diff --git a/doc/pub/week48/html/._week48-bs006.html b/doc/pub/week48/html/._week48-bs006.html index 5a086a8d8..703810d96 100644 --- a/doc/pub/week48/html/._week48-bs006.html +++ b/doc/pub/week48/html/._week48-bs006.html @@ -261,9 +261,9 @@ MathJax.Hub.Config({

The equations

-Suppose we define a polynomial transformation of degree two only (we continue to live in a plane with \( x_i \) and \( y_i \) as variables) +Suppose we define a polynomial transformation of degree two only (we continue to live in a plane with \( x_i \) and \( y_i \) as variables). We define a vector \( \boldsymbol{x}_i=[x_i,y_i] \) and have $$ -z = \phi(x_i) =\left(1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i\right). +\boldsymbol{z}_i = \phi(\boldsymbol{x}_i)^T =\left(1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i\right). $$

@@ -291,10 +291,10 @@ $$

We note that this dot product can be rewritten as $$ -K(\boldsymbol{x}_i,\boldsymbol{x}_j)=[1+\boldsymbol{x}^T\boldsymbol{x}']^d, +K(\boldsymbol{x}_i,\boldsymbol{x}_j)=[1+\boldsymbol{x}_i^T\boldsymbol{x}_j]^d, $$ -where \( d=2 \) in our case and \( \boldsymbol{x}=[x_i,y_i] \) and \( \boldsymbol{x}=[x_j,y_j] \). +where \( d=2 \) in our case and \( \boldsymbol{x}_i=[x_i,y_i] \) and \( \boldsymbol{x}_j=[x_j,y_j] \). To compute the last equation is however inefficient from a computational stand. Instead of computing the last equation for the kernel, we simply compute the dot product \( (\boldsymbol{x}_i^T\boldsymbol{x}_j)^2 \). diff --git a/doc/pub/week48/html/._week48-bs015.html b/doc/pub/week48/html/._week48-bs015.html index fa35d4a8c..2f2a0ee84 100644 --- a/doc/pub/week48/html/._week48-bs015.html +++ b/doc/pub/week48/html/._week48-bs015.html @@ -273,14 +273,11 @@ $$

  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. -
  4. The \( \boldsymbol{q} \) is zero.
  5. +
  6. The vector \( \boldsymbol{q} \) is zero.
  7. The constraint \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \) leads to \( f=0 \) and \( \boldsymbol{A}=\boldsymbol{y} \).
  8. -
  9. To set up the matrix \( \boldsymbol{G} \) we note that the inequalities \( 0\leq \lambda_i \leq C \) can be split up into
  10. +
  11. 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 \( \boldsymbol{h} \).
-\( 0\leq \lambda_i \) and \( \lambda_i \leq C \). These two inequalities define then the matrix \( \boldsymbol{G} \) and the vector \( \boldsymbol{h} \). - -