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January 15th, 2015, 05:49 AM  #1 
Newbie Joined: May 2013 Posts: 5 Thanks: 0  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}{11/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}{11/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}{11/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}{11/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}{11/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}{11/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. 
January 15th, 2015, 12:49 PM  #2 
Newbie Joined: May 2013 Posts: 5 Thanks: 0 
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. Thread will be closed. Last edited by skipjack; January 15th, 2015 at 02:08 PM. 

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flaws, hypothesis, proof, res>1 or 2, riemann, riemann hypothesis, zetas 
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