【转载请注明出处】http://www.cnblogs.com/mashiqi
2017/06/16
适合于自己的关于Jacobi-Anger expansion的推导方法，这里记下来，方便以后查阅。
现记住下面四个关系式：
\begin{align*}
& (1)~ |x-y|=|x| -\hat{x} \cdot y + \mathcal{O}\left(\frac{1}{|x|}\right), ~|x| \to +\infty. \\
& (2)~ \sum_{m=-n}^{n} Y_n^m(\hat{x})\overline{Y_n^m(\hat{y})} = \frac{2n+1}{4\pi} P_n(\cos\theta). \\
& (3)~ \Phi (x,y) \triangleq \frac{e^{ik|x-y|}}{4\pi|x-y|} = ik \sum_{n=-\infty}^{\infty}\sum_{m=-n}^{n} h_n^{(1)}(k|x|)Y_n^m(\hat{x}) j_n(k|y|)\overline{Y_n^m(\hat{y})}, \forall~ |x| > |y|. \\
&(4)~ h_n^{(1)}(t) = \frac{1}{i^{n+1}t} e^{it} \left\{1 + \mathcal{O}\left(\frac{1}{t}\right)\right\}, ~t \to +\infty.
\end{align*}
于是当$|x|$充分大时，我们可以得到
\begin{align*}
\frac{e^{ik|x-y|}}{4\pi|x-y|} & = \frac{e^{ik|x|}}{4\pi|x|} \left\{ e^{-ik\hat{x} \cdot y} + \mathcal{O}\left(\frac{1}{|x|}\right) \right\} \\
& = ik \sum_{n=-\infty}^{\infty}\sum_{m=-n}^{n} h_n^{(1)}(k|x|)Y_n^m(\hat{x}) j_n(k|y|)\overline{Y_n^m(\hat{y})} \\
& = ik \sum_{n=-\infty}^{\infty} \left\{ j_n(k|y|)h_n^{(1)}(k|x|) \left[ \sum_{m=-n}^{n} Y_n^m(\hat{x}) \overline{Y_n^m(\hat{y})} \right] \right\} \\
& = ik \sum_{n=-\infty}^{\infty} \left\{ j_n(k|y|)h_n^{(1)}(k|x|) \frac{2n+1}{4\pi} P_n(\cos\theta) \right\} \\
& = ik \sum_{n=-\infty}^{\infty} \frac{2n+1}{4\pi} j_n(k|y|) P_n(\cos\theta) h_n^{(1)}(k|x|) \\
& = ik \sum_{n=-\infty}^{\infty} \frac{2n+1}{4\pi} j_n(k|y|) P_n(\cos\theta) \frac{e^{ik|x|}}{i^{n+1}k|x|} \left\{1 + \mathcal{O}\left(\frac{1}{|x|}\right)\right\} \\
& = \frac{e^{ik|x|}}{4\pi |x|} \sum_{n=-\infty}^{\infty} \frac{2n+1}{i^n} j_n(k|y|) P_n(\cos\theta) \left\{1 + \mathcal{O}\left(\frac{1}{|x|}\right)\right\} \\
& = \frac{e^{ik|x|}}{4\pi |x|} \left\{ \sum_{n=-\infty}^{\infty} \frac{2n+1}{i^n} j_n(k|y|) P_n(\cos\theta) + \mathcal{O}\left(\frac{1}{|x|}\right)\right\}.
\end{align*}
于是$$e^{-ik\hat{x} \cdot y} = \sum_{n=-\infty}^{\infty} \frac{2n+1}{i^n} j_n(k|y|) P_n(\cos\theta).$$将$\hat{x}$换做$-d$，$y$换做$x$，可得：
\begin{align*}
e^{ikd \cdot x} & = \sum_{n=-\infty}^{\infty} \frac{2n+1}{i^n} j_n(k|x|) P_n(\cos(\pi-\theta)) \\
& = \sum_{n=-\infty}^{\infty} \frac{2n+1}{i^n} j_n(k|x|) (-1)^n P_n(\cos\theta) \\
& = \sum_{n=-\infty}^{\infty} i^n(2n+1) j_n(k|x|) P_n(\cos\theta).
\end{align*}
原文：http://www.cnblogs.com/mashiqi/p/7029226.html