Bessel Functions of the First Kind $J_n(x)$ I: Generating Function, Recurrence Relation, Bessel’s Equation

Let us begin with the generating function

$$g(x,t) = e^{\frac{x}{2}\left(t-\frac{1}{t}\right)}.$$
Expanding this function in a Laurent series, we obtain
$$e^{\frac{x}{2}\left(t-\frac{1}{t}\right)} = \sum_{n=-\infty}^\infty J_n(x)t^n.$$
The coefficient of $t^n$, $J_n(x)$, is defined to be a Bessel function of the first kind of order $n$.
Now, we determine $J_n(x)$.
\begin{align*}
e^{\frac{x}{2}t}e^{-\frac{x}{2t}}&=\sum_{r=0}^\infty\left(\frac{x}{2}\right)^r\frac{t^r}{r!} \sum_{s=0}^\infty(-1)^s\left( \frac{x}{2}\right)^s \frac{t^{-s}}{s!}\\
&=\sum_{r=0}^\infty\sum_{s=0}^\infty\frac{(-1)^s}{r!s!}\left(\frac{x}{2}\right)^{r+s}t^{r-s}.
\end{align*}
Set $r=n+s$. Then for $n\ge 0$ we obtain
$$J_n(x)=\sum_{s=0}^\infty \frac{(-1)^s}{s!(n+s)!}\left(\frac{x}{2}\right)^{n+2s}.$$

Bessel Functions

Now we find $J_{-n}(x)$. In the above series for $J_n(x)$, we obtain
$$J_{-n}(x)=\sum_{s=0}^\infty\frac{(-1)^s}{s!(s-n)!} \left(\frac{x}{2}\right)^{2s-n}.$$
However, $(s-n)!\rightarrow\infty$ for $s=0,1,\cdots,(n-1)$. So the series may be considered to begin at $s=n$. Replacing $s$ by $s+n$, we obtain
$$J_{-n}(x)=\sum_{s=0}^ \infty\frac{(-1)^{s+n}}{s!(s+n)!}\left( \frac{x}{2} \right)^{n+2s}.$$ Note that $J_n(x)$ and $J_{-n}(x)$ satisfy the relation
$$J_{-n}=(-1)^nJ_n(x).$$

Let us differentiate the generating function $g(x,t)$ with respect to $t$:
\begin{align*}
\frac{\partial g(x,t)}{\partial t} &=\frac{x}{2}\left(1+ \frac{1}{t^2}\right) e^{\frac{x}{2}\left(t-\frac{1}{t}\right)}\\
&=\sum_{n=-\infty}^\infty n J_n(x) t^{n-1}.
\end{align*}
Replace $e^{\frac{x}{2}\left(t-\frac{1}{t}\right)}$ by $\sum_{n=-\infty}^\infty J_n(x) t^n$.
Then
\begin{align*}
\sum_{n=-\infty}^\infty \frac{x}{2} (1+ \frac{1}{t^2}) J_n(x) t^{n}&=\sum_{n=-\infty}^\infty \frac{x}{2} [J_n(x) t^n + J_n(x) t^{n-2}]\\
&=\sum_{n=-\infty}^\infty \frac{x}{2} [J_{n-1}(x) + J_{n+1}(x)] t^{n-1}.
\end{align*}
Thus,
$$\sum_{n=-\infty}^\infty \frac{x}{2} [J_{n-1}(x) + J_{n+1}(x)] t^{n-1}=\sum_{n=-\infty}^\infty n J_n(x) t^{n-1}$$
or we obtain the recurrence relation,
\begin{equation}J_{n-1}(x) + J_{n+1}(x) = \frac{2n}{x} J_n(x).\end{equation}
Now we differentiate $g(x,t)$ with respect to $x$:
\begin{align*}
\frac{\partial g(x,t)}{\partial x}&=\frac{1}{2}\left(1-\frac{1}{t}\right)e^{\frac{x}{2}\left(t-\frac{1}{t}\right)}\\
& =\sum_{n=-\infty}^\infty J_n'(x)t^n.
\end{align*}
This leads to the recurrence relation
\begin{equation}J_{n-1}(x) – J_{n+1}(x) = 2 J_n'(x).\end{equation}
As a special Case of this recurrence relation, we obtain,
$$J_{0}'(x)=-J_1(x).$$
Adding (1) and (2), we have
\begin{equation}J_{n-1}(x)=\frac{n}{x}J_n(x) + J_n'(x).\end{equation}
Multiplying (3) by $x^n$:
\begin{align*}
x^n J_{n-1}(x) & = n x^{n-1} J_n(x) + x^n J_n'(x)\\
& = \frac{d}{dx}[ x^n J_n(x)].
\end{align*}
Subtracting (2) from (1), we have
\begin{equation}J_{n+1}(x) = \frac{n}{x} J_n(x) – J_n'(x).\end{equation}
Multiplying (4) by $-x^{-n}$:
\begin{align*}
-x^{-n} J_{n+1}(x) & = -n x^{-n-1} J_n(x) + x^{-n} J_n'(x)\\
& = \frac{d}{dx}[x^{-n} J_n(x)].
\end{align*}

Using recurrence relations, we can show that the Bessel functions $J_n(x)$ are the solutions of the Bessel’s differential equation. The recurrence relation (3) can be written as
\begin{equation}x J_n'(x) = x J_{n-1}(x) – n J_n(x).\end{equation}
Differentiating this equation with respect to $x$, we obtain
$$J_n'(x) + x J_n^{\prime\prime}(x) = J_{n-1}(x) + x J_{n-1}'(x) – n J_n'(x)$$
or
\begin{equation}x J_n^{\prime\prime}(x) + (n+1) J_n'(x) – x J_{n-1}'(x) – J_{n-1}(x) = 0.\end{equation}
Subtracting (5) times $n$ from (6) times $x$ results the equation
\begin{equation}x^2 J_n^{\prime\prime}(x) + x J_n'(x) – n^2 J_n(x) + x(n-1) J_{n-1}(x) – x^2 J_{n-1}'(x) = 0.\end{equation}
Replace $n$ by $n-1$ in (4) and multiply the resulting equation by $x^2$ to get the equation
\begin{equation}x^2 J_n(x) = x (n-1) J_{n-1}(x) – x^2 J_{n-1}'(x).\end{equation}
With the equation (8), the equation (7) can be written as
\begin{equation}\label{eq:bessel9}x^2 J_n^{\prime\prime}(x) + x J_n'(x) + (x^2 – n^2) J_n(x) = 0.\end{equation}
This is Bessel’s equation. Hence the Bessel functions $J_n(x)$ are the solutions of Bessel’s equation.

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5 thoughts on “Bessel Functions of the First Kind $J_n(x)$ I: Generating Function, Recurrence Relation, Bessel’s Equation

  1. Pingback: Bessel Functions of the First Kind $J_n(x)$ II: Orthogonality | MathPhys Archive

  2. John Whitman

    The equation before equation 1 is wrong. The t should be raised to the n-1 power, in order to make the recurrence relation correct. Good work overall though 🙂

    Reply
    1. John Lee Post author

      John Whitman,

      Yes, you are correct and I fixed it. Thanks to your comment, I also noticed other typos. Thanks a lot!

      John Lee

      Reply
  3. Pingback: Neumann Functions, Bessel Function of the Second Kind $N_nu(X)$ | MathPhys Archive

  4. Pingback: Bessel Functions of the First Kind $J_n(x)$ II: Orthogonality | MathPhys Archive

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