[math-fun] For lovers of Pi
Hello, For all x, sum((2*n)!/(2^(3*n+1/2))/n!^2/(2*n+1)*(1/cosh(2*x)^(1/2))^(2*n+1)*cosh((2*n+1)*x),n = 0 .. infinity) = 1/4*Pi FME...
Dr. E, Mathematica thinks you need x real: In[429]:= %424[[1, 1]] /. x -> I \[Pi] Out[429]= (2^(-(1/2) - 3 n) Cos[(1 + 2 n) \[Pi]] (2 n)!)/((1 + 2 n) (n!)^2) Similarly for t in your ln 2 formula. For π, best convergence is 1 bit/term at x=0, rapidly decaying toward 0 bits/term away from x=0. —rwg On Mon, Sep 24, 2018 at 10:45 AM françois mendzina essomba2 < m_essob@yahoo.fr> wrote:
Hello,
For all x,
sum((2*n)!/(2^(3*n+1/2))/n!^2/(2*n+1)*(1/cosh(2*x)^(1/2))^(2*n+1)*cosh((2*n+1)*x),n = 0 .. infinity) = 1/4*Pi
FME...
another of this type sum((2^(-4*n-1)*(2*n)!*(2*cosh(2*x)+sqrt(3))^((-2*n-1)/2)*cosh((2*n+1)*x))/((2*n+1)*n!^2),n=0..infinity)=Pi/12 Le lundi 24 septembre 2018 à 18:45:10 UTC+1, françois mendzina essomba2 <m_essob@yahoo.fr> a écrit : Hello, For all x, sum((2*n)!/(2^(3*n+1/2))/n!^2/(2*n+1)*(1/cosh(2*x)^(1/2))^(2*n+1)*cosh((2*n+1)*x),n = 0 .. infinity) = 1/4*Pi FME...
participants (2)
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Bill Gosper -
françois mendzina essomba2