On the power tower $exp(x-exp(x-cdots))$
$begingroup$
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, $$exp(x-exp(x-cdots))=kimplies exp(x-exp(x-k))=k$$ not $exp(x-k)=k$.
But we face a problem, how do we write $k$ purely in terms of $x$?
exponential-function power-towers
asked Nov 10 '18 at 15:55
TheSimpliFireTheSimpliFire
12.6k62461
$endgroup$
add a comment |
$begingroup$
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, $$exp(x-exp(x-cdots))=kimplies exp(x-exp(x-k))=k$$ not $exp(x-k)=k$.
But we face a problem, how do we write $k$ purely in terms of $x$?
exponential-function power-towers
asked Nov 10 '18 at 15:55
TheSimpliFireTheSimpliFire
12.6k62461
$endgroup$
1
$begingroup$
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
add a comment |
$begingroup$
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, $$exp(x-exp(x-cdots))=kimplies exp(x-exp(x-k))=k$$ not $exp(x-k)=k$.
But we face a problem, how do we write $k$ purely in terms of $x$?
exponential-function power-towers
asked Nov 10 '18 at 15:55
TheSimpliFireTheSimpliFire
12.6k62461
$endgroup$
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, $$exp(x-exp(x-cdots))=kimplies exp(x-exp(x-k))=k$$ not $exp(x-k)=k$.
But we face a problem, how do we write $k$ purely in terms of $x$?
exponential-function power-towers
exponential-function power-towers
asked Nov 10 '18 at 15:55
TheSimpliFireTheSimpliFire
12.6k62461
asked Nov 10 '18 at 15:55
TheSimpliFireTheSimpliFire
12.6k62461
asked Nov 10 '18 at 15:55
TheSimpliFireTheSimpliFire
12.6k62461
asked Nov 10 '18 at 15:55
TheSimpliFireTheSimpliFire
12.6k62461
asked Nov 10 '18 at 15:55
TheSimpliFireTheSimpliFire
12.6k62461
12.6k62461
1
$begingroup$
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
add a comment |
1
$begingroup$
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
1
1
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@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
$begingroup$
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
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
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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 $f(x)= lim_{nto infty} f_n(x)$
(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).
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 $err_{501}(x)$
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.
answered Dec 21 '18 at 22:54
Gottfried HelmsGottfried Helms
23.5k24599
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add a comment |
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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.
answered Dec 21 '18 at 23:08
J.G.J.G.
28.4k22844
$endgroup$
$begingroup$
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
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– TheSimpliFire
Dec 22 '18 at 9:01
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
(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 $f(x)= lim_{nto infty} f_n(x)$
(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).
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 $err_{501}(x)$
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.
answered Dec 21 '18 at 22:54
Gottfried HelmsGottfried Helms
23.5k24599
$endgroup$
add a comment |
$begingroup$
(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 $f(x)= lim_{nto infty} f_n(x)$
(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).
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 $err_{501}(x)$
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.
answered Dec 21 '18 at 22:54
Gottfried HelmsGottfried Helms
23.5k24599
$endgroup$
add a comment |
$begingroup$
(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 $f(x)= lim_{nto infty} f_n(x)$
(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).
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 $err_{501}(x)$
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.
answered Dec 21 '18 at 22:54
Gottfried HelmsGottfried Helms
23.5k24599
$endgroup$
(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 $f(x)= lim_{nto infty} f_n(x)$
(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).
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 $err_{501}(x)$
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.
answered Dec 21 '18 at 22:54
Gottfried HelmsGottfried Helms
23.5k24599
edited Dec 24 '18 at 4:25
answered Dec 21 '18 at 22:54
Gottfried HelmsGottfried Helms
23.5k24599
answered Dec 21 '18 at 22:54
Gottfried HelmsGottfried Helms
23.5k24599
answered Dec 21 '18 at 22:54
Gottfried HelmsGottfried Helms
23.5k24599
23.5k24599
add a comment |
add a comment |
$begingroup$
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.
answered Dec 21 '18 at 23:08
J.G.J.G.
28.4k22844
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 21 '18 at 23:08
J.G.J.G.
28.4k22844
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 21 '18 at 23:08
J.G.J.G.
28.4k22844
$endgroup$
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.
answered Dec 21 '18 at 23:08
J.G.J.G.
28.4k22844
answered Dec 21 '18 at 23:08
J.G.J.G.
28.4k22844
answered Dec 21 '18 at 23:08
J.G.J.G.
28.4k22844
answered Dec 21 '18 at 23:08
J.G.J.G.
28.4k22844
28.4k22844
$begingroup$
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
add a comment |
$begingroup$
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
$begingroup$
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
$begingroup$
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
add a comment |
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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