# Lagrange inversion theorem application

Can someone give me an example of where the Lagrange inversion theorem is applied in such a way it inverts a formal series? For example, say I have

$$\sum_{i>-1} a_it^i = u.$$

Can someone show me the step by step process by which

$$\sum_{i>-1}b_iu^i = t$$

is obtained. I can seem to find any links which "dumb" it down for me, or deal with series inversion.

We consider the series \begin{align*} f(t)=\sum_{n=1}^{\infty}nt^n \end{align*} and look for a function $$g(u)$$ with \begin{align*} g(u)=t\qquad\text{ and }\qquad f(t)=f(g(u))=u \end{align*}

Note the index $$n\geq 1$$. In order to find a compositional inverse we have to check that $$f(0)=0$$ and $$f^\prime(0)\ne 0$$. This is the case and we can proceed.

The following is often helpful, when applying the Lagrange Inversion Formula. If there is a function $$\phi(t)$$ with \begin{align*} f(t)=\frac{t}{\phi(t)} \end{align*} then the coefficients of the series expansion of the compositional inverse $$g(u)=t$$ with $$f(g(u))=u$$ are given by \begin{align*} [u^n]g(u)=\frac{1}{n}[t^{n-1}]\phi(t)^n\tag{1}\\ \end{align*}

We find a representation of $$f(t)$$ from which we can derive $$\phi(t)$$. \begin{align*} f(t)&=\sum_{n=1}^\infty nt^n=t\sum_{n=0}^{\infty}(n+1)t^n=t\sum_{n=0}^{\infty}\binom{-2}{n}(-t)^n\\ &=\frac{t}{(1-t)^2} \end{align*} We conclude $$\phi(t)=(1-t)^2$$ and get according to (1) \begin{align*} [u^n]g(u)&=\frac{1}{n}[t^{n-1}]\phi(t)^n\\ &=\frac{1}{n}[t^{n-1}](1-t)^{2n}\\ &=\frac{(-1)^{n-1}}{n}\binom{2n}{n-1}\\ &=\frac{(-1)^{n-1}}{n+1}\binom{2n}{n}\\ &=(-1)^{n-1}C_n \end{align*} with $$C_n$$ the well known Catalan numbers. Since the generating function of $$C_n$$ is \begin{align*} \sum_{n=0}^{\infty}C_nu^n=\frac{1}{2u}\left(1-\sqrt{1-4u}\right)\tag{2} \end{align*} we obtain from (2) by respecting $$g(0)=0$$ \begin{align*} g(u)&=\sum_{n=1}^{\infty}(-1)^{n-1}C_nu^n\\ &=-\sum_{n=1}^{\infty}C_n(-u)^n\\ &=\frac{1}{2u}\left(1-\sqrt{1+4u}\right)+1 \end{align*}

We finally conclude the compositional inverse of \begin{align*} f(t)&=\sum_{n=1}^{\infty}nt^n=\frac{t}{(1-t)^2}\\ \end{align*} is the function \begin{align*} g(u)&=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n+1}\binom{2n}{n}u^n =\frac{1}{2u}\left(1-\sqrt{1+4u}\right)+1 \end{align*}

Hint: Another variation of the Lagrange inversion theorem can be used to derive from the series expansion of the Catalan numbers $$\sum_{n=0}^{\infty}C_nu^n$$ the generating function $$\frac{1}{2u}\left(1-\sqrt{1-4u}\right)$$ according to (2). This is shown here.

If $a_{0}=0=b_{0}$ you can use the series reversion. See http://mathworld.wolfram.com/SeriesReversion.html.

Well, a really typical example for using this Method is The Keppler's equation.

                    M = E - e Sin E


As you can see, It is a transcendental function, because it involves a trigonometric function, as well, you shall fin two methods to solve it.

The first one is by using approximatios for M, the most common one is M = E; and the second one is using the Lagrange inversion for Taylor Series.

A very simple and interesting example: the Lambert W function, implcitly defined by $$W(z)e^{W(z)} = z.$$ Using the version of the formula quoted by @MarkusScheuer, their power series around the zero is: $$W(z) = \sum_{n=1}^\infty\left(\frac{d^{n-1}}{ds^{n-1}}e^{-ns}\Bigg|_{s=0}\right)\frac{z^n}{n!} = \sum_{n=1}^\infty(-n)^{n-1}\frac{z^n}{n!},$$ with radius of convergence $$1/e$$ (¿why?).