Methods of Mathematical Physics(2016. 11)Chapter 8 Series solutions oflinear difference equations and some special functions YLMaa Phys FDU 将(8.3)和(8.1)代入方程,我们得到 (212+o2a)+(3,2a+o2a)+…+[(n+2)( ∑[(n+2)(n+1 (84) 如果使方程有非零解,全部系数应为零,因此 2.1a,+2a=0 3·2a2+o2a1=0 5·4a5+O2a3=0 5·4 (n+2)n+ (2k2k-)“2(2k2k-1(2k-2(2k (-1)yo2 (-1)ya (2k)(2k-1X2k-2 (2k) a2k+1= (2k+1)(2k)(2k+1X2k)2k-1X2 (-1)a2 (-1)y (2k+1)(2k)X2k-1(2k-2)…32(2k+1) y=ao +a,I (2k) (- (-1) k+1 1-些t2 (2k)Methods of Mathematical Physics (2016.11) Chapter 8 Series solutions of linear difference equations and some special functions YLMa@Phys.FDU 10 将（8.3）和（8.1）代入方程，我们得到 ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 0 3 1 2 2 2 0 2 1 3 2 2 1 2 1 0. n n n n n n n a a a a t n n a a t n n a a t     +  + =  + +  + + + +  + + +     = +  + + =     （8.4） 如果使方程有非零解，全部系数应为零，因此 2 1 0 0 2  a2 + a = ， 即 0 2 0 2 2 2 1 2! a a a   = −  = − 3 2 1 0 2  a3 + a = ， 即 1 2 1 2 3 3 2 3! a a a   = −  = − 4 3 2 0 2  a4 + a = ， 即 0 4 2 2 4 4 3 4! a a a   =  = − 5 4 3 0 2  a5 + a = ， 即 1 4 3 2 5 5 4 5! a a a   =  = − … … ( 2) ( 1) 0 2 n +  n + an+2 + an = ，即 ( )( ) n an n n a 2 1 2 2 + + + = −  ( )( ) ( )( )( )( ) ( ) ( )( )( )( ) ( ) ( ) 0 2 0 2 2 4 4 2 2 2 2 2 ! 1 2 2 1 2 2 2 3 2 1 1 2 2 1 2 2 1 2 2 2 3 a k a k k k k a k k k k a k k a k k k k k k k     − = − − −  − = − − − = − = − − −  ( )( ) ( )( )( )( ) ( ) ( )( )( )( ) ( ) ( ) 1 2 1 2 2 3 4 2 1 2 2 1 2 1 ! 1 2 1 2 2 1 2 2 3 2 1 2 1 2 2 1 2 2 1 2 2 a k a k k k k a k k k k a k k a k k k k k k k + − = + − −  − = + − − = + + = − − −      ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 3 2 2 1 0 1 0 1 0 1 2 2 2 2 2 2 3 2 1 0 1 0 0 2 3 2 2 2 1 0 0 1 1 2! 3! 2 ! 2 1 ! 1 1 1 2! 2 ! 3! 2 1 ! 1 1 2! 2 ! k k k k k k k k k k k k k k k k k k y a a t a t a t a t a t k k a t t a t t t k k a a t t t k              +   + = =  = − − = + − − + + + + +     − − = − + + + + − + + +     +       − = − + + + + −          ( ) ( ) 2 1 3 2 1 0 1 3! 2 1 ! k k k k t t k   + + =   −   + + +   +   