Page 361 - Schaum's Outline of Theory and Problems of Advanced Calculus

352 FOURIER SERIES [CHAP. 13

The required result follows on dividing both sides of (1)by L. Parseval’s identity is valid under less
restrictive conditions than that imposed here.

13.14. (a)Write Parseval’s identity corresponding to the Fourier series of Problem 13.12(b).
1 1 1 1
(b) Determine from (a) the sum S of the series þ þ þ þ þ .
1 4 2 4 3 4 n 4
4
ðcos n 1Þ; n 6¼ 0; b n ¼ 0.
n
(a)Here L ¼ 2; a 0 ¼ 2; a n ¼ 2 2
Then Parseval’s identity becomes
1 ð 2 2 1 ð 2 2 ð2Þ 2 X 16 2
1
f f ðxÞg dx ¼ x dx ¼ þ 4 4 ðcos n 1Þ
2 2 2 2 2 n
n¼1
8 64 1 1 1 1 1 1 4
or þ ; i.e.,
3 ¼ 2 þ 4 1 4 þ 3 4 þ 5 4 1 4 þ 3 4 þ 5 4 þ ¼ 96:
1 1 1 1 1 1 1 1 1
ðbÞ S ¼ 4 þ 4 þ 4 þ ¼ 4 þ 4 þ 4 þ þ 4 þ 4 þ 4 þ
1 2 3 1 3 5 2 4 6

1 1 1 1 1 1 1
1 3 5 2 1 2 3
¼ 4 þ 4 þ 4 þ þ 4 4 þ 4 þ 4 þ
4 S 4
;
96 16 90
¼ þ from which S ¼
13.15. Prove that for all positive integers M,
2 M ð L
a 0 X 2 2 1 2
ða n þ b n Þ @ f f ðxÞg dx
2 þ L L
n¼1
where a n and b n are the Fourier coefficients corresponding to f ðxÞ, and f ðxÞ is assumed piecewise
continuous in ð L; LÞ.
M
X n x n x
a 0
Let S M ðxÞ¼ þ a n cos þ b n sin (1)
2 L L
n¼1
For M ¼ 1; 2; 3; ... this is the sequence of partial sums of the Fourier series corresponding to f ðxÞ.
We have
ð L
2
f f ðxÞ S M ðxÞg dx A 0 ð2Þ
L
since the integrand is non-negative. Expanding the integrand, we obtain
L L L
ð ð ð
2
2
2 f ðxÞ S M ðxÞ dx S M ðxÞ dx @ f f ðxÞg dx ð3Þ
L L L
Multiplying both sides of (1)by2 f ðxÞ and integrating from L to L,using equations (2)ofProblem
13.13, gives
( )
ð L 2 M
a 0 X 2 2
2 f ðxÞ S M ðxÞ dx ¼ 2L þ ða n þ b n Þ ð4Þ
L 2 n¼1
Also, squaring (1) and integrating from L to L,using Problem 13.3, we find
( )
M
ð L 2 X
2
2
2
S M ðxÞ dx ¼ L a 0 þ ða n þ b n Þ ð5Þ
L 2 n¼1
Substitution of (4) and (5)into (3) and dividing by L yields the required result.