corrected typos

doc/pub/week35/html/._week35-bs010.html

@@ -381,10 +381,10 @@ $$
 
 <p>which means that (using our previous example and keeping track of our definition of the derivative of a scalar) we have</p>
 $$
-\frac{\partial \alpha}{\partial \boldsymbol{x}} = \frac{\partial \boldsymbol{z}^T\boldsymbol{x}}{\partial \boldsymbol{x}}=\boldsymbol{z}^T=\boldsymbol{A}^T\boldsymbol{y}.
+\frac{\partial \alpha}{\partial \boldsymbol{x}} = \frac{\partial \boldsymbol{z}^T\boldsymbol{x}}{\partial \boldsymbol{x}}=\boldsymbol{z}^T.
 $$
 
-<p>Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one is just the transpose of the other.</p>
+<p>Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one is just the transpose of the other. We have the transposed here since we have used that the inner product of two vectors is a scalar.</p>
 
 <p>Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that</p>
 $$

doc/pub/week35/html/week35-reveal.html

@@ -540,11 +540,11 @@ $$
 <p>which means that (using our previous example and keeping track of our definition of the derivative of a scalar) we have</p>
 <p>&nbsp;<br>
 $$
-\frac{\partial \alpha}{\partial \boldsymbol{x}} = \frac{\partial \boldsymbol{z}^T\boldsymbol{x}}{\partial \boldsymbol{x}}=\boldsymbol{z}^T=\boldsymbol{A}^T\boldsymbol{y}.
+\frac{\partial \alpha}{\partial \boldsymbol{x}} = \frac{\partial \boldsymbol{z}^T\boldsymbol{x}}{\partial \boldsymbol{x}}=\boldsymbol{z}^T.
 $$
 <p>&nbsp;<br>
 
-<p>Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one is just the transpose of the other.</p>
+<p>Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one is just the transpose of the other. We have the transposed here since we have used that the inner product of two vectors is a scalar.</p>
 
 <p>Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that</p>
 <p>&nbsp;<br>

doc/pub/week35/html/week35-solarized.html

@@ -604,10 +604,10 @@ $$
 
 <p>which means that (using our previous example and keeping track of our definition of the derivative of a scalar) we have</p>
 $$
-\frac{\partial \alpha}{\partial \boldsymbol{x}} = \frac{\partial \boldsymbol{z}^T\boldsymbol{x}}{\partial \boldsymbol{x}}=\boldsymbol{z}^T=\boldsymbol{A}^T\boldsymbol{y}.
+\frac{\partial \alpha}{\partial \boldsymbol{x}} = \frac{\partial \boldsymbol{z}^T\boldsymbol{x}}{\partial \boldsymbol{x}}=\boldsymbol{z}^T.
 $$
 
-<p>Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one is just the transpose of the other.</p>
+<p>Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one is just the transpose of the other. We have the transposed here since we have used that the inner product of two vectors is a scalar.</p>
 
 <p>Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that</p>
 $$

doc/pub/week35/html/week35.html

@@ -681,10 +681,10 @@ $$
 
 <p>which means that (using our previous example and keeping track of our definition of the derivative of a scalar) we have</p>
 $$
-\frac{\partial \alpha}{\partial \boldsymbol{x}} = \frac{\partial \boldsymbol{z}^T\boldsymbol{x}}{\partial \boldsymbol{x}}=\boldsymbol{z}^T=\boldsymbol{A}^T\boldsymbol{y}.
+\frac{\partial \alpha}{\partial \boldsymbol{x}} = \frac{\partial \boldsymbol{z}^T\boldsymbol{x}}{\partial \boldsymbol{x}}=\boldsymbol{z}^T.
 $$
 
-<p>Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one is just the transpose of the other.</p>
+<p>Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one is just the transpose of the other. We have the transposed here since we have used that the inner product of two vectors is a scalar.</p>
 
 <p>Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that</p>
 $$

doc/src/week35/week35.do.txt

@@ -318,11 +318,11 @@ $\bm{z}^T=\bm{y}^T\bm{A}$, a vector of length $n$. We have then, using the defin
 which means that (using our previous example and keeping track of our definition of the derivative of a scalar) we have
 !bt
 \[
-\frac{\partial \alpha}{\partial \bm{x}} = \frac{\partial \bm{z}^T\bm{x}}{\partial \bm{x}}=\bm{z}^T=\bm{A}^T\bm{y}.
+\frac{\partial \alpha}{\partial \bm{x}} = \frac{\partial \bm{z}^T\bm{x}}{\partial \bm{x}}=\bm{z}^T.
 \]
 !et
 
-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.
+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. We have the transposed here since we have used that the inner product of two vectors is a scalar.
 
 Since $\alpha$ is a scalar we have $\alpha =\alpha^T=\bm{x}^T\bm{A}^T\bm{y}$. Defining now $\bm{z}^T=\bm{x}^T\bm{A}^T$ we find that
 !bt