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 Transseries, nest-series, and other exotic series representations for tetration mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/14/2009, 09:22 PM (12/14/2009, 03:16 PM)bo198214 Wrote: Mike, I have another bad news, I found out that not even the Faulhaber sum of $e^{\frac{x^2}{2}}$ is convergent, contrary to what I said before because it looks convergent up to n=100 terms; so your transseries sum $e^{e^x}$ can not be convergent. How does that refute the summation I gave with the transseries? That involves continuum-summing on terms of the form $ae^{nx}$, not $ae^{nx^2}$ or $ae^{x^n}$ or something like that. bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 12/15/2009, 01:17 AM (This post was last modified: 12/15/2009, 01:17 AM by bo198214.) (12/14/2009, 09:22 PM)mike3 Wrote: How does that refute the summation I gave with the transseries? That involves continuum-summing on terms of the form $ae^{nx}$, not $ae^{nx^2}$ or $ae^{x^n}$ or something like that. Its because $\frac{x^2}{2}. Thatswhy both Faulhaber sums are also in the < relation. Thatswhy the sum of $e^{\frac{x^2}{2}}$ and of $e^{e^x}$ are also in this relation. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/15/2009, 06:18 AM (This post was last modified: 12/15/2009, 06:41 AM by mike3.) (12/15/2009, 01:17 AM)bo198214 Wrote: Its because $\frac{x^2}{2}. Thatswhy both Faulhaber sums are also in the < relation. Thatswhy the sum of $e^{\frac{x^2}{2}}$ and of $e^{e^x}$ are also in this relation. So then how come the series given in the original post $\sum_{n=0}^{x-1} e^{e^n} = x + \sum_{n=1}^{\infty} \frac{e^{nx} - 1}{n! \left(e^n - 1\right)} = -\left(\sum_{n=1}^{\infty} \frac{1}{n! \left(e^n - 1\right)}\right) + x + \left(\sum_{n=1}^{\infty} \frac{1}{n! \left(e^n - 1\right)} e^{nx}\right)$ looks to converge? In fact, the two series on the right can be proven to converge by applying the direct comparison test between these and the original series without its initial term, at a given x-value for the second series (the one that depends on x) and at $x = 0$ for the first (the one that sums to a constant independent of x). The original is known to converge because the radius of convergence of the exponential function's Taylor series is infinity and it is a simple substitution of a finite value (for finite x) into that series. So can you shoot down this proof? Unless you can claim $\sum_{n=0}^{x-1} e^{nx} = \frac{e^{nx} - 1}{e^n - 1}$ is wrong or inconsistent with the Faulhaber formula. How would you do that? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/15/2009, 06:56 AM Hmm. I think I may have figured out what's going on. It seems like Faulhaber's formula fails to sum $\sum_{n=0}^{x-1} e^{qn}$ for what appears to be $q \ge 2\pi$. Below that it works and sums it to, apparently $\frac{e^{qx} - 1}{e^q - 1}$. So that would provide an analytic continuation to higher q-values, thus the formula given is valid in a sense, but direct application of the Faulhaber formula will not work. It seems that a combination of analytic continuation and divergent summation theory applied to Faulhaber and transseries theory is needed to put together a comprehensive theory of continuum sums of general analytic functions. I find the appearance of $2\pi$ interesting, considering it is the magnitude of the imaginary period of the exponential function. I don't know if that is significant or not. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/15/2009, 07:47 AM (This post was last modified: 12/15/2009, 09:11 AM by mike3.) That's interesting, Ansus. What's $\prod_x x^{x^x}$ ? According to your hypothesis, it should be $\prod_x x^{x^x} = C e^{\psi(-3, x) + P(x)}$ for some polynomial $P(x)$. Does that work? bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 12/15/2009, 12:46 PM (12/15/2009, 06:18 AM)mike3 Wrote: In fact, the two series on the right can be proven to converge by applying the direct comparison test between these and the original series Actually that seems right too. Hm, I have to look where the error is then. bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 12/15/2009, 01:00 PM (This post was last modified: 12/15/2009, 01:13 PM by bo198214.) (12/15/2009, 12:46 PM)bo198214 Wrote: Hm, I have to look where the error is then. It appears that the Faulhaber sum is not monotonic (which I assumed), i.e. we dont have that from $f(x) follows that $\sum_{x} f(x) < \sum_{x} g(x)$. This puzzles me a bit, shouldnt one expect that from a sum operator? But no, one shouldnt. The simplest example is perhaps $f(x)=1$ and $g(x)=1+x^2$ with the Faulhaber sums $x$ and $\frac{1}{3} x^{3} - \frac{1}{2} x^{2} + \frac{7}{6} x$; the latter cuts $x$ at 0 because the first derivative there is different from 1. « Next Oldest | Next Newest »

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