Update 1D_tensor_parallel.md

docs/source/en/features/1D_tensor_parallel.md

@@ -19,9 +19,16 @@ An efficient 1D tensor parallelism implementation was introduced by [Megatron-LM
 
 Let's take a linear layer as an example, which consists of a GEMM $Y = XA$. Given 2 processors, we split the columns of $A$ into $[A_1 ~ A_2]$, and calculate $Y_i = XA_i$ on each processor, which then forms $[Y_1 ~ Y_2] = [XA_1 ~ XA_2]$. This is called a column-parallel fashion.
 
-When a second linear layer $Z=YB$ follows the column-parallel one, we split $B$ into $\left[\begin{matrix} B_1 \\ B_2 \end{matrix} \right]$,
+When a second linear layer $Z=YB$ follows the column-parallel one, we split $B$ into 
+```math
+\left[\begin{matrix} B_1 \\ B_2 \end{matrix} \right]
+```
 which is called a row-parallel fashion.
-To calculate $Z = [Y_1 ~ Y_2] \left[\begin{matrix} B_1 \\ B_2 \end{matrix} \right]$, we first calculate $Y_iB_i$ on each processor, then use an all-reduce to aggregate the results as $Z=Y_1B_1+Y_2B_2$.
+To calculate 
+```math
+Z = [Y_1 ~ Y_2] \left[\begin{matrix} B_1 \\ B_2 \end{matrix} \right]
+```
+we first calculate $Y_iB_i$ on each processor, then use an all-reduce to aggregate the results as $Z=Y_1B_1+Y_2B_2$.
 
 We also need to note that in the backward pass, the column-parallel linear layer needs to aggregate the gradients of the input tensor $X$, because on each processor $i$ we only have $\dot{X_i}=\dot{Y_i}A_i^T$.
 Thus, we apply an all-reduce across the processors to get $\dot{X}=\dot{Y}A^T=\dot{Y_1}A_1^T+\dot{Y_2}A_2^T$.