Page 868 - Advanced_Engineering_Mathematics o'neil

848 Answers to Selected Problems

Section 15.2 Legendre Polynomials
1. With n = 5 in the recurrence relation,
1
P 6 (x) = (11xP 5 (x) − 5P 4 (x))
6
1
4
2
6
= (231x − 315x + 105x − 5)
16
Similarly,
1
7
5
3
P 7 (x) = (429x − 693x + 315x − 35x),
16
6435 8 3003 6 3465 4 315 2 35
P 8 (x) = x − x + x − x +
128 32 64 32 128
3. For n = 2, this gives
1 d 2 1
2
2
2
((x − 1) ) = (3x − 1) = P 2 (x).
4(2!) dx 2 2
5. Use the binomial series to write
∞
1 (−1/2)(−3/2)···(−1/2 − n + 1) 2 n
√ = (t − 2at)
1 − 2at + t 2 n!
n=0
∞ k j−k j−k j−k
(−1) 1 · 3···(2 j − 1) j!(−1) 2 a

n
= t .
j
2 j!k!( j − k)!
n=0 j+k=n,k≤ j
n
Show that the coefficient of t is P n (a).For r < d,let a = cos(θ) and t =r/d.For r > d,use a = cos(θ) and t = d/r.
8. (a)
2 2
2
1 + 2x − x = P 0 (x) + 2P 1 (x) − P 2 (x)
3 3
(c)
37 34 32
2 4
2 − x + 4x = P 0 (x) + P 2 (x) + P 4 (x)
15 21 35
9. For −1 < x < 1,
2
12 π − 10
sin(πx/2) = x + 168 P 3 (x)
π 2 π 4
2 4
−112π + π + 1008
+ 660 P 5 (x) +···
π 6
11. For −1 < x < 1,
1 1
2
sin (x) =− cos(1)sin(1) +
2 2

5 15 15
2
+ − cos(1) + − cos (1) P 2 (x)
8 8 4

531 585 585
2
+ cos(1)sin(1) − + cos (1) P 4 (x) +···
32 32 16
13. For −1 < x < 0and 0 < x < 1,
3 7 11
f (x) = P 1 (x) − P 3 (x) + P 5 (x) +···
2 8 16
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October 14, 2010 17:50 THM/NEIL Page-848 27410_25_Ans_p801-866