Lagrange Bürmann Inversion Series Example

Solution 1:

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