Page 479 - Advanced engineering mathematics

13.6 Complex Fourier Series 459

Then
1 L 1 L
d n = f (x)e −inω 0 x dx = f (x)e inω 0 x dx = d −n .
2L −L 2L −L
With this, the expansion of equation (13.14) becomes
∞ ∞

d 0 + d n e inω 0 x + d −n e −inω 0 x
n=1 n=1
∞
inω 0 x
= d n e .
n=−∞

This leads us to define the complex Fourier series of f on [−L, L] to be
∞
inω 0 x
d n e ,
n=−∞
with coefficients
1 L
d n = f (x)e −inω 0 x dx
2L −L
for n = 0,±1,±2,···.

Because of the periodicity of f , the integral defining the coefficients can be carried out over
any interval [α,α + 2L] of length 2L. The Fourier convergence theorem applies to this complex
Fourier expansion, since it is just the Fourier series in complex form.

EXAMPLE 13.17
Let f (x) = x for −1 ≤ x < 1 and suppose f has fundamental period 2, so f (x + 2) = f (x) for
all x. Figure 13.23 is part of a graph of f .Now ω 0 = π.
Immediately d 0 = 0 because f is an odd function. For n = 0,
1 1
d n = xe −inπx dx
2 −1
1
= inπe inπ − e inπ + inπe −inπ + e −inπ
2
2n π 2
1
= inπ e inπ + e −inπ − e inπ − e −inπ .
2
2n π 2
The complex Fourier series of f is
∞
1
inπ −inπ inπ −inπ inπx
inπ e + e − e − e e .
2n π 2
2
n=−∞,n =0
This converges to x for −1 < x < 1. In this example we can simplify the series. For n = 0,
1
d n = [2inπ cos(nπ) − 2i sin(nπ)]
2n π 2
2
i
n
= (−1) .
nπ
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October 14, 2010 14:57 THM/NEIL Page-459 27410_13_ch13_p425-464