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+--- Thread: Quick way to get the repelling fixed point from the attracting fixed point? (/showthread.php?tid=1567)
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Quick way to get the repelling fixed point from the attracting fixed point? - JmsNxn - 07/06/2022
So, this may seem like an odd question, but can anyone describe for me the function that takes the attracting fixed point to the repelling fixed point? I currently have a subpar manner of calculating this, and I'm wondering if there is a quicker way.
Now, to explain, let \(|\log(y)| < 1\), so for example \(y=2\). I want a function \(f\) such that \(f(2) = 4\), and similarly for the other fixed point pairs.
That is:
\[
f(y)^{1/f(y)} = y^{1/y}\\
\]
or more specifically: \(\log(f(y))/f(y) = \log(y)/y\) for the principal branch of the log.
While \(|\log(y) | < 1\) and \(|\log f(y) | > 1\)?
I feel there must be some literature on this function. So far, I've been running the protocol:
\[
f(y) = \lim_{n\to\infty} \log^{\circ n}_{y^{1/y}}(e)\\
\]
Which works pretty well, but I'm not even sure works in the general sense. Is \(e\) always in the attracting orbit of \(\log_{y^{1/y}}(z)\) of the repelling fixed point \(f(y)\)? I feel like there may be counter examples, but I'm too lazy, and my code seems to be working
This problems been annoying me. And it's either something gnawingly stupid that I missed, or it's a rather deep problem.
To reiterate the question. I'm looking for a function:
\[
\begin{align}
f : \{y \in \mathbb{C}\,|\,|\log(y)| < 1\} &\to \{ y \in \mathbb{C}\,|\, |\log(y)| > 1\}\\
f(y)^{1/f(y)} &= y^{1/y}\\
\end{align}
\]
I might ask this on MO, but god knows they're all a bunch of c**** there.
Regards, James
EDIT:
here's the MO question link https://mathoverflow.net/questions/426150/finding-the-repelling-fixed-point-of-an-exponential-knowing-only-its-attracting
I'm expecting a bunch of c***** to discredit my question by tomorrow. I hate MO.
RE: Quick way to get the repelling fixed point from the attracting fixed point? - Gottfried - 07/09/2022
Hmm, my mind is not well to follow today, sorry James.
I remember to have worked with the problem to find the fixpoints by an analysis different from iterating the function.
This was one of my earliest contributions here and I had not much experience in putting things in established mathematical terms and derivation-process. Anyway, perhaps there is an idea in it that you've overlooked so far. In short I looked at the conditions on \(a\) and \(b\) where \(t=a+bî \) and \(\alpha\) and \(\beta\) where \( \log(t)=u=\alpha+\beta î \) to write this as function on \( \beta \) alone and if I remember correctly had a better/faster access to the fixpoints than by the iteration, however didn't use it much. See here: tetration-function (stupid name, I simply had no better ideas...)
Gottfried
RE: Quick way to get the repelling fixed point from the attracting fixed point? - JmsNxn - 07/10/2022
Ooooooooooooo
This looks really promising. It sucks the paper is structured in the exact opposite direction, lol. I know how to go from repelling to attracting fast, that's not too hard, as iterating the exponential will always converge, but for the repelling case this isn't so clear. I'll have to experiment with this. This analysis is a little adjacent to what I was looking for, but I think there might be something I can unearth from here.
MO, suggested that I look at \(f(y)\) about \(y = e\) because there should be a branching point centered around that point. By this you can define a function holomorphic on \(|\log(y)| \neq 1\) with a fixed/branch point at \(e\) (that may be holomorphic there), such that:
\[
f(y) : \{|\log(y)| < 1\} \to \{|\log(y)| > 1\}\\
\]
Up to here everything is possible. Someone on MO suggests I try to develop a series about \(e\) from here. Which honestly sounds like a closed form using W-Lambert and one of its branches is really possible from here. (See how the inverse of \(w^{1/w}\) is expressible in W-lambert.) I'd just have to choose a different branch of the inverse. This sounds troubling, because I hate W-Lambert, lol. I'll see if I can go on a deepdive using W-lambert.
let:
\[
F(y^{1/y}) = y\,\,\text{for}\,|\log(y)| > 1\\
\]
Where \(F\) is some W-lambert atrocity. This would actually solve my problem, and I wouldn't need \(f\). If I have an attracting fixed point, take the self root and run an inverse W-lambert trick. YA, this should work.
Now to read up how to run W-Lambert. Skipped all those lessons, lmao.
EDIT:
I answered my own question on MO at https://mathoverflow.net/questions/426150/finding-the-repelling-fixed-point-of-an-exponential-knowing-only-its-attracting
This entirely solves the problem. But I don't know what branch of the Lambert function to use as of now.
Does anyone have a good formula for how the various branches of Lambert relate to the Lambert formula for the inverse of:
\[
g(y) = y^{1/y}\\
\]
RE: Quick way to get the repelling fixed point from the attracting fixed point? - bo198214 - 07/19/2022
I thought you just would calculate the base from the fixed point via \(b=z^{\frac{1}{z}}\) and then apply \(\text{LambertW}(-\log(b))/(-\log(b))\) with the right branch?
When I made the animated picture of the primary fixed points, I used these branches \(k\) of the LambertW:
For the attracting fixed point \(z\) in the upper half plane excluding the real line (left of \(e\)) use \(k=1\). Otherwise use \(k=-1\).
For example for \(z_1=2\) is \(b=\sqrt{2}\) then \(z_2 = \text{LambertW}_{-1}(-\log(b))/(-\log(b)) = 4\)
RE: Quick way to get the repelling fixed point from the attracting fixed point? - Gottfried - 07/19/2022
(07/10/2022, 02:24 AM)JmsNxn Wrote: Now to read up how to run W-Lambert. Skipped all those lessons, lmao.
(...)
This entirely solves the problem. But I don't know what branch of the Lambert function to use as of now.
Does anyone have a good formula for how the various branches of Lambert relate to the Lambert formula for the inverse of:
\[
g(y) = y^{1/y}\\
\]
Hmm, just to make sure: you know that LambertW (though without branch-index so far) is in Pari/GP? (I think in the 2.14-versions (alpha,experimental) they have it with branchindex) . And even better, the implementation in Pari/GP by our (former?) member mike3 - a real good one as it seems to me, allowing branch-indexing.
Gottfried
(Late corrections due to JmsNxn)
RE: Quick way to get the repelling fixed point from the attracting fixed point? - JmsNxn - 07/19/2022
(07/19/2022, 08:36 AM)Gottfried Wrote:(07/10/2022, 02:24 AM)JmsNxn Wrote: Now to read up how to run W-Lambert. Skipped all those lessons, lmao.
(...)
This entirely solves the problem. But I don't know what branch of the Lambert function to use as of now.
Does anyone have a good formula for how the various branches of Lambert relate to the Lambert formula for the inverse of:
\[
g(y) = y^{1/y}\\
\]
Hmm, just to make sure: you know that LambertW (though without branch-index so far) is in Pari/GP? (I think in the 2.14-versions (alpha,experimental) they have it with branchindex) . And even better, the implementation in Pari/GP by our (former?) member mike4 - a real good one as it seems to me, allowing branch-indexing.
Gottfried
Hey, Gottfried.
All the pari-gp literature says that lambertW only works for the \(0\)-th branch on the real positive line. Are you saying I can call \(W_1\), because if I can call that in the complex plane, I'd dance around. That will save me fkn hours in CPU time, if it's just a taylor expansion built into pari.
Also, it was mike3, not mike4, who was the member here. I still use his graphing program....
Please tell me I can call:
\[
\frac{W_1(-\log(b))}{-\log(b)} = y\\
\]
Such that:
\[
|\log(y)| > 1
\]
And \(b = y^{1/y}\).
If you can point me to some pari-gp code which does this, Gottfried. THANK YOU!
I'll fucking s*** your d***
PLEASE TELL ME THIS IS POSSIBLE!!!
RE: Quick way to get the repelling fixed point from the attracting fixed point? - Gottfried - 07/19/2022
(07/19/2022, 09:01 AM)JmsNxn Wrote: If you can point me to some pari-gp code which does this, Gottfried. THANK YOU!
PLEASE TELL ME THIS IS POSSIBLE!!!
See this Pari/GP-source of Mike3 in the attachment. I've slightly adapted to fit my environment of Pari/GP-functions, and hope I'd removed that references to my private environment properly.
Please tell me if something does not work.
- - - - - - - - - -
Additionally, this is what Pari/GP (alpha version 2.14) tells us about "lambertw":
Code:
lambertw(y,{branch=0}): solution of the implicit equation x*exp(x)=y. In the
p-adic case, gives a solution of x*exp(x)=y if x has positive valuation, of
x+log(x)=log(y) otherwise.using version
Code:
\v
GP/PARI CALCULATOR Version 2.14.0 (development 27026-bc08af1825)
amd64 running mingw (x86-64/GMP-6.1.2 kernel) 64-bit version
compiled: Oct 25 2021, gcc version 8.3-win32 20190406 (GCC)
threading engine: single
(readline v8.0 disabled, extended help enabled)- - - - - - - -
One more point
I see you're discussing \(b = y^{1/y} \) in context of multiple fixpoints.
Are you aware that for multiple fixpoints \(t_k\) with property \( e^{t_k} = t_k \) we must write \( e= t_k^{1/(t_k+k\cdot 2 \pi î)} \) ? (where the reference to the branch index must occur in the exponent in the rhs) See the protocol using Mike's LambertW: (update: Pari/GP V 2.14 woks identically, calling "lambertw()" instead)
Code:
bb = exp(1)
k=-1; tk=(exp(-LambertW(-log(bb),k))); [bb^tk,tk ; bb,tk^(1/(tk+k*2*Pi*I))]
%298 =
[0.318131505205 + 1.33723570143*I 0.318131505205 + 1.33723570143*I] \\ OK
[ 2.71828182846 0.764563417922 + 0.0623544878074*I] \\ ERR: at k=-1 the branchindex in exponent must be chose to be zero!!
k=-1; tk=(exp(-LambertW(-log(bb),k))); [bb^tk,tk ; bb,tk^(1/(tk+0*k*2*Pi*I))] \\OK: at k=-1 the branchindex in exponent must be chose to be zero!!
%300 =
[0.318131505205 + 1.33723570143*I 0.318131505205 + 1.33723570143*I] \\OK
[ 2.71828182846 2.71828182846 + 9.21798806441 E-810*I] \\OK: at k=-1 the branchindex in exponent must be chose to be zero!!
k=0; tk=(exp(-LambertW(-log(bb),k))); [bb^tk,tk ; bb,tk^(1/(tk+k*2*Pi*I))]
%302 =
[0.318131505205 - 1.33723570143*I 0.318131505205 - 1.33723570143*I] \\ OK
[ 2.71828182846 2.71828182846 - 9.21798806441 E-810*I] \\ OK
k=1; tk=(exp(-LambertW(-log(bb),k))); [bb^tk,tk ; bb,tk^(1/(tk+k*2*Pi*I))]
%304 =
[2.06227772960 - 7.58863117847*I 2.06227772960 - 7.58863117847*I] \\ OK
[ 2.71828182846 2.71828182846 + 0.E-809*I] \\ OK
k=2; tk=(exp(-LambertW(-log(bb),k))); [bb^tk,tk ; bb,tk^(1/(tk+k*2*Pi*I))]
%306 =
[2.65319197404 - 13.9492083345*I 2.65319197404 - 13.9492083345*I] \\ OK
[ 2.71828182846 2.71828182846 + 4.79335379349 E-808*I] \\ OK- - - - - - - - - - -
FInal remark: it comes out that the following Pari/GP-call is always correct:
Code:
k=-12 \\ init k with wanted branch-index
t_k= exp( - lambertw(-log(bb),k) ) \\ "lambertw()" with branching only available in Pari/GP v. 2.14 - otherwise use Mike's implementation
print ( [bb, k, t_k, t_k^(1/ ( t_k+ (k + (k<0))*2*Pi*I ) ) ])
[2.71828182846, -12, 4.25920785594 + 70.6256008021*I, 2.71828182846 + 1.36057503831 E-806*I]For other bases, for instance \( bb=12 \) we first compute \( \lambda = \log(bb) \) and then
Code:
[bb=12, lam = log(bb)]
{for(k=-3,3,
tt=exp(-LambertW(-lam,k));
print("--- ",k," ---");
printp( mytrunc( [bb^tt ; tt] ) ); \\ "mytrunc()" is userdefined truncation-procedure, to set spurious real or imag values to zero
printp( mytrunc( [bb ; tt^(1/(tt+(k+(k<0))*2*Pi*I/lam)) ]) );
)}Gottfried
------------------------------------------------
How to get Pari/GP v 2.14 (development version)
https://pari.math.u-bordeaux.fr/download.html
"Daily Snapshots (32-bit, 64-bit): basic GP binaries built daily from the master branch. Bleeding edge ! "
https://pari.math.u-bordeaux.fr/pub/pari/windows/snapshots/gp-gmp-git-latest.exe
https://pari.math.u-bordeaux.fr/pub/pari/windows/snapshots/gp64-gmp-git-latest.exe
RE: Quick way to get the repelling fixed point from the attracting fixed point? - JmsNxn - 07/20/2022
Hmm, I'm wondering if you have a rundown for me, because I'm confused by how you are referencing your indexes of LAmbertW. I absolutely deplore Lambert's W function, and it always gives me a headache, so I've successfully avoided it my entire mathematical career.
The function:
\[
y^{1/y} : \mathbb{R}_{y\ge 1} \to [1,e^{1/e}]\\
\]
There is a critical point at \(y=e\) and this critical point is of the second order (the second derivative is non-zero). So two branches spawn from here. There exists firstly the function:
\[
f(b) : (1,e^{1/e}) \to (1,e)\\
\]
And secondly the function:
\[
g(b) : (1,e^{1/e}) \to (e, \infty)\\
\]
The first function \(f\) is expressible as:
\[
f(b) = \frac{W_0(-\log(b))}{-\log(b)}
\]
For the principle branch. Since \(W_1\) is the only other real valued branch of Lambert W (correct me if I'm wrong), I was under the impression:
\[
g(b) = \frac{W_1(-\log(b))}{-\log(b)}\\
\]
Or something similar. Maybe I've made a grave error in my assumptions. I am primarily interested in the inverse of \(y^{1/y}\) which is the other branch than the typical infinite tetration branch with the Lambert W formula.
I may have gotten ahead of myself. To set things straight, I'll ask a question:
Would I just run:
Code:
-LambertW(-log(sqrt(2)),1)/log(sqrt(2))To spit out the value \(4\)? I was under the impression this would work, as the different branch of Lambert W (at the critical point \(1/e\)) would just choose the different real valued branch of the inverse of \(y^{1/y}\).
RE: Quick way to get the repelling fixed point from the attracting fixed point? - JmsNxn - 07/21/2022
(07/19/2022, 06:51 AM)bo198214 Wrote: I thought you just would calculate the base from the fixed point via \(b=z^{\frac{1}{z}}\) and then apply \(\text{LambertW}(-\log(b))/(-\log(b))\) with the right branch?
When I made the animated picture of the primary fixed points, I used these branches \(k\) of the LambertW:
For the attracting fixed point \(z\) in the upper half plane excluding the real line (left of \(e\)) use \(k=1\). Otherwise use \(k=-1\).
For example for \(z_1=2\) is \(b=\sqrt{2}\) then \(z_2 = \text{LambertW}_{-1}(-\log(b))/(-\log(b)) = 4\)
Oh my God I totally missed your comment BO!
YES THAT IS WHAT I WANT!
I want that in PARI-GP, and everything I've read says that pari doesn't support complex Lambert W, and additionally only has the principal branch.
Thank you though, I want \(W_{-1}\) not \(W_1\), I knew I was screwing something up.
You answered my question, but I'm still confused how to run this in Pari-GP. That's still confusing me.
But yes, that's what I want! Thank you, that's the mathematical problem explained.
How do I write
Code:
LambertW(sqrt(2) + O(y),-1) = 4 + O(y)\\Where we can gather taylor terms in the same manner any transcendental function works in pari. (I'm kind of a stickler for this--I don't want a numerical value, I want a polynomial value).
So I want, LambertW(sqrt(2) + x,-1) to be a polynomial expansion about the point \(\sqrt{2}\) in the variable x. If I can get an efficient way of doing that, that isn't my hacked together way now. Omg, I wouldn't know how to say thank you.
Tell me how to program LambertW(.,-1); ffs. Lambert has always been a handicap of mine.
RE: Quick way to get the repelling fixed point from the attracting fixed point? - Gottfried - 07/21/2022
(07/21/2022, 06:09 AM)JmsNxn Wrote: I want that in PARI-GP, and everything I've read says that pari doesn't support complex Lambert W, and additionally only has the principal branch.
Thank you though, I want \(W_{-1}\) not \(W_1\), I knew I was screwing something up.
You answered my question, but I'm still confused how to run this in Pari-GP. That's still confusing me.
How do I write
Code:LambertW(sqrt(2) + O(y),-1) = 4 + O(y)\\
Hi James -
two ways: either use Mike3's implementation (I'd attached the Pari/GP-code to my earlier answer) or download the latest Pari/GP version 2.14 (which firms as "development version" at the "bleeding edge" ;-) )from the Pari-group's website. With Mike3's implementation he didn't call it "LambertW()" but simply "lambw()" or so; I renamed it in the downloaded Pari/GP-source simply for my own convenience.
Gottfried