[注]:
泰勒公式才是以后解题的主流方法,洛必达(买来的)是辅助的;

8个泰勒展开公式熟稔于心:

x0 x \to 0 时:

sinx=x13!x3+o(x3)=x16x3+o(x3) sinx = x-\frac{1}{3!}x^3+o(x^3) = x-\frac{1}{6}x^3+o(x^3)

arcsinx=x+13!x3+o(x3)=x+16x3+o(x3) arcsinx = x+\frac{1}{3!}x^3+o(x^3) = x+\frac{1}{6}x^3+o(x^3)

tanx=x+13x3+o(x3) tanx = x+\frac{1}{3}x^3+o(x^3)

arctanx=x13x3+o(x3) arctanx = x-\frac{1}{3}x^3+o(x^3)

cosx=112!x2+14!x4+o(x4)=112x2+124x4+o(x4) cosx = 1-\frac{1}{2!}x^2+\frac{1}{4!}x^4+o(x^4) = 1-\frac{1}{2}x^2+\frac{1}{24}x^4+o(x^4)

ln(1+x)=x12x2+13x314x4+o(x4) ln(1+x) = x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+o(x^4)

ex=1+x+12!x2+13!x3+o(x3) e^x = 1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+o(x^3)

(1+x)α=1+αx+α(α1)2x2+o(x2) (1+x)^\alpha = 1 + \alpha x + \frac{\alpha (\alpha - 1)}{2}x^2 + o(x^2)

再补充2个:

11x=1+x+x2+x3+o(x3)   (x<1) \frac{1}{1-x} = 1+x+x^2+x^3+o(x^3) \space\space\space (|x|<1)

11+x=1x+x2x3+o(x3)   (x<1) \frac{1}{1+x} = 1-x+x^2-x^3+o(x^3) \space\space\space (|x|<1)

记忆技巧:

1、需要阶乘的: sinx,cosx,exsinx, cosx, e^x
小结:因为上面的sinxsinxcosxcosx直接把阶乘给运算了,所以只记exe^x带阶乘即可;
2、记住sinxsinx的,然后直接对sinxsinx求导即可得到conxconx的展开式;
3、ln(1+x)ln(1+x)x开头,正负相间
4、exe^x:e与1同音,所以记为1开头,全为正,需阶乘
5、记住11x\frac{1}{1-x}全正,然后将x-x带入即得11+x\frac{1}{1+x}的展开式;

笔记:

1、主部:上面9个泰勒展开式的头1个人就叫做主部,也是等价替换的人;
2、泰勒 (Taylor)

文档更新时间: 2019-10-06 15:23   作者:数学公式