Commit 40598751 authored by Conor McCoid's avatar Conor McCoid
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Extrap: overview of how to get to Shanks transformation, Sidis (b) process

parent 41c59b00
......@@ -357,4 +357,63 @@ Let $\vec{r}(\vec{x}_i) = \vec{x}_{i+1} - \vec{x}_i$ be the residual function.
Then MPE may be expressed as equation (\ref{eq:uqn}) with $\fxi{i} = \vec{r}(\vec{x}_i)$ and $\vec{v}_i = \vec{r}(\vec{x}_{n+i-1})$.
RRE and MMPE may also be expressed as such, using $\vec{v}_i = \vec{r}(\vec{x}_{n+i}) - \vec{r}(\vec{x}_{n+i-1})$ and $\vec{v}_i$ some fixed vector independent of $\vec{r}(\vec{x})$, respectively.
\section{Generalized Shanks Transformation}
Suppose that we have a sequence $\set{x_n}$ that tends towards $s$ as
$$ x_n \tilde s + \sum_{i=1}^\infty a_i \lambda_i^n. $$
Then we say $s_{n,k}$ is an approximation of $s$ such that
\begin{equation}
x_{n+j} = s_{n,k} + \sum_{i=1}^k a_i \lambda_i^{n+j}
\end{equation}
for all $0 \leq j \leq 2k$.
This gives $2k+1$ equations to solve for the $2k+1$ unknowns ($s_{n,k}$, $a_i$ and $\lambda_i$).
We know that the final result will be a sum of $x_{n+j}$.
Suppose the solution is
\begin{align*}
s_{n,k} = & \sum_{j=0}^m \gamma_j x_{n+j} \\
= & \sum_{j=0}^m \gamma_j \left ( s_{n,k} + \sum_{i=0}^k a_i \lambda_i^{n+j} \right ) \\
= & \left ( \sum_{j=0}^m \gamma_j \right ) s_{n,k} + \sum_{j=0}^m \sum_{i=0}^k a_i \lambda_i^{n+j}.
\end{align*}
Naturally this means $\sum_{j=0}^m \gamma_j = 1$.
Suppose also that
\begin{equation*}
s_{n,k} = \sum_{j=0}^m \gamma_j x_{n+j+i}
\end{equation*}
for $0 \leq i \leq 2k-m-1$.
Then
\begin{align*}
s_{n,k} - s_{n,k} = & \sum_{j=0}^m \gamma_j \left ( x_{n+j+i+1} - x_{n+j+i} \right ) \\
0 = & \sum_{j=0}^m \gamma_j r_{n+j+i}
\end{align*}
for the same set of $i$.
This gives the following linear equation:
\begin{equation*}
\begin{bmatrix} 1 & \dots & 1 \\ r_n & \dots & r_{n+m} \\ \vdots & & \vdots \\ r_{n+2k-m-1} & \dots & r_{n+2k} \end{bmatrix} \begin{bmatrix} \gamma_0 \\ \vdots \\ \gamma_m \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}.
\end{equation*}
For this equation to be solvable one requires that $2k-m = m$, or $m=k$.
The solution is then
\begin{align*}
\gamma_i = & \frac{\vmat{
1 & \dots & 1 & 1 & 1 & \dots & 1 \\
r_n & \dots & r_{n+i-1} & 0 & r_{n+i+1} & \dots & r_{n+k} \\
\vdots & & \vdots & \vdots & \vdots & & \vdots \\
r_{n+k-1} & \dots & r_{n+i+k-2} & 0 & r_{n+i+k} & \dots & r_{n+2k-1}
}}{\vmat{
1 & \dots & 1 \\ r_n & \dots & r_{n+k} \\ \vdots & & \vdots \\ r_{n+k-1} & \dots & r_{n+2k-1} }} \\
= & \frac{\vmat{
r_n & \dots & r_{n+i-1} & r_{n+i+1} & \dots & r_{n+k} \\
\vdots & \vdots & \vdots & \vdots & & \vdots \\
r_{n+k-1} & \dots & r_{n+i+k-2} & r_{n+i+k} & \dots & r_{n+2k-1}
}}{\vmat{
1 & \dots & 1 \\ r_n & \dots & r_{n+k} \\ \vdots & & \vdots \\ r_{n+k-1} & \dots & r_{n+2k-1} }}
\end{align*}
which gives
\begin{equation}
s_{n,k} = \frac{\vmat{
x_n & \dots & x_{n+k} \\ r_n & \dots & r_{n+k} \\ \vdots & & \vdots \\ r_{n+k-1} & \dots & r_{n+2k-1} }}{\vmat{
1 & \dots & 1 \\ r_n & \dots & r_{n+k} \\ \vdots & & \vdots \\ r_{n+k-1} & \dots & r_{n+2k-1} }}.
\end{equation}
\end{document}
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