f(x)=1/x在x=3的泰勒展开式为?
问题描述：
f(x)=1/x在x=3的泰勒展开式为?
答
f(x)=1/x=1/(x-3+3)=(1/3)/[1+(x-3)/3]
=(1/3)*{1-(x-3)/3+(x-3)^2/3^2+...+(-1)^n*(x-3)^n/3^n};n=0,1,...
答
∑(-1)^n*(x-3)^n/3^(n+1)+o((x-3)^n)
答
f(x)=1/x
= 1/[(x-3)+3]
= 1/3*1/[1+(x-3)/3]
= 1/3*∑(n=0,+∞) (-1)^n*[(x-3)/3]^n
= ∑(n=0,+∞) (-1)^n/3^(n+1)*(x-3)^n