# chpp06 - 6 5 ~ ~ I O d 2 w d z 2 p z d w d z q z w = 0(1...

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Unformatted text preview: 6 5 ~ ? { ~ I O / d 2 w d z 2 + p ( z ) d w d z + q ( z ) w = 0 (1) 6.1 5 ~ ~ : : ~ : : p ( z ) q ( z ) 5 ~ d 2 w d z 2 + p ( z ) d w d z + q ( z ) w = 0 X . X J p ( z ) , q ( z ) 3 z : , K z : ~ : . X J p ( z ) , q ( z ) k 3 z : , K z : . Example 6.1 A z (1- z ) d 2 w d z + [ γ- (1 + α + β ) z ] d w d z- αβw = 0 Solution d 2 w d z + γ- (1 + α + β ) z z (1- z ) d w d z- αβ z (1- z ) w = 0 X p ( z ) = γ- (1 + α + β ) z z (1- z ) q ( z ) =- αβ z (1- z ) p ( z ), q ( z ) k : : z = 0 , 1. z = 0 , 1 : , : ~ : . Example 6.2 Legendre (1- x 2 ) d 2 y d x 2- 2 x d y d x + l ( l + 1) y = 0 Solution d 2 y d x 2- 2 x 1- x 2 d y d x + l ( l + 1) 1- x 2 y = 0 d X : x = ± 1 : . 6.2 ~ : S ~ : S Theorem 6.1. X J p ( z ) q ( z ) 3 | z- z | < R S , K3 d S ~ K d 2 w d z 2 + p ( z ) d w d z + q ( z ) w = 0 w ( z ) = C , w ( z ) = C 1 C , C 1 ? ~ , k w ( z ) , w ( z ) 3 S . 1 n , ^ ? { ~ : S . k p ( z ) q ( z ) 3 | z- z | < R S m Taylor ? p ( z ) = ∞ X k =0 a k ( z- z ) k (2) q ( z ) = ∞ X k =0 b k ( z- z ) k (3) n , q r w ( z ) 3 | z- z | < R S m Taylor ? w ( z ) = ∞ X k =0 c k ( z- z ) k (4) d ^ c = w ( z ) = C , c 1 = w ( z ) = C 1 ? m , = ^ X { X c k . Example 6.3 Legendre d 2 y d x 2- 2 x 1- x 2 d y d x + l ( l + 1) 1- x 2 y = 0 3 x = 0 : S , l . Solution x = 0 ~ : . x = ± 1 : , p ( x ), q ( x ) 3 | x | < 1 S , u 3 | x | < 1 S .- y ( x ) = ∞ X k =0 c k x k e e X p ( x ), q ( x ) m Taylor ? , K L ' . E (1- x 2 ) d 2 y d x 2- 2 x d y d x + l ( l + 1) y = 0 ? , k (1- x 2 ) ∞ X k =0 c k k ( k- 1) x k- 2- 2 x ∞ X k =0 c k kx k- 1 + l ( l + 1) ∞ X k =0 c k x k = 0 > 1 ∞ X k =2 c k k ( k- 1) x k- 2- ∞ X k =0 c k k ( k- 1) x k = ∞ X k =0 c k +2 ( k + 2)( k + 1) x k- ∞ X k =0 c k k ( k- 1) x k , n ! , ∞ X k =0 { ( k + 2)( k + 1) c k +2- [ k ( k + 1)- l ( l + 1)] c k } x k = 0 2 ' x g X ( k + 2)( k + 1) c k +2- [ k ( k + 1)- l ( l + 1)] c k = 0 X m 4 ' X c k +2 = ( k- l )( k + l + 1) ( k + 2)( k + 1) c k E | ^ 4 ' X c 2 n = c (2 n )! (2 n- l- 2)(2 n- l- 4) ··· (- l ) (2 n + l- 1)(2 n + l- 3) ··· ( l + 1) c 2 n +1 = c 1 (2 n + 1)! (2 n- l- 1)(2 n- l- 3) ··· (- l + 1) (2 n + l )(2 n + l- 2) ··· ( l + 2) | ^ Γ 5 Γ( z + 1) = z Γ( z ) Γ( z + n ) = ( z + n- 1)( z + n- 2) ··· ( z + 1) z Γ( z ) ^ Γ L n ( z + n- 1)( z + n- 2) ··· ( z + 1) z = Γ( z + n ) Γ( z ) (5) u c 2 n = 2 2 n (2 n )!...
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## This note was uploaded on 05/05/2010 for the course PHYSICS 122 taught by Professor Weizhen during the Spring '06 term at Peking Uni..

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chpp06 - 6 5 ~ ~ I O d 2 w d z 2 p z d w d z q z w = 0(1...

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