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520    CHAPTER 15  Special Functions and Eigenfunction Expansions

                                 With n = 4 we obtain
                                                           20 − λ    −λ(6 − λ)(20 − λ)
                                                       a 6 =     a 4 =              a 0 ,
                                                           (5)(6)           6!
                                 and so on, with each even-indexed a 2n in terms of λ, n and a 0 . Similarly, with n = 3,
                                                            12 − λ    (2 − λ)(12 − λ)
                                                        a 5 =     a 3 =            a 1 ,
                                                             (4)(5)         5!
                                 and with n = 7,
                                                         30 − λ    (2 − λ)(12 − λ)(30 − λ)
                                                     a 7 =     a 5 =                   a 1
                                                          (6)(7)            7!
                                 and so on. Each odd-indexed a 2n+1 can be written in terms of λ, n and a 1 .
                                    Now we can write the solution
                                           ∞
                                                          λ     λ(6 − λ)   λ(6 − λ)(20 − λ)
                                                n            2          4                 6
                                    y(x) =   a n x = a 0 1 − x −       x −               x + ···
                                                          2       4!             6!
                                           n=0

                                                   2 − λ  3  (2 − λ)(12 − λ)  5  (2 − λ)(12 − λ)(30 − λ)  7
                                          + a 1 x +     x +              x +                     x + ··· .
                                                     3!           5!                  7!
                                 These two series solutions in large parentheses are linearly independent, one containing only odd
                                 powers of x, the other only even powers. This expression therefore gives the general solution of
                                 Legendre’s equation, with a 0 and a 1 arbitrary constants.
                                    Now observe that we obtain polynomial solutions by choosing λ of the form n(n + 1), and
                                 either a 0 or a 1 zero. For example:
                                    With n = 0 and a 1 = 0, we have λ = 0 and

                                                              y(x) = a 0 = constant ;
                                 with n = 1 and a 0 = 0, we have λ = 2 and
                                                                  y(x) = a 1 x;
                                 with n = 2 and a 1 = 0, we have λ = 6 and
                                                                             2
                                                               y(x) = a 0 (1 − 3x );
                                 with n = 3 and a 0 = 0, we have λ = 12 and

                                                                           5
                                                              y(x) = a 1 x − x 3  ;
                                                                           3
                                 with n = 4 and a 1 = 0, we have λ = 20 and

                                                                             35
                                                                          2     4
                                                          y(x) = a 0 1 − 10x +  x  ;
                                                                             3
                                 and so on. These polynomial solutions are bounded on [−1,1]. If the constant is always chosen
                                 so that y(1) = 1, we obtain the Legendre polynomials P n (x), the first six of which are
                                                                         1
                                                                             2
                                                P 0 (x) = 1, P 1 (x) = x, P 2 (x) = (3x − 1),
                                                                         2
                                                       1                 1
                                                                              4
                                                           3
                                                                                    2
                                                P 3 (x) = (5x − 3x), P 4 (x) = (35x − 30x + 3), and
                                                       2                 8
                                                       1
                                                            5
                                                                  3
                                                P 5 (x) = (63x − 70x + 15x).
                                                       8
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                                   October 14, 2010  15:20  THM/NEIL   Page-520        27410_15_ch15_p505-562
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