洛朗级数 f(=) ze 0<z<∞0 e==2 ∑=∑=∑ 1= (k+2) f()=sin 0 ● 2 f(z=sin =sin( 1=sin()cos 1-cos(sin 1 2 2k+ 2k 2k+1 2k COS sIn k=0 (2k+1) 2 k=6(2k)!1
洛朗级数 1 2 ( ) 0 z f z z e z = 1 2 2 2 2 0 0 0 0 1 1( ) ! ! ! ( 2)! n n k z n n n n k z z z z e z z n z n n k − − = = = = = = = = + ( ) sin 0 1 1 z f z z z = − − 1 1 1 ( ) sin sin( 1) sin( ) cos1 cos( ) sin1 1 1 1 1 z f z z z z z = = − = − − − − − 2 1 2 2 1 2 0 0 ( 1) 1 ( 1) 1 cos1 ( ) sin1 ( ) (2 1)! 1 (2 )! 1 k k k k k k k z k z + + = = − − = − + − −
确定收敛环 ∑2K( ∑2K(-1)+∑ k k: 一0 k=1 ∑ 1)+∑2-(二-1) k=0 正幂部分 ∑2(=-1 =m-1 lim(2)K=lim 2 k→>∞ (2 k/kk→>o 负幂部分 R2=1/im-_yk1/∞o=0 k→>∞(2
确定收敛环 2 2 ( 1) k k k z − =− − 2 2 0 1 2 ( 1) 2 ( 1) k k k k k k z z − − =− = = − + − 2 2 0 1 2 ( 1) 2 ( 1) k k k k k k z z − − − = = = − + − 正幂部分 2 1 2 ( 1) k k k z − = − 2 2 1/ 1 1/ 1 lim lim(2 ) lim 2 (2 ) k k k k k k k k R → → → − = = = = 负幂部分 2 2 1/ 1 1/ lim 1/ 0 (2 ) k k k R → − = = =
计算 de 3(=+1)(z+4) dz 二(二+1)(+4)4z3(二+1)12(2+4) 4z3z(+ 48(1+ 4 4-32(1)(+4(-)G k+1 4z3z48 3∑ 4 4312 dz =2ia l=(二+1)(z+4) 6
3 ( 1)( 4) z dz z z z = + + 计算 1 1 1 ( 1)( 4) 4 3( 1) 12( 4) dz z z z z z z = − + + + + + 1 1 1 4 1 3 (1 ) 48(1 ) 4 z z z z = − + + + 0 0 1 1 1 1 ( 1) ( ) ( 1) ( ) 4 3 48 4 k k k k k k z z z z = = = − − + − 1 1 1 1 1 1 1 1 1 ( 1) ( ) ( 1) ( ) 4 3 48 3 48 4 k k k k k k z z z z + = = = − + + − + − 1 1 1 1 4 3 12 a− = − = − 1 1 2 z ( 1)( 4) 6 dz i ia z z z − = = = − + +
