KaTeX测试

{\KaTeX}

\iiint\limits_{V} \left( \frac{\partial^2 P}{\partial x^2} + \frac{\partial^2 Q}{\partial y^2} + \frac{\partial^2 R}{\partial z^2} \right) \, dV = \oint_{\partial V} \left( P\, dy \wedge dz + Q\, dz \wedge dx + R\, dx \wedge dy \right)

\det \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix} = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma(i)}

\lim_{n \to \infty} \left( \sum_{k=1}^n \frac{1}{k^2} \right) = \frac{\pi^2}{6} \quad \text{且} \quad e^{ix} = \cos x + i\sin x = \sum_{n=0}^\infty \frac{(ix)^n}{n!}

\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt \quad \text{满足} \quad \frac{\partial^2 u}{\partial t^2} = c^2 \nabla^2 u

P(A \mid B) = \frac{P(B \mid A) P(A)}{P(B)} \quad \text{其中} \quad P(B) = \sum_{i=1}^n P(B \mid A_i) P(A_i)

\nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) \mathbf{i} + \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \mathbf{j} + \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \mathbf{k}

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \quad \text{且} \quad \sqrt{\frac{a + \sqrt{b}}{c}} + \int_0^1 \frac{\ln(1 + x^2)}{\sqrt{x}} \, dx

\hat{H} \psi = \left( -\frac{\hbar^2}{2m} \nabla^2 + V(\mathbf{r}) \right) \psi = i\hbar \frac{\partial \psi}{\partial t}

z_{n+1} = z_n^2 + c \quad \text{其中} \quad z, c \in \mathbb{C} \quad \text{且} \quad |z_n| \to \infty \text{ 或收敛}

Z = \sum_{i} g_i e^{-\beta E_i} \quad \text{且} \quad S = -k_B \sum_{i} P_i \ln P_i