求(sinx)的平方在x=0处的幂级数展开式,并确定它收敛于该函数的区间

2025-04-04 15:27:16
推荐回答(1个)
回答1:

(sinx)^2=(1-cos2x)/2
而cos2x=1-(2x)^2/2!+(2x)^4/4!-(2x)^6/6!+...., 收敛域为R
故(sinx)^2=1/2[4x^2/2!-2^4x^4/4!+2^6x^6/6!-..]
=x^2-2^3x^4/4!+2^5x^6/6!-....