Page 867 - Advanced_Engineering_Mathematics o'neil

Answers to Selected Problems 847

Section 14.7 DFT Approximation of the Fourier Transform
1.

4 − 4iω 1
ˆ f (ω) = , ˆ f (4) = (1 − i),
2
ω + 16 8
511
3π 3π j
DFT approximation = f e −3πij/64 = 0.143860 − 0.124549i
256 256
j=0
3.

4 − ω 2 4ωi
ˆ f (ω) = − , ˆ f (12) ≈−0.006392 − 0.002191i,
2
(ω + 4) 2 (ω + 4) 2
2
511
3π 3π j
DFT approximation = f e −9πij/64 =−0.006506 − 0.002191i
256 256
j=0
CHAPTER FIFTEEN SPECIAL FUNCTIONS AND EIGENFUNCTION EXPANSIONS
Section 15.1 Eigenfunction Expansions
2
1. The problem is regular on [0, L] with eigenvalues ((2n − 1)π/2L) for n = 1,2,···. The functions
sin((2n − 1)πx/2L) are eigenfunctions.
2
3. Regular on [0,4], ((2n − 1)π/8) ,cos((2n − 1)πx/8)
2
5. Periodic on [−3π,3π], n /9for n = 0,1,2,···, a n cos(nx/3) + b n sin(nx/3) with not both a n and b n equal to 0
√ √
1
7. Regular on [0,1], eigenvalues are positive solutions of tan( λ) = λ.If λ is an eigenvalue, an eigenfunction is
√ √ √ 2
2 λcos( λx) + sin( λx).
2
9. Regular on [0,π],1 + n and e −x sin(nx) for n = 1,2,···
11. For 0 < x < 1,
∞
2
n
1 − x = (1 + (−1) (L − 1))sin(nπx).
nπ
n=1
13. The expansion is
√ √
∞ n
4 2sin(nπ/2) − (−1) 2n − 1
2cos(nπ/2) −
cos πx .
π 2n − 1 8
n=1
This converges to −1for0 < x < 2, to 1 for 2 < x < 4,andto0at x = 0.
15. For −3π< x < 3π,
∞ n
(−1)
2
2
x = 3π + 36 cos(nx/3).
n 2
n=1
17. Compute both sides of Bessel’s inequality. With some rearrangement, obtain
∞ n 2
4(−1) + (2n − 1)π 512 2 1

≤ =
(2n − 1) π 3 15 256 2 960
3
n=1
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October 14, 2010 17:50 THM/NEIL Page-847 27410_25_Ans_p801-866