March 2005 - mathfun@mailman.xmission.com

[math-fun] decagonal gamma product
by R. William Gosper 18 Mar '05

18 Mar '05

One might hope to generalize the factorial reflection formula, z! (-z)! = pi z/(sin pi z) to include things like z! (wz)! (z/w)!, with w^3 = 1, but these have unilaterally infinite rows of singularities, whereas finite combinations of elementary functions have their singularities bilaterally infinite. So the only oddgons we can do are the special cases where pairs of vertices sum to integers. E.g. for triangles of integer radius centered at the origin, 2 even! (even w)! (even w )! even + 1 %pi even 2 even! = -----------------------------------------------------, even ---- 2 /===\ sqrt(3) %pi even | | 2 2 sqrt(3) sinh(----------------) | | (4 k + 3 even ) 2 | | k = 1 with a similar expression for odd!. We can also slide the triangle left or right by an integer. For a 2n-gon of any phase and radius, we can just combine opposite vertices in pairs: %i %pi n 2 n - 1 %i %pi k -------- /===\ -------- n 2 n | | n %i %pi %e z | | (%e z)! = - ---------------------, | | f (%pi z) k = 0 n n - 1 %i %pi k /===\ -------- | | n f (z) = | | sin(%e z). n | | k = 0 f_n(z) is fairly interesting. For the decagon, f_5(z) = i pi 2 i pi 3 i pi 4 i pi ---- ------ ------ ------ 5 5 5 5 sin(z) sin(e z) sin(e z) sin(e z) sin(e z) 15 25 5 z 521 z = z - ----- + ---------------- + . . . 93555 4593988395871875 inf ==== j 10 j + 5 \ (- 1) (fib(10 j + 6) + fib(10 j + 4) + 1) (2 z) 5 > -------------------------------------------------------- / (10 j + 5)! ==== j = 0 = ---------------------------------------------------------------- 16 (after strenuous simplification) where the fibby factor can be written as Lucas(10j+5) + 1, since Lucas(n) = fib(n-1) + fib(n+1) = fib(2n)/fib(n). The powers of 2 in the coefficients exactly cancel in terms 5,15,25,45,65,75,105,... (What is the largest n for which the numerator of 2^n/n! is 2? 4? 8? ...) For the hexagon, f_3(z) = %i sqrt(3) 1 %i sqrt(3) 1 sin(z) sin((---------- - -) z) sin((---------- + -) z) 2 2 2 2 inf ==== i 6 i + 3 \ 3 (- 1) (2 z) = > (- ---------------------) / 4 (6 i + 3)! ==== i = 0 9 15 21 3 z 4 z 2 z = - z + --- - --------- + -------------- + . . ., 945 212837625 64965492466875 and for the octagon, f_4(z) = (%i - 1) z (%i + 1) z %i sin(z) sinh(z) sin(----------) sin(----------) = sqrt(2) sqrt(2) inf ==== 4 i + 2 4 i + 2 i 8 i + 4 \ %i ((2 sqrt(2) + 4) + (4 - 2 sqrt(2)) ) (- 1) z > (- --------------------------------------------------------------------) / 2 (8 i + 4)! ==== i = 0 12 20 4 %i z (59 %i) z = - %i z + ------ - ------------- + . . . 9450 1302566265000 --rwg PS, "square" tree rings: Replace one side of a square with a semicircle to make a figure with two corners and three straight sides. Near me is a large, sawn-off tree stump whose outer rings are routinely round, but about halfway in gradually deform to the aforementioned shape, with sides quite straight and corners quite sharp. Explain.

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[math-fun] ~s highly miscellaneous hypergeometric identity
by R. William Gosper 17 Mar '05

17 Mar '05

Just to shake off some rust, I fired up an old Macsyma notebook containing some matrix product rearrangements. A variable dropped out from an unusual spot, producing oo n 5 ==== (n + a) (11 n + 6 a + 1) 4 (n + a - -)! (n + 2 a - 2)! (2 n + a - 1)! \ 6 > ---------------------------------------------------------------------- / 2 ==== (n + a - -)! (3 n + 2 a + 1)! n = 0 3 1 3 3 (-)! (a - -)! 3 2 = ---------------, 2/3 4 an ungainly 7F6[16/27]. This is a fairly unusual z for a closed form. Also strange is the single a on the right, and the two linear factors in the summand. But oddest is the absence of n!, or (2n)!, or ..., from the summand's denominator, thanks to the dropped variable. Normally, closed forms like this require "termination to the left". So prevalent is that denominator factor that pFq notation supplies it implicitly, yet here it must be artificially (and ungainly) canceled out. A similar but lesser curiosity is oo ==== 2 2 \ 3 k + (3 a + 1) k + a > --------------------------------- = 1, / 2 k + a + 1 ==== (k + a) (k + a + 1) ( ) k = 0 k where the k! is in the numerator instead of the denominator. A slightly uglier cousin gives pi for all a (from rational terms). --rwg Holey mackerel, scombroid cobordism. PPS, the history of the Cal\Nevada border (MKleber> http://laws.findlaw.com/us/447/125.html) is indeed fascinating, The surveying uncertainties dwarfed the mathematical ones. Besides the inevitable problem with longitude, the south end was defined to be where the Colorado River crossed the 35th parallel, which turned out to be highly variable. The town of Aurora, claimed by both states, sent representatives to both legislatures, who simultaneously became house speaker (1862).

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[math-fun] better posed problem? (small surprise)
by Meeussen Wouter (bkarnd) 17 Mar '05

17 Mar '05

it is a small surprise that, for w = Exp[ 2 Pi I /30 ] alias 1^(1/30) , w+w^7+w^8+w^18+w^19+w^25 = 0 no cancellation of symmetric pairs, triples or quintuples in sight, but w+w^7+w^19+w^25 = -w^13 (* four out of five *) and w^8+w^18 = -w^28 = +w^13 (* two out of three *) could be called "cancellation by missing parts" ;-)) Wouter Meeussen Senior Scientist nv Vandemoortele Izegem R&D Center Albertlaan 79, B8870 Izegem tel +32 (0)51 33 21 24 fax +32 (0)51 33 21 75 =============================== This email is confidential and intended solely for the use of the individual to whom it is addressed. If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited. You are explicitly requested to notify the sender of this email that the intended recipient was not reached.

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[math-fun] click the little books as they pop up to spell "number theory"
by Thane Plambeck 16 Mar '05

16 Mar '05

http://amaztype.tha.jp/US/Books/Title?q=number+theory -- Thane Plambeck http://www.plambeck.org/ehome.htm

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RE: [math-fun] CA/NV border dispute
by Henry Baker 16 Mar '05

16 Mar '05

I have to drive to Las Vegas in a couple of weeks. Any chance that I could fall through a hole at the CA/NV border? Is there a "no man's land", where neither the CA nor the NV highway patrol will give tickets? Does this have anything to do with Area 51? At 12:18 PM 3/16/2005, Schroeppel, Richard wrote: >PPS, the history of the Cal\Nevada border (MKleber> >http://laws.findlaw.com/us/447/125.html) >is indeed fascinating, The surveying uncertainties dwarfed the >mathematical ones. Besides the inevitable problem with longitude, the >south end was defined to be where the Colorado River crossed the 35th >parallel, which turned out to be highly variable. The town of Aurora, >claimed by both states, sent representatives to both legislatures, who >simultaneously became house speaker (1862).

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RE: [math-fun] ~s highly miscellaneous hypergeometric identity
by Schroeppel, Richard 16 Mar '05

16 Mar '05

For just a moment I thought you'd busted the Curse of Gamma that prevents us from finding a good series for ! that will give a T4 computing time for real! . But it does appear that you've found a polynomial diffeqn for !, which I thought had been proved impossible. Rich PS: Utah-Colorado border has a glitch. I noticed it on a large map & thought "printing error", but a web search turned up the story of a surveying error. The UT-CO glitch is a moderately long segment that isn't quite due North. --R -----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 R. William Gosper Sent: Wednesday, March 16, 2005 11:18 AM To: math-fun(a)mailman.xmission.com Subject: [math-fun] ~s highly miscellaneous hypergeometric identity Just to shake off some rust, I fired up an old Macsyma notebook containing some matrix product rearrangements. A variable dropped out from an unusual spot, producing oo n 5 ==== (n + a) (11 n + 6 a + 1) 4 (n + a - -)! (n + 2 a - 2)! (2 n + a - 1)! \ 6 > ---------------------------------------------------------------------- / 2 ==== (n + a - -)! (3 n + 2 a + 1)! n = 0 3 1 3 3 (-)! (a - -)! 3 2 = ---------------, 2/3 4 an ungainly 7F6[16/27]. This is a fairly unusual z for a closed form. Also strange is the single a on the right, and the two linear factors in the summand. But oddest is the absence of n!, or (2n)!, or ..., from the summand's denominator, thanks to the dropped variable. Normally, closed forms like this require "termination to the left". So prevalent is that denominator factor that pFq notation supplies it implicitly, yet here it must be artificially (and ungainly) canceled out. A similar but lesser curiosity is oo ==== 2 2 \ 3 k + (3 a + 1) k + a > --------------------------------- = 1, / 2 k + a + 1 ==== (k + a) (k + a + 1) ( ) k = 0 k where the k! is in the numerator instead of the denominator. A slightly uglier cousin gives pi for all a (from rational terms). --rwg Holey mackerel, scombroid cobordism. PPS, the history of the Cal\Nevada border (MKleber> http://laws.findlaw.com/us/447/125.html) is indeed fascinating, The surveying uncertainties dwarfed the mathematical ones. Besides the inevitable problem with longitude, the south end was defined to be where the Colorado River crossed the 35th parallel, which turned out to be highly variable. The town of Aurora, claimed by both states, sent representatives to both legislatures, who simultaneously became house speaker (1862). _______________________________________________ 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] better posed problem?
by wouter meeussen 12 Mar '05

12 Mar '05

of all k-subsets (including the empty subset) of the n-th roots of 1, say n= 0 to 20, 0>= k>=n, how many add to zero? 1, 1, 2, 2, 4, 2, 10, 2, 16, 8, 34, 2, 100, 2, 130, 38, 256, 2, 1000, 2, 1156 ... {1} {1, 0}, {1, 0, 1}, {1, 0, 0, 1}, {1, 0, 2, 0, 1}, {1, 0, 0, 0, 0, 1}, {1, 0, 3, 2, 3, 0, 1}, {1, 0, 0, 0, 0, 0, 0, 1}, {1, 0, 4, 0, 6, 0, 4, 0, 1}, {1, 0, 0, 3, 0, 0, 3, 0, 0, 1}, {1, 0, 5, 0, 10, 2, 10, 0, 5, 0, 1}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {1, 0, 6, 4, 15, 12, 24, 12, 15, 4, 6, 0, 1}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {1, 0, 7, 0, 21, 0, 35, 2, 35, 0, 21, 0, 7, 0, 1}, {1, 0, 0, 5, 0, 3, 10, 0, 0, 10, 3, 0, 5, 0, 0, 1}, {1, 0, 8, 0, 28, 0, 56, 0, 70, 0, 56, 0, 28, 0, 8, 0, 1}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {1, 0, 9, 6, 36, 36, 96, 90, 162, 128, 162, 90, 96, 36, 36, 6, 9, 0, 1}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {1, 0, 10, 0, 45, 4, 120, 20, 210, 40, 256, 40, 210, 20, 120, 4, 45, 0, 10, 0, 1}} ?? it feels funny to take the empty subset of the zero'th (?) roots of 1. Superseek comes up empty handed on the rowsum 2,2,4,2,10,...,1156 Thanks to Daniel Asimov, Franklin T. Adams-Watters and Mike Speciner for their kind tutoring. Will submit to OEIS unless anyone still sees bugs. W. ----- Original Message ----- From: "wouter meeussen" <wouter.meeussen(a)pandora.be> To: "Seqfan (E-mail)" <seqfan(a)ext.jussieu.fr>; "math-fun" <math-fun(a)mailman.xmission.com> Sent: Friday, March 11, 2005 11:21 PM Subject: well posed problem? of all subsets of the n-th roots of -1, n= 1 to 20, how many add to zero? ?? 0, 1, 1, 3, 1, 6, 1, 11, 4, 8, 1, 50, 1, 10, 9, 107, 1, 240, 1, 316 ?? where are the 'groupies' when you need them? W.

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RE: [math-fun] better posed problem?
by Daniel Asimov 12 Mar '05

12 Mar '05

<< of all k-subsets (including the empty subset) of the n-th roots of 1, say n= 0 to 20, 0>= k>=n, how many add to zero? 1, 1, 2, 2, 4, 2, 10, 2, 16, 8, 34, 2, 100, 2, 130, 38, 256, 2, 1000, 2, 1156 ... >> Some patterns here look intriguing. Like f(2^n) = 2^(2^(n-1)) for one, and f(6n) = 10^n for another. Do these extend? --Dan

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[math-fun] Provable semiprime, probably unfactorable.
by Phil Carmody 12 Mar '05

12 Mar '05

After last week's http://www.graysage.com/djr/isp.txt http://physics.open.ac.uk/~dbroadhu/cert/semgpch.gp After (DB) finding and fixing (me) a bug in Pari/GP, this was only a fraction of a day's work. Larger ones would be easy (for DB, not me) to churn out. Phil When inserting a CD, hold down shift to stop the AutoRun feature In the Device Manager, disable the SbcpHid device. http://www.cs.princeton.edu/~jhalderm/cd3/ __________________________________ Do you Yahoo!? Yahoo! Sports - Sign up for Fantasy Baseball. http://baseball.fantasysports.yahoo.com/

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Re: [math-fun] Keen approximations
by Phil Carmody 12 Mar '05

12 Mar '05

From: ed pegg <ed(a)mathpuzzle.com> > I recently did a column on approximations. > http://www.maa.org/editorial/mathgames/mathgames_02_14_05.html > > I've gotten a lot of results, including an article on class > polynomials: > http://www.geocities.com/titus_piezas/Approximations.htm I'm perturbed that multiplication, division, addition, subtraction, and exponentiation aren't considered functions when calculating complexity. Just noodling on the train yesterday, I looked at the CF representations of n^(1/m), n^(Pi/m), and n^(e/m) for n<=9999, m<=9. I was surprised that n^(e/m) showed a noticable paucity of close approximations compared to the other two families. None were particularly impressive though, with no CF term seen over ~1.5e6 (sthg^(Pi/9)). The e family did provide the most fun with its deliberately lengthened (which could be decrease complexity in other languages) (62+101)*(62*101)^(e/4)=61989.99999996... Phil When inserting a CD, hold down shift to stop the AutoRun feature In the Device Manager, disable the SbcpHid device. http://www.cs.princeton.edu/~jhalderm/cd3/ __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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