Commit f4376ef4 authored by Conor McCoid's avatar Conor McCoid
Browse files

Extrap: fixed some notation in equivalence section

parent 7adfa993
......@@ -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 $\bbr^d$.
......
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