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I am trying to understand how one applies Lagrange Bürmann Inversion to solve an implicit equation in real variables(given that the equation satisfies the needed conditions). I have tried looking for examples of this, but all I have found is the wikipedia article for the topic and the examples there were too rushed(or requiring too much knowledge of an outside topic) for me to understand.

Could someone please walk me through an example of how to use this (beautiful) theorem so that I may use it for myself?

QTHalfTau
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1 Answers1

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An example of solution of a transcendental equation by means of Lagrange inversion can be the following. Consider the transcendental equation: $$(x-a)(x-b) = l e^ x $$

You can rewrite as: $$x = a+ \frac{l e^ x}{(x-b)} $$

Applying Lagrange inversion:

$$x = a+ \sum_{n=1}\frac{l^n}{n!}\left[\left(\frac{d}{dx}\right)^{n-1}\frac{e^{nx}}{(x-b)^n} \right]_{x=a}$$ and developing the derivative, you can find an explicit solution:

$$x=a+\sum_{n=1}^{n=\infty} \frac{1}{n n!}\left( \frac{nle^a}{a-b}\right)^n B_{n-1}\left( \frac{-2}{n(a-b)}\right)$$

where $B_n(x)$ are the Bessel polynomials are defined as:

$$B_n(x)=\sum_{k=0}^n\frac{(n+k)!}{(n-k)!k!}\left(\frac{x}{2}\right)^k$$ See: http://en.wikipedia.org/wiki/Bessel_polynomials

Another solution can be obtained swapping $a$ with $b$:

$$x=b+\sum_{n=1}^{n=\infty} \frac{1}{n n!}\left( \frac{nle^b}{b-a}\right)^n B_{n-1}\left( \frac{-2}{n(b-a)}\right)$$

A numerical example:

$$x^2-1=e^{x-1}$$

gives as solutions:

$$x=1+\sum_{n=1}^{n=\infty} \frac{1}{n n!}\left( \frac{n}{2}\right)^n B_{n-1}\left( \frac{-1}{n}\right) = 1.78947...$$

$$x=-1+\sum_{n=1}^{n=\infty} \frac{1}{n n!}\left( \frac{-ne^{-2}}{2}\right)^n B_{n-1}\left( \frac{1}{n}\right) = -1.0617135...$$

For details, see :

"Generalization of Lambert W-function, Bessel polynomials and transcendental equations" http://arxiv.org/abs/1501.00138

giorgiomugnaini
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