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The intention is to find the maximum of the power tower $\exp(x-\exp(x-\cdots))$. From here, we see that it is around $0.965$ or possibly even higher. The approximate value of its integral is also given for interest.

Since $\exp(x-\exp(x-\cdots))$ has this shape in the link for every two iterations, $$\exp(x-\exp(x-\cdots))=k\implies \exp(x-\exp(x-k))=k$$ not $\exp(x-k)=k$.

But we face a problem, how do we write $k$ purely in terms of $x$?

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    $\begingroup$ For $x=1$ you can see that $e^{x-e^{x-k}}=k$ has a solution for $k=1$. So the maximum should be 1 or higher. $\endgroup$ – garondal Nov 10 '18 at 16:28
  • $\begingroup$ It seems that the integral from $0$ to $1$ is approx $0.77203095$ ? (Based on my proposal for the powerseries) Desmos gives some smaller value, but the difference to my value si surely based on insufficient iteration in the representation of the function in Desmos (try higher iterations to see approximations to my value). $\endgroup$ – Gottfried Helms Dec 23 '18 at 21:40
  • $\begingroup$ @GottfriedHelms Thank you for your attempt. But as I have said to J.G., the iteration $f_{k+1}(x)=\exp(x-f_k(x))$ is not the same as $$f_{k+1}(x)=\exp(x-\exp(x-f_k(x)))$$ as the integral fluctuates for odd iterations. $\endgroup$ – TheSimpliFire Dec 24 '18 at 8:54
  • $\begingroup$ Well, doing the expansion $u_0=\exp(x)$, $t_0=\exp(x-u_0)$,$u_1 = \exp(x-t_0)$ and so on for $n$ iterations it shows, that for $n \to \infty$ not only $u_n - u_{n+1}$ and $t_n-t_{n+1}$ converge to zero, but also $u_n-t_n$ converge to zero. For $x=0.8$ I found improvement of one significant digit for $k \approx 12$ steps, so for this range $0<x<=1$ we have for $n \to \infty$ that in the limit indeed $u_n=t_n$ so we can identify the short version of the iteration $\lim_{n \to \infty} f_{n}=\exp(x-f_n)=\exp(x-\exp(x-f_n))$ . The purpose of 3. picture in my answer was to visualize just this. $\endgroup$ – Gottfried Helms Dec 24 '18 at 9:14
  • $\begingroup$ For more derivation of the problem of convergence $t_n \to u_n$ you might like to look at math.stackexchange.com/q/3050819/1714 espec. the answer of Henning Makholm $\endgroup$ – Gottfried Helms Dec 24 '18 at 9:43
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(0) $\qquad$ Let basically $f_0(x)=\exp(x)$ and then iterate $f_{k+1}(x)=\exp(x-f_k(x))$ a couple of times and assume $f(x)= \lim_{n\to \infty} f_n(x)$


(I) $\qquad$ First we assume all that in terms of a formal powerseries. Then this converges to something like $$ f_\infty(x) \approx +0.567143290410 \\ +0.361896256635 x^{1} \\ +0.0736778051764 x^{2} \\ -0.00134285965499 x^{3} \\ -0.00163606514791 x^{4} \\ +0.000232149655570 x^{5} \\ +0.0000474223203353 x^{6} \\ -0.0000189444233824 x^{7} \\ -0.0000000208785458195 x^{8} \\ +0.00000117699067908 x^{9} \\ -0.000000179633602646 x^{10} \\ -0.0000000510936764494 x^{11} \\ +0.0000000206028966316 x^{12} \\ +0.000000000306827812731 x^{13} \\ -0.00000000154503877033 x^{14} \\ + O(x^{15}) $$ using Pari/GP. Setting $x=1$ it approximates nicely the value $f(1)=1$. Moreover, for $x=0$ it gives immediately the value $\omega=0.5671432...=W(1)$

(II) $\qquad$ If we say $t:=f_\infty(x) $ and $t=\exp(x-t)$ then we can derive $t \exp(t) = \exp(x)$ and thus $ t=W(\exp(x)) = f_\infty(x) $ and this is $1$ for $x=1$ and $\omega$ for $x=0$ as before.

(III) $\qquad$ If we use the basic definition (but not as powerseries but as evaluated values) and take the mean $g_k = (f_k(x) + f_{k+1}(x))/2$ for some (high) iteration $k$ then $err_k(x)= g_k(x) -f_\infty(x) $ shows a small difference-curve increasing with $x \to 1$ but accordingly and in concurrence decreasing with $k \to \infty$ , so it seems also by this limiting-process that the definition of the limit using the Lambert-W-definition makes sense.
For (III) see the following pictures. The first picture shows $f_{101}(x),f_{102}(x),f_\infty(x),err_{101}(x)$ It makes visually that it makes sense to look at the mean of the alternating values $f_{101}(x)$ and $f_{102}$, and also, that the difference of the mean and the $f_\infty(x)$ is small but increasing somewhat when $x \to 1$ (the scale for the err-curve is at the rhs of the picture).
picture101

The second picture shows $f_{501}(x),f_{502}(x),f_\infty(x),err_{501}(x)$ and we see, that the two curves approximate the $f_\infty(x)$ curve much more which is displayed in the smaller $err_{501}(x)$
picture501

The third picture shows the rate of convergence improving by iterations. I use the value $x_0=0.8$ and document $t_n(x_0)=f_{2n}(x_0)$ and $u_n(x_0)=f_{2n+1}(x_0)$ for $1$ to $128$ iterations. We see that $t_n()$ and $u_n()$ converge well and the difference decreases by about one significant decimal digits by ca 12 iterations.
picture

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You've noted $k=e^{x-k}$ so $ke^k=e^x$. In terms of the Lambert $W$ function, $k=W(e^x)$. The main question is which branch choice we take as our definition of the tower.

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  • $\begingroup$ No, it's not $k=\exp(x-k)$, it's $k=\exp(x-\exp(x-k))$! As I have said, the integral only converges for every two iterations, that's why I am asking $\endgroup$ – TheSimpliFire Dec 22 '18 at 9:01

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