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 Simplified regular tetration andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 05/02/2009, 12:13 AM Regular iteration of exponentials gives a power series of $\exp_a^{y}(z)$ about z. In order to obtain regular tetration from this we must evaluate this at (z=1), but in doing so, it is no longer a power series. In order to make this a power series again we have to re-expand it about 'y' or 'a', which makes really messy power series. This messy power series is of the form ${}^{y}a = \sum_{i \ge 0} \sum_{j \ge 0} \sum_{k \ge 0} C_{ijk} (a - 1)^i \ln(a - 1)^j y^k$ This is not a power series in 'a', because it involves both (a - 1) and $\ln(a - 1)$. If we are to compare this with other methods, then it would be beneficial to have a true power series in 'a'. To this end, we can define the following function $X(u, v) = (e^{ue^{-u}}) \uparrow\uparrow \log_u(v)$ and define an an inverse function such that $Z(u, X(u, v)) = v$. There are numerous benefits to defining the function this way. Since all of the functions used are invertible, we can express tetration in terms of X: ${}^{y}a = X(\ln({}^{\infty}a), \ln({}^{\infty}a)^y)$. We can also express the superlog in terms of Z: $\text{slog}_a(z) = \log_{\ln({}^{\infty}a)}(Z(\ln({}^{\infty}a), z))$. Superroots cannot be expressed with these functions, however. To show that all of the logarithms are gone, here is the resulting power series of X: $ \begin{tabular}{rl} -X(u, v+1) & = -1 + uv + \frac{u^2v}{2} \\ & + {u^3}\left(\frac{2}{3}v + \frac{1}{2}v^2\right) \\ & + {u^4}\left(\frac{13}{24}v + \frac{1}{2}v^2\right) \\ & + {u^5}\left(\frac{43}{60}v + \frac{25}{24}v^2 + \frac{1}{3}v^3\right) \\ & + {u^6}\left(\frac{331}{720}v + \frac{19}{24}v^2 + \frac{1}{3}v^3\right) \\ & + {u^7}\left(\frac{1999}{2520}v + \frac{1231}{720}v^2 + \frac{7}{6}v^3 + \frac{1}{4}v^4\right) \\ & + {u^8}\left(\frac{17977}{40320}v + \frac{811}{720}v^2 + \frac{67}{72}v^3 + \frac{1}{4}v^4\right) \\ & + {u^9}\left(\frac{25705}{36288}v + \frac{85681}{40320}v^2 + \frac{281}{120}v^3 + \frac{9}{8}v^4 + \frac{1}{5}v^5\right) \\ & + {u^{10}}\left(\frac{296113}{518400}v + \frac{14143}{8064}v^2 + \frac{583}{288}v^3 + \frac{25}{24}v^4 + \frac{1}{5}v^5\right) \\ & + \cdots \end{tabular}$ as you can see, some of the coefficients display a pattern, like the $\frac{1}{n}v^n$ terms, but I don't know if this pattern continues. Even so, I think this form is much easier to analyze than picking a base and sticking with it for all calculations. For example, the Julia function of exponentials (evaluated at 1) can be expressed with this function as well. $\mathcal{J}[\exp_a](1) = \ln(\ln({}^{\infty}a))\left[\frac{\partial}{\partial v} X(\ln({}^{\infty}a), v)\right]_{v=1}$ so to summarize, these are the benefits of this simplified view of regular tetration: The coefficients of (-X) are always positive (except for the -1 in the first term), which means X itself has negative coefficients (except for the 1 in the first term). All of the logarithms in the direct expansion are gone, meaning they must have come from the sub-expansions of ${}^{\infty}a$ in the power series. We can express tetration in terms of X, without any loss of generality. We can express superlogs in terms of the inverse function Z. We can express the Julia function of exponentials in terms of $\frac{\partial X}{\partial v}$ Some other properties of this function include $X(u, 0) = e^{u}$ $X(u, 1/u) = 0$ $X(u, 1) = 1$ $X(u, u) = e^{ue^{-u}}$ $X(u, \infty) = e^{u}$ $X(u, uv) = e^{ue^{-u}} \uparrow X(u, v)$ Andrew Robbins 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 ${^xb}=\eta(c^x)+a$. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 05/03/2009, 08:49 AM So is X the inverse of the Schroeder function? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/03/2009, 09:33 AM (05/03/2009, 08:49 AM)andydude Wrote: So is X the inverse of the Schroeder function?Actually I am also somewhat puzzled. I would suggest to always use variable $b$ for the base. In this terminology you wrote: ${}^{y}b = X(\ln({}^{\infty}b), \ln({}^{\infty}b)^y)$ Now ${^\infty b}$ is the fixed point which I denoted with $a$. And $\ln(a)$ is the derivative at the fixed point which I denoted with $c$. So your formula is: ${}^{x}b = X(c, c^x)$ while my formula is: ${^xb}=\eta(c^x)+a$ Your coefficients are polynomials in $c$, while my coefficients are rational functions in $c$ and also use $\ln(b)$. Though both must be equal up to translation. *headscratch* « Next Oldest | Next Newest »

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