typo week48

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 \\