Recurrence Formula for Bessel Function of the First Kind
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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:
- $\map {J_{n + 1} } x = -2 \map {J_n'} x + \map {J_{n - 1} } x$
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Proof
From Generating Function for Bessel Function of the First Kind of Order n of x we have:
- $\ds \map \exp {\dfrac x 2 \paren {t - \dfrac 1 t} } = \sum_{n \mathop = - \infty}^\infty \map {J_n} x t^n$
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 | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {J_{n - 1} } x - \map {J_{n + 1} } x\) | \(=\) | \(\ds 2 \map {J_n'} x\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {J_{n + 1} } x\) | \(=\) | \(\ds - 2 \map {J_n'} x + \map {J_{n - 1} } x\) |
$\blacksquare$
Sources
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Some Special Functions: $\text {II}$. Bessel functions: $2$