Page 470 - Advanced engineering mathematics

450 CHAPTER 13 Fourier Series

We therefore have
L N N
2 L 2
2
0 ≤ (g(x)) dx − L b + b ,
n
n
0 n=1 2 n=1
and this gives us
N L
2
2 2
b ≤ ( f (x)) dx.
n
L
n=1 0
Since this is true for all positive integers N, we can let N →∞ to obtain
∞ L
2
2 2
b ≤ ( f (x)) dx.
n
L 0
n=1
EXAMPLE 13.14
We will use Bessel’s inequality to derive an upper bound for the sum of a series. Let f (x) = x 2
on [−π,π]. The Fourier series is
∞ n
1 2 (−1)
f (x) = π + 4 cos(nx)
3 n 2
n=1
n
2
2
2
for −π ≤ x ≤π.Here a 0 =2π /3 and a n =4(−1) /n , while each b n =0(x is an even function).
By Bessel’s inequality,
2 ∞ n 2 π
1 2π 4(−1) 1 4 2 4

+ ≤ x dx = π .
2 3 n 2 π 5
n=1 −π
Then
∞ 4
1 2 2 8π
4
16 ≤ − π = .
n 4 5 9 45
n=1
Then
∞ 4
1 π

≤ ,
n 4 90
n=1
which is approximately 1.0823. Infinite series are generally difficult to sum, so it is sometimes
useful to be able to derive an upper bound.
With stronger assumptions than just existence of the integral over the interval, we can derive
an important equality satisfied by the Fourier coefficients of a function on [−L, L] or by the
Fourier sine or cosine coefficients of a function on [0, L]. We will state the result for f (x)
defined on [−L, L].
THEOREM 13.7 Parseval’s Theorem

Let f be continuous on [−L, L] and let f be piecewise continuous. Suppose that f (−L) =
f (L). Then the Fourier coefficients of f on [−L, L] satisfy
∞
1 2 2 2 1 L 2
a + (a + b ) = f (x) dx.
2 0 n n L −L
n=1

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October 14, 2010 14:57 THM/NEIL Page-450 27410_13_ch13_p425-464