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Chem Differential Eq HW Solutions Fall 2011 88

# Chem Differential Eq HW Solutions Fall 2011 88 - 88 Chapter...

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88 Chapter 5 Partial Diﬀerential Equations in Spherical Coordinates Solutions to Exercises 5.5 1. Putting n = 0 in (9), we obtain P 0 ( x )= 1 2 0 0 ± m =0 ( - 1) m (0 - 2 m )! m !(0 - m )!(0 - 2 m )! x 0 - 2 m . The sum contains only one term corresponding to m = 0. Thus P 0 ( x )=( - 1) 0 0! 0!0!0! x 0 =1 , because 0! = 1. For n = 1, formula (9) becomes P 1 ( x 1 2 1 M ± m =0 ( - 1) m (2 - 2 m )! m !(1 - m )!(1 - 2 m )! x 1 - 2 m , where M = 1 - 1 2 = 0. Thus the sum contains only one term corresponding to m =0 and so P 1 ( x 1 2 1 ( - 1) 0 2! 0!1!1! x 1 = x. For n = 2, we have M = 2 2 = 1 and (9) becomes P 2 ( x 1 2 2 1 ± m =0 ( - 1) m (4 - 2 m )! m !(2 - m )!(2 - 2 m )! x 2 - 2 m = 1 2 2 m =0 ² ³´ µ ( - 1) 0 4! 0!2!2! x 2 + 1 2 2 m =1 ² ³´ µ ( - 1) 1 (4 - 2)! 1!1!0! x 0 = 1 4 6 x 2 + 1 4 ( - 1)2 = 3 2 x 2 - 1 2 . n = 3, we have M = 3 - 1 2 = 1 and (9) becomes P 3 ( x 1 2 3 1 ± m =0 ( - 1) m (6 - 2 m )! m !(3 - m )!(3 - 2 m )! x 3 - 2 m = 1 2 3 ( - 1) 0 6! 0!3!3!
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