If the Bessel function of integer order n is defined as $J_{n} (x) = \sum_{k = 0}^{\infty} \frac{(- 1)^{k}}{k! (n + k)!} {(\frac{x}{2})}^{2 k + n}$ then $\frac{d}{d x} [x^{- n} J_{n} (x)]$ is

Question

Question

Answer 1

Explanation:

The recurrence relation is: $(n) J_{n} (x) - x J_{n}^{'} (x) = x J_{n - 1} (x)$
Rearranging and multiplying the relation by $x^{- n}$ , we get: $- x^{- n} J_{n}^{'} (x) = x^{- (n - 1)} J_{n - 1} (x) - x^{- n} (n) J_{n} (x)$ ,
Which simplifies to: $- x^{- n} J_{n}^{'} (x) = x^{- n} (J_{n - 1} (x) - n J_{n} (x))$
But another recurrence relation involves Bessel functions, which is $J_{n - 1} (x) + J_{n + 1} (x) = 2 n J_{n} (x) / x$ ,
From above we can replace $(J_{n - 1} (x) = x [J_{n + 1} (x) - 2 n J_{n} (x) / x])$ in our expression for $- x^{- n} J_{n}^{'} (x)$
This results in: $- x^{- n} J_{n}^{'} (x) = x^{- n} (x [J_{n + 1} (x) - 2 n J_{n} (x) / x] - n J_{n} (x))$
After simplifying, it becomes: $- x^{- n} J_{n}^{'} (x) = - x^{- n} J_{n + 1} (x)$
Solving for the derivative of interest, we get: $[\frac{d}{d x} [x^{- n} J_{n} (x)] = - x^{- n} J_{n + 1} (x)]$