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548 CHAPTER 15 Special Functions and Eigenfunction Expansions

With a bit more analysis, we can observe the skin effect in this mathematical model of current
flow in the wire. The entire current flowing through a cylinder of radius r within the wire, and
having the same central axis as the wire, is
r
b
Re Ce iωt 2πξ I 0 (bξ)dξ .
2πaI (ba) 0

0
Again using equation (15.9), this is
rI (br) iωt

0
Re Ce .

aI (ba)
0
Then
current in the cylinder of radius r r I (br)

0
= .

total current in the wire a I (ba)
0
When the frequency ω is large, b is large in magnitude and we can use the asymptotic expansion
(15.13) to write
√
r I (br) r e br ba b −b(a−r)

0
≈ √ = e .

a I (ba) a br e ba a
0
For any r with 0 < r < a, the quantity on the right in this equation can be made arbitrarily small
by choosing ω sufficiently large. This means that, for large frequencies of the current, “most” of
the current flows in a thin layer near the outer surface of the wire. This is the skin effect.
15.3.8 A Generating Function for J n (x)
Thus far, we have defined Bessel and modified Bessel functions as solutions to Bessel’s equation,
and examined some applications. Now we will carry out a program like that for Legendre poly-
nomials. We will develop a generating function, recurrence relations, zeros of Bessel functions,
and eigenfunction expansions involving Bessel functions.
THEOREM 15.9 Generating Function for J n (x)

∞

n
e x(t−1/t)/2 = J n (x)t . (15.19)
n=−∞
This means that, if we expand the exponential e x(t−1/t)/2 in an infinite series, then the coef-
ficient of t is J n (x), for any integer n. This is in the same spirit that P n (x) is the coefficient of
n
√
n
t in the expansion of L(x,t) = 1/ 1 − 2xt + t . One difference is that the expansion of L(x,t)
2
involves only nonnegative powers of t, while this expansion of e x(t−1/t)/2 involves negative powers
of t because of the 1/t term in the exponent.
To understand why equation (15.19) is true, begin with the familiar Maclaurin expansions
of the exponential function:
e
e x(t−1/t)/2 = e xt/2 −x/2t
m
∞
∞ k
1 xt 1 x
k
= (−1)
m! 2 k! 2t
m=0 k=0

2 2 3 3
2 3
xt 1 x t 1 x t x 1 x 1 x
= 1 + + + + ··· 1 − + − + ··· .
2 2
3 3
2 2! 2 2 3! 2 3 2t 2! 2 t 3! 2 t
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October 14, 2010 15:20 THM/NEIL Page-548 27410_15_ch15_p505-562