LaTeX Test

$\huge\rm\LaTeX$ \[\] A test collection for use on this website.
Accents and Diacritics: $$\acute{a} \grave{a} \hat{a} \tilde{a} \breve{a}$$ $$\check{a} \bar{a} \ddot{a} \dot{a}$$ $$x’, y”$$
Superscript and subscript: $$a_i$$ $$b_{ij}$$ $$C_{m,n}$$ $$\delta_{j+k}$$ $$x^y$$ $$a^{j2\pi}$$ $$x^2_3$$ $$C^k_{\mu,\nu}$$ $${}_1^2 \Psi_3^4$$
Underlines, overlines and stackings: \[\hat a \ \bar b \ \vec c\] \[\overrightarrow{a b} \ \overleftarrow{c d} \ \widehat{d e f}\] $$\overbrace{ 1+2+\cdots+100 }^{5050}$$ $$\underbrace{ a+b+\cdots+z }_{26}$$ $$A \xleftarrow{n+\mu-1} B \xrightarrow[T]{n\pm i-1} C$$ $$\overset{\alpha}{\omega} \underset{\mu}{\nu} \overset{\beta}{\underset{\Delta}{\tau}} \stackrel{\zeta}{\eta}$$
Trigonometric functions: \begin{align*} \cos^2 x +\sin^ 2 x=1 \end{align*}
Radicals: \[ \sqrt{\matrix{a & b\cr c & d}} \sqrt{ \rlap{\smash{\rm Hi!}} \phantom{\matrix{a & b\cr c & d}} } \]
Arrows: \begin{align*} A\iff B\\ \implies\\ \impliedby\\ \iff\\ \mapsto\\ \to\\ \gets\\ \rightarrow\\ \leftarrow\\ \Rightarrow\\ \Leftarrow\\ \hookrightarrow\\ \hookleftarrow \end{align*}
Cases: \begin{align*} \det(A)&=1+(-1)^{n+1} \\ &= \begin{cases} 2 & \text{ if } n \text{ is odd}\\ 0 & \text{ if } n \text{ is even}. \end{cases} \end{align*} \[ |x| = \begin{cases} x & \text{ if } x\ge 0 \\ -x & \text{ if } x \lt 0 \end{cases} \]
Providing reasons for each line in align: \begin{align*} f(ab)&=(ab)^2 && (\text{by definition of $f$})\\ &=(ab)(ab)\\ &=a^2 b^2 && (\text{since $G$ is abelian})\\ &=f(a)f(b) && (\text{by definition of $f$}). \end{align*} \begin{gather} a = a \tag{$*$}\\ \text{if } a=b \text{ then } b=a \tag{$\dagger$}\\ \text{if } a=b \text{ and } b=c \text{ then } a=c \tag{3.1} \end{gather}
System of equations: \begin{align*} \left\{ \begin{array}{c} a_1x+b_1y+c_1z=d_1 \\ a_2x+b_2y+c_2z=d_2 \\ a_3x+b_3y+c_3z=d_3 \end{array} \right. \end{align*}
Brackets: \[ \left\langle \matrix{a & b\cr c & d} \right\rangle \]
Braces: \[ \left\lbrace \matrix{a & b\cr c & d} \right\rbrace \]
Matrix: \[ A = \pmatrix{ a_{11} & a_{12} & \ldots & a_{1n} \cr a_{21} & a_{22} & \ldots & a_{2n} \cr \vdots & \vdots & \ddots & \vdots \cr a_{m1} & a_{m2} & \ldots & a_{mn} \cr } \] Above is an $m \times n$ matrix, where $a_{ij}$ are elements of the matrix $A$, $i$ = the $i_{th}$ row and $j$ = the $j_{th}$ column. $\\\\$ The sequence of numbers: $$A_{(i)} = (a_{i1}, \dotsc, a_{in})$$ is the $i_{th}$ row of A, and the sequence of numbers: $$A_{(j)} = (a_{1j}, \dotsc, a_{mj})$$ is the $j_{th}$ column of A. Displaying a matrix: $$\begin{bmatrix} 17 & 18 & 5 & 5 & 45 & 1 \\ 42 & 28 & 30 & 15 & 115 & 3 \\ 10 & 10 & 10 & 21 & 51 & 2 \\ 28 & 5 & 65 & 39 & 132 & 5 \\ 24 & 26 & 45 & 21 & 116 & 4 \end{bmatrix}$$
Augmented matrix: \begin{align*} \left[\begin{array}{rrr|rrr} 1 & 0 & 0 & 1 &1 & 1 \\ 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 0 & 1 \\ \end{array} \right] \end{align*}
Block matrix: \begin{align*} M= \left[\begin{array}{c|c} A & B\\ \hline C & D \end{array} \right] \end{align*}
Matrix with fractions: \begin{align*} A=\begin{bmatrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{2}{3} &\frac{-1}{3} &\frac{-1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{-2}{3} \end{bmatrix} \end{align*}
Elementary row operations (Gauss-Jordan elimination): \begin{align*} \left[\begin{array}{rrrr|r} 1 & 1 & 1 & 1 &1 \\ 0 & 1 & 2 & 3 & 5 \\ 0 & -2 & 0 & -2 & 2 \\ 0 & 1 & -2 & 3 & 1 \\ \end{array}\right] \xrightarrow{\substack{R_1-R_2 \\ R_3-R_2\\R_4-R_2}} \left[\begin{array}{rrrr|r} 1 & 0& -1 & -2 &-4 \\ 0 & 1 & 2 & 3 & 5 \\ 0 & 0 & 4 & 4 & 12 \\ 0 & 0 & -4 & 0 & -4 \\ \end{array}\right] \xrightarrow[\frac{-1}{4}R_4]{\frac{1}{4}R_3} \left[\begin{array}{rrrr|r} 1 & 0& -1 & -2 &-4 \\ 0 & 1 & 2 & 3 & 5 \\ 0 & 0 & 1 & 1 & 3 \\ 0 & 0 & 1 & 0& 1 \\ \end{array}\right] \end{align*}
Use array for tabular: \begin{array}{ |c|c|c| } \hline a & a^2 \pmod{5} & 2a^2 \pmod{5} \\ \hline 0 & 0 & 0 \\ 1& 1 & 2 \\ 2& 4 & 3 \\ 3 & 4 & 3\\ 4 & 1 & 2\\ \hline \end{array}
Sums: $$\sum_{k=1}^n a_k$$ \[ \sum_{ \substack{ 1\lt i\lt 3 \\ 1\le j\lt 5 }} a_{ij} \]
Sub-arrays: \[ \prod_{\begin{subarray}{rl} i\lt 5\quad & j\gt 1 \\ k\ge2,\,k\ne 5 \quad & \ell\le 5,\,\ell\ne 2 \end{subarray}} x_{ijk\ell} \]
Integration: $$\int\limits_a^b f(x)\,dx$$ \begin{align*} \int_{a}^{b} \! f(x)\,\mathrm{d}x \end{align*} \[ \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-\frac{x^2}{2}}dx \]
Second derivative: \begin{align*} f^{\prime\prime} \end{align*}
Explanations under equations: \begin{align*} n=\underbrace{1+1+\cdots+1}_{\text{$n$ times}} \end{align*}
Explanations over equations: \[ \overbrace{x+\cdots+x}^{n\text{ times}} \] \[ \overparen a \quad \overparen ab \quad \overparen{ab} \quad \overparen{abc} \quad \overparen{abcdef} \quad \overparen{\underparen{abcd}} \]
Inline: $\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$
Displayed: $$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$ \[ (x+1)^2 % original expression = (x+1)(x+1) % definition of exponent = x^2 + 2x + 1 % FOIL, combine like terms \\ \{x\,|\,x\in\Bbb Z\} \]
Parentheses: $$\Biggl(\biggl(\Bigl(\bigl((x)\bigr)\Bigr)\biggr)\Biggr)$$
Fractions: $$\frac{100}{20} = 5$$ $${a+1\over b+1}$$ $$\displaystyle\frac ab + {\textstyle \frac cd + \frac ef} + \frac gh$$ $$\left\{\frac ab,c\right\}$$ $$|\frac ab|$$ $$\left(\frac{a+b}{\dfrac{c}{d}}\right)$$ $$\left(\vcenter{\frac{a+b}{\dfrac{c}{d}}}\right)$$ \[ \displaystyle \frac{2x+3y-\phantom{5}z} {\phantom{2}x+\phantom{3}y+5z} \] $${a+1 \overwithdelims \{ \} b+2}+c$$ $$\left(\frac ab,c\right)$$ $${a+1 \above 1.5pt b+2}+c$$ $$\cfrac{2}{1+\cfrac{2}{1+\cfrac{2}{1}}}$$ \[ \def\specialFrac#1#2{\frac{x + #1}{y + #2}} \specialFrac{7}{z+3} \] $$\sqrt[3]{\frac xy}$$
Partial differentials: \[ i\hbar\frac{\partial}{\partial t}\left|\Psi(t)\right>=H\left|\Psi(t)\right> \]
Graphics:
cayley graph blurred
From the image we can see…
Definitions, Remarks, lemmas, etc.:
Definition 1. Suppose that $(X,\mathcal M)$ and $(Y,\mathcal N)$ are measurable spaces, and $f:X\to Y$ is a map. We call $f$ is measurable if for every $B\in\mathcal N$ the set $f^{-1}(B)$ is in $\mathcal M$.

Remark 1. If $Y$ is a topological space, and $\mathcal N$ is the $\sigma$-algebra of Borel sets, then $f$ is measurable if and only if the following condition satisfied:
  • For every open set $V$ in $Y$, the inverse image $f^{-1}(V)$ is measurable.
Lemma 2 (fundamental lemma of integration). Let $\set{f_n}$ be a Cauchy sequence of step mappings. Then there exists a subsequence which converges pointwise almost everywhere, and satisfies the additional property: given $\eps$ there exists a set $Z$ of measure $< \eps$ such that this subsequence converges absolutely and uniformly outside $Z$.
Lemma 2 (fundamental lemma of integration). Let $\set{f_n}$ be a Cauchy sequence of step mappings. Then there exists a subsequence which converges pointwise almost everywhere, and satisfies the additional property: given $\eps$ there exists a set $Z$ of measure $< \eps$ such that this subsequence converges absolutely and uniformly outside $Z$.

Referencing equations: \[ \] At first, we sample $f(x)$ in the $N$ ($N$ is odd) equidistant points around $x^*$: \[ f_k = f(x_k),\: x_k = x^*+kh,\: k=-\frac{N-1}{2},\dots,\frac{N-1}{2} \] where $h$ is some step. Then we interpolate points $\{(x_k,f_k)\}$ by polynomial \[\begin{equation} \label{eq:poly} P_{N-1}(x)=\sum_{j=0}^{N-1}{a_jx^j} \end{equation}\] Its coefficients $\{a_j\}$ are found as a solution of system of linear equations: \[\begin{equation} \label{eq:sys} \left\{ P_{N-1}(x_k) = f_k\right\},\quad k=-\frac{N-1}{2},\dots,\frac{N-1}{2} \end{equation}\] Here are references to existing equations: (\[eq:poly\]), (\[eq:sys\]). Here is reference to non-existing equation (\[eq:unknown\]).
Math scripting: $$\mathscr{ABCDEFGHIJKLMNOPQRSTUVWXYZ}$$ $$\Bbb{ABCDEFGHIJKLMNOPQRSTUVWXYZ}$$ $$\cal ABCDEFGHIJKLMNOPQRSTUVWXYZ$$ $$\frak ABCDEFGHIJKLMNOPQRSTUVWXYZ$$ $$\huge AaBb\alpha\beta123\frac ab\sqrt x$$
In-line Mathematics: While display equations look good for a page of samples, the ability to mix math and text in a paragraph is also important. This expression \(\sqrt{3x-1}+(1+x)^2\) is an example of an inline equation. As you see, equations can be used this way as well, without unduly disturbing the spacing between lines.
A Cross Product Formula: \[\mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\ \frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0 \\ \end{vmatrix}\]
The probability of getting \(k\) heads when flipping \(n\) coins is: \[P(E) = {n \choose k} p^k (1-p)^{ n-k}\]
$$\pmb{\mathrm{An \ Identity \ of \ Ramanujan:}}$$ \[\frac{1}{(\sqrt{\phi \sqrt{5}}-\phi) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\ldots} } } }\]
$$\pmb{\mathrm{A \ Rogers-Ramanujan \ Identity:}}$$ \[ 1 + \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots = \prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})}, \quad\quad \text{for $|q|<1$}.\]
$$\pmb{\mathrm{Maxwell's \ Equations:}}$$ \[\begin{align} \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{align}\]
$$\pmb{\mathrm{The \ Lorenz \ Equations:}}$$ \[\begin{align} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x - y - xz \\ \dot{z} & = -\beta z + xy \end{align}\]
$$\pmb{\mathrm{The \ Cauchy-Schwarz \ Inequality:}}$$ \[\left( \sum_{k=1}^n a_k b_k \right)^{\!\!2} \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)\]

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