bessel function maximizer

Question

I try to find global maximum for $ \frac{J_2(x)}{x^2} $ I suspect it happens at x=0 ( plotting the graph) where the value of the function is $ \frac{1}{8} $ I know local maximizers are at zeros of $ J_3(x) $ since $ \frac{d}{dx}\frac{J_2(x)}{x^2}=-\frac{J_3(x)}{x^2} $. My problem is to evaluate $ \frac{J_2(x)}{x^2} $ at the zeros of $ J_3(x) $ or compare it to $ \frac{1}{8} $ . thanks.

Answer 1

From Abramowitz/Stegun 9.1.62 or http://dlmf.nist.gov/10.14#E4 we know that $$|J_\nu(z)| \le \frac{|\frac{1}{2}z|^{\nu}e^{|\Im z|}}{\Gamma(\nu+1)}$$ for $\nu \ge -\frac{1}{2}\cdot$ Assuming that your $x$ is real this implies

$$|J_2(x)| \le \frac{|\frac{1}{2}x|^2e^{0}}{\Gamma(2+1)}= \frac{x^2}{8},$$ from which your result follows with division by $x^2$ and the continuity of your function at $x=0$.