I'm going to start working through it and compile their test programs.
+\subsection{A sequence that seems to converge to a wrong limit - pgs 9-10, \cite{HFP}}
+
+\begin{align*}
+ u_n &= \left\{ \begin{array}{c} u_0 = 2 \\ u_1 = -4 \\ u_n = 111 - \frac{1130}{u_{n-1}} + \frac{3000}{u_{n-1}u_{n-2}}\end{array}\right.
+\end{align*}
+
+The limit of the series should be $6$ but when calculated with IEEE floats it is actually $100$
+The authors show that the limit is actually $100$ for different starting values, and the error in floating point arithmetic causes the series to go to that limit instead.
+
+\begin{figure}[H]
+ \centering
+ \includegraphics[width=0.8\textwidth]{figures/handbook1-1.pdf}
+ \caption{Output of Program 1.1 from \emph{Handbook of Floating-Point Arithmetic}\cite{HFP} for various IEEE types}
+ \label{HFP-1-1}
+\end{figure}
+
+\subsection{Mr Gullible and the Chaotic Bank Society pgs 10-11 \cite{HFP}}
+
+This is an example of a sequence involving $e$. Since $e$ cannot be represented exactly with FP, even though the sequence should go to $0$ for $a_0 = e - 1$, the representation of $a_0 \neq e - 1$ so the sequence goes to $\pm \infty$.
+
+To eliminate these types of problems we'd need an \emph{exact} representation of all real numbers.
+For \emph{any} FP representation, regardless of precision (a finite number of digits) there will be numbers that can't be represented exactly hence you could find a similar sequence that would explode.
+
+IE: The more precise the representation, the slower things go wrong, but they still go wrong, {\bf even with errorless operations}.
+
+
+\subsection{Rump's example pg 12 \cite {HFP}}
+
+This is an example where the calculation of a function $f(a,b)$ is not only totally wrong, it gives completely different results depending on the CPU. Despite the CPU conforming to IEEE.
+
\chapter{General Notes}
+\section{Floating-Point \cite{HFP,goldberg1991whatevery,goldberg1992thedesign,priest1991algorithms}}
+
+A set of FP numbers is characterised by:
+\begin{enumerate}
+ \item Radix (base) $\beta \geq 2$
+ \item Precision %$p \req 2$ ``number of sig digits''
+ \item Two ``extremal`` exponents $e_min < 0 < e_max$ (generally, don't have to have the $0$ in there)
+\end{enumerate}
+
+Numbers are represented by {\bf integers}: $(M, e)$ such that $x = M \times \beta^{e - p + 1}$
+
+Require: $|M| \leq \beta^{p}-1$ and $e_min \leq e \leq e_max$.
+
+Representations are not unique; set of equivelant representations is a cohort.
+
+$\beta^{e-p+1}$ is the quantum, $e-p+1$ is the quantum exponent.
+
+Alternate represetnation: $(s, m, e)$ such that $x = (-1)^s \times m \times \beta^{e}$
+$m$ is the significand, mantissa, or fractional part. Depending on what you read.
+
+
+
+
+
\section{Rounding Errors}
+
They happen. There is ULP and I don't mean a political party.
TODO: Probably say something more insightful. Other than "here is a graph that shows errors and we blame rounding".
Tests with \verb/calculatepi/ show it's not quite as simple as just blindly replacing all your additions with Fast2Sum from Dekker\cite{dekker1971afloating}.
ie: The graph looks exactly the same for single precision. \verb/calculatepi/ obviously also has multiplication ops in it which I didn't change. Will look at after sleep maybe.
-\subsection{A sequence that seems to converge to a wrong limit - pgs 9-10, \cite{HFP}}
-
-\begin{align*}
- u_n &= \left\{ \begin{array}{c} u_0 = 2 \\ u_1 = -4 \\ u_n = 111 - \frac{1130}{u_{n-1}} + \frac{3000}{u_{n-1}u_{n-2}}\end{array}\right.
-\end{align*}
-
-The limit of the series should be $6$ but when calculated with IEEE floats it is actually $100$
-The authors show that the limit is actually $100$ for different starting values, and the error in floating point arithmetic causes the series to go to that limit instead.
-
-\begin{figure}[H]
- \centering
- \includegraphics[width=0.8\textwidth]{figures/handbook1-1.pdf}
- \caption{Output of Program 1.1 from \emph{Handbook of Floating-Point Arithmetic}\cite{HFP} for various IEEE types}
-\end{figure}
\pagebreak
\bibliographystyle{unsrt}
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