常用公式¶
Wallis 公式¶
\[I_n = \int_0^{\frac{\pi}{2}}\sin^n xdx = \int_0^{\frac{\pi}{2}}\cos^n xdx=\begin{cases} \frac{(n-1)!!}{n!!}\frac{\pi}{2} & n \text{为偶数} \\ \frac{(n-1)!!}{n!!} & n \text{为奇数} \end{cases}\]
\[I_{m,n} = \int_0^{\frac{\pi}{2}}\sin^m x\cos^n xdx = \begin{cases} \frac{(m-1)!!(n-1)!!}{(m+n)!!}\frac{\pi}{2} & m,n \text{均为偶数} \\ \frac{(m-1)!!(n-1)!!}{(m+n)!!} & \text{其他} \end{cases}\]
积分,代换¶
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\[\int_0^\infty \frac{1}{1+x^4}dx = \int_0^\infty \frac{x^2}{1+x^4}dx = \frac{\pi}{2\sqrt{2}}\]
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\[d\left(x-\frac{1}{x} \right) = \left(1+\frac{1}{x^2}\right)dx\]
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高斯积分
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\[\int_{-\infty}^\infty e^{-ax^2+bx}dx = \sqrt{\frac{\pi}{a}}e^{\frac{b^2}{4a}}\]
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\[\int_{-\infty}^{\infty} \sin x^2dx=\int_{-\infty}^{\infty} \cos x^2 dx=\sqrt{\frac{\pi}{2}}\]
- \(\sin x = \frac{e^{ix}-e^{-ix}}{2i}\) 和 高斯积分结合
级数, 代换¶
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\[a_n+2a_{n-1}+\cdots+na_1 =\sum \begin{pmatrix} a_1 & 0 & 0 & \cdots & 0 \\ a_1 & a_2 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_1 & a_2 & a_3 & \cdots & a_n \end{pmatrix} = S_1 + S_2 + \cdots +S_n\]
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\[1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n} = \ln n + \gamma + o(1)\]
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若 \(a_n\to 0\), 则 \(S_{2n}\) 收敛到 \(S\), 意味着 \(S_{2n+1}\) 收敛到 \(S\) , 也就意味着 \(S_n\) 收敛到 \(S\)
- 可推广为 \(S_{pn}\) 收敛到 \(S\) 意味着 \(S_n\) 收敛到 \(S\)
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\[\sqrt[n]{n}-1\sim \ln[1+(\sqrt[n]{n}-1)]\sim \frac{\ln n}{n}\]
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\[1^k+2^k+\cdots+n^k\sim\frac1{k+1}n^{k+1}\]
其他代换¶
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\[\arctan x - \arctan y = \arctan \frac{x-y}{1+xy}\]
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\[\arctan\frac{1}{2k^2} = \arctan\frac{1}{2k-1}-\arctan\frac{1}{2k+1}\]
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\[\sqrt[n]{a_n} = \frac{a_{n+1}}{a_n}\]
等价代换¶
\[x\sim \sin x\sim \tan x\sim \arcsin x\sim \ln(x+1)\sim e^x-1\sim \frac{a^x-1}{\ln a}\sim \frac{(1+x)^b-1}{b}\]
Stirling 公式¶
\[n!\sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^ne^\frac{\theta_n}{12}\]
Vieta 公式¶
\[
\frac\pi2=\frac1{\sqrt{\frac12}\sqrt{\frac12+\frac12\sqrt{\frac12}}\sqrt{\frac12+\frac12\sqrt{\frac12+\frac12\sqrt{\frac12}}\cdots}}
\]
\[\frac{\sin x}{x}=\cos\frac{x}{2}\cos\frac{x}{4}\cos\frac{x}{8}\cdots\]
令 \(x=\frac{\pi}{2}\), 得到
\[\frac2\pi=\frac{\sqrt2}{2}\cdot\frac{\sqrt{2+\sqrt2}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt2}}}{2}\cdots\]
中值公式¶
[!NOTE] 需要连续性和可导性
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Lagrange 中值定理: \(f(b)-f(a)=f'(\xi)(b-a)\)
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Cauchy 中值定理: \(\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(\xi)}{g'(\xi)}\)
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Rolle 定理: 若\(f(a)=f(b)\), 则\(\exists \xi\in(a,b),s.t.f'(\xi)=0\)
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积分中值定理
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\(\int_a^bf(x)g(x)dx=f(\xi)\int_a^bg(x)dx\)
- \(\xi\in (a,b)\) 书上是取闭区间,但是可以强化成开区间
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\(g(x)\) 单增且 \(g(a)\geq 0\) ,则 \(\int_a^bg(x)f(x)dx=g(b)\int_\eta^bf(x)dx\)
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\(g(x)\) 单减且 \(g(b)\geq 0\) ,则 \(\int_a^bg(x)f(x)dx=g(a)\int_a^\eta f(x)dx\)
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\(g(x)\) 单调,则 \(\int_a^bg(x)f(x)dx=g(a)\int_a^\xi f(x)dx+g(b)\int_\xi^bf(x)dx\)
- 记忆方法,总之 \(g\geq 0\) 单调,取 \(g\) 的最大值点和靠近那个点的 \(f\) 的积分
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Cauchy 不等式¶
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\[\left(a_1^2+a_2^2+\cdots+a_n^2\right)\left(b_1^2+b_2^2+\cdots+b_n^2\right)\geq \left(a_1b_1+a_2b_2+\cdots+a_nb_n\right)^2\]
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\[\left(\sum\limits_{i=1}^{n} a_ib_i\right)^2\leq \left(\sum\limits_{i=1}^{n} a_i^2\right)\left(\sum\limits_{i=1}^{n} b_i^2\right)\]
Schwarz 不等式¶
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\[\left(\int_a^b f(x)g(x)dx\right)^2\leq \int_a^b f^2(x)dx\int_a^b g^2(x)dx\]
[!NOTE] 就是 Cauchy 不等式的积分形式
Holder 不等式¶
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\(\frac1p+\frac1q=1,p,q>1\)
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\[\sum\limits_{i=1}^{n} a_ib_i \leq \left(\sum\limits_{i=1}^{n} a_i^p\right)^{\frac1p}\left(\sum\limits_{i=1}^{n} b_i^q\right)^{\frac1q}\]
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\[\int_a^b f(x)g(x)dx\leq \left(\int_a^b f^p(x)dx\right)^{\frac1p}\left(\int_a^b g^q(x)dx\right)^{\frac1q}\]
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[!NOTE] \(p,q<1\) 的时候不等号反向
Minkowski 不等式¶
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\(r>1\)
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\[\left(\sum\limits_{i=1}^{n} (a_i+b_i)^r\right)^{\frac1r}\leq \left(\sum\limits_{i=1}^{n} a_i^r\right)^{\frac1r}+\left(\sum\limits_{i=1}^{n} b_i^r\right)^{\frac1r}\]
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\[\left(\int_a^b (f(x)+g(x))^r dx\right)^{\frac1r}\leq \left(\int_a^b f^r(x)dx\right)^{\frac1r}+\left(\int_a^b g^r(x)dx\right)^{\frac1r}\]
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\(r<1\)
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\[\left(\sum\limits_{i=1}^{n} (a_i+b_i)^r\right)^{\frac1r}\geq \left(\sum\limits_{i=1}^{n} a_i^r\right)^{\frac1r}+\left(\sum\limits_{i=1}^{n} b_i^r\right)^{\frac1r}\]
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推广到 \(n\) 元, \(r>1\)
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\[\left(\sum\limits_{i=1}^{n} \left(\sum\limits_{j=1}^{m} a_{ij}\right)^r\right)^{\frac1r}\leq \sum\limits_{j=1}^{m}\left(\sum\limits_{i=1}^{n} a_{ij}^r\right)^{\frac1r}\]
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Young 不等式¶
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\(f(x)\) 在 \((a,b)\) 连续且单增, \(f(0)=0,a,b>0\), 则
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\[ab\leq \int_0^a f(x)dx+\int_0^b f^{-1}(x)dx\]
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当且仅当 \(f(a)=b\) 时取得等号,其中 \(f^{-1}(x)\) 为 \(f(x)\) 的反函数
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