Page 451 - Advanced engineering mathematics

13.2 The Fourier Series of a Function 431

4 4
=− cos(nπ) =− (−1) n
n 2 n 2
4(−1) n+1
= ,
n 2
and
1 π 2
b n = (x − x )sin(nx)dx
π −π
2sin(nπ) − 2nπ cos(nπ)
=
πn 2
2 2 n
=− cos(nπ) =− (−1)
n n
2(−1) n+1
= .
n
n
We have used the facts that sin(nπ) = 0 and cos(nπ) = (−1) if n is an integer.
2
The Fourier series of f (x) = x − x on [−π,π] is
1 4(−1) n+1 2(−1) n+1
∞
2
− π + cos(nx) + sin(nx) .
3 n 2 n
n=1
This example illustrates a fundamental issue. We do not know what this Fourier series con-
verges to. We need something that establishes a relationship between the function and its Fourier
series on an interval. This will require some assumptions about the function.
Recall that f is piecewise continuous on [a,b] if f is continuous at all but perhaps finitely
many points of this interval, and, at a point where the function is not continuous, f has finite
limits at the point from within the interval. Such a function has at worst jump discontinuities, or
finite gaps in the graph, at finitely many points. Figure 13.1 shows a typical piecewise continuous
function.
If a < x 0 < b, denote the left limit of f (x) at x 0 as f (x 0 −), and the right limit of f (x) at x 0
as f (x 0 +):
f (x 0 −) = lim f (x 0 − h) and f (x 0 +) = lim f (x 0 + h).
h→0+ h→0+
If f is continuous at x 0 , then these left and right limits both equal f (x 0 ).

y

x

FIGURE 13.1 A piecewise con-
tinuous function.

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-431 27410_13_ch13_p425-464