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[math-fun] Algebra 2 exercise
by Bill Gosper 15 Aug '16

15 Aug '16

If a regular *n*-gon with unit circumradius has side s, the 2*n*-gon has side Sqrt[2 - Sqrt[4 - s^2]]. Shew that this denests to (-2 + s + Sqrt[4 - s^2])/Sqrt[4 - 2 s] when 0 ≤ s ≤ 2. --rwg

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[math-fun] Feynman and the Thinking Machine
by Jason Holt 14 Aug '16

14 Aug '16

This is a delightful reminiscence of Danny Hillis about working with Feynman at Thinking Machines: http://longnow.org/essays/richard-feynman-connection-machine/

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Re: [math-fun] Explain the 710.
by Keith F. Lynch 14 Aug '16

14 Aug '16

Bill Gosper <billgosper(a)gmail.com> wrote: > Ultrahard: Change 39999999 to ?. Will that change the answer? --rwg > Do you think the ? answer will *ever* be known? Not to those of us with the ASCII digest option. Please avoid using non-ASCII characters, as the digest renders them as question marks, making the post incomprehensible. Thanks.

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[math-fun] Brains patterns for solving math problems
by Hilarie Orman 14 Aug '16

14 Aug '16

I hope this line of research leads to improvements in education: http://www.nytimes.com/2016/07/29/science/brain-scans-math.html Looks like "encoding" really heats up the grey matter. Hilarie

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[math-fun] Explain the 710.
by Bill Gosper 12 Aug '16

12 Aug '16

Easy. In[136]:= TakeLargest[Table[Csc@n/n/n, {n, 999}], 7] // Timing Out[136]= {0.08285, {Csc[1], Csc[3]/9, Csc[2]/4, Csc[710]/504100, Csc[7]/49, Csc[9]/81, Csc[44]/1936}} Change 999 to 39999999: In[135]:= TakeLargest[Table[Csc@n/n/n, {n, 39999999}], 7] // Timing Out[135]= {3135.54, {Csc[1], Csc[3]/9, Csc[2]/4, Csc[710]/504100, Csc[7]/49, Csc[9]/81, Csc[44]/1936}} It took 38 thousand times longer to get the same answer. Ultrahard: Change 39999999 to ∞. Will that change the answer? --rwg Do you think the ∞ answer will *ever* be known?

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[math-fun] something I just noticed about helices
by Bill Gosper 11 Aug '16

11 Aug '16

Viewed side-on in perspective, e.g., gosper.org/helix.png there are *two* points where its tangent vector points almost straight at you. In this case z ~ 1.7 (ascending) and 3.7 (descending). --rwg

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[math-fun] Draft of my August 2016 blog post
by James Propp 10 Aug '16

10 Aug '16

I started writing a new draft titled "Bertrand's Ballot Problem" and would love to get your feedback. I plan on publishing it on the 17th. It's still quite rough (and I haven't made the figures yet), but I thought I'd share it now, given that August 17th is only a week away. I'll probably update it now and then during the week ahead. I'm hoping to learn Omnigraffle by using it to prepare the eight illustrations. Are any of you Omnigraffle users, and hence potentially available to offer me assistance if I have an "Oh Mister Wizard! Mister Wizard!" moment? Suggestions for references, and comments of all kinds, are welcome. Note that all comments submitted through Wordpress will be relayed to me anonymously. If you want me acknowledge your contribution, or if you think I might have a question for you in response to your comment, please sign your comment, or else comment by email. Please leave your feedback here: https://mathenchant.wordpress.com?p=879&shareadraft=57a549f09f22d Title: Bertrand's Ballot Problem Beginning: An early lead in an election doesn't necessarily mean much. Remember Republican presidential candidate Rudy Giuliani? Back in 2008, the outspoken New Yorker polled so well that some pundits predicted he'd win the nomination, but his lead fizzled out before the convention. The same thing has happened... Read more: https://mathenchant.wordpress.com?p=879&shareadraft=57a549f0 9f22d Thanks, jamespropp

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[math-fun] An infinite sequence of Langford pairs... with words
by Eric Angelini 09 Aug '16

09 Aug '16

Hello Math-Fun, Start with this: THREE, TEN, THREE, SEVEN, TEN, FIVE, SEVEN, FIVE ... where there are indeed: 3 letters inside the pair of "THREE" 10 letters inside the pair of "TEN" 7 letters inside the pair of "SEVEN" 5 letters inside the pair of "FIVE". This "core" is now the seed of an infinite such sequence: 1) Measure the length (in letters) of the "core" (here 34); 2) write this length before and after the core; 3) repeat ad lib. CORE (34 letters) THIRTY-FOUR -----------------THIRTY-FOUR FIFTY-FOUR ---------------------------------------- FIFTY-FOUR etc. Will you find more such "cores" (where each number-name has exactly one copy)? And a way to enter them into the OEIS? Best, É.

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[math-fun] Those weird Borwein integrals
by Bill Gosper 08 Aug '16

08 Aug '16

https://en.wikipedia.org/wiki/Borwein_integral Mma 10.4 gets In[562]:= Integrate[Product[Sinc[x/(2 k - 1)], {k, 8}], {x, 0,∞}] Out[562]= (467807924713440738696537864469 π/ 935615849440640907310521750000 surprisingly (suspiciously?) quickly, but has run all night on Integrate[2 Cos[x] Product[Sinc[x/(2 k - 1)], {k, 18}], {x, 0,∞}] // tim which is only a quarter of the way to where it gets interesting. In[421]:= ContinuedFraction[467807924713440738696537864469/ 935615849440640907310521750000] Out[421]= {0, 2, 33999077543, 5, 16, 1, 2, 1, 2, 1, 1, 1, 5, 1, 2, 1, 15, 1, 13, 2, 16, 1, 2, 3, 1, 7, 1, 10, 3, 1, 1, 2, 1, 1, 3, 3, 1, 11, 10, 1, 1, 3, 1, 19} AFAI can tell, http://schmid-werren.ch/hanspeter/publications/ 2014elemath.pdf mentioned by Wikipedia tells *where* (the Cos[x] case) gets interesting, but not *how* interesting. Has anybody actually computed that huge rational? We wonder how far from π/2 the integral manages to stray as we tip over the 8, It's a bit surprising, though well known, that Product[Cos[x/(2*k)], {k, Infinity}] == Product[Sinc[x/(2*k-1)], {k, Infinity}] Might the left infinite product be any easier to integrate? Not likely! All the integrals of the finite cosine products diverge! Mathematica is so far unhelpful finding the integral. I may need my homebrew Euler-Maclaurin function. --rwg In fact, I've so far failed to qualitatively accelerate the infinite product for a given x. This gets it from zeta 2 to zeta 8 type convergence, In[323]:= sncprd[x_?NumericQ] := Block[{p = 1/(\[Pi]^6 x^6) 30 Sqrt[ 42] (Cos[(\[Pi] x)/30^(1/4)] - Cosh[(\[Pi] x)/30^(1/4)]) (Cos[(3^(1/3) \[Pi] x)/( Sqrt[2] 35^(1/6))] - Cosh[(\[Pi] x)/(Sqrt[2] 105^(1/6))]) Sin[(\[Pi] x)/Sqrt[ 6]] Sinh[(\[Pi] x)/(Sqrt[2] 105^(1/6))], n = 1}, While[p != (p *= Sinc[x/n]/((1 - x^2/(6 n^2)) (1 + x^4/(120 n^4)) (1 + x^6/( 840 n^6)))), ++n]; Print@n; p] In[740]:= sncprd[\[Pi]/2`22] During evaluation of In[740]:= 87 Out[740]= 0.48690721697316073847 In[741]:= sncprd[\[Pi]/2`33] // tim During evaluation of In[741]:= 1454 During evaluation of In[741]:= 0.095577,0 Out[741]= 0.48690721697316075167428625082 In[742]:= sncprd[\[Pi]/2`39] // tim During evaluation of In[742]:= 6751 During evaluation of In[742]:= 0.302979,0 Out[742]= 0.48690721697316075167428628848113831 In[743]:= sncprd[\[Pi]/2`44] // tim During evaluation of In[743]:= 24261 During evaluation of In[743]:= 1.299224,0 Out[743]= 0.486907216973160751674286288481949287233 but that's still zero bits/term. The integer printouts are number of terms until the cumulative product stops changing. Note that this is several digits short of the truth. The plot strongly resembles some band-limited, interpolated row of Pascal's triangle. It sucks that I can't attach a plot.

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[math-fun] Strange(?) Gamma identities
by Bill Gosper 06 Aug '16

06 Aug '16

Mathematica failed to telescope an infinite product and produced a big mess. Plugging in small integers, Gamma[Root[-17 + 18 #1 - 7 #1^2 + #1^3 &, 1]] Gamma[ Root[-17 + 18 #1 - 7 #1^2 + #1^3 &, 2]] Gamma[ Root[-17 + 18 #1 - 7 #1^2 + #1^3 &, 3]] == 5 Gamma[Root[-5 + 7 #1 - 4 #1^2 + #1^3 &, 1]] Gamma[ Root[-5 + 7 #1 - 4 #1^2 + #1^3 &, 2]] Gamma[ Root[-5 + 7 #1 - 4 #1^2 + #1^3 &, 3]], Different integers: Gamma[Root[-25 + 21 #1 - 7 #1^2 + #1^3 &, 1]] Gamma[ Root[-25 + 21 #1 - 7 #1^2 + #1^3 &, 2]] Gamma[ Root[-25 + 21 #1 - 7 #1^2 + #1^3 &, 3]] == 10 Gamma[Root[-10 + 10 #1 - 4 #1^2 + #1^3 &, 1]] Gamma[ Root[-10 + 10 #1 - 4 #1^2 + #1^3 &, 2]] Gamma[ Root[-10 + 10 #1 - 4 #1^2 + #1^3 &, 3]] etc., apparently inexhaustibly. Products and sums of a function over a complete set of algebraic conjugates often come out nice. E.g., Out[654]= (E^(-((2 I Pi)/3)) z)! (E^((2 I Pi)/3) z)! == (Pi Csc[E^(-((I Pi)/3)) Pi z])/Pochhammer[1 + E^(-((2 I Pi)/3)) z, -1 + z] In[655]:= Table[ N[Divide @@ %], {z, {-1, 1/3, 1/2, 1, 6, 9, E, Pi, I}}] Out[655]= {0.9999999999999925 - 9.025031411218348*10^-18 I, 1.000000000000002 - 2.272641409138545*10^-16 I, 1. + 0. I, 0.9999999999999911 + 3.885780586188048*10^-16 I, 0.9999999999999996 + 4.33801051959268*10^-15 I, 0.9999999999999991 + 4.773959005888173*10^-15 I, 0.9999999999999993 + 8.129524651428513*10^-16 I, 0.9999999999999978 + 6.339440984328298*10^-16 I, 1.000000000000002 + 2.775557561562891*10^-17 I} Similarly for Gammas of cubic surds? --rwg

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