以《改变世界的17个方程》测试『神马笔记』方程编辑器功能。
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a 2 1 + 1 b 2 1 = 1 c 2 a^2\phantom{1}+\phantom{1}b^2\phantom{1}=\phantom{1}c^2 a21+1b21=1c2
l o g 1 x y = l o g 1 x + l o g 1 y log\phantom{1}xy = log\phantom{1}x + log\phantom{1}y log1xy=log1x+log1y
d f d t = lim h → 0 f ( t + h ) 1 − 1 f ( t ) h \frac{df}{dt}=\lim_{h \to 0}\frac{f(t+h)\phantom{1}-\phantom{1}f(t)}{h} dtdf=h→0limhf(t+h)1−1f(t)
F g = G m 1 m 2 r 2 F_g=\frac{Gm_1m_2}{r^2} Fg=r2Gm1m2
i 2 = − 1 i^2 = -1 i2=−1
F − E + V = 2 F-E+V=2 F−E+V=2
Φ ( x ) = 1 2 π σ e ( x − μ ) 2 2 σ 2 \Phi(x)=\frac{1}{\sqrt{2\pi\sigma}}e^{\frac{{(x-\mu)}^2}{2\sigma^2}} Φ(x)=2πσ 1e2σ2(x−μ)2
∂ 2 u ∂ t 2 = c 2 ∂ 2 u ∂ x 2 \frac{\partial^2u}{\partial{t^2}}=c^2\frac{\partial^2u}{\partial{x^2}} ∂t2∂2u=c2∂x2∂2u
f ^ ( ζ ) = ∫ − ∞ − ∞ f ( x ) e − 2 π i x ζ d x \hat{f}(\zeta)=\int_{-\infty}^{-\infty}f(x)e^{-2\pi{ix}\zeta}dx f^(ζ)=∫−∞−∞f(x)e−2πixζdx
ρ ( ∂ v ∂ t + v ⋅ ∇ v ) = − ∇ p + ∇ ⋅ T + f \rho\big(\frac{\partial{v}}{\partial{t}}+v\cdot\nabla{v}\big)=-\nabla{p}+\nabla\cdot{T}+f ρ(∂t∂v+v⋅∇v)=−∇p+∇⋅T+f
∇ ⋅ E = 0 ∇ × E = − 1 c ∂ H ∂ t ∇ ⋅ H = 0 ∇ × H = 1 c ∂ E ∂ t \begin{array}{ll} \nabla\cdot{E} = 0 & \nabla\times{E}=-\frac{1}{c}\frac{\partial{H}}{\partial{t}} \\ \nabla\cdot{H} = 0 & \nabla\times{H}=\frac{1}{c}\frac{\partial{E}}{\partial{t}} \end{array} ∇⋅E=0∇⋅H=0∇×E=−c1∂t∂H∇×H=c1∂t∂E
d S ≥ 0 dS\ge0 dS≥0
E = m c 2 E=mc^2 E=mc2
i ℏ ∂ Ψ ∂ t = − ℏ 2 2 m ∂ 2 Ψ ∂ x 2 i\hbar\frac{\partial{\Psi}}{\partial{t}}=-\frac{\hbar{^2}}{2m}\frac{\partial{^2}\Psi}{\partial{x^2}} iℏ∂t∂Ψ=−2mℏ2∂x2∂2Ψ
H = − ∑ p ( x ) l o g p ( x ) H=-\sum{p}(x)logp(x) H=−∑p(x)logp(x)
X t + 1 = k x t ( 1 − x t ) X_{t+1}=kx_t(1-x_t) Xt+1=kxt(1−xt)
1 2 σ 2 S 2 ∂ 2 V ∂ S 2 + r S ∂ V ∂ S + ∂ V ∂ t − r V = 0 \frac{1}{2}\sigma{^2}S^2\frac{\partial{^2}V}{\partial{S^2}}+ rS\frac{\partial{V}}{\partial{S}}+ \frac{\partial{V}}{\partial{t}}- rV= 0 21σ2S2∂S2∂2V+rS∂S∂V+∂t∂V−rV=0