Texコードの数式を表示してみる。
\(\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}\) \[ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} \tag{3} \] \begin{align*} A & = B \\ & = C \\ & = D. \end{align*} $$ \newcommand\C{{\mathbb C}} \newcommand\np[2]{{#1}#2{#1}} \C[y_1,\ldots,y_n]\to {\mathcal A}, \quad a\mapsto\np{:}{a} $$ \begin{eqnarray} \delta S & = & \int_{t_{1}}^{t_{2}}dt\left\{ L(q+\delta q,\dot{q}+\delta\dot{q})-L(q,\dot{q})\right\} \nonumber \\ & = & \int_{t_{1}}^{t_{2}}dt\left(\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial\dot{q}}\delta\dot{q}\right)\nonumber \\ & = & \int_{t_{1}}^{t_{2}}dt\left\{ \frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial\dot{q}}\left(\dfrac{d}{dt}\delta q\right)\right\} \nonumber \\ & = & \int_{t_{1}}^{t_{2}}dt\left\{ \frac{\partial L}{\partial q}\delta q+\dfrac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\delta q\right)-\dfrac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right)\delta q\right\} \nonumber \\ & = & \left[\frac{\partial L}{\partial\dot{q}}\delta q\right]_{t_{1}}^{t_{2}}+\int_{t_{1}}^{t_{2}}dt\left\{ \frac{\partial L}{\partial q}-\dfrac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right)\right\} \delta q\nonumber \\ & = & \int_{t_{1}}^{t_{2}}dt\left\{ \frac{\partial L}{\partial q}-\dfrac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right)\right\} \delta q \end{eqnarray}最近、数式の通し番号が自動処理できるようになった。また、自動処理で得た番号を文章中で参照もできる。