论坛 › LaTeX/ScienceWord论坛 › MadioLaTex--网页公式编辑器使用教程

suolunga 发表于 2009-10-9 09:47

http://latex.codecogs.com/gif.latex?\begin{cases} \frac{di }{dt}=\lambda si-\mu i, & i(0)=i_0 \\ \frac{ds}{dt}=-\lambda si, & s(0)=s_0 \end{cases}

suolunga 发表于 2009-10-9 10:01

呵呵~复杂一点儿的数学公式!!!!!!!!!!!!!!!
http://latex.codecogs.com/gif.latex?\begin{cases} f(x_1,x_2)\equiv r_1x_1(1-\frac{x_1}{N_1}-\sigma _1\frac{x_2}{N_2})=0 \\ g(x_1,x_2)\equiv r_2x_2(1-\sigma _2\frac{x_1}{N_1}-\frac{x_2}{N_2})=0 \end{cases}\\ \\ \\ \\ A=\begin{vmatrix} f_{x_1} & f_{x_2}\\ g_{x_1} & g_{x_2} \end{vmatrix}=\begin{vmatrix} r_1(1-\frac{2x_1}{N_1}-\frac{\sigma _1x_2}{N_2}) & -\frac{r_1\sigma_1x_1}{N_2}\\ -\frac{r_2\sigma_2x_2}{N_1} & r_2(1-\frac{\sigma _2x_1}{N_1}-\frac{2x_2}{N_2}) \end{vmatrix}

suolunga 发表于 2009-10-9 10:03

http://latex.codecogs.com/gif.latex?\begin{cases} f(x_1,x_2)\equiv r_1x_1(1-\displaystyle\frac{x_1}{N_1}-\displaystyle\sigma _1\frac{x_2}{N_2})=0 \\ g(x_1,x_2)\equiv r_2x_2(1-\displaystyle\sigma _2\frac{x_1}{N_1}-\displaystyle\frac{x_2}{N_2})=0 \end{cases}\\ \\ \\ \\ A=\begin{vmatrix} f_{x_1} & f_{x_2}\\ g_{x_1} & g_{x_2} \end{vmatrix}=\begin{vmatrix} r_1(1-\displaystyle\frac{2x_1}{N_1}-\displaystyle\frac{\sigma _1x_2}{N_2}) & -\displaystyle\frac{r_1\sigma_1x_1}{N_2}\\ -\displaystyle\frac{r_2\sigma_2x_2}{N_1} & r_2(1-\displaystyle\frac{\sigma _2x_1}{N_1}-\displaystyle\frac{2x_2}{N_2}) \end{vmatrix}

suolunga 发表于 2009-10-9 10:28

平衡点 http://latex.codecogs.com/gif.latex?p http://latex.codecogs.com/gif.latex?q 稳定条件 http://latex.codecogs.com/gif.latex?p_1(N_1,0) http://latex.codecogs.com/gif.latex?r_1-r_2(1-\sigma_2) http://latex.codecogs.com/gif.latex?-r_1r_2(1-\sigma_2) 1" name=equationview>http://latex.codecogs.com/gif.latex?p_2(0,N_2) http://latex.codecogs.com/gif.latex?-r_1(1-\sigma_1)+r_2 http://latex.codecogs.com/gif.latex?-r_1r_2(1-\sigma_1) 1,\sigma_2 http://latex.codecogs.com/gif.latex?p_3\Big(\frac{N_1(1-\sigma_1)}{1-\sigma_1\sigma_2},\frac{N_2(1-\sigma_2)}{1-\sigma_1\sigma_2}\Big)   http://latex.codecogs.com/gif.latex?\frac{r_1(1-\sigma_1)+r_2(1-\sigma_2)}{1-\sigma_1\sigma_2} http://latex.codecogs.com/gif.latex?\frac{r_1r_2(1-\sigma_1)(1-\sigma_2)}{1-\sigma_1\sigma_2} http://latex.codecogs.com/gif.latex?\sigma_1<1,\sigma_2<1 http://latex.codecogs.com/gif.latex?p_4(0,0) http://latex.codecogs.com/gif.latex?-(r_1+r_2) http://latex.codecogs.com/gif.latex?r_1r_2 不稳定

sdoerhang 发表于 2009-10-9 11:15

http://latex.codecogs.com/gif.latex?\sum_{i=1}^{100}{}a_{i}=b_{}

dongdong3344 发表于 2009-10-9 12:08

哈哈 应该不错吧

dongdong3344 发表于 2009-10-9 12:09

非常感谢楼主分享哈

icymirror 发表于 2009-10-9 17:49

1# huashi3483 http://latex.codecogs.com/gif.latex?1.%20%5Cln%28x+1%29%3Dx-%5Cfrac%7Bx%5E%7B2%7D%7D%7B2%7D%20+%20%5Cfrac%7Bx%5E3%7B3%7D%7D%7B3%7D%20-%20...%28%7Cx%7C%20%3C%201%29%202.%20%7Bd%20e%5Ex%5Cover%20dx%7D%20%3D%20e%5Ex

pinquion 发表于 2009-10-9 20:45

http://latex.codecogs.com/gif.latex?%5Csqrt%5B3%5D%7B27%7D

pinquion 发表于 2009-10-9 20:46

看到成功了，好玩，搞个复杂的看看，

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