small typo

doc/pub/week35/html/._week35-bs068.html

@@ -298,7 +298,7 @@ $$
 
 <p>Taking the derivative with respect to \( \boldsymbol{\beta} \) and recalling that the derivative of the absolute value is (we drop the boldfaced vector symbol for simplicty)</p>
 $$
-\frac{d \vert \beta\vert}{d \boldsymbol{\beta}}=\mathrm{sgn}(\boldsymbol{\beta})=\left\{\begin{array}{cc} 1 & \beta > 0 \\ 0 & \beta =0\\-1 & \beta < 0, \end{array}\right.
+\frac{d \vert \beta\vert}{d \boldsymbol{\beta}}=\mathrm{sgn}(\boldsymbol{\beta})=\left\{\begin{array}{cc} 1 & \beta > 0 \\-1 & \beta < 0, \end{array}\right.
 $$
 
 <p>we have that the derivative of the cost function is</p>

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

@@ -3465,7 +3465,7 @@ $$
 <p>Taking the derivative with respect to \( \boldsymbol{\beta} \) and recalling that the derivative of the absolute value is (we drop the boldfaced vector symbol for simplicty)</p>
 <p>&nbsp;<br>
 $$
-\frac{d \vert \beta\vert}{d \boldsymbol{\beta}}=\mathrm{sgn}(\boldsymbol{\beta})=\left\{\begin{array}{cc} 1 & \beta > 0 \\ 0 & \beta =0\\-1 & \beta < 0, \end{array}\right.
+\frac{d \vert \beta\vert}{d \boldsymbol{\beta}}=\mathrm{sgn}(\boldsymbol{\beta})=\left\{\begin{array}{cc} 1 & \beta > 0 \\-1 & \beta < 0, \end{array}\right.
 $$
 <p>&nbsp;<br>
 

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

@@ -3309,7 +3309,7 @@ $$
 
 <p>Taking the derivative with respect to \( \boldsymbol{\beta} \) and recalling that the derivative of the absolute value is (we drop the boldfaced vector symbol for simplicty)</p>
 $$
-\frac{d \vert \beta\vert}{d \boldsymbol{\beta}}=\mathrm{sgn}(\boldsymbol{\beta})=\left\{\begin{array}{cc} 1 & \beta > 0 \\ 0 & \beta =0\\-1 & \beta < 0, \end{array}\right.
+\frac{d \vert \beta\vert}{d \boldsymbol{\beta}}=\mathrm{sgn}(\boldsymbol{\beta})=\left\{\begin{array}{cc} 1 & \beta > 0 \\-1 & \beta < 0, \end{array}\right.
 $$
 
 <p>we have that the derivative of the cost function is</p>

doc/pub/week35/html/week35.html

@@ -3386,7 +3386,7 @@ $$
 
 <p>Taking the derivative with respect to \( \boldsymbol{\beta} \) and recalling that the derivative of the absolute value is (we drop the boldfaced vector symbol for simplicty)</p>
 $$
-\frac{d \vert \beta\vert}{d \boldsymbol{\beta}}=\mathrm{sgn}(\boldsymbol{\beta})=\left\{\begin{array}{cc} 1 & \beta > 0 \\ 0 & \beta =0\\-1 & \beta < 0, \end{array}\right.
+\frac{d \vert \beta\vert}{d \boldsymbol{\beta}}=\mathrm{sgn}(\boldsymbol{\beta})=\left\{\begin{array}{cc} 1 & \beta > 0 \\-1 & \beta < 0, \end{array}\right.
 $$
 
 <p>we have that the derivative of the cost function is</p>

doc/src/week35/week35.do.txt

@@ -2495,7 +2495,7 @@ C(\bm{X},\bm{\beta})=\left\{(\bm{y}-\bm{X}\bm{\beta})^T(\bm{y}-\bm{X}\bm{\beta})
 Taking the derivative with respect to $\bm{\beta}$ and recalling that the derivative of the absolute value is (we drop the boldfaced vector symbol for simplicty)
 !bt
 \[
-\frac{d \vert \beta\vert}{d \bm{\beta}}=\mathrm{sgn}(\bm{\beta})=\left\{\begin{array}{cc} 1 & \beta > 0 \\ 0 & \beta =0\\-1 & \beta < 0, \end{array}\right.
+\frac{d \vert \beta\vert}{d \bm{\beta}}=\mathrm{sgn}(\bm{\beta})=\left\{\begin{array}{cc} 1 & \beta > 0 \\-1 & \beta < 0, \end{array}\right.
 \]
 !et
 we have that the derivative of the cost function is