测试一下KaTex #1

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2023-01-23T22:24:53+08:00

skyone-wzw commented

2023-01-23 22:24:53 +08:00

测试数学公式 S=\pi r^2


\cos(x)


f(x)=\frac{1}{1+e^x}

测试数学公式 $S=\pi r^2$ $$ \cos(x) $$ $$ f(x)=\frac{1}{1+e^x} $$

skyone-wzw commented

2023-07-22 09:50:35 +08:00


\LaTeX


\KaTeX

$$ \LaTeX $$ $$ \KaTeX $$

skyone-wzw commented

2023-12-20 20:57:30 +08:00


D_n=\left|\begin{array}{}

   1+x_{1}^2   &   x_1x_2             &   x_1x_3      &   ...   &   x_1x_n      \\
   x_2x_1      &   1+x_{2}^2          &   x_2x_3      &   ...   &   x_2x_n      \\
   x_3x_1      &   x_3x_2             &   1+x_{3}^2   &   ...   &   x_3x_n      \\
   ...         &   ...                &   ...         &   ...   &   ...         \\
   x_nx_1      &   x_nx_2             &   x_nx_3      &   ...   &   1+x_{n}^2

\end{array}\right|

解法：加边升阶，得：


D_n=\left|\begin{array}{}

   1   &   x_1         &   x_2         &   x_3         &   ...   &   x_n         \\
   0   &   1+x_{1}^2   &   x_1x_2      &   x_1x_3      &   ...   &   x_1x_n      \\
   0   &   x_2x_1      &   1+x_{2}^2   &   x_2x_3      &   ...   &   x_2x_n      \\
   0   &   x_3x_1      &   x_3x_2      &   1+x_{3}^2   &   ...   &   x_3x_n      \\
   0   &   ...         &   ...         &   ...         &   ...   &   ...         \\
   0   &   x_nx_1      &   x_nx_2      &   x_nx_3      &   ...   &   1+x_{n}^2

\end{array}\right|

再将第 i,i=2...n+1 行都减去第一行的 x_i,i=1...n 倍，得：


D_n=\left|\begin{array}{}

   1      &   x_1   &   x_2   &   x_3   &   ...   &   x_n   \\
   -x_1   &   1     &   0     &   0     &   ...   &   0     \\
   -x_2   &   0     &   1     &   0     &   ...   &   0     \\
   -x_3   &   0     &   0     &   1     &   ...   &   0     \\
   0      &   ...   &   ...   &   ...   &   ...   &   ...   \\
   -x_n   &   0     &   0     &   0     &   ...   &   1

\end{array}\right|

即又化成了箭型行列式，可得通式：


D_n=1+\sum_{i=1}^{n}x_{i}^{2}

$$ D_n=\left|\begin{array}{} 1+x_{1}^2 & x_1x_2 & x_1x_3 & ... & x_1x_n \\ x_2x_1 & 1+x_{2}^2 & x_2x_3 & ... & x_2x_n \\ x_3x_1 & x_3x_2 & 1+x_{3}^2 & ... & x_3x_n \\ ... & ... & ... & ... & ... \\ x_nx_1 & x_nx_2 & x_nx_3 & ... & 1+x_{n}^2 \end{array}\right| $$ 解法：加边升阶，得： $$ D_n=\left|\begin{array}{} 1 & x_1 & x_2 & x_3 & ... & x_n \\ 0 & 1+x_{1}^2 & x_1x_2 & x_1x_3 & ... & x_1x_n \\ 0 & x_2x_1 & 1+x_{2}^2 & x_2x_3 & ... & x_2x_n \\ 0 & x_3x_1 & x_3x_2 & 1+x_{3}^2 & ... & x_3x_n \\ 0 & ... & ... & ... & ... & ... \\ 0 & x_nx_1 & x_nx_2 & x_nx_3 & ... & 1+x_{n}^2 \end{array}\right| $$ 再将第 $i,i=2...n+1$ 行都减去第一行的 $x_i,i=1...n$ 倍，得： $$ D_n=\left|\begin{array}{} 1 & x_1 & x_2 & x_3 & ... & x_n \\ -x_1 & 1 & 0 & 0 & ... & 0 \\ -x_2 & 0 & 1 & 0 & ... & 0 \\ -x_3 & 0 & 0 & 1 & ... & 0 \\ 0 & ... & ... & ... & ... & ... \\ -x_n & 0 & 0 & 0 & ... & 1 \end{array}\right| $$ 即又化成了箭型行列式，可得通式： $$ D_n=1+\sum_{i=1}^{n}x_{i}^{2} $$

skyone-wzw commented

2023-12-20 20:58:48 +08:00


D_n=\left|\begin{array}{}

a_{1}^n&a_{1}^{n-1}b_1&...&a_1b_1^{n-1}&b_1^n\\
a_{2}^n&a_{2}^{n-1}b_2&...&a_2b_2^{n-1}&b_2^n\\
...&...&...&...&...\\
a_{n}^n&a_{n}^{n-1}b_n&...&a_nb_n^{n-1}&b_n^n\\
a_{n+1}^n&a_{n+1}^{n-1}b_{n+1}&...&a_{n+1}b_{n+1}^{n-1}&b_{n+1}^n

\end{array}\right|

解法：将每行都提出 a_i^{n},i=1...n+1 倍，得：


D_n=\prod_{i=1}^{n+1}a_i^n\left|\begin{array}{}

1&\frac{b_1}{a_1}&...&(\frac{b_1}{a_1})^{n-1}&(\frac{b_1}{a_1})^{n}\\
1&\frac{b_2}{a_2}&...&(\frac{b_2}{a_2})^{n-1}&(\frac{b_2}{a_2})^{n}\\
...&...&...&...&...\\
1&\frac{b_n}{a_n}&...&(\frac{b_n}{a_n})^{n-1}&(\frac{b_n}{a_n})^{n}\\
1&\frac{b_{n+1}}{a_{n+1}}&...&(\frac{b_{n+1}}{a_{n+1}})^{n-1}&(\frac{b_{n+1}}{a_{n+1}})^{n}

\end{array}\right|

上式即为范德蒙德行列式，所以通式为：


D_n=\prod_{1\le+i<j\le+n+1}(a_ib_j-b_ia_j)

$$ D_n=\left|\begin{array}{} a_{1}^n&a_{1}^{n-1}b_1&...&a_1b_1^{n-1}&b_1^n\\ a_{2}^n&a_{2}^{n-1}b_2&...&a_2b_2^{n-1}&b_2^n\\ ...&...&...&...&...\\ a_{n}^n&a_{n}^{n-1}b_n&...&a_nb_n^{n-1}&b_n^n\\ a_{n+1}^n&a_{n+1}^{n-1}b_{n+1}&...&a_{n+1}b_{n+1}^{n-1}&b_{n+1}^n \end{array}\right| $$ 解法：将每行都提出 $a_i^{n},i=1...n+1$ 倍，得： $$ D_n=\prod_{i=1}^{n+1}a_i^n\left|\begin{array}{} 1&\frac{b_1}{a_1}&...&(\frac{b_1}{a_1})^{n-1}&(\frac{b_1}{a_1})^{n}\\ 1&\frac{b_2}{a_2}&...&(\frac{b_2}{a_2})^{n-1}&(\frac{b_2}{a_2})^{n}\\ ...&...&...&...&...\\ 1&\frac{b_n}{a_n}&...&(\frac{b_n}{a_n})^{n-1}&(\frac{b_n}{a_n})^{n}\\ 1&\frac{b_{n+1}}{a_{n+1}}&...&(\frac{b_{n+1}}{a_{n+1}})^{n-1}&(\frac{b_{n+1}}{a_{n+1}})^{n} \end{array}\right| $$ 上式即为范德蒙德行列式，所以通式为： $$ D_n=\prod_{1\le+i<j\le+n+1}(a_ib_j-b_ia_j) $$

skyone-wzw closed this issue

2023-12-20 21:03:53 +08:00

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Reference: skyone-wzw/SiteTest#1