题目

求e^x*sinx展开为x的幂级数
如题,要详解

提问时间：2020-10-09

答案

如果学了Euler恒等式e^(ix) = cos(x)+isin(x),那么可以有sin(x) = (e^(ix)-e^(-ix))/(2i).
于是e^x·sin(x) = (e^((1+i)x)-e^((1-i)x))/(2i).
e^x = ∑{n ≥ 0} x^n/n!.
故e^((1+i)x) = ∑{n ≥ 0} (1+i)^n·x^n/n!.
e^((1-i)x) = ∑{n ≥ 0} (1-i)^n·x^n/n!.
注意到1+i = √2·e^(πi/4),1-i = √2·e^(-πi/4).
于是(1+i)^n-(1-i)^n = √2^n·(e^(nπi/4)-e^(-nπi/4)) = √2^n·2i·sin(nπ/4).
代回得到e^x·sin(x) = ∑{n ≥ 0} √2^n·sin(nπ/4)·x^n/n!.
没学Euler恒等式就直接求导.
设f(x) = e^x·sin(x),则f'(x) = e^x·sin(x)+e^x·cos(x) = √2e^x·sin(x+π/4).
可得f'(x) = √2·e^(-π/4)·f(x+π/4).
求导得f"(x) = √2·e^(-π/4)·f‘(x+π/4) = √2^2·e^(-2π/4)·f(x+2π/4).
依次类推得f^(n)(x) = √2^n·e^(-nπ/4)·f(x+nπ/4) = √2^n·e^x·sin(x+nπ/4) (也可直接用归纳法证明).
于是f^(n)(0) = √2^n·sin(nπ/4).
e^x·sin(x) = ∑{n ≥ 0} √2^n·sin(nπ/4)·x^n/n!.