Power series around $w = 0$ of the implicit function $z(w) = w(1+z(w))^t$

Question

I have to determine the power series around $w = 0$ of the implicit function $z(w) = w(1+z(w))^t, \ t \in \mathbb N$ and calculate the radius of convergence.

I tried to compute $\frac{dz}{dw}$ because I wanted to "tackle" the problem by trying to determine the Taylor series. Here, I get $\frac{dz}{dw} = \frac{(1+z)^t}{1-wt(1+z)^{t-1}}$. For $w = 0$, this becomes $\frac{dz}{dw} = (1+z)^t$. But already calculating the second derivative is a "tedious" calculation.

EDIT: I asked this question yesterday and got a response. I should assume $z(w) = \sum_{n=0}^{p} a_n w^n$ and consider $z(w) - w(1+z(w))^t$ and then expand and cancel out the coefficients. If I expand $(1+z(w))^t$ with the binomial expression, I get for the first 4 terms $(1+z(w))^t = 1 + tz(w) + \frac{t(t-1)z^2}{2} + \frac{t(t-1)(t-2)z^3}{6} + ...$. But the first 4 coefficients $a_n$ should be, as per the hint from yesterday, $a_0 = 0$, $a_1 = 1$, $a_2 = t$, $a_3 = \frac{1}{2}3t(t-1)$. But I don't know how to get these coefficients.

Answer 1

We can derive a series expansion of $z(w)$ at $w=0$ by applying the Lagrange–Bürmann formula.

We show the following is valid \begin{align*} \color{blue}{z(w)}&\color{blue}{=w+\sum_{k=1}^{\infty}\frac{t}{k}\binom{(k+1)t-1}{k-1}w^{k+1}}\tag{1}\\ &\,\,\color{blue}{=w+tw^2+\frac{3}{2}t(t-1)w^3+\cdots} \end{align*}

Following the Wiki-page we set \begin{align*} \phi(z)=(1+z)^t\qquad\qquad H(z)=(1+z)^t\tag{2} \end{align*}

Using the coefficient of operator $[w^n]$ to denote the coefficient of $w^n$ of a series we obtain \begin{align*} \color{blue}{[w^k](1+z)^t}&=\frac{1}{k}[z^{k-1}]\left(H^{\prime}(z)\phi(z)^k\right)\\ &=\frac{1}{k}[z^{k-1}]t(1+z)^{t-1}(1+z)^{kt}\\ &=\frac{t}{k}[z^{k-1}](1+z)^{(k+1)t-1}\\ &=\,\,\color{blue}{\frac{t}{k}\binom{(k+1)t-1}{k-1}} \end{align*} and the claim (1) follows.

Note: An answer to a slightly more general problem together with an additional solution path is presented here by @MarkoRiedel.