Super convergent cos(x)

Question

could you find where it come from? $$\cos (x)=\sum _{n=0}^{\infty } \frac{\left((-1)^n+1\right) e^{\frac{i \pi n}{2}} (n+1)^2 J_{n+1}(x)}{x}$$ it is seem as bessel series?

Answer 1

Hint: You can consider the Fourier series for $\cos(x \cos\theta)$ given by the Jacobi-Anger expansion, differentiate it twice with respect to $\theta$, then evaluate it in $\theta=\pi$. You'll get your identity.

Answer 2

Here's some if the simplifications Jack D'Aurizio mentioned in his comment (this is really a comment, but the math is too complicated for me to enter it as such):

$\begin{array}\\ \sum _{n=0}^{\infty } \frac{\left((-1)^n+1\right) e^{\frac{i \pi n}{2}} (n+1)^2 J_{n+1}(x)}{x} &=\sum _{n=0}^{\infty } \frac{2 e^{\frac{i \pi 2 n}{2}} (2n+1)^2 J_{2n+1}(x)}{x} \quad\text{ (terms 0 for odd n)}\\ &=\frac{2}{x}\sum _{n=0}^{\infty } e^{i \pi n} (2n+1)^2 J_{2n+1}(x) \quad (e^{i \pi n} = (-1)^n)\\ &=\frac{2}{x}\sum _{n=0}^{\infty } (-1)^{ n} (2n+1)^2 J_{2n+1}(x) \text{ }\\ \end{array} $