Lagrange inversion theorem applied on $f(x)=x-x^p$

Question

I was looking at the Wikipedia page of the Lagrange inversion theorem and stumbled upon the example. It says that to find the root of the polynomial $x^p-x+z=0$ we can find the inverse of the function $f(x)=x-x^p$ using the theorem. So we have $f(x)=z$, therefore the inverse is $x=g(z)$ where $$ g(z)=x_0+\sum_{n=1}^\infty \frac{g_n}{n!}(z-f(x_0))^n \qquad g_n=\lim_{w\to x_0}\frac{\partial^{n-1}}{\partial w^{n-1}}\left[\left(\frac{w-x_0}{f(w)-f(x_0)}\right)^n\right] $$ The result should be, according to Wikipedia: $$ g(z)=\sum_{n=1}^\infty\binom{pk}{k}\frac{z^{(p-1)k+1}}{(p-1)k+1} $$ So, I decided to try it myself. First I put $x_0=0$, because $f'(0)=1$, and calculated $g_n$ $$ g_n=\lim_{w\to0}\frac{\partial^{n-1}}{\partial w^{n-1}}\left[(1-w^{p-1})^{-n}\right] $$ Using the negative binomial theorem we obtain $$ g_n=\lim_{w\to0}\frac{\partial^{n-1}}{\partial w^{n-1}}\left[\sum_{k=0}^\infty(-1)^k\binom{n+k-1}{k}w^{-(n+k)(p-1)}\right]= \qquad |w|<1\\ =\lim_{w\to0}\sum_{k=0}^\infty(-1)^k\binom{n+k-1}{k}(-(n+k)(p-1))...(-(n+k)(p-1)-n-2)w^{-(n+k)(p-1)-n-1} $$ Now, this implies that $g_n=0$ $\forall n\geq2$, which isn't right. What did I do wrong? Thanks.

Answer 1

The binomial series expansion needs to be revised a little. We consider the algebraic equation \begin{align*} x^p-x+z=0\qquad\qquad\mathrm{with}\quad\qquad f(x)=x-x^p \end{align*} and calculate $g_n$ in \begin{align*} g(z)=\sum_{n=1}^{\infty}g_n\frac{z^n}{n!}\tag{1} \end{align*} according to the Lagrange inversion formula. We obtain \begin{align*} {\color{blue}{g_n}}&\color{blue}{=\lim_{w\to 0}\frac{d^{n-1}}{dw^{n-1}}\left[\left(\frac{w}{w-w^p}\right)^n\right]}\\ &=\lim_{w\to 0}\frac{d^{n-1}}{dw^{n-1}}\left[\left(1-w^{p-1}\right)^{-n}\right]\\ &=\lim_{w\to 0}\frac{d^{n-1}}{dw^{n-1}}\sum_{k=0}^{\infty}\binom{-n}{k}\left(-w^{p-1}\right)^k\\ &=\lim_{w\to 0}\frac{d^{n-1}}{dw^{n-1}}\sum_{k=0}^{\infty}\binom{n+k-1}{k}w^{(p-1)k}\\ \end{align*} When we differentiate $n-1$ times all terms with $(p-1)k<n-1$ cancel out and we get \begin{align*} g_n=\lim_{w\to 0}\sum_{k=\left\lfloor\frac{n-1}{p-1}\right\rfloor}^{\infty} \binom{n+k-1}{k}\left((p-1)k\right)^{\underline{n-1}}w^{(p-1)k-(n-1)}\tag{2} \end{align*} where we use the falling factorial notation $q^{\underline{n}}=q(q-1)\cdots (q-n+1)$. Thanks to the limit all non-constant terms cancel out and for the constant term we need \begin{align*} (p-1)k-(n-1)=0\qquad \mathrm{which\ gives}\qquad k=\frac{n-1}{p-1}\tag{3} \end{align*} This implies that $p-1$ has to be a divisor of $n-1$. We obtain from (2) and (3) \begin{align*} \color{blue}{g_n}&=\binom{n+\frac{n-1}{p-1}-1}{\frac{n-1}{p-1}}(n-1)^{\underline{n-1}}\\ &\color{blue}{=\binom{\frac{p}{p-1}(n-1)}{n-1}(n-1)!} \end{align*} where we simplified the upper index of the binomial coefficient and also use $\binom{p}{q}=\binom{p}{p-q}$.

We therefore can write (1) as \begin{align*} g(z)&=\sum_{n=1}^{\infty}g_n\frac{z^n}{n!}\\ &=\sum_{{n=1}\atop{p-1|n-1}}^{\infty}\binom{\frac{p}{p-1}(n-1)}{n-1}(n-1)!\frac{z^n}{n!}\\ &=\sum_{{n=1}\atop{p-1|n-1}}^{\infty}\binom{\frac{p}{p-1}(n-1)}{n-1}\frac{z^n}{n}\\ \end{align*} Noting that $p-1|n-1$ we can change the index $n$ by letting \begin{align*} n-1&=k(p-1)\qquad\qquad k=1,2,\ldots\\ \end{align*} and we obtain \begin{align*} \color{blue}{g(z)}&=\sum_{k=1}^{\infty}\binom{\frac{p}{p-1}k(p-1)}{k(p-1)}\frac{z^{(p-1)k+1}}{k(p-1)+1}\\ &\color{blue}{=\sum_{k=1}^{\infty}\binom{pk}{k}\frac{z^{(p-1)k+1}}{(p-1)k+1}} \end{align*} in accordance with the Wikipedia example. In the final step we use again $\binom{p}{q}=\binom{p}{p-q}$.