typo

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

@@ -381,10 +381,10 @@ $$
 
 <p>which means that (using our previous example) we have</p>
 $$
-\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=bm{A}^T\boldsymbol{y}.
+\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=\boldsymbol{A}^T\boldsymbol{y}.
 $$
 
-<p>Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one if 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.</p>
 
 <p>Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that</p>
 $$

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

@@ -539,11 +539,11 @@ $$
 <p>which means that (using our previous example) we have</p>
 <p>&nbsp;<br>
 $$
-\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=bm{A}^T\boldsymbol{y}.
+\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=\boldsymbol{A}^T\boldsymbol{y}.
 $$
 <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 if 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.</p>
 
 <p>Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that</p>
 <p>&nbsp;<br>

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

@@ -603,10 +603,10 @@ $$
 
 <p>which means that (using our previous example) we have</p>
 $$
-\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=bm{A}^T\boldsymbol{y}.
+\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=\boldsymbol{A}^T\boldsymbol{y}.
 $$
 
-<p>Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one if 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.</p>
 
 <p>Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that</p>
 $$

doc/pub/week35/html/week35.html

@@ -680,10 +680,10 @@ $$
 
 <p>which means that (using our previous example) we have</p>
 $$
-\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=bm{A}^T\boldsymbol{y}.
+\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=\boldsymbol{A}^T\boldsymbol{y}.
 $$
 
-<p>Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one if 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.</p>
 
 <p>Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that</p>
 $$