The Maclaurin expansion of the Lambert function $y=y(x)$ defined as $y(x)\cdot e^{y(x)}=x, y(0)=0.$

Question

I'm trying to find the Maclaurin expansion of the Lambert function $y=y(x)$ defined as $$y(x)\cdot e^{y(x)}=x, y(0)=0.$$

I find it easier to think of the Lambert function as the solution of the equation $ye^y=x$. So $y=y(x)$.

Maybe it's more clear if we denote it $w(x)$. So the solution of $ye^y=x$ is $y=w(x)$. It gives us $y$ in terms of $x$.

We know that $ye^y=x$. Let's differentiate both sides to get $$y'e^y+yy'e^y=1\\e^yy'(1+y)=1\\y'=\dfrac{y}{x(1+y)}$$

How do I find the next derivatives?

A Maclaurin series is a Taylor series expansion of a function about $0$, so it will be something like this: $$y(0)+\dfrac{y'(0)}{1!}x+\dfrac{y''(0)}{2!}x^2+\dfrac{y'''(0)}{3!}x^3+\dots$$

Okay but isn't $y'(0)$ undefined as we plug $x=0$ which is in the denominator?

THIS DOES NOT ANSWER MY QUESTION: How to derive the Lambert W function series expansion?. Thank you!

Answer 1

From my old cookbook.

The $n^{\text{th}}$ derivative of $W_0(x)$ is given by $$\frac {d^n\,W_0(x)}{dx^n}=\frac{e^{-n W_0(x)}\,P_n\big(W_0(x)\big)}{ \big(1+W_0(x)\big)^{2n-1}}$$ where the very first polynomials are $$\left( \begin{array}{cc} n & P_n(w) \\ 1 & 1 \\ 2 & -(w+2) \\ 3 & 2 w^2+8 w+9 \\ 4 & -(6 w^3+36 w^2+79 w+64) \\ 5 & 24 w^4+192 w^3+622 w^2+974 w+625 \\ \end{array} \right)$$

These polynomials satisfy the recurrence relation $$P_{n+1}(w)=-\big(n (w + 3) - 1\big)P_{n}(w)+(1+w)P'_{n}(w)$$