This is inline: $E=mc^2$.

This is display: $$E=mc^2$$

A display with a tag: $$E=mc^2\tag{2.1}$$

A wide display: \[ 1 + \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots = \prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})}, \quad\quad \text{for $|q|<1$}. \]

A tall display: \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}

Some actual mathematical text:

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This procedure is simply a generalization of the method used in Sects. 1-3 and 1-4 to obtain the equations of the osculating plane and the osculating circle. Let $f(u)$ near $P(u=u_0)$ have finite derivatives $f^{(i)}(u_0)$, $i = 1, 2, \ldots, n+1$. Then if we take $u=u_1$ at $A$ and write $h = u_1 - u_0$, then there exists a Taylor development of $f(u)$ of the form (compare Eq. (1-5)): $$ f(u_1) = f(u_0) + hf'(u_0)+{h^2\over 2!}f''(u_0) + \cdots + {h^{n+1}\over (n+1)!}f^{(n+1)}(u_0) + o(h^{n+1}). $$ Here, $f(u_0)=0$ since $P$ lies on $\Sigma_2$, and $h$ is of order $AP$ (see theorem Sec. 1-2); $f(u_1)$ is of order $AD$. Hence necessary and sufficient conditions that the surface has a contact of order $n$ at $P$ with the curve are that at $P$ the relations hold: $$ f(u) = f'(u) = f''(u) = \cdots = f^{(n)}(u) = 0;\quad f^{(n+1)}(u) \ne 0. $$

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If $P(u,v)$ and $Q(u,v)$ are two functions of $u$ and $v$ on a surface, then according to Green's theorem and the expression in Chapter 2, Eq. (3-4) for the element area: $$ \int_C P\,du + Q\, dv = \int\!\!\!\int_A \left({\partial Q\over \partial u} - {\partial P\over \partial v}\right) {1\over \sqrt{EG-F^2}}\,dA, $$ where $dA$ is the element of area of the region $R$ enclosed by the curve $C$. With the aid of this theorem we shall evaluate $$ \int_C \kappa_g\,ds, $$ where $\kappa_g$ is the geodesic curvature of the curve $C$. If $C$ at a point $P$ makes the angle $\theta$ with the coordinate curve $v = {\rm constant}$ and if the coordinate curves are orthogonal, then, according to Liouville's formula (1-13): $$ \kappa_g\,ds = d\theta + \kappa_1(\cos\theta)\,ds + \kappa_2(\sin\theta)\,ds. $$ Here, $\kappa_1$ and $\kappa_2$ are the geodesic curvatures of the curves $v = {\rm constant}$ and $u = {\rm constant}$ respectively. Since $$ \cos\theta\,ds = \sqrt{E}\,du, \qquad \sin\theta\,ds = \sqrt{G}\,dv, $$ we find by application of Green's theorem: $$ \int_C\kappa_g\,ds = \int_C d\theta + \int\!\!\!\int_A\left({\partial\over\partial u} \left(\kappa_2\sqrt{G}\,\right) - {\partial\over \partial v}\left(\kappa_1\sqrt{E}\,\right)\right)\,du\,dv. $$

The Gaussian curvature can be written, according to Chapter 3, Eq. (3-7), $$ K = -{1\over 2\sqrt{EG}} \left[{\partial\over\partial u}{G_u\over \sqrt{EG}} + {\partial\over\partial v}{E_v\over\sqrt{EG}}\right] ={1\over\sqrt{EG}}\left[ -{\partial\over\partial u} \left(\kappa_2\sqrt{G}\,\right) + {\partial\over\partial v} \left(\kappa_1\sqrt{E}\,\right)\right], $$ so we obtain the formula $$ \int_C\kappa_g\,ds = \int_C d\theta - \int\!\!\!\int_A K\,dA. $$ The integral $\int\!\!\int_A K\,dA$ is known as the total or integral curvature, or curvature integra, of the region $R$, the name by which Gauss introduced it.

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