 My Math Forum riemann hypothesis proof?: zeta(s) is never 0 for Re(s)>1/2, flaws? thanks!
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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}{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. 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. Tags flaws, hypothesis, proof, res>1 or 2, riemann, riemann hypothesis, zetas Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Vincenzo Oliva Number Theory 12 November 27th, 2014 03:14 PM Nick Vandaele Number Theory 4 October 1st, 2014 07:43 AM eddybob123 Number Theory 18 May 21st, 2013 06:10 PM mathbalarka Number Theory 9 February 12th, 2013 07:24 AM joexian Number Theory 5 January 16th, 2013 06:13 AM

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