Extrap: overview of how to get to Shanks transformation, Sidis (b) process

Research/Extrapolation methods/notes.tex

@@ -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}
\ No newline at end of file