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January 15th, 2015, 05:49 AM   #1
jlb
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riemann hypothesis proof?: zeta(s) is never 0 for Re(s)>1/2, flaws? thanks!

The Riemann Zeta function can be represented by the Euler product \begin{align}
\zeta \left( s \right) = \prod^{\infty}_{p=primes} \frac {1}{1-1/p^{s}}
\end{align}
for $ s = a + ib \in C$ and $a=Re\left( s \right) > 1$.
Let
\begin{align}
F \left( s \right) = \prod^{\infty}_{p=primes} e^{- \frac {1}{p^{s}}}=e^{- P \left( s \right)}
\end{align}
for $ s = a + ib \in C $ and $a=Re\left( s \right) > 1$, where $P \left( s \right)$ is the Riemann Prime Zeta function. $P \left( s \right), F \left( s \right)$ and $\zeta \left( s \right)$ are analytic and can be extended to the complex plane for $0 < Re\left( s \right) < 1$.
The norm (squared) of the product $F \left( s \right)\zeta \left( s \right)$ is defined for $a=Re\left( s \right) > 1$ by
\begin{align}
|F \left( s \right)\zeta \left( s \right) |^2 =
\prod^{\infty}_{p=primes} | \frac {1}{1-1/p^{s}} |^2
\prod^{\infty}_{p=primes} | e^{- \frac {1}{p^{s}}} |^2
\end{align}
\begin{align}
|F \left( s \right)\zeta \left( s \right) |^2 =
\prod^{\infty}_{p=primes} \left( \frac {1}{1-1/p^{2a}} \right)
\left(\frac {p^{2a}-1}{p^{2a}-2 p^{a}\cos(b\ln{p})+1} \right)
\prod^{\infty}_{p=primes} e^{- \frac {2 \cos(b\ln{p}) }{p^{a}}}
\end{align}
\begin{align}
|F \left( s \right)\zeta \left( s \right) |^2 =
\prod^{\infty}_{p=primes} \left( \frac {1}{1-1/p^{2a}} \right)
\prod^{\infty}_{p=primes} \left(\frac {p^{2a}-1}{p^{2a}-2 p^{a}\cos(b\ln{p})+1} \right)
e^{- \frac {2 \cos(b\ln{p}) }{p^{a}}} (1)
\end{align}
The first product $ \prod^{\infty}_{p=primes} \left( \frac {1}{1-1/p^{2a}} \right)$ obviously converges absolutely for $a=Re\left( s \right) > \frac {1}{2}$ and since
\begin{align*}
\frac {p^{2a}-1}{p^{2a}-2 p^{a}\cos(b\ln{p})+1} = 1 + 2 \sum^{\infty}_{j=1} \frac {\cos \left( jb \ln {p} \right)}{p^{ja}} &
\end{align*}
for $a=Re\left( s \right) > 0$, then the second product in $(1)$
\begin{align}
\prod^{\infty}_{p=primes} \left(\frac {p^{2a}-1}{p^{2a}-2 p^{a}\cos(b\ln{p})+1} \right)
e^{- \frac {2 \cos(b\ln{p}) }{p^{a}}}
\end{align}
also converges absolutely for $a=Re\left( s \right) > \frac {1}{2}$ and $(1)$ can be extended in that range.
Since all terms in the products of $(1)$ are positive and real, then the norm
\begin{align*}
|F \left( s \right)\zeta \left( s \right) |^2 =
|e^{- P \left( s \right)}\zeta \left( s \right) |^2 =
\prod^{\infty}_{p=primes} \left( \frac {1}{1-1/p^{2a}} \right)
\prod^{\infty}_{p=primes} \left(\frac {p^{2a}-1}{p^{2a}-2 p^{a}\cos(b\ln{p})+1} \right)
e^{- \frac {2 \cos(b\ln{p}) }{p^{a}}}
\end{align*}
is always greater than 0 for $a=Re\left( s \right) > \frac {1}{2}$, and the Riemann hypothesis would be correct.
I would appreciate, if you have time, that you let me know what you think of this. If I have made a false assumption regarding analycity, continuation of the norm inside the critical strip, or if the products were to be only semi convergent (possibly allowing exceptions to the argument?), I would still be at a loss to explain why the partial products agree so well with the actual result. I have checked the solution numerically for several points of $a=Re\left( s \right) > \frac {1}{2}$ and compared it with mathematical software output.
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January 15th, 2015, 12:49 PM   #2
jlb
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I finally figured out that I have just proved 1=1 ...
all that I needed was to substitute the analytic formula for P(s)=sum mu(k) ln(zeta(ks))/k.

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Last edited by skipjack; January 15th, 2015 at 02:08 PM.
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