Merge pull request #126 from xhqing/master

fix math formulas render problem

README.md

@@ -80,4 +80,4 @@ PDF版本是我们寄送出版社的全书初稿，经由人民邮电出版社
 </div>
 
 ## LICENSE
-<a rel="license" href="http://creativecommons.org/licenses/by-nc-sa/4.0/"><img alt="知识共享许可协议" style="border-width:0" src="https://img.shields.io/badge/license-CC%20BY--NC--SA%204.0-lightgrey" /></a><br />本作品采用<a rel="license" href="http://creativecommons.org/licenses/by-nc-sa/4.0/">知识共享署名-非商业性使用-相同方式共享 4.0 国际许可协议</a>进行许可。
+<a rel="license" href="http://creativecommons.org/licenses/by-nc-sa/4.0/"><img alt="知识共享许可协议" style="border-width:0" src="https://img.shields.io/badge/license-CC%20BY--NC--SA%204.0-lightgrey" /></a><br />本作品采用<a rel="license" href="http://creativecommons.org/licenses/by-nc-sa/4.0/">知识共享署名-非商业性使用-相同方式共享 4.0 国际许可协议</a>进行许可。

docs/chapter10/chapter10.md

@@ -48,7 +48,7 @@ $$
 ## 10.4
 
 $$
-\sum^m_{i=1}dist^2_{ij}=tr(\boldsymbol B)+mb_{jj}
+\sum^m_{i=1}dist^2_{ij}=\operatorname{tr}(\boldsymbol B)+mb_{jj}
 $$
 
 [解析]：首先根据式10.3有
@@ -119,7 +119,7 @@ $$
 由公式（10.6）和（10.9）可得
 $$
 \begin{aligned}
-tr(\boldsymbol B)&=\frac{1}{2m}\sum^m_{i=1}\sum^m_{j=1}dist^2_{ij}\\
+\operatorname{tr}(\boldsymbol B)&=\frac{1}{2m}\sum^m_{i=1}\sum^m_{j=1}dist^2_{ij}\\
 &=\frac{m}{2}dist^2_{\cdot}
 \end{aligned}
 $$
@@ -133,7 +133,7 @@ $$
 由公式（10.5）和（10.7）可得
 $$
 \begin{aligned}
-b_{ii}&=\frac{1}{m}\sum^m_{j=1}dist^2_{ij}-\frac{1}{m}tr(\boldsymbol B)\\
+b_{ii}&=\frac{1}{m}\sum^m_{j=1}dist^2_{ij}-\frac{1}{m}\operatorname{tr}(\boldsymbol B)\\
 &=dist^2_{i\cdot}-\frac{1}{2}dist^2_{\cdot}
 \end{aligned}
 $$
@@ -379,7 +379,7 @@ $$
 ## 10.31
 $$
 \begin{aligned}
-&\min\limits_{\boldsymbol Z}tr(\boldsymbol Z \boldsymbol M \boldsymbol Z^T)\\
+&\min\limits_{\boldsymbol Z}\operatorname{tr}(\boldsymbol Z \boldsymbol M \boldsymbol Z^T)\\
 &s.t. \boldsymbol Z^T\boldsymbol Z=\boldsymbol I. 
 \end{aligned}
 $$
@@ -393,13 +393,12 @@ $$
 &=\sum^m_{i=1}\|\boldsymbol Z(\boldsymbol I_i-\boldsymbol W_i)\|^2_2\\
 &=\sum^m_{i=1}(\boldsymbol Z(\boldsymbol I_i-\boldsymbol W_i))^T\boldsymbol Z(\boldsymbol I_i-\boldsymbol W_i)\\
 &=\sum^m_{i=1}(\boldsymbol I_i-\boldsymbol W_i)^T\boldsymbol Z^T\boldsymbol Z(\boldsymbol I_i-\boldsymbol W_i)\\
-&=tr((\boldsymbol I-\boldsymbol W)^T\boldsymbol Z^T\boldsymbol Z(\boldsymbol I-\boldsymbol W))\\
-&=tr(\boldsymbol Z(\boldsymbol I-\boldsymbol W)(\boldsymbol I-\boldsymbol W)^T\boldsymbol Z^T)\\
-&=tr(\boldsymbol Z\boldsymbol M\boldsymbol Z^T)
+&=\operatorname{tr}((\boldsymbol I-\boldsymbol W)^T\boldsymbol Z^T\boldsymbol Z(\boldsymbol I-\boldsymbol W))\\
+&=\operatorname{tr}(\boldsymbol Z(\boldsymbol I-\boldsymbol W)(\boldsymbol I-\boldsymbol W)^T\boldsymbol Z^T)\\
+&=\operatorname{tr}(\boldsymbol Z\boldsymbol M\boldsymbol Z^T)
 \end{aligned}
 $$
 
-
 其中，$\boldsymbol M=(\boldsymbol I-\boldsymbol W)(\boldsymbol I-\boldsymbol W)^T$。
 [解析]：约束条件$\boldsymbol Z^T\boldsymbol Z=\boldsymbol I$是为了得到标准化（标准正交空间）的低维数据。
 

docs/index.html

@@ -28,12 +28,10 @@
     }
   </script>
   <!-- CDN files for docsify-katex -->
-  <script src="//cdn.jsdelivr.net/npm/docsify-katex@latest/dist/docsify-katex.js"></script>
-  <!-- or <script src="//cdn.jsdelivr.net/gh/upupming/docsify-katex/dist/docsify-katex.js"></script> -->
-  <link
-    rel="stylesheet"
-    href="//cdn.jsdelivr.net/npm/katex@0.10.2/dist/katex.min.css"
-  />
+  <!-- script src="//cdn.jsdelivr.net/npm/docsify-katex@latest/dist/docsify-katex.js"></script> -->
+  <script src="//cdn.jsdelivr.net/gh/upupming/docsify-katex@latest/dist/docsify-katex.js"></script>
+  <link rel="stylesheet" href="//cdn.jsdelivr.net/npm/katex@latest/dist/katex.min.css"/>
+
   <!-- Put them above docsify.min.js -->
   <script src="//cdn.jsdelivr.net/npm/docsify@latest/lib/docsify.min.js"></script>
 </body>