Processing math: 100%

Rexdf

The devil is in the Details.

测试数学公式

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[latex]r=rF+β(rMrF)+ϵ[/latex] [latex] ˙x=σ(yx) ˙y=ρxyxz ˙z=βz+xy [/latex] [latex] x^2+y1+z{12}^{34} \ \sin^{-1} (x) \ \lim\limits{h->0} \frac{f(x+h)-f(x)}{h} \ \lim\nolimits{h\to 0} \frac{f(x+h)-f(x)}{h} \ f(x)=\sum\limits{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \ \int0^1 f(x)dx \ \left[ 123 456 789 \right] \ \left[ ab cd \right] \left( n k \right) [/latex] [latex] \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) \ \mathbf{V}1 \times \mathbf{V}2 = |ijk XuYu0 XvYv0| \ P(E) = {n \choose k} p^k (1-p)^{ n-k} \ \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) 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 + \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 <1$}. \ ×B1cEt=4πcj E=4πρ ×E+1cBt=0 B=0 [/latex]\begin{equation} \label{eq1}r = r_F+ \beta (r_M - r_F) + \epsilon \end{equation}˙x=σ(yx) ˙y=ρxyxz ˙z=βz+xy\[ \mathbf{V}1 \times \mathbf{V}2 = |ijk XuYu0 XvYv0| \]x2+y1+z3412 sin1(x) limh>0f(x+h)f(x)h limh0f(x+h)f(x)h f(x)=n=0f(n)(a)n!(xa)n 10f(x)dx [123 456 789] [ab\cd](n\k)$$\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)\\mathbf{V}1 \times \mathbf{V}2 = |ijkfracXuYu0fracXvYv0|\P(E) = {n \choose k} p^k (1-p)^{ n-k}\\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) 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 + \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 <1}.\\begin{aligned}\nabla \times \vec{\mathbf{B}} -\, \frac1c\, \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}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{aligned}$

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rexdf:

#include <stdio.h>
#include <stdlib.h>
int main()
{
printf("Hello\n");
return 0;
}
i=i=0xii!

rexdf: i=i=0xii!

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