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Markdown+itex2MML Sandbox

Play around below. Your changes will, periodically, be rolled back.

Some examples

(1)minw hp h+w rp r+w lp l\mathop{min} w_h p_h + w_r p_r + w_l p_l

Here’s an equation

(2) e x 2 /2 dx=2 π\int_{-\infty}^\infty e^{-x^2/2} \mathrm{d}x = \sqrt{2\pi}

which we can later refer1 back to as (2).

The Dirac equation:

(iD+m)ψ=0 (i\slash{D}+m)\psi =0

Here’s the table of Clifford2 algebras over :

j0 1 2 3 4 5 6 7 8
𝒞 j (2 )(4 )(8 )(8 )(8 )(16 )
𝒞 j +(2 )(2 )(2 )(2 )(2 )(4 )(8 )(16 )

where the generators of 𝒞 j ± satisfy

γ iγ j+γ jγ i=±2 δ ij\gamma_i\gamma_j +\gamma_j \gamma_i =\pm 2\delta_{i j}

and 𝒞 n+8 ±=𝒞 n ±(16 ).

(3)lim n k=1 n1 k 2 =π 2 6 \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6}

More Examples

(4)×E=Bt\nabla \times \vec{E} = - \frac {\partial \vec{B}}{\partial t}
(5)Bdl=μ 0 I enc\oint \mathbf{B}\cdot \mathrm{d}\mathbf{l} = \mu_0 I_\text{enc}

H 1 (𝒵,𝒪(k)) Let G=(V,E) be a graph, with w:V[0,1 ] a weight function.

(6){Q i,Q j}=δ ij.\{Q_i, Q_j\} = \delta_{ij}\mathcal{H}.

Here is a groupoid 𝒢. 3 (mod5 ).

Theorems

Definition

Let H be a subgroup of a group G. A left coset of H in G is a subset of G that is of the form xH, where xG and xH={xh:hH}.

Similarly a right coset of H in G is a subset of G that is of the form Hx, where Hx={hx:hH}.

Lemma

Let H be a subgroup of a group G, and let x and y be elements of G. Suppose that xHyH is non-empty. Then xH=yH.

Proof

Let z be some element of xHyH. Then z=xa for some aH, and z=yb for some bH. If h is any element of H then ahH and a 1 hH, since H is a subgroup of G. But zh=x(ah) and xh=z(a 1 h) for all hH. Therefore zHxH and xHzH, and thus xH=zH. Similarly yH=zH, and thus xH=yH, as required.

Lemma

Let H be a finite subgroup of a group G. Then each left coset of H in G has the same number of elements as H.

Proof

Let H={h 1 ,h 2 ,,h m}, where h 1 ,h 2 ,,h m are distinct, and let x be an element of G. Then the left coset xH consists of the elements xh j for j=1,2 ,,m. Suppose that j and k are integers between 1 and m for which xh j=xh k. Then h j=x 1 (xh j)=x 1 (xh k)=h k, and thus j=k, since h 1 ,h 2 ,,h m are distinct. It follows that the elements xh 1 ,xh 2 ,,xh m are distinct. We conclude that the subgroup H and the left coset xH both have m elements, as required.

Theorem

(Lagrange’s Theorem). Let G be a finite group, and let H be a subgroup of G. Then the order of H divides the order of G.

Proof

Each element x of G belongs to at least one left coset of H in G (namely the coset xH), and no element can belong to two distinct left cosets of H in G (see Lemma 1). Therefore every element of G belongs to exactly one left coset of H. Moreover each left coset of H contains H elements (Lemma 2). Therefore G=nH, where n is the number of left cosets of H in G. The result follows.

Corollary

Let x be an element of a finite group G. Then the order of x divides the order of G.

Problems?

Maybe the notation should be different than the Latex counterparts, but these do not seem to be rendering correctly (JD: They look fine to me. A font problem in your browser?) (JB: That was exactly the problem, thanks for the suggestion. The braces weren’t stretching correctly with my old fonts.):

  • SVG:
Box diagram d s¯ u, c, t s d¯ u, c, t W W+
K 0 K¯ 0 Mixing
  • Complicated commutative diagrams (equations in SVG)
Complicated commutative diagram, realized in SVG 1 1 1 1 Id Id A B ρ H H K K ϕ 1 ϕ 2 N A N B N A N B
  • SVG in equations.

In SU(3 ), Rank-2 Symmetric Tensor Representation Fundamental Representation = Adjoint Representation Rank-3 Symmetric Tensor Representation .

  • Cases:
r a+1 ={0 with prob.exp(θr a) max{δr a,z} with prob.1 exp(θr a) r_{a+1} = \begin{cases} 0 & \text{with prob.}\quad \exp(-\theta r_a) \\ \max \lbrace \delta r_a, z \rbrace & \text{with prob.}\quad 1 - \exp(-\theta r_a) \\ \end{cases}
  • Stretchy Brackets:
q a(z)=σ a 1 exp[γ+zσ a]q_a(z) = \sigma_a^{-1} \exp{\left[ -\frac{\gamma + z}{\sigma_a} \right]}
  • Linearity of Quadrature Rules
i=1 N(αf(x i)+βg(x i))w i=α i=1 Nf(x i)w i+β i=1 Ng(x i)w i\sum_{i = 1}^N {\left( {\alpha f(x_i ) + \beta g(x_i )} \right)w_i } = \alpha \sum_{i = 1}^N {f(x_i )w_i } + \beta \sum_{i = 1}^N {g(x_i )w_i }
  • Linearity of Integrals
a b(αf(x)+βg(x))dx=α a bf(x)dx+β a bg(x)dx{\int_a^b {\left( {\alpha f(x)\, + \beta g(x)} \right)dx = } \alpha \int_a^b {f(x)\,dx} + \beta \int_a^b {g(x)\,dx} }
  • Can we talk about x i 2 inline? What about a bx 2 dx? Inline fractions xx 2 x 1 x 2 ?

  • Big fractions

p 3 (x)=(1 2 )(x1 2 )(x3 4 )(x1 )(1 4 1 2 )(1 4 3 4 )(1 4 1 )+(1 2 )(x1 2 )(x3 4 )(x1 )(1 4 1 2 )(1 4 3 4 )(1 4 1 )+(1 2 )(x1 2 )(x3 4 )(x1 )(1 4 1 2 )(1 4 3 4 )(1 4 1 )+(1 2 )(x1 2 )(x3 4 )(x1 )(1 4 1 2 )(1 4 3 4 )(1 4 1 )p_3 (x) = \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}}
  • Diagram
P 1 (Y) P 1 (X) T T\array{ P_1(Y) &\to& P_1(X) \\ \downarrow &\Downarrow^\sim& \downarrow \\ T' &\to& T }

Will indented code work

This should be code
And this

yep

(7) A_n Quiver v1 v2 vn1 (U(k) n 1 ,{v i})\array{\arrayopts{\rowalign{center}}\begin{svg} <svg xmlns="http://www.w3.org/2000/svg" width="10em" height="10em" viewBox="0 0 120 120"> <desc>A_n Quiver</desc> <g fill="none" stroke="black" stroke-width="1"> <path d="M 60 25 l 25 20 l 0 30 l -25 20 l -25 -20 l 0 -30 z" /> <g stroke-dasharray="2"> <path d="M 60 0 l 0 25" /> <path d="M 85 45 l 25 -20" /> <path d="M 85 75 l 25 20" /> <path d="M 60 95 l 0 25" /> <path d="M 35 75 l -25 20" /> <path d="M 35 45 l -25 -20" /> </g> </g> <g fill="red" stroke="none"> <circle cx="60" cy="25" r="4" /> <circle cx="85" cy="45" r="4" /> <circle cx="85" cy="75" r="4" /> <circle cx="60" cy="95" r="4" /> <circle cx="35" cy="75" r="4" /> <circle cx="35" cy="45" r="4" /> </g> <foreignObject x="45" y="0" width="16" height="20"><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><msub><mi>v</mi><mn>1</mn></msub></math></foreignObject> <foreignObject x="83" y="15" width="16" height="20"><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><msub><mi>v</mi><mn>2</mn></msub></math></foreignObject> <foreignObject x="5" y="25" width="20" height="25"><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><msub><mi>v</mi><mrow><msub><mi>n</mi><mn>1</mn></msub></mrow></msub></math></foreignObject> </svg> \end{svg}& \equiv \left({U(k)}^{n_1},\{v_i\}\right)}

I want to cut and paste a bunch of stuff in here, just to see how well it works. Unfortunately, this doesn’t recognise ‘\(’ and ‘\)’, which I always use instead of single dollar signs, so it will all break rather trivially. I don’t know why ‘\’ and ‘\’ caught on (as a replacement for double dollar signs) while ‘\(’ and ‘\)’ didn’t catch on, but it means that nobody ever supports the latter (unless they’re actually doing LaTeX, of course). —Toby Bartels

foo


  1. You can also refer to it as (2). Chacun à son goût!.

  2. For more information, see Wikipedia.