Page 358 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 13] FOURIER SERIES 349

1 ð L n x 1 ð L n x
f ðxÞ sin f ðxÞ sin dx ¼ 0
b n ¼ dx þ
L 0 L L 0 L
X n x n x
1
a 0
Method 2: Assume f ðxÞ¼ þ a n cos þ b n sin :
2 L L
n¼1
X n x n x
1
a 0
Then f ð xÞ¼ þ a n cos b N sin :
2 L L
n¼1
If f ðxÞ is even, f ð xÞ¼ f ðxÞ. Hence,
1
1
a 0 X n x n x a 0 X n x n x
a n cos þ b n sin a n cos b n sin
2 þ L L ¼ 2 þ L L
n¼1 n¼1
1 1
X n x a 0 X n x
and so b n sin ¼ 0; a n cos
L i.e., f ðxÞ¼ 2 þ L
n¼1 n¼1
and no sine terms appear.
In a similar manner we can show that an odd function has no cosine terms (or constant term) in its
Fourier expansion.
2 ð L n x
13.10. If f ðxÞ is even, show that (a) a n ¼ f ðxÞ cos dx; ðbÞ b n ¼ 0.
L 0 L
1 ð L n x 1 ð 0 n x 1 ð L n x
f ðxÞ cos f ðxÞ cos f ðxÞ cos dx
ðaÞ a n ¼ L dx ¼ L dx þ L
L L L L L 0
Letting x ¼ u,
1 ð 0 n x 1 ð L n u 1 ð L n u
f ðxÞ cos dx ¼ f ð uÞ cos du ¼ f ðuÞ cos du
L L L L 0 L L 0 L
since by definition of an even function f ð uÞ¼ f ðuÞ. Then
ð L ð L ð L
1 n u 1 n x 2 n x
f ðuÞ cos f ðxÞ cos f ðxÞ cos dx
a n ¼ du þ dx ¼
L 0 L L 0 L L 0 L
(b) This follows by Method 1 of Problem 13.9.
13.11. Expand f ðxÞ¼ sin x; 0 < x < ,ina Fourier cosine series.
A Fourier series consisting of cosine terms alone is obtained only for an even function. Hence, we
extend the definition of f ðxÞ so that it becomes even (dashed part of Fig. 13-11 below). With this extension,
f ðxÞ is then defined in an interval of length 2 . Taking the period as 2 ,we have 2L ¼ 2 so that L ¼ .

f (x)

x
_ _ O
2p p p 2p
Fig. 13-11

By Problem 13.10, b n ¼ 0 and

2 ð L n x 2 ð
f ðxÞ cos sin x cos nx dx
a n ¼ L dx ¼
L 0 0