## New method for calculating zeta zeroes

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• 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
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