Page 478 - Advanced engineering mathematics

458 CHAPTER 13 Fourier Series

and by replacing θ with −θ, we get
e −iθ = cos(θ) − i sin(θ).
Solve these equations for cos(θ) and sin(θ) to obtain the complex exponential forms of the
trigonometric functions:
1 iθ −iθ 1 iθ −iθ
cos(θ) = e + e ,sin(θ) = e − e .
2 2i
ix
We will also use the fact that, if x is a real number, then the conjugate of e is
e = e −ix .
ix
This follows from Euler’s formula, since
ix
e = cos(x) + i sin(x) = cos(x) − i sin(x) = e −ix .
Now let f be a piecewise smooth periodic function with fundamental period 2L. To derive a
complex Fourier expansion of f (x) on [−L, L], begin with the Fourier series of f (x). With
ω 0 = π/L, this series is
∞
1
a 0 + (a n cos(nω 0 x) + b n sin(nω 0 x))
2
n=1
Put the complex forms of cos(nω 0 x) and sin(nω 0 x) into this expansion:
∞
1 1 1
a 0 + a n e inω 0 x + e −inω 0 x + b n e inω 0 x − e −inω 0 x
2 2 2i
n=1
∞
1 1 1
= a 0 + (a n − ib n )e inω 0 x + (a n + ib n )e −inω 0 x ,
2 2 2
n=1
in which we used the fact that 1/i =−i. In this series, let
1
d 0 = a 0
2
and, for n = 1,2,··· ,
1
d n = (a n − ib n ).
2
The Fourier series on [−L, L] becomes
∞ ∞

d 0 + d n e inω 0 x + d n e −inω 0 x (13.14)
n=1 n=1
Now
1 1 L
d 0 = a 0 = f (x)dx
2 L −L
and for n = 1,2,···,
1
d n = (a n − ib n )
2
1 L i L
= f (x)cos(nω 0 x)dx − f (x)sin(nω 0 x)dx
2L −L 2L −L
1 L
= f (x)[cos(nω 0 x) − i sin(nω 0 x)]dx
2L −L
1 L −inω 0 x
= f (x)e dx.
2L −L

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

October 14, 2010 14:57 THM/NEIL Page-458 27410_13_ch13_p425-464