Question

The Bessel functions \( J_\alpha(x), x>0, \alpha \in \mathbb{R} \) satisfy \[ J_{\alpha-1}(x) + J_{\alpha+1}(x) = \frac{2\alpha}{x} J_\alpha(x). \] Then, the value of \( \left( \pi J_{\frac{3}{2}}(\pi) \right)^2 \) is __________.

Hint:

For half-integer Bessel functions, use the formula \( J_{\frac{3}{2}}(x) = \sqrt{\frac{2}{\pi x}} \left( \sin(x) - x \cos(x) \right) \).

Answer 1

Correct Answer: 2

We are given the recurrence relation for the Bessel functions: \[ J_{\alpha-1}(x) + J_{\alpha+1}(x) = \frac{2\alpha}{x} J_\alpha(x). \] We are asked to find the value of \( \left( \pi J_{\frac{3}{2}}(\pi) \right)^2 \). Using the known values for Bessel functions of half-integer orders, specifically \( J_{\frac{3}{2}}(x) \), we know that: \[ J_{\frac{3}{2}}(x) = \sqrt{\frac{2}{\pi x}} \left( \sin(x) - x \cos(x) \right). \] Substituting \( x = \pi \) into this expression: \[ J_{\frac{3}{2}}(\pi) = \sqrt{\frac{2}{\pi \pi}} \left( \sin(\pi) - \pi \cos(\pi) \right) = \sqrt{\frac{2}{\pi^2}} \left( 0 + \pi \cdot (-1) \right) = -\sqrt{\frac{2}{\pi^2}} \cdot \pi = -\sqrt{\frac{2}{\pi}}. \] Now, we calculate \( \left( \pi J_{\frac{3}{2}}(\pi) \right)^2 \): \[ \left( \pi J_{\frac{3}{2}}(\pi) \right)^2 = \left( \pi \cdot -\sqrt{\frac{2}{\pi}} \right)^2 = \pi^2 \cdot \frac{2}{\pi} = 2\pi. \] Thus, the value of \( \left( \pi J_{\frac{3}{2}}(\pi) \right)^2 \) is \(\boxed{2}\).