WebKit MathML Demos (2016)

Mathematical Fonts

γ=limn→∞(−lnn+∑k=1n1k)=∫1∞(1⌊x⌋−1x)dx≈0.5772156649{\displaystyle \gamma = {\lim_{n\to \infty }\left(-\operatorname{ln} n+\sum _{k=1}^{n}{\frac {1}{k}}\right)} = {\int _{1}^{\infty }\left({\frac {1}{\lfloor x\rfloor }}-{\frac {1}{x}}\right)\,dx} \approx 0.5772156649}

γ=limn→∞(−lnn+∑k=1n1k)=∫1∞(1⌊x⌋−1x)dx≈0.5772156649{\displaystyle \gamma = {\lim_{n\to \infty }\left(-\operatorname{ln} n+\sum _{k=1}^{n}{\frac {1}{k}}\right)} = {\int _{1}^{\infty }\left({\frac {1}{\lfloor x\rfloor }}-{\frac {1}{x}}\right)\,dx} \approx 0.5772156649}

γ=limn→∞(−lnn+∑k=1n1k)=∫1∞(1⌊x⌋−1x)dx≈0.5772156649{\displaystyle \gamma = {\lim_{n\to \infty }\left(-\operatorname{ln} n+\sum _{k=1}^{n}{\frac {1}{k}}\right)} = {\int _{1}^{\infty }\left({\frac {1}{\lfloor x\rfloor }}-{\frac {1}{x}}\right)\,dx} \approx 0.5772156649}

a2+b2=c2\href{https://en.wikipedia.org/wiki/Pythagorean_theorem}{a^2 + b^2 = c^2}

(qp)(pq)=(−1)p−12q−12\href{https://en.wikipedia.org/wiki/Legendre_symbol#Definition}{\left({\frac{q}{p}}\right)} \left({\frac {p}{q}}\right) = {(-1)}^{{\frac{p-1}{2}}{\frac{q-1}{2}}}

det(expA)=exp(Tr(A))\href{https://en.wikipedia.org/wiki/Determinant}{\det}(\href{https://en.wikipedia.org/wiki/Matrix_exponential}{\exp{A}}) = \exp{(\href{https://en.wikipedia.org/wiki/Trace_(linear_algebr}{\mathrm{Tr}{(A)}})}

Mathvariants

f:g→g′,f([x,y])=[f(x),f(y)]f:{\mathfrak {g}}\to {\mathfrak {g'}},\quad f([x,y])=[f(x),f(y)]

SLn(R)={A∊Mn(R):detA=1}{\mathrm{SL}}_n(\mathbb{R}) = \left\{ A \in \mathscr{M}_n(\mathbb{R}) : \det{A} = 1 \right\}

Z/NZ≅Z/n1Z×⋯×Z/nkZ{\mathbb {Z}}/N{\mathbb {Z}}\cong {\mathbb {Z}}/n_{1}{\mathbb {Z}}\times \cdots \times {\mathbb {Z}}/n_{k}{\mathbb {Z}}

Operators

(23)N=23×23×23×…×23⏟Ntimes=2N/3\left(\sqrt[3]{2}\right)^N = \underset{N\,\text{times}}{\underbrace{\sqrt[3]{2} \times \sqrt[3]{2} \times \sqrt[3]{2} \times \dots \times \sqrt[3]{2}}} = 2^{N/3}

Γ(t)=limn→∞n!ntt(t+1)⋯(t+n)=1t∏n=1∞(1+1n)t1+tn=e−γtt∏n=1∞(1+tn)−1etn{\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}}

Displaystyle

∑k=1+∞9k4=π410=π410=∑k=1+∞9k4{{\sum_{k=1}^{+\infty} \frac{9}{k^4}} = \frac{\pi^4}{10}} = {\displaystyle \frac{\pi^4}{10} = {\sum_{k=1}^{+\infty} \frac{9}{k^4}}}

B(t)=∑i=0nbi,n(t)Piwi∑i=0nbi,n(t)wi{\displaystyle \mathbf {B} (t)={\frac {\sum _{i=0}^{n}b_{i,n}(t)\mathbf {P} _{i}w_{i}}{\sum _{i=0}^{n}b_{i,n}(t)w_{i}}}}

(∏i=1nAi)2=(A∑i=1ni)2=An(n−1)\left(\prod_{i=1}^n A^i\right)^2 = \left(A^{\sum_{i=1}^n i}\right)^2 = A^{n{(n-1)}}

OpenType MATH Parameters

π=41+122+322+522+⋱=∑n=0∞4(−1)n2n+1=41−43+45−47+−⋯{\displaystyle \pi ={\displaystyle \frac{4}{1+{\displaystyle \frac {1^{2}}{2+{\displaystyle \frac {3^{2}}{2+{\displaystyle \frac {5^{2}}{2+\ddots }}}}}}}}=\sum _{n=0}^{\infty }{\frac {4(-1)^{n}}{2n+1}}={\frac {4}{1}}-{\frac {4}{3}}+{\frac {4}{5}}-{\frac {4}{7}}+-\cdots }

ΦE=∯SE⋅dA=Qϵ0{\mathrm \Phi}_E = {\oiint_S \mathbf {E} \cdot \mathrm{d} \mathbf{A}} = \frac{Q}{\epsilon_0}

Arabic Mathematics

ل 𞸟 𞸝 = 𞸝 𞸟 ل 𞸟 − ١ 𞸝 − ١

س=−ب±ب٢−٤اج٢اس = \frac{-ب\pm\sqrt{ب^٢-٤اج}}{٢ا}