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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.

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  • $\begingroup$ Are $a_k$ known? Is $z$ a constant? If no, how is $z$ related to $x$? $\endgroup$
    – Szeto
    Oct 25 '19 at 12:31
  • $\begingroup$ $a_k$ is to be determined and $z$ is a constant. $\endgroup$ Oct 25 '19 at 12:39
  • $\begingroup$ How did you get from equation $(1)$ to equation $(2)$? $\endgroup$
    – Somos
    Oct 25 '19 at 12:55
  • $\begingroup$ 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) $\endgroup$ Oct 28 '19 at 14:53
  • $\begingroup$ @user90369 Hi there, yes definitely, I have been on holiday and haven't had any attention on this question when I posted a bounty :( $\endgroup$ Nov 11 '19 at 21:12
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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.

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