
New Users Post up here and introduce yourself! 
 LinkBack  Thread Tools  Display Modes 
August 23rd, 2014, 03:02 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(1s) (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^((1s)) =η(1s) R>0 = 1/1^s  1/2^s + 1/3^s  1/4^s … = 1/1^((1s))  1/2^((1s)) + 1/3^((1s))  1/4^((1s)) … Subtracting … 0 = ∑_(n=1)^∞▒〖〖(1)〗^(n+1)/n^((1s)) 〗 〖(1)〗^(n+1)/n^s = ∑_(n=1)^∞▒〖(〖(1)〗^(n+1) n^s)/(n^((1s) ) n^s )  〗 (〖(1)〗^(n+1) n^((1s)))/(n^((1s) ) n^s ) 0 = ∑_(n=1)^∞▒〖(〖(1)〗^(n+1) 〖(n〗^s n^((1s))))/n 〗 = ( (0))/1  (( 2^s2^(1s)))/2 + ( ( 3^s3^(1s)))/3 … $ Now this is true for all positive n so if 0 = n^s n^((1s)) => n^s= n^((1s)) 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. 
August 23rd, 2014, 03:11 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, 04:10 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(1s) (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^((1s)) =η(1s) R>0 = 1/1^s  1/2^s + 1/3^s  1/4^s … = 1/1^((1s))  1/2^((1s)) + 1/3^((1s))  1/4^((1s)) … Subtracting … 0 = \sum_(n=1)^∞▒〖〖(1)〗^(n+1)/n^((1s)) 〗 〖(1)〗^(n+1)/n^s = \sum_(n=1)^∞▒〖(〖(1)〗^(n+1) n^s)/(n^((1s) ) n^s )  〗 (〖(1)〗^(n+1) n^((1s)))/(n^((1s) ) n^s ) 0 = ∑_(n=1)^∞▒〖(〖(1)〗^(n+1) 〖(n〗^s n^((1s))))/n 〗 = ( (0))/1  (( 2^s2^(1s)))/2 + ( ( 3^s3^(1s)))/3 … $ Now this is true for all positive n so if 0 = n^s n^((1s)) => n^s= n^((1s)) 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. 

Tags 
hypothesis, riemann 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Riemann Hypothesis.  mathbalarka  Number Theory  0  October 31st, 2013 01:54 AM 
Book on Riemann hypothesis  madromanoff  Math Books  1  September 16th, 2012 08:34 AM 
Do you believe the Riemann hypothesis?  Eureka  Number Theory  1  September 12th, 2011 02:07 PM 
The Riemann Hypothesis and what it say  MyNameIsVu  Complex Analysis  2  April 16th, 2009 08:51 AM 
Riemann Hypothesis  Whoever  Number Theory  72  February 11th, 2009 07:03 AM 