Merge branch 'master' of https://github.com/CompPhysics/MachineLearning

doc/pub/week48/html/._week48-bs007.html

@@ -268,7 +268,7 @@ MathJax.Hub.Config({
 <p>
 Using our definition of the kernel, we can rewrite again the Lagrangian
 $$
-{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{z}_j,
+{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{z}_i^T\boldsymbol{z}_j,
 $$
 
 subject to the constraints \( \lambda_i\geq 0 \), \( \sum_i\lambda_iy_i=0 \) in terms of a convex optimization problem

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 \( \boldsymbol{q} \) is zero.</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-reveal.html

@@ -355,7 +355,7 @@ kernel trick.
 Using our definition of the kernel, we can rewrite again the Lagrangian
 <p>&nbsp;<br>
 $$
-{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{z}_j,
+{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{z}_i^T\boldsymbol{z}_j,
 $$
 <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 \( \boldsymbol{q} \) is zero.</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>

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

@@ -394,7 +394,7 @@ kernel trick.
 <p>
 Using our definition of the kernel, we can rewrite again the Lagrangian
 $$
-{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{z}_j,
+{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{z}_i^T\boldsymbol{z}_j,
 $$
 
 subject to the constraints \( \lambda_i\geq 0 \), \( \sum_i\lambda_iy_i=0 \) in terms of a convex optimization problem
@@ -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 \( \boldsymbol{q} \) is zero.</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.html

@@ -399,7 +399,7 @@ kernel trick.
 <p>
 Using our definition of the kernel, we can rewrite again the Lagrangian
 $$
-{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{z}_j,
+{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{z}_i^T\boldsymbol{z}_j,
 $$
 
 subject to the constraints \( \lambda_i\geq 0 \), \( \sum_i\lambda_iy_i=0 \) in terms of a convex optimization problem
@@ -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 \( \boldsymbol{q} \) is zero.</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/ipynb/week48.ipynb

@@ -252,7 +252,7 @@
    "metadata": {},
    "source": [
     "$$\n",
-    "{\\cal L}=\\sum_i\\lambda_i-\\frac{1}{2}\\sum_{ij}^n\\lambda_i\\lambda_jy_iy_j\\boldsymbol{x}_i^T\\boldsymbol{z}_j,\n",
+    "{\\cal L}=\\sum_i\\lambda_i-\\frac{1}{2}\\sum_{ij}^n\\lambda_i\\lambda_jy_iy_j\\boldsymbol{z}_i^T\\boldsymbol{z}_j,\n",
     "$$"
    ]
   },
@@ -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 $\\boldsymbol{q}$ is zero.\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",

doc/src/week48/week48.do.txt

@@ -178,7 +178,7 @@ kernel trick.
 Using our definition of the kernel, we can rewrite again the Lagrangian
 !bt
 \[
-{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\bm{x}_i^T\bm{z}_j,
+{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\bm{z}_i^T\bm{z}_j,
 \]
 !et
 subject to the constraints $\lambda_i\geq 0$, $\sum_i\lambda_iy_i=0$ in terms of a convex optimization problem
@@ -582,7 +582,7 @@ 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 $\bm{q}$ is zero.
+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 $\bm{h}$.