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.

  • $\begingroup$ 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

Your series $(2)$ looks like:


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:



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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.