small typo

doc/src/week35/week35.do.txt

@@ -317,7 +317,7 @@ multiplications
 !et
 with $\bm{y}$ a vector of length $m$, $\bm{A}$ an $m\times n$ matrix and $\bm{x}$ a vector of length $n$. We assume also that $\bm{A}$ does not depend on any of the two vectors.
 In order to find the derivative of $\alpha$ with respect to the two vectors, we define an intermediate vector $\bm{z}$. We define first
-$\bm{z}^T=\bm{y}^T\bm{A}$, a vector of length $n$. We have then
+$\bm{z}^T=\bm{y}^T\bm{A}$, a vector of length $n$. We have then, using the definition of the Jacobian,
 !bt
 \[
 \alpha = \bm{z}^T\bm{x},
@@ -326,10 +326,12 @@ $\bm{z}^T=\bm{y}^T\bm{A}$, a vector of length $n$. We have then
 which means that (using our previous example) we have
 !bt
 \[
-\frac{\partial \alpha}{\partial \bm{x}} = \bm{z}^T=\bm{y}^T\bm{A}.
+\frac{\partial \alpha}{\partial \bm{x}} = \bm{z}=bm{A}^T\bm{y}.
 \]
 !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.
+
 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
 \[