Hello, here is one that I like a lot : 1304*sum(1/(4*n^2-4*n+4*163^2+1),n=1..infinity); is equal to Pi at 443 digits. if 163 is replaced by 1000 : the precision is 2727 Digits. The trick is the expression : 8*sum(1/(4*n^2-4*n+4*k^2+1),n=1..infinity) = = 1/k*Pi*tanh(Pi*k) and since tanh(X) is very close to 1 when X >> 1 then the approximation is very good. The underlying explanation is that the sum is complex but the imaginary parts cancel each other and gives birth to the tanh( ) argument. (classical result). Simon Plouffe Le 2017-10-06 à 19:32, Bill Gosper a écrit :
Gack. So you're saying someone actually needs to _prove_ Sum[Floor[n GoldenRatio]/2^n,{n,0,∞}] = 1 + Sum[2^-Floor[n GoldenRatio],{n,0,∞}] .
Imponderable: How can a mirage have solid angles? In any case, I worked out that the solid angle at the apex of a regular n-gon pyramid with vertex angles a: pyramidSolidAngle[n_, a_] := n 2 ArcSin[Sec[a/2] Sin[2π/n] (1 - √(1 - Csc[π/n]^2 Sin[a/2]^2))/2]
along with three sadistically dissimilar alternatives, to frustrate FullSimplify and trigonometry students. --rwg Another imponderable: If radishes grow underground, why do they take the trouble to be red?
On Fri, Oct 6, 2017 at 8:24 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
Bill,
Define the function f(x) as follows:
f(x) := Sum[2^-k Sinc[x] Product[ If[PrimeQ[n], Sinc[x/n], 1], {n, 1, k, 8}], {k, 0, Infinity}]
Then the integral between 0 and infinity of f(x) agrees to pi for slightly more than 10^2000 digits.
[Regarding the discussion of pyramids, you really want every face to be an equilateral triangle. Then, from a distance, the reflection in the sand will give a mirage of a floating octahedron.]
Best wishes,
Adam P. Goucher
Sent: Friday, October 06, 2017 at 1:05 PM From: "Bill Gosper" <billgosper@gmail.com> To: math-fun@mailman.xmission.com Subject: Re: [math-fun] Borweins' high precision fraud
%479, below, is Sum[Floor[n GoldenRatio]/2^n,{n,0,∞}] = 0 + 1/2 + 3/4 + 1/ 2 + 3/8 + 1/4 + 9/64 + 11/128 + 3/64 + . . . = 2.709803442861291314629 G,Knuth,&P, , Exercise 6.49, gives Sum[2^-Floor[n GoldenRatio],{n,0,∞}] = 1+1/2+1/8+1/16+1/64+1/256+1/512+1/2048+1/4096+...= 1.709803442861291314642 ! --rwg
On Sat, Sep 30, 2017 at 12:31 PM, Bill Gosper <billgosper@gmail.com> wrote:
From http://www.math.grinnell.edu/~chamberl/courses/444/ worksheets/high-precision-fraud.pdf
[CHOP]
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