• 如何在博客园中插入公式


    egin{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 Ccos(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*}

    egin{align}a^2+b^2&=c^2\eta^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}

    egin{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}

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

     egin{align}

    E(S^2) &=Eleft(frac{1}{n} sum_{i=1}^n (X_i-ar{X})^2 ight) \ & =Eleft(frac{1}{n}sum_{i=1}^n X_i^2 ight) - Eleft(frac{1}{n}sum_{i=1}^n 2ar{X}X_i ight) + Eleft(frac{1}{n}sum_{i=1}^n ar{X}^2 ight) \ & =EX^2 -E(ar{X}^2) \ & =DX + (EX)^2 - Dar{X} - (Ear{X})^2 \ & =frac{n-1}{n}DX end{align}

    egin{equation*}
      egin{rcase}
        B' &= -partial imes E          \
        E' &=  partial imes B - 4pi j \,
      end{rcase}
      quad ext {Maxwell's equations}
    end{equation*}

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

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

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

    egin{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 Ccos(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*}
    
     
    
    egin{align}a^2+b^2&=c^2\eta^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
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  • 原文地址:https://www.cnblogs.com/dapeng-bupt/p/7628904.html
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