32.8. EXERCISES 633

The interchange of the integral with the summation follows from noting that the sumsof the form ∑

km=−k cos(nθ −mθ)Jm (x) converge uniformly on [0,π] to the infinite sum

thanks to the 1/m! in the estimates of 32.22. Thus, from the fact that the integral is linear,∫π

0

∞

∑m=−∞

cos(nθ −mθ)Jm (x)dθ =∫

π

0limk→∞

k

∑m=−k

cos(nθ −mθ)Jm (x)dθ

= limk→∞

k

∑m=−k

∫π

0cos(nθ −mθ)Jm (x)dθ =

∞

∑m=−∞

∫π

0cos(nθ −mθ)Jm (x)dθ

Theorem 32.7.2 Let n be a positive integer. Then

Jn (x) =1π

∫π

0cos(nθ − xsin(θ))dθ

How do you compute Jn (x)? You can’t get it the usual way very conveniently becausethe leading term vanishes at 0. This integral will give an easy way to do it. For example,

J4 (6) =1π

∫π

0cos(4θ −6sin(θ))dθ = 0.35764

I just did the integral numerically in Scientific Notebook and got the answer easily. Onecan also produce a graph of x→ J4 (x) very easily in this software by graphing the functionof x given by 1

π

∫π

0 cos(4θ − xsin(θ))dθ . To do this, you simply type the expression inmath mode and then select plot 2d. It has to work at it a little but will produce the graph. Itknows that the variable is x and acts accordingly. In the exercises is a problem on how todo this in MATLAB. It is more elaborate.

There are whole books written on Bessel functions, [17].

32.8 Exercises1. The Hermite equation is

y′′− xy′+ny = 0

Verify that if n = 0 or a positive integer, then this equation always has a polynomialsolution. These are called Hermite polynomials. Hint: This is easier than the caseof a regular singular point. Just look for a solution of the form y = ∑

∞n=0 anxn and

choose the an in such a way that the series satisfies the equation using the fact thatyou can differentiate a power series term by term. In this case, there should be twosolutions.

2. If you have two polynomial solutions to the Hermite equation above, pm (x) corre-sponding to m in the equation and pn (x) corresponding to n in the equation, n ̸= m,show that ∫

∞

−∞

e−x2pm (x) pn (x)dx = 0

3. The equation (1− x2)y′′−2xy′+n(n+1)y = 0

is Legendre’s equation. Note that 0 is an ordinary point for this equation. Show thatfor n a non-negative integer, this equation has polynomial solutions. Also explainwhy this equation has a regular singular point at 1,−1.