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 Superlog with exact coefficients andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 06/11/2008, 05:46 AM (This post was last modified: 06/11/2008, 05:52 AM by andydude.) We have investigated the coefficients of the super-logarithm for quite a long time now, and so far, all my attempts have been met with approximations of approximations. Finally, I may have found a super-logarithm with exact coefficients. You be the judge. First lets start with some recent realizations about Abel functions, Julia functions, and topological conjugacy. So, I'll use the $\text{dxp}_b(x) = b^x - 1$ notation, and thus the topological conjugacy between "exp" and "dxp" can be expressed as: $\exp_{(b^{1/b})}^{\circ t}(x) = b(\text{dxp}_b^{\circ t}(x/b-1)+1)$ which can also be found in this thread. The Abel functions and Julia functions can be related as: $\mathcal{J}[f](x) = \frac{1}{\frac{\partial}{\partial x} \mathcal{A}[f](x)}$ which, as they apply to exp/dxp, can be proven (and is proven later) to imply: $\mathcal{A}[\exp_{(b^{1/b})}](x) = \mathcal{A}[\text{dxp}_b](x/b-1)$ $\mathcal{J}[\exp_{(b^{1/b})}](x) = b\mathcal{J}[\text{dxp}_b](x/b-1)$ Second, I think the process of finding iterated-dxp is well understood by now, so I'll start with that. One recent observation in this thread has been that Szekeres' Julia functions and Jabotinsky's L-functions (or iterative logarithm) are actually the same functions, which has opened my eyes to a whole new approach to iteration. With this in mind, not only can we express the Abel function as: $\mathcal{A}[f](x) = \lim_{n\to\infty}(\log_a(f^{\circ n}(x)) - n)$ or the logarithm of the Schroeder function, but we can also express it as: $\mathcal{A}[f](x) = \int\frac{dx}{\left[\frac{\partial}{\partial t} f^{\circ t}(x)\right]_{t=0}}$ which also serves to emphasize the fact that Abel functions are only determined up to a constant, and that a solution $\alpha(x)$ (to the Abel functional equation) can always be generalized to $\alpha(x) + C$, which is also true of integrals. Iterated-dxp can be expressed as the hyperbolic iteration of $e^{ax}-1$ as: $\text{dxp}_{(e^a)}^{\circ t}(x) = a^tx + \frac{a^{t+1}(a^t-1)x^2}{2(a-1)} + \frac{a^{t+2}(a^t-1)(-1-2a+a^t(a+2))x^3}{6(a-1)^2(a+1)}$ and since the Julia function of dxp is also the iterative logarithm of dxp: $\left[\frac{\partial}{\partial t}\text{dxp}_{(e^a)}^{\circ t}(x)\right]_{t=0} = \mathcal{J}[\text{dxp}_{(e^a)}](x)$ which evaluates to the power series: $\mathcal{J}[\text{dxp}_{(e^a)}](x) = \ln(a)\left( x + \frac{ax^2}{2(a-1)} + \frac{a^2x^3}{6(1-a^2)} - \frac{a^3(1+2a)x^4}{24(-1-a+a^3+a^4)} + \cdots \right)$ We can now find the Abel function of dxp by integrating its Julia function, but we need to find the reciprocal first. Finding the reciprocal of a power series can be a tedious task, so I've done the work for you: $\frac{1}{f'(x)} = \frac{1}{f'(0)} - \frac{f''(0)}{f'(0)^2}x + \frac{2f''(0)^2-f'(0)f'''(0)}{2f'(0)^3}x^2 + \cdots$ and since $\mathcal{J}[f](x) = \frac{1}{\frac{\partial}{\partial x}\mathcal{A}[f](x)}$, we solve for $f^{(k)}(0)$ by equating the coefficients of x, and the solution to these equations is: $\mathcal{A}[\text{dxp}_{(e^a)}](x) = \frac{1}{\ln(a)}\left( x + \frac{ax^2}{4(1-a)} + \frac{a^2(1+5a)x^3}{36(a-1)^2(a+1)} - \frac{a^4(2+a+3a^2)x^4}{32(a-1)^3(a+1)(1+a+a^2)} + \cdots \right)$ Lastly, we can relate these findings back to the super-logarithm. Let $\alpha(x) = \mathcal{A}[\text{dxp}_{b}](x)$ be the Abel function of dxp. Let $\beta(x) = \alpha(x/b - 1)$. Then $\beta(x)$ is an Abel function of exp. Proof. $\alpha(b^x-1) = \alpha(x) + 1$ $\alpha(b^{(x/b - 1)}-1) = \alpha(x/b - 1) + 1$ $\alpha(b^{(x/b)}/b-1) = \alpha(x/b - 1) + 1$ $\alpha((b^{1/b})^x/b-1) = \alpha(x/b - 1) + 1$ $\beta((b^{1/b})^x) = \beta(x) + 1$. [] This means that $\beta(x) = \text{slog}_{(b^{1/b})}(x)$ which relates back to the super-logarithm as follows: $\text{slog}_{(e^{a(e^{-a})})}(x) = C + \frac{1}{\ln(a)}\left( (x(e^{-a})-1) + \frac{a(x(e^{-a})-1)^2}{4(1-a)} + \frac{a^2(1+5a)(x(e^{-a})-1)^3}{36(a-1)^2(a+1)} - \frac{a^4(2+a+3a^2)(x(e^{-a})-1)^4}{32(a-1)^3(a+1)(1+a+a^2)} + \cdots \right)$ Andrew Robbins « Next Oldest | Next Newest »

 Messages In This Thread Superlog with exact coefficients - by andydude - 06/11/2008, 05:46 AM RE: Superlog with exact coefficients - by andydude - 06/11/2008, 06:02 AM RE: Superlog with exact coefficients - by Gottfried - 06/11/2008, 07:18 AM RE: Superlog with exact coefficients - by Gottfried - 06/13/2008, 06:38 AM RE: Superlog with exact coefficients - by bo198214 - 06/20/2008, 01:26 PM RE: Superlog with exact coefficients - by Gottfried - 03/10/2009, 09:59 PM

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