# Thread: Bessel function J1

1. ## Bessel function J1

2. You have the Bessel function of the first kind

$\displaystyle J_1(x) = \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{n!(n+1)!2^{2n+1}}$

The first derivative is

$\displaystyle J_1'(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!(n+1)!2^{2n+1}}\frac{d}{dx}x^{2n+1} = \sum_{n=0}^{\infty}\frac{(-1)^n(2n+1)}{n!(n+1)!2^{2n+1}}x^{2n}$

Analogously, the second derivative is

$\displaystyle J_1''(x) = \sum_{n=0}^{\infty}\frac{(-1)^n(2n+1)}{n!(n+1)!2^{2n+1}}\frac{d}{dx}x^{2n} = \sum_{n=0}^{\infty}\frac{(-1)^n(2n+1)(2n)}{n!(n+1)!2^{2n+1}}x^{2n-1}$

Plugging those in in the equation you will get

$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n(2n+1)(2n)}{n!(n+1)!2^{2n+1}}x^{2n+1} + \sum_{n=0}^{\infty}\frac{(-1)^n(2n+1)}{n!(n+1)!2^{2n+1}}x^{2n+1} +(x^2-1)\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{n!(n+1)!2^{2n+1}}$

All you need to do now is multiply out the x^2-1 on the third series, equate the summing ranges for all three series (you will have to take out the first 2 terms of the first two series) and then you should get some cancellation and eventually 0.