求arctan(2x/2-x^2)的麦克劳林级数,求详解
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/24 16:56:21
求arctan(2x/2-x^2)的麦克劳林级数,求详解
![求arctan(2x/2-x^2)的麦克劳林级数,求详解](/uploads/image/z/5409382-22-2.jpg?t=%E6%B1%82arctan%282x%2F2-x%5E2%29%E7%9A%84%E9%BA%A6%E5%85%8B%E5%8A%B3%E6%9E%97%E7%BA%A7%E6%95%B0%2C%E6%B1%82%E8%AF%A6%E8%A7%A3)
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.
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.