xe^x展开为x-1的幂级数

2025-03-11 00:01:57
推荐回答(1个)
回答1:

Let x-1=t. Then we have x e^x =(t+1) e^(t+1)=e(t+1)e^t=e(t+1)\sum t^n/n! =e\sum (1/n!+1/(n-1)!) t^n.