Show that the real zeros of the Bessel function $J_0(x) \stackrel{\text{def}}{=} \sum_{n=0}^{\infty} \frac{(-1)^n}{(n!)^2} \left(\frac{x}{2}\right)^{2n}$ are simple, that is, if $J_0(\alpha) = 0$, then $J_0'(\alpha) \neq 0$.
Show that the real zeros of the Bessel function $J_0(x) \stackrel{\text{def}}{=} \sum_{n=0}^{\infty} \frac{(-1)^n}{(n!)^2} \left(\frac{x}{2}\right)^{2n}$ are simple, that is, if $J_0(\alpha) = 0$, then $J_0'(\alpha) \neq 0$.