(10/03/2021, 09:17 PM)tommy1729 Wrote: Again , nice question !

OMG this forum has upgraded so hugelyyyyyyyy my final exams just ended i'm back and find me unable to catch up on

anyway

Happy new year, Tommy!

This question is exactly not the same as original ones that do map the fixed point at infinity to local one and then solvable,

you see, given that Gamma function has this asymptotic expansion at infinity with a_n coefficients:

$$\Gamma(z)=\sqrt{\frac{2\pi}{z}}\bigg(\frac{z}{e}\bigg)^z\sum_{n\ge0}{\frac{a_n}{z^n}}$$

And recall back on the technique we used in the case

$$\begin{align}f(z)=z+e^z\\g(z)=ln(z)\\g^{-1}(f(g(z)))=ze^z\\\text{The fixed point at }-\infty\text{ is mapped to }g^{-1}(-\infty)\end{align}$$

And we notice that since Gamma function has zeroes at directed infinity $$e^{i\theta}\infty$$ where $$\frac{\pi}{2}\le|\theta|\le\pi$$

And then the function f(z)=z+Gamma(z) must have fixed point at these infinities*

We may pick some function g to map that fixed point to local ones, 0 as preferred.

First attempt, let $$g(z)=-log(z)e^{i \theta}, g^{-1}(z)=e^{-e^{-i \theta}z}$$ and hence the infinity in the direction at e^(i \theta) will be mapped to 0.

However this leads to such asymptotic expansion at z=0:

$$\begin{align}\text{Let sgnL(z)=}(-1)^{2+\left\lfloor \frac{\arg \left(-e^{i \theta } \log (z)\right)+\pi }{2 \pi }\right\rfloor } 2^{\frac{1}{2}-\left\lfloor \frac{\arg \left(-e^{i \theta } \log (z)\right)+\pi }{2 \pi }\right\rfloor }\sqrt{\pi }\\\times\left(e^{i \theta } \log (z) (-1)^{1+\left\lfloor \frac{\arg \left(-e^{i \theta } \log (z)\right)+\pi }{2 \pi }\right\rfloor }\right)^{\frac{1}{2}-e^{i \theta } \log (z)} \left(-\csc \left(\pi e^{i \theta } \log (z)\right)\right)^{\left\lfloor \frac{\arg \left(-e^{i \theta } \log (z)\right)+\pi }{2 \pi }\right\rfloor}\\g^{-1}(f(g(z)))=ze^{\frac{\text{sgnL(z)}e^{i\theta}}{log(z)}+P_\theta\small(\frac{1}{log(z)}\small)}\end{align}$$

where P is another Taylor series

Even if its series can be calculated at a specific direction, we won't be able to apply multiplier method:

Assuming z is positive and theta is not \pi, then $$\begin{align}g^{-1}(f(g(z)))=ze^{-\sqrt{2\pi}e^{-log(z)P\small(\frac{1}{log(z)}\small)}}\\=z e^{-\sqrt{2 \pi } z^{-P\left(\frac{1}{\log (z)}\right)}}\end{align}$$

It doesn't have a formal series at 0 at all

Second attempt, let g(z)=1/z, this method will map all infinities* to 0:

$$g^{-1}(f(g(z)))=z\frac{1}{1+e^{-\frac{1+log(z)}{z}}\sqrt{z^3}(1+O(z))}$$

It has no computable expansion at z=0 either.

So we have to find a map g(z) that has these very properties:

g has to map the fixed point at infinity to a local point, say, 0

$$g^{-1}(e^{i\theta}\infty)=0$$

After the mapping, we can find a series at z=0 only containing z (or abracadabra we have a much more complicated issue than ever!) That is, we can find the second derivative, third derivative at 0 and so on. (And they can't be all zeroes)

$$F(z)=g^{-1}(f(g(z))),\forall{k\in\mathbb{N}_+},\exists{\mathcal{D}_z^k{F}(z)}|_{z=0}$$

Let's calculate the second derivative where p is some function of z

$$g(0)=e^{i\theta}\infty,g^{-1}(e^{i\theta}\infty)=0,f'(e^{i\theta}\infty)=1,\mathcal{D}_z^2{F}(z)|_{z=0}=\lim_{z\to 0,p\to e^{i\theta}\infty}{f''(p)g'(z)}$$

And the third derivative

$$\mathcal{D}_z^3{F}(z)|_{z=0}=\lim_{z\to 0,p\to e^{i\theta}\infty}{f'''(p)g'(z)^2}$$

Forth derivative

$$\mathcal{D}_z^4{F}(z)|_{z=0}=\lim_{z\to 0,p\to e^{i\theta}\infty}{f^{(4)}(p)g'(z)^3-2f''(p)g'''(z)+2g'(z)g''(z)f'''(p)+3f''(p)g''(z)^2g'(z)^{-1}-3g'(z)f''(p)^2g''(z)}$$

You can see that all derivatives of F would contain some information from f, but not about its first derivative=1, and since all derivatives of f at very infinities* are all 0 (Easily proved), it is almost impossible to find such function g.

Also it tests why the two maps above are no-use, the log map leads that all derivatives oscillates at z=0, thus has no specific derivatives, even if assumed z>0, whether the derivative diverges, or all derivatives are 0. The 1/z map makes all derivatives are 0.

The reason why this is so hard is that the fundamental term, or leading term in the Gamma(z) expansion is

$$\Gamma(z)\sim\Gamma_0(z)=\sqrt{\frac{2\pi}{z}}\bigg(\frac{z}{e}\bigg)^z$$

*maybe it's easier to deal with this function?

How will you solve iterations of

$$f^*(z)=z+\Gamma_0(z)=z+\sqrt{\frac{2\pi}{z}}\bigg(\frac{z}{e}\bigg)^z$$

without using fixed point 0 (since it is excluded in the z+gamma case)?

*and maybe equally hard to use z=0 looooooooooooool?

Again, happy new year!!!

Leo