Quick way to get the repelling fixed point from the attracting fixed point?
#9
(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.
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Messages In This Thread
RE: Quick way to get the repelling fixed point from the attracting fixed point? - by JmsNxn - 07/21/2022, 06:09 AM

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