Extrap: fixed some notation in equivalence section

Research/Extrapolation methods/notes.tex

@@ -460,7 +460,7 @@ x_n & \dots & x_{n+k} \\ r_n & \dots & r_{n+k} \\ \vdots & & \vdots \\ r_{n+k-1}
 
 Suppose we have a method such that
 \begin{equation}
-\begin{bmatrix} 1 & \dots & 1 \\ \fxi{n} & \dots & \fxi{n+k} \end{bmatrix} \vec{u} = F_{n,k} \vec{u} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \\ \hat{\vec{x}} = \begin{bmatrix} \vec{x}_n & \dots & \vec{x}_{n+k} \end{bmatrix} \vec{u} = X_{n,k} \vec{u}
+\begin{bmatrix} 1 & \dots & 1 \\ \fxi{n} & \dots & \fxi{n+k} \end{bmatrix} \vec{u} = \begin{bmatrix} \vec{1}^\top \\ F_{n,k} \end{bmatrix} \vec{u} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \\ \hat{\vec{x}} = \begin{bmatrix} \vec{x}_n & \dots & \vec{x}_{n+k} \end{bmatrix} \vec{u} = X_{n,k} \vec{u}
 \end{equation}
 then one can write an equivalent method:
 \begin{equation*}
@@ -468,17 +468,17 @@ then one can write an equivalent method:
 \end{equation*}
 where $\hat{\vec{u}}$ satisfies
 \begin{equation*}
-F_{n,k} \hat{\vec{u}} = \begin{bmatrix} 0 \\ \fxi{n+i} \end{bmatrix}.
+\begin{bmatrix} \vec{1}^\top \\ F_{n,k} \end{bmatrix} \hat{\vec{u}} = \begin{bmatrix} 0 \\ \fxi{n+i} \end{bmatrix}.
 \end{equation*}
 Replace $\hat{\vec{u}}$ with $\Delta \tilde{\vec{u}}$ where $\Delta$ is some matrix representing a differentiation operator.
-The conditions on $\Delta$ are that it has $k-1$ rows and $k$ columns, and its columns sum to zero.
+The conditions on $\Delta$ are that it has $k+1$ rows and $k$ columns, and its columns sum to zero.
 Then an equivalent method is
 \begin{equation*}
 \hat{\vec{x}} = \vec{x}_{n+i} - X_{n,k} \Delta \tilde{\vec{u}}
 \end{equation*}
 where
 \begin{equation*}
-F_{n,k} \Delta \tilde{\vec{u}} = \begin{bmatrix} 0 \\ \fxi{n+i} \end{bmatrix}.
+\begin{bmatrix} \vec{1}^\top \\ F_{n,k} \end{bmatrix} \Delta \tilde{\vec{u}} = \begin{bmatrix} 0 \\ \fxi{n+i} \end{bmatrix} \implies F_{n,k} \Delta \tilde{\vec{u}} = \fxi{n+i}.
 \end{equation*}
 Some standard choices of $\Delta$:
 \begin{equation*}
@@ -490,9 +490,8 @@ For example, if $\vec{f} : \bbr^d \to \bbr^d$ and $k< d$ then each of the presen
 If $k>d$ then they are overdetermined.
 To solve the system(s) one can multiply by a matrix $A^\top$:
 \begin{equation*}
-A^\top F_{n,k} \vec{u} = A^\top \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad A^\top F_{n,k} \Delta \tilde{\vec{u}} = A^\top \begin{bmatrix} 0 \\ \fxi{n+i} \end{bmatrix}.
+\begin{bmatrix} \vec{1}^\top \\ A^\top F_{n,k} \end{bmatrix} \vec{u} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad A^\top F_{n,k} \Delta \tilde{\vec{u}} = A^\top \fxi{n+i} .
 \end{equation*}
-The first column of $A^\top$ should be $\begin{bmatrix} 1 & 0 \end{bmatrix}^\top$ to preserve the first equation of the system.
 
 Note that $\hat{\vec{x}}$ is always in the span of $\set{\vec{x}_{n+i}}_{i=0}^k$.
 Therefore, if $k<d$ then $\hat{\vec{x}}$ is in a restricted subspace of $\bbr^d$.