series approximation by iteration

Question

Given the following transcendental generating function $H(x) = x\sum_ka_k[H(x)]^k$, I want to approximate $zH(x)$. I do this by setting the zeroth term to $e^{-z}x$.

How can I generate higher order terms? My attempt is to iteratively insert the zeroth solution as $$ zH(x) = xze^{-z} + \dots \tag{1} $$

$$ zH(x) = xze^{-z} + (xze^{-z})^2 + \frac{3}{2}(xze^{-z})^3 + \dots \tag{2} $$

However, I'm not sure how to continue this to the next term, or if a closed form can be found for the $n$th term.

Answer 1

Your series $(2)$ looks like:

$$\sum\limits_{k=1}^\infty\frac{k^{k-1}}{k!}(xze^{-z})^k$$

Please have a look at Lambert W-function, here: $W_0(x)$ . It's $~W_0(x)e^{W_0(x)}=x~$ .

Or in the way as you've written it in the first line:

$$W_0(x)=xe^{-W_0(x)} ~~ => ~~ a_k:=\frac{(-1)^k}{k!}$$

It follows:

$$zH(x)=W_0(-xze^{-z})$$

$~$

We have $~W_0(x) = x h(e^{-x})~$ where $~h(x)~$ is the infinite power tower.

$h(x)~$ exists for $~e^{-1/e}<x<e^{1/e}~$ and therefore $~W_0(x)~$ for $~-1/e<x<1/e~$ .

Iteration with the start $~x_1:=x_0~$:

$$x_{n+1}=x_n^{x_{n+1}}~~ , ~~~~ \lim\limits_{n\to\infty}x_n = h(x_0)$$

This is not the best iteration, but good if $~x_0~$ is near by $~1~$ .

You can also try to construct an iteration directly from $~W_0(x)e^{W_0(x)}=x~$, it's not complicate.