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 Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 09/08/2007, 09:07 AM (This post was last modified: 09/10/2007, 04:28 PM by Gottfried.) Hmm, I'm already losing the structure of the current threads, which focus the same things from different views. May be it's a good place here. I was searching for a method, to find t where t^(1/t) = s, given s, where s is an arbitrary real number outside the range e^(-e)..e^(1/e). The h()-function derived from the python-implementation for Lambert-W in wikipedia seems to be valid only for the "inside bounds" case. In another thread I have presented a graph, which shows some of such real values s for complex t: but my "tracing" idea is not yet implemented. For a another quick view into it I asked someone to help me to plot a graph for complex t in the complex unit square indicating the contour where t^(1/t) is purely real-valued. The plot shows this (in principle, it is far too granular!) by colors for imag(s) where t = x + I*y. Where the color is black, purely real values occur for s, because black means imag(s)~0. The real value of s is not visible from that plot, but I showed in my other graph, that it ranges over a wide range up to sinh(10) and above for some complex t, taken from coordinates in that unit square. Here is the rough plot (only 129 x 64 points in the right half of the unit-square. The black regions should collapse into black lines only and their curved forms resemble roughly the bold red interpolation-lines in my other graph. The granularity of this plot is far too bad; unfortunately I don't own mathematica&co, and wouldn't feel good when asking the correspondent for more than one or the other favor. Gottfried [update] plot removed; improvement see second-next post Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 09/09/2007, 07:04 AM (This post was last modified: 09/09/2007, 05:45 PM by Gottfried.) Gottfried Wrote:The plot shows this (in principle, it is far too granular!) by colors for imag(s) where t = x + I*y. Where the color is black, purely real values occur for s, because black means imag(s)~0. The real value of s is not visible from that plot, but I showed in my other graph, that it rangesI've just got another image with a higher resolution and stronger rescaling of imag(s) $u = asinh(imag(s)) \hspace{24} y_{row,col} = u_{row,col} / \sqrt{max(abs(u_{,col})) }$ The black's are still areas in the plot, and I'd like to see them as lines. The closer the argument real(t) is at zero, the more solutions for real(s)>e^(1/e) are available, and also the absolute values real(s) increase strongly(which again cannot be seen in the graph) The tick-marks for x and y-axes are not correct in the plot; the range is the right half of the complex half-unit-square. I'd like to understand, whether the imaginary nulls of the function f(x)=x^(1/x) form continuous lines... Gottfried [update] plot removed, improved plot see next post Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 09/09/2007, 04:54 PM (This post was last modified: 09/09/2007, 05:34 PM by Gottfried.) A better view of the contours of the imaginary zeros of z=u^(1/u). This is a plot of imag(z) for u in the right half complex (half-unit)-square. The black lines are interpolations for the zeros of imag(z), where the sign of imag(z) changes (as far as the given computation accuracy allows). x-axis is the real part of u, y-axis is the imaginary part of u, color is a scaling of imag(z). The plot is a bit rough; the black lines shouldn't be dotted this way.     Gottfried     Gottfried Helms, Kassel andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 09/10/2007, 06:58 AM Nice graph! I think Scott Draves once said that fractals are only half-way between math and art, the other half is coloring algorithms Anyways, I wanted to give some pointers to info about the Lambert W-function. Mostly anything published by Corless/Jeffery et.al. or Galidakis is a great place to start. I also have done research on it, and one of the series expansions I found for it could only be found on OEIS, but I'm assuming it has to be published somewhere, so I'm including it below. I'm not sure if series expansions are really what you need, it sounds like you need Maple or Mathematica, but if it helps, here are the two series expansions I know for the solution to $b = h^{1/h}$: $ h = {}^{\infty}{b} = \sum^{\infty}_{k=0} \frac{\log(b)^k}{k!} (k+1)^{(k-1)}$ found in (A000272) and using the Taylor-Puiseux conversion detailed in Abel Functional Equation, the other one is relatively easy to derive: $ h = {}^{\infty}{b} = \sum^{\infty}_{k=0} \frac{(b - 1)^k}{k!} \sum^k_{j=0} \left[{k \atop j}\right] (j+1)^{(j-1)}$ found in (A033917). Andrew Robbins bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 09/10/2007, 01:33 PM It is well known that the (lower real) solution of $x=y^{1/y}$ is $h(x)=y=\frac{W(-\ln(x))}{-\ln(x)}$ where $W$ is the Lambert W function which is the solution of $ye^y=x$ where $W(x)=y$. Short derivation: $x^y=y$ $e^{\ln(x)y}=y=y\ln(x)/\ln(x)$ $\ln(x)e^{\ln(x)y}=y\ln(x)$ $\ln(x)=y\ln(x)e^{-\ln(x)y}$ $-\ln(x)=-y\ln(x)e^{-\ln(x)y}$ $W(-\ln(x))=-y\ln(x)$ $y=\frac{W(-\ln(x))}{-\ln(x)}$ Now the series development of the Lambert W function is well known to be: $W(z)=\sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!} z^n$. Then $h(z)=\frac{W(-\ln(z))}{-\ln(z)}=\sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!} (-\ln(z))^{n-1}=\sum_{n=0}^\infty \frac{(n+1)^n}{(n+1)!} \log(z)^n$ with $\frac{(n+1)^n}{(n+1)!}=\frac{(n+1)^{n-1}}{n!}$ your formula is confirmed However the radius of convergence of the $W$ series is $1/e$ and hence the radius of convergence of $h(z)$ is $e^{1/e}$ but Gottfried was just interested in a solution for $z>e^{1/e}$. The radius of convergence is limited because the LambertW function has a singularity at $-1/e$. It can however analytically continued past that singularity to values below $-1/e$ which are then complex. Such an analytic continuation is somewhat cumbersome. You start with the development point $a_0=0$. Then you compute the development at a point $a_1$ with $|a_1-(-1/e)|<1/e$ then you compute the development at a point $a_2$ with $|a_2-a_1|<|a_1-(-1/e)|$ and so on until you can reach an $a_n<-1/e$. See the attachment for a depiction. A development $\sum_{k=0}^\infty f_{n,k} (x-a_n)^k$ at point $a_n$ turns into the development $\sum_{k=0}^\infty f_{n+1,k} (x-a_{n+1})^k$ at point $a_{n+1}$ via $f_{n+1,k} = \sum_{j=k}^\infty \left(j\\k\right) f_{n,j} (a_{n+1}-a_n)^{j-k}$. For our task it should suffice to have 4 steps to reach a development at $a_4=-\frac{2}{e}$. $a_n=-\frac{1}{e} + \frac{1}{e}e^{i\frac{\pi}{4}n}, n=0..4$ Taking 3 steps (each 60 degrees) would result in $|a_{n+1}-a_n|=1/e$ which is not allowed. So the four step solution would result in $f_{4,k} = \sum_{j_4=k }^\infty\sum_{j_3=j_4}^\infty \sum_{j_2=j_3}^\infty\sum_{j_1=j_2}^\infty \left(j_4\\k\right)\left(j_3\\j_4\right)\left(j_2\\j_3\right) \left( j_1 \\ j_2 \right) f_{0,j_4}a_1^{j_1-j_2}(a_2-a_1)^{j_2-j_3} (a_3-a_2)^{j_3-j_4} (a_4-a_3)^{j_4-k}$ But nobody wants to use this *lol* Attached Files   analytic continuation.pdf (Size: 23.44 KB / Downloads: 432) Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 09/10/2007, 01:59 PM (This post was last modified: 09/10/2007, 02:21 PM by Gottfried.) I've just prepared a plot (resized here) which focuses the contours of the imag(b)=imag(x^(1/x))=0 - coordinates of x. I'm trying to hack my "tracer" now, if Andrew's reference shouldn't work for real values b>e^(1/e) Gottfried     A bigger version is at contour Some "fun"-versions are relief 2 (usable as tiling-background, for instance at the wall of my bathroom ;-) ) and for instance inverted some coefficients surface and as relief again reflief 3 Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 09/11/2007, 12:18 AM (This post was last modified: 09/15/2007, 12:24 PM by Gottfried.) Hmm. Manually I get a satisfying result today. First recall my matrix-formula: $\hspace{24} V(x)\sim * B_s = V(s^x)\sim$ with the hypothesis that the eigensystem-decomposition $\hspace{24} B_s = W * E * W^{-1}$ holds for $\hspace{24} 1/e^e Assuming a value $\hspace{24} t$ so that $\hspace{24} s = t^{1/t} \hspace{24} tl=\log(t) \hspace{24} -1 I use for the following my analytical result for the construction of B_s from the Eigensystem. The basic definition is simply $\hspace{24} B_s =\phantom{a}_dV(log(s)) * B$ where B = matrix(c^r/r!), c=colindex, r=rowindex, c,r>=0 and there is no complex value involved here, if only s is real and s>0. But for the eigensystem, the needed parameters t and log(t) are conventionally defined only for the above mentioned range of s. What if s is beyond the above bounds? We don't have then immediately a valid t available. The empirical eigensystems of truncated matrices Bs of such parameters have partly erratic structure and instability even when increasing the dimension, and powers of its diagonal can only be used in a very limited range for the exponents and unknown approximation quality. Backed by my previous plots I searched manually for the example $\hspace{24} s=7$ and found one possible t by binary search and approximation as solution $\hspace{24} \begin{eqnarray} t &&=&& 0.108080950260 + 0.243817409271*I \\ tl &&=&& -1.32163414553 + 1.15353940987*I \end{eqnarray}$ I can insert these values for the composition of W, E and W^-1 due to my analytical description for their entries. Let's denote the above eigensystem-decomposition and indicate the used parameters t and tl : $\hspace{24} \begin{eqnarray} B_t &&=&& W_t * E_t * W_t^{-1} \\ &&=&& (\phantom{a}_dV(1/t) * P^{-1}\sim * XI )* \phantom{a}_dV(tl) * (X * P\sim * \phantom{a}_dV(t)) \end{eqnarray}$ and the basic formula $\hspace{24} V(x)\sim * B_t = V(s^x)\sim$ is expanded $\hspace{24} V(x)\sim* (\phantom{a}_dV(1/t) * P^{-1}\sim * XI )* \phantom{a}_dV(tl) * (X * P\sim * \phantom{a}_dV(t)) = V(s^x)\sim$ It should be omputed by parts, using associativity: $ 1)\hspace{24} V(x')\sim = (V(x)\sim* \phantom{a}_dV(1/t) * P^{-1}\sim) = V(x/t-1)\sim \\ \\ 2) \hspace{24} M = ( XI * \phantom{a}_dV(tl) * X) * (P\sim * \phantom{a}_dV(t))$ I computed the direct sums of the final matrix-product and its expected result $\hspace{24} V(x')\sim * M = V(s^x)\sim = [1, s^1, s^2, s^3 ,....]$ giving the following result in the first four columns (dim=64, rounded to six digits) Code:[1.000000, 7.000000, 49.000000, 343.000000 + 8.0307505 E-11*I]which agrees beautifully with the expected result So the search for solutions t=h(s) where s out of the bounds [1/e^e .. e^(1/e)], as indicated in the previous graphs, seems to be useful. Next step would either be, to make some more numerical results available, or to prove the appropriateness of defining B_s via a complex eigensystem, using the real values of complex h(x)-parameters. The graph gives an impression, where the imaginary zeros of h(x) are located; it does not show, that the real values on the traces seem continuous and to cover all values (possibly below a bound) on each of the "circles". The graph shows some repetitions; I assume, there are infinitely many solutions t for each real s and they are correct values for the eigensystem. [update] The beautiful result is not so beautiful, if we change some of the parameters; then the problem of branches of logarithm occurs. Using the formula V(x)~ * Bs = V(s^x) ~ where Bs was created by the eigensystem-approach, then the simplest example is, using s=7, and the above values for t and tl, V(1)~ * Bs = V(s)~ which worked fine. However, trying V(0.5)~ * Bs = Y~ gave the negative root -sqrt(s) in Y[1], and generally [update] V(1/k)~ * Bs = V(y)~ where y is s^^(1/k)*exp(2*Pi*I/k), so this problem may be cured by rotating the result appropriately. [/update] Trying any integer or fractional iterations m V(1)~ * Bs^m = Y~ then relating y=Y[1] to the expected result was intractable at a first glance. So there are some more considerations needed... Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 09/19/2007, 05:17 PM (This post was last modified: 09/19/2007, 05:29 PM by Gottfried.) I've sort of a solution for the problem of finding complex t, where t^(1/t) =b and b is real, b>eta, by 2-dimensional newton-like approximation. A plot for t = h(b), b=2..8192 is appended. The line seems to be that one in my previous plot, which leaves the displayed area at the top nearest the imaginary axis. Gottfried     Gottfried Helms, Kassel GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 09/21/2007, 02:01 PM Dear Gottfried, Andrew and Henryk, I was impressed by the last plot posted by you (Gottfried) and I am pleased to tell you that I agree completely with your conclusions. Please find attached some notes concerning that matter. Actually, the same result can be reached from another end, by using the two real branches of the product-logarithm (Lambert) function. Therefore, I presume that the plot is correct. Nevertheless, by using the same method, it seems that for b -> oo the complex infinite towers don't vanish, as it could be felt from a first look at your graph. Is it correct? Can that be proved? Best wishes. Gianfranco Attached Files   Complex towers.pdf (Size: 71.97 KB / Downloads: 538) Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 09/21/2007, 04:57 PM GFR Wrote:Dear Gottfried, Andrew and Henryk, I was impressed by the last plot posted by you (Gottfried) and I am pleased to tell you that I agree completely with your conclusions. Please find attached some notes concerning that matter. Actually, the same result can be reached from another end, by using the two real branches of the product-logarithm (Lambert) function. Therefore, I presume that the plot is correct. Nevertheless, by using the same method, it seems that for b -> oo the complex infinite towers don't vanish, as it could be felt from a first look at your graph. Is it correct? Can that be proved? Best wishes. GianfrancoDear Gianfranco - thanks for that comment! I just read your pdf. In fact I found the values being symmetric w.r. to the real line. The first image adressing this was [attachment=67] in this thread, where the black lines are the different branches. Each of these branches seem to contain *all* real valued b's, so it shows for each b a multitude of solutions (possibly infinitely many). The complex conjugacy can possibly better be seen in the other graph which focuses the lines better. The contour, which I used in my last post, is that one, which escapes in the above grey contourplot at the top of the image; it curves then to the right. It would be interesting to see its contour down to b->1.4446.... (eta). I don't think it escapes to infinity, but let's see... It is also nice to see your graphs of the productlog; unfortunately I don't have the possibility to obtain them using Pari/Gp, so I had to hack my own procedure. Tonight I'l read your post a second time, I just came back home from a walk. Kind regards - Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »

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