fix rendering

Author: yzh119

_posts/2025-03-10-sampling.md

@@ -253,15 +253,15 @@ The sum of the coefficient $\sum_{\substack{r \notin T \\ p_r > \tau}} \frac{p_r
 
 $$0 \leq \frac{W(\tau)}{S(\tau)} < 1$$
 
-Suppose $\tau^*$ is the pivot where $|\Delta_j(\tau)|$ reach its maximum, if it's positive, we have:
+Suppose $\tau^*$ is the pivot where $\left|\Delta_j(\tau)\right|$ reach its maximum, if it's positive, we have:
 
 $$
 |\Delta_j(\tau^*)| \leq \sum_{\substack{r \notin T \\ p_r > \tau^*}}
         \frac{p_r}{S(\tau)} |\Delta_j(p_r)| \leq \sum_{\substack{r \notin T \\ p_r > \tau^*}}
         \frac{p_r}{S(\tau)} |\Delta_j(\tau^*)| = \frac{W(\tau^*)}{S(\tau^*)} |\Delta_j(\tau^*)| < |\Delta_j(\tau^*)|
 $$
 
-which is contradiction, which means $\Delta_j(\tau^*) = 0$, and our solution is unique.
+which leads to a contradiction, which means $\Delta_j(\tau^*) = 0$, and our solution is unique.
 
 The algorithm starts with $\tau = 0$; therefore