typo week48

doc/pub/week48/html/._week48-bs013.html

@@ -289,7 +289,7 @@ $$
 
 The minimization problem can be rewritten in terms of vectors and matrices as (with \( x \) and \( y \) being the unknowns)
 $$
-\frac{1}{2}\begin{bmatrix} x\\ y \end{bmatrix}^T   \begin{bmatrix} 1 & 0\\ 0 & 0  \end{bmatrix}  \begin{bmatrix} x \\ y \end{bmatrix}  + \begin{bmatrix}3\\ 4  \end{bmatrix}^T \begin{bmatrix}x \\ y  \end{bmatrix}.  
+\frac{1}{2}\begin{bmatrix} x\\ y \end{bmatrix}^T   \begin{bmatrix} 1 & 0\\ 0 & 0  \end{bmatrix}  \begin{bmatrix} x \\ y \end{bmatrix}  + \begin{bmatrix}5\\ 3  \end{bmatrix}^T \begin{bmatrix}x \\ y  \end{bmatrix}.  
 $$
 
 Similarly, we can now set up the inequalities (we need to change \( \geq \) to \( \leq \) by multiplying with \( -1 \) on bot sides) as the following matrix-vector equation

doc/pub/week48/html/._week48-bs015.html

@@ -278,7 +278,7 @@ $$
 <ol>
 <li> With a given kernel we can thus define the matrix \( \boldsymbol{P} \).</li>
 <li> The matrix \( \boldsymbol{P} \) has matrix elements \( p_{ij}=y_iy_jK(\boldsymbol{x}_i,\boldsymbol{x}_j) \). Given a kernel \( K \) and the targets \( y_i \) this matrix is easy to set up.</li>
-<li> The vector \( \boldsymbol{q} \) has all elements equal 1.</li>
+<li> The vector \( \boldsymbol{q} \) has all elements equal -1.</li>
 <li> The constraint \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \) leads to \( f=0 \) and \( \boldsymbol{A}=\boldsymbol{y} \).</li>
 <li> To set up the matrix \( \boldsymbol{G} \) we note that the inequalities \( 0\leq \lambda_i \leq C \) can be split up into \( 0\leq \lambda_i \) and \( \lambda_i \leq C \). These two inequalities define then the matrix \( \boldsymbol{G} \) and the vector \( \boldsymbol{h} \).</li>
 </ol>

doc/pub/week48/html/._week48-bs016.html

@@ -286,7 +286,7 @@ $$
 \lambda_3 \\
 \dots \\
 \lambda_n \\
-\end{bmatrix}=
+\end{bmatrix}\wedge 
 \begin{bmatrix} 0 \\
 0 \\
 0 \\

doc/pub/week48/html/week48-reveal.html

@@ -720,7 +720,7 @@ $$
 The minimization problem can be rewritten in terms of vectors and matrices as (with \( x \) and \( y \) being the unknowns)
 <p>&nbsp;<br>
 $$
-\frac{1}{2}\begin{bmatrix} x\\ y \end{bmatrix}^T   \begin{bmatrix} 1 & 0\\ 0 & 0  \end{bmatrix}  \begin{bmatrix} x \\ y \end{bmatrix}  + \begin{bmatrix}3\\ 4  \end{bmatrix}^T \begin{bmatrix}x \\ y  \end{bmatrix}.  
+\frac{1}{2}\begin{bmatrix} x\\ y \end{bmatrix}^T   \begin{bmatrix} 1 & 0\\ 0 & 0  \end{bmatrix}  \begin{bmatrix} x \\ y \end{bmatrix}  + \begin{bmatrix}5\\ 3  \end{bmatrix}^T \begin{bmatrix}x \\ y  \end{bmatrix}.  
 $$
 <p>&nbsp;<br>
 
@@ -809,7 +809,7 @@ $$
 <ol>
 <p><li> With a given kernel we can thus define the matrix \( \boldsymbol{P} \).</li>
 <p><li> The matrix \( \boldsymbol{P} \) has matrix elements \( p_{ij}=y_iy_jK(\boldsymbol{x}_i,\boldsymbol{x}_j) \). Given a kernel \( K \) and the targets \( y_i \) this matrix is easy to set up.</li>
-<p><li> The vector \( \boldsymbol{q} \) has all elements equal 1.</li>
+<p><li> The vector \( \boldsymbol{q} \) has all elements equal -1.</li>
 <p><li> The constraint \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \) leads to \( f=0 \) and \( \boldsymbol{A}=\boldsymbol{y} \).</li>
 <p><li> To set up the matrix \( \boldsymbol{G} \) we note that the inequalities \( 0\leq \lambda_i \leq C \) can be split up into \( 0\leq \lambda_i \) and \( \lambda_i \leq C \). These two inequalities define then the matrix \( \boldsymbol{G} \) and the vector \( \boldsymbol{h} \).</li>
 </ol>
@@ -841,7 +841,7 @@ $$
 \lambda_3 \\
 \dots \\
 \lambda_n \\
-\end{bmatrix}=
+\end{bmatrix}\wedge 
 \begin{bmatrix} 0 \\
 0 \\
 0 \\

doc/pub/week48/html/week48-solarized.html

@@ -743,7 +743,7 @@ $$
 
 The minimization problem can be rewritten in terms of vectors and matrices as (with \( x \) and \( y \) being the unknowns)
 $$
-\frac{1}{2}\begin{bmatrix} x\\ y \end{bmatrix}^T   \begin{bmatrix} 1 & 0\\ 0 & 0  \end{bmatrix}  \begin{bmatrix} x \\ y \end{bmatrix}  + \begin{bmatrix}3\\ 4  \end{bmatrix}^T \begin{bmatrix}x \\ y  \end{bmatrix}.  
+\frac{1}{2}\begin{bmatrix} x\\ y \end{bmatrix}^T   \begin{bmatrix} 1 & 0\\ 0 & 0  \end{bmatrix}  \begin{bmatrix} x \\ y \end{bmatrix}  + \begin{bmatrix}5\\ 3  \end{bmatrix}^T \begin{bmatrix}x \\ y  \end{bmatrix}.  
 $$
 
 Similarly, we can now set up the inequalities (we need to change \( \geq \) to \( \leq \) by multiplying with \( -1 \) on bot sides) as the following matrix-vector equation
@@ -820,7 +820,7 @@ $$
 <ol>
 <li> With a given kernel we can thus define the matrix \( \boldsymbol{P} \).</li>
 <li> The matrix \( \boldsymbol{P} \) has matrix elements \( p_{ij}=y_iy_jK(\boldsymbol{x}_i,\boldsymbol{x}_j) \). Given a kernel \( K \) and the targets \( y_i \) this matrix is easy to set up.</li>
-<li> The vector \( \boldsymbol{q} \) has all elements equal 1.</li>
+<li> The vector \( \boldsymbol{q} \) has all elements equal -1.</li>
 <li> The constraint \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \) leads to \( f=0 \) and \( \boldsymbol{A}=\boldsymbol{y} \).</li>
 <li> To set up the matrix \( \boldsymbol{G} \) we note that the inequalities \( 0\leq \lambda_i \leq C \) can be split up into \( 0\leq \lambda_i \) and \( \lambda_i \leq C \). These two inequalities define then the matrix \( \boldsymbol{G} \) and the vector \( \boldsymbol{h} \).</li>
 </ol>
@@ -850,7 +850,7 @@ $$
 \lambda_3 \\
 \dots \\
 \lambda_n \\
-\end{bmatrix}=
+\end{bmatrix}\wedge 
 \begin{bmatrix} 0 \\
 0 \\
 0 \\

doc/pub/week48/html/week48.html

@@ -748,7 +748,7 @@ $$
 
 The minimization problem can be rewritten in terms of vectors and matrices as (with \( x \) and \( y \) being the unknowns)
 $$
-\frac{1}{2}\begin{bmatrix} x\\ y \end{bmatrix}^T   \begin{bmatrix} 1 & 0\\ 0 & 0  \end{bmatrix}  \begin{bmatrix} x \\ y \end{bmatrix}  + \begin{bmatrix}3\\ 4  \end{bmatrix}^T \begin{bmatrix}x \\ y  \end{bmatrix}.  
+\frac{1}{2}\begin{bmatrix} x\\ y \end{bmatrix}^T   \begin{bmatrix} 1 & 0\\ 0 & 0  \end{bmatrix}  \begin{bmatrix} x \\ y \end{bmatrix}  + \begin{bmatrix}5\\ 3  \end{bmatrix}^T \begin{bmatrix}x \\ y  \end{bmatrix}.  
 $$
 
 Similarly, we can now set up the inequalities (we need to change \( \geq \) to \( \leq \) by multiplying with \( -1 \) on bot sides) as the following matrix-vector equation
@@ -825,7 +825,7 @@ $$
 <ol>
 <li> With a given kernel we can thus define the matrix \( \boldsymbol{P} \).</li>
 <li> The matrix \( \boldsymbol{P} \) has matrix elements \( p_{ij}=y_iy_jK(\boldsymbol{x}_i,\boldsymbol{x}_j) \). Given a kernel \( K \) and the targets \( y_i \) this matrix is easy to set up.</li>
-<li> The vector \( \boldsymbol{q} \) has all elements equal 1.</li>
+<li> The vector \( \boldsymbol{q} \) has all elements equal -1.</li>
 <li> The constraint \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \) leads to \( f=0 \) and \( \boldsymbol{A}=\boldsymbol{y} \).</li>
 <li> To set up the matrix \( \boldsymbol{G} \) we note that the inequalities \( 0\leq \lambda_i \leq C \) can be split up into \( 0\leq \lambda_i \) and \( \lambda_i \leq C \). These two inequalities define then the matrix \( \boldsymbol{G} \) and the vector \( \boldsymbol{h} \).</li>
 </ol>
@@ -855,7 +855,7 @@ $$
 \lambda_3 \\
 \dots \\
 \lambda_n \\
-\end{bmatrix}=
+\end{bmatrix}\wedge 
 \begin{bmatrix} 0 \\
 0 \\
 0 \\

doc/pub/week48/ipynb/week48.ipynb

@@ -675,7 +675,7 @@
    "metadata": {},
    "source": [
     "$$\n",
-    "\\frac{1}{2}\\begin{bmatrix} x\\\\ y \\end{bmatrix}^T   \\begin{bmatrix} 1 & 0\\\\ 0 & 0  \\end{bmatrix}  \\begin{bmatrix} x \\\\ y \\end{bmatrix}  + \\begin{bmatrix}3\\\\ 4  \\end{bmatrix}^T \\begin{bmatrix}x \\\\ y  \\end{bmatrix}.\n",
+    "\\frac{1}{2}\\begin{bmatrix} x\\\\ y \\end{bmatrix}^T   \\begin{bmatrix} 1 & 0\\\\ 0 & 0  \\end{bmatrix}  \\begin{bmatrix} x \\\\ y \\end{bmatrix}  + \\begin{bmatrix}5\\\\ 3  \\end{bmatrix}^T \\begin{bmatrix}x \\\\ y  \\end{bmatrix}.\n",
     "$$"
    ]
   },
@@ -816,7 +816,7 @@
     "\n",
     "2. The matrix $\\boldsymbol{P}$ has matrix elements $p_{ij}=y_iy_jK(\\boldsymbol{x}_i,\\boldsymbol{x}_j)$. Given a kernel $K$ and the targets $y_i$ this matrix is easy to set up.\n",
     "\n",
-    "3. The vector $\\boldsymbol{q}$ has all elements equal 1.\n",
+    "3. The vector $\\boldsymbol{q}$ has all elements equal -1.\n",
     "\n",
     "4. The constraint $\\boldsymbol{y}^T\\boldsymbol{\\lambda}=0$ leads to $f=0$ and $\\boldsymbol{A}=\\boldsymbol{y}$.\n",
     "\n",
@@ -850,7 +850,7 @@
     "\\lambda_3 \\\\\n",
     "\\dots \\\\\n",
     "\\lambda_n \\\\\n",
-    "\\end{bmatrix}=\n",
+    "\\end{bmatrix}\\wedge \n",
     "\\begin{bmatrix} 0 \\\\\n",
     "0 \\\\\n",
     "0 \\\\\n",

doc/src/week48/week48.do.txt

@@ -498,7 +498,7 @@ Let us show how to perform the optmization using a simple case. Assume we want t
 The minimization problem can be rewritten in terms of vectors and matrices as (with $x$ and $y$ being the unknowns)
 !bt
 \[
-\frac{1}{2}\begin{bmatrix} x\\ y \end{bmatrix}^T   \begin{bmatrix} 1 & 0\\ 0 & 0  \end{bmatrix}  \begin{bmatrix} x \\ y \end{bmatrix}  + \begin{bmatrix}3\\ 4  \end{bmatrix}^T \begin{bmatrix}x \\ y  \end{bmatrix}.  
+\frac{1}{2}\begin{bmatrix} x\\ y \end{bmatrix}^T   \begin{bmatrix} 1 & 0\\ 0 & 0  \end{bmatrix}  \begin{bmatrix} x \\ y \end{bmatrix}  + \begin{bmatrix}5\\ 3  \end{bmatrix}^T \begin{bmatrix}x \\ y  \end{bmatrix}.  
 \]
 !et
 Similarly, we can now set up the inequalities (we need to change $\geq$ to $\leq$ by multiplying with $-1$ on bot sides) as the following matrix-vector equation
@@ -575,7 +575,7 @@ We have the general problem
 
 o With a given kernel we can thus define the matrix $\bm{P}$.
 o The matrix $\bm{P}$ has matrix elements $p_{ij}=y_iy_jK(\bm{x}_i,\bm{x}_j)$. Given a kernel $K$ and the targets $y_i$ this matrix is easy to set up.
-o The vector $\bm{q}$ has all elements equal 1.
+o The vector $\bm{q}$ has all elements equal -1.
 o The constraint $\bm{y}^T\bm{\lambda}=0$ leads to $f=0$ and $\bm{A}=\bm{y}$.
 o To set up the matrix $\bm{G}$ we note that the inequalities $0\leq \lambda_i \leq C$ can be split up into $0\leq \lambda_i$ and $\lambda_i \leq C$. These two inequalities define then the matrix $\bm{G}$ and the vector $\bm{h}$.
 
@@ -603,7 +603,7 @@ can be written as
 \lambda_3 \\
 \dots \\
 \lambda_n \\
-\end{bmatrix}=
+\end{bmatrix}\wedge 
 \begin{bmatrix} 0 \\
 0 \\
 0 \\