Consider the coordinates $(x, y)$ of a point, let us introduce the following quantity:
\begin{equation*}
\forall p \in \mathbb Z, \quad \alpha_p = \sum_{n=-\infty}^{+\infty} a_n J_{n-p}(R)e^{i(n-p)\Phi},
\end{equation*}
where $a_n=(-i)^n=i^{-n}$, $R=\sqrt{x^2+y^2}$, $\tan(\Phi)=\frac{y}{x}$ and $J_n$ is the Bessel function of the first kind of order $n$.
It seems that whenever $R\ll 1$, we have:
\begin{equation*}
\forall p \in \mathbb Z, \quad \alpha_p \approx a_p e^{-ix}.
\end{equation*}
I verified numerically that this is true. However, I failed to demonstrate it mathematically. Here is my attempt:
\begin{align}
\alpha_p &= \sum_{n=-\infty}^{+\infty} a_n J_{n-p}(R)e^{i(n-p)\Phi}\\
&= \sum_{n=-\infty}^{p} a_n J_{n-p}(R)e^{i(n-p)\Phi} + \sum_{n=p}^{+\infty} a_n J_{n-p}(R)e^{i(n-p)\Phi} – a_p J_{0}(R)\\
&= \sum_{n=-\infty}^{p} a_n (-1)^{p-n} J_{p-n}(R)e^{-i(p-n)\Phi} + \sum_{n=p}^{+\infty} a_n J_{n-p}(R)e^{i(n-p)\Phi} – a_p J_{0}(R) &J_{-n}(z) = (-1)^nJ_n(z)\\
&= \sum_{q=+\infty}^{0} a_{p-q} (-1)^{q} J_{q}(R)e^{-iq\Phi} + \sum_{q=0}^{+\infty} a_{p+q} J_{q}(R)e^{iq\Phi} – a_p J_{0}(R) &q=p-n, \quad q=n-p\\
&= a_p\left[\sum_{q=0}^{+\infty} (-i)^{q} J_{q}(R)\left(e^{-iq\Phi} + e^{iq\Phi}\right) – J_{0}(R)\right] & a_{p-q}=a_pi^q, \quad a_{p+q}=a_p(-i)^q\\
\end{align}
From here, I would be tempted to use the asymptotic behaviour of Bessel function, which is:
\begin{equation*}
z\to 0, \quad J_n(z) \approx \frac{1}{n!}\left(\frac{z}{2}\right)^n
\end{equation*}
However, I know that asymptotic behaviour does not work great with sum, and numerically it does not work great either.
Best Answer
The proposed series can be converted into a generating function for the Bessel function by shifting the summation index ($n=p+k$): \begin{align} \alpha_p &= \sum_{n=-\infty}^{+\infty} (-i)^n J_{n-p}(R)e^{i(n-p)\Phi}\\ &= (-i)^p\sum_{k=-\infty}^{+\infty} (-i)^k J_k(R)e^{ik\Phi}\\ &=(-i)^p\sum_{k=-\infty}^{+\infty} J_k(R)e^{ik(\Phi-\pi/2)} \end{align} The generating function of the Bessel function reads \begin{equation} e^{\frac{1}{2}z(t-t^{-1})}=\sum_{k=-\infty}^{\infty}t^{k}J_{k}\left(z\right) \end{equation} It is valid for $z\in C$ and $t\in C\setminus \{0\}$. By taking $z=R$ and $t=\exp\left(i(\Phi-\pi/2)\right)$, it comes \begin{align} \alpha_p&=(-i)^pe^{iR\sin(\Phi-\pi/2)}\\ &=a_pe^{-iR\cos\Phi}\\ &=a_pe^{-ix} \end{align} as expected. It is however valid for all $R$.