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[CHAP. 13
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1 2 2 X 1 cos n cos nx   n 2 n2 1   2 4 cos 2x cos 4x cos 6x 2 2   22 1 4 1 6 1
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13.12. Expand f x x; 0 < x < 2, in a half range
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(a) sine series,
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(b) cosine series.
(a) Extend the de nition of the given function to that of the odd function of period 4 shown in Fig. 13-12 below. This is sometimes called the odd extension of f x . Then 2L 4; L 2.
f (x)
_6 _4 _2 2 4 6
Fig. 13-12
Thus an 0 and bn 2 L nx 2 2 nx f x sin x sin dx dx L 0 L 2 0 2 &    '2 2 nx 4 nx   4 cos n cos 1 2 2 sin x  n 2 2 n n  0
Then
f x
1 X 4 n 1
  4 x 1 2x 1 3x sin sin sin  2 2 2 3 2
cos n sin
nx 2
(b) Extend the de nition of f x to that of the even function of period 4 shown in Fig. 13-13 below. This is the even extension of f x . Then 2L 4; L 2.
CHAP. 13]
FOURIER SERIES
f (x)
2 4 6
Fig. 13-13
Thus bn 0, an 2 L nx 2 2 nx dx dx f x cos x cos L 0 L 2 0 2 &    ' 2 nx 4 nx 2  x sin 1 2 2 cos  n 2 2 n  0 4 2 2 cos n 1 If n 6 0 n 
If n 0; a0
x dx 2:
1 X 4 nx cos n 1 cos 2 n2 2 n 1   8 x 1 3x 1 5x 1 2 cos 2 cos 2 cos 2 2 2  3 5
Then
f x 1
It should be noted that the given function f x x, 0 < x < 2, is represented equally well by the two di erent series in (a) and (b).
PARSEVAL S IDENTITY 13.13. Assuming that the Fourier series corresponding to f x converges uniformly to f x in L; L , prove Parseval s identity 1 L a2 f f x g2 dx 0 a2 b2 n n L L 2 where the integral is assumed to exist.
1 a0 X nx nx , then multiplying by f x and integrating term by term bn sin an cos L L 2 n 1 from L to L (which is justi ed since the series is uniformly convergent) we obtain ' L L 1 X& L a L nx nx dx bn dx f f x g2 dx 0 f x dx an f x cos f x sin L L 2 L L L L n 1
If f x
1 X 2 a2 0 L L an b2 n 2 n 1
where we have used the results L nx f x cos dx Lan ; L L obtained from the Fourier coe cients.
f x sin
nx dx Lbn ; L
f x dx La0
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 4 4 4 4 . 1 2 3 n
(a) Here L 2; a0 2; an 4 cos n 1 ; n 6 0; bn 0. n2 2 Then Parseval s identity becomes 1 1 2 1 2 2 2 2 X 16 f f x g2 dx x dx cos n 1 2 2 2 2 2 2 n4 4 n 1
  8 64 1 1 1 1 1 1 4 2 4 4 4 4 ; i.e., 4 4 4 3 96:  1 3 5 1 3 5     1 1 1 1 1 1 1 1 1 4 4 4 4 4 4 4 4 14 2 3 1 3 5 2 4 6     1 1 1 1 1 1 1 4 4 4 4 4 4 4 1 3 5 2 1 2 3 4 S ; 96 16 from which S 4 90
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