Show that these identities are true.

Question

$\dfrac{\text{d}}{\text{d}x}[x^{-n}J_n(x)]=-x^{-n}J_{n+1}(x)$

$\dfrac{\text{d}}{\text{d}x}[x^{n}J_n(x)]=x^{n}J_{n-1}(x)$

$xJ_n'(x)=nJ_n(x)-xJ_{n+1}(x)$

Where $J_n(x)=\sum\limits_{j=0}^\infty\dfrac{(-1)^j}{j!(n+j)!}\left(\dfrac{x}{2}\right)^{n+2j}$

I'm having some trouble with the first one in particular. Can someone show me the steps?

Could you give some context? – Tim Ratigan Dec 16 '13 at 01:47 — Tim Ratigan, Dec 16 '13 at 01:47
$J_n$ is the $n$-th Bessel function of the first kind. – Chuck Franklin Dec 16 '13 at 01:50 — Chuck Franklin, Dec 16 '13 at 01:50

score 0 · Accepted Answer · answered Dec 16 '13 at 01:57

Here's $1$:$$\begin{align} \frac{\text d}{\text dx}x^{-n}J_n(x)&=\frac{\text d}{\text dx}\sum_{j=0}^\infty \frac{(-1)^jx^{2j}}{j!(n+j)!2^{n+2j}}\\ &=\sum_{j=1}^\infty\frac{2j(-1)^jx^{2j-1}}{j!(n+j)!2^{n+2j}}\\ &=\sum_{j=1}^\infty\frac{(-1)^jx^{2j-1}}{(j-1)!(n+j)!2^{n+2j-1}}\\ &=\sum_{j=0}^\infty\frac{(-1)^{j+1}x^{2j+1}}{j!(n+j+1)!2^{n+2j+1}}\\ &=-x^{-n}\sum_{j=0}^\infty\frac{(-1)^{j}}{j!(n+1+j)!}\left(\frac x2\right)^{n+1+2j}\\ &=-x^{-n}J_{n+1}(x)\end{align}$$

Show that these identities are true.

1 Answers1