Page 360 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 13] FOURIER SERIES 351

f (x)

x
_ _ _ O
6 4 2 2 4 6
Fig. 13-13

Thus b n ¼ 0,
2 ð L n x 2 ð 2 n x
f ðxÞ cos x cos dx
a n ¼ dx ¼
L 0 L 2 0 2
2
2 n x 4 n x
sin cos
n 2 n 2 0
¼ðxÞ ð1Þ 2 2
4
If n 6¼ 0
¼ 2 2 ðcos n 1Þ
n
ð 2
xdx ¼ 2:
If n ¼ 0; a 0 ¼
0
1
X 4 n x
Then f ðxÞ¼ 1 þ ðcos n 1Þ cos
2 2
n 2
n¼1
8 x 1 3 x 1 5 x
cos cos cos
2 3 2 5 2
¼ 1 2 þ 2 þ 2 þ
It should be noted that the given function f ðxÞ¼ x,0 < x < 2, is represented equally well by the
two different series in (a) and (b).
PARSEVAL’S IDENTITY
13.13. Assuming that the Fourier series corresponding to f ðxÞ converges uniformly to f ðxÞ in ð L; LÞ,
prove Parseval’s identity
1 ð L 2 a 0 2 2 2
f f ðxÞg dx ¼ þ ða n þ b n Þ
L L 2
where the integral is assumed to exist.

1
X n x n x
a 0
a n cos þ b n sin ,then multiplying by f ðxÞ and integrating term by term
2 L L
If f ðxÞ¼ þ
n¼1
from L to L (which is justified since the series is uniformly convergent) we obtain
L L 1 L n x L n x
ð ð ð ð
2 a 0 X
a n f ðxÞ cos dx þ b n f ðxÞ sin dx
f f ðxÞg dx ¼ f ðxÞ dx þ
L 2 L n¼1 L L L L
2 1
a 0 X 2 2
L þ L
2
¼ ða n þ b n Þ ð1Þ
n¼1
where we have used the results
ð L n x ð L n x ð L
f ðxÞ cos dx ¼ La n ; f ðxÞ sin dx ¼ Lb n ; f ðxÞ dx ¼ La 0 ð2Þ
L L L L L
obtained from the Fourier coefficients.