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 [repost] A nowhere analytic infinite sum for tetration. tommy1729 Ultimate Fellow     Posts: 1,358 Threads: 330 Joined: Feb 2009 03/20/2018, 12:16 AM In the title I mentioned that this is a repost. Indeed these ideas are not completely new and have been mentioned before. However I Will reinvent my own wheel , by a different presentation of  the/my ideas ... The basic principle is very old : infinite Sum and telescoping. We want to solve t(exp(x)) = e t(x). Define  0 < a < 1. f(x,a) = exp(x) + a x. f^[n+1](x,a) = f( f^[n](x,a) ,a). —- Now consider for Every real y there is a solution x to f(x,a) = y. Nomatter What a is , but still 0 < a < 1. —— Now define the Sum from m = - Oo to Oo : t(z,a) = ... + exp(z + 2)/f^[-2](1/2,a) + exp(z + 1)/f^[-1](1/2,a) + ... So t(z,a) = sum_m exp(z - m)/f^[m](1/2,a). Now let Lim a —> 0 : z = x then : Lim t(z,0) = t(z) = t(x) t(exp(x)) = e t(x). Qed ___ For practical reasons I suggest  Compute for x < 1 : t(x) by computing t(exp(x))/e. —— We can continue since 1/z + 1/exp(z) + 1/exp(exp(z)) + ... Is nowhere analytic since exp iterations are chaotic we have : t(x) is nowhere analytic. Also t(exp(x)) = t(x) e. ln t(  exp(x)  ) = ln t(x) + 1 Thus ln t(x) = slog(x). A nowhere analytic slog(x) ! ——- Due to the telescoping infinite Sum , this mzthod is probably the simplest nowhere analytic solution in the spirit of calculus. I assume it was worthy of Being repeated. Regards Tommy1729 « Next Oldest | Next Newest »

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