第三章无穷级数习题课 小结。见书 展开: 1e在=0泰勒展开,预期结果;∑a2,zk1(二=1为奇 方法一:若利用已知级数 则e:当|zk1 1+z+z2+.+ ∑ k! m=o m!
2 0 0 1 1- 1 1- 1 1- 1 ,| | 1( 1 ) | | . 0 1 k k k k k z k z z z z z z e z a z z z e z e e e ¥ = ¥ = å + + +¼¼+ å < = = = < = 见书 预 结 开 为 则 当 结 数 开 级 ， 期 果： 奇 一、小 。 二、展 ： 1 在 泰勒 方法一： 展 若利用已知 = × × × × × = × × ×¼ ¼ × × × å å ¥ = ¥ = 0 2 0 ! ! 2 m m k k z z z m z k z e e e e e k 第三章无穷级数习题课
方法二:直接用公式 f()=e,f(0)=e f'(=)=e f'(0) 1-z)2f'(z)=f(z) (1-=)2f"(x)-2(1-)f(2)=f(2) f"(=)=(a)(3+23),r"0)=3c f(z)=e1+z+ 间:若1=1=1能否这样展开?
1 1 - 1 1 - 2 2 2 2 ( ) , ( 0 ) 1 ( ) , ( 0 ) (1 - ) (1 - ) ( ) ( ) (1 - ) ( ) - 2 (1 - ) ( ) ( ) ( ) ( 3 2 ) ( ) , ( 0 ) 3 2 (1 - ) . .. .. . 3 ( ) 1 ... 2 ! | | 1 z z f z e f e f z e f e z z f z f z z f z z f z f z f z z f z f e z f z e z z z 方法二：直 接 用公式： ： 若 能 否 展 ？ = = ¢ ¢ = × \ = \ = ¢ ¢ ¢¢ ¢ ¢ = ¢ + ¢¢ = = ¢¢ \ é ù = + + + ë û 问 = 这 样 开
2.cosz,在z=1,泰勒展开 结果∑a(-12-1k COS z=COS(2-1+1)=cos(z-1)cosl-sin(z-1)sin1 2k+1 = COS 2n/(-1yks1(-y k20(2k+1) |1+(y CS1+(-1)2 121+(-y sin1(z-1)
0 ( -1) ,| -1| cos cos( -1 1) cos( -1)cos1-sin 2.cos , 1 ( -1)sin1 , k k k a z z z z z z z z ¥ = å < ¥ = + = = 结 开 果 在 泰勒展 ： 2 2 1 0 0 1 1 2 2 0 ( 1) ( 1) cos1 ( 1) sin1 ( 1) 2 ! (2 1)! 1 ( 1) 1 ( 1) ( 1) cos1 ( 1) sin1 ( 1) 2 2 ! k k k k k k n n n n n z z k k z n ¥ ¥ + = = + ¥ = - - = å å - - - + é ù + - + - - = å + - - ê ú ë û
z=1泰勒展开 解:结果:∑a1(-1,z-1k2 首先应分项分式 (+11=21.2 (z+1)2(z+1 + z+12-1+22 ∑(-1)(三 又 (z+1) 2+ 2 dz ∑(-1) 2∑(1)k(2 k=0 n+l n
0 2 2 2 2 2 0 2 2 2 ( -1) ,| -1| 2 ( 1-1) 2 1 1- ( 1 3. 1) ( 1) ( 1) 1 1 1 1 1 -1 (-1) ( , 1, ) 1 -1 2 2 1 -1 2 2 2 1 1 - ( 1 1) 1 ( ) k k k k k k a z z z z z z z z z z z z z z z z z ¥ = ¥ = å < + = = + + + + + = = = å + + + ¢ é ù = ê ú ë + + = + û Q 解： 果： 首先 泰勒展 分 。 分式： 又 开 结 应 项 å å å ¥ = + ¥ = - ¥ = - - = - - - = - - × - - - = 0 1 0 1 0 ) 2 1 ( 2 1 ( 1 ) 2 1 ) 2 1 ( 2 1 ( 1 ) 2 1 ) 2 1 ( 1 ) ( 2 1 n n n k k k k k k n z z k z dz d
1 f(=)=1 z+l(x+1)2 =l∑(2y+∑ (-y(k+1) k+2 k=0 k=0 1-54 -2(-)+2-k(=-lyl k=0 k=0 ∑ye=y 1-20k=3 k+2 k=0
2 ( 1) 1 1 2 ( ) 1 + + + = - z z f z 2 0 0 2 2 0 0 0 2 0 ( 1) ( 1) ( 1) 1 ( 1) ( 1) 2 2 ( 1) ( 1) 1 4 ( 1) ( 1) 2 2 ( 1) ( 1) 2 ( 1) 1 ( 1) ( 3) 2 k k k k k k k k k k k k k k k k k k k k k k k k k z z k z z z z k ¥ ¥ + = = ¥ ¥ + + = = ¥ = ¥ + = - - + = - - + - - - × = - × - + - - + - - = - - - å å å å å å