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