arctanx的麦克劳林展开式是什么?还有tanx的呢?

2024-11-29 04:01:36
推荐回答(5个)
回答1:

任意函数的迈克劳林展开式为

据此可以求得:

arctanx(x)=x-1/3*x^3+1/5*x^5-1/7*x^7+1/9*x^9+...+(-1)^(n+1)/(2n-1)*x^(2n-1)

tan(x)=x+1/3*x^3+2/15*x^5+17/315*x^7+62/2835*x^9+...+[2^(2n)*(2^(2n)-1)*B(2n-1)*x^(2n-1)]/(2n)!

回答2:

  1. arctanx(x)前五项是:x-1/3*x^3+1/5*x^5-1/7*x^7+1/9*x^9;

    第n项是 [(-1)^(n+1)*x^(2*n-1)]/(2*n-1);

    拉格朗日余项是:

    第n项是 (-1)^(n+1)*x^(2*n-1)/(2*n-1);

  2. tan(x)前五项是:x+1/3*x^3+2/15*x^5+17/315*x^7+62/2835*x^9; 


        拉格朗日的余项只要把最后一项f(x)的n+1次方导数换成f(ξ)的n+1次方导数就行了,其它的不变

回答3:

arctanx(x)=x-1/3*x^3+1/5*x^5-1/7*x^7+1/9*x^9

tan(x)=x+1/3*x^3+2/15*x^5+17/315*x^7+62/2835*x^9

回答4:

arctanx(x)前五项是:x-1/3*x^3+1/5*x^5-1/7*x^7+1/9*x^9;
tan(x)前五项是:x+1/3*x^3+2/15*x^5+17/315*x^7+62/2835*x^9;

回答5:

arctanx(x)=x-1/3*x^3+1/5*x^5-1/7*x^7+1/9*x^9
tan(x)=x+1/3*x^3+2/15*x^5+17/315*x^7+62/2835*x^9