Page 362 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 13] FOURIER SERIES 353

Taking the limit as M !1,we obtain Bessel’s inequality
2 1 ð L
a 0 X 2 2 1 2
ða n þ b n Þ @ f f ðxÞg dx
2 þ L L ð6Þ
n¼1
If the equality holds, we have Parseval’s identity (Problem 13.13).
We can think of S M ðxÞ as representing an approximation to f ðxÞ, while the left-hand side of (2), divided
by 2L, represents the mean square error of the approximation. Parseval’s identity indicates that as M !1
the mean square error approaches zero, while Bessels’ inequality indicates the possibility that this mean
square error does not approach zero.
The results are connected with the idea of completeness of an orthonormal set. If, for example, we were
to leave out one or more terms in a Fourier series (say cos 4 x=L,for example), we could never get the mean
square error to approach zero no matter how many terms we took. For an analogy with three-dimensional
vectors, see Problem 13.60.

DIFFERENTIATION AND INTEGRATION OF FOURIER SERIES
2
13.16. (a) Find a Fourier series for f ðxÞ¼ x ; 0 < x < 2, by integrating the series of Problem 13.12(a).
n 1
1
X ð 1Þ
(b) Use (a)to evaluate the series .
n 2
n¼1
(a)From Problem 13.12(a),
4 x 1 2 x 1 3 x
sin sin sin
2 2 2 3 2
x ¼ þ ð1Þ
Integrating both sides from 0 to x (applying the theorem of Page 339) and multiplying by 2, we find

2 16 x 1 2 x 1 3 x
2 2 2 3 2
x ¼ C 2 cos 2 cos þ 2 cos ð2Þ

16 1 1 1
where C ¼ 2 1 2 þ 2 2 þ :
2 3 4
2
(b)Todetermine C in another way, note that (2) represents the Fourier cosine series for x in 0 < x < 2.
Then since L ¼ 2in thiscase,
1 ð L 1 ð 2 4
a 0 2
C ¼ ¼ f ðxÞ¼ x dx ¼
2 L 0 2 0 3
Then from the value of C in (a), we have

n 1 2 2
1
X 1 1 1 4
ð 1Þ
n 2 ¼ 1 2 2 þ 3 2 ¼ 4 2 þ ¼ 16 3 ¼ 12

n¼1
13.17. Show that term by term differentiation of the series in Problem 13.12(a)is not valid.

x 2 x 3 x

Term by term differentiation yields 2 cos cos þ cos :
2 2 2
Since the nth term of this series does not approach 0, the series does not converge for any value of x.