# mathml.css

## Basic Examples

Inline equation: x

Display equation: x

Fraction: x 2

Binom: ( nk )

Subcript and Superscripts: x 2 + y 3 + a 12

Multiscripts: BASE 1 2 3 4 5 6 7 8

Underscripts and Overscripts: x 2 + y 3 + a 12

Roots: x + x 3

menclose element: top bottom left right madruwb actuarial box roundedbox circle horizontalstrike top left horizontalstrike

table: | 1 2 3 4 5 6 7 8 9 | = 0

Token elements: x such that x ≤10 ; x ; x

mathvariants: normal bold italic bold-italic sans-serif monospace bold-italic bold-sans-serif sans-serif-italic sans-serif-bold-italic

phantom element: x+phantom y

merror element: missing argument

annotation: xy \frac{x}{y}

## More complex examples

x 2a 2+y 2b 2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 ∑n=1+∞1 n 2=π 2 6\sum_{n=1}^{+\infty} \frac{1}{n^2} = \frac{\pi^2}{6} x=-b±b 2-4 ac2 a x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} f(x)=∑n=-∞∞cn e 2 πi(n/T)x=∑n=-∞∞f^(ξn)e 2 πi ξn x Δξ f(x)=\sum_{n=-\infty}^\infty c_n e^{2\pi i(n/T) x} =\sum_{n=-\infty}^\infty \hat{f}(\xi_n) e^{2\pi i\xi_n x}\Delta\xi Γ(t)=lim n→∞n!ntt(t+1)⋯(t+n)=1 t∏n=1∞(1+1 n)t 1+tn=e-γt t∏n=1∞(1+tn)-1e tn\Gamma(t) = \lim_{n \to \infty} \frac{n! \; n^t}{t \; (t+1)\cdots(t+n)}= \frac{1}{t} \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^t}{1+\frac{t}{n}} = \frac{e^{-\gamma t}}{t} \prod_{n=1}^\infty \left(1 + \frac{t}{n}\right)^{-1} e^{\frac{t}{n}} sl(n,𝔽)={A∊ℳn(𝔽):Tr(A)=0}\mathfrak{sl}(n, \mathbb{F}) = \left\{ A \in \mathscr{M}_n(\mathbb{F}) : \operatorname{Tr}(A) = 0 \right\} x 2y 2 x^2 y^2 F 3 2\multiscripts{_2}{F}{_3} x+y 2k+1\frac{x+y^2}{k+1} x+y 2 k+1 x+y^{\frac 2 {k+1}} a b/2\frac{a}{b/2} a 0+1 a 1+1 a 2+1 a 3+1 a 4\displaystyle a_0 + \frac{1}{\displaystyle a_1+\frac{1}{\displaystyle a_2+\frac{1}{\displaystyle a_3+\frac{1}{\displaystyle a_4}}}} (n k/2)\binom{n}{k/2} (p 2)x 2y p-2-1 1-x1 1-x 2\binom{p}{2} x^2 y^{p-2} - \frac{1}{1-x} \frac{1}{1-x^2} ∑0≤i≤m0<j<n P(i,j)\sum_{\substack{ 0 \le i \le m \\ 0 < j < n }} {P(i,j)} x 2 y x^{2y} ∑i=1 p∑j=1 q∑k=1 r a ijb jkc ki\sum_{i=1}^p \sum_{j=1}^q \sum_{k=1}^r {a_{i j} b_{j k} c_{k i}} 1+1+1+1+1+1+1+x\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+x}}}}}}} (∂2∂x 2+∂2∂y 2)|φ(x+iy)|2=0\left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \right) {| \varphi(x+ i y)|}^2 = 0 2 2 2 x 2^{2^{2^x}} ∫1 x dt t\int_1^x \frac{dt}{t} ∬D dxdy\iint_D {dx dy} f(x)={1/3if 0≤x≤1;2/3if 3≤x≤4;0elsewhere.f(x) = \begin{cases} 1/3 & \text{if} \quad 0 \leq x \leq 1; \\ 2/3 & \text{if} \quad 3 \leq x \leq 4; \\ 0 & \text{elsewhere}. \end{cases} y x 2 y_{x^2} ((abcd)(efgh)0(ijkl))\begin{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} & \begin{pmatrix} e & f \\ g & h \end{pmatrix} \\ 0 & \begin{pmatrix} i & j \\ k & l \end{pmatrix} \end{pmatrix} det|c 0c 1c 2…cnc 1c 2c 3…c n+1c 2c 3c 4…c n+2⋮⋮⋮⋮cnc n+1c n+2…c 2 n|>0\det \begin{vmatrix} c_0 & c_1 & c_2 & \dots & c_n \\ c_1 & c_2 & c_3 & \dots & c_{n+1} \\ c_2 & c_3 & c_4 & \dots & c_{n+2} \\ \vdots & \vdots & \vdots & & \vdots \\ c_n & c_{n+1} & c_{n+2} & \dots & c_{2n} \end{vmatrix} > 0 y x 2 y_{x_2} x 9231415+π x^31415_92 + \pi x ybazcd x^{z^d_c}_{y_b^a} y 3‴y_3''' x˙=σ(y-x)y˙=ρx-y-xzz˙=-βz+xy\begin{aligned} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x - y - xz \\ \dot{z} & = -\beta z + xy \end{aligned} x˙=σ(y-x)y˙=ρx-y-xzz˙=-βz+xy\begin{aligned} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x - y - xz \\ \dot{z} & = -\beta z + xy \end{aligned} 𝐕 1×𝐕 2=|𝐢𝐣𝐤∂X∂u∂Y∂u0∂X∂v∂Y∂v0|\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} P(E)=(nk)pk(1-p)n-k P(E) = {n \choose k} p^k (1-p)^{ n-k} 1(ϕ 5-ϕ)e 25 π=1+e-2 π 1+e-4 π 1+e-6 π 1+e-8 π 1+…\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) 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} } } } 1+q 2(1-q)+q 6(1-q)(1-q 2)+⋯=∏j=0∞1(1-q 5 j+2)(1-q 5 j+3),for|q|<1.1 + \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} \quad |q| < 1. ∇×𝐁⇀-1 c∂𝐄⇀∂t=4 π c 𝐣⇀∇⋅𝐄⇀=4 πρ∇×𝐄⇀+1 c∂𝐁⇀∂t=0⇀∇⋅𝐁⇀=0\begin{aligned} \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{aligned}