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August 23rd, 2014, 03:02 PM   #1
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Help with Riemann's Hypothesis

I'm working on Riemann;s Hypothesis. See any mistakes so far?
Since the analytic continuation function η(s) has the exact same zeroes as ϛ(s) then we expect η(s) = η(1-s) (Hardy.1999)
$\displaystyle 0=η(s)≡(1- 2/2^s )ϛ(s)=∑_(n=1)^∞▒〖(-1)〗^(n+1)/n^s =∑_(n=1)^∞▒〖(-1)〗^(n+1)/n^((1-s)) =η(1-s) R>0
= 1/1^s - 1/2^s + 1/3^s - 1/4^s … = 1/1^((1-s)) - 1/2^((1-s)) + 1/3^((1-s)) - 1/4^((1-s)) …
Subtracting …
0 = ∑_(n=1)^∞▒〖〖(-1)〗^(n+1)/n^((1-s)) -〗 〖(-1)〗^(n+1)/n^s = ∑_(n=1)^∞▒〖(〖(-1)〗^(n+1) n^s)/(n^((1-s) ) n^s ) - 〗 (〖(-1)〗^(n+1) n^((1-s)))/(n^((1-s) ) n^s )
0 = ∑_(n=1)^∞▒〖(〖(-1)〗^(n+1) 〖(n〗^s- n^((1-s))))/n 〗 = ( (0))/1 - (( 2^s-2^(1-s)))/2 + ( ( 3^s-3^(1-s)))/3 … $
Now this is true for all positive n so
if 0 = n^s- n^((1-s)) => n^s= n^((1-s))
and this is possible only if s = (1 – s) => s = 1/2 or R[s] = 1/2
0 < s < 1 (the critical strip)
There are no other zeroes to be found in the equation, so I believe we have them all. Otherwise we disprove Riemann’s Hypothesis.

Last edited by robinlrandall; August 23rd, 2014 at 03:19 PM.
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August 23rd, 2014, 03:11 PM   #2
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Does My Math Forum have a proprietary Math formatter similar to TEX?
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RR
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August 23rd, 2014, 04:10 PM   #3
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robinlrandall;204010]I'm working on Riemann;s Hypothesis. See any mistakes so far?
Since the analytic continuation function η(s) has the exact same zeroes as ϛ(s) then we expect η(s) = η(1-s) (Hardy.1999)
$\displaystyle 0=η(s)≡(1- 2/2^s )ϛ(s)=\sum_n=1_∞〖(-1)〗^(n+1)/n^s =\sum_(n=1)^∞▒〖(-1)〗^(n+1)/n^((1-s)) =η(1-s) R>0
= 1/1^s - 1/2^s + 1/3^s - 1/4^s … = 1/1^((1-s)) - 1/2^((1-s)) + 1/3^((1-s)) - 1/4^((1-s)) …
Subtracting …
0 = \sum_(n=1)^∞▒〖〖(-1)〗^(n+1)/n^((1-s)) -〗 〖(-1)〗^(n+1)/n^s = \sum_(n=1)^∞▒〖(〖(-1)〗^(n+1) n^s)/(n^((1-s) ) n^s ) - 〗 (〖(-1)〗^(n+1) n^((1-s)))/(n^((1-s) ) n^s )
0 = ∑_(n=1)^∞▒〖(〖(-1)〗^(n+1) 〖(n〗^s- n^((1-s))))/n 〗 = ( (0))/1 - (( 2^s-2^(1-s)))/2 + ( ( 3^s-3^(1-s)))/3 … $
Now this is true for all positive n so
if 0 = n^s- n^((1-s)) => n^s= n^((1-s))
and this is possible only if s = (1 – s) => s = 1/2 or R[s] = 1/2
0 < s < 1 (the critical strip)
There are no other zeroes to be found in the equation, so I believe we have them all. Otherwise we disprove Riemann’s Hypothesis.[/QUOTE]

Last edited by robinlrandall; August 23rd, 2014 at 04:14 PM.
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