November 2nd, 2015, 12:22 PM  #21  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,659 Thanks: 2635 Math Focus: Mainly analysis and algebra  Quote:
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I also have been looking at those functions and integrals, but they aren't all of the answer because these results are the ones that come from the analytic continuation of the Riemann Zeta function, and not from techniques such as Generic Summation. Other results remain unexplained by the integrals: $$11+11+\ldots = \tfrac12 \\ 1+01+10+1+\ldots = \tfrac23 \\ 1+2+4+8+16+\ldots = 1$$ Note the the latter can be obtained by analytic continuation, and the two results agree. Last edited by v8archie; November 2nd, 2015 at 12:26 PM.  
November 2nd, 2015, 12:34 PM  #22 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  
November 2nd, 2015, 06:01 PM  #23  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,659 Thanks: 2635 Math Focus: Mainly analysis and algebra  Quote:
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November 3rd, 2015, 06:40 PM  #24  
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry  Quote:
 
November 3rd, 2015, 06:47 PM  #25  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,160 Thanks: 879 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan Last edited by skipjack; November 3rd, 2015 at 10:59 PM.  
November 3rd, 2015, 07:09 PM  #26 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond  
November 13th, 2015, 06:37 PM  #27 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,160 Thanks: 879 Math Focus: Wibbly wobbly timeywimey stuff. 
I realize this topic is basically done and buried but I spent a few idle moments of research and though I'd tell y'all my results. As the topic has come up frequently in the last few months I thought it might be helpful. First: I have no idea how to derive this formula. But it is the simplest I could find that analytically continues the (Riemann) zeta function to negative s. I am uncertain of the region of convergence but $\displaystyle \zeta$ is meromorphic and it converges everywhere in the complex s plain with the exception of s = 1. The formula below provides a form that converges for all but even negative integer s (and, of course, s = 1) so far as I can tell. I also can't find a way to use this definition to produce a value for s = 0. $\displaystyle \zeta (s) = 2^s \pi ^{s  1} ~ sin \left ( \frac{ \pi s}{2} \right ) ~ \Gamma (1  s) ~ \zeta (1  s)$ (This is called the "Riemann function equation.") For those who don't know, the Gamma function ( $\displaystyle \Gamma(n)$) is closely related to the factorial function and for our purposes all we need to know is that $\displaystyle \Gamma (n) = (n  1)!$ for integer $\displaystyle n \geq 1$. The idea here is that we know the values of $\displaystyle \zeta (s)$ for any positive integer greater than 1. So to get $\displaystyle \zeta (1)$: $\displaystyle \zeta (1) = 2^{1} \pi ^{1  1} ~ sin \left ( \frac{ \pi (1)}{2} \right ) ~ \Gamma (1  1) ~ \zeta (1  1)$ $\displaystyle = \frac{1}{2 \pi ^2} (1) \Gamma (2) ~ \zeta (2)$ and since $\displaystyle \Gamma (2) = (2  1)! = 1! = 1$ and $\displaystyle \zeta (2) = \sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{ \pi ^2}{6}$ we get $\displaystyle \zeta (1) = \frac{1}{2 \pi ^2} (1) \frac{\pi ^2}{6}$ $\displaystyle \zeta (1) =  \frac{1}{12}$ This has nothing to do with the sum 1 + 2 + 3 + ... Dan Last edited by topsquark; November 13th, 2015 at 07:05 PM. 
November 13th, 2015, 07:14 PM  #28 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,659 Thanks: 2635 Math Focus: Mainly analysis and algebra  
November 13th, 2015, 07:21 PM  #29  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,160 Thanks: 879 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  

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1/12, 1 or 12, infinity, ramanujan, relationship, series, zeta 
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