f(z)=z/(z+1)(z+2)在z0=2处展开成泰勒级数,要详细步骤
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/20 00:31:41
f(z)=z/(z+1)(z+2)在z0=2处展开成泰勒级数,要详细步骤f(z)=z/(z+1)(z+2)在z0=2处展开成泰勒级数,要详细步骤f(z)=z/(z+1)(z+2)在z0=2处展开成泰勒
f(z)=z/(z+1)(z+2)在z0=2处展开成泰勒级数,要详细步骤
f(z)=z/(z+1)(z+2)在z0=2处展开成泰勒级数,要详细步骤
f(z)=z/(z+1)(z+2)在z0=2处展开成泰勒级数,要详细步骤
先裂项
f(z)=z/(z+1)(z+2)=-1/(1+z)+2/(2+z)
再根据需要变项
f(z)=-1/(3+z-2)+2/(4+z-2)=(-1/3){1/[1-[(-1)(z-2)/3]}+(1/2){1/[1-[(-1)(z-2)/4]}
再展开,z-2的绝对值小于3(因为-1是奇点)
f(z)=sigma{(-1/3)[(-1)(z-2)/3]^n}+(1/2)[(-1)(z-2)/4]^n}
sigma是求和号,从n=0到无穷
f(z)=z/(z+1)(z+2)在z0=2处展开成泰勒级数,要详细步骤
f(z)=2z+(共轭z)-3i,f(z0共轭+i)=6-3i,求f(-z0)注:共轭z表示z的共轭复数
1、 求1/z(4-3z)在z0=1+i展开成泰勒级数的收敛半径.2、z=0是f(z)=1/(e^z-1)-1/z的何种类型的奇点?
求f(z)=1/z^2,在z0=-1处的泰勒展开式,及收敛半径.
已知复数z0=3+2i,复数z满足z+z0 =3z+z0,则复数z=
已知复数z0=3+2i,复数z满足z·z0=3z+z0,求复数z
泰勒级数,z/(z+2),z0=1,按z0展开,写出收敛半径
F(Z)=1/(Z-1)(z-2) 在Z=1处的泰勒展开式
f(Z)=1/z(z+1)(z+4)在2
将函数f(z)= 1/[(z-1)(z-2)]在|z|
将函数f(z)= 1/[(z-1)(z-2)]在|z|
把F(z)=1/z(z-1)在1
求函数f(z)=z/(z-1)(z+3)^2在z=1处的留数.
复数x0=3+2i,复数z满足z*z0 =3z+z0,则复数z=
f'(z)=(z+1)'(2-z)+(z+1)(2-z)'如何计算
将函数f(z)=1/(z^3+1),在Z0=0展开成泰勒级数
f(z)=z^2/{(z^2+1)*(z^2+9)}求Res(z=i)f(z)和Res(z=3i)f(z)
如果f(z)与g(z)是以z0为零点的两个不恒为0的解析函数,证明 lim(z->z0)f(z)/g(z)=lim(z->zo)f'(z)/g'(z)或两端均为∞