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 August 23rd, 2014, 02:02 PM #1 Newbie   Joined: Aug 2014 From: Mountain View Posts: 10 Thanks: 0 Math Focus: Number Theory 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 02:19 PM. August 23rd, 2014, 02:11 PM #2 Newbie   Joined: Aug 2014 From: Mountain View Posts: 10 Thanks: 0 Math Focus: Number Theory Does My Math Forum have a proprietary Math formatter similar to TEX? My summations are not looking good. Or do people save screenshots and post as JPG or PNG files? RR August 23rd, 2014, 03:10 PM #3 Newbie   Joined: Aug 2014 From: Mountain View Posts: 10 Thanks: 0 Math Focus: Number Theory 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 03:14 PM. Tags hypothesis, riemann 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 mathbalarka Number Theory 0 October 31st, 2013 12:54 AM madromanoff Math Books 1 September 16th, 2012 07:34 AM Eureka Number Theory 1 September 12th, 2011 01:07 PM MyNameIsVu Complex Analysis 2 April 16th, 2009 07:51 AM Whoever Number Theory 72 February 11th, 2009 06:03 AM

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