This essay demonstrates how to write technical content with full LaTeX math support.
Write inline math with single dollar signs: f ( x ) = ∑ n = 0 ∞ f ( n ) ( a ) n ! ( x − a ) n f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n f ( x ) = ∑ n = 0 ∞ n ! f ( n ) ( a ) ( x − a ) n .
Greek letters work naturally: α , β , γ , δ , ϵ , θ , λ , μ , σ , ϕ , ψ , ω \alpha, \beta, \gamma, \delta, \epsilon, \theta, \lambda, \mu, \sigma, \phi, \psi, \omega α , β , γ , δ , ϵ , θ , λ , μ , σ , ϕ , ψ , ω .
Use double dollar signs for display mode:
F { f ( t ) } = ∫ − ∞ ∞ f ( t ) e − 2 π i ξ t d t \mathcal{F}\{f(t)\} = \int_{-\infty}^{\infty} f(t) e^{-2\pi i \xi t} \, dt F { f ( t )} = ∫ − ∞ ∞ f ( t ) e − 2 π i ξ t d t
For important equations you want to reference:
∇ 2 ϕ = ∂ 2 ϕ ∂ x 2 + ∂ 2 ϕ ∂ y 2 + ∂ 2 ϕ ∂ z 2 = 0 (1) \nabla^2 \phi = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2} = 0 \tag{1} ∇ 2 ϕ = ∂ x 2 ∂ 2 ϕ + ∂ y 2 ∂ 2 ϕ + ∂ z 2 ∂ 2 ϕ = 0 ( 1 )
Laplace’s equation (1) appears throughout physics.
For multi-line derivations:
d d x ln ( x ) = lim h → 0 ln ( x + h ) − ln ( x ) h = lim h → 0 1 h ln ( x + h x ) = lim h → 0 1 h ln ( 1 + h x ) = 1 x \begin{aligned}
\frac{d}{dx} \ln(x) &= \lim_{h \to 0} \frac{\ln(x+h) - \ln(x)}{h} \\
&= \lim_{h \to 0} \frac{1}{h} \ln\left(\frac{x+h}{x}\right) \\
&= \lim_{h \to 0} \frac{1}{h} \ln\left(1 + \frac{h}{x}\right) \\
&= \frac{1}{x}
\end{aligned} d x d ln ( x ) = h → 0 lim h ln ( x + h ) − ln ( x ) = h → 0 lim h 1 ln ( x x + h ) = h → 0 lim h 1 ln ( 1 + x h ) = x 1
det ( A − λ I ) = det ( a 11 − λ a 12 ⋯ a 1 n a 21 a 22 − λ ⋯ a 2 n ⋮ ⋮ ⋱ ⋮ a n 1 a n 2 ⋯ a n n − λ ) = 0 \det(A - \lambda I) = \det \begin{pmatrix}
a_{11} - \lambda & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} - \lambda & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn} - \lambda
\end{pmatrix} = 0 det ( A − λ I ) = det a 11 − λ a 21 ⋮ a n 1 a 12 a 22 − λ ⋮ a n 2 ⋯ ⋯ ⋱ ⋯ a 1 n a 2 n ⋮ a nn − λ = 0
Theorem (Fundamental Theorem of Calculus). Let f f f be continuous on [ a , b ] [a,b] [ a , b ] and let F F F be any antiderivative of f f f . Then:
∫ a b f ( x ) d x = F ( b ) − F ( a ) \int_a^b f(x) \, dx = F(b) - F(a) ∫ a b f ( x ) d x = F ( b ) − F ( a )
Proof. Let G ( x ) = ∫ a x f ( t ) d t G(x) = \int_a^x f(t) \, dt G ( x ) = ∫ a x f ( t ) d t . By the first part of the FTC, G ′ ( x ) = f ( x ) G'(x) = f(x) G ′ ( x ) = f ( x ) .
Since F F F is also an antiderivative, F ( x ) = G ( x ) + C F(x) = G(x) + C F ( x ) = G ( x ) + C for some constant.
Then F ( b ) − F ( a ) = G ( b ) − G ( a ) = ∫ a b f ( t ) d t − 0 F(b) - F(a) = G(b) - G(a) = \int_a^b f(t) \, dt - 0 F ( b ) − F ( a ) = G ( b ) − G ( a ) = ∫ a b f ( t ) d t − 0 . ∎
Maxwell’s equations in differential form:
∇ ⋅ E = ρ ε 0 (Gauss’s law) ∇ ⋅ B = 0 (No magnetic monopoles) ∇ × E = − ∂ B ∂ t (Faraday’s law) ∇ × B = μ 0 J + μ 0 ε 0 ∂ E ∂ t (Amp e ˋ re-Maxwell) \begin{aligned}
\nabla \cdot \mathbf{E} &= \frac{\rho}{\varepsilon_0} & \text{(Gauss's law)} \\
\nabla \cdot \mathbf{B} &= 0 & \text{(No magnetic monopoles)} \\
\nabla \times \mathbf{E} &= -\frac{\partial \mathbf{B}}{\partial t} & \text{(Faraday's law)} \\
\nabla \times \mathbf{B} &= \mu_0 \mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t} & \text{(Ampère-Maxwell)}
\end{aligned} ∇ ⋅ E ∇ ⋅ B ∇ × E ∇ × B = ε 0 ρ = 0 = − ∂ t ∂ B = μ 0 J + μ 0 ε 0 ∂ t ∂ E (Gauss’s law) (No magnetic monopoles) (Faraday’s law) (Amp e ˋ re-Maxwell)
The cross-entropy loss for classification:
L ( θ ) = − 1 N ∑ i = 1 N ∑ c = 1 C y i , c log ( y ^ i , c ) \mathcal{L}(\theta) = -\frac{1}{N} \sum_{i=1}^{N} \sum_{c=1}^{C} y_{i,c} \log(\hat{y}_{i,c}) L ( θ ) = − N 1 i = 1 ∑ N c = 1 ∑ C y i , c log ( y ^ i , c )
Backpropagation gradient:
∂ L ∂ w j k ( l ) = ∂ L ∂ a k ( l ) ⋅ ∂ a k ( l ) ∂ z k ( l ) ⋅ ∂ z k ( l ) ∂ w j k ( l ) = δ k ( l ) ⋅ a j ( l − 1 ) \frac{\partial \mathcal{L}}{\partial w_{jk}^{(l)}} = \frac{\partial \mathcal{L}}{\partial a_k^{(l)}} \cdot \frac{\partial a_k^{(l)}}{\partial z_k^{(l)}} \cdot \frac{\partial z_k^{(l)}}{\partial w_{jk}^{(l)}} = \delta_k^{(l)} \cdot a_j^{(l-1)} ∂ w j k ( l ) ∂ L = ∂ a k ( l ) ∂ L ⋅ ∂ z k ( l ) ∂ a k ( l ) ⋅ ∂ w j k ( l ) ∂ z k ( l ) = δ k ( l ) ⋅ a j ( l − 1 )
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