diff --git a/doc/pub/week48/html/._week48-bs007.html b/doc/pub/week48/html/._week48-bs007.html index 53618c612..cba270fd8 100644 --- a/doc/pub/week48/html/._week48-bs007.html +++ b/doc/pub/week48/html/._week48-bs007.html @@ -268,7 +268,7 @@ MathJax.Hub.Config({

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 diff --git a/doc/pub/week48/html/._week48-bs015.html b/doc/pub/week48/html/._week48-bs015.html index 598ecf1af..d78bf5373 100644 --- a/doc/pub/week48/html/._week48-bs015.html +++ b/doc/pub/week48/html/._week48-bs015.html @@ -278,7 +278,7 @@ $$

  1. With a given kernel we can thus define the matrix \( \boldsymbol{P} \).
  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.
    3. -
  3. The vector \( \boldsymbol{q} \) is zero.
    4. +
  4. The vector \( \boldsymbol{q} \) has all elements equal 1.
  5. The constraint \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \) leads to \( f=0 \) and \( \boldsymbol{A}=\boldsymbol{y} \).
  6. 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 \( \boldsymbol{h} \).
diff --git a/doc/pub/week48/html/week48-reveal.html b/doc/pub/week48/html/week48-reveal.html index 2d65f4154..a6d8cd3fb 100644 --- a/doc/pub/week48/html/week48-reveal.html +++ b/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

 
$$ -{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{z}_j, +{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{z}_i^T\boldsymbol{z}_j, $$

 
@@ -809,7 +809,7 @@ $$

  1. With a given kernel we can thus define the matrix \( \boldsymbol{P} \).

  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.
    3. -

  3. The vector \( \boldsymbol{q} \) is zero.
    4. +

  4. The vector \( \boldsymbol{q} \) has all elements equal 1.

  5. The constraint \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \) leads to \( f=0 \) and \( \boldsymbol{A}=\boldsymbol{y} \).

  6. 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 \( \boldsymbol{h} \).
diff --git a/doc/pub/week48/html/week48-solarized.html b/doc/pub/week48/html/week48-solarized.html index a342cc370..225b53558 100644 --- a/doc/pub/week48/html/week48-solarized.html +++ b/doc/pub/week48/html/week48-solarized.html @@ -394,7 +394,7 @@ kernel trick.

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 @@ $$

  1. With a given kernel we can thus define the matrix \( \boldsymbol{P} \).
  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.
    3. -
  3. The vector \( \boldsymbol{q} \) is zero.
    4. +
  4. The vector \( \boldsymbol{q} \) has all elements equal 1.
  5. The constraint \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \) leads to \( f=0 \) and \( \boldsymbol{A}=\boldsymbol{y} \).
  6. 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 \( \boldsymbol{h} \).
diff --git a/doc/pub/week48/html/week48.html b/doc/pub/week48/html/week48.html index 199b62748..32f5f77ef 100644 --- a/doc/pub/week48/html/week48.html +++ b/doc/pub/week48/html/week48.html @@ -399,7 +399,7 @@ kernel trick.

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 @@ $$

  1. With a given kernel we can thus define the matrix \( \boldsymbol{P} \).
  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.
    3. -
  3. The vector \( \boldsymbol{q} \) is zero.
    4. +
  4. The vector \( \boldsymbol{q} \) has all elements equal 1.
  5. The constraint \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \) leads to \( f=0 \) and \( \boldsymbol{A}=\boldsymbol{y} \).
  6. 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 \( \boldsymbol{h} \).
diff --git a/doc/pub/week48/ipynb/ipynb-week48-src.tar.gz b/doc/pub/week48/ipynb/ipynb-week48-src.tar.gz index 93c9e0356..990f429ba 100644 Binary files a/doc/pub/week48/ipynb/ipynb-week48-src.tar.gz and b/doc/pub/week48/ipynb/ipynb-week48-src.tar.gz differ diff --git a/doc/pub/week48/ipynb/week48.ipynb b/doc/pub/week48/ipynb/week48.ipynb index 600b2ed39..4b000801a 100644 --- a/doc/pub/week48/ipynb/week48.ipynb +++ b/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", diff --git a/doc/src/week48/week48.do.txt b/doc/src/week48/week48.do.txt index 66084c5ac..fb4411189 100644 --- a/doc/src/week48/week48.do.txt +++ b/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 $\bm{h}$.