Recurrence Formula for Bessel Function of the First Kind

Theorem

Let $\map {J_n} x$ denote the Bessel function of the first kind of order $n$.

Then:

$\map {J_{n + 1} } x = \dfrac {2 n} x \map {J_n} x - \map {J_{n - 1} } x$

And:

$\map {J_{n + 1} } x = -2 \map {J_n'} x + \map {J_{n - 1} } x$

Proof

$\ds \map \exp {\dfrac x 2 \paren {t - \dfrac 1 t} } = \sum_{n \mathop = - \infty}^\infty \map {J_n} x t^n$

Differentiating both sides of the equation with respect to $t$:

 $\ds \dfrac x 2 \paren {1 + \dfrac 1 {t^2} } \map \exp {\dfrac x 2 \paren {t - \dfrac 1 t} }$ $=$ $\ds \sum_{m \mathop = - \infty}^\infty n \map {J_n} x t^{n - 1}$ $\ds \leadsto \ \$ $\ds \dfrac x 2 \paren {1 + \dfrac 1 {t^2} } \sum_{n \mathop = - \infty}^\infty \map {J_n} x t^n$ $=$ $\ds \sum_{n \mathop = - \infty}^\infty n \map {J_n} x t^{n - 1}$ Generating Function for Bessel Function of the First Kind of Order n of x $\ds \leadsto \ \$ $\ds \dfrac x 2 \paren {\sum_{n \mathop = - \infty}^\infty \map {J_n} x t^n + \sum_{n \mathop = - \infty}^\infty \map {J_n} x t^{n - 2} }$ $=$ $\ds \sum_{n \mathop = - \infty}^\infty n \map {J_n} x t^{n - 1}$ $\ds \leadsto \ \$ $\ds \dfrac x 2 \paren {\sum_{n \mathop = - \infty}^\infty \map {J_{n - 1} } x t^{n - 1} + \sum_{n \mathop = - \infty}^\infty \map {J_{n + 1} } x t^{n - 1} }$ $=$ $\ds \sum_{n \mathop = - \infty}^\infty n \map {J_n} x t^{n - 1}$ Translation of Index Variable of Summation $\ds \leadsto \ \$ $\ds \dfrac x 2 \paren {\map {J_{n - 1} } x + \map {J_{n + 1} } x}$ $=$ $\ds n \map {J_n} x$ by comparing coefficients $\ds \leadsto \ \$ $\ds \map {J_{n - 1} } x + \map {J_{n + 1} } x$ $=$ $\ds \dfrac {2n} x \map {J_n} x$ $\ds \leadsto \ \$ $\ds \map {J_{n + 1} } x$ $=$ $\ds \dfrac {2n} x \map {J_n} x - \map {J_{n - 1} } x$

This is the first recurrence formula.

$\Box$

We prove the second recurrence formula by differentiating both sides of the original equation with respect to $x$:

 $\ds \dfrac 1 2 \paren {t - \dfrac 1 t} \map \exp {\dfrac x 2 \paren {t - \dfrac 1 t} }$ $=$ $\ds \sum_{m \mathop = - \infty}^\infty \map {J_n'} x t^n$ $\ds \leadsto \ \$ $\ds \dfrac 1 2 \paren {t - \dfrac 1 t} \sum_{m \mathop = - \infty}^\infty \map {J_n} x t^n$ $=$ $\ds \sum_{m \mathop = - \infty}^\infty \map {J_n'} x t^n$ $\ds \leadsto \ \$ $\ds \dfrac 1 2 \paren {\sum_{m \mathop = - \infty}^\infty \map {J_n} x t^{n + 1} - \sum_{m \mathop = - \infty}^\infty \map {J_n} x t^{n - 1} }$ $=$ $\ds \sum_{m \mathop = - \infty}^\infty \map {J_n'} x t^n$ $\ds \leadsto \ \$ $\ds \dfrac 1 2 \paren {\sum_{m \mathop = - \infty}^\infty \map {J_{n - 1} } x t^n - \sum_{m \mathop = - \infty}^\infty \map {J_{n + 1} } x t^n}$ $=$ $\ds \sum_{m \mathop = - \infty}^\infty \map {J_n'} x t^n$ Translation of Index Variable of Summation $\ds \leadsto \ \$ $\ds \dfrac 1 2 \paren {\map {J_{n - 1} } x - \map {J_{n + 1} } x}$ $=$ $\ds \map {J_n'} x$ by comparing coefficients $\ds \leadsto \ \$ $\ds \map {J_{n - 1} } x - \map {J_{n + 1} } x$ $=$ $\ds 2 \map {J_n'} x$ $\ds \leadsto \ \$ $\ds \map {J_{n + 1} } x$ $=$ $\ds - 2 \map {J_n'} x + \map {J_{n - 1} } x$

$\blacksquare$