Question

xe^x展开为x-1的幂级数

Answer 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.

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最后一步怎么合并的

喔喔 懂了 3q