October 2007 - mathfun@mailman.xmission.com

Re: [math-fun] radio math
by James Propp 31 Oct '07

31 Oct '07

Dan writes: >Darn, can't they broadcast it later? This means I'll have to wake up at 5 a.m! They're recording the conversation, and a few days from now it'll be possible to listen to the conversation via podcast. Jim

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Re: [math-fun] radio math
by Dan Asimov 31 Oct '07

31 Oct '07

Jim wrote: << Thane asks: >what timezone? Oops, sorry about that. It's 8:05 a.m. Eastern time. >> Darn, can't they broadcast it later? This means I'll have to wake up at 5 a.m! --Dan

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Re: [math-fun] radio math
by James Propp 31 Oct '07

31 Oct '07

Thane asks: >what timezone? Oops, sorry about that. It's 8:05 a.m. Eastern time. Jim

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Re: [math-fun] radio math
by James Propp 30 Oct '07

30 Oct '07

My "morning math" gig will be at 8:05 a.m. tomorrow, on WUML Sunrise (91.5 FM or on the web at http://www.wuml.org/webcast.php) I've been told that the conversation will last about half an hour. Jim Propp

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P.S. Re: Correction Re: [math-fun] random walk
by Dan Asimov 30 Oct '07

30 Oct '07

he probabiity that a random walk (with probabiity p of stepping to the left) will, starting form 0, ever reach a negative number is L = (1-sqrt(1-4pq)) / 2q, which simplifies to L = (1-|2p-1|) / 2(1-p). Considering the cases p >= 1/2 and p <= 1/2 separately: ------------------------------- L(p) = 1 if p >= 1/2, and L(p) = p/(1-p) if p <= 1/2. ------------------------------- I find this highly bipartite behavior rather amusing. --Dan

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Correction Re: [math-fun] random walk
by Dan Asimov 30 Oct '07

30 Oct '07

Having first solved this with the meanings of p & q interchanged (since p is traditionally the prob. of stepping to the right in a random walk on Z), I flubbed the denominator. The following is corrected: ------------------------------------------------------------------------ Bill Cordwell asks for a derivation of the probability that a random walk on Z, starting from 0, ever reaches a negative number, where the probability of any one step's going to the left = p. Let L be the desired probability: that the walk ever reaches -1. Then L = p + q*L^2 (q = 1-p), since if the first step is to the right, the probability of then ever reaching -1 requires first ever reaching 0 (prob = L by translation-invariance) and then ever reaching -1 from there (L again). Which gives L = (1+-sqrt(1-4pq)) / 2q. The + sign would give probabilities > 1, so the answer must be L = (1-sqrt(1-4pq)) / 2q. ------------------------------------------------------------------------

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Re: [math-fun] random walk
by Dan Asimov 30 Oct '07

30 Oct '07

Bill Cordwell asks: << I'm looking for a simple derivation of the probability that a random walk starting at 0, with unit steps, will ever end up to the left of 0. Probability that the person goes to the left (towards the negative) at any step is p. >> Here's a stab at it: Let L be the desired probability: that the walk ever reaches -1. Then L = p + q*L^2 (q = 1-p), since if the first step is to the right, the probability of then ever reaching -1 requires first ever reaching 0 (prob = L by translation-invariance) and then ever reaching -1 from there (L again). Which gives L = (1+-sqrt(1-4pq))/(2p). The + sign would give probabilities > 1, so the answer must be L = (1-sqrt(1-4pq))/(2p). --Dan

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[math-fun] Re: radio math
by Phil Carmody 30 Oct '07

30 Oct '07

From: James Propp <jpropp(a)cs.uml.edu> > I've been invited to speak on a college morning radio program next > week, on the topic of mathematical proof and infinity. It'll be a > conversation with two interviewers (no call-ins). I've not notice anyone mention proofs of the infinitude of primes yet. That should be simple enough for anyone reasonably awake to understand, and introduces both important techniques used in mathematical proof as well as an infinity. Phil () ASCII ribbon campaign () Hopeless ribbon campaign /\ against HTML mail /\ against gratuitous bloodshed [stolen with permission from Daniel B. Cristofani] __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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Re: [math-fun] New York Times Joke of the Day for Sunday
by Henry Baker 29 Oct '07

29 Oct '07

Supposedly, Feynmann would sign in at Los Alamos, sneak out & come back & sign in again: http://oakridgevisitor.com/history/behindfence1.html Richard Stallman, of GNU fame, used to do the same thing at Project MAC, MIT. Even though Project MAC was never a "secure" facility like Los Alamos, this behavior still drove the guards crazy. At 11:53 AM 10/29/2007, Dan Asimov wrote: >>From the N.Y. Times, October 28: > ><< >Joke of the Day > >A mathematician is someone who thinks that if there are supposed to be three people in a room and five walk out, then two must enter the room in order for it to be empty. > >Â Found on an mitadmissions.org blog. >>> > >--Dan

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[math-fun] New York Times Joke of the Day for Sunday
by Dan Asimov 29 Oct '07

29 Oct '07

>From the N.Y. Times, October 28: << Joke of the Day A mathematician is someone who thinks that if there are supposed to be three people in a room and five walk out, then two must enter the room in order for it to be empty. — Found on an mitadmissions.org blog. >> --Dan

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