Taylor series for tetration

Question

I would like to know what is known about Taylor series for tetration (and other hyper-exponentiations).

Surprisingly, such information is rare on internet. Numerical values for expansion in hyperexponent for tetration with basis $e$ can be found here:

http://en.citizendium.org/wiki/Tetration#Taylor_expansion_at_zero

However I am interested in expansion in the basis of tetration. I look for

$$ x \uparrow \uparrow m = \sum_{n=0}^{\infty} c_n(m)x^n $$

Wolfram mathematica gives me wired result (obviously a bug) at $x=0$ for $x \uparrow 3$:

https://www.wolframalpha.com/input/?i=taylor+series+x%5E(x%5Ex)

where in the output $log(x)$ appears (not a polynomial).

It is better at $x=1$

https://www.wolframalpha.com/input/?i=taylor+series+x%5E(x%5Ex)+at+x%3D1

I imagine the point $x=0$ may be "peculiar", but at least expansion at $x=1$ should be possible.

List of my questions:

1) Is tetration analytic at $x=0$? At $x=1$?

2) If yes, is an explicit closed-form formula for $c_n(m)$ known? (at any of these points)

3) If not, is an explicit closed-form formula for $c_n(m)$ know for some specific values of $m$?

4) Same questions for extension to higher hyper-exponentiations...

Thank you.

Answer 1

$x \uparrow \uparrow m$ is never analytic at $x = 0$ for $m > 1$. For example, consider $m = 2$, then you have the function $x \uparrow \uparrow 2 = x^x = e^{x \ln x}$ which clearly has a branch point at $x = 0$ and hence is not analytic. You can see this if we substitute $\ln x = 2\pi i n + \log x$ where $\log$ represents the principal branch of the logarithm, in which case we have $x^x = e^{x (2\pi i n + \log x)} = e^{x 2 \pi i n} e^{x\log x}$, which certainly depends on $n$ as long as $x$ is not an integer. You have similar problems for all $m > 1$.

The function $x \uparrow \uparrow m$ is always analytic at $1$, which can be seen inductively: Clearly for $m=1$ it is analytic at $1$. If $x \uparrow \uparrow m$ is analytic at 1, then $x \uparrow \uparrow{(m+1)} = e^{(\ln x)(x\uparrow \uparrow m)}$ is analytic at $1$ because $\exp$ is analytic everwhere and $\ln$ is analytic at 1. The Taylor series of $x \uparrow \uparrow m$ at $1$ is not very nice, but the terms can be computed explicitly. Wolfram Mathworld tells us that $$ (e^x) \uparrow\uparrow m = \sum_{n=0}^m \frac{(n+1)^n}{(n+1)!} x^n + \sum_{n = m+1}^\infty a_{m n} x^n$$ where $a_{m n}$ have a rather complicated recursive formula. To compute the Taylor series of $x \uparrow\uparrow m$ at $1$, you could compose the above series with the series for $\ln(x)$ at $1$ using Faa di Bruno's formula. Not very simple, but it probably as close to a closed form as you will get.

Answer 2

Consider the tetration of the function $e^x$ $$^n(e^x)=(e^x)_1^{(e^x)_2^{(e^x)_3^{_..^{.e^x_n}}}}$$

For a natural number n, the taylor series of that function is

$$^n(e^x)=\sum_{k}^{\infty}\frac{1}{k!}*T(n,k)*x^k$$

where $T(n,k)$ is the OEIS A210725 (if k<n, $T(n,k)=T(k,k)$)

then, for $n\in\mathbb{N}$:

$$^n(x)=\sum_{k}^{\infty}\frac{1}{k!}*T(n,k)*(ln(x))^k$$

On the limit for $n\to\infty$, the T(n,k)=T(k,k) are the coefficients of the Lambert w function.

There are other series, that you can find here.

For code able to calculate tetrations with high precision, (for a limited set of bases, and $n\in\mathbb{R}$) check Sheldonison's fatou.gp code

In Wolframalpha, you can get the series of $^n(x+1)$, and it will produce the Stirling transform of $^n(e^x)$. The series has power of x, and no logarithms.

Answer 3

I think really the extension of the progression of operations by introducing a new one that is the $n$-fold iteration of the thus-far highest one is a process that on the face of it kindof looks like it ought to be susceptible of being plied indefinitely ... but at the end of the day isn't! (Except in this case to the natural numbers only.) You get it with complex-quaternion-octonion-etc also - that progression seems also to fizzle-out rather disappointingly; and a similar principle seems to be abroad in very many areas of mathematics. Sometimes I so want this or that progression to be extensible without limit ... but then I find eventually that I need to resign myself to the first two or three items of the progression being beautiful special cases. I must admit, though ... sometimes it's actually quite a relief!

However - I do think that business of relating the tetration by $n$ of the exponential function to the Lambert w function as $n\rightarrow\infty$, and the way there is a taylor series for the function at finite $n$ with coefficients that can be found by a recursion algorithm - and the way they 'peel-away' term-by-term to leave the coefficients of the lambert w function: I do find that to be an extraordinarily sweet little oasis in the midst of a parched desert ... and it is certainly an item that I treasure amongst my crown jewels!

Answer 4

Just for your curiosity.

Expanding as Taylor series aound $x=1$ as $$f_n=\sum_{k=0}^\infty c_k\,(x-1)^k$$ see how fast converge the coefficients $$f_1=\left\{1,1,1,\frac{1}{2},\frac{1}{3},\frac{1}{12},\frac{3}{40},-\frac{1}{120},\frac{59} {2520},-\frac{71}{5040},\frac{131}{10080}\right\}$$ $$f_2=\left\{1,1,1,\frac{3}{2},\frac{4}{3},\frac{3}{2},\frac{53}{40},\frac{233}{180},\frac{56 27}{5040},\frac{2501}{2520},\frac{8399}{10080}\right\}$$ $$f_3=\left\{1,1,1,\frac{3}{2},\frac{7}{3},3,\frac{163}{40},\frac{1861}{360},\frac{33641}{504 0},\frac{8363}{1008},\frac{22391}{2160}\right\}$$ $$f_4=\left\{1,1,1,\frac{3}{2},\frac{7}{3},4,\frac{243}{40},\frac{3421}{360},\frac{71861}{504 0},\frac{54371}{2520},\frac{69281}{2160}\right\}$$ $$f_5=\left\{1,1,1,\frac{3}{2},\frac{7}{3},4,\frac{283}{40},\frac{4321}{360},\frac{102941}{50 40},\frac{85871}{2520},\frac{61333}{1080}\right\}$$ $$f_6=\left\{1,1,1,\frac{3}{2},\frac{7}{3},4,\frac{283}{40},\frac{4681}{360},\frac{118061}{50 40},\frac{106661}{2520},\frac{81583}{1080}\right\}$$ $$f_7=\left\{1,1,1,\frac{3}{2},\frac{7}{3},4,\frac{283}{40},\frac{4681}{360},\frac{123101}{50 40},\frac{115481}{2520},\frac{93013}{1080}\right\}$$ $$f_8=\left\{1,1,1,\frac{3}{2},\frac{7}{3},4,\frac{283}{40},\frac{4681}{360},\frac{123101}{50 40},\frac{118001}{2520},\frac{97333}{1080}\right\}$$ $$f_9=\left\{1,1,1,\frac{3}{2},\frac{7}{3},4,\frac{283}{40},\frac{4681}{360},\frac{123101}{50 40},\frac{118001}{2520},\frac{97333}{1080}\right\}$$