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 Simplified regular tetration bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/02/2009, 08:47 AM (This post was last modified: 05/03/2009, 09:25 AM by bo198214.) We know that we can express the Abel function $\alpha$ of $f$ at $a$ by $\alpha = \log_c \circ \chi$ where $\chi$ is the Schröder function of $f$ at $a$, where $c=f'(a)$. This means that the inverse of the Abel function which is actually the superexponential can be expressed by $F(x)=\chi^{-1}(c^x)$ with appropriate translation along the x-axis choosen such that $F(0)=1$. Here $c=\exp_b'(a)=\ln(b) \exp_b(a)=\ln(b)a=\ln(b^a)=\ln(a)$ The coefficients of $\eta+a=\chi^{-1}$, $\eta_0=0$, $\eta_1=1$, can be recursively computed from the equation (*) $\eta(cx)=f(\eta(x))$, $f(x)=b^{x+a}-a = a (b^x-1)$ by the composition formula $(f\circ g)_n = \sum_{k=1}^n f_k (g^k)_n$ where $(g^k)_n = \sum_{n_1+\dots+n_k = n} f_{n_1} \dots f_{n_k}$. Now we put formula (*) in: $c^n \eta_n = \sum_{k=1}^n f_k (\eta^k)_n$ On the right side $\eta_n$ only occurs for $k=1$ in $\eta^k$, namely in the summand $f_1 (\eta^1)_n = c \eta_n$. Thatswhy we have the recursive formula: $\eta_n = \frac{1}{c^n - c} \sum_{k=2}^n f_k (\eta^k)_n$ So each coefficient of $\eta$ is a polynomial in $\ln(b)$ ($b$ base) and a rational function in $c=\ln(a)$ ($a$ fixed point). and upto translation along the x-Axis we have ${^xb}=\eta(c^x)+a$. « Next Oldest | Next Newest »

 Messages In This Thread Simplified regular tetration - by andydude - 05/02/2009, 12:13 AM RE: Simplified regular tetration - by bo198214 - 05/02/2009, 08:47 AM RE: Simplified regular tetration - by andydude - 05/03/2009, 08:49 AM RE: Simplified regular tetration - by bo198214 - 05/03/2009, 09:33 AM

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