CSDN中编辑数学公式的几种代码,根据代码去分析每种符号的表示方法吧,只能找到这些了^-^

注意$$之间的是另起一行显示
$在当前位置显示
这里是在$$代码$$显示的情况
\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt\,.
Γ ( z ) = 0 t z 1 e t d t . \Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt\,.
\ f(x)=\int_0^\infty t{z-1}dt\,.
  f ( x ) = 0 t z 1 d t . \ f(x)=\int_0^\infty t{z-1}dt\,.
\ f{x} = \int_{-\infty}^\infty \hat f\xi\,e^{2 \pi i \xi x} \,d\xi
  f x = f ^ ξ e 2 π i ξ x d ξ \ f{x} = \int_{-\infty}^\infty \hat f\xi\,e^{2 \pi i \xi x} \,d\xi
这里就是$代码$显示
$\Gamma(n) = (n-1)!\quad\forall n\in\mathbb N
Γ ( n ) = ( n 1 ) ! n N \Gamma(n) = (n-1)!\quad\forall n\in\mathbb N

2.度为m的树中第i层上至多有 m i 1 m^{i-1} 这里就是$ $显示
\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+\cdots} } } }

1 ( ϕ 5 ϕ ) e 2 5 π = 1 + e 2 π 1 + e 4 π 1 + e 6 π 1 + e 8 π 1 + \displaystyle \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+\cdots} } } }
可以不用\displaystyle
\displaystyle \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)
( k = 1 n a k b k ) 2 ( k = 1 n a k 2 ) ( k = 1 n b k 2 ) \displaystyle \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)
\displaystyle {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 }\lvert q\rvert<1.
1 + q 2 ( 1 q ) + q 6 ( 1 q ) ( 1 q 2 ) + = j = 0 1 ( 1 q 5 j + 2 ) ( 1 q 5 j + 3 ) , for  q < 1. \displaystyle {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 }\lvert q\rvert<1.

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转载自blog.csdn.net/qq_43625764/article/details/103915722