Bảng cheat toán học
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main.tex
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\begin{document}
\begin{center}
{\large\bfseries\color{accent} Calculus and Linear Algebra Cheat Sheet}
\quad {\small MATH 2410 Final \ | \ one card allowed}
\end{center}
\vspace{-4pt}
\begin{multicols}{4}
\footnotesize
\block{Trigonometric identities}
$\sin^2\theta + \cos^2\theta = 1$\\
$1 + \tan^2\theta = \sec^2\theta$\\
$\sin(a \pm b) = \sin a\cos b \pm \cos a\sin b$\\
$\cos(a \pm b) = \cos a\cos b \mp \sin a\sin b$\\
$\sin 2\theta = 2\sin\theta\cos\theta$\\
$\cos 2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$\\
$\sin^2\theta = \tfrac{1 - \cos 2\theta}{2}$, \ $\cos^2\theta = \tfrac{1 + \cos 2\theta}{2}$
\block{Derivatives}
$(x^n)' = nx^{n-1}$, \ $(e^{x})' = e^{x}$, \ $(\ln x)' = \tfrac1x$\\
$(\sin x)' = \cos x$, \ $(\cos x)' = -\sin x$\\
$(\tan x)' = \sec^2 x$, \ $(\sec x)' = \sec x\tan x$\\
$(\arcsin x)' = \tfrac{1}{\sqrt{1-x^2}}$, \ $(\arctan x)' = \tfrac{1}{1+x^2}$\\
$(fg)' = f'g + fg'$, \quad $\bigl(\tfrac{f}{g}\bigr)' = \tfrac{f'g - fg'}{g^2}$\\
$(f\circ g)'(x) = f'(g(x))\,g'(x)$
\block{Integrals}
$\int x^n\,dx = \tfrac{x^{n+1}}{n+1} + C \ (n \neq -1)$\\
$\int \tfrac{1}{x}\,dx = \ln|x| + C$, \ $\int e^{x}\,dx = e^{x} + C$\\
$\int \sin x\,dx = -\cos x + C$\\
$\int \sec^2 x\,dx = \tan x + C$\\
$\int \tfrac{dx}{1+x^2} = \arctan x + C$\\
$\int \tfrac{dx}{\sqrt{1-x^2}} = \arcsin x + C$\\
Parts: $\int u\,dv = uv - \int v\,du$\\
$\int_0^\infty e^{-x^2}dx = \tfrac{\sqrt{\pi}}{2}$
\block{Limits and series}
$\lim_{x\to 0}\tfrac{\sin x}{x} = 1$, \ $\lim_{x\to 0}\tfrac{1-\cos x}{x^2} = \tfrac12$\\
$\lim_{n\to\infty}\bigl(1 + \tfrac{x}{n}\bigr)^n = e^{x}$\\
$e^{x} = \sum_{k\ge 0} \tfrac{x^k}{k!}$, \quad
$\tfrac{1}{1-x} = \sum_{k\ge 0} x^k$ ($|x|<1$)\\
$\ln(1+x) = x - \tfrac{x^2}{2} + \tfrac{x^3}{3} - \cdots$ ($|x|<1$)\\
$\sin x = x - \tfrac{x^3}{3!} + \tfrac{x^5}{5!} - \cdots$\\
$\cos x = 1 - \tfrac{x^2}{2!} + \tfrac{x^4}{4!} - \cdots$
\block{Taylor with remainder}
$f(x) = \sum_{k=0}^{n} \tfrac{f^{(k)}(a)}{k!}(x-a)^k + R_n$,\\
$R_n = \tfrac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}$ for some $\xi$ between $a$ and $x$.
\block{Matrices ($2\times 2$)}
$\det\begin{pmatrix} a & b\\ c & d \end{pmatrix} = ad - bc$\\
$\begin{pmatrix} a & b\\ c & d\end{pmatrix}^{-1} = \tfrac{1}{ad-bc}\begin{pmatrix} d & -b\\ -c & a \end{pmatrix}$\\
$\det(AB) = \det A \det B$, \ $\det A^{\top} = \det A$\\
$\operatorname{tr} A = \sum \lambda_i$, \quad $\det A = \prod \lambda_i$
\block{Eigenvalues}
$Av = \lambda v$, $v \neq 0$; solve $\det(A - \lambda I) = 0$.\\
$A$ diagonalizable iff it has $n$ independent eigenvectors: $A = PDP^{-1}$.\\
Symmetric real $A$: real eigenvalues, orthonormal eigenbasis, $A = QDQ^{\top}$.
\block{Vector geometry}
$u\cdot v = \lVert u\rVert\,\lVert v\rVert\cos\theta$\\
$\operatorname{proj}_v u = \tfrac{u\cdot v}{v\cdot v}\,v$\\
$\lVert u\times v\rVert = \lVert u\rVert\,\lVert v\rVert\sin\theta$ (area of parallelogram)\\
Cauchy-Schwarz: $|u\cdot v| \le \lVert u\rVert\,\lVert v\rVert$
\block{Probability}
$P(A\cup B) = P(A) + P(B) - P(A\cap B)$\\
$P(A\mid B) = \tfrac{P(A\cap B)}{P(B)}$\\
Bayes: $P(A\mid B) = \tfrac{P(B\mid A)P(A)}{P(B)}$\\
$\mathbb{E}[aX+bY] = a\mathbb{E}X + b\mathbb{E}Y$\\
$\operatorname{Var} X = \mathbb{E}[X^2] - (\mathbb{E}X)^2$\\
Indep.: $\operatorname{Var}(X+Y) = \operatorname{Var}X + \operatorname{Var}Y$
\block{Distributions}
Binomial$(n,p)$: $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$, mean $np$, var $np(1-p)$\\
Poisson$(\lambda)$: $P(X=k) = e^{-\lambda}\tfrac{\lambda^k}{k!}$, mean and var $\lambda$\\
Normal$(\mu,\sigma^2)$: $f(x) = \tfrac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma^2}$\\
CLT: $\tfrac{\bar X_n - \mu}{\sigma/\sqrt n} \Rightarrow N(0,1)$
\end{multicols}
\end{document}
Trong ứng dụng: mở thư viện Dự án mới, cài đặt gói {nhãn} trong "Nhận thêm mẫu" và mẫu này xuất hiện cùng với bản xem trước trực tiếp và tạo dự án chỉ bằng một cú nhấp chuột. Quá trình biên dịch chạy cục bộ trên công cụ đi kèm.