求e^2x in (x-1)的泰勒级数?
1个回答
关注
![]()
展开全部
为方便，令t=x-1
e^2x in (x-1)=(t-1)e^(-t-1)
=1/e[te^(-t)-e^(-t)]
=1/e[t(1-t+t^2/2!-t^3/3!+...)-(1-t+t^2/2!-t^3/3!-..)]
=1/e[ -1+2t-t^2(1+1/2!)+t^3(1/2!+1/3!)-....-(-1)^nt^n(1/(n-1)!+1/n!)+...]
咨询记录 · 回答于2021-12-11
求e^2x in (x-1)的泰勒级数?
为方便，令t=x-1e^2x in (x-1)=(t-1)e^(-t-1)=1/e[te^(-t)-e^(-t)]=1/e[t(1-t+t^2/2!-t^3/3!+...)-(1-t+t^2/2!-t^3/3!-..)]=1/e[ -1+2t-t^2(1+1/2!)+t^3(1/2!+1/3!)-....-(-1)^nt^n(1/(n-1)!+1/n!)+...]
为方便，令t=x-1e^2x in (x-1)=(t-1)e^(-t-1)=1/e[te^(-t)-e^(-t)]=1/e[t(1-t+t^2/2!-t^3/3!+...)-(1-t+t^2/2!-t^3/3!-..)]=1/e[ -1+2t-t^2(1+1/2!)+t^3(1/2!+1/3!)-....-(-1)^nt^n(1/(n-1)!+1/n!)+...]
已赞过
评论
收起
你对这个回答的评价是？