2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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?).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.