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[math-fun] What's the cross product of two mathematicians?
by Schroeppel, Richard 17 Dec '05

17 Dec '05

Congratulations to Scott & Cheryl Beaver on their early Christmas present! --Rich ----- Cheryl writes We are happy to announce the arrival of baby Evan Walter Beaver on Wednesday December 14, 2005 at 8:51 am. Evan weighed in at 7 lbs. 5 oz and 20 1/4 inches long. Everything went great and the whole family is doing fine.

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[math-fun] 2F1(golden ratio)
by R. William Gosper 16 Dec '05

16 Dec '05

I have the weird feeling I once knew this and forgot it, but since I no longer have a decent message massager (or "masausage", as a young friend recently misspoke) we'll have to risk repetition. 1 - sqrt(5) hyper_2f1(a, 5 a + 1, 3 a + 1, -----------) = 2 5 a --- 2 3 2 pi 5 (a - -)! (a - -)! 4 %pi cos(--) 5 5 10 (3 a)! (---------------------- + ----------------------------) pi 5 a (5 a)! 4 %pi cos(--) --- 10 2 3 2 5 a! 5 (a - -)! (a - -)! 5 5 Note G/(5a)! + 1/G/5a!. I think this is almost, but not quite derivable from the formulas in A&S Chapter 15, starting with 15.1.31, and then various linear and quadratic transformations. The obstacle seems to be A&S's negligence to provide contiguous pairs both for 15.1.31 and the quadratic transformations. The latter can be somewhat tediously derived from the three-term recurrences, but 15.1.31 seems to be cursed with some sort of singularity that thwarts population of the contiguity grid starting with F(a) and F(a+-3). (Of course, the good way to do things is with 3x3 matrices that have the contiguous pair built in. I'm belatedly nagging the DLMF people about this.) There are also similar identities for 2F1(phi^-2). Once all this goes into the HYPERSIMP facility, we'll see whether 2F1(a,5a,3a,1/phi) or some contiguous neighbor has a monomial rhs. --rwg

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[math-fun] integral identities for Pi and Log[2]
by Pacher Christoph 14 Dec '05

14 Dec '05

Hi, I found (redevelopped ?) some integral-identities for Pi and Log[2]. Starting from Eq. 34 of http://mathworld.wolfram.com/PiFormulas.html, i.e. Pi = 22/7 - Integrate[x^4(1 - x)^4/(1 + x^2), {x, 0, 1}], (in this special case 22/7 is a continued fraction of Pi) I found the more general relations: Pi = p_n/q_n -(-1)^n Integrate[x^(2n)(1 - x)^4/(1 + x^2), {x, 0, 1}], n>=0 where p_n/q_n is not a cont. fraction of Pi for n != 2. For increasing n the integral approaches quite slowly 0. The sequence p_n/q_n starts with 10/3, 47/15, 22/7, 1979/630, 10886/3465, 141511/45045, 141514/45045, 9622853/3063060, 45708802/14549535, 45708659/14549535, 150185878/47805615, 10512998863/3346393050, 3153902146/1003917915, 457315566013/145568097675, 14176787858138/4512611027925, 113414272443349/36100888223400, 14176786835558/4512611027925, 524541036307361/166966608033225, 40349314946642/12843585233325, 43012366079184317/13691261858724450, 924765931862392498/294362129962575675 Note the adjacent terms with the denominators 45045 and 14549535... -- For Log[2] I found the following (pls. note the strong similarity to the Pi integrals) Log[2] = 1 - Integrate[(1 - x)^2/(1 + x^2), {x, 0, 1}] Log[2] = 2/3 + Integrate[x^2(1 - x)^2/(1 + x^2), {x, 0, 1}] Log[2] = 7/10 - Integrate[x^4(1 - x)^2/(1 + x^2), {x, 0, 1}] Interestingly, 1, 2/3 and 7/10 are continued fractions of Log[2]. The general form is Log[2] = r_n/s_n -(-1)^n Integrate[x^(2n)(1 - x)^2/(1 + x^2), {x, 0, 1}], n>=0 but for n>2 r_n/s_n is not a continued fraction (at least I have not found any further). The sequence r_n/s_n starts with 1, 2/3, 7/10, 29/42, 25/36, 457/660, 541/780, 97/140, 9901/14280, 33181/47880, 1747/2520, 441871/637560, 96079/138600, 749293/1081080, 7244119/10450440, 1548577/2234232, 99917/144144, 8492321/12252240, 62847065/90666576, 32271119/46558512, 6615877847/9544494960 it looks that (2n+1)|s_n, and very often 6|s_n... And again, for increasing n the integral approaches quite slowly 0. I am quite sure that one could derive direct expressions for p_n,q_n,r_n,s_n through recurrence relations obtained by partial integration... -- In general Integrate[x^n(1 - x)^m/(1 + x^2)^p, {x, 0, 1}] , n,m,p>=0 gives a linear combination with rational coefficients of 1, Pi and Log[2]. The Pi and Log[2] come from the denominator in the integrals, so other polynomials (orthogonal ones?) in the numerator might give further interesting identities... Christoph

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RE: [math-fun] Erdos-Bacon number
by Schroeppel, Richard 13 Dec '05

13 Dec '05

It's a dubious distinction, but my Fischer number is 2. The intermediary is a friend from high school days, Kate Sillars, who managed to spring an opening surprise on him in a simultaneous exhibition. This would likely translate into a Morphy number of 6 or 7. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Jud McCranie Sent: Tuesday, December 13, 2005 11:06 AM To: math-fun Cc: math-fun Subject: Re: [math-fun] Erdos-Bacon number At 12:48 PM 12/13/2005, Jud McCranie wrote: >There are also Morphy numbers - chessmaster Paul Morphy had Morphy >Number 0. If you played a game with Morphy, you have Morphy number 1, >etc. The sum of Erdos number and Morphy number was discussed on a >chess newsgroup about a year ago. See http://www.oakland.edu/enp/related.html _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Erdos-Bacon number
by Mike Stay 13 Dec '05

13 Dec '05

http://en.wikipedia.org/wiki/Erdos-Bacon_number -- Mike Stay metaweta(a)gmail.com http://math.ucr.edu/~mike

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Re: [math-fun] Origamic Number Theory
by Michael Reid 13 Dec '05

13 Dec '05

> A Pythagorean triple is any integer solution of x^2+y^2=z^2 and, as we > know, one gets all solutions by choosing any integers n>m. and defining > T(m,n) by, > x=n^2-m^2n, > y=2mn, > z=m^2+n^2. > (T stands for triple and/or triangle) > > CHEERFUL FACT > To see a given solution lay an mxn sheet of paper on the desk > A_____ m ____ B > > > n n > > > C_____m_____D > > Fold and crease so that corner D falls on corner A. The triangle bounded by > AB and the image of BD is (up to similarity) T(m,n). this is not a coincidence. given a pythagorean triple, (a, b, c) , we get an element in the field Q(i) of norm 1 , namely (a + bi) / c . a consequence of hilbert's "theorem 90" is that any element of norm 1 is of the form z / z^bar for some z in the field Q(i) . multiplying z by a rational number does not change z / z^bar , so we may even take z to be in the ring Z[i] , i.e. a gaussian integer. also, pi / 2 > arg((a + bi) / c > 0 , and arg(z / z^bar) = 2 arg(z) . therefore we may take z = n + mi , where m and n are integers satisfying n > m > 0 . this gives the parametrization above. let P be the point on BD which is one of the endpoints of the diagonal crease, so that ABP is the right triangle of interest. simple geometry shows that angle APB = 2 (angle ADB) . since angle ADB = arg(n + mi) , it follows that angle APB = arg((n + mi) / (n - mi)) , which shows that triangle APB is similar to the pythagorean triangle given by parameters m and n . mike

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RE: [math-fun] Erdos-Bacon number
by Jud McCranie 13 Dec '05

13 Dec '05

At 01:23 PM 12/13/2005, Schroeppel, Richard wrote: > It's a dubious distinction, but my Fischer number is 2. >The intermediary is a friend from high school days, Kate Sillars, >who managed to spring an opening surprise on him in a >simultaneous exhibition. This would likely translate into >a Morphy number of 6 or 7. It is probably 6. I played Norman Whitaker, who is a 3, and I *think* Fisher played them. I know they knew each other, so I assume they played )although there is no official match or tournament game listed).

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RE: [math-fun] Erdos-Bacon number
by Schroeppel, Richard 12 Dec '05

12 Dec '05

Did Euler or Fermat ever have coauthors? Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of victor(a)idaccr.org Sent: Monday, December 12, 2005 3:40 PM To: math-fun(a)mailman.xmission.com Subject: Re: [math-fun] Erdos-Bacon number Very interesting. My Erdos Number is 2, and I was an extra in "A Beautiful Mind" (unfortunately not credited). So if I had gotten credit I would have 4 :-). Close, but no cigar. Victor _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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RE: [math-fun] Lie (circle) geometry
by dasimov＠earthlink.net 11 Dec '05

11 Dec '05

Fred writes: << Is anybody interested in the Laguerre group of equilong transformations --- nowadays apparently subsumed, along with the Moebius group, into something called Lie geometry? I've looked in books by Cecil and by Coolidge, and at numerous web sites, but I'm not much nearer understanding what a Laguerre involution --- let alone a general equilong transformation --- actually _looks_ like! I could say (much) more, but I'll refrain until I find out if there's any takers out there. >> The Moebius group PSL(2,C) of conformal automorphism of S^2 is pretty darn fascinating, so whatever this is about sounds promising. Can you tell us at least a bit more about what these equilong thingies are (even if the def. is somewhat opaque)? --Dan

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Re: [math-fun] Origamic Number Theory
by David Gale 11 Dec '05

11 Dec '05

> >Moving on a little, if D is superposed onto A, then its image D' = A lies >on AB, and the "triangle" referred to appears to have zero area. >Might this submission have benefitted from more careful presentation? Gosh, I was hoping someone would grab a piece of 8.5 by 11 and actually do it. The image of BD (under the folding map) is the broken line forming the hypotenuse and one side of the triangle. This of course only gives the "shape" of the Pyth. triple, which is all one could expect. dg >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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