Page 859 - Advanced_Engineering_Mathematics o'neil

Answers to Selected Problems 839

n
8 (−1) n 2(1 − (−1) )
∞
− + sin(nπx/2),
3
π n n π 2
n=1
2
converging to x if 0 < x < 2and to 0for x = 0and for x = 2.
7. The cosine series is
1
2
4 12 6
∞

n
+ sin(2nπ/3) + cos(2nπ/3) − (1 + (−1) ) cos(nπx/3),
2
2
nπ n π 2 n π 2
n=1
converging to x if 0 ≤ x < 2,to1if x = 2 and to 2 − x if 2 < x ≤ 3. The sine series is
12 4 2
∞
n
sin(2nπ/3) − cos(2nπ/3) + (−1) sin(nπx/3),
2
n π 2 nπ nπ
n=1
converging to x if 0 ≤ x < 2to1if x = 2, to 2 − x if 2 < x < 3and to0if x = 3.
9. The cosine series is
5 16 1 nπ 4 nπ nπx
∞
+ cos − sin cos
6 π 2 n 2 4 n π 4 4
3
n=1
2
converging to x for 0 ≤ x ≤ 1andto 1for 1 < x ≤ 4.
The sine series is
16 nπ 64 nπ
2(−1) nπx
∞
n

sin + cos − 1 − sin
3
2
n π 2 4 n π 3 4 nπ 4
n=1
2
converging to x for 0 ≤ x ≤ 1, 1 for 1 < x < 4, and 0 for x = 4.
11. The series converges to 1/2 − π/4.
13. If f is both even and odd, then f (x) = f (−x) =− f (x),so f (x) = 0.
Section 13.4 Integration and Differentiation of Fourier Series
1. The Fourier series of f on [−π,π] is
n
1 (−1) − 1 (−1) n+1
∞
π + cos(nx) + sin(nx) .
4 πn 2 n
n=1
This converges to 0 for −π< x < 0and to x for 0 < x <π. Because f is continuous, its Fourier series can be
integrated term by term, yielding the integral of the sum of the series. Term by term integration yields
x
π
f (t)dt = (x + π)
−π 4
1 (−1) 1
∞ n

n
+ ((−1) − 1)sin(nx) + cos(nx) −
πn 3 n 2 n 2
n=1
3. For −π ≤ x ≤ π,
∞ n+1
1 (−1)
x sin(x) = 1 − cos(x) + 2 cos(nx).
2
2 n − 1
n=2
f is continuous with continuous first and second derivatives on [−π,π],and f (−π) = f (π), so we can differentiate
theseriestermbytermtoobtain
∞
1 n(−1) n
x cos(x) + sin(x) = sin(x) + 2 sin(nx)
2 n − 1
2
n=2
for −π< x <π.
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October 14, 2010 17:50 THM/NEIL Page-839 27410_25_Ans_p801-866