Chem Differential Eq HW Solutions Fall 2011 92

Chem Differential Eq HW Solutions Fall 2011 92 - 92 Chapter...

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Unformatted text preview: 92 Chapter 5 Partial Differential Equations in Spherical Coordinates By Example 1, we have Pn+1 (1) − Pn−1(1) = 0. So for n = 1, 2, . . . 1 Pn(t) dt = x 1 [Pn−1(x) − Pn+1 (x)] . 2n + 1 (b) First let us note that because Pn is even when is even and odd when is odd, it follows that Pn(−1) = (−1)n Pn(1) = (−1)n . Taking x = −1 in (a), we get 1 Pn(t) dt = −1 1 [Pn−1(−1) − Pn+1(−1)] = 0, 2n + 1 because n − 1 and n +1 are either both even or both odd, so Pn−1(−1) = Pn+1 (−1). (c) We have 1 0= x 1 Pn(t) dt = Pn(t) dt + −1 −1 Pn(t) dt. x So x 1 Pn(t) dt = − Pn (t) dt −1 x 1 [Pn−1(x) − Pn+1 (x)] 2n + 1 = − = 1 [Pn+1(x) − Pn−1(x)] 2n + 1 13. We will use Dn f to denote the nth derivative of f . Using Exercise 12, 1 (1 − x2) P13(x) dx = −1 1 (−1)13 213(13)! D13 [(1 − x2 )] (x2 − 1)13 dx = 0 −1 because D13[(1 − x2 )] = 0. 17. Using Exercise 12, 1 ln(1 − x) P2(x) dx = −1 = = 1 (−1)2 22 2! 1 1 8 −1 D2 [ln(1 − x)] (x2 − 1)2 dx −1 −1 (x − 1)2(x + 1)2 dx (1 − x)2 1 −1 8 (x + 1)2 dx = −1 −1 (x + 1)3 24 1 = −1 −1 . 3 21. For n > 0, we have Dn [ln(1 − x)] = −(n − 1)! (1 − x)n . So 1 ln(1 − x) Pn(x) dx = −1 = = = (−1)n 2nn! 1 Dn [ln(1 − x)] (x2 − 1)n dx 1 2nn! (−1) −1 n+1 −1 (n − 1)! (x + 1)n (x − 1)n dx (1 − x)n −1 1 −1 (x + 1)n dx = n (x + 1)n+1 2nn −1 2 n(n + 1) −2 n(n + 1) 1 −1 ...
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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