Test: rst

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(ℱf)(y)=12πn∫ℝnf(x)e−iy⋅xdx(\mathcal{F}f)(y) = \frac{1}{\sqrt{2\pi}^{\ n}} \int_{\mathbb{R}^n} f(x)\, e^{-\mathrm{i} y \cdot x} \,\mathrm{d} x

Complex math example, the Fourier transform.

Inline: ab&\sqrt{a}^{b} \&.

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Test: rst

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Very complex math

x0GL𝔻xbx0GL𝔻xaf(x)=limδ→01δb∑k=0⌊x−x0δ⌋(−1)k(bk)x0GL𝔻xaf(x−kδ)expand x0GL𝔻xb=limδ→01δb∑k=0⌊x−x0δ⌋(−1)k(bk)limε→01εa∑j=0⌊x−kδ−x0ε⌋(−1)j(aj)f(x−kδ−jε)expand x0GL𝔻xa=limδ→0limε→01δb1εa∑k=0⌊x−x0δ⌋∑j=0⌊x−kδ−x0ε⌋(−1)k(−1)j(bk)(aj)f(x−kδ−jε)push constants into lim and ∑=limε→01εb+a∑k=0⌊x−x0ε⌋∑j=0⌊x−x0ε⌋−k(−1)k+j(bk)(aj)f(x−(k+j)ε)(1), unify δ and ε=limε→01εb+a∑i=0⌊x−x0ε⌋∑k=0i(−1)i(bk)(ai−k)f(x−iε)rearrange triangle-sum, over i=k+j=limε→01εb+a∑i=0⌊x−x0ε⌋(−1)i(b+ai)f(x−iε)Vandermonde’s identity=x0GL𝔻xb+af(x)\begin{align} & \quad\,\, {}^{GL}_{x_0}\mathbb{D}^b_x {}^{GL}_{x_0}\mathbb{D}^a_x f(x) \\ & = \lim_{\delta \to 0} \frac{1}{\delta^b} \sum_{k=0}^{\lfloor \frac{x-x_0}{\delta} \rfloor} (-1)^k {b \choose k} {}^{GL}_{x_0}\mathbb{D}^a_x f(x-k\delta) & \text{expand ${}^{GL}_{x_0}\mathbb{D}^b_x$} \\ & = \lim_{\delta \to 0} \frac{1}{\delta^b} \sum_{k=0}^{\lfloor \frac{x-x_0}{\delta} \rfloor} (-1)^k {b \choose k} \lim_{\varepsilon \to 0} \frac{1}{\varepsilon^a} \sum_{j=0}^{\lfloor \frac{x-k\delta-x_0}{\varepsilon} \rfloor} (-1)^j {a \choose j} f(x-k\delta-j\varepsilon) & \text{expand ${}^{GL}_{x_0}\mathbb{D}^a_x$} \\ & = \lim_{\delta \to 0} \lim_{\varepsilon \to 0} \frac{1}{\delta^b} \frac{1}{\varepsilon^a} \sum_{k=0}^{\lfloor \frac{x-x_0}{\delta} \rfloor} \sum_{j=0}^{\lfloor \frac{x-k\delta-x_0}{\varepsilon} \rfloor} (-1)^k (-1)^j {b \choose k} {a \choose j} f(x-k\delta-j\varepsilon) & \text{push constants into $\lim$ and $\sum$} \\ & = \lim_{\varepsilon \to 0} \frac{1}{\varepsilon^{b+a}} \sum_{k=0}^{\lfloor \frac{x-x_0}{\varepsilon} \rfloor} \sum_{j=0}^{\lfloor \frac{x-x_0}{\varepsilon} \rfloor - k} (-1)^{k+j} {b \choose k} {a \choose j} f(x-(k+j)\varepsilon) & \text{(1), unify $\delta$ and $\varepsilon$} \\ & = \lim_{\varepsilon \to 0} \frac{1}{\varepsilon^{b+a}} \sum_{i=0}^{\lfloor \frac{x-x_0}{\varepsilon} \rfloor} \sum_{k=0}^{i} (-1)^i {b \choose k} {a \choose i-k} f(x-i\varepsilon) & \text{rearrange triangle-sum, over $i = k+j$} \\ & = \lim_{\varepsilon \to 0} \frac{1}{\varepsilon^{b+a}} \sum_{i=0}^{\lfloor \frac{x-x_0}{\varepsilon} \rfloor} (-1)^i {b+a \choose i} f(x-i\varepsilon) & \text{Vandermonde's identity} \\ & = {}^{GL}_{x_0}\mathbb{D}^{b+a}_x f(x) \end{align}