如何在博客园中插入公式

Posted 2020-10-09 皇家大鹏鹏

\\begin{align*}\\sin^2 A+\\sin^2B+\\sin^2C &=\\frac{1-\\cos 2A}{2}+\\frac{1-\\cos 2B}{2}+\\frac{1-\\cos 2C}{2}\\\\ &=\\frac{3}{2}-\\frac{1}{2}(\\cos 2A+\\cos 2B+\\cos 2C)\\\\ &=\\frac{3}{2}-\\cos(A+B)\\cos(A-B)-\\cos^2C+\\frac{1}{2}\\\\ &=2+\\cos C\\cos(A-B)-\\cos^2C \\\\ &\\leq 2+|\\cos C|-\\cos^2C \\\\ &=-(|\\cos C|-\\frac{1}{2})^2+\\frac{9}{4}\\\\ &\\leq \\frac{9}{4}.\\end{align*}

 

\\begin{align}a^2+b^2&=c^2\\\\\\beta^2&=\\sum_{n=0}^{\\infty}{\\frac{1}{n}}+\\int_0^1{\\frac{x}{e^x-1}dx}\\end{align}

 {\\sigma}_{k}(n)=\\sum_{d|n}^{}{d}^{k}

\\begin{align}
  (a + b)^3  &= (a + b) (a + b)^2        \\\\
             &= (a + b)(a^2 + 2ab + b^2) \\\\
             &= a^3 + 3a^2b + 3ab^2 + b^3
\\end{align}

 \\begin{flalign}
& q=
  \\begin{cases}
    0 & \\mbox{ if $(A+B)\\bmod 2 =0$}\\\\  第一个分支
    1 & \\mbox{ if $(A+B)\\bmod 4 =0$};\\\\ 第二个分支
    -1 & \\mbox{ if $(A-B)\\bmod 2=0$}\\\\  第三个分支
  \\end{cases}  & \\nonumber
\\end{flalign}

 \\begin{align}

E(S^2) &=E\\left(\\frac{1}{n} \\sum_{i=1}^n (X_i-\\bar{X})^2\\right) \\\\ & =E\\left(\\frac{1}{n}\\sum_{i=1}^n X_i^2\\right) - E\\left(\\frac{1}{n}\\sum_{i=1}^n 2\\bar{X}X_i\\right) + E\\left(\\frac{1}{n}\\sum_{i=1}^n \\bar{X}^2\\right) \\\\ & =EX^2 -E(\\bar{X}^2) \\\\ & =DX + (EX)^2 - D\\bar{X} - (E\\bar{X})^2 \\\\ & =\\frac{n-1}{n}DX \\end{align}

\\begin{equation*}
  \\begin{rcase}
    B\' &= -\\partial\\times E          \\\\
    E\' &=  \\partial\\times B - 4\\pi j \\,
  \\end{rcase}
  \\quad \\text {Maxwell\'s equations}
\\end{equation*}

 

\\begin{equation} \\begin{aligned}
  V_j &= v_j                      &
  X_i &= x_i - q_i x_j            &
      &= u_j + \\sum_{i\\ne j} q_i \\\\
  V_i &= v_i - q_i v_j            &
  X_j &= x_j                      &
  U_i &= u_i
\\end{aligned} \\end{equation}

 

\\begin{align*} f(x) & = (x+a)(x+b) \\\\ & = x^2 + (a+b)x + ab \\end{align*}

\\begin{gather*} E(X)=\\lambda \\qquad D(X)=\\lambda \\\\ E(\\bar{X})=\\lambda \\\\ D(\\bar{X})=\\frac{\\lambda}{n} \\\\ E(S^2)=\\frac{n-1}{n}\\lambda \\\\ \\end{gather*}

\\begin{align*}\\sin^2 A+\\sin^2B+\\sin^2C &=\\frac{1-\\cos 2A}{2}+\\frac{1-\\cos 2B}{2}+\\frac{1-\\cos 2C}{2}\\\\ &=\\frac{3}{2}-\\frac{1}{2}(\\cos 2A+\\cos 2B+\\cos 2C)\\\\ &=\\frac{3}{2}-\\cos(A+B)\\cos(A-B)-\\cos^2C+\\frac{1}{2}\\\\ &=2+\\cos C\\cos(A-B)-\\cos^2C \\\\ &\\leq 2+|\\cos C|-\\cos^2C \\\\ &=-(|\\cos C|-\\frac{1}{2})^2+\\frac{9}{4}\\\\ &\\leq \\frac{9}{4}.\\end{align*}

 

\\begin{align}a^2+b^2&=c^2\\\\\\beta^2&=\\sum_{n=0}^{\\infty}{\\frac{1}{n}}+\\int_0^1{\\frac{x}{e^x-1}dx}\\end{align}

 

参考链接如下：
http://www.cnblogs.com/cmt/p/3279312.html#!comments
http://www.cnblogs.com/apprenticeship/p/4215755.html
http://www.cnblogs.com/cmt/p/markdown-latex.html
http://blog.sina.com.cn/s/blog_5e16f1770100gror.html