typo

doc/src/week35/week35.do.txt

@@ -317,11 +317,11 @@ $\bm{z}^T=\bm{y}^T\bm{A}$, a vector of length $n$. We have then, using the defin
 which means that (using our previous example) we have
 !bt
 \[
-\frac{\partial \alpha}{\partial \bm{x}} = \bm{z}=bm{A}^T\bm{y}.
+\frac{\partial \alpha}{\partial \bm{x}} = \bm{z}=\bm{A}^T\bm{y}.
 \]
 !et
 
-Note that the resulting vector elements are the same for $\bm{z}^T$ and $\bm{z}$, the only difference is that one if just the transpose of the other.
+Note that the resulting vector elements are the same for $\bm{z}^T$ and $\bm{z}$, the only difference is that one is just the transpose of the other.
 
 Since $\alpha$ is a scalar we have $\alpha =\alpha^T=\bm{x}^T\bm{A}^T\bm{y}$. Defining now $\bm{z}=\bm{x}^T\bm{A}^T$ we find that
 !bt