Question

ezsinz展开为z的幂级数

您可不可以帮我把e^(z/z-1)展开成z的幂级数?
我只会解e^z或者sinz这种问题,对于我刚才提的问题及e^(z^2)*sinz^2这种难度的就不会解了,

Answer 1

(1)e^(z/(z-1))无法给出通式
1.
e^(z/(z-1))=e^(1+1/(z-1))可以按照泰勒展开
令[e^(1+1/(z-1))](n)'代表n次导数
那么[e^(1+1/(z-1))](1)'=[e^(1+1/(z-1))]*[-1/(z-1)^2]
[e^(1+1/(z-1))](2)'=[e^(1+1/(z-1))]*[1/(z-1)^4]+[e^(1+1/(z-1))]*[2/(z-1)^3]=
=[e^(1+1/(z-1))]*[(2z-1)/(z-1)^4]
[e^(1+1/(z-1))](3)'= (e^(z/(z-1)) (-6 z^2+6 z-1))/(z-1)^6
...
求出每个在x=0的值
得到
1-z-z^2/2-z^3/6+z^4/24+(19 z^5)/120+O(z^6)
2.
e^x=1+x+x^2/2+x^3/3!+...
e^(z/(z-1))=1+z/(z-1)+(z/(z-1))^2/2+(z/(z-1))^3/3+...
又z/(z-1)= -z/(1-z)=-z-z^2-z^3-z^4+...
[z/(z-1)]^n=(-z)^n * (1+c(n,1)z+(c(n,1)+c(n,2))z^2+
(c(n,1)+c(n,2)+c(n,3))z^3+...)
令bni=c(n,1)+c(n,2)+...+c(n,i)
对应的有[z/(z-1)]^n=(-z)^n * (1+bn1*z+bn2*z^2+...)
e^(z/(z-1))=1+z/(z-1)+(z/(z-1))^2/2+(z/(z-1))^3/3+...=
=1-z(1+b11*z+b12*z^2+...)+z^(1+b21*z+b22*z^2+...)+...
对应的z^n的系数是(+b1n+b2n+b3n+...+bnn)/n!
这里只是抽象的写出系数的解析式,并不具有实际意义.
那么还是通过求值得
1-z-z^2/2-z^3/6+z^4/24+(19 z^5)/120+O(z^6)
(2)
e^(z^2)*sinz^2反而比较容易
按照e^x=1+x+x^2/2+...展开得
e^(z^2)=1+z^2+z^4/2+...
sin(z^2)=z^2-z^6/3!+z^10/5!...
e^(z^2)*sin(z^2)=z^2+z^4+z^6/3-z^10/30-z^12/90-z^14/630...
想不出有什么好的方法.