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I was playing around looking at the Taylor series of x^(x^x). This is what the first few terms looks like: http://i.imgur.com/QbRNp.png

Each term consists of (and forgive me if I don't parse it nicely) (x*ln(x))^(n+1)/n!, and then a polynomial of natural logs. If you write out the coefficients of each term in the polynomial in order, and put 1 in the first few terms with no polynomial, you get the following sequence:

1,1,1,1,1,3,1,1,7,6,1

I plugged that into Sloane's and found out that those are Stirling numbers of the second kind.

This is interesting because there have been some papers (http://projecteuclid.org/euclid.rmjm/1181072076, http://projecteuclid.org/euclid.rmjm/1250127221) that show that higher order derivatives of x^x can be described using Stirling numbers. I guess this shows that third order tetration can be described similarly.

The coefficients in the polynomials for higher order tetrations don't bring up anything in Sloane's.

Is this of interest to anyone? Also, can someone figure out the general expression for the nth term? It's just beyond my grasp right now.

Edit: After doing some more math, the Taylor series is:

Edit 2: I now realize that the coefficient in the inner summation is just S(n,m) where m goes from 1 to n. That simplifies it. And upon some more tinkering, I found out that the entire thing inside the sum is called the Bell polynomial.

Has anyone worked out the Taylor series for nth order tetration?
(03/15/2010, 04:52 PM)Ztolk Wrote: [ -> ]Has anyone worked out the Taylor series for nth order tetration?

yes.

http://mathworld.wolfram.com/PowerTower.html

formula ( 6 ) , ( 7 ) and ( 8 ).

a classic.

regards

tommy1729
(06/24/2010, 08:23 PM)tommy1729 Wrote: [ -> ]http://mathworld.wolfram.com/PowerTower.html

formula ( 6 ) , ( 7 ) and ( 8 ).

a classic.

Hm, that is a development in $\ln(x)$, Not really a Taylor devlopment of x^^n at some point $x_0$.
I wonder whether we have formulas for the powerseries development of $x\^\^n$ at $x_0=1$ (and not at 0 because the powertower is not analytic there).

And indeed Andrew pointed it out in his tetration-reference formula (4.17-4.19) (or in the Andrew's older Tetration FAQ 20080112: (4.23-25)):

$\begin{equation}
{}^{n}{x} = \sum^\infty_{k=0} t_{n,k} (x-1)^k
\end{equation}$

where:
$\begin{equation}
t_{n,k} = \begin{cases}
1 & \text{if } n \ge 0 \text{ and } k = 0, \\
0 & \text{if } n = 0 \text{ and } k > 0, \\
1 & \text{if } n = 1 \text{ and } k = 1, \\
0 & \text{if } n = 1 \text{ and } k > 1,
\end{cases}
\end{equation}
$

otherwise:
$\begin{equation}
t_{n,k} = \frac{1}{k} \sum^k_{j=1} \frac{1}{j} \sum^k_{i=j} i {(-1)^{j-1}} t_{n,k-i} t_{n-1,i-j}
\end{equation}
$

I think it has convergence radius 1 because the substituted logarithm has convergence radius 1 and also because of the singularity at 0.