# series approximation by iteration

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.

• Are $a_k$ known? Is $z$ a constant? If no, how is $z$ related to $x$? Oct 25 '19 at 12:31
• $a_k$ is to be determined and $z$ is a constant. Oct 25 '19 at 12:39
• How did you get from equation $(1)$ to equation $(2)$? Oct 25 '19 at 12:55
• Let $f(t)$ be the power series $f(t)=\sum_{t=1}^\infty a_tx^t$. Then what you're looking for is the solution to the equation $H(x)=xf(H(x)).$ I'm going to write $H(x)=xG(x)$; then this is equivalent to $xG(x)=xf(xG(x))$. Dividing out on both sides we get $G(x)=f(G(x))$. But this equation isn't formally well-determined; in particular, setting $x=0$ we have $G(0)=f(G(0))$, so that $G(0)$ is a fixed point of $f()$. But without further information, there's no way of determining which fixed point of $f$ we have. (All these manipulations are formal, BTW; dividing by '0' shouldn't matter) Oct 28 '19 at 14:53
• @user90369 Hi there, yes definitely, I have been on holiday and haven't had any attention on this question when I posted a bounty :( Nov 11 '19 at 21:12

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