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 Binary partition at oo ? tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 10/07/2014, 07:22 PM (This post was last modified: 10/07/2014, 07:23 PM by tommy1729.) (10/06/2014, 07:17 PM)jaydfox Wrote: $ \frac{\sum_{k=-\infty}^{\infty} \left((-1)^k\, 2^{-k^2/2}\right)}{\sum_{k=-\infty}^{\infty} \left(2^{-k^2/2}\right)}$ This evaluates to approximately (0.004872868560797)/(3.01076739115959), which is approximately 0.001618480582427. In general $\sum_{k=-\infty}^{\infty} \left((-1)^k\, f(-k^2)\right)$ equals $2 \sum_{k=1}^{\infty} \left((-1)^k\, f(-k^2)\right)-f(0)$. You probably already know that , but maybe it simplifies matters ? Are you claiming that there are no zero's off the real line ? Very intresting stuff. I think we are onto a general thing ; No zero's in the upper complex plane Good entire approximations ( fake function , J(x) for the binary p ) ... with all derivatives positive. It all seems connected. Id like to see how you arrived at these things. Thanks. regards tommy1729 « Next Oldest | Next Newest »

 Messages In This Thread Binary partition at oo ? - by tommy1729 - 10/03/2014, 09:11 PM RE: Binary partition at oo ? - by jaydfox - 10/06/2014, 07:17 PM RE: Binary partition at oo ? - by tommy1729 - 10/07/2014, 07:22 PM

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