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 04 Dec '06

04 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 04 Dec '06

04 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 01 Dec '06

01 Dec '06

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

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Re: [math-fun] [0, 1] to (0, 1)?
by Daniel Asimov 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)? ... 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? >> For a function f with just one discontinuity, one can map the half-open interval (0,1] continuously onto (0,1) using f: [0,1] -> (0,1) defined by f(x) = (1/2)(1 + sin(1/x)/(1+x)) for x > 0 and f(0) = 1. --Dan

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Re: [math-fun] [0, 1] to (0, 1)?
by Daniel Asimov 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? >> Any continuous function f: [0,1] -> (0,1) cannot be *onto*, just because the continuous image of a compact set is compact. Since a continuous map also preserves connectedness, f([0,1]) will be a closed interval [a,b] with 0 < a <= b < 1 (or a point if a = b). -------------------------------------------------------------------------- There is an interesting curiosity, however, that there *does* exist such a function, in fact a *homeomorphism*, from the rational numbers in [0,1] to the rational numbers in (0,1): h: Q \int [0,1] -> Q \int (0,1) where these spaces are topologized as subspaces of R. I'll leave it as a puzzle how to define such an h (for those who haven't already seen it). --Dan

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