高中/笔记

$$P\left(\frac{X_{1}+X_{2}+\cdots+X_{n}-n \mu}{\sigma \sqrt{n}} \leq x\right)\to \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{x} e^{-t^{2} / 2} d t$$ $$F=G \frac{m_{1} m_{2}}{r^{2}}$$ $$\frac{\partial^{2} y}{\partial t^{2}}=c^{2} \frac{\partial^{2} y}{\partial x^{2}}$$ $$V-E+F=2$$ $$\Delta_{\psi}(P) \Delta_{\psi}(X) \geq \hbar$$ $$H(S) \leq m(S) \leq H(S)+1$$ $$P \stackrel{?}{=} N P$$ $$i \hbar \frac{\partial \psi}{\partial t}=\frac{-\hbar^{2}}{2 m} \nabla^{2} \psi+V \psi$$ $$\frac{1}{\pi}=\frac{\sqrt{8}}{99^{2}} \sum_{n=0}^{\infty} \frac{(4 n) !}{(n !)^{4}} \frac{(1103+26390 n)}{396^{4 n}}$$ $$l^{\prime}=\frac{l}{\sqrt{1-\frac{v^{2}}{c^{2}} }}$$ $$\frac{d N}{d t}=r N\left(1-\frac{N}{K}\right)$$ $$P(A)=(k/N)^n p(B|A)=1-\sum_{i=1}^{k-1}\binom{k}{i}(1-i/k)^n(-1)^{i+1} P\{D=k\}=\binom{N}{k}P(AB)$$ $$\left\{ \begin{array}{l} U = \frac { Q } { C } = \frac { n q } { C } \\ { a = g - \frac { E q } { m } = g - \frac { U q } { d m } = g - \frac { n q ^ { 2 } } { d m C } } \\ { \frac { d } { 2 } = \frac { 1 } { 2 } a t ^ { 2 } = \frac { 1 } { 2 } ( g - \frac { n q ^ { 2 } } { d m C } ) \cdot ( \frac { L } { v _ { 0 } } ) ^ { 2 } } \end{array} \right .\Rightarrow n = ( g - \frac { d v _ { 0 } ^ { 2 } } { L ^ { 2 } } ) \cdot \frac { d m C } { q ^ { 2 } }$$ $$\oint IR=0$$ $$\left \{\begin{array}{l}x=x_0+t\cos\theta,\\y=y_0+t\sin\theta \end{array}\right .(\theta\in [0,\pi])$$ $$\rho =\frac{ep}{1-e\cos\theta}$$ $$\frac {x^2} {a^2}\pm \frac {y^2} {b^2}=1$$ $$\left\{ \begin{array}{l}m\frac{v^2}{r}=Bqv\\T=\frac {2\pi r} v \end{array}\right .\Rightarrow T=\frac{2\pi m} {Bq}\Rightarrow t=\frac{\theta m} {Bq}$$ $$\sum_{i = 1}^{k} i ^ {2} = \frac{k {(k + 1)} {(2 k + 1)}}{6}\\$$ $$\begin{array} {}\sum_\limits{i = 1}^{k+1}i^{2}&= \sum_\limits{i = 1}^{k} i ^ {2}+(k+1)^2\\&= \frac{k {(k + 1)} {(2 k + 1)+6(k+1)^2}}{6}\\&=\frac{(k + 1)(2k^2+7k+6)}{6}\\&=\frac{(k + 1)[(k+1)+1][2(k+1)+1]}{6}\end{array}$$ $$t=(1+x)^n-1\rightarrow0\\ \begin{array} {aligned}\lim_\limits{x \to 0} \frac{(1+x)^n-1}{x}(n\in R)&=\lim_\limits{x \to 0}[\frac{(1+x)^n-1}{\ln(1+x)^n}\cdot\frac{n\ln(1+x)}{x}]\\&=\lim_\limits{x \to 0} \frac{t}{\ln(1+t)}\cdot\lim_\limits{x \to 0}\frac{n\ln(1+x)}{x}\\&=\frac 1 e\cdot e\cdot n\\&=n\end{array}\\ (1+x)^\alpha\thicksim 1+\alpha x(x\rightarrow0)$$ $$(1+x)^n> 1+n x\;(n>1,n\in N^*)\\ ( 1 + x ) ^ { k } > 1 + k x\\ \begin{array} {aligned} ( 1 + x ) ^ { k + 1 } &= ( 1 + x ) ( 1 + x ) ^ { k } \\&>{ ( 1 + x ) ( 1 + k x ) }\\ &={ 1 + x + k x + k x ^ { 2 } }\\ &> {1 + ( k + 1 ) x }\end{array}$$ $$\int^{t}_{0}\mathrm{d}t =-\int_{v_0}^{v}\frac{mR}{B^2L^2v}\mathrm{d}v$$ $$\int \textrm{efforts}+\texttt{immaturity}+\textrm{d}\textbf{lucky}=Success$$ $$\frac{d}{dx}(f(x)+g(x))=\frac {df}{dx}+\frac{dg}{dx}$$ $$\int_{a}^{b}f(x)dx=F(b)-F(a)$$ $$-2^{w-1}\le x<2^{w-1}-1\qquad \textrm{and}\qquad-2^{w-1}\le y<2^{w-1}-1\qquad\\ x+y=\left\{\begin{array}{l}x+y-2^w,\quad x+y\ge 2^{w-1}\\x+y,\quad 2^{w-1}\le x+y<2^{w-1}\\x+y-2^w,\quad x+y<-2^{w-1}\end{array}\right.$$ $$\textrm{binary to unsigned}\\ B2U_w(\overrightarrow{x})=x_{w-1}\cdot 2^{w-1}+x_{w-2}\cdot 2^{w-2}+\cdot\cdot\cdot+x_0\cdot2^0=\sum_{i=0}^{w-1}x_i\cdot 2^i$$ $$\textrm{binary to two's complement representation}\\ B2U_w(\overrightarrow{x})=x_{w-1}\cdot -2^{w-1}+x_{w-2}\cdot 2^{w-2}+\cdot\cdot\cdot+x_0\cdot2^0=x_{w-1}\cdot -2^{w-1}+\sum_{i=0}^{w-2}x_i\cdot 2^i$$ $$x\cdot2^k=x<<k$$ $$x=d_md_{m-1}…d_1d_0.d_{-1}…d_{-(n-1)}d_{-n}\\ x=\sum_{i=-n}^{m}d_i\cdot10^i$$ $$float=>V=(-1)^s\times M\times2^E\\ E=e-bias(bias=2^{8-1}-1=127) \\e\in[1,255]\\E\in[-126,127]\\M=1+f=1.f_{22}f_{21…}f_1$$ $$\bar{F} =\frac{m_0}{l_1}\sum_{i=1}^{N}v_{ix}^2$$ $$p=n_0kTe^{-\frac{m_0gh}{kT}}$$ $$\rm Youth\;passes,but\;with\;luck\;immaturity\;can\;last\;a\;lifetime.$$