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New method for calculating zeta zeroes

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  • Jon Perry
    Although not in it s final stages yet. zeta(s) = sigma{n=1,oo,1/n^s} azeta(s) = sigma{n=1,oo,(-1)^(n-1)/n^s} and of course zeta(s) = 1/[1-2^(1-s)].azeta(s) A
    Message 1 of 1 , May 27, 2002
      Although not in it's final stages yet.

      zeta(s) = sigma{n=1,oo,1/n^s}

      azeta(s) = sigma{n=1,oo,(-1)^(n-1)/n^s}

      and of course zeta(s) = 1/[1-2^(1-s)].azeta(s)

      A single term in azeta(s) is (-1)^(n-1).[coslnn^t + i.sinlnn^t]/n^s

      where s=s+it.

      Consider zeta(s)=zeta(s')=0

      Summing the two expansions of these, we get single terms of:

      (-1)^(n-1).[coslnn^t + coslnn^t'+ i.sinlnn^t + i.sinlnn^t]/n^s

      As the sum is zero, the sum of this expansion must be zero,

      therefore there exists some t'' such that coslnn^t'' = coslnn^t + coslnn^t'

      for all n.

      Jon Perry
      perry@...
      http://www.users.globalnet.co.uk/~perry/maths
      BrainBench MVP for HTML and JavaScript
      http://www.brainbench.com
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