Question

求arctan(2x/2-x^2)的麦克劳林级数,高手帮帮忙啊，求详解

Answer 1

arctan(2x/2-x^2)求导得：
1/[1+(4x^2/(2-x^2)^2]=(2-x^2)^2/[(2-x^2)^2+4x^2]=(4-4x^2+x^4)/(4+x^4)
=4/(4+x^4)-4x^2/(4+x^4)+x^4/(4+x^4)
=1/(1+x^4/4)-x^2/(1+x^4/4)+(1/4)*x^4/(1+x^4/4)
=Σ(-1)^nx^(4n)/4^n-x^2Σ(-1)^nx^(4n)/4^n+(1/4)*x^4Σ(-1)^nx^(4n)/4^n n从0到∞
=Σ(-1)^nx^(4n)/4^n-Σ(-1)^nx^(4n+2)/4^n+(1/4)Σ(-1)^nx^(4n+4)/4^n n从0到∞
然后从0到x作积分得：
arctan(2x/2-x^2)=Σ(-1)^nx^(4n+1)/[(4n+1)4^n]-Σ(-1)^nx^(4n+3)/[(4n+3)4^n]+(1/4)Σ(-1)^nx^(4n+5)/[(4n+5)4^n]
然后把三个级数合成一个级数就行了。