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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 andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 06/11/2008, 06:02 AM Also, I wonder if this is the same as regular slog (rslog)? Gottfried Ultimate Fellow Posts: 765 Threads: 119 Joined: Aug 2007 06/11/2008, 07:18 AM (This post was last modified: 06/11/2008, 12:44 PM by Gottfried.) andydude Wrote:This means that $\beta(x) = \text{slog}_{(b^{1/b})}(x)$ which relates back to the super-logarithm as follows: $\text{slog}_{(e^{a(e^{-a})})}(x) = C + \frac{1}{\ln(a)}\left( (x(e^{-a})-1) + \frac{a(x(e^{-a})-1)^2}{4(1-a)} + \frac{a^2(1+5a)(x(e^{-a})-1)^3}{36(a-1)^2(a+1)} - \frac{a^4(2+a+3a^2)(x(e^{-a})-1)^4}{32(a-1)^3(a+1)(1+a+a^2)} + \cdots \right)$ Hmm, at least it looks somehow similar to the "regular"-formula. If I replace a by u, and introduce t for exp(u) so that $\text{slog}_{(e^{a(e^{-a})})}(x) = \text{slog}_{(e^{u/t})}(x)$ and $(x(e^{-a})-1) = \frac{x}{t}-1 = x'$ and factorize the denominators differently, for instance $\hspace{24} (a-1)^3(a+1)(1+a+a^2) = (a-1)(a-1)(a-1)*(a+1)*(1+a+a^2) = (a-1)(a^2-1)(a^3-1)$ to get a more familiar looking formula for me, this is then, using "rsdxplog" as rslog for the dxp-function: $\text{slog}_{e^{u/t}}(x) = \text{rsdxplog}_t(x') = C + \frac{1}{\ln(u)}\left( (x' - \frac{ux'^2}{4(u-1)} \hspace{12} + \frac{u^2(1+5u)x'^3}{36(u-1)(u^2-1)} \hspace{12} - \frac{u^4(2+u+3u^2)x'^4}{32(u-1)(u^2-1)(u^3-1)} \hspace{12} + \cdots \right)$ where especially the denominators-products are the same as in my Ut-formulae, and also the numerators look very familiar. I'll see, whether we have the same coefficients later today. I had the rsdxplog as logarithm of the Schroeder-function, assuming x' as h'th (continuous) iteration of $\text{dxp}_t^{{^o}h}(1)$ then $ \\ \vspace{12}\hspace{24} x' = \text{dxp}_t^{{^o}h}(1) = \sigma_t^{-1}(u^h \sigma_t(1)) \\ \vspace{12}\hspace{24} \sigma_t(x') = u^h \sigma_t(1) \\ \vspace{12}\hspace{24} \frac {\sigma_t(x')}{\sigma_t(1)} = u^h \\ \vspace{12}\hspace{24} \text{rsdxplog}(x')= h = \log_u(\sigma_t(x')) - \log_u(\sigma_t(1)) \\ \vspace{12}\hspace{24} \text{rsdxplog}(x')= h = C + \log_u(\sigma_t(x')) =C + \frac1{ln(u)}\log(\sigma_t(x'))$ where I got the coefficients of the sigma-function by the eigenmatrices of Ut - and the structure of these coefficients look very similar to yours above Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 765 Threads: 119 Joined: Aug 2007 06/13/2008, 06:38 AM andydude Wrote:This means that $\beta(x) = \text{slog}_{(b^{1/b})}(x)$ which relates back to the super-logarithm as follows: $\text{slog}_{(e^{a(e^{-a})})}(x) = C + \frac{1}{\ln(a)}\left( (x(e^{-a})-1) + \frac{a(x(e^{-a})-1)^2}{4(1-a)} + \frac{a^2(1+5a)(x(e^{-a})-1)^3}{36(a-1)^2(a+1)} - \frac{a^4(2+a+3a^2)(x(e^{-a})-1)^4}{32(a-1)^3(a+1)(1+a+a^2)} + \cdots \right)$ andydude Wrote:Also, I wonder if this is the same as regular slog (rslog)? Hmm, I give up here - I couldn't reproduce these coefficients yet; I had to determine the coefficients of $ln(\frac{\sigma(x)}{\sigma(1)})$ ... I'll try another time. So, if your coefficients give the correct powerseries (and values): congratulation! (and I put it aside for a later reconsideration) Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/20/2008, 01:26 PM There is an error in your calculation of the Abel function of dxp: andydude Wrote: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: $\mathcal{A}[\text{dxp}_{(e^a)}](x) = \frac{1}{\ln(a)}\left( x + \frac{ax^2}{4(1-a)} + \frac{a^2(1+5a)x^3}{36(a-1)^2(a+1)} - \frac{a^4(2+a+3a^2)x^4}{32(a-1)^3(a+1)(1+a+a^2)} + \cdots \right)$ The regular Abel function (for a function with fixed point at 0) always has a singularity at 0. You can not expand it into a powerseries at 0. In the hyperbolic case the Abel function is the logarithm of the Schroeder function (as Gottfried also pointed out in his post), here you can also see that it is not developable at 0, because log is not. What you however can do is to express the Abel function as $c\log(z)+M(z)$ where M is a function, such that $z^m M(z)$ is holomorphic and so developable at at 0. This is an extension to holomorphic functions (so called meromorphic functions) which also allow a finite number of negative powers in the power series development, i.e. $M(z) = \sum_{n=-m}^\infty M_n z^n$. I explained here how this comes. To compute the inverse of a powerseries $f$ - which is a meromorphic function - we determine the first index $m$ such that $f_m\neq 0$ then we divide $f$ by $z^m$ and get a powerseries with $g$ with $g_0\neq 0$. We can then compute the reciprocal of this powerseries and the inverse of $f$ is then $\frac{1}{f(z)} = z^{-m} \frac{1}{f(z)/z^m}$ is a powerseries with $m$ negative powers. In our case $f(z)=\mathcal{J}[f](z)$, $f_0=0$ and $m=1$. $\frac{1}{f(z)}=z^{-1}\frac{1}{f(z)/z} = \frac{1}{\log(a)} z^{-1} + \frac{-{a}^{2} }{{2 \left( {a}^{2} - a \right)}}z^0 - \frac{\frac{{-{a}^{5} }}{{2 \left( {a}^{2} - a \right)}} + \frac{{{a}^{4} }}{{2 \left( {a}^{2} - a \right)}} + \frac{{{a}^{3} }}{3}}{{\left( {a}^{3} - a \right) }} z + \dots$ If we now integrate this: $\alpha(z)=\mathcal{A}[\text{dxp}_{e^a}](z)=\int{\frac{dz}{f(z)}=C+\log_a(z)+\frac{-{a}^{2} }{{2 \left( {a}^{2} - a \right)}}z^1 - \frac{\frac{{-{a}^{5} }}{{2 \left( {a}^{2} - a \right)}} + \frac{{{a}^{4} }}{{2 \left( {a}^{2} - a \right)}} + \frac{{{a}^{3} }}{3}}{{\left( {a}^{3} - a \right) }} \frac{z^2}{2} + \dots$ Quote: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. $\alpha(b^x-1) = \alpha(x) + 1$ $\alpha(b^{(x/b - 1)}-1) = \alpha(x/b - 1) + 1$ $\alpha(b^{(x/b)}/b-1) = \alpha(x/b - 1) + 1$ $\alpha((b^{1/b})^x/b-1) = \alpha(x/b - 1) + 1$ $\beta((b^{1/b})^x) = \beta(x) + 1$. [] Yes this is correct and this is equal to the rslog because of the following: You put the formula $\left(\exp_{p^{1/p}}\right)^{\circ t}=\mu_p\circ \tau_1\circ \text{dxp}_p^{\circ t} \circ \tau_1^{-1}\circ \mu_p^{-1}$ where $p$ is the lower fixed point, $\mu_c(z)=cz$ and $\tau_c(z)=z+c$. If $\alpha$ is the (principal) Abel function of $\text{dxp}_p$ we write this is as: $\left(\exp_{p^{1/p}}\right)^{\circ t}=\mu_p\circ \tau_1\circ \alpha^{-1}\circ \tau_t \circ \alpha \circ \tau_1^{-1}\circ \mu_p^{-1}=\beta^{-1}\circ \tau_t \circ \beta$ with $\beta = \alpha\circ \tau_1^{-1}\circ \mu_p^{-1}$ as you already pointed out with $\beta(z)=\alpha(z/p-1)$. Now the rslog computation is slightly different, we start with $h=\tau_p^{-1}\circ \exp_{b}\circ \tau_p=\mu_p\circ \text{dxp}_b$ i.e. $\exp_b^{\circ t} = \tau_p \circ h^{\circ t} \circ \tau_p^{-1} = \tau_p \circ \left(\mu_p\circ \text{dxp}_b\right)^{\circ t} \circ \tau_p^{-1}$ to show that both Abel functions of $\exp_b$ are identical we just need to show that $\mu_p\circ \tau_1\circ \text{dxp}_p^{\circ t} \circ \tau_1^{-1}\circ \mu_p^{-1} = \tau_p \circ \left(\mu_p\circ \text{dxp}_b\right)^{\circ t} \circ \tau_p^{-1}$ This is be done by the following equivalences:  \begin{align*} \tau_p\circ\mu_p \circ \text{dxp}_p^{\circ t} \circ \mu_p^{-1}\circ \tau_p^{-1} &= \tau_p \circ \left(\mu_p\circ \text{dxp}_b\right)^{\circ t} \circ \tau_p^{-1}\\ \mu_p \circ \text{dxp}_p^{\circ t} \circ \mu_p^{-1} &= \left(\mu_p\circ \text{dxp}_b\right)^{\circ t}=\left(\mu_p \circ \text{dxp}_p \circ \mu_p^{-1}\right)^{\circ t} \end{align*} And the last line is automatically satisfied by the regular iteration. Quote:This means that $\beta(x) = \dots$ If you now redevelop $\beta(z)=\alpha(z/b - 1)$ with the correct $\alpha$ - note that the logarithmic term $\log_a(z/b-1)$ in $\alpha(z/b-1)$ is now developable at 0 - you see that the coefficients become infinite sums, which is again no finite description (except you find some closed expression). Gottfried Ultimate Fellow Posts: 765 Threads: 119 Joined: Aug 2007 03/10/2009, 09:59 PM (This post was last modified: 03/12/2009, 08:19 PM by Gottfried.) andydude Wrote: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)}( (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)} + \dots ) $ Andrew Robbins One year late... and only partial progress. I express this with my toolbox of matrices. First we agree, that we can use for exp_b(x) the dxp_t(x/t-1) (while, however the full equivalence is then (dxp_t(x/t-1) + 1)*t ) For exp_b(x) I use the transposed Carleman-matrix Bb V(x) ~ * Bb = V(b^x)~ For dxp_t(x) I use t^(1/t) = b ,u = log(t) V(x/t-1) ~ *Ut = V( b^x/t- 1) ~ where Ut is the matrix-operator for dxp_t having the coefficients in its second column Create it as "Utsym" symbolically with the parameter a (from your notation = log(t) = u) to dimension, say 5: Utsym = dV(a,5) * VE(fS2F,5); Now the iterative logarithm can be found by finding the matrix-log of Utsym. Since the diagonal is not the Id-matrix, I take the eigen-decomposition (efficiently configured for triangular matrices) UtsymKenn = triEigSys(Utsym) ; and this has in the three components UtsymKenn[2] =W, UtsymKenn[3]=D ,UtsymKenn[4] = W^-1 where D is diagonal and contains the eigenvalues [1,a,a^2,a^3,a^4], sucht that Utsym = W * D * W^-1 The log of Utsym is then the re-composition with the log of the diagonal, (Log(D)) and this is the diagonalmatrix Log(D) = diag([0,1,2,3,4])*lna where I use the symbol lna for the ln(a) (since we want to keep it symbolically) UtsymLog = W * lna* diag(0,1,2,3,4) * W^-1 ... = lna* W * diag(0,1,2,3,4) * W^-1 The result is not a matrix-operator because the eigenvalues are not vandermonde; so we have to construct the matrixopertator from its second column: UtJulia = matfromser(UtsymLog[,2]) $ \begin{matrix} {rrrr} & lna* (x & -\frac{a}{2(1-a)}*x^2 & + \frac{a^2(1-a)}{3!(1-a)(1-a^2)}*x^3 & - \frac{((a^3(2a+1)(1-a)}{4!(1-a^2)(1-a^3)}*x^4 & \dots ) \end{matrix}$ This agrees with your representation except of a bit straightening of denominators, so far. Now to proceed I needed the integral-representation, but unfortunately I don't have the appropriate "toolbox-matrix" for this at the moment... So I'll try the remaining part another day Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 765 Threads: 119 Joined: Aug 2007 03/13/2009, 02:20 PM Ansus Wrote:For the base e sexp'(-1)=sexp'(0) so slog'(0)=slog'(1), can you verify this for your formula? Don't think so. I've not even completed the full reformulation of Andrew's formula... Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/13/2009, 06:14 PM Gottfried Wrote:Ansus Wrote:For the base e sexp'(-1)=sexp'(0) so slog'(0)=slog'(1), can you verify this for your formula? Don't think so. I've not even completed the full reformulation of Andrew's formula... Gottfried Perhaps one should add that the superlogarithm for base $b>e^{1/e}$ by this formula is not a real function, i.e. has complex values for real arguments. While the superlogarithm with Andrew's original formula is a real function. « Next Oldest | Next Newest »

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