Derivative of x^n by Bessel Function of the First Kind of Order n of x
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Theorem
Let $\map {J_n} x$ denote the Bessel function of the first kind of order $n$.
Then:
- $\map {\dfrac \d {\d x} } {x^n \map {J_n} x} = x^n \map {J_{n - 1} } x$
Proof
\(\ds \map {\frac \d {\d x} } {x^n \map {J_n} x}\) | \(=\) | \(\ds n x^{n - 1} \map {J_n} x + x^n \map {J_n'} x\) | Product Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds n x^{n - 1} \map {J_n} x - x^n \paren {\frac {\map {J_{n + 1} } x - \map {J_{n - 1} } x} 2}\) | Recurrence Formula for Bessel Function of the First Kind | |||||||||||
\(\ds \) | \(=\) | \(\ds n x^{n - 1} \map {J_n} x - x^n \paren {\frac {\frac {2 n} x \map {J_n} x - 2 \map {J_{n - 1} } x} 2}\) | Recurrence Formula for Bessel Function of the First Kind | |||||||||||
\(\ds \) | \(=\) | \(\ds n x^{n - 1} \map {J_n} x - x^n \paren {\frac n x \map {J_n} x - \map {J_{n - 1} } x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n x^{n - 1} \map {J_n} x - n x^{n - 1} \map {J_n} x + x^n \map {J_{n - 1} } x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^n \map {J_{n - 1} } x\) |
$\blacksquare$
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: $3$