幂级数的展开
发布网友
发布时间：2022-04-25 22:12
共1个回答
热心网友
时间：2022-06-18 00:52
f(x)= 1/(x+1) =>f(1)=1/2
f'(x)= -1/(x+1)^2 =>f'(1)/1!=-1/4
f''(x)= 2/(x+1)^3 =>f''(1)/2!=1/8
...
f^(n)(x) = (-1)^n.n!/(x+1)^(n+1) => f^(n)(1)/n! = (-1)^n/2^(n+1)
1/(1+x)
=f(x)
=f(1) +[f'(1)/1!](x-1) +[f''(1)/2!](x-1)^2 +...+[f^(n)(1)/n!](x-1)^n+...
=1/2 -(1/4)(x-1) +(1/8)(x-1)^2+....+[(-1)^n/2^(n+1)](x-1)^n +....
