Review Note
Last Update: 10/05/2024 09:24 AM
Current Deck: 25考研高等数学公式+概念+定理+典例【5.0版】【数二版】【latex精制版】::02函数，极限，连续::02极限::06求极限的基本方法::05利用泰勒公式求极限
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问题
| 泰勒公式表格 |
| 定理 (带 Peano 余项的泰勒公式) 设 \(f(x)\) 在 \(x=x_{0}\) 处 \(n\) 阶可导, 则 \[f(x)=\text{____}\] 特别是当\(x_{0}=0\) 时, \[f(x)=\text{____}\] |
| \[e^{x} \stackrel{\text { 展开 } }= \text{____}\stackrel{\text { 求和 } }=\text{____}\] |
| \[\sin x \stackrel{\text { 展开 } } =\text{____} \stackrel{\text { 求和} }=\text{____}\] |
| \[\cos x \stackrel{\text { 展开 } } =\text{____} \stackrel{\text { 求和} } =\text{____} \] |
| \[\ln (1+x) \stackrel{\text { 展开} }=\text{____} \stackrel{\text { 求和 } }=\text{____}\] |
| \(\begin{align}\frac{1}{1-x}\stackrel{\text{展开} }=\text{____}\stackrel{\text { 求和 } }=\text{____}\end{align}\)|x|<1(等比数列，公比为x) |
| \(\begin{align} \frac{1}{1+x}\stackrel{\text{展开} }=\text{____}\stackrel{\text { 求和 } }=\text{____}\end{align}\) |x|<1（等比数列，公比为-x） |
| \(\begin{align}(1+x)^{\alpha}=\text{____}\end{align}\) |
| \(\tan x=\text{____}\) |
| \(\arcsin x=\text{____}\)做题一般也是用到三阶 |
| \(\arctan x=\text{____}\) |
| \(\ln \left( x+\sqrt{1+x^2} \right) =\text{____} \) |
| \((1+x)^{\frac{1}{x} }= \text{____} \left( x\rightarrow 0 \right) \) |
| \(\begin{aligned}tan(tan x)=\text{____}\end{aligned}\) |
\(\begin{aligned} sin(sinx)=\text{____}\end{aligned}\) |
答案
| 泰勒公式表格 |
| 定理 (带 Peano 余项的泰勒公式) 设 \(f(x)\) 在 \(x=x_{0}\) 处 \(n\) 阶可导, 则 \[f(x)=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)+\frac{f^{\prime \prime}\left(x_{0}\right)}{2 !}\left(x-x_{0}\right)^{2}+\cdots+\frac{f^{(n)}\left(x_{0}\right)}{n !}\left(x-x_{0}\right)^{n}+o\left(\left(x-x_{0}\right)^{n}\right)\] 特别是当\(x_{0}=0\) 时, \[f(x)=f(0)+f^{\prime}(0) x+\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\cdots+\frac{f^{(n)}(0)}{n !} x^{n}+o\left(x^{n}\right)\] |
| \[e^{x} \stackrel{\text { 展开 } }= 1+x+\frac{x^{2} }{2 !}+\frac{x^{3} }{3 !}+\cdots+\frac{x^{n} }{n !}+o\left(x^{n}\right) \stackrel{\text { 求和 } }=\displaystyle\sum_{n=0}^{\infty} \frac{x^{n} }{n !} \left( -\infty <x<+\infty \right) \] |
| \[\sin x \stackrel{\text { 展开 } } =x-\frac{1}{3 !} x^{3}+\frac{1}{5 !} x^{5}-\cdots+(-1)^{n} \frac{1}{(2 n+1) !} x^{2 n+1}+o\left(x^{2 n+1}\right) \stackrel{\text { 求和} }=\displaystyle\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n+1} }{(2 n+1) !}\left( -\infty <x<+\infty \right) \] |
| \[\cos x \stackrel{\text { 展开 } } =1-\frac{1}{2 !} x^{2}+\frac{1}{4 !} x^{4}-\cdots+(-1)^{n} \frac{1}{(2 n) !} x^{2 n}+o\left(x^{2 n}\right) \stackrel{\text { 求和} } =\displaystyle\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n} }{(2 n) !}\left( -\infty <x<+\infty \right) \] |
| \[\ln (1+x) \stackrel{\text { 展开} }=x-\frac{1}{2} x^{2}+\frac{1}{3} x^{3}-\cdots+(-1)^{n-1} \frac{x^{n} }{n}+o\left(x^{n}\right) \stackrel{\text { 求和 } }=\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1} \frac{x^{n} }{n} , \left( -1<x\leqslant 1 \right) \] |
| \(\begin{align}\frac{1}{1-x}\stackrel{\text{展开} }=1+x+x^{2}+\cdots+x^{n}+o\left(x^{n}\right) \stackrel{\text { 求和 } }=\displaystyle\sum_{n=0}^{\infty} x^{n}\left( -1<x<1 \right)\end{align}\)|x|<1(等比数列，公比为x) |
| \(\begin{align} \frac{1}{1+x}\stackrel{\text{展开} }=1-x+x^2-x^3+\cdots +\left( -1 \right) ^nx^n+o\left( x^n \right) \stackrel{\text { 求和 } }=\displaystyle\sum_{n=0}^{\infty}{\left( -1 \right) ^nx^n}\,\end{align}\) |x|<1（等比数列，公比为-x） |
| \(\begin{align}(1+x)^{\alpha}=1+\alpha x+\frac{\alpha(\alpha-1)}{2} x^{2}+\cdots+\frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{n !} x^{n}+o\left(x^{n}\right) \end{align}\) |
| \(\tan x=x+\frac{x^3}{3}+o\left( x^3 \right) \) |
| \(\arcsin x=x+\frac{1}{6} x^{3}+o\left(x^{3}\right)\)做题一般也是用到三阶 |
| \(\arctan x=x-\frac{x^{3} }{3}+o\left(x^{3}\right)\) |
| \(\ln \left( x+\sqrt{1+x^2} \right) =x-\frac{1}{6}x^3+o\left( x^3 \right) \left( x\rightarrow 0 \right) \) |
| \((1+x)^{\frac{1}{x} }= e-\frac{e}{2}x+\frac{11e}{24}x^2-\frac{7e}{16}x^3+o\left( x^3 \right) \left( x\rightarrow 0 \right) \) |
| \(\begin{aligned}tan(tan x)=&\underbrace{\tan x}_{\text {低阶展开 } }+\underbrace{\frac{\tan ^{3} x}{3} }_{\text {高阶等价 } }+o\left(\tan ^{3} x\right)=x+\frac{x^{3} }{3}+\frac{x^{3} }{3}+o\left(x^{3}\right) \\&=x+\frac{2}{3} x^{3}+o\left(x^{3}\right)\end{aligned}\) |
\(\begin{aligned} sin(sinx)=&\sin x-\frac{\sin ^{3} x}{6}+o\left(\sin ^{3} x\right)=x-\frac{x^{3} }{6}-\frac{x^{3} }{6} \\ &=x-\frac{1}{3} x^{3}+o\left(x^{3}\right) \end{aligned}\) |
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