修订9.35符号问题

docs/chapter9/chapter9.md

@@ -49,8 +49,8 @@ $$\begin{aligned}
 &=\sum_{j=1}^m\frac{\partial}{\partial\boldsymbol\Sigma_{i}}ln\Bigg(\sum_{i=1}^k \alpha_{i}\cdot p(\boldsymbol{x_{j}}|\boldsymbol\mu_{i},\boldsymbol\Sigma_{i})\Bigg) \\
 &=\sum_{j=1}^m \frac{\alpha_{i}\cdot \frac{\partial}{\partial\boldsymbol\Sigma_{i}}p(\boldsymbol x_{j}|\boldsymbol \mu_{i},\boldsymbol\Sigma_{i})}{\sum_{l=1}^k\alpha_{l}\cdot p(\boldsymbol x_{j},|\boldsymbol \mu_{l},\boldsymbol\Sigma_{l})} \\
 &=\sum_{j=1}^m \frac{\alpha_{i}\cdot \frac{\partial}{\partial\boldsymbol\Sigma_{i}}\frac{1}{(2\pi)^\frac{n}{2}\left| \boldsymbol\Sigma_{i}\right |^\frac{1}{2}}e^{-\frac{1}{2}(\boldsymbol{x_{j}}-\boldsymbol\mu_{i})^T\boldsymbol\Sigma_{i}^{-1}(\boldsymbol{x_{j}-\mu_{i}})}}{\sum_{l=1}^k\alpha_{l}\cdot p(\boldsymbol x_{j},|\boldsymbol \mu_{l},\boldsymbol\Sigma_{l})}\\
-&=\sum_{j=1}^m \frac{\alpha_{i}\cdot \frac{\partial}{\partial\boldsymbol\Sigma_{i}}e^{ln\left(\frac{1}{(2\pi)^\frac{n}{2}\left| \boldsymbol\Sigma_{i}\right |^\frac{1}{2}}e^{-\frac{1}{2}(\boldsymbol{x_{j}}-\boldsymbol\mu_{i})^T\boldsymbol\Sigma_{i}^{-1}(\boldsymbol{x_{j}-\mu_{i}})}\right)}}{\sum_{l=1}^k\alpha_{l}\cdot p(\boldsymbol x_{j},|\boldsymbol \mu_{l},\boldsymbol\Sigma_{l}} \\
-&=\sum_{j=1}^m \frac{\alpha_{i}\cdot \frac{\partial}{\partial\boldsymbol\Sigma_{i}}e^{-\frac{n}{2}ln\left(2\pi\right)-\frac{1}{2}ln\left(|\boldsymbol\Sigma_{i}|\right)-\frac{1}{2}\left(\boldsymbol{x_{j}-\mu_{i}}\right)^T\boldsymbol\Sigma_{i}^{-1}\left(\boldsymbol{x_{j}-\mu_{i}}\right)}}{\sum_{l=1}^k\alpha_{l}\cdot p(\boldsymbol x_{j},|\boldsymbol \mu_{l},\boldsymbol\Sigma_{l}} \\
+&=\sum_{j=1}^m \frac{\alpha_{i}\cdot \frac{\partial}{\partial\boldsymbol\Sigma_{i}}e^{ln\left(\frac{1}{(2\pi)^\frac{n}{2}\left| \boldsymbol\Sigma_{i}\right |^\frac{1}{2}}e^{-\frac{1}{2}(\boldsymbol{x_{j}}-\boldsymbol\mu_{i})^T\boldsymbol\Sigma_{i}^{-1}(\boldsymbol{x_{j}-\mu_{i}})}\right)}}{\sum_{l=1}^k\alpha_{l}\cdot p(\boldsymbol x_{j},|\boldsymbol \mu_{l},\boldsymbol\Sigma_{l})} \\
+&=\sum_{j=1}^m \frac{\alpha_{i}\cdot \frac{\partial}{\partial\boldsymbol\Sigma_{i}}e^{-\frac{n}{2}ln\left(2\pi\right)-\frac{1}{2}ln\left(|\boldsymbol\Sigma_{i}|\right)-\frac{1}{2}\left(\boldsymbol{x_{j}-\mu_{i}}\right)^T\boldsymbol\Sigma_{i}^{-1}\left(\boldsymbol{x_{j}-\mu_{i}}\right)}}{\sum_{l=1}^k\alpha_{l}\cdot p(\boldsymbol x_{j},|\boldsymbol \mu_{l},\boldsymbol\Sigma_{l})} \\
 &=\sum_{j=1}^m \frac{\alpha_{i}\cdot p(\boldsymbol x_{j}|\boldsymbol \mu_{i},\boldsymbol\Sigma_{i})}{\sum_{l=1}^k\alpha_{l}\cdot p(\boldsymbol x_{j},|\boldsymbol \mu_{l},\boldsymbol\Sigma_{l})}\frac{\partial}{\partial\boldsymbol\Sigma_{i}}\left(-\frac{n}{2}ln\left(2\pi\right)-\frac{1}{2}ln\left(|\boldsymbol\Sigma_{i}|\right)-\frac{1}{2}\left(\boldsymbol{x_{j}-\mu_{i}}\right)^T\boldsymbol\Sigma_{i}^{-1}\left(\boldsymbol{x_{j}-\mu_{i}}\right)\right) \\
 &=\sum_{j=1}^m \frac{\alpha_{i}\cdot p(\boldsymbol x_{j}|\boldsymbol \mu_{i},\boldsymbol\Sigma_{i})}{\sum_{l=1}^k\alpha_{l}\cdot p(\boldsymbol x_{j},|\boldsymbol \mu_{l},\boldsymbol\Sigma_{l})}\left(-\frac{1}{2}\left(\boldsymbol\Sigma_{i}^{-1}\right)^T-\frac{1}{2}\frac{\partial}{\partial\boldsymbol\Sigma_{i}}\left(\boldsymbol{x_{j}-\mu_{i}}\right)^T\boldsymbol\Sigma_{i}^{-1}\left(\boldsymbol{x_{j}-\mu_{i}}\right)\right)
 \end{aligned}$$