we know that \(e^{t}=1+t+\frac{t^{2}}{2!}+\frac{t^{3}}{3!} +\cdots\)
\begin{equation*}
\therefore e^{\frac{xz}{2}}=1+\frac{xz}{2}+\frac{1}{2!}\left(\frac{xz}{2}\right)^{2}+\frac{1}{3!}\left(\frac{xz}{2}\right)^{3}+\cdots
\end{equation*}
\begin{equation}
= \sum\limits_{r=0}^{\infty}\frac{x^{r}z^{r}}{2^{r}r!}\tag{4.7.10}
\end{equation}
and
\begin{equation*}
e^{-\frac{x}{2z}}=1-\left(\frac{x}{2z}\right)+\frac{1}{2!}\left(\frac{x}{2z}\right)^{2}-\frac{1}{3!}\left(\frac{x}{2z}\right)^{3}+\cdots
\end{equation*}
\begin{equation}
= \sum\limits_{s=0}^{\infty}(-1)^{s}\frac{x^{s}}{2^{s}z^{s}s!}\tag{4.7.11}
\end{equation}
\begin{equation}
e^{\frac{x}{2}(z-\frac{1}{z})}=\sum\limits_{r=0}^{\infty}\frac{x^{r}z^{r}}{2^{r}r!}\sum\limits_{s=0}^{\infty}(-1)^{s}\frac{x^{s}}{2^{s}z^{s}s!}\tag{4.7.12}
\end{equation}
Replacing \(r\) by \(n+s\text{,}\) we get the coefficient of \(z^{n}\) as
\begin{equation*}
e^{\frac{x}{2}(z-\frac{1}{z})} = \sum\limits_{s=-n}^{-1}\{\}z^{n+s}+\sum\limits_{s=0}^{\infty}\frac{x^{n+s}}{2^{n+s}(n+s)!}z^{n+s}\frac{(-1)^{s}x^{s}}{2^{s}z^{s}s!}
\end{equation*}
\begin{equation*}
=\{\}z^{0}+\cdots\{\}z^{n-1}+\sum\limits_{s=0}^{\infty}\frac{x^{n+s}}{2^{n+s}(n+s)!}z^{n+s}\frac{(-1)^{s}x^{s}}{2^{s}z^{s}s!}
\end{equation*}
\begin{equation*}
=\sum\limits_{s=0}^{\infty}\frac{(-1)^{s}x^{n+2s}}{s!(n+s)!2^{n+2s}}z^{n}
\end{equation*}
\begin{equation}
=\sum\limits_{s=0}^{\infty}\frac{(-1)^{s}}{s!(n+s)!}\left(\frac{x}{2}\right)^{n+2s}z^{n}=J_{n}(x)z^{n}\tag{4.7.13}
\end{equation}
The coefficient of \(z^{-n}\) is obtained by putting \(s=n+r\text{,}\) so we have
\begin{equation*}
\sum\limits_{r=0}^{\infty}\frac{x^{r}z^{r}}{2^{r}r!}\cdot\frac{(-1)^{n+r}x^{n+r}}{2^{n+r}(n+r)!z^{n+r}}
\end{equation*}
\begin{equation*}
=(-1)^{n}\sum\limits_{r=0}^{\infty}\frac{(-1)^{r}}{r!(n+r)!}\left(\frac{x}{2}\right)^{n+2r}z^{-n}
\end{equation*}
\begin{equation}
=(-1)^{n}J_{n}(x)z^{-n}=J_{-n}(x)z^{-n}\tag{4.7.14}
\end{equation}
\begin{equation}
e^{\frac{x}{2}(z-\frac{1}{z})}=\sum\limits_{n=-\infty}^{\infty}J_{n}(x)z^{n}\tag{4.7.15}
\end{equation}
This is the reason that \(e^{\frac{x}{2}(z-\frac{1}{z})}\) is said to be generating function of Bessel’s functions. The generating function also express the trigonometric functions as expansions involving Bessel’s functions. For this purpose put \(z=e^{i\theta}\) in the generating function. That is,
\begin{equation*}
e^{\frac{x}{2}(z-\frac{1}{z})}=\sum\limits_{n=-\infty}^{\infty}J_{n}(x)z^{n}
\end{equation*}
or,
\begin{equation*}
e^{\frac{x}{2}(e^{i\theta}-e^{-i\theta})}=e^{ix\sin\theta}=\sum\limits_{n=-\infty}^{\infty}J_{n}(x)e^{ni\theta}
\end{equation*}
or,
\begin{equation*}
\cos(x\sin\theta)+i\sin(x\sin\theta)
\end{equation*}
\begin{equation*}
=J_{o}(x)+\{J_{1}(x)e^{i\theta}+J_{-1}(x)e^{-i\theta}\}+\{J_{2}(x)e^{2i\theta}+J_{-2}(x)e^{-2i\theta}\}+\cdots
\end{equation*}
\begin{equation*}
=J_{o}(x)+\{J_{1}(x)e^{i\theta}-e^{-i\theta}\}+\{J_{2}(x)e^{2i\theta}+e^{-2i\theta}\}+\cdots
\end{equation*}
\([\because J_{-n}(x)=(-1)^{n}J_{n}(x)]\)
\begin{equation*}
=J_{o}(x)+2i\sin\theta J_{1}(x)+2\cos2\theta J_{2}(x)+\cdots
\end{equation*}
Equating real and imaginary parts, we get -
\begin{equation}
\cos(x\sin\theta)=J_{o}(x)+2 J_{2}(x)\cos2\theta +2 J_{4}(x)\cos4\theta+\cdots\tag{4.7.16}
\end{equation}
and
\begin{equation}
\sin(x\sin\theta)=2J_{1}(x)\sin\theta+2J_{3}(x)\sin3\theta+\cdots\tag{4.7.17}
\end{equation}
The above two series (4.7.16) and (4.7.17) are known as Jacobian series. Replacing \(\theta\) by \((\frac{\pi}{2}-\theta)\text{,}\) we have
\begin{equation}
\cos(x\cos\theta)=J_{o}(x)-2 J_{2}(x)\cos2\theta +2 J_{4}(x)\cos4\theta-\cdots\tag{4.7.18}
\end{equation}
and
\begin{equation}
\sin(x\cos\theta)=2J_{1}(x)\cos\theta-2J_{3}(x)\cos3\theta+\cdots \tag{4.7.19}
\end{equation}
