2015 Continuum sum conjecture tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 05/25/2015, 11:29 PM I noticed a possible pattern. $\sum_{k=0}^{z-1} e^{nk} = \frac{e^{nz} - 1}{e^{n} - 1}$ next $\sum_{k=0}^{z-1} e^{e^k} = \sum_{k=0}^{z-1} f(k) = \sum_{n=0}^{\infty} \frac{1}{n!} \left(\sum_{k=0}^{z-1} e^{nk}\right) = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{z-1} e^{nk} = \sum_{n=0}^{\infty} \frac{e^{nz} - 1}{n! \left(e^{n} - 1\right)}$ next define : $e^{e^{e^x}} = \sum_{n=0}^{\infty} a_n e^{nx}$ then \begin{align}\sum_{n=0}^{x-1} e^{e^{e^n}} &= a_0 x + \sum_{n=1}^{\infty} \frac{a_n}{e^{n} - 1} \left(e^{nx} - 1\right) \\ &= \left(a_0 - \sum_{n=1}^{\infty} \frac{a_n}{e^{n} - 1}\right) x + \sum_{n=1}^{\infty} \frac{a_n}{e^{n} - 1} e^{nx}\end{align} Now it appears The continuum sum up to z-1 of exp^[n](x) is given by CS [ exp^[n](x) , z-1] = P_n(x) + F_n(x) where F_n is 2pi i periodic and P_n is a polynomial of degree (at most) n. I guess there is a simple reason for it. Right ?? What if n is not an integer ? say n = 3/2 ? Does that imply P_{3/2}(x) =< O( x^{3/2} ) ?? regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 05/26/2015, 12:10 PM The answer appears to be simple. That is considering we work with the same continuum sum. Within an analytic region : Let r > 0. exp^[1+r](x) = exp^[r](0) + F(exp(x)) where F(0) = 0 and is analytic. Therefore the continuum sum CS is given by T_r(x) = CS exp^[1+r](x) = exp^[r](0) x + G(exp(x)). Therefore T(x) - T(x - 2pi i) = exp^[r](0) * 2 pi i. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 05/26/2015, 12:18 PM One wonders what happens with r < 0. I will post a related thread. I have some ideas but little time. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 05/26/2015, 12:24 PM (This post was last modified: 05/26/2015, 12:26 PM by tommy1729.) Notice the counterintuitive nature sum [ f(x) - f(x - period f) ] =/= sum [ f(x) ] - sum[ f(x - period f) =/= 0. even if f(x) - f(x - period f) = 0 and sum [ 0 ] = 0. I have to think more about this. Any ideas ? I considered using tommysum[f(x)] := continuum sum [f(x) - f(0)]. regards tommy1729 « Next Oldest | Next Newest »

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