KaTeX test

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KaTeX vs MathJax



KaTeX test s privzetimi nastavitvami. Primerjaj z [MathJax test].
[MathJax vs KaTeX - povzetek testa]



Inline ϕn(κ)=14π2κ2∫0∞sin⁡(κR)κR∂∂R[R2∂Dn(R)∂R],dR \phi_n(\kappa) = \frac{1}{4\pi^2\kappa^2} \int_0^\infty \frac{\sin(\kappa R)}{\kappa R} \frac{\partial}{\partial R} \left [ R^2\frac{\partial D_n( R )}{\partial R} \right ] ,dR



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G:=∐i=1∞X^{j<i}\mathfrak G \mathrel{:}= \coprod_{i=1}^{\infty} \widehat{\mathbb X}_{\{ j \lt i \}}



(∑k=1nakbk)2≤(∑k=1nak2)(∑k=1nbk2) \left( \sum_{k=1}^n a_k b_k \right)^{2} \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)



V1×V2=∣ijk∂X∂u∂Y∂u0∂X∂v∂Y∂v0∣\mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\ \frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0 \\ \end{vmatrix}



1(ϕ5−ϕ)e25π=1+e−2π1+e−4π1+e−6π1+e−8π1+… \frac{1}{(\sqrt{\phi \sqrt{5}}-\phi) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\ldots} } } }



1+q2(1−q)+q6(1−q)(1−q2)+⋯=∏j=0∞1(1−q5j+2)(1−q5j+3),for ∣q∣<1. 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 $|q| \lt 1$}.



∇×B⃗−,1c,∂E⃗∂t=4πcj⃗∇⋅E⃗=4πρ∇×E⃗,+,1c,∂B⃗∂t=0⃗∇⋅B⃗=0\begin{aligned} \nabla \times \vec{\mathbf{B}} -, \frac{1}{c}, \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}}, +, \frac{1}{c}, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{aligned}



x(t)=e∫t0tp(s)ds(∫t0t(q(s)e−∫t0sp(τ)dτ)ds+x0). x(t) = e^{\int_{t_0}^tp(s)ds}\Bigg(\int_{t_0}^t\Big(q(s)e^{-\int_{t_0}^sp(\tau)d\tau}\Big)ds + x_0\Bigg).



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