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Re: Fibonacci-Lucas-Jacobi-Gauss identity

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  • David Broadhurst
    You also get the log of the golden mean from an agm. Let S2=sum_{n 0} 1/F(2*n-1) S3=sum_{n 0} 1/L(2*n) then agm(S2/sqrt(5),1/4+S3) =Pi/8/log((1+sqrt(5))/2)
    Message 1 of 4 , Oct 23, 2002
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      You also get the log of the
      golden mean from an agm.

      Let
      S2=sum_{n>0} 1/F(2*n-1)
      S3=sum_{n>0} 1/L(2*n)
      then
      agm(S2/sqrt(5),1/4+S3)
      =Pi/8/log((1+sqrt(5))/2)

      Proof: As before, but now use
      the evaluations of the nome
      q=exp(-Pi*K(k')/K(k))=(3-sqrt(5))/2

      Check:
      F(n)=fibonacci(n)
      L(n)=if(n,F(2*n)/F(n),2)
      S2=suminf(n=1,1./F(2*n-1));
      S3=suminf(n=1,1./L(2*n));
      print(agm(S2/sqrt(5),1/4+S3)-Pi/8/log((1+sqrt(5))/2));
      -2.5860875718090325175134436270000000000 E-37
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