January 2009 - mathfun@mailman.xmission.com

[math-fun] Oliver Atkin passed away December 28th
by rcs＠xmission.com 01 Jan '09

01 Jan '09

(from the Number Theory list --Rich) ----- Forwarded message from henriettaatkin(a)yahoo.com ----- Date: Tue, 30 Dec 2008 11:57:51 -0500 From: Henrietta Atkin <henriettaatkin(a)yahoo.com> Reply-To: Henrietta Atkin <henriettaatkin(a)yahoo.com> Subject: My father, Dr. Oliver Atkin, passed away December 28th. To: NMBRTHRY(a)LISTSERV.NODAK.EDU My father had a fall down the stairs of his home on December 23rd. Although he had a fractured vertebra, and other injuries, he was doing very well until he contracted hospital pneumonia. He weakened rapidly after that, and passed away December 28th. We wanted to invite everyone from the math community at UIC to his wake and funeral service. Visitation will be held at Hurston Funeral Home from 3-9 p.m. on Friday, January 2. It's located at 4001 W. Mannheim Road, Hillside-Westchester (corner of Roosevelt and Mannheim roads). His funeral will be held at Grace Episcopal Church, 924 Lake Street, Oak Park on Saturday at 11 a.m. He had fought a battle with prostate cancer about a year ago, and was dealing with other health issues. Naturally, my brother and I are saddened at his unexpected passing, but the end was very peaceful. Henrietta Atkin ----- End forwarded message -----

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Re: [math-fun] math-fun Digest, Vol 71, Issue 1
by Steve Witham 01 Jan '09

01 Jan '09

>From: rwg(a)sdf.lonestar.org > >PS, Veit Elser's difference map algorithm, >http://en.wikipedia.org/wiki/Difference_map_algorithm , has become only >the second entity to solve the 82% Arnold Dozenegger disk packing puzzle >completely unaided. (Not counting Emma Cohen, who got massive clues from > Emma Cohen.) Also, it's clear that Rod Stephenson's clustering algorithm >will do it, probably in ~1 hr --way longer than Veit's, who clearly has >something dangerous. "Searching with Iterated Maps", free at: http://www.pnas.org/content/104/2/418.full.pdf Yes, scary. --Steve

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Re: [math-fun] http://gosper.org/Snowflake.xml
by rwg＠sdf.lonestar.org 01 Jan '09

01 Jan '09

Uh oh, it looks like we need some kind of Mathematica Player to read the formulae on that page. Anybody know how to win here? > On 1/1/09, rwg(a)sdf.lonestar.org <rwg(a)sdf.lonestar.org> wrote: >> is an addendum to http://mathworld.wolfram.com/KochSnowflake.html . >> Summary: Exact Fourier series and exact spacefill. >> --rwg > > The pictures are intriguing --- once the URL has been disinterred from its lair > in the subject line! > > But (using Firefox) something goes awry with the displayed equations --- in > several places (e.g. the first such) I see the square root of a black rectangle > above a digit 3. > > WFL >

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Re: [math-fun] Correlation puzzle
by Eugene Salamin 01 Jan '09

01 Jan '09

From: Dan Asimov <dasimov(a)earthlink.net> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Thursday, January 1, 2009 12:35:00 PM Subject: [math-fun] Correlation puzzle Phil wrote in the "recent illogical conflation" thread: << It's rather "the enemy of my enemy is my friend"-like, which is also commonly used. >> This reminds me of an interesting statistics thang: (I don't recall posting this here, but if I did please forgive me.) PUZZLE ------ Given real-valued random variables X,Y,Z in a joint distribution, suppose that the correlation coefficient* between any two of them is the same number C. QUESTION: What is the minimum value, over all such joint distributions, that C can take? (We assume all means and variances are finite.) --Dan _____________________________________________________________ * There are several types of correlation coefficient, but this refers to what is by far the commonest one: CC(U,V) = E(U_s * V_s) where W_s denotes the standardized version of the random variable W: W_s := (W - E(W)) / sd(W) where E is expectation and sd is standard deviation. _____________________________________________________________________ The minimum C is -1/2. Without changing C, we may shift x, y, and z so that each has zero mean, and then scale them so that each has unit variance. Then 0 <= E((x + y + z)^2) = 3 + 6 C, so C >= -1/2. To actually achieve this value, let the distribution consist of the three points (2, -1, -1), (-1, 2, -1), and (-1, -1, 2). We have E(x) = 0, etc., E(x^2) = 2, etc., E(xy) = -1, etc. Then C = E(xy)/(sqrt(E(x^2)) sqrt(E(y^2))) = -1/2. Gene

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[math-fun] Correlation puzzle
by Dan Asimov 01 Jan '09

01 Jan '09

Phil wrote in the "recent illogical conflation" thread: << It's rather "the enemy of my enemy is my friend"-like, which is also commonly used. >> This reminds me of an interesting statistics thang: (I don't recall posting this here, but if I did please forgive me.) PUZZLE ------ Given real-valued random variables X,Y,Z in a joint distribution, suppose that the correlation coefficient* between any two of them is the same number C. QUESTION: What is the minimum value, over all such joint distributions, that C can take? (We assume all means and variances are finite.) --Dan _____________________________________________________________ * There are several types of correlation coefficient, but this refers to what is by far the commonest one: CC(U,V) = E(U_s * V_s) where W_s denotes the standardized version of the random variable W: W_s := (W - E(W)) / sd(W) where E is expectation and sd is standard deviation. _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] recent illogical conflation
by Phil Carmody 01 Jan '09

01 Jan '09

From: Henry Baker <hbaker1(a)pipeline.com> > In simplest terms, the logical fallacy is essentially the > following: > > A is a subset of (not C) > B is a subset of (not C) > > Therefore, A is somehow "equivalent" to B. It's rather "the enemy of my enemy is my friend"-like, which is also commonly used. Phil WILLIAM RALPH GOSPER ~ PROGRAMS WILL HELP A.I.

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