Special Functions – Bessel functions – Generating functions

The Bessel function \(J_n(z)\) of the first kind of integer order \(n\in\mathbb{Z}\) can also be defined by means of the generating function

\[\exp\left(\frac{1}{2}z\left(t-t^{-1}\right)\right)=\sum_{n=-\infty}^{\infty}J_n(z)t^n.\tag1\]

In fact, the series on the right-hand side is a so-called Laurent series at \(t=0\) for the function at the left-hand side. Using the Taylor series for the exponential function we obtain

\[\exp\left(\frac{1}{2}z\left(t-t^{-1}\right)\right)=\exp\left(\frac{zt}{2}\right)\cdot\exp\left(-\frac{z}{2t}\right) =\sum_{j=0}^{\infty}\frac{1}{j!}\left(\frac{zt}{2}\right)^j\cdot\sum_{k=0}^{\infty}\frac{(-1)^k}{k!}\left(\frac{z}{2t}\right)^k =\sum_{n=-\infty}^{\infty}a_nt^n.\]

For \(n\in\{0,1,2,\ldots\}\) we have

\[a_n=\sum_{k=0}^{\infty}\frac{1}{(n+k)!}\left(\frac{z}{2}\right)^{n+k}\cdot\frac{(-1)^k}{k!}\left(\frac{z}{2}\right)^k =\left(\frac{z}{2}\right)^n\sum_{k=0}^{\infty}\frac{(-1)^k}{(n+k)!\,k!}\left(\frac{z}{2}\right)^{2k}=J_n(z)\]

and

\[a_{-n}=\sum_{j=0}^{\infty}\frac{1}{j!}\left(\frac{z}{2}\right)^j\cdot \frac{(-1)^{j+n}}{(n+j)!}\left(\frac{z}{2}\right)^{n+j}=(-1)^nJ_n(z)=J_{-n}(z).\]

This proves (1).

If \(t=e^{i\theta}\), then we have

\[\frac{1}{2}\left(t-t^{-1}\right)=\frac{e^{i\theta}-e^{-i\theta}}{2}=i\sin\theta\quad\Longrightarrow\quad \exp\left(\frac{1}{2}x\left(t-t^{-1}\right)\right)=e^{i x\sin\theta}=\cos(x\sin\theta)+i\sin(x\sin\theta).\]

Hence we have

\[\cos(x\sin\theta)+i\sin(x\sin\theta)=e^{i x\sin\theta}=\sum_{n=-\infty}^{\infty}J_n(x)e^{i n\theta} =\sum_{n=-\infty}^{\infty}J_n(x)\left[\cos(n\theta)+i\sin(n\theta)\right].\]

Since \(J_{-n}(x)=(-1)^nJ_n(x)\) this implies that

\[\cos(x\sin\theta)=\sum_{n=-\infty}^{\infty}J_n(x)\cos(n\theta)=J_0(x)+2\sum_{k=1}^{\infty}J_{2k}(x)\cos(2k\theta)\]

and

\[\sin(x\sin\theta)=\sum_{n=-\infty}^{\infty}J_n(x)\sin(n\theta)=2\sum_{k=0}^{\infty}J_{2k+1}(x)\sin(2k+1)\theta.\]

For \(\theta=\pi/2\) this implies that

\[\cos x=J_0(x)+2\sum_{k=1}^{\infty}(-1)^kJ_{2k}(x)\quad\text{and}\quad\sin x=2\sum_{k=0}^{\infty}(-1)^kJ_{2k+1}(x).\]

For \(\theta=0\) we also have

\[1=J_0(x)+2\sum_{k=1}^{\infty}J_{2k}(x).\]

The generating function (1) can be used to prove that

\[J_n(x+y)=\sum_{k=-\infty}^{\infty}J_k(x)J_{n-k}(y).\]

The proof is as follows:

\begin{align*} \sum_{n=-\infty}^{\infty}J_n(x+y)t^n&=\exp\left(\frac{1}{2}(x+y)\left(t-t^{-1}\right)\right) =\exp\left(\frac{1}{2}x\left(t-t^{-1}\right)\right)\cdot\exp\left(\frac{1}{2}y\left(t-t^{-1}\right)\right)\\[2.5mm] &=\sum_{k=-\infty}^{\infty}J_k(x)t^k\cdot\sum_{m=-\infty}^{\infty}J_m(y)t^m =\sum_{n=-\infty}^{\infty}\left(\sum_{k=-\infty}^{\infty}J_k(x)J_{n-k}(y)\right)t^n. \end{align*}

We can also find an integral representation for the Bessel function \(J_n(x)\) of the first kind of integral order \(n\) starting from the generating function

\[e^{i x\sin\theta}=\sum_{n=-\infty}^{\infty}J_n(x)e^{i n\theta}.\]

We use the orthogonality property of the exponential function, id est

\[\int_{-\pi}^{\pi}e^{i k\theta}\,d\theta =\left\{\begin{array}{ll}0, & k\in\mathbb{Z},\quad k\ne 0\\[2.5mm]2\pi, & k=0.\end{array}\right.\]

Hence we have

\[\int_{-\pi}^{\pi}e^{i x\sin\theta}\cdot e^{-in\theta}\,d\theta =\sum_{k=-\infty}^{\infty}J_k(x)\int_{-\pi}^{\pi}e^{i k\theta}\cdot e^{-i n\theta}\,d\theta=2\pi\cdot J_n(x).\]

This implies that

\[J_n(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i\left(x\sin\theta-n\theta\right)}\,d\theta =\frac{1}{\pi}\int_0^{\pi}\cos\left(x\sin\theta-n\theta\right)\,d\theta.\]

The special case \(n=0\) reads

\[J_0(x)=\frac{1}{\pi}\int_0^{\pi}\cos\left(x\sin\theta\right)\,d\theta =\frac{2}{\pi}\int_0^1\frac{\cos(xt)}{\sqrt{1-t^2}}\,dt.\]

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Last modified on September 30, 2021