February 2011 - mathfun@mailman.xmission.com

Re: [math-fun] Big bug in W&W?
by Bill Gosper 18 Feb '11

18 Feb '11

* Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> [Feb 14. 2011 09:51]: > http://mathworld.wolfram.com/EllipticIntegralSingularValue.html leads off > with > "Abel (quoted in Whittaker and Watson 1990, p. 525) proved that whenever > (K'(k))/(K(k))=(a+b sqrt(n))/(c+d sqrt(n)), > where a, b, c, d, and n are integers, K(k) is a complete elliptic integral > of the first kind, and K'(k)=K(sqrt(1-k^2)) is the complementary complete > elliptic integral of the first kind, then the elliptic modulus k is the root > of an algebraic equation with integer coefficients." > (Note these are old style K(k) := EllipticK[k^2] .) > E.g., choosing a:=0, b:=1, n:=5, c:=2, and d:=0, so that K'(k)/K(k) = > Sqrt[5]/2, > then k^2 -> 4*Sqrt[2]*(-(3491/2) - 1234*Sqrt[2] - (1561*Sqrt[5])/2 - > 552*Sqrt[10] + (1/22)* > Sqrt[38078 + 17029*Sqrt[5]]*(285 - 3*Sqrt[5] + 88*Sqrt[10])) > > (probably undersimplified). OK, so change a:=1 to get K'/K = GoldenRatio. > I cannot to save my life PSLQ a polynomial satisfied by > LambdaStar[GoldenRatio^2]. (E.g., degree = 100, $MaxExtraPrecision = > 69999.) > Is the theorem false? Misquoted? I'm spazzing? It's true, but the > polynomial > is stupendous?? Joerg>Same here, I can only obtain a minpoly for sqrt of any rational >(which can be a square). >So the mathworld statement appears to be incorrect (or incomplete). >I'd be very interested what the correction might be. Here it is: In[105]:= FindIntegerNullVector[ Table[ModularLambda[(1 + Sqrt[-5])/2]^(2*k), {k, 0, 4}]] Out[105]= {-256 I, -1792 I, -3168 I, 5008 I, -I} I.e., K'/K must have the form (a + b sqrt(-n))/(c + d sqrt(-n)). --rwg > > So what do W&W say? "The theorem is beyond the scope of this book."! > --rwg

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[math-fun] An improper integral equals to log(\pi/2)-\gamma
by Warut Roonguthai 18 Feb '11

18 Feb '11

Can anyone tell me how to prove that \int_0^\infty log(x)/(1+cosh(x)) dx = log(\pi/2)-\gamma? Any info related to this integral is also appreciated. I could find nothing except that Wolfram|Alpha knew it. Hope that this isn't too trivial for math-fun. Thanks, Warut

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Re: [math-fun] An improper integral equals to log(\pi/2)-\gamma
by Henry Baker 17 Feb '11

17 Feb '11

I tried this in maxima using risch(log(x)/(1+cosh(x)),x): Maxima encountered a Lisp error: Error in PROGN [or a callee]: Caught fatal error [memory may be damaged] Automatically continuing. To reenable the Lisp debugger set *debugger-hook* to nil. Whoops! At 12:10 PM 2/17/2011, Warut Roonguthai wrote: >Can anyone tell me how to prove that > >\int_0^\infty log(x)/(1+cosh(x)) dx = log(\pi/2)-\gamma? > >Any info related to this integral is also appreciated. I could find >nothing except that Wolfram|Alpha knew it. > >Hope that this isn't too trivial for math-fun. > >Thanks, >Warut

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[math-fun] LambdaStar (singular value) as eta
by Bill Gosper 15 Feb '11

15 Feb '11

LambdaStar[r]== 4*eta[E^(-4*r*Pi)]^4/Sqrt[eta[E^(-r*Pi)]^8+16*eta[E^(-4*r*Pi)]^8]== 4*eta[E^(-4*r*Pi)]^8*eta[E^(-r*Pi)]^4/eta[E^(-2*r*Pi)]^12 where eta[q] := EllipticEta[q] := DedekindEta[Log[q]/2/I/Pi] --rwg

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[math-fun] JSBach as hacker: "Well" Tempered /= "Equal Tempered
by Henry Baker 14 Feb '11

14 Feb '11

Bach apparently encoded his "well-tempered" tuning formula into a squiggle of loops & knots on the cover page of his "Well-Tempered Clavier" music. The subsequent music in the book is itself is a "proof" that his tuning formula works, by running through the various chords & progressions. Bach's diagram/cryptogram was apparently overlooked as a mere ornament for the past nearly 300 years, until musician/software engineer Bradley Lehman decoded this tuning diagram. Yes, equal tempering has the advantage of removing the differences of different key signatures, but also the disadvantage (for the composer) of taking advantage of different keys for different stylistic effects (including removing boredom) -- following the practise of the medieval "modes". I'm sure some mathematician has studied music tuning formulas in terms of continued fractions -- does anyone know any references? Some interesting fractions from the paper: 531441/524288 = [1, 73, 3, 2, 1, 1, 1, 23, 2, 5] 81/80 7/11 = [0, 1, 1, 1, 3] 128/125 = [1, 41, 1, 2] 1/11 http://em.oxfordjournals.org/content/33/1.toc

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[math-fun] Fwd: [xbbn] New Knuth books
by Allan Wechsler 14 Feb '11

14 Feb '11

This announcement went by on a mailing list for former BBN employees. ---------- Forwarded message ---------- From: Richard M Koolish <koolish(a)theworld.com> Date: Sun, Feb 13, 2011 at 8:46 PM Subject: [xbbn] New Knuth books To: xbbn(a)xbbn.org The Art of Computer Programming, Vol 4A: Combinatorial Algorithms http://www.amazon.com/gp/product/0201038048/ Selected Papers on Fun and Games http://www.amazon.com/FunandGames/dp/157586584X/ -- You received this message because you are subscribed to the Google Groups "xBBN" group. To post to this group, send email to xbbn(a)googlegroups.com. To unsubscribe from this group, send email to xbbn+unsubscribe(a)googlegroups.com. For more options, visit this group at http://groups.google.com/group/xbbn?hl=en.

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[math-fun] Big bug in W&W?
by Bill Gosper 14 Feb '11

14 Feb '11

http://mathworld.wolfram.com/EllipticIntegralSingularValue.html leads off with "Abel (quoted in Whittaker and Watson 1990, p. 525) proved that whenever (K'(k))/(K(k))=(a+b sqrt(n))/(c+d sqrt(n)), where a, b, c, d, and n are integers, K(k) is a complete elliptic integral of the first kind, and K'(k)=K(sqrt(1-k^2)) is the complementary complete elliptic integral of the first kind, then the elliptic modulus k is the root of an algebraic equation with integer coefficients." (Note these are old style K(k) := EllipticK[k^2] .) E.g., choosing a:=0, b:=1, n:=5, c:=2, and d:=0, so that K'(k)/K(k) = Sqrt[5]/2, then k^2 -> 4*Sqrt[2]*(-(3491/2) - 1234*Sqrt[2] - (1561*Sqrt[5])/2 - 552*Sqrt[10] + (1/22)* Sqrt[38078 + 17029*Sqrt[5]]*(285 - 3*Sqrt[5] + 88*Sqrt[10])) (probably undersimplified). OK, so change a:=1 to get K'/K = GoldenRatio. I cannot to save my life PSLQ a polynomial satisfied by LambdaStar[GoldenRatio^2]. (E.g., degree = 100, $MaxExtraPrecision = 69999.) Is the theorem false? Misquoted? I'm spazzing? It's true, but the polynomial is stupendous?? So what do W&W say? "The theorem is beyond the scope of this book."! --rwg

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[math-fun] The three quartersth singular value
by Bill Gosper 14 Feb '11

14 Feb '11

Mathworld gives as an unsolved biquartic. LambdaStar[3/4]=2 2^(1/4) Sqrt[-147 - 104 Sqrt[2] + 85 Sqrt[3] + 60 Sqrt[6]] --rwg

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[math-fun] Euler's crazy pi product
by Bill Gosper 14 Feb '11

14 Feb '11

http://math.ucr.edu/home/baez/week127.html finds in Lennart Berggren, Jonathan Borwein and Peter Borwein, π: A Source Book, Springer-Verlag, New York, 1997, "the following weirdly beautiful formula due to Euler, which unfortunately is not explained:" (Then how can they call it a sourcebook?--rwg) 3 5 7 11 13 17 19 π/2 = - × - × - × -- × -- × -- × -- × ... 2 6 6 10 14 18 18 "Here the numerators are the odd primes, and the denominators are the closest numbers of the form 4n+2." E.g., for 10, 100, ..., 10^6 terms, In[209]:= Table[Block[{p=2},Nest[N[#*(p=NextPrime[p])/(2+4*Round[(p-2)/4]),9]&,2,10^n]],{n,6}] Out[209]= {3.10152045, 3.13398462, 3.13772561, 3.14073685, 3.14143290, 3.14157195} Why the heck?? Was Euler a preincarnation of Ramanujan? --rwg

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[math-fun] New End-User License Agreement for OEIS
by N. J. A. Sloane 12 Feb '11

12 Feb '11

Dear Math Fun, New End-User License Agreement for OEIS After two months of discussions, the OEIS Foundation has adopted the Creative Commons Attribution Non-Commercial 3.0 license (CC-BY-NC) as its End-User License Agreement. The Contributor's License Agreement will remain unchanged. Neil Sloane Here is the announcement about the new OEIS, in case anyone missed it. Announcement, Nov 17 2010: New Version of OEIS! =============================================== The OEIS is now running on the web. There are two parts: the main OEIS page is oeis.org, which has all the sequence data, and a wiki, oeis.org/wiki, which has discussion pages. To contribute a new sequence or edit a sequence, you must first register at oeis.org/wiki - click Register, enter your real name (use John M. Smith, NOT john smith, NOT johnny33), enter your email address, and a brief bio. You will be sent an automatic email (from admin(a)oeis.org, I think), which you must answer to get your password. When you get the password, go to the main page, oeis.org, and login with the same user name and password. Neil Sloane President, OEIS Foundation Inc (see http://oeisf.org/)

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