Question

Use the formula obtained in Examplealong with part (a) of Problem$\mathbf{27}$to derive the recurrence relation$\mathbf{2}{\mathbf{vJ}}_{\mathbf{v}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{xJ}}_{\mathbf{v}\mathbf{+}\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}{\mathbf{xJ}}_{\mathbf{v}\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.

Short answer

The derived recurrence relation is $2{\mathrm{vJ}}_{\mathrm{v}}\left(\mathrm{x}\right)={\mathrm{xJ}}_{\mathrm{v}-1}\left(\mathrm{x}\right)+{\mathrm{xJ}}_{\mathrm{v}+1}\left(\mathrm{x}\right)$.

Step-by-step solution

Define differential recurrence relation.

Recurrence formulas that relate Bessel functions of different orders are important in theory and in applications.

$\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{x}}^{\mathrm{v}}{\mathrm{J}}_{\mathrm{v}}\left(\mathrm{x}\right)\right]={\mathrm{x}}^{\mathrm{v}}{\mathrm{J}}_{\mathrm{v}-1}\left(\mathrm{x}\right)$

Derive the function xJv'(x)=-vJv(x)+xJv-1(x)

(a)

Let the function be, ${\mathrm{xJ}}_{\mathrm{v}}^{\text{'}}={\mathrm{vJ}}_{\mathrm{v}}\left(\mathrm{x}\right)-{\mathrm{xJ}}_{\mathrm{v}+1}\left(\mathrm{x}\right)$… (1)

and${\mathrm{xJ}}_{\mathrm{v}}^{\text{'}}={\mathrm{xJ}}_{\mathrm{v}-1}\left(\mathrm{x}\right)-{\mathrm{vJ}}_{\mathrm{v}}\left(\mathrm{x}\right)$ …(2)

Substitute the equation (1) into (2) yields,

$\begin{array}{c}{\mathrm{vJ}}_{\mathrm{v}}\left(\mathrm{x}\right)-{\mathrm{xJ}}_{\mathrm{v}+1}\left(\mathrm{x}\right)={\mathrm{xJ}}_{\mathrm{v}-1}\left(\mathrm{x}\right)-{\mathrm{vJ}}_{\mathrm{v}}\left(\mathrm{x}\right)\\ 2{\mathrm{vJ}}_{\mathrm{v}}\left(\mathrm{x}\right)={\mathrm{xJ}}_{\mathrm{v}-1}\left(\mathrm{x}\right)+{\mathrm{xJ}}_{\mathrm{v}+1}\left(\mathrm{x}\right)\end{array}$