I noticed that the following expression holds true for $k \in \left[-1,1\right]$:

$$ \operatorname{J}_{0}\left(x\right) + 2\sum_{n = 1}^{\infty}{\rm i}^{n}\operatorname{J}_n\left(x\right)\cos\left(n \cos^{-1}\left(k\right)\right) = \operatorname{J}_{0}\left(kx\right) + 2\sum_{n = 1}^{\infty}{\rm i}^{n} \operatorname{J}_{n}\left(kx\right), $$ where $\operatorname{J}_{n}$ is the Bessel function of the first kind and ${\rm i} = \sqrt{-1}$ is the imaginary unit.

It is related to the Jacobi-Anger expansion. I stumbled upon this equality, it fascinates me, and I don't understand how it is equal. I have checked numerically for many values of $x$.

Appreciate any advice or explanation $!$.


1 Answer 1


This is not a complete answer, as I don't recognize the exact equality. However, I can point out a few things that you could look into.

Both sides are Neumann series of Bessel functions.

$\cos(n\cos^{-1}(k))$ is a Chebyshev polynomial evaluated at $k$.

It looks like one side uses a scaled version of the other. So the multiplication theorem may come into play: \begin{equation*} \lambda^{-\nu}J_\nu(\lambda z) = \sum_{n=0}^\infty \frac{1}{n!} \left(\frac{(1-\lambda)^2z}{2} \right)^n J_{\nu+n}(z) \end{equation*}

Using the the multiplication theorem on the RHS, you may recognize a series representation of the Chebyshev polynomials appearing.

Note: If you find this equality interesting, you should check out Watson's A Treatise on the Theory of Bessel Functions


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