Page 450 - Advanced engineering mathematics

430 CHAPTER 13 Fourier Series

And, if n = 0, then
L L
2 2
cos (nπx/L)dx = sin (nπx/L)dx = L. (13.5)
−L −L
Now let n be any positive integer. To solve for a n , multiply equation (13.1) by cos(nπx/L) and
integrate the resulting equation to get
1
L L
f (x)cos(nπx/L)dx = a 0 cos(nπx/L)dx
−L 2 −L
∞ L L

+ a k cos(kπx/L)cos(nπx/L)dx + b k sin(kπx/L)cos(nπx/L)dx .
−L −L
k=1
Because of equations (13.3) and (13.4), all of the terms on the right are zero except the coefficient
of a n , which occurs in the summation when k = n. The last equation reduces to
L L
2
cos (nπx/L)dx = a n L
f (x)cos(nπx/L)dx = a n
−L −L
by equation (13.5). Therefore
1 L
a n = f (x)cos(nπx/L)dx. (13.6)
L L
This expression contains a 0 if we let n = 0.
Similarly, if we multiply equation (13.1) by sin(nπx/L) instead of cos(nπx/L) and
integrate, we obtain
1 L
b n = f (x)sin(nπx/L)dx. (13.7)
L −L

The numbers

1 L
a n = f (x)cos(nπx/L)dx for n = 0,1,2,··· (13.8)
L −L
1 L
b n = f (x)sin(nπx/L)dx for n = 1,2,··· (13.9)
L −L
are called the Fourier coefficients of f on [L, L]. When these numbers are used, the series
(13.1) is called the Fourier series of f on [L, L].

EXAMPLE 13.1
2
Let f (x) = x − x for −π ≤ x ≤ π.Here L = π. Compute
1 π 2
2
2
a 0 = (x − x )dx =− π ,
π −π 3
1 π
2
a n = (x − x )cos(nx)dx
π −π
2
2
4sin(nπ) − 4nπ cos(nπ) − 2n π sin(nπ)
=
πn 3

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October 14, 2010 14:57 THM/NEIL Page-430 27410_13_ch13_p425-464