泰勒展开公式意义（泰勒展开公式是什么）

几个常见的泰勒公式 (x→0)(x\rightarrow0) ：

sinx=x−x36+o(x3)arcsinx=x+x36+o(x3)sinx = x -\frac{x^3}{6} +o(x^3)\qquad \qquad \quad \ \ arcsinx=x+\frac{x^3}{6}+o(x^3)

cosx=1−x22+x424+o(x4)arccosx=?cosx=1-\frac{x^2}{2}+\frac{x^4}{24}+o(x^4)\qquad \quad arccosx=? [1]

tanx=x+x33+o(x3)arctanx=x−x33+o(x3)tanx = x +\frac{x^3}{3}+o(x^3)\qquad \qquad \quad \ arctanx=x-\frac{x^3}{3}+o(x^3)

ex=1+x+x22!+x33!+o(x3)ln(1+x)=x−x22+x33+o(x3)e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+o(x^3) \qquad ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3)

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

另外

对于当，则当，则对于(1+x)α=1+αx+α(α−1)2x2+o(x2)当α=12，则1+x=1+12x−18x2+o(x2)当α=13，则1+x3=1+13x−19x2+o(x2)\begin{align} &对于 (1+x)^{\alpha}=1+\alpha x+\frac{\alpha(\alpha-1)}{2}x^2+o(x^2) \\ &\text{当}\alpha =\frac{1}{2}\text{，则}\sqrt{1+x}=1+\frac{1}{2}x-\frac{1}{8}x^2+o\left( x^2 \right) \\ &\text{当}\alpha =\frac{1}{3}\text{，则}\sqrt[3]{1+x}=1+\frac{1}{3}x-\frac{1}{9}x^2+o\left( x^2 \right) \end{align}

习题中常见 (x→0)(x \rightarrow 0) ：

tanx−sinx=12x3+o(x3)x−sinx=16x3+o(x3)arcsinx−x=16x3+o(x3)tanx−x=13x3+o(x3)x−arctanx=13x3+o(x3)\begin{align} tanx - sinx &= \frac{1}{2}x^3+o(x^3)\\ x - sinx &= \frac{1}{6}x^3+o(x^3)\\ arcsinx - x &= \frac{1}{6}x^3+o(x^3)\\ tanx - x &= \frac{1}{3}x^3+o(x^3)\\ x-arctanx &=\frac{1}{3}x^3+o(x^3) \end{align}

即有

tanx−sinx∼12x3x−sinx∼16x3arcsinx−x∼16x3tanx−x∼13x3x−arctanx∼13x3\begin{align*} tanx - sinx &\sim \frac{1}{2}x^3\\ x - sinx &\sim \frac{1}{6}x^3\\ arcsinx - x &\sim \frac{1}{6}x^3\\ tanx - x &\sim \frac{1}{3}x^3\\ x-arctanx &\sim\frac{1}{3}x^3 \end{align*}

还可以得到 (x→0)(x\rightarrow0) ：

当且x−ln⁡(1+x)∼x22ex−1−x∼x221−cosa⁡x∼ax22f(x)g(x)−1∼g(x)[f(x)−1](当f(x)→1且f(x)g(x)→1)\begin{align} x-\ln \left( 1+x \right) \,&\sim \frac{x^2}{2} \\ e^x-1-x\,&\sim \frac{x^2}{2} \\ 1-\cos ^ax\ &\sim \frac{ax^2}{2} \\ f\left( x \right) ^{g\left( x \right)}-1 &\sim g\left( x \right) \left[ f\left( x \right) -1 \right] \qquad \left( 当f\left( x \right) \rightarrow 1\text{且}f\left( x \right) ^{g\left( x \right)}\rightarrow 1 \right) \end{align}

注：上述四结论来自：

Takatomon：利用等价无穷小解决一些简单极限116 赞同 · 21 评论文章

有时还会用到

(1+x)1x=e−e2x+11e24x2+o(x2)\left( 1+x \right) ^{\frac{1}{x}}=e-\frac{e}{2}x+\frac{11e}{24}{x^2}+o\left( x^2 \right) [2]

一般地

ex=∑n=0∞xnn!=1+x+x22!+⋯+xnn!xn+⋯sinx=∑n=0∞(−1)n(2n+1)!x2n+1=x−x33!+x55!−⋯+(−1)n(2n+1)!x2n+1+⋯cosx=∑n=0∞(−1)n(2n)!x2n=1−x22!+x44!−⋯+(−1)n(2n)!x2n+⋯ln(1+x)=∑n=0∞(−1)nn+1xn+1=x−12x2+13x3−⋯+(−1)nn+1xn+1+⋯,x∈(−1,1]11−x=∑n=0∞xn=1+x+x2+x3+⋯+xn+⋯,x∈(−1,1)11+x=∑n=0∞(−1)nxn=1−x+x2−x3+⋯+(−1)nxn+⋯,x∈(−1,1)(1+x)α=1+∑n=1∞α(α−1)⋯(α−n+1)n!xn=1+αx+α(α−1)2!x2+⋯+α(α−1)…(α−n+1)n!xn+⋯,x∈(−1,1)arctan⁡x=∑n=0∞(−1)n2n+1x2π+1=x−13x3+15x5+⋯+(−1)n2n+1x2n+1+⋯,x∈[−1,1]\begin{align} e^{x}&=\sum_{n=0}^{\infty} \frac{x^{n}}{n!} =1+x+\frac{x^{2}}{2 !} +\cdots+\frac{x^{n}}{n!} x^{n}+\cdots \\ \ sin x&=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1) !} x^{2 n+1}=x-\frac{x^{3}}{3 !} +\frac{x^{5}}{5!} -\cdots+\frac{(-1)^{n}}{(2 n+1) !} x^{2 n+1}+\cdots\\ \ cos x&=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n) !} x^{2 n}=1-\frac{x^{2}}{2!} +\frac{x^{4}}{4!} -\cdots+\frac{(-1)^{n}}{(2n)!} x^{2n}+\cdots \\ \ ln (1+x)&=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+1} x^{n+1}=x-\frac{1}{2} x^{2}+\frac{1}{3} x^{3}-\cdots+\frac{(-1)^{n}}{n+1} x^{n+1}+\cdots, x \in(-1,1] \\ \frac{1}{1-x}&=\sum_{n=0}^{\infty} x^{n}=1+x+x^{2}+x^{3}+\cdots+x^{n}+\cdots, x \in(-1,1) \\ \frac{1}{1+x} &= \sum_{n = 0}^{\infty}(-1)^{n} x^{n} = 1-x+x^{2}-x^{3}+\cdots+(-1)^{n} x^{n}+\cdots, x \in(-1,1) \\ (1+x)^{\alpha} &= 1+\sum_{n = 1}^{\infty} \frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{n !} x^{n} = 1+\alpha x+\frac{\alpha(\alpha-1)}{2 !} x^{2}+\cdots+\frac{\alpha(\alpha-1) \ldots(\alpha-n+1)}{n !} x^{n}+\cdots, x \in(-1,1) \\ \arctan x &= \sum_{n = 0}^{\infty} \frac{(-1)^{n}}{2 n+1} x^{2 \pi+1} = x-\frac{1}{3} x^{3}+\frac{1}{5} x^{5}+\cdots+\frac{(-1)^{n}}{2 n+1} x^{2 n+1}+\cdots, x \in[-1,1] \\ \end{align}

arcsin⁡x=∑n=0∞(2n!)x2n+14n(n!)2(2n+1)=x+16x3+340x5+5112x7+351152x2+⋯+(2n)!4n(n!)2(2n+1)x2n+1+⋯,x∈(−1,1)tan⁡x=∑n=1∞B2n4n(4n−1)(2n)!x2n−1=x+13x3+215x5+17315x7+622835x9+1382155925x11+218446081075x13+929569638512875x15+⋯,x∈(−1,1)sec⁡x=∑π=0∞(−1)nE2nx2n(2n)!=1+12x2+524x4+61720x6+⋯,x∈(−π2,π2)csc⁡x=∑n=0∞(−1)n+12(22n−1−1)B2n(2n)!x2x−1=1x+16x+7360x3+3115120x5+127604800x7+733421440x2+141447765383718400x11+⋯,x∈(0,π)cot⁡x=∑n=0∞(−1)n22nB2n(2n)!x2n−1=1x−13x−145x3−2945x5−⋯,x∈(0,π){\LARGE \begin{align} \arcsin x &= \sum_{n = 0}^{\infty} \frac{(2 n!)x^{2n+1}}{4^{n}(n !)^{2}(2 n+1)} = x+\frac{1}{6} x^{3}+\frac{3}{40} x^{5}+\frac{5}{112} x^{7}+\frac{35}{1152} x^{2}+\cdots+\frac{(2 n) !}{4^{n}(n !)^{2}(2 n+1)} x^{2 n+1}+\cdots, x \in(-1,1) \\ \tan x &= \sum_{n = 1}^{\infty} \frac{B_{2n}4^{n}(4^{n}-1)}{(2 n) !} x^{2 n-1} = x+\frac{1}{3} x^{3}+\frac{2}{15} x^{5}+\frac{17}{315} x^{7}+\frac{62}{2835} x^{9}+\frac{1382}{155925} x^{11}+\frac{21844}{6081075} x^{13}+\frac{929569}{638512875} x^{15}+\cdots ,x \in(-1,1) \\ \sec x &= \sum_{\pi = 0}^{\infty} \frac{(-1)^{n}E_{2n} x^{2 n}}{(2 n) !} = 1+\frac{1}{2} x^{2}+\frac{5}{24} x^{4}+\frac{61}{720} x^{6}+\cdots, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\\ \csc x &= \sum_{n = 0}^{\infty} \frac{(-1)^{n+1} 2\left(2^{2 \mathrm{n}-1}-1\right) B_{2n}}{(2 n) !} x^{2 x-1} = \frac{1}{x}+\frac{1}{6} x+\frac{7}{360} x^{3}+\frac{31}{15120} x^{5}+\frac{127}{604800} x^{7}+\frac{73}{3421440} x^{2}+\frac{1414477}{65383718400} x^{11}+\cdots, x \in(0, \pi)\\ \cot x &= \sum_{n = 0}^{\infty} \frac{(-1)^{n} 2^{2n} B_{2n}}{(2 n) !} x^{2 n-1} = \frac{1}{x}-\frac{1}{3} x-\frac{1}{45} x^{3}-\frac{2}{945} x^{5}-\cdots, x \in(0, \pi) \end{align}}

相关链接：

sinX：如何记忆泰勒展开式41 赞同 · 2 评论文章tan(x)的泰勒展开有通项公式吗？848 赞同 · 0 评论回答arcsinX和tanX的泰勒展开有没有通项公式？434 赞同 · 15 评论回答请问这四个展开式是怎么来的？69 赞同 · 7 评论回答

参考

^利用arccosx = pi/2 - arcsinx即可得出。^https://zhuanlan.zhihu.com/p/427747801