Cheat Sheet Template
Author
stanlrt
Last Updated
6 месяцев назад
License
Creative Commons CC BY 4.0
Аннотация
A 3-column cheat sheet template
Improved version of: https://www.overleaf.com/articles/130-cheat-sheet/ntwtkmpxmgrp
\documentclass{article}
\usepackage{dependencies, formatting, cheatsheet}
% To be used for normal cheatbox content
\newcommand{\textStrech}{1.4}
% To be used with environments (e.g. amsmath's align) that have spacing issues
\newcommand{\mathStrech}{0.1}
\begin{document}
\graphicspath{{figures/}}
% Arg 1: The title of the cheat sheet. Can be empty.
% Arg 2: The spacing between boxes. Must be a distance unit.
\begin{cheatsheet}{Cheat Sheet Template}{1pt}
% Arg 1: The cheat box title. Can be empty.
% Arg 2: The stretch (used to fix padding uses with packages).
% Arg 3: The box's contents
\cheatbox{Physical Layer}{\textStrech}{
$\frac{dT}{dt} = -k(T-T_o)$ \\
$ T_o =$ outside temperature
% Arg 1: The width
% Arg 2: The filename
% Arg 3: The caption. Can be empty.
\cheatboxfigure{0.4}{colorful-design-with-spiral-design_188544-9588.jpg}{A colourful spiral}
}
\cheatbox{Inner Product Spaces}{\textStrech}{
1. $ \langle v,v \rangle \geq 0$ Furthermore, $ \langle v,v \rangle = 0 \leftrightarrow v = 0 $ \\
2. $\langle v,u \rangle = \langle u,v \rangle $ \\
3. $ \langle ku,v \rangle =k\langle u,v \rangle $ \\
4. $\langle u+v,w\rangle = \langle u,w \rangle + \langle v,w\rangle $ \\
$ ||v|| = \langle v, v \rangle $ \\
$ \cos^{-1}(\frac{\langle v,u \rangle}{||v|| ||u||}) $
}
\cheatbox{Gram-Schmidt}{\mathStrech}{
\begin{align*}
v_{1} &= x_{1} \\
v_{2} &= x_{2} - \frac{\langle x_2, v_1\rangle}{||v_{1}||^2}v_{1}\\
&\shortvdotswithin{=}
v_{n} &= x_{m} - \sum_{k=1}^{m-1} \frac{\langle x_{m},v_{k} \rangle}{||v_{k}||^{2}}v_{k}
\end{align*}
}
\cheatbox{Variation of Parameters}{\mathStrech}{
\begin{align*}
F(x) &= y'' + y' \\
y_h &= b_1y_1(x) + b_2y_2(x), y_1 y_2 \text{ are L.I.} \\
y_p &= u_1(x)y_1(x) + u_2(x)y_2(x) \\
u_1 &= \int^t -\frac{y_2F(t)dt}{w[y_1,y_2](t)} \\
u_2 &= \int^t \frac{y_1F(t)dt}{w[y_1,y_2](t)} \\
y &= y_h + y_p
\end{align*}
}
\cheatbox{Systems}{\textStrech}{
\begin{tabular}{lp{4cm} l}
$\vec{x}' = A\vec{x}$ \\
\textit{A is diagonalizable} & $\vec{x}(t)=a_{1}e^{\lambda_1 t}\vec{v_1}+\cdots+ a_{n}e^{\lambda_n t}\vec{v_n}$ \\ \hline
\textit{A is not diagonalizable} & $\vec{x}(t)=a_1e^{\lambda_1 t}\vec{v_1} + a_2e^{\lambda t}(\vec{w} + t\vec{v} )$ \\
& where $(A - \lambda I)\vec{w} = \vec{v} $\\
& $\vec{v}$ is an Eigenvector w/ value $\lambda$ \\
& i.e. $\vec{w}$ is a generalized Eigenvector \\ \hline
$\vec{x}' = A\vec{x} + \vec{B}$ &Solve $y_h$ \\
& $\vec{x_1} = e^{\lambda_1t}\vec{v_1}, \vec{x_2} = e^{\lambda_2t}\vec{v_2}$ \\ & $\vec{X} = [\vec{x_1},\vec{x_2}]$ \\
& $\vec{X}\vec{u}'=\vec{B}$ \\
& $y_p = \vec{X}\vec{u}$ \\
& $y = y_h + y_p$
\end{tabular}
}
\cheatbox{ODEs}{\textStrech}{
\tiny{
\begin{tabular}{lp{4cm} l}
\textit{1st Order Linear} & Use integrating factor,
\\ & $I = e^{\int P(x) dx}$ \\ \hline
\textit{Separable:} & $ \int P(y) dy/dx = \int Q(x) $ \\ \hline
\textit{HomogEnEous:} & $ dy/dx = f(x,y) = f(xt,yt) $ \\ &
sub $ y = xV $ solve, then sub $ V = y/x $ \\ \hline
\textit{Exact:} & If $ M(x,y) + N(x,y)dy/dx = 0 $ and $ M_y = N_x $ i.e. $ \langle M,N \rangle = \nabla F $ then $ \int_x M + \int_y N = F $ \\ \hline
\textit{Order Reduction} & Let $ v = dy/dx $ then check other types \\
&\textit{If purely a function of y, }$\frac{dv}{dx} = v\frac{dv}{dy}$\\
\hline
\textit{Variation of Parameters:} & When $y''+a_1y'+a_2y = F(x)$ \\
& $F$ contains $\ln x$, $\sec x$, $\tan x$, $\div$ \\ \hline
\textit{Bernoulli} & $y' + P(x)y = Q(x)y^n$ \\
& $\div y^n$ \\
&$y^{-n}y'+P(x)y^{1-n}=Q(x)$ \textit{Let }$U(x) = y^{1-n}(x)$ \\
&$\frac{dU}{dx}=(1-n)y^{-n}\frac{dy}{dx}$ \\
&$\frac{1}{1-n}\frac{du}{dx} + P(x)U(x) = Q(x)$ \textit{solve as a 1st order} \\ \hline
\textit{Cauchy-Euler} &$x^ny^n + a_1x^{n-1}y^{n-1} + \cdots + a_{n-1}y^{n-2}+a_ny = 0$ \\
&guess $y = x^r$ \\
\textit{3 Cases:} \\
\textit{1) Distinct real roots} &$y = ax^{r_1}+bx^{r_2}$ \\
\textit{2) Repeated real roots} &$y = Ax^r + y_2$ \\
&\textit{Guess} $y_2 = x^ru(x)$ \\
&\textit{Solve for $u(x)$ and choose one ($A=1, C=0$)} \\
\textit{3) Distinct complex roots} &$y=B_1x^a \cos (b \ln x) + B_2x^a\sin (b \ln x)$
\end{tabular}}
}
\cheatbox{Matrix Exponentiation}{\textStrech}{
\begin{tabular}{lp{4cm} l}
$A^n = SD^nS^{-1}$ \\
\textit{D is the diagonalization of A}
\end{tabular}
}
\cheatbox{Complex Numbers}{\textStrech}{
\begin{tabular}{lp{4cm} l}
\textit{Systems of equations} & If $\vec{w_1} = \vec{u(t)} + i\vec{v(t)}$ is a solution, $\vec{x_1} = \vec{u(t)}, \vec{x_2} = \vec{v(t)}$ are solutions \\
& i.e. $\vec{x_h} = c_1 \vec{x_1} + c_2 \vec{x_2}$ \\
\hline
\textit{Euler's Identity} &$e^{ix} = \cos x + i \sin x$
\end{tabular}
}
\cheatbox{Laplace Transforms}{\textStrech}{
$L[f](s) = \int_0^{\infty} e^{-sx}f(x)dx $\\
\begin{tabular}{lp{4cm} l}
$f(t) = t^n, n \geq 0 $ &$F(s) = \frac{n!}{s^{n+1}}, s > 0 $ \\
$f(t) = e^{at}, a \textit{ constant}$ & $ F(s) = \frac{1}{s-a}, s > a$ \\
$f(t) = \sin{bt}, b \textit{ constant}$ & $ F(s) = \frac{b}{s^2 + b^2}, s > 0$ \\
$f(t) = \cos{bt}, b \textit{ constant}$ & $ F(s) = \frac{s}{s^2 + b^2}, s > 0$ \\
$f(t) = t^{-1/2}$ & $F(s) = \frac{\pi}{s^{1/2}}, s > 0$ \\
$f(t) = \delta(t-a)$ & $F(s) = e^{-as}$ \\
$f'$ & $L[f'] = sL[f] - f(0)$ \\
$f''$ & $L[f''] = s^2 L[f] - sf(0) - f'(0)$ \\
$L[e^{at}f(t)]$ & $L[f](s-a)$ \\
$L[u_a(t)f(t-a)]$ & $L[f]e^{-as}$ \\
\end{tabular}
}
\columnbreak
\cheatbox{Gaussian Integral}{\textStrech}{
$\int_{-\infty}^{+\infty} e^{-1/2(\vec{x}^TA\vec{x})} = \frac{\sqrt{2\pi}^n}{\sqrt{\det A}}$
}
\cheatbox{Vector Spaces}{\textStrech}{
$v_1, v_2 \in V$\\
1. $v_1 + v_2 \in V$ \\
2. $k \in \mathbb{F}, kv_1 \in V $ \\
3. $ v_1 + v_2 = v_2 + v_1 $ \\
4. $(v_1 + v_2) + v_3 = v_1 + (v_2 + v_3) $ \\
5. $\forall v \in V, 0 \in V \mid 0 + v_1 = v_1 + 0 = v_1$ \\
6. $\forall v \in V, \exists -v \in V \mid v + (-v) = (-v) + v = 0 $ \\
7. $\forall v \in V, 1 \in \mathbb{F} \mid 1*v = v$ \\
8. $\forall v \in V, k,l \in \mathbb{F}, (kl)v = k (lv)$ \\
9. $\forall k \in \mathbb{F}, k(v_1 + v_2) = kv_1 + kv_2$ \\
10. $\forall v \in V, k,l \in \mathbb{F}, (k+l)v = kv + lv$
}
\end{cheatsheet}
\end{document}