Adding notes on Sidis proof of MPE=Arnoldi

Research/Extrapolation methods/notes.tex

@@ -20,6 +20,7 @@
 \newcommand{\Set}[2]{\left \{ #1 \ \middle \vert \ #2 \right \}}
 \newcommand{\vmat}[1]{\begin{vmatrix} #1 \end{vmatrix}}
 \DeclareMathOperator{\sign}{sign}
+\DeclareMathOperator{\Span}{span}
 
 \newcommand{\bbn}{\mathbb{N}}
 \newcommand{\bbz}{\mathbb{Z}}
@@ -143,7 +144,7 @@ T_n = S \ \forall n \in \bbn \iff (S_n) \in \kernel.
 (Somehow this already defines the kernel of the transformation; is this definition then vacuous?
 It is stated in Brezinski that because of this definition any sequence transformation can be viewed as an extrapolation method.)
 
-This relation may be expressed as
+For some methods this relation may be expressed as
 \begin{equation}
 R(S_n, \dots, S_{n+q}, S) =0 \iff S = T_n = \sum_{i=0}^q a_i S_{n+i}.
 \end{equation}
@@ -171,4 +172,70 @@ T_n = \frac{S_n S_{n+2} - S_{n+1}^2}{S_{n+2} - 2 S_{n+1} + S_n}.
 Note that $dR/dS = -(a_1 + a_2)$.
 \end{example}
 
+\section{E-algorithm and derivatives}
+
+For the E-algorithm it is assumed that the relation $R$ has the form
+\begin{equation}
+S_n - S - \sum_{i=1}^k a_i g_i(n) = 0
+\end{equation}
+where $g_i(n)$ are some functions that depend on the indices $i$ and $n$ and the elements $\set{S_{n+j}}_{j=0}^k$.
+By solving the system created by repeating this equation for $n$ to $n+k$ in the same manner as the previous example one arrives at the solution
+\begin{equation}
+S = T_k^{(n)} = \frac{\vmat{S_n & \dots & S_{n+k} \\ g_1(n) & \dots & g_1(n+k) \\ \vdots & & \vdots \\ g_k(n) & \dots & g_k(n+k)}}{\vmat{1 & \dots & 1 \\ g_1(n) & \dots & g_1(n+k) \\ \vdots & & \vdots \\ g_k(n) & \dots & g_k(n+k)}} .
+\end{equation}
+The kernel of this transformation is
+\begin{equation}
+S_n = S + \sum_{i=1}^k a_i g_i(n)
+\end{equation}
+which may be readily obtained by rearranging the relation $R$.
+
+\section{Summary of Sidi proof of MPE=Arnoldi}
+
+Let $(x_n)$ be a sequence to be accelerated, then we produce approximations
+\begin{equation}
+T_n^{(k)} = \sum_{j=0}^k \gamma_j^{(n,k)} x_{n+j}
+\end{equation}
+where
+\begin{equation*}
+\sum_{j=0}^k \gamma_j^{(n,k)} = 1 .
+\end{equation*}
+The $\gamma_j^{(n,k)}$ also satisfy
+\begin{equation}
+\sum_{j=0}^k \langle \Delta x_{n+i}, \Delta x_{n+j} \rangle \gamma_j^{(n,k)} = 0, \quad 0 \leq i \leq k-1
+\end{equation}
+where $\langle \cdot, \cdot \rangle$ is the L2 inner product.
+The $\Delta$ operator has been defined previously but we reiterate it here:
+\begin{equation*}
+\Delta x_{n+i} = x_{n+i+1} - x_{n+i}.
+\end{equation*}
+We've seen previously that $T_n^{(k)}$ can be written as
+\begin{equation*}
+T_n^{(k)} = \frac{ \vmat{ x_n & \dots & x_{n+k} \\ \langle \Delta x_n, \Delta x_n \rangle & \dots & \langle \Delta x_n, \Delta x_{n+k} \rangle \\ \vdots & & \vdots \\ \langle \Delta x_{n+k-1}, \Delta x_n \rangle & \dots & \langle \Delta x_{n+k-1}, \Delta x_{n+k} \rangle }}{ \vmat{ 1 & \dots & 1 \\ \langle \Delta x_n, \Delta x_n \rangle & \dots & \langle \Delta x_n, \Delta x_{n+k} \rangle \\ \vdots & & \vdots \\ \langle \Delta x_{n+k-1}, \Delta x_n \rangle & \dots & \langle \Delta x_{n+k-1}, \Delta x_{n+k} \rangle}} .
+\end{equation*}
+The notion of the determinant is generalized here to allow for a vector result.
+That is, we expand along the first row so that each $x_{n+i}$ is multiplied by the determinant of a submatrix.
+That means each $\gamma_j^{(n,k)}$ is equal to
+\begin{equation}
+\gamma_j^{(n,k)} = \frac{\vmat{\langle \Delta x_n, \Delta x_n \rangle & \dots & \langle \Delta x_n, \Delta x_{n+j-1} \rangle & \langle \Delta x_n, \Delta x_{n+j+1} \rangle & \dots & \langle \Delta x_n, \Delta x_{n+k} \rangle \\
+				 \vdots & & \vdots & \vdots & & \vdots \\ 
+				 \langle \Delta x_{n+k-1}, \Delta x_n \rangle & \dots & \langle \Delta x_{n+k-1}, \Delta x_{n+j-1} \rangle & \langle \Delta x_{n+k-1}, \Delta x_{n+j+1} \rangle & \dots & \langle \Delta x_{n+k-1}, \Delta x_{n+k} \rangle 
+				 }}{ 
+				 \vmat{ 1 & \dots & 1 \\ 
+				 \langle \Delta x_n, \Delta x_n \rangle & \dots & \langle \Delta x_n, \Delta x_{n+k} \rangle \\ 
+				 \vdots & & \vdots \\ 
+				 \langle \Delta x_{n+k-1}, \Delta x_n \rangle & \dots & \langle \Delta x_{n+k-1}, \Delta x_{n+k} \rangle} 
+				 }
+\end{equation}
+
+Suppose the sequence $(x_n)$ is defined as $x_{n+1} = A x_n + b$ for some matrix $A$ and vector $b$.
+The residual $r(x) = Ax + b - x$ satisfies $r(x_i) = \Delta x_i$ and $r(s)=0$ where $s$ is the limit of the sequence, $s=As+b$.
+Moreover,
+\begin{equation}
+\Delta x_{i+1} = A x_i + b - x_i = A x_i + b - A x_{i-1} - b = A \Delta x_i = A^{i+1} \Delta x_0.
+\end{equation}
+Thus, 
+\begin{equation}
+r(T_n^{(k)}) = \sum \gamma_i^{(n,k)} \Delta x_{n+i} \in \Span \set{\Delta x_n, A \Delta x_n, \dots , A^k \Delta x_n}.
+\end{equation}
+
 \end{document}
\ No newline at end of file