I am doing phased array antenna research and am trying to understand a final step in the Taylor circular aperture distribution solution. The solution Taylor presents has removable singularities at:
$$\frac{J_1(\pi x)}{1-(x/\mu)^2}$$
when $x=\mu$, where as stated in the questions $\mu$ is a solution of $J_1(\pi x) = 0$. Applying L'Hôpital to this, I get:
$$\frac{2}{\pi \mu}(\lim_{x \to \mu} \frac{J_1(\pi x)}{1-(x/\mu)^2})=\frac{-J_0(\pi \mu)+J_2(\pi \mu)}{2}$$
but Taylor has:
$$\frac{2}{\pi \mu}(\lim_{x \to \mu} \frac{J_1(\pi x)}{1-(x/\mu)^2})=-J_0(\pi \mu)$$
implying that $J_2(\pi \mu) = -J_0(\pi \mu)$.
How can I prove that this is true?
I have checked and both these solutions are equal, and I have checked that $-J_0(\pi \mu) = J_2(\pi \mu)$ using C++ and it is indeed true for all $\mu$ which I have listed below:
$\mu$ = 1.2196699, 2.2331306, 3.2383154, 4.2410628, 5.2427643, 6.2439216, 7.2447598, 8.2453948, 9.2458927, 10.2462933;
For reference, here is a plot of the two functions which should be equal for all $\mu$:
Recall the recurrence relation
$$J_{\nu-1}(z)+J_{\nu+1}(z)=\frac{2\nu}{z}J_\nu(z)$$
Let $\nu=1$ and $z=\pi\mu$, and use the fact that $J_1(\pi\mu)=0$.