§1.1 Legendre多项式 一、 Legendre方程的级数解 1-x2)y2xy+(+1)y=0 对照常微分方程的级数解法 此处y(小/2x(321x2 1(1+1) 它们在x=0点邻域均解析,故x=0为该方程 的常点,从而可令
§1.1Legendre多项式 一、Legendre方程的级数解 ( ) ( ) ( ) 2 1- x y ¢¢ ¢ - 2xy + l l y + = 1 0 1 对照常微分方程的级数解法 ( ) ( ) ( ) 2 2 -2 1 , , 1- 1- x l l p x q x x x + 此处 = = 它们在x=0点邻域均解析，故x=0为该方程 的常点，从而可令
2) k=0 代入(1)得 ∑k(k-)lcx2-∑k(-1)cx2-2∑kcx2 k=1 1(7+1)∑cx=0 x0:2·l2+(1+1)cn=0 (+1) 2·1 0
0 2 k k k y c x ¥ = = å （） 代入(1),得 ( ) ( ) -2 2 2 1 -1 - -1 - 2 k k k k k k k k k k k c x k k c x kc x ¥ ¥ ¥ = = = å å å ( ) 0 1 0 k k k l l c x ¥ = + + = å 0 2 0 x : 2 × lc +（l l c + = 1) 0 2 0 ( 1) 2 1 l l c c + \ = ×
x1:3,2c3-2c1+l(l+1l1=0 (+1)-2 3.2 [(+1)-k(k+1 k+2 (k+2)(+1) 2+l-2.3 2(-2)(+1)(1+ (-1)
1 3 1 1 x : 3× 2c - 2c + l(l c + = 1) 0 3 1 ( 1)-2 - 3 2 l l c c + \ = × [ ] ( ) 2 ( 1) - ( 1) : - 3 ( 2) ( 1) k k k l l k k x c c k l + + + = + × + 2 4 2 - 2 3 - 4 3 l l c c + × \ = × ( ) 2 ( ) ( )( ) 0 - 2 1 3 -1 4! l l l l c + + =
(2n) 2n+1 yn+(-2+3)(-(+2(+4)(+2x (2n+1)!
( ) 2 ( )( )( )( ) 5 1 - 3 -1 2 4 -1 5! l l l l c c + + = 2n c = ( ) ( )( ) ( )( ) ( ) ( ) ( ) 0 - 2 2 - 2 4 ... 1 3 ... 2 -1 -1 4 2 ! n l n l n l l l l n c n + + + + + 2n+1 c = ( ) ( )( ) ( )( )( ) ( ) ( ) 1 ( ) - 2 1 - 2 3 ... -1 2 4 ... 2 -1 5 2 1 ! n l n l n l l l l n c n + + + + + +
故y=∑c k=0 C1x+∑C =y(x)+y1(x) 式中y(x)=c+2cnx(6) n=1 V(x=cxi cent/2n+ n-
k k k=0 y = c x ∞ 故 å 2 2 0 2 1 2 1 1 1 0 1 ( ) ( ) n n l n n n n c c x c x c x y x y x ¥ ¥ + + = = = + å å + + = + ( ) ( ) 2 0 0 2 1 6 n n n y x c c x ¥ = 式中 = + å ( ) ( ) 2 1 1 1 2 1 1 7 n n n y x c x c x ¥ + + = = + å
二、解的敛散性 1.由达氏判 R=lim k→∞ k+1 k+2 lim (k+2)(k+1) (+1)-k(k+1 x1发散 ④)当 x=1收敛?发散?
二、解的敛散性 1.由达氏判 1 2 lim lim k k k k k k a c R ®¥ a c ®¥ + + = = ( ) ( )( ) ( ) ( ) 3 2 1 lim 1 k 1 - 1 k k ®¥ l l k k + + = = + + ∴y(x)当 x 1 发散 x = 1 收敛 发? 散?
2.由高斯判 ReH>1收敛 则Σ/当 ReH≤1发散 将x=代入(6)和(7)得: y(1)=c2+cn();cn c2/—常数
2.由高斯判 1 k k f ¥ = 则å 当 Re 1 m > 收敛 Re 1 m £ 发散 将x = ±1代入 和 (6) (7 : )得 ( ) ( ) 2 0 0 2 2 0 1 1 1 , n n n n y c c c c ¥ = ± = + å ± ¬ 0 0 n n n c f f ¥ = = å 一常数
类似y(±1)=c∑8n 2n+2)(2n+ 1+-+ fm12n(2n+1)-1(1+1)n42 +_+0 类似的有: 8n=1+-+02 n+1
1 1 1 ( 1) n n y c g ¥ = 类似 ± = å ( )( ) ( ) ( ) ( ) 2 1 2 2 2 1 1 1 1 2 2 1 - 1 4 n n f n n l l f n n l l n n + + + + \ = = + + ¼ + + 2 1 1 1 o n n æ ö = + + ç ÷ è ø 类似的有： 2 1 1 1 1 n n g o g n n + æ ö = + + ç ÷ è ø
(x)y(x)在x=1发散 本征值问题 -)y2xy+1(+1)y=01(+1)常数(8) y→有限 由(3)可看出 若取l=0,12,则当l=时 k+2 1+2 C1三
\ y0 1 ( x), 1 y ( x x )在 = ± 发散 三、本征值问题 ( ) ( ) ( ) ( ) 2 1- x y ¢¢ ¢ - 2xy + l l +1 y = + 0,l l 1 8 常数 x 1 y =± ® 有限, 由(3)可看出 若取l = = 0,1,2,...,则当l k时 2 2 0 0 k l l c c c + + = = ´ =
从而有C4=0,c1=0 即y或→次多项式 1若l=k=2n,n=0,1,2 则a2=C2=Cn12=0 n+4 n+6 =..=0 yo(x) C +cx+.+C, x n n =cn+c,x2+…+cx→>次多项式 y,(x) 2n+1 2n+3 Cx+cx+.+ 2n+1 + C →无穷级数
4 6 0, 0... k k c c 从而有 + + = = 0 1 即y 或y l → 项 次多 式 1.若l = k = = 2n n, 0,1,2,... 2 2 2 2 0 k l n c c c 则 + = + + = = 2 4 2 6 ... 0 n n c c \ + + = = = ( ) 2 2 0 0 2 2 ... n n y x = c + c x + + c x 2 0 2 ... l l = c + c x + + ® c x l次多项式 ( ) 3 2 1 2 3 1 1 3 2 1 2 3 ... ... n n n n y x c x c x c x c x + + = + + + + + + + → 无穷级数