fix 9.35

docs/chapter9/chapter9.md

@@ -211,7 +211,13 @@ $$
 \sum_{j=1}^m\sum_{i=1}^k\frac{\alpha_{i}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{i},\mathbf\Sigma_{i})}{\sum_{l=1}^k\alpha_{l}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{l},\mathbf\Sigma_{l})}&=-\lambda\sum_{i=1}^k\alpha_{i}
 \end{aligned}
 $$
-由$$m=-\lambda$$有
+因为
+$$
+\sum_{i=1}^k\frac{\alpha_{i}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{i},\mathbf\Sigma_{i})}{\sum_{l=1}^k\alpha_{l}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{l},\mathbf\Sigma_{l})}=\frac{\sum_{i=1}^k\alpha_{i}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{i},\mathbf\Sigma_{i})}{\sum_{l=1}^k\alpha_{l}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{l},\mathbf\Sigma_{l})}=1
+$$
+且 $\sum_{i=1}^k\alpha_{i}=1$，所以有$$m=-\lambda$$
+
+因此
 $$
 \sum_{j=1}^m\frac{\alpha_{i}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{i},\mathbf\Sigma_{i})}{\sum_{l=1}^k\alpha_{l}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{l},\mathbf\Sigma_{l})}=-\lambda\alpha_{i}=m\alpha_{i}
 $$