1
$\begingroup$

I deleted this at StackExchange as it seems advanced or not very relevant for there and I want to ask here.

Let z be a complex number, $Im(z)\neq 0$ or $Re(z)<4$, m be a natural number, let $f(z,m)$ be defined recursively;

$f(z, 0) = z$

$f(z, 1) = z^{z/2}$

$f(z,2)=\left(z^{z/2}\right)^{\frac{z^{z/2}}{2}}$

and so on,

$$f(z,n)=f(z,n-1)^{f(z,n-1)/2}$$

with

$$g(z,m)=\prod _{n=0}^m \left(1-f\left(\sqrt{z},n\right)\right)$$

then,

$$\sum _{m=0}^{\infty} f\left(\sqrt{z},m\right)g(z,m)=1-z-\sum _{m=0}^{\infty}(f\left(\sqrt{z},m+1\right)){}^{2}g(z,m)$$ with the first three terms of the sum on the left given by

$\left(1-\sqrt{z}\right) \sqrt{z}+\left(1-\sqrt{z}\right) z^{\frac{\sqrt{z}}{4}} \left(1-z^{\frac{\sqrt{z}}{4}}\right)+\left(1-\sqrt{z}\right) \left(z^{\frac{\sqrt{z}} {4}}\right)^{\frac{z^{\frac{\sqrt{z}}{4}}}{2}} \left(1-z^{\frac{\sqrt{z}}{4}}\right) \left(1-\left(z^{\frac{\sqrt{z}}{4}}\right)^{\frac{z^{\frac{\sqrt{z}}{4}}}{2}}\right) etc$

and the values at zero by taking the appropriate limit of each term.

What are the usual/standard ways to approach a problem of this kind? Is there a known method to establish it?

What is known about the convergence of such series?

Is there any immediate way to prove, or analytically establish the convergence of either identity?

$\endgroup$
9
  • $\begingroup$ what is the problem? you have only made a definition. is it to determine convergence? $\endgroup$ Jun 19, 2014 at 12:33
  • $\begingroup$ To determine convergence. And to prove this identity. $\endgroup$
    – Tisego
    Jun 19, 2014 at 12:53
  • $\begingroup$ From your post, it looks like you state the identity as a fact... $\endgroup$ Jun 19, 2014 at 12:59
  • $\begingroup$ I can prove it myself, but I want to do so analytically,and rigorously. So I ask help and advice. $\endgroup$
    – Tisego
    Jun 19, 2014 at 13:00
  • 1
    $\begingroup$ While experimenting a bit, I have found that the equations somewhat simplify, if one replaces the variable $z \rightarrow x^2/W(x)^2$ ($W(x)$ the Lambert-W-Function). Calling the new function $\phi_{n}(x):=f(n,x/W(x))$ I found (without proof): $\phi_{n}(x)=\exp\{\frac{x}{2^n}\prod_{k=1}^{n-1}\phi_{k}(x)\}$ might be true. Maybe this helps. $\endgroup$ Jul 21, 2014 at 11:10

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.