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 Something interesting about Taylor series bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/29/2010, 06:32 AM (This post was last modified: 06/29/2010, 06:42 AM by bo198214.) (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. « Next Oldest | Next Newest »

 Messages In This Thread Something interesting about Taylor series - by Ztolk - 03/13/2010, 10:55 PM

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