diff options
author | Karl Berry <karl@freefriends.org> | 2021-12-12 22:31:41 +0000 |
---|---|---|
committer | Karl Berry <karl@freefriends.org> | 2021-12-12 22:31:41 +0000 |
commit | 9d4b1fda1b468db18f8f70ece19f42e4e1a69f58 (patch) | |
tree | 2b3abf211a8f8436b254a10308f37329323c5d71 | |
parent | cf5e51a222298c80ebd729561b06698c259aa1ca (diff) |
numerica-plus (12dec21)
git-svn-id: svn://tug.org/texlive/trunk@61289 c570f23f-e606-0410-a88d-b1316a301751
-rw-r--r-- | Master/texmf-dist/doc/latex/numerica-plus/README.txt | 31 | ||||
-rw-r--r-- | Master/texmf-dist/doc/latex/numerica-plus/numerica-plus.pdf | 7174 | ||||
-rw-r--r-- | Master/texmf-dist/doc/latex/numerica-plus/numerica-plus.tex | 2060 | ||||
-rw-r--r-- | Master/texmf-dist/tex/latex/numerica-plus/numerica-plus.sty | 916 | ||||
-rw-r--r-- | Master/tlpkg/tlpsrc/numerica-plus.tlpsrc | 0 |
5 files changed, 10181 insertions, 0 deletions
diff --git a/Master/texmf-dist/doc/latex/numerica-plus/README.txt b/Master/texmf-dist/doc/latex/numerica-plus/README.txt new file mode 100644 index 00000000000..e0ee344ea4b --- /dev/null +++ b/Master/texmf-dist/doc/latex/numerica-plus/README.txt @@ -0,0 +1,31 @@ +numerica-plus: a package to iterate and find fixed points of +functions, to find zeros and extrema of functions and to +calculate recurrence relations. + +Andrew Parsloe (ajparsloe@gmail.com) + +This work may be distributed and/or modified under the conditions +of the LaTeX Project Public License, either version 1.3c of this +license or any later version; see +http://www.latex-project.org/lppl.txt + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + +This is version 2.0.0 of numerica-plus, the first stand-alone +version. (Version 1 of numerica-plus was invoked from the +numerica package by means of a package option.) The packages +numerica, l3kernel, l3packages, amsmath and mathtools are +required. numerica-plus defines three commands: \nmcIterate to +iterate and find fixed points of functions, \nmcSolve to find +zeros and extrema of functions, and \nmcRecur to calculate the +terms of recurrence relations. See numerica-plus.pdf for +details on how to use the package. + +Manifest +%%%%%%%% +README.txt this document + +numerica-plus.sty LaTeX .sty file +numerica-plus.pdf documentation for numerica-plus.sty +numerica-plus.tex documentation source file + diff --git a/Master/texmf-dist/doc/latex/numerica-plus/numerica-plus.pdf b/Master/texmf-dist/doc/latex/numerica-plus/numerica-plus.pdf new file mode 100644 index 00000000000..325537a64b7 --- /dev/null +++ b/Master/texmf-dist/doc/latex/numerica-plus/numerica-plus.pdf @@ -0,0 +1,7174 @@ +%PDF-1.5 +% +1 0 obj +<< /S /GoTo /D (chapter.1) >> +endobj +4 0 obj +(\376\377\0001\000\040\000I\000n\000t\000r\000o\000d\000u\000c\000t\000i\000o\000n) +endobj +5 0 obj +<< /S /GoTo /D (section.1.1) >> +endobj +8 0 obj +(\376\377\0001\000.\0001\000\040\000E\000x\000a\000m\000p\000l\000e\000\040\000o\000f\000\040\000u\000s\000e\000:\000\040\000t\000h\000e\000\040\000r\000o\000t\000a\000t\000i\000n\000g\000\040\000d\000i\000s\000k) +endobj +9 0 obj +<< /S 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For more info, see https://www.lyx.org/. +%% Do not edit unless you really know what you are doing. +\documentclass[english,tableposition=top]{report} +\usepackage{lmodern} +\renewcommand{\sfdefault}{lmss} +\renewcommand{\ttdefault}{lmtt} +\usepackage[T1]{fontenc} +\usepackage{textcomp} +\usepackage[latin9]{inputenc} +\setcounter{secnumdepth}{3} +\usepackage{color} +\definecolor{shadecolor}{rgb}{0.667969, 1, 1} +\usepackage{babel} +\usepackage{array} +\usepackage{wrapfig} +\usepackage{booktabs} +\usepackage{framed} +\usepackage{url} +\usepackage{amsmath} +\usepackage[unicode=true,pdfusetitle, + bookmarks=true,bookmarksnumbered=true,bookmarksopen=true,bookmarksopenlevel=2, + breaklinks=true,pdfborder={0 0 1},backref=section,colorlinks=true] + {hyperref} + +\makeatletter + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% LyX specific LaTeX commands. +%% Because html converters don't know tabularnewline +\providecommand{\tabularnewline}{\\} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Textclass specific LaTeX commands. +\newenvironment{centred}% + {\begin{center}\baselineskip=13pt\parskip=1pt}{\end{center}} +\newenvironment{lyxcode} + {\par\begin{list}{}{ + \setlength{\rightmargin}{\leftmargin} + \setlength{\listparindent}{0pt}% needed for AMS classes + \raggedright + \setlength{\itemsep}{0pt} + \setlength{\parsep}{0pt} + \normalfont\ttfamily}% + \item[]} + {\end{list}} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands. +\usepackage{numerica-plus} +\usepackage{upquote} +\newcommand\rel{\,\varrho\;} +\DeclareMathOperator{\erf}{erf} +\DeclareMathOperator{\gd}{gd} + +\reuse{} +\constants{ c=30,\omega=0.2 } + +\makeatother + +\begin{document} +\title{\texttt{numerica-plus}~\\ +} +\author{Andrew Parsloe\\ +(\url{ajparsloe@gmail.com})} +\maketitle +\begin{abstract} +The \verb`numerica-plus` package defines commands to iterate and +find fixed points of functions of a single variable, to find the zeros +or extrema of such functions, and to calculate the terms of recurrence +relations.\\ +\\ +\noindent\begin{minipage}[t]{1\columnwidth}% +\begin{shaded}% + +\paragraph*{Note:} +\begin{itemize} +\item {\normalsize This document applies to version 2.0.0 of }{\normalsize\texttt{numerica-plus}}{\normalsize .def.}{\small\par} +\item {\normalsize A version of }{\normalsize\texttt{numerica}}{\normalsize{} +from or later than 2021/12/07 is required; (}{\normalsize\texttt{numerica}}{\normalsize{} +requires }{\normalsize\texttt{amsmath}}{\normalsize , }{\normalsize\texttt{mathtools}}{\normalsize{} +and the \LaTeX 3 bundles }{\normalsize\texttt{l3kernel}}{\normalsize{} +and }{\normalsize\texttt{l3packages}}{\normalsize ).}{\small\par} +\item {\normalsize I refer a number of times in this document to }{\normalsize\emph{Handbook +of Mathematical Functions}}{\normalsize , edited by Milton Abramowitz +and Irene A. Segun, Dover, 1965. This is abbreviated to }{\normalsize\emph{HMF}}{\normalsize .}{\small\par} +\item {\normalsize Version 2 of }{\normalsize\texttt{numerica-plus}}{\small\par} +\begin{itemize} +\item {\normalsize is the first stand-alone version; (in v.1 of }{\normalsize\texttt{numerica}}{\normalsize{} +the commands }{\normalsize\texttt{\textbackslash nmcIterate}}{\normalsize , +}{\normalsize\texttt{\textbackslash nmcSolve}}{\normalsize{} and }{\normalsize\texttt{\textbackslash nmcRecur}}{\normalsize{} +were loaded with the }{\normalsize\texttt{plus}}{\normalsize{} package +option);}{\small\par} +\item {\normalsize makes some small code adjustements;}{\small\par} +\item {\normalsize amends documentation.}{\small\par} +\end{itemize} +\end{itemize} +\end{shaded}% +\end{minipage} +\end{abstract} +\tableofcontents{} + +\chapter{Introduction} + +Entering +\begin{lyxcode} +~\textbackslash usepackage\{numerica-plus\} +\end{lyxcode} +in the preamble of your document makes available the commands +\begin{itemize} +\item \verb`\nmcIterate`, a command to iterate a function (apply it repeatedly +to itself), including finding fixed points (values $x$ where $f(x)=x$); +\item \verb`\nmcSolve`, a command to find the zeros of functions of a single +variable (values $x$ for which $f(x)=0$) or, failing that, local +maxima or minima of such functions; +\item \verb`\nmcRecur`, a command to calculate the values of terms in recurrence +relations in a single (recurrence) variable (like the terms of the +Fibonacci sequence or Legendre polynomials). +\end{itemize} +\verb`numerica-plus` requires a version of \verb`numerica` from +or later than 2021/11/26. If found, \verb`numerica` is loaded automatically, +making available the \verb`\nmcEvaluate`, \verb`\nmcInfo`, \verb`\nmcMacros`, +\verb`\nmcConstants`, and \verb`\nmcReuse` commands; see the \verb`numerica` +documentation for details on the use of these commands. + +The commands of the present package all share the syntax of \verb`\nmcEvaluate`. +I will discuss them individually in later chapters but turn first +to something more than a `toy' example that illustrates their use +and gives a sense of `what they are about'. + +\section{Example of use: the rotating disk} + +\label{sec:introExampleOfUse}Consider a disk rotating uniformly with +angular velocity $\omega$ in an anticlockwise sense in an inertial +system in which the disk's centre \textbf{0} is at rest. Three distinct +points \textbf{1}, \textbf{2}, \textbf{3} are fixed in the disk and, +in a co-rotating polar coordinate system centred at \textbf{0}, have +polar coordinates $(r_{i},\theta_{i})$ ($i,j=1,2,3$). Choose \textbf{01} +as initial line so that $\theta_{1}=0$. + +The cosine rule for solving triangles tells us that the time $t_{ij}$ +in the underlying inertial system for a signal to pass from \textbf{i} +to \textbf{j} satisfies the equation +\[ +t_{ij}=c^{-1}\sqrt{r_{i}^{2}+r_{j}^{2}-2r_{i}r_{j}\cos(\theta_{j}-\theta_{i}+\omega t_{ij})}\equiv f(t_{ij}), +\] +where $c$ is the speed of light. (Equally, we could be describing +an acoustic signal between points on a disk rotating uniformly in +a still, uniform atmosphere \textendash{} in which case $c$ would +be the speed of sound.) Although the equation doesn't solve algebraically +for the time $t_{ij},$ it does tell us that $t=t_{ij}$ is a \emph{fixed +point} of the function $f(t)$. To calculate fixed points we use the +command \verb`\nmcIterate`, or its short-name form \verb`\iter`, +with the star option, \verb`\iter*`. For \verb`\iter` the star option +means: continue iterating until a fixed point has been reached and, +as with the \verb`\eval` command, suppress all elements from the +display save for the numerical result. + +First, though, values need to be assigned to the various parameters. +Suppose we use units in which $c=30,$ and $\omega=0.2$ radians per +second. To avoid having to write these values in the vv-list every +time, I have put in the preamble to this document the statement +\begin{verbatim} + \constants{ c=30,\omega=0.2 } +\end{verbatim} +For the polar coordinates of \textbf{1 }and \textbf{3 }I have chosen +$r_{1}=10$, $r_{3}=20$ and $\theta_{3}=0.2$ radians (remember $\theta_{1}=0$). +To find a fixed point $t_{13}$ I give $t$ an initial trial value +$1$ (plucked from the air). Its position as the rightmost item in +the vv-list tells \verb`\iter` that $t$ is the iteration variable: +\begin{verbatim} + \iter*{ c^{-1}\sqrt{r_1^2+r_3^2-2r_1 r_3 + \cos(\theta_3+\omega t)} + }[ r_1=10,r_3=20,\theta_3=0.2,t=1 ], + \quad\info{iter}. +\end{verbatim} +$\Longrightarrow$ \iter*{ c^{-1}\sqrt{r_1^2+r_3^2-2r_1 r_3 + \cos(\theta_3+\omega t)} + }[ r_1=10,r_3=20,\theta_3=0.2,t=1 ], + \quad\info{iter}. The short-name form of the \verb`\nmcInfo` command from \verb`numerica` + has been used to display the number of iterations required to attain +the fixed-point value. + +To six figures, only five iterations are needed, which seems rapid +but we can check this by substituting $t=0.356899$ back into the +formula and \verb`\eval`-uating it: +\begin{verbatim} + \eval*{ c^{-1}\sqrt{r_1^2+r_3^2-2r_1 r_3 + \cos(\theta_3+\omega t)} + }[ r_1=10,r_3=20,\theta_3=0.2,t=0.356899 ] +\end{verbatim} +$\Longrightarrow$ \eval*{ c^{-1}\sqrt{r_1^2+r_3^2-2r_1 r_3 + \cos(\theta_3+\omega t)} + }[ r_1=10,r_3=20,\theta_3=0.2,t=0.356899 ], confirming that we have indeed calculated a fixed point. That it +did indeed take only $5$ iterations can be checked by omitting the +asterisk from the \verb`\iter` command and specifying the total number +of iterations to perform. I choose \texttt{do=}7 to show not just +the $5$th iteration but also the next two just to confirm that the +result is stable. We shall view all $7$: \texttt{see=7}. Because +of the length of the formula I have suppressed display of the vv-list +by giving the key \texttt{vvd}\emph{ }an empty value:\footnote{See the associated document \texttt{numerica.pdf}, the chapter on +settings.} +\begin{verbatim} + \iter[do=7,see=7,vvd=] + {\[ c^{-1}\sqrt{r_1^2+r_3^2-2r_1 r_3 + \cos(\theta_3+\omega t)} \]} + [ r_1=10,r_3=20,\theta_3=0.2,t=1 ] +\end{verbatim} +$\Longrightarrow$ \iter[do=7,see=7,vvd=] + {\[ c^{-1}\sqrt{r_1^2+r_3^2-2r_1 r_3 + \cos(\theta_3+\omega t)} \]} + [ r_1=10,r_3=20,\theta_3=0.2,t=1 ] +\noindent \begin{flushleft} +The display makes clear that on the $5$th iteration, the $6$-figure +value has been attained. +\par\end{flushleft} + +Alternatively, we could use the \verb`\nmcRecur` command, or its +short-name form \verb`\recur`, to view the successive iterations, +since an iteration is a first-order recurrence: $f_{n+1}=f(f_{n})$: +\begin{verbatim} + \recur[do=8,see1=0,see2=5,vvd={,\\(vv)\\},*] + {\[ f_{n+1}=c^{-1}\sqrt{r_1^2+r_3^2-2r_1 r_3 + \cos(\theta_3+\omega f_{n})} \]} + [ r_1=10,r_3=20,\theta_3=0.2,f_{0}=1 ] +\end{verbatim} +$\Longrightarrow$ \recur[do=8,see1=0,see2=5,vvd={,\\(vv)\\},*] + {\[ f_{n+1}=c^{-1}\sqrt{r_1^2+r_3^2-2r_1 r_3 + \cos(\theta_3+\omega f_{n})} \]} + [ r_1=10,r_3=20,\theta_3=0.2,f_{0}=1 ] + +\noindent I have specified \texttt{do=8} terms rather than $7$ since +the zero-th term ($f_{0}=1$) is included in the count. I've chosen +to view the last $5$ of them but none prior to those by writing \texttt{see1=0,see2=5}. +Notice the \texttt{vvd} setting, pushing display of the vv-list and +result to new lines and suppressing equation numbering with the \texttt{{*}} +setting (which turns the \verb`multline` environment into a \verb`multline*` +environment). + +Another and perhaps more obvious way to find the value of $t_{13}$, +is to look for a zero of the function $f(t)-t$. That means using +the command \verb`\nmcSolve` or its short-name form \verb`\solve`. +I shall do so with the star option \verb`\solve*` which suppresses +display of all but the numerical result. A trial value for $t$ is +required. I have chosen \texttt{t=0}: +\begin{verbatim} + \solve*{ c^{-1}\sqrt{r_1^2+r_3^2-2r_1 r_3 + \cos(\theta_3+\omega t)} - t } + [ r_1=10,r_3=20,\theta_3=0.2,t=0 ], + \quad\nmcInfo{solve}. +\end{verbatim} +$\Longrightarrow$ \solve*{ c^{-1}\sqrt{r_1^2+r_3^2-2r_1 r_3 + \cos(\theta_3+\omega t)} - t } + [ r_1=10,r_3=20,\theta_3=0.2,t=0 ], + \quad\nmcInfo{solve}. + +Nearly the same answer as before is attained but this time many more +steps have been required. This is to be expected. The \verb`\solve` +command uses the bisection method. Since $1/2^{10}\approx1/10^{3}$, +about $10$ bisections are needed to determine $3$ decimal places. +Hence we can expect about $20$ bisections for a $6$-decimal-place +answer. The particular form of the \verb`\nmcInfo` command display, +`$1+20$ steps', indicates that it took $1$ search step to find +an interval in which the function vanished and, having found that +interval, $20$ bisections to narrow the position of the zero to $6$-figures. +I will discuss the discrepancy in the final figure in Chapter~\ref{chap:solveSolve}; +see \ref{subsec:solveExtraRounding}. + +\subsection{Circuits} + +Okay, so we can calculate the time taken in the underlying inertial +system for a signal to pass from one point of the rotating disk to +another. How long does it take to traverse the circuit \textbf{1} +to \textbf{2} to \textbf{3} and back to \textbf{1}? That means forming +the sum $t_{12}+t_{23}+t_{31}$, hence calculating the separate $t_{ij}$ +and then using \verb`\eval` to calculate their sum. + +To simplify things, I assume a little symmetry. Let the (polar) coordinates +of \textbf{1} be $(a,0),$ of \textbf{2} be $(r,-\theta)$, and of +\textbf{3} be $(r,\theta)$: \textbf{2} and \textbf{3} are at the +same radial distance from the centre \textbf{0} and at the same angular +distance from the line \textbf{01} but on opposite sides of it, \textbf{3} +ahead of the line, \textbf{2} behind it. The rotation is in the direction +of positive $\theta$. Rather than just calculate $t_{12}+t_{23}+t_{31}$ +for the circuit \textbf{1231}, I also calculate the time $t_{13}+t_{32}+t_{21}$ +for a signal to traverse the same circuit but in the opposite sense, +\textbf{1321}, and compare them (form the difference). + +Note that with \textbf{2} and \textbf{3} positioned as they are relative +to \textbf{1}, a signal against the rotation from \textbf{3} to \textbf{1} +takes the same time as a signal from \textbf{1} to \textbf{2} and, +in the sense of rotation, a signal from \textbf{2} to \textbf{1} takes +the same time as a signal from \textbf{1} to \textbf{3}. To see this, +suppose the signal from \textbf{2} to \textbf{1} starts at time $t=0$; +it reaches \textbf{1} at a later time $t=t'$ when the disk has rotated +an angle $\omega t'$. Viewed from the underlying inertial system, +the signal path is a straight line from a point on a circle of radius +$r$ to a point on a concentric circle of radius $a$, the points +subtending an angle at the centre \textbf{0} of $\theta+\omega t'$. +But \textbf{3} at time $t'$ and \textbf{1} at time $t=0$ also subtend +an angle $\theta+\omega t'$ at \textbf{0}, and also lie on circles +of radii $r$ and $a$ respectively. In the underlying inertial system +the line segments \textbf{1}$(0)$\textbf{3}$(t')$ and \textbf{2}$(0)$\textbf{1(}$t')$ +are of equal length. Similarly, if a signal from \textbf{3} at time +$t=0$ reaches \textbf{1} at time $t=t''$ then \textbf{3}$(0)$\textbf{1}$(t'')$ +and \textbf{1}$(0)$\textbf{2}$(t'')$ are of equal length. Hence +the round trip times are $2t_{12}+t_{23}$ and $2t_{13}+t_{32}$. + +\subsubsection{Nesting commands} + +Analytically, both $t_{21}$ and $t_{13}$ are the same fixed point +of the function +\[ +c^{-1}\sqrt{r^{2}+a^{2}-2ra\cos(\theta+\omega t)} +\] + and $t_{31}$ and $t_{12}$ are the same fixed point of the function +\[ +c^{-1}\sqrt{r^{2}+a^{2}-2ra\cos(\theta-\omega t).} +\] +To calculate $2t_{12}+t_{23}$ therefore means calculating +\begin{verbatim} + 2\iter*{ c^{-1}\sqrt{a^2+r^2-2ar + \cos(\theta-\omega t)} } + + \iter*{ c^{-1}\sqrt{2r^2-2r^2 + \cos(2\theta+\omega t)} } +\end{verbatim} +with the analogous expression for $2t_{13}+t_{32}$. But we can do +the comparison of round trip times `in one go' by nesting the \verb`\iter*` +commands inside an \verb`\eval*` command: +\begin{verbatim} + + \eval*{ % circuit 1231 + 2\iter*[var=t]{ c^{-1}\sqrt{a^2+r^2-2ar + \cos(\theta-\omega t)} }[8] + + \iter*[var=t]{ c^{-1}\sqrt{2r^2-2r^2 + \cos(2\theta+\omega t)} }[8] + % circuit 1321 + - 2\iter*[var=t]{ c^{-1}\sqrt{a^2+r^2-2ar + \cos(\theta+\omega t)} }[8] + - \iter*[var=t]{ c^{-1}\sqrt{2r^2-2r^2 + \cos(2\theta-\omega t)} }[8] + }[ a=10,r=20,\theta=0.2,t=1 ] +\end{verbatim} +$\Longrightarrow$ \eval*{ % circuit 1231 + 2\times\iter*[var=t]{ c^{-1}\sqrt{a^2+r^2-2ar + \cos(\theta-\omega t)} }[8] + + \iter*[var=t]{ c^{-1}\sqrt{2r^2-2r^2 + \cos(2\theta+\omega t)} }[8] + % circuit 1321 + - 2\times\iter*[var=t]{ c^{-1}\sqrt{a^2+r^2-2ar + \cos(\theta+\omega t)} }[8] + - \iter*[var=t]{ c^{-1}\sqrt{2r^2-2r^2 + \cos(2\theta-\omega t)} }[8] + }[ a=10,r=20,\theta=0.2,{t}=1 ] . + +By itself this result is of little interest beyond seeing that \verb`numerica-plus` +can handle the calculation. What \emph{is} interesting is to find +values of our parameters for which the time difference vanishes \textendash{} +say values of $\theta$, given the other parameters, especially the +value of $r$. Is there a circuit such that it takes a signal the +same time to travel in opposite senses around the circuit, despite +the rotation of the disk? Rather than nesting the \verb`\iter*` commands +inside an \verb`\eval`, we need to nest them in a \verb`\solve` +command: +\begin{verbatim} + \solve[p=.,var=\theta,+=1,vvd=\\,*,+=1] + {\[ % circuit 1231 + 2\times\iter*[var=t,+=1]{ c^{-1}\sqrt{a^2+r^2-2ar + \cos(\theta-\omega t)} } + + \iter*[var=t,+=1]{ c^{-1}\sqrt{2r^2-2r^2 + \cos(2\theta+\omega t)} } + % circuit 1321 + - 2\times\iter*[var=t,+=1]{ c^{-1}\sqrt{a^2+r^2-2ar + \cos(\theta+\omega t)} } + - \iter*[var=t,+=1]{ c^{-1}\sqrt{2r^2-2r^2 + \cos(2\theta-\omega t)} } + \]}[ a=10,r=20,\theta=0.1,t=1 ] +\end{verbatim} +$\Longrightarrow$ \solve[p=.,var=\theta,+=1,vvd=\\,*] + {\[ % circuit 1231 + 2\times\iter*[var=t,+=1]{ c^{-1}\sqrt{a^2+r^2-2ar + \cos(\theta-\omega t)} } + + \iter*[var=t,+=1]{ c^{-1}\sqrt{2r^2-2r^2 + \cos(2\theta+\omega t)} } + % circuit 1321 + - 2\times\iter*[var=t,+=1]{ c^{-1}\sqrt{a^2+r^2-2ar + \cos(\theta+\omega t)} } + - \iter*[var=t,+=1]{ c^{-1}\sqrt{2r^2-2r^2 + \cos(2\theta-\omega t)} } + \]}[ a=10,r=20,\theta=0.1,t=1 ] + +One point to note here is the use of \verb`\times` (in \verb`2\times\iter*`). +In this example the formula is displayed (\verb`\solve` wraps around +math delimiters). Without the \verb`\times` the result would have +been the same but the display of the formula would have juxtaposed +the `$2$'s against the following decimals, making it look as if +signal travel times were $20.537778$ and $20.61442$ (and no doubt +causing perplexity). Also note the \texttt{vvd=\textbackslash\textbackslash} +to place the result on a new line and suppress display of the vv-list. + +So this expression gives a value of $\theta_{\Delta t=0}$ for one +value of $r$. The obvious next step is to create a table of such +values. I show how that is done in the document \verb`numerica-tables.pdf` +using the command \verb`\nmcTabulate` defined in the associated package +\verb`numerica-tables`. But this is not a research paper on the rotating +disk. I wished to show how the different commands of \verb`numerica-plus` +can be used to explore a meaningful problem. And although it looks +as if a lot of typing is involved, once $c^{-1}\sqrt{r^{2}+a^{2}-2ra\cos(\theta-\omega t)}$ +has been formed in \LaTeX{} and values specified in the vv-list, much +of the rest is copy-and-paste with minor editing. + +\section{Shared syntax of the new commands} + +\texttt{numerica-plus} offers three new commands for three processes: +\verb`\nmcIterate` (short-name form \verb`\iter`) for iterating +functions, \verb`\nmcSolve` (short-name form \verb`\solve`) for +finding the zeros or (local) extrema of functions, and \verb`\nmcRecur` +(short-name form \verb`\recur`) for calculating terms of recurrence +relations. + +All three commands share the syntax of the \verb`\nmcEvaluate` (or +\verb`\eval`) command detailed in the associated document \texttt{numerica.pdf}. +When all options are used the command looks like, for instance, +\begin{centred} +\noindent \verb`\nmcIterate*[settings]{expr.}[vv-list][num. format]` +\end{centred} +You can substitute \verb`\nmcSolve`, or \verb`\nmcRecur` for \verb`\nmcIterate` +here. The arguments are similar to those for \verb`\nmcEvaluate`. +\begin{enumerate} +\item \verb`*` optional switch; if present ensures a single number output +with no formatting, or an appropriate error message if the single +number cannot be produced; +\item \verb`[settings]` optional comma-separated list of \emph{key=value +}settings for this particular command and calculation; +\item \verb`{expr.}` the only mandatory argument; the mathematical expression +in \LaTeX{} form that is the object of interest; +\item \verb`[vv-list]` optional comma-separated list of \emph{variable=value +}items; for \verb`\iter` and \verb`\solve` the \emph{rightmost} +(or innermost) variable in the vv-list may have special significance; +\item \verb`[num. format]` optional format specification for presentation +of the numerical result (rounding, padding with zeros, scientific +notation); boolean output is suppressed for these commands. +\end{enumerate} +Like \verb`\nmcEvaluate`, for all three commands the way the result +is displayed depends on whether the command wraps around math delimiters, +or is used between math delimiters or in the absence of math delimiters. +These distinctions are relevant\emph{ only if the optional star {*} +is absent.} +\begin{itemize} +\item When the star option is used, the \emph{result} is a number only, +without any formatting or vv-list display, or an error message is +displayed. +\item When the star option is not used and one of the following is the case +\begin{itemize} +\item the command wraps around math delimiters, e.g. \verb`\iter{$ expr. $}`, +then +\begin{itemize} +\item the result is displayed in the form \emph{formula~=~result, (vv-list)} +or the form \emph{formula~$\rightarrow$~result, (vv-list)} as appropriate; +\end{itemize} +\item the command is used within math delimiters, e.g. \verb`\[\iter...\]`, +then +\begin{itemize} +\item the result is displayed in the form \emph{result, (vv-list)} (without +reference to the formula); +\end{itemize} +\item the command is used in the absence of delimiters, then +\begin{itemize} +\item the result is presented as if it had been used between \verb`\[` +and \verb`\]`. +\end{itemize} +\end{itemize} +\end{itemize} +Looking at the various examples in the preceding section on the rotating +disk you will see illustrations of all these situations. + +\subsection{Settings} + +Nearly all the settings available to the \verb`\eval` command are +available to these other commands. To save switching between documents +I reproduce in Table~\ref{tab:introSettingsInherited} the options +found in \texttt{numerica.pdf}, although for discussion of the options +you will need to refer to that document. In addition, each of the +present commands also has settings of its own, discussed at the relevant +parts of the following chapters. + +\subsection{Nesting} + +In v.1 of \verb`numerica`, for commands to be nested one within an +another, it was necessary for the inner command to be starred (and +thus produce a purely numerical result). With v.2 of \verb`numerica` +this is no longer the case. A nested command is detected as such and +the star automatically set, whether the user has explicitly starred +the command or not. Provided the starred form of a command actually +does produce a numerical result and not an error message then it can +be nested within the main argument of any one of the other commands, +including itself. The example of use, \ref{sec:introExampleOfUse} +above, shows several examples of this. The starred form can also be +used in the vv-list of any one of the commands, including itself. +The associated document \texttt{numerica.pdf} shows examples of an +\verb`\eval*` command being used in the vv-list of an \verb`\eval` +command. + +\begin{table} +\noindent \centering{}\caption{\protect\label{tab:introSettingsInherited}Inherited settings options} +\noindent \begin{center} +\begin{tabular}{ll>{\raggedright}p{4cm}>{\raggedright}p{4cm}} +\toprule +{\small key} & {\small type} & {\small meaning} & {\small default}\tabularnewline +\midrule +{\small\texttt{dbg}} & {\small int} & {\small debug `magic' integer} & {\small\texttt{0}}\tabularnewline +{\small\texttt{\textasciicircum}} & {\small char} & {\small exponent mark for sci. notation input} & {\small\texttt{e}}\tabularnewline +{\small\texttt{xx}} & {\small int (0/1)} & {\small multi-token variable switch} & {\small\texttt{1}}\tabularnewline +{\small\texttt{()}} & {\small int (0/1/2)} & {\small trig. function arg. parsing} & {\small\texttt{0}}\tabularnewline +{\small\texttt{o}} & & {\small degree switch for trig. funcions} & \tabularnewline +{\small\texttt{log}} & {\small num} & {\small base of logarithms for }{\small{\small\verb`\log`}} & {\small\texttt{10}}\tabularnewline +{\small\texttt{vvmode}} & {\small int (0/1)} & {\small vv-list calculation mode} & {\small\texttt{0}}\tabularnewline +{\small\texttt{vvd}} & {\small tokens} & {\small vv-list display-style spec.} & {\small\texttt{\{,\}\textbackslash mskip 12mu plus 6mu minus 9mu(vv)}}\tabularnewline +{\small\texttt{vvi}} & {\small token(s)} & {\small vv-list text-style spec.} & {\small\texttt{\{,\}\textbackslash mskip 36mu minus 24mu(vv)}}\tabularnewline +{*} & & {\small suppress equation numbering if }{\small\texttt{\textbackslash\textbackslash}}{\small{} +in }{\small\texttt{vvd}} & \tabularnewline +{\small\texttt{p}} & char(s) & {\small punctuation (esp. in display-style)} & {\small\texttt{, (comma)}}\tabularnewline +{\small\texttt{S+}} & {\small int} & {\small extra rounding for stopping criterion for sums} & {\small\texttt{2}}\tabularnewline +{\small\texttt{S?}} & {\small$\text{int}\ge0$} & {\small stopping criterion query terms for sums} & {\small\texttt{0}}\tabularnewline +{\small\texttt{P+}} & {\small int} & {\small extra rounding for stopping criterion for products} & {\small\texttt{2}}\tabularnewline +{\small\texttt{P?}} & {\small$\text{int}\ge0$} & {\small stopping criterion query terms for products} & {\small\texttt{0}}\tabularnewline +\bottomrule +\end{tabular} +\par\end{center} +\end{table} + + +\chapter{Iterating functions: \texttt{\textbackslash nmcIterate}} + +\label{chap:Iterating-functions}Only in desperation would one try +to evaluate a continued fraction by stacking fraction upon fraction +upon fraction like so: +\begin{verbatim} + \eval{\[ 1+\frac{1}{1+\frac{1}{1+\frac{1} + {1+\frac{1}{1+\frac{1}{1+\frac{1} + {1+\frac{1}{1+\frac{1}{1+\frac{1} + {1+\frac{1}{1+\frac{1} + {1+\frac{1}{1}}}}}}}}}}}} \]} +\end{verbatim} +$\Longrightarrow$ \eval{\[ 1+\frac{1}{1+\frac{1}{1+\frac{1} + {1+\frac{1}{1+\frac{1}{1+\frac{1} + {1+\frac{1}{1+\frac{1}{1+\frac{1} + {1+\frac{1}{1+\frac{1} + {1+\frac{1}{1}}}}}}}}}}}} \]} + +\noindent \verb`numerica-plus` provides a command for tackling problems +like this sensibly. In such problems a function is repeatedly applied +to itself (iterated). This is done through the command \verb`\nmcIterate` +or (short-name form) \verb`\iter`. Thus to evaluate this continued +fraction write (for instance), +\begin{centred} +\verb`\iter[do=15,see=5]{\[ 1+1/x \]}[x=1]` $\Longrightarrow$ \iter[do=15,see=5]{\[ 1+1/x \]}[x=1] +\end{centred} +The \verb`\iter` command evaluates $1+1/x$ when $x=1$ and then +uses this value as a new $x$-value to substitue into $1+1/x$, to +once again evaluate and use as a new $x$-value, and so on. It looks +as if the repeated iterations are approaching \verb`\eval{$ \tfrac{\sqrt{5}+1}2 $}` +$\Longrightarrow$ \eval{$ \tfrac{\sqrt{5}+1}2 $}. Increasing the +number of iterations in the example from \texttt{do=15} to, say, \texttt{do=18}, +shows that this is indeed the case. +\begin{centred} +\verb`\iter[do=18,see=5]{\[ 1+1/x \]}[x=1]` $\Longrightarrow$ \iter[do=18,see=5]{\[ 1+1/x \]}[x=1] +\end{centred} +But iteration of functions is not limited to continued fractions. +Particularly since the emergence of chaos theory, iteration has become +an important study in its own right. Any function with range within +its domain can be iterated \textendash{} repeatedly applied to itself +\textendash{} like the cosine: +\begin{centred} +\verb`\iter[do=20,see=4]{\[ \cos x \]}[x=\pi/2]` $\Longrightarrow$ +\iter[do=20,see=4]{\[ \cos x \]}[x=\pi/2] +\end{centred} +which displays the first one and last four of 20 iterations of $\cos x$ +when $x=\tfrac{\pi}{2}$. It looks as if the cosine is `cautiously' +approaching a limit, perhaps around $0.738$ or $0.739$. You need +to nearly double the number of iterations (\texttt{do=40}) to confirm +that this is so. + +The logistic function $kx(1-x)$ exhibits a variety of behaviours +depending on the value of $k$. For instance, with $k=3.5$ we get +a period-4 cycle: +\begin{centred} +\verb`\iter[do=12,see=8]{\[ kx(1-x) \]}[k=3.5,x=0.5]` $\Longrightarrow$ + \iter[do=12,see=8]{\[ kx(1-x) \]}[k=3.5,x=0.5] +\end{centred} +and with $k=3.1$ we get a period-2 cycle, although it takes many +more iterations to stabilize there: +\begin{centred} +\verb`\iter[do=42,see=4]{\[ kx(1-x) \]}[k=3.1,x=0.5]` $\Longrightarrow$ +\iter[do=42,see=4]{\[ kx(1-x) \]}[k=3.1,x=0.5] +\end{centred} + +\section{Star (\texttt{{*}}) option: fixed points} + +In the first two of these examples, iteration eventually ended at +a \emph{fixed point}. This is a point $x$ where $f(x)=x$. Appending +a star (asterisk) to the \verb`\iter` command is the signal for iteration +to continue until a fixed point has been reached at the specified +rounding value: +\begin{centred} +\verb`\iter*{ 1+a/x }[a=n(n+1),n=1,x=1]` $\Longrightarrow$ \iter*{ 1+a/x }[a=n(n+1),n=1,x=1] +\end{centred} +(with the default rounding value $6$).\footnote{For your own interest try also putting $n=2,3,4,\ldots$ in the vv-list +of this expression.} The star overrides any value for the number of iterations to perform +(the \texttt{do} key) that may have been entered in the settings option. +It also overrides any elements of the display other than the numerical +result. With the star option math delimiters are irrelevant \textendash{} +other than displaying minus signs correctly when \verb`\iter*` is +between them. + +A function may not approach a fixed point when iterated \textendash{} +see the examples with the logistic function above. To prevent an infinite +loop \verb`\iter*` counts the number of iterations performed and +when that number reaches a certain cut-off value \textendash{} the +default is $100$ \textendash{} the loop terminates and a message +is displayed: +\begin{centred} +\verb`\iter*{kx(1-x)}[k=3.5,x=0.5]` $\Longrightarrow$ \iter*{kx(1-x)}[k=3.5,x=0.5] +\end{centred} +In this case we \emph{know} that a fixed point does not exist, but +that may not always be the case. One response to a message like this +is to change parameter values or starting value of the iteration variable. +For instance, changing the parameter value to $k=1.5$, +\begin{centred} +\verb`\iter*{kx(1-x)}[k=1.5,x=0.5]` $\Longrightarrow$ \iter*{kx(1-x)}[k=1.5,x=0.5], +\end{centred} +means a fixed point is now attained. It is easy to check that $1/3$ +is indeed a fixed point (but that makes the $4$ in the last decimal +place a concern; see the extra rounding setting, \ref{subsec:iterExtra-rounding}). + +But should a fixed point still not eventuate after `fiddling' with +parameter and start values, there are two general adjustments one +might try: either +\begin{enumerate} +\item reduce the rounding value, from the default $6$ (or the one specified), +to a smaller value, or +\item increase the cut-off figure from the default 100 to some higher value. +\end{enumerate} +The former is done via the trailing number format optional argument +of the \verb`\iter` command; the latter is done via the settings +option, see \ref{sec:iterSettings-option}, specifically \ref{subsec:iterMaximum-iteration-count}. + +\subsection{Use with \texttt{\textbackslash nmcInfo}} + +It is of interest to know how many iterations are required to reach +a fixed point at a particular rounding value. That knowledge allows +a good guess as to whether a fixed point will be attained at a greater +rounding value. Thus when iterating the function +\[ +f(t_{ij})=c^{-1}\sqrt{r_{i}^{2}+r_{j}^{2}-2r_{i}r_{j}\cos(\theta_{j}-\theta_{i}+\omega t_{ij})} +\] +in \ref{sec:introExampleOfUse} only $5$ iterations were required +to attain $6$-figure accuracy for the fixed point. That information +came by following the \verb`\iter*` command with \verb`\nmcInfo` +(or \verb`\info`) with the argument \verb`iter`. And generally, +for any `infinite' process, follow the command with an \verb`\info` +command if you want to know how many `steps' \textendash{} in the +present case iterations \textendash{} are required to achieve the +result. So, if $5$ iterations achieve $6$-figure accuracy, presumably +something like $10$ iterations will achieve $12$-figure accuracy: +\begin{verbatim} + \iter*{ c^{-1}\sqrt{r_i^2+r_j^2-2r_i r_j + \cos(\theta_{ij}+\omega t)} + }[ r_i=10,r_j=20,\theta_{ij}=0.2,t=1 ][12] + ,\quad\info{iter}. +\end{verbatim} +$\Longrightarrow$ \iter*{ c^{-1}\sqrt{r_i^2+r_j^2-2r_i r_j + \cos(\theta_{ij}+\omega t)} + }[ r_i=10,r_j=20,\theta_{ij}=0.2,t=1 ][12] + ,\quad\info{iter}. (Remember, \verb`numerica-plus` knows the values of $c$ and $\omega$ +from a \verb`\constants` statement in the preamble.) And indeed only +$9$ iterations suffice to achieve $12$-figure accuracy: +\begin{verbatim} + \iter[do =11,see=4] + { c^{-1}\sqrt{r_i^2+r_j^2-2r_i r_j + \cos(\theta_{ij}+\omega t)} + }[ r_i=10,r_j=20,\theta_{ij}=0.2,t=1 ][12] +\end{verbatim} +$\Longrightarrow$ \iter[do =11,see=4] + { c^{-1}\sqrt{r_i^2+r_j^2-2r_i r_j + \cos(\theta_{ij}+\omega t)} + }[ r_i=10,r_j=20,\theta_{ij}=0.2,t=1 ][12] + +Or again, with another example from earlier, +\begin{centred} +\verb`$ \iter*{\cos x}[x=\pi/2] $,\ \info{iter}.` $\Longrightarrow$ +$\iter*{\cos x}[x=\pi/2] $,\ \info{iter}. +\end{centred} +That suggests that around $2\times37=74$ iterations will give a $2\times6=12$-figure +answer, well within the cut-off figure of $100$: +\begin{centred} +\verb`$ \iter*{\cos x}[x=\pi/2][12] $,\ \info{iter}.` $\Longrightarrow$ +$\iter*{\cos x}[x=\pi/2][12] $,\ \info{iter}. +\end{centred} + +\section{Settings option} + +\label{sec:iterSettings-option}The settings option is a comma-separated +list of items of the form \emph{key~=~value}. + +\subsection{Inherited settings} + +Nearly all of the keys discussed in the settings\emph{ }option for +\verb`\nmcEvaluate` are available for \verb`\nmcIterate`. Table~\ref{tab:introSettingsInherited} +above lists these, repeating a table from \texttt{numerica.pdf}. Thus +should a quantity in the vv-list depend on the iteration variable, +forcing an implicit mode calculation, simply enter, as with \verb`\eval`, +\verb`vv@=1` (alternatively, \verb`vvmode=1`) in the settings option: +\begin{centred} +\verb`\iter*[vv@=1]{$ f(x) $}[f(x)=1+a/x,a=12,x=1]` $\Longrightarrow$ +\iter*[vv@=1]{$ f(x) $}[f(x)=1+a/x,a=12,x=1]. +\end{centred} +Implicit in the example is the default multi-token setting \texttt{xx=1} +inherited from \verb`\eval` and ensuring that the multi-token variable +$f(x)$ is treated correctly. + +Let's add \verb`dbg=1` to the example: +\begin{centred} +\verb`\iter*[dbg=1,vv@=1]{$ f(x) $}[f(x)=1+a/x,a=12,x=1]` $\Longrightarrow$ +\iter*[dbg=1,vv@=1]{$ f(x) $} [f(x)=1+a/x,a=12,x=1] +\end{centred} +The multi-token variable \verb`f(x)` has been changed to a single-token. +The values shown under `stored' and `fp-form' are those of the +\emph{final} iteration. + +\subsection{\texttt{\textbackslash nmcIter}ate-specific settings} + +In addition to the inherited settings there are some specific to \verb`\nmcIterate`. +These are listed in Table~\ref{tab:iterSettings}. + +\subsubsection{Iteration variable} + +In nearly all of the examples so far, the iteration variable has been +the rightmost variable in the vv-list and has not needed to be otherwise +specified. However it is sometimes not feasible to indicate the variable +in this way. In that case, entering +\begin{lyxcode} +var~=~<variable~name> +\end{lyxcode} +in the settings option enables the variable to be specified, irrespective +of what the rightmost variable in the vv-list is. Here, \texttt{<variable +name>} will generally be a character like \texttt{x} or \texttt{t} +or a token like \verb`\alpha`, but it could also be a multi-token +name like {\ttfamily\verb`x'`}\texttt{ }or \texttt{\textbackslash beta\_\{ij\}} +(or even \texttt{Fred} if you so chose). Although the iteration variable +can be independently specified like this, it must still be given an +initial \emph{value} in the vv-list \textendash{} only it need not +be the rightmost variable. + +In the following example the rightmost variable is $a$ which is clearly +\emph{not} the iteration variable: +\begin{centred} +\verb`\iter[var=x,do=40,see=5]{$ 1+a/x $}[x=a/6,a=6][*]` $\Longrightarrow$ +\iter[var=x,do=40,see=5]{$ 1+a/x $}[x=a/6,a=6][*] +\end{centred} + +\subsubsection{Extra rounding for fixed-point calculations} + +\label{subsec:iterExtra-rounding} +\begin{table} +\centering{}\caption{\protect\label{tab:iterSettings}Settings for \texttt{\textbackslash nmcIterate}} +\begin{center} +\begin{tabular}{llll} +\toprule +{\small key} & {\small type} & {\small meaning} & {\small default}\tabularnewline +\midrule +{\small\texttt{var}} & {\small token(s)} & {\small iteration variable} & \tabularnewline +{\small\texttt{+}} & {\small int} & {\small fixed point extra rounding} & {\small\texttt{0}}\tabularnewline +{\small\texttt{max}} & {\small int > 0} & {\small max. iteration count (fixed points)} & {\small\texttt{100}}\tabularnewline +{\small\texttt{do}} & {\small int > 0} & {\small number of iterations to perform} & {\small\texttt{5}}\tabularnewline +{\small\texttt{see}} & {\small int > 0} & {\small number of final iterations to view} & {\small\texttt{4}}\tabularnewline + & {\small int ($\mathtt{0}/\mathtt{1}/\mathtt{2}$)} & {\small form of result saved with }{\small{\small\verb`\`}} & {\small\texttt{0}}\tabularnewline +\bottomrule +\end{tabular} +\par\end{center} +\end{table} +\verb`numerica-plus` determines that a fixed point has been reached +when the difference between successive iterations vanishes when rounded +to the current rounding value. One might want reassurance that this +really is the correct value by seeking a fixed point at a higher rounding +value than that displayed. This extra rounding is achieved by entering +\begin{lyxcode} ++~=~<integer> +\end{lyxcode} +in the settings option. By default this extra rounding is set to zero. + +We have seen before that $\cos x$ starting at $x=\tfrac{1}{2}\pi$ +takes $37$ iterations to reach a $6$-figure fixed point $0.739085$, +about $6$ iterations per decimal place. By entering \texttt{+=1} +in the settings option the number of iterations is increased to $43$, +$6$ more than $37$ but, reassuringly, the $6$-figure result that +is displayed remains unchanged: +\begin{centred} +\verb`$ \iter*[+=1]{\cos x}[x=\pi/2] $,\ \info{iter}.` $\Longrightarrow$ +$ \iter*[+=1]{\cos x}[x=\pi/2] $,\ \info{iter}. +\end{centred} + +\subsubsection{Maximum {\small iteration count for fixed-point searches}} + +\label{subsec:iterMaximum-iteration-count}To prevent a fixed-point +search from continuing indefinitely when no fixed point exists, there +needs to be a maximum number of iterations specified after which point +the search is called off. By default this number is $100$. To change +it enter +\begin{lyxcode} +max~=~<positive~integer> +\end{lyxcode} +in the settings option. + +\subsubsection{Number of iterations to perform} + +To specify the number of iterations to perform enter +\begin{lyxcode} +do~=~<positive~integer> +\end{lyxcode} +in the settings option. Note that if the \verb`*` option is present +this value will be ignored and iteration will continue until either +a fixed point or the maximum iteration count is reached. By default +\texttt{do} is set to $5$. (Note that \texttt{do} can be set to a +greater number than \texttt{max}; \texttt{max} applies only to \verb`\iter*`.) + +\subsubsection{Number of iterations to show} + +To specify the number of final iterations to show enter +\begin{lyxcode} +see~=~<positive~integer> +\end{lyxcode} +in the settings option. By default \texttt{see} is set to $4$. Always +it is the \emph{last} \texttt{see} iterations that are displayed. +If \texttt{see} is set to a greater value than \texttt{do}, all iterations +are shown. If the star option is used the \texttt{see} value is ignored. + +\subsubsection{Form of result saved by \texttt{\textbackslash nmcReuse}} + +By entering +\begin{lyxcode} +reuse~=~<integer> +\end{lyxcode} +in the settings option of the \verb`iter` command it is possible +to specify the form of result that is saved when using \verb`\nmcReuse`. +(This setting has no effect when the star option is used with \verb`\nmcIterate`. +In that case only the numerical result of the fixed point calculation +\textendash{} if successful \textendash{} is saved.) The possibilities +are: +\begin{itemize} +\item \texttt{int=0} (or any integer $\mathtt{\ne1,2}$) saves the display +resulting from the \verb`\iter` command (the default); +\item \texttt{int=1} saves a comma-separated list of braced pairs of the +form:\texttt{ }~\\ +\texttt{\{$k$, value-of-$k$-th-iterate\}}; +\item \texttt{int=2} saves a comma-separated list of iterate values. +\end{itemize} +Note that the number and content of the items in the lists are those +resulting from the \texttt{see} setting (the number of iterations +to view). +\begin{verbatim} + \iter[reuse=1,do=12,see=4] + {\[ kx(1-x) \]}[k=3.5,x=0.5] + \reuse{logistic} +\end{verbatim} +$\Longrightarrow$ \iter[reuse=1,do=12,see=4]{\[ kx(1-x) \]}[k=3.5,x=0.5] \reuse[renew]{logistic} + +\noindent whence \verb`\logistic` $\Longrightarrow$ \logistic{}. +As you can see the control sequence \verb`\logistic` displays as +a comma-separated list of numbers, alternating between the iterate +ordinal and the iterate value. That these are stored as braced pairs +can be seen by using \TeX 's \verb`\meaning` command: +\begin{verbatim} + \meaning \logistic +\end{verbatim} +$\Longrightarrow$ \meaning \logistic + +\subsection{Changing default values} + +\begin{wraptable}[11]{o}{0.5\columnwidth}% +\centering{}\vspace{-2.7ex} +\caption{\protect\label{tab:iterSettingsDefaults}Defaults for \texttt{\textbackslash nmcIterate}} +\begin{center} +\begin{tabular}{ll} +\toprule +{\small key} & {\small default}\tabularnewline +\midrule +{\small iter-extra-rounding} & {\small\texttt{0}}\tabularnewline +{\small iter-max-iterations} & {\small\texttt{100}}\tabularnewline +{\small iter-do} & {\small\texttt{5}}\tabularnewline +{\small iter-see-last} & {\small\texttt{4}}\tabularnewline +{\small iter-reuse} & {\small\texttt{0}}\tabularnewline +\bottomrule +\end{tabular} +\par\end{center}\end{wraptable}% +If you wish to change the default values of the various settings for +\verb`\nmcIterate` this can be done by entering new values in a configuration +file \texttt{numerica-plus.cfg} as described in the chapter on settings +in the associated document \texttt{numerica.pdf}. The relevant keys +are listed in Table~\ref{tab:iterSettingsDefaults}, corresponding +to the \verb`+`, \verb`max`, \verb`do`, \verb`see` and \verb`reuse` +settings of the \verb`\iter` command. (Obviously it makes no sense +to have a default setting for the iteration variable. That will change +from case to case.) + +\section{Errors} + +By errors I refer to \verb`numerica-plus` errors rather than \LaTeX{} +errors. We have already met one in the discussion of fixed points: +\begin{centred} +\verb`\iter*{kx(1-x)}[k=3.5,x=0.5]` $\Longrightarrow$ \iter*{kx(1-x)}[k=3.5,x=0.5] +\end{centred} +For a function to be iterated indefinitely, its range must lie within +or be equal to its domain. If even part of the range of a function +lies outside its domain, then on repeated iteration there is a chance +that a value will eventually be calculated which lies in this `outside' +region. Iteration cannot continue beyond this point and an error message +is generated. As an example consider the inverse cosine, \verb`\arccos`. +This can be iterated only so far as the iterated values lie between +$\pm1$ inclusive. If we try to iterate \verb`\arccos` at 0 for example, +since $\cos\frac{1}{2}\pi=0$, $\arccos0=\eval{0.5\pi}[4]$ (which +is $\tfrac{1}{2}\pi$) so only a first iterate is possible. But we +could choose an initial value more carefully; $37$ iterations of +the cosine at $\tfrac{1}{2}\pi$ led to a fixed point $0.739085$, +so let's choose $0.739085$ as initial point and perform $37$ iterations: +\begin{centred} +\verb`\iter[do=37,see=4]{\[ \arccos x \]}[x=0.739085]` $\Longrightarrow$ +\iter[do=37,see=4]{\[ \arccos x \]}[x=0.739085] +\end{centred} +The result of the $37$th iteration is greater than $1$. Thus increasing +the number of iterations to 38 should generate an error message: +\begin{centred} +\verb`\iter[do=38,see=4]{\[ \arccos x \]}[x=0.739085]` $\Longrightarrow$\iter[do=38,see=4]{\[ \arccos x \]} [x=0.739085] +\end{centred} +\verb`l3fp` objects when asked to find the inverse cosine of a number +greater than $1$. + +\chapter{Finding zeros and extrema: \texttt{\textbackslash nmcSolve}} + +\label{chap:solveSolve}\verb`numerica-plus` provides a command,\textbf{ +}\verb`\nmcSolve` (short-name form \verb`\solve`), for finding a +zero of a function, should it have one. In the following example, +\begin{centred} +\verb`\solve[p]{\[ e^{ax}-bx^2 \]}[a=2,b=3,{x}=0]` $\Longrightarrow$ +\solve[p]{\[ e^{ax}-bx^2 \]}[a=2,b=3,{x}=0] +\end{centred} +I have sought and found a solution $x$ to the equation $e^{ax/2}-bx^{2}=0$ +when $a=2$ and $b=3$, starting with a trial value $x=0$, entered +as the \emph{rightmost} variable in the vv-list (and em-braced since +I don't want this trial value displaying in the presentation of the +result). Although $x$ has been found to the default six-figure accuracy, +it is evident that the function vanishes only to five figures. Let's +check: +\begin{centred} +\verb`\eval{$ bx^2 $}[b=3,x=x=-0.390647]` $\Longrightarrow$ \eval{$ bx^2 $}[b=3,x=-0.390647], + +\verb`\eval{$ e^{ax} $}[a=2,x=-0.390647]` $\Longrightarrow$ \eval{$ e^{ax} $}[a=2,x=-0.390647]; +\end{centred} +the values agree save in the final digit. + +This discrepancy in the final decimal place or places is a general +feature of solutions found by \verb`\solve`. It is the value of $x$, +not the value of $f(x)$, that is being found (in this case) to six +figures. If the graph of a function crosses the $x$-axis steeply +then the $x$ value (the zero) may be located to a higher precision +than the function value. Conversely, if the graph of a function crosses +the $x$-axis gently (at a shallow angle) then the function value +will vanish to a greater number of decimal places than the zero (the +$x$ value) is found to. + +A second example, which we can check against values tabulated in \emph{HMF}, +is to find a value of $x$ that satisfies $\tan x=\lambda x$. In +other words, find a zero of $\tan x-\lambda x$. In the example $\lambda$ +is negative, so a trial value for $x$ greater than $\pi/2$ seems +like a good idea. I've chosen $x=2$. +\begin{centred} +\verb`\solve{$ \tan x - \lambda x $}[\lambda=-1/0.8,{x}=2][5]` $\Longrightarrow$ +\solve{$ \tan x - \lambda x $}[\lambda=-1/0.8,{x}=2][5]. +\end{centred} +Table 4.19 of \emph{HMF }lists values of $x$ against $\lambda$ and +this is the value tabulated there. + +\section{Extrema} + +A function may not have a zero; or, for the given initial trial value +and initial step in the search for a zero, there may be a local extremum +in the way. In that case \verb`numerica-plus` may well locate the +local extremum (maximum or minimum but not a saddle point). For example +for the quadratic $(2x-1)^{2}+3x+1$ the \verb`\solve` command gives +the result +\begin{centred} +\verb`\solve[vvi=]{$ (2x-1)^2+3x+1 $}[x=2]` $\Longrightarrow$\solve[vvi=]{$ (2x-1)^2+3x+1 $}[{x}=2]. +\end{centred} +Since $(2x-1)^{2}+3x+1\ne0$ for any (real number) $x$, we deduce +that the quadratic takes a minimum value $1.9375$ at $x=0.125$ \textendash{} +easily confirmed analytically. This particular minimum is a global +minimum but in general any extremum found is only \emph{local}. The +function may well take larger or smaller values (or vanish for that +matter) further afield. + +It is also worth noting in this example the \verb`vvi=` in the settings +option which suppresses display of the vv-list. (The only member of +the vv-list is the trial value \verb`x=2` which we do not want to +display.) + +\noindent\begin{minipage}[t]{1\columnwidth}% +\begin{shaded}% +Note that the function for which a zero is being sought is \emph{not} +equated to zero when entered in the \verb`\solve` command. It is +\verb`\solve{ f(x) }`, not \verb`\solve{ f(x)=0 }`. This is precisely +because it may be an extremum that is found rather than a zero (if +extremum or zero is found at all \textendash{} think $e^{x}$). The +display of the result makes clear which is which, equating $f(x)$ +to its value, zero or extremum depending on what has been found, as +you can see in the preceding examples.\end{shaded}% +\end{minipage} + +\subsection{The search strategy} + +\label{subsec:solveSearch-strategy}If you have some sense of where +a function has a zero, then choose a trial value in that vicinity. +\verb`\solve` uses a bisection method to home in on the zero. It +therefore needs \emph{two} initial values. For the first it uses the +trial value you specify, call it $a$ and for the second, by default, +it uses $a+1$. (The default value $1$ for the initial step from +the trial value can be changed in the settings option; see \ref{sec:solveSettings-option}.) +If $f(a)$ and $f(a+1)$ have opposite signs then that is good. Bisection +of the interval $[a,a+1]$ can begin immediately in order to home +in on the precise point where $f$ vanishes. Write $b=a+1$. +\begin{itemize} +\item Let $c=\tfrac{1}{2}(a+b)$; if $f(c)=0$ the zero is found; otherwise +either $f(a),f(c)$ are of opposite signs or $f(c),f(b)$ are of opposite +signs. In the former case write $a_{1}=a,$ $b_{1}=c$; in the latter +case write $a_{1}=c$, $b_{1}=b$ and then redefine $c=\tfrac{1}{2}(a_{1}+b_{1})$. +Continue the bisection process, either until an exact zero $c$ of +$f$ is reached ($f(c)=0$) or a value $c$ is reached where the difference +between $a_{n+1}$ and $b_{n+1}$ is zero at the specified rounding +value. (But note, $f(c)$ may not vanish at that rounding value \textendash{} +the zero might be elsewhere in the interval and $f$ might cross the +axis at a steep slope.) +\end{itemize} +However $f(a)$ and $f(b)=f(a+1)$ may not have opposite signs. If +we graph the function $y=f(x)$ and suppose $f(a),f(b)$ are distinct +but of the same sign, then the line through the points $(a,f(a))$, +$(b,f(b))$ will intersect the $x$-axis to the left of $a$ or the +right of $b$ depending on its slope. We search always \emph{towards +the $x$-axis} in steps of $b-a$ ($=1$ with default values). +\begin{itemize} +\item If the line intersects the axis to the left of $a$ then $c=a-(b-a)$ +and we set $a_{1}=c,b_{1}=a$; if the line intersects the axis to +the right of $b$ then $c=b+(b-a)$ and we set $b_{1}=c,a_{1}=b$. +The hope is that by always taking steps in the direction towards the +$x$-axis that eventually $f(c)$ will be found to lie on the \emph{opposite} +side of the axis from $f(a_{n})$ or $f(b_{n})$, at which point the +bisection process begins. +\item Of course this may not happen. At some point $c$ may lie to the left +of $a_{n}$ but $\left|f(c)\right|>\left|f(a_{n})\right|$, or $c$ +may lie to the right of $b_{n}$ but $\left|f(c)\right|>\left|f(b_{n})\right|$. +The slope has reversed. In that case we halve the step value to $\tfrac{1}{2}(b-a)$ +and try again in the same direction as before from the same point +as before ($a_{n}$ or $b_{n}$ as the case may be). +\item Should we find at some point that $f(a_{n})=f(b_{n})$ then the previous +strategy does not apply. In this case we choose $a_{n+1}$ and \textbf{$b_{n+1}$} +at the quarter and three-quarter marks between $a_{n}$ and $b_{n}$. +Either $f(a_{n+1})$ and $f(b_{n+1})$ will differ and the previous +search strategy can start again or we are on the way to finding an +extremum of $f$. +\end{itemize} +As already noted it is also possible that our function has neither +zeros nor extrema. To prevent the search continuing indefinitely, +\texttt{numerica} uses a cut-off value for the maximum number of steps +pursued \textendash{} by default set at 100. + +\subsubsection{Elusive extrema} + +The strategy `search always towards the $x$-axis' has a consequence: +it means that a local maximum above the $x$-axis will almost certainly +not be found, since `towards the $x$-axis' pulls the search away +from the maximum. Similarly a local minimum below the $x$-axis will +also not be found since `towards the $x$-axis' pulls the search +away from the minimum. + +One way of countering this elusiveness is to add a constant value +(possibly negative) to the function whose zeros and extrema are being +sought. The zeros of the function will change but the abscissae ($x$ +values) of the extrema remain unchanged. If the constant is big enough +it will push a local minimum above the axis where it can be found +or, for a negative constant, push a local maximum below the axis where +it can be found. + +For example $f(x)=x^{3}-x$ has roots at $-1,0,1$, a local maximum +at $-\tfrac{1}{\surd3}$ and a local minimum at $\tfrac{1}{\surd3}$. +To locate the minimum, I have added an unnecessarily large constant +$k$ to $f(x)$. ($k=1$ would have sufficed, but note, $k=0$ fails.) +\begin{centred} +\verb`\solve{$ x^3-x+k $}[k=5,{x}=0.5]` $\Longrightarrow$ \solve{$ x^3-x+k $}[k=5,{x}=0.5]. +\end{centred} +Checking, \verb`\eval{$\tfrac1{\surd 3}$}` $\Longrightarrow$ \eval{$\tfrac1{\surd 3}$}. +There is a discrepancy in the $6$th decimal place which can be eliminated +by using the extra rounding setting; see \ref{subsec:solveExtraRounding}. + +\subsubsection{False extrema} + +A function which `has an infinity' at a particular value can result +in a false extremum being found: +\begin{centred} +\verb`\solve{$ 1/x $}[x=-1/3]` $\Longrightarrow$ \solve{$ 1/x $}[x=-1/3]. +\end{centred} +One needs to look for extrema with some awareness, a general sense +of how the function behaves. `Searching blind' may lead to nonsense +results. In this particular example, changing the rounding value will +show the supposed extremum jumping from one large value to another +and not settling at a particular value. + +\section{Star (\texttt{{*}}) option} + +A starred form of the\textbf{ }\verb`\nmcSolve` command suppresses +all elements of display of the result apart from the numerical value. +In v.1 of \verb`numerica`, when the commands of \verb`numerica-plus` +were invoked with a package option, nesting a \verb`\solve*` command +within another command was the form to use. Now that \verb`numerica-plus` +is a separate package (but labelled v.2), the star is no longer necessary. +\verb`numerica` and associated packages understand a nested command +to be the starred form, whether the star is explicitly present or +not. + +With the `elusive' extremum example above, we can find the actual +value of the minimum by nesting \verb`\solve*` or \verb`\solve` +within the vv-list of an \verb`\eval` command: +\begin{centred} +\verb`\eval{$ x^3-x $}[x={\solve{y^3-y+k}[k=5,y=0.5]}]` $\Longrightarrow$ +\eval{$ x^3-x $}[x={\solve{y^3-y+k}[k=5,y=0.5]}]. +\end{centred} +(Note the braces around the \verb`\solve` and its vv-list to hide +\emph{its} square-brackets from the parsing of the vv-list of the +\verb`\eval` command.) The result is to be compared with \verb`$\eval*{x^3-x}[x=\tfrac1{\surd3}]$` +$\Longrightarrow$ $\eval*{x^3-x}[x=\tfrac1{\surd3}]$. + +\section{Settings option} + +\label{sec:solveSettings-option}The settings option is a comma-separated +list of items of the form \emph{key~=~value}. + +\subsection{Inherited settings} + +The keys discussed in the settings\emph{ }option for \verb`\nmcEvaluate` +are also available for \verb`\nmcSolve`. The very first example in +this chapter used the punctuation option \texttt{p} (\verb`\solve[p]{\[... `) +inherited from the \verb`\eval` command to ensure a comma after the +display-style presentation of the result. We also saw in the quadratic +example illustrating extrema the use of \texttt{vvi} with no value +to suppress display of the vv-list: \verb`\solve[vvi=]{$ ...`. + +Putting \verb`dbg=1` produces a familiar kind of display. Using the +function +\[ +ct-\sqrt{a^{2}+b^{2}-2ab\cos(\beta+\omega t)} +\] +from the rotating disk problem, +\begin{verbatim} + \solve[dbg=1,var=t] + {$ ct-\sqrt{a^{2}+b^{2}-2ab\cos(\beta+\omega t)} + $}[a=10,b=20,\beta=1,{t}=0][4] +\end{verbatim} +$\Longrightarrow$ \solve[dbg=1,var=t] + {$ ct-\sqrt{a^{2}+b^{2}-2ab\cos(\beta+\omega t)} + $}[a=10,b=20,\beta=1,{t}=0][4] + +\subsubsection{Multi-line display of the result} + +\label{subsec:solveMulti-line-display}By default the result is presented +on a single line. Unless the star option is being used, this can be +of the form \emph{function = function value, (vv-list) $\rightarrow$ +result}. It takes only a slightly complicated formula and only a few +variables in the vv-list before this becomes a crowded line, likely +to exceed the line width and extend into the margin. To split the +display over two lines choose a \texttt{vvd} specification in the +vv-list like, for instance, \texttt{vvd=\{,\}\textbackslash\textbackslash (vv)}. +The \texttt{\textbackslash\textbackslash} is a trigger for \texttt{numerica} +to replace whatever environment the \texttt{\textbackslash eval} +command is wrapped around with a \texttt{multline} environment. An +asterisk in the vv-list replaces \texttt{multline} with \texttt{multline{*}} +so that no equation number is used: +\begin{verbatim} + \solve[p=.,vvd={,}\\(vv),*] + {$ ct-\sqrt{a^{2}+b^{2}-2ab\cos(\beta+\omega t)} + $}[a=10,b=20,\beta=1,{t}=0][4] +\end{verbatim} +$\Longrightarrow$ \solve[p=.,vvd={,}\\(vv),*] + {$ ct-\sqrt{a^{2}+b^{2}-2ab\cos(\beta+\omega t)} + $}[a=10,b=20,\beta=1,{t}=0][4] +\begin{verbatim} + +\end{verbatim} +You could introduce a third line if you wished to display the result +on a line of its own by using the spec. \texttt{vvd=\{,\}\textbackslash\textbackslash (vv)\textbackslash\textbackslash}. + +The function evaluates to $-0.0007$. Is this a zero that has been +found or an extremum? To find out, the calculation needs to be carried +out to a higher rounding value which is the reason why \verb`\nmcSolve` +has an extra rounding setting; see \ref{subsec:solveExtraRounding} +below. + +\subsection{\texttt{\textbackslash nmcSolve}-specific settings} + +In addition there are some settings specific to \verb`\nmcSolve`. +These are listed in Table~\ref{tab:solveSettings}. + +\begin{table} +\centering{}\caption{\protect\label{tab:solveSettings}Settings for \texttt{\textbackslash nmcSolve}} +\begin{center} +\begin{tabular}{llll} +\toprule +{\small key} & {\small type} & {\small meaning} & {\small default}\tabularnewline +\midrule +{\small\texttt{var}} & {\small token(s)} & {\small equation variable} & \tabularnewline +{\small\texttt{dvar}} & {\small real $\ne0$ } & {\small initial step size} & {\small\texttt{1}}\tabularnewline +{\small\texttt{+}} & {\small int} & {\small extra rounding} & {\small\texttt{0}}\tabularnewline +{\small\texttt{max}} & {\small int > 0} & {\small max. number of steps before cut off} & {\small\texttt{100}}\tabularnewline +{\small\texttt{reuse}} & {\small int ($\mathtt{0}/\mathtt{1}$)} & {\small form of result saved with }{\small{\small\verb`\reuse`}} & {\small\texttt{0}}\tabularnewline +\bottomrule +\end{tabular} +\par\end{center} +\end{table} + + +\subsubsection{Equation variable} + +By default the equation variable is the \emph{rightmost} variable +in the vv-list. This may not always be convenient. A different equation +variable can be specified by entering +\begin{lyxcode} +var~=~<variable~name> +\end{lyxcode} +in the vv-list. \texttt{<variable name>} will generally be a single +character or token \textendash{} $x$, $t$, $\alpha$, $\omega$ +\textendash{} but is not necessarily of this kind. Multi-token names +are perfectly acceptable (with the default multi-token setting; see +the associated document \verb`numerica.pdf` about this). + +\subsubsection{Initial step size} + +The vv-list contains the equation variable set to a trial value. But +\verb`\solve` needs \emph{two} initial values to begin its search +for a zero or extremum; see \ref{subsec:solveSearch-strategy}. Ideally, +these values will straddle a zero of the function being investigated. +By default, the second trial value is $1$ more than the first: if +the equation variable is set to a trial value $a$ then the second +value defaults to $a+1$. The `$+1$' here can be changed by entering +in the settings option +\begin{lyxcode} +dvar~=~<non-zero~real~number> +\end{lyxcode} +For instance, \texttt{dvar=-1}, or \texttt{dvar=\textbackslash pi} +are two valid specifications of initial step size. The notation is +prompted by the use of expressions like $x$ and $x+dx$ for two nearby +points in calculus. + +An example where the default step value is too big and a smaller one +needs to be specified is provided by Planck's radiation function (\emph{HMF +}Table 27.2), +\[ +f(x)=\frac{1}{x^{5}(e^{1/x}-1)}. +\] +From the (somewhat coarse-grained) table in \emph{HMF }it is clear +that there is a maximum of about 21.2 when $x$ is a little more than +$0.2$. This is a maximum above the $x$-axis and hence `elusive'. +To find it, substract $100$ (say) from the formula and again use +the ability to nest commands to display the result. In the example, +I find in the vv-list of the \verb`\eval` command the value of $x$ +which maximizes the Planck radiation function, then calculate the +maximum in the main argument of the \verb`\eval` command. Note the +\verb`dvar=0.1` in the settings option of the \verb`\solve*` command: +\begin{verbatim} + \eval[p=.]{\[ \frac1{x^5(e^{1/x}-1)} \]} + [ x={ \solve*[dvar=0.1] + { \frac1{y^5(e^{1/y}-1)}-100 }[y=0.1] + } ] +\end{verbatim} +$\Longrightarrow$ \eval[p=.]{\[ \frac1{x^5(e^{1/x}-1)} \]} + [ x={ \solve*[dvar=0.1] + { \frac1{y^5(e^{1/y}-1)}-100 }[y=0.1] + } ] + +\noindent The maximum is indeed a little over $21.2$ and the $x$ +value a little more than $0.2$. + +The default \verb`dvar=1` is too big for this problem. From the table +in \emph{HMF},\emph{ }$f(0.1)=4.540$ and $f(1.1)=0.419$. Thus for +$f(x)-100$ the `towards the $x$-axis' search strategy would lead +to negative values of $x$ with the default \verb`dvar` setting. + +\subsubsection{Extra rounding} + +\label{subsec:solveExtraRounding}\verb`\solve` determines that a +zero or an extremum has been reached when the difference between two +successive bisection values vanishes at the specified rounding value +(the value in the final trailing optional argument of the \verb`\solve` +command; $6$ by default). If our function is $f(x)$ then $\abs{x_{n+1}-x_{n}}=0$ +to the specified rounding value and $f(x_{n})$, $f(x_{n+1})$ have +opposite signs or at least one vanishes. Then (assuming $x_{n+1}>x_{n}$ +and continuity) there must be a critical value $x_{c}\in[x_{n},x_{n+1}]$ +such that $f(x_{c})=0$ exactly. But in general the critical value +$x_{c}$ will not coincide with $x_{n}$ or $x_{n+1}$. If $f(x)$ +crosses the $x$-axis at a steep angle it may well be that although +$f(x_{c})$ vanishes to all $16$ figures, $f(x_{n})$ and $f(x_{n+1})$ +do not, not even at the (generally smaller) specified rounding value. +For instance, suppose $f(x)=1000x-3000$ and that our trial value +is $x=e$: +\begin{centred} +\verb`\solve[vvi=]{$ 1000x-3000 $}[x=e][4*]` $\Longrightarrow$ \solve[vvi=]{$ 1000x-3000 $}[x=e][4*]. +\end{centred} +Although the difference between successive $x$ values vanishes to +$4$ places of decimals, $f(x)$ does not, not even to $2$ places. +If we want the function to vanish at the specified rounding value +\textendash{} $4$ in the example \textendash{} then we will need +to locate the zero more precisely than that. + +This is the purpose of the extra rounding key in the settings option. +Enter +\begin{lyxcode} ++~=~<integer> +\end{lyxcode} +in the settings option of the \verb`\solve` command to add \texttt{<integer>} +to the rounding value determining the conclusion of the calculation. +By default, \texttt{+=}0. + +With this option available it is easy to check that \verb`+=3` suffices +in the example to ensure that both $x$ and $f(x)$ vanish to $4$ +places of decimals, +\begin{centred} +\verb`\solve[+=3]{$ 1000x-3000 $}[x=e][4*]` $\Longrightarrow$ \solve[+=3]{$ 1000x-3000 $}[x=e][4*], +\end{centred} +and that \texttt{+=2} does not, i.e., we need to locate the zero to +$4+3=7$ figures to ensure the function vanishes to $4$ figures. + +There is no need for the \texttt{<integer>} to be positive. In fact +negative values can illuminate what is going on. In the first of the +following, the display is to $10$ places but (\verb`+=-4`) the calculation +is only to $10-4=6$ places. In the second, the display is again to +$10$ places, but (\verb`+=-3`) the calculation is to $10-3=7$ places. +\begin{centred} +\verb`\solve[+=-4]{$ 1000x-3000 $}[x=e][10*]` $\Longrightarrow$ +\solve[+=-4]{$ 1000x-3000 $}[x=e][10*], + +\verb`\solve[+=-3]{$ 1000x-3000 $}[x=e][10*]` $\Longrightarrow$ +\solve[+=-3]{$ 1000x-3000 $}[x=e][10*]. +\end{centred} +Only in the second does $f(x)=1000x-3000$ vanish when rounded to +$4$ figures. + +Returning to an earlier example (\ref{subsec:solveMulti-line-display}) +in which it was not entirely clear whether a zero or an extremum had +been found, we can now resolve the confusion. Use the extra rounding +setting (and pad with zeros to emphasize the $4$-figure display by +adding an asterisk in the trailing optional argument): +\begin{verbatim} + \solve[+=2,vvd={,}\\(vv),*] + {$ ct-\sqrt{a^{2}+b^{2}-2ab\cos(\beta+\omega t)} + $}[a=10,b=20,\beta=1,{t}=0][4*] +\end{verbatim} +$\Longrightarrow$ \solve[+=2,vvd={,}\\(vv),*] + {$ct-\sqrt{a^{2}+b^{2}-2ab\cos(\beta+\omega t)} + $}[a=10,b=20,\beta=1,{t}=0][4*] + +\subsubsection{Maximum number of steps before cut-off} + +Once two function values have been found of different sign, bisection +is guaranteed to arrive at a result. The problem is the \emph{search} +for two such values. This may not terminate \textendash{} think of +a function like $e^{x}$ which lacks both zeros and extrema. To prevent +an infinite loop, \verb`\solve` cuts off the search after $100$ +steps. This cut-off value can be changed for a calculation by entering +\begin{lyxcode} +max~=~<positive~integer> +\end{lyxcode} +in the settings option. + +To illustrate, we know that $1/x$ has neither zero nor extremum, +but we do not get an infinite loop; we get an error message if we +attempt to `solve' $1/x$: +\begin{centred} +\verb`\solve{ 1/x }[x=1]` $\Longrightarrow$ \solve{ 1/x }[x=1] +\end{centred} + +\subsubsection{Form of result saved by \texttt{\textbackslash nmcReuse}} + +As wth \verb`\eval` and \verb`\iter` it is possible to specify to +some extent what is saved to file when using \verb`\reuse` after +a \verb`\solve` command. The form of entry in the settings option +is +\begin{lyxcode} +reuse~=~<integer> +\end{lyxcode} +If the star option is used with the \verb`\solve` command the numerical +result is the only thing saved, but in the absence of the star option, +\begin{itemize} +\item \texttt{reuse=0} saves\emph{ the form that is displayed}. For example, +if the display is of the form \emph{function = function value, (vv-list) +$\rightarrow$ result} then that is what is saved; this is the default +behaviour; +\item \texttt{reuse=1} (or any non-zero integer) saves only the numerical +result. +\end{itemize} + +\subsection{Changing default values} + +\begin{wraptable}[12]{o}{0.5\columnwidth}% +\centering{}\caption{\protect\label{tab:solveSettingsDefaults}Defaults for \texttt{\textbackslash nmcSolve}} +\begin{center} +\begin{tabular}{ll} +\toprule +{\small key} & {\small default}\tabularnewline +\midrule +{\small solve-first-step} & 1\tabularnewline +{\small solve-extra-rounding} & {\small\texttt{0}}\tabularnewline +{\small solve-max-steps} & {\small\texttt{100}}\tabularnewline +{\small solve-reuse} & {\small\texttt{0}}\tabularnewline +\bottomrule +\end{tabular} +\par\end{center}\end{wraptable}% +If you wish to change the default values of the various settings for +\verb`\nmcSolve` this can be done by entering new values in a configuration +file \texttt{numerica-plus.cfg} as described in the chapter on settings +in the associated document \texttt{numeric.pdf}. The relevant keys +are listed in Table~\ref{tab:solveSettingsDefaults}, corresponding +to the \verb`dvar`, \verb`+`, \verb`max` and \verb`reuse` settings +of the \verb`\solve` command. (Obviously it makes no sense to have +a default setting for the solution variable. That will change from +case to case.) + +\chapter{Recurrence relations: \texttt{\textbackslash nmcRecur}} + +One of the simplest recurrence relations is that determining the Fibonacci +numbers, $f_{n+2}=f_{n+1}+f_{n}$, with initial values $f_{0}=f_{1}=1$. +The command \verb`\nmcRecur`, short-name form \verb`\recur`, allows +calculation of the terms of this sequence: +\begin{verbatim} + $ \nmcRecur[do=8,see1=8,...] + { f_{n+2}=f_{n+1}+f_{n} } + [f_{1}=1,f_{0}=1] $ +\end{verbatim} +$\Longrightarrow$ $\nmcRecur[do=8,see1=8,...] + { f_{n+2}=f_{n+1}+f_{n} } + [f_{1}=1,f_{0}=1]$ + +The recurrence relation is entered in the main argument (between braces), +the initial values in the vv-list trailing the main argument, and +the display specification is placed in the settings option: \texttt{do=8} +terms to be calculated, all $8$ to be viewed (\texttt{see1=8}), and +the display to be concluded by an ellipsis to indicate that the sequence +continues (those are three dots/periods/full stops in the settings +option, not an ellipsis glyph). + +A more complicated recurrence relation determines the Legendre polynomials: +\[ +(n+2)P_{n+2}(x)-(2n+3)xP_{n+1}(x)+(n+1)P_{n}(x)=0. +\] +For the purposes of \verb`\recur` we need $P_{n+2}$ expressed in +terms of the lower order terms: +\[ +P_{n+2}(x)=\frac{1}{n+2}\left((2n+3)xP_{n+1}(x)-(n+1)P_{n}(x)\right). +\] + It is this standard form \textendash{} the term to be calculated +on the left, equated to an expression involving a fixed number of +lower-order terms on the right \textendash{} that \verb`numerica-plus` +works with. For $P_{0}(x)=1,~P_{1}(x)=x$ and $x=0.5$, the terms +are calculated thus: +\begin{verbatim} + \recur[p,do=11,see1=4,see2=2,vvd={,}\\(vv)\\,*] + {\[ P_{n+2}(x)=\frac{1}{n+2} + \Bigl((2n+3)xP_{n+1}(x)-(n+1)P_{n}(x)\Bigr) + \]}[P_{1}(x)=x,P_{0}(x)=1,x=0.5] +\end{verbatim} +$\Longrightarrow$ \recur[p,do=11,see1=4,see2=2,vvd={,}\\(vv)\\,*] + {\[ P_{n+2}(x)=\frac{1}{n+2} + \Bigl((2n+3)xP_{n+1}(x)-(n+1)P_{n}(x)\Bigr) + \]}[P_{1}(x)=x,P_{0}(x)=1,x=0.5] + +\noindent where $P_{9}(0.5)$ and $P_{10}(0.5)$ are the last two +displayed values (and to $6$-figures are the values listed in \emph{HMF +}Table 8.1). + +These examples also illustrate a common behaviour of the commands +in \verb`numerica` and associated packages: when wrapped around math +delimiters: the display is of the \emph{expression=result} form, and +when placed between math delimiters the display is simply of the \emph{result}. +When used without math delimiters, \verb`numerica-plus` treats the +command as if it had been placed between \texttt{\textbackslash{[} +\textbackslash{]}}. + +\section{Notational niceties} + +More than the other commands in {\ttfamily\verb`numerica`} +and associated packages, \verb`\nmcRecur` depends on getting the +notation into a standard form. +\begin{itemize} +\item The terms of the recurrence must be \emph{subscripted}: $f_{n}$, +$P_{n}(x)$ are examples. +\item The recurrence relation is placed in the main (mandatory) argument +of \verb`\nmcRecur` in the form: \emph{high-order term=function of +lower-order terms}. +\item The initial-value terms in the vv-list must occur left-to-right in +the order \emph{high }to \emph{low} order. +\item The recurrence variable changes by $1$ between successive terms. +\end{itemize} +The example for Legendre polynomials in particular shows what is required. +The Fibonacci example is simpler, since the recurrence variable does +not occur independently in the recurrence relation as it does with +the Legendre polynomials. In both cases though the recurrence variable +is absent from the vv-list. + +\subsection{Vv-list and recurrence variable} + +The recurrence variable is required in the vv-list only when an implicit +mode calculation is undertaken. Suppose we write $A$ and $B$ for +the coefficients $2n+3$ and $n+1$ respectively in the Legendre recurrence. +$A$ and $B$ will now need entries in the vv-list which means the +recurrence variable will need a value assigned to it there too, and +we will need to add \verb`vv@=1` (or \verb`vvmode=1`) to the settings +option. +\begin{verbatim} + \recur[p,vvmode=1,do=11,see1=4,see2=2,vvd={,}\\(vv)\\,*] + {\[ P_{n+2}(x)=\frac{1}{n+2} + \Bigl(AxP_{n+1}(x)-BP_{n}(x)\Bigr) + \]}[P_{1}(x)=x,P_{0}(x)=1,x=0.5,A=2n+3,B=n+1,n=0] +\end{verbatim} +$\Longrightarrow$ \recur[p,vvmode=1,do=11,see1=4,see2=2,vvd={,}\\(vv)\\,*] + {\[ P_{n+2}(x)=\frac{1}{n+2} + \Bigl(AxP_{n+1}(x)-BP_{n}(x)\Bigr) + \]}[P_{1}(x)=x,P_{0}(x)=1,x=0.5,A=2n+3,B=n+1,n=0] + +Since the vv-list is evaluated from the right, the left-to-right high-to-low +ordering of the initial-value terms means the value of the lowest +order term is read first. Although \verb`numerica-plus` depends on +this order of occurrence of the terms, they do not need to be \emph{consecutive} +as in the examples so far (although it is natural to enter them in +this way). \verb`numerica-plus` reads the value of the subscript +of only the right-most term (the lowest order term), increments it +by $1$ when reading the next recurrence term to the left, and so +on. The reading of the subscript of the lowest order term in the vv-list +provides the initial value of the recurrence variable. + +In the following example I have placed other items between $P_{1}(x)$ +and $P_{0}(x)$ in the vv-list (but maintained their left-to-right +order) and given the recurrence variable $n$ a ridiculous initial +value $\pi^{2}/12$. (Because of the order in which things get done +`behind the scenes', \emph{some} value is necessary so that the +$n$ in `$B=n+1$' does not generate an `unknown token' message.) +The result is unchanged. +\begin{verbatim} + \recur[p,vvmode=1,do=11,see1=4,see2=2,vvd={,}\\(vv)\\,*] + {\[ P_{n+2}(x)=\frac{1}{n+2} + \Bigl(AxP_{n+1}(x)-BP_{n}(x)\Bigr) + \]}[A=2n+3,P_{1}(x)=x,B=n+1,n=\pi^2/12,P_{0}(x)=1,x=0.5] +\end{verbatim} +$\Longrightarrow$ \recur[p,vvmode=1,do=11,see1=4,see2=2,vvd={,}\\(vv)\\,*] + {\[ P_{n+2}(x)=\frac{1}{n+2} + \Bigl(AxP_{n+1}(x)-BP_{n}(x)\Bigr) + \]}[A=2n+3,P_{1}(x)=x,B=n+1,n=\pi^2/12,P_{0}(x)=1,x=0.5] + +\subsection{Form of the recurrence relation} + +As noted earler, the form of the recurrence must be entered in the +main argument in the form: \emph{highest order term = function of +consecutive lower order terms}. The number of lower\emph{ }order terms +is the order of the recurrence. The Fibonacci and Legendre polynomial +recurrences are both second order and presented in the form: \emph{$n+2$-th +term = function of $n+1$-th term and $n$-th term}. We could equally +have done +\begin{verbatim} + \nmcRecur[p,do=8,see1=8,...] + {$ f_{n}=f_{n-1}+f_{n-2} $} + [f_{1}=1,f_{0}=1] +\end{verbatim} +$\Longrightarrow$ \nmcRecur[p,do=8,see1=8,...] + {$ f_{n}=f_{n-1}+f_{n-2} $} + [f_{1}=1,f_{0}=1] where now the recurrence is of the form $n$\emph{-th term = function +of $n-1$-th term and $n-2$-th term}, or (adjusting the coefficients +as well as the recurrence terms), +\begin{verbatim} + \recur[p=.,do=10,see1=4,see2=2,vvd={,}\\(vv)\\,*]{\[ + P_{n+1}(x)=\frac{1}{n+1} + \Bigl((2n+1)xP_{n}(x)-nP_{n-1}(x)\Bigr) + \]}[P_{2}(x)=-0.125,P_{1}(x)=x,x=0.5] +\end{verbatim} +$\Longrightarrow$ \recur[p=.,do=10,see1=4,see2=2,vvd={,}\\(vv)\\,*]{\[ + P_{n+1}(x)=\frac{1}{n+1} + \Bigl((2n+1)xP_{n}(x)-nP_{n-1}(x)\Bigr) + \]}[P_{2}(x)=-0.125,P_{1}(x)=x,x=0.5] + +\noindent The recurrence here is of the form $n+1$\emph{-th term += function of $n$-th term and $n-1$-th term}. This last example +has one further `wrinkle'. I've made $P_{1}(x)$ the lowest order +term and decreased the number of terms to calculate by $1$ accordingly. + +\subsection{First order recurrences (iteration)} + +The recurrence relations for both the Fibonacci sequence and Legendre +polynomials are second order. There is no reason why the recurrence +should not be of third or higher order or, indeed, lower. A first +order recurrence provides an alternative means of iterating functions. +\verb`\recur` therefore provides a means to display the results of +an iteration in a different form from \verb`\iter`. + +Iterating $1+a/x$ in this way, $16$ terms gives the sequence +\begin{verbatim} + \recur[do=16,see1=0,see2=3,...]{$ + x_{n+1}=1+a/x_{n} + $}[x_{0}=1,a=1] +\end{verbatim} +$\Longrightarrow$ \recur[do=16,see1=0,see2=3,...]{$ + x_{n+1}=1+a/x_{n} + $}[x_{0}=1,a=1] + +\noindent to be compared with the example near the start of Chapter~\ref{chap:Iterating-functions}. +(\emph{That} effected $15$ iterations; \emph{this} uses $16$ terms +because of the extra $x_{0}=1$ term.) + +\section{Star (\texttt{{*}}) option} + +When the star option is used with the \verb`\nmcRecur` command, only +a single term, the \emph{last}, is presented as the result. Repeating +the last calculation, but with the star option produces +\begin{verbatim} + \recur*[p=.,do=10]{\[ + P_{n+1}(x)=\frac{1}{n+1} + \Bigl((2n+1)xP_{n}(x)-nP_{n-1}(x)\Bigr) + \]}[P_{2}(x)=-0.125,P_{1}(x)=x,x=0.5] +\end{verbatim} +$\Longrightarrow$ \recur*[p=.,do=10]{\[ + P_{n+1}(x)=\frac{1}{n+1} + \Bigl((2n+1)xP_{n}(x)-nP_{n-1}(x)\Bigr) + \]}[P_{2}(x)=-0.125,P_{1}(x)=x,x=0.5] + +Although punctuation (a full stop) was specified in the settings, +it has been ignored in the display of the result. Other settings would +also have been ignored with the exception of the \verb`do` key which +is required to know exactly which term to calculate. The star option +produces a purely numerical answer without any trimmings. + +\noindent\begin{minipage}[t]{1\columnwidth}% +\begin{shaded}% +This seems something of a waste of the star option since it gives +much the same result as choosing \texttt{do=10,see1=0,see2=1}. Not +\emph{exactly} the same, since math delimiters are involved now, but +sufficiently similar to make me wonder if I should change the starred +form to apply only to those recurrences which approach a limit. The +starred form would then produce the limiting value as its result (like +\verb`\iter*`). This is a possible change for future versions of +\verb`numerica-plus` and should be borne in mind if using \verb`\recur*`.\end{shaded}% +\end{minipage} + +\section{Settings} + +The settings option is a comma-separated list of items of the form +\emph{key~=~value}. + +\subsection{Inherited settings} + +Because recurrence terms are necessarily multi-token, the multi-token +key is hard-coded in \verb`\recur` to \texttt{xx=1}. + +\subsubsection{Multi-line formatting of result} + +When the \verb`\recur` command wraps around math delimiters, the +\texttt{vvd} setting is available to split display of the result over +two or more lines. For example, \texttt{vvd=\{,\}\textbackslash\textbackslash (vv)}pushes +the vv-list and sequence of calculated values to a second line; or, +\texttt{vvd=\{,\}\textbackslash qquad(vv)\textbackslash\textbackslash} +pushes only the sequence of calculated values to a second line; or +\texttt{vvd=\{,\}\textbackslash\textbackslash (vv)\textbackslash\textbackslash} +pushes the vv-list, centred, to a second line and the sequence of +values, right aligned, to a third line. The \texttt{{*}} setting is +available to suppress equation numbering (by substituting \verb`multline*` +for \verb`multline`). +\begin{verbatim} + \nmcRecur[do=8,see1=8,...,vvd={,}\qquad(vv)\\,*] + {$ f_{n+2}=f_{n+1}+f_{n} $} + [f_{1}=1,f_{0}=1] +\end{verbatim} +$\Longrightarrow$ \nmcRecur[do=8,see1=8,...,vvd={,}\qquad(vv)\\,*] + {$ f_{n+2}=f_{n+1}+f_{n} $} + [f_{1}=1,f_{0}=1] + +\subsection{\texttt{\textbackslash nmcRecur}-specific settings} + +\label{subsec:recurSpecific-settings}In addition to the inherited +settings there are some specific to \verb`\nmcRecur`. These are listed +in Table~\ref{tab:recurSettings} below. + +\subsubsection{Number of terms to calculate} + +By entering +\begin{lyxcode} +do~=~<integer> +\end{lyxcode} +in the settings option you can specify how many terms of a recurrence +to calculate. The default is set to $7$ (largely to show a sufficient +number of terms of the Fibonacci series to begin to be interesting). +Note that \texttt{<integer>} will generally \emph{not} correspond +to the subscript on the last term calculated since that also depends +on the value of the subscript of the lowest order term in the vv-list. + +\subsubsection{Number of terms to display} + +\begin{table}[b] +\centering{}\caption{\protect\label{tab:recurSettings}Settings for \texttt{\textbackslash nmcRecur}} +\begin{center} +\begin{tabular}{llll} +\toprule +{\small key} & {\small type} & {\small meaning} & {\small default}\tabularnewline +\midrule +{\small\texttt{do}} & {\small int$\ge0$} & {\small number of terms to calculate} & {\small\texttt{7}}\tabularnewline +{\small\texttt{see1}} & {\small int$\ge0$} & {\small number of initial terms to display} & {\small\texttt{3}}\tabularnewline +{\small\texttt{see2}} & {\small int$\ge0$} & {\small number of final terms to display} & {\small\texttt{2}}\tabularnewline +{\small\texttt{...}} & {\small chars} & {\small follow display of values with an ellipsis} & \tabularnewline +{\small\texttt{reuse}} & {\small int ($\mathtt{0}/\mathtt{1}/\mathtt{2}$)} & {\small form of result saved with }{\small{\small\verb`\reuse`}} & {\small\texttt{0}}\tabularnewline +\bottomrule +\end{tabular} +\par\end{center} +\end{table} +By entering +\begin{lyxcode} +see1~=~<integer1>,~see2=<integer2> +\end{lyxcode} +in the settings option, you can specify how many initial terms of +the recurrence and how many of the final terms calculated you want +to view. If the sum of these settings is less than the \texttt{do} +setting, then the terms are displayed with an intervening ellipsis. +If the sum is greater than the \texttt{do} setting, then the values +are adjusted so that their sum equals the \texttt{do} setting and +all terms are displayed. + +The adjustment is preferentially to \texttt{see1}. Suppose \texttt{do=}7, +\texttt{see1=5}, \texttt{see2=4}. Then \texttt{see2} is left unchanged +but \texttt{see1} is reduced to \texttt{7-4=3}. If, say, \texttt{do=}7, +\texttt{see1=5}, \texttt{see2=8}, then \texttt{see2} is reduced to +\texttt{7} and \texttt{see1} to \texttt{-1} (rather than zero, for +technical reasons). The reason for preserving \texttt{see2} over \texttt{see1} +is for the functioning of the \texttt{reuse} setting (see below). + +The default value for \texttt{see1} is $3$; the default value for +\texttt{see2} is $2$. + +\subsubsection{Ellipsis} + +Including three dots in the settings option +\begin{lyxcode} +... +\end{lyxcode} +ensures that a (proper) ellipsis is inserted after the final term +is displayed. An example is provided by the display of the Fibonacci +sequence at the start of this chapter. By default this option is turned +off. + +\subsubsection{Form of result saved by \texttt{\textbackslash nmcReuse}} + +By entering +\begin{lyxcode} +reuse~=~<integer> +\end{lyxcode} +it is possible to specify the form of result that is saved when using +\verb`\nmcReuse`. (This setting has no effect when the star option +is used with \verb`\nmcRecur`. In that case only the numerical result +of the final term calculated is saved.) There are three different +outputs possible: +\begin{itemize} +\item \texttt{int=0} (or any integer $\mathtt{\ne1,2}$) saves the full +display (the default); +\item \texttt{int=1} saves a comma-separated list of braced pairs of the +form: \texttt{\{$k$, value-of-term-$k$\}} for the last \texttt{see2} +terms calculated; +\item \texttt{int=2} saves a comma-separated list of the values of the last +\texttt{see2} terms calculated. +\end{itemize} +As an example, using \texttt{reuse=1}, +\begin{verbatim} + \recur[reuse=1,p=.,vvmode=1,do=11,see1=4,see2=2, + vvd={,}\\(vv)\\,*] + {\[ P_{n+2}(x)=\frac{1}{n+2} + \Bigl(kxP_{n+1}(x)-(n+1)P_{n}(x)\Bigr) + \]}[k=2n+3,n=1,P_{1}(x)=x,P_{0}(x)=1,x=0.5] + \reuse[legendre] +\end{verbatim} +$\Longrightarrow$ \recur[reuse=1,p=.,vvmode=1,do=11,see1=4,see2=2, + vvd={,}\\(vv)\\,*] + {\[ P_{n+2}(x)=\frac{1}{n+2} + \Bigl(kxP_{n+1}(x)-(n+1)P_{n}(x)\Bigr) + \]}[k=2n+3,n=0,P_{1}(x)=x,P_{0}(x)=1,x=0.5] + \reuse[renew]{ legendre} + +\noindent Now check to see what has been saved: +\begin{centred} +\verb`$\legendre$` $\Longrightarrow$ $ \legendre $. +\end{centred} +As you can see, the final two (because of \texttt{see2=2}) of the +$11$ Legendre polynomials calculated ($P_{0}(x)$ is the first) have +been saved, each value preceded by its index value. If the setting +had been \texttt{reuse=2}, only the two values without the index values +would have been saved. To see that they are saved as braced pairs, +use \TeX 's \verb`\meaning` command: +\begin{verbatim} + \meaning \legendre +\end{verbatim} +$\Longrightarrow$ \meaning \legendre + +\subsection{Changing default values} + +\begin{wraptable}[12]{o}{0.5\columnwidth}% +\centering{}\caption{\protect\label{tab:recurSettingsDefaults}Defaults for \texttt{\textbackslash nmcRecur}} +\begin{center} +\begin{tabular}{ll} +\toprule +{\small key} & {\small default}\tabularnewline +\midrule +{\small recur-do} & {\small\texttt{7}}\tabularnewline +{\small recur-see-first} & {\small\texttt{3}}\tabularnewline +{\small recur-see-last} & {\small\texttt{2}}\tabularnewline +{\small recur-reuse} & {\small\texttt{0}}\tabularnewline +\bottomrule +\end{tabular} +\par\end{center}\end{wraptable}% +If you wish to change the default values of the various settings for +\verb`\nmcRecur` this can be done by entering new values in a configuration +file \texttt{numerica-plus.cfg} as described in the chapter on settings +in the associated document \texttt{numerica.pdf}. The relevant keys +are listed in Table~\ref{tab:recurSettingsDefaults}, corresponding +to the \verb`do`, \verb`see1`, \verb`see2` and \verb`reuse` settings +of the \verb`\recur` command. + +\subsection{Orthogonal polynomials} + +I've used Legendre polynomials in examples above, but orthogonal polynomials +generally lend themselves to the \verb`\recur` treatment. Quoting +from \emph{HMF} 22.7, orthogonal polynomials $f_{n}$ satisfy recurrence +relations of the form +\[ +a_{1n}f_{n+1}(x)=(a_{2n}+a_{3n}x)f_{n}(x)-a_{4n}f_{n-1}(x), +\] +or in the standard form required by \verb`\recur`, +\[ +f_{n+1}(x)=\frac{a_{2n}+a_{3n}x}{a_{1n}}f_{n}(x)-\frac{a_{4n}}{a_{1n}}f_{n-1}(x). +\] +\emph{HMF} 22.7 provides a listing of the coefficients $a_{in}$ for +the polynomials of Jacobi, Chebyshev, Legendre, Laguerre, Hermite +and others, and tables for these polynomials. + +For example, Laguerre polynomials satisfy the recurrence +\[ +L_{n+1}(x)=\frac{2n+1-x}{n+1}L_{n}(x)-\frac{n}{n+1}L_{n-1}(x). +\] +with initial values $L_{0}(x)=1$ and $L_{1}(x)=1-x$. So let's calculate +the first $13$ Laguerre polynomials for, say, $x=0.5$: +\begin{verbatim} + \recur[do=13,see1=4,see2=2,vvd={,}\\(vv)\\,*]{\[ + L_{n+1}(x)=\frac{2n+1-x}{n+1}L_{n}(x)- + \frac{n}{n+1}L_{n-1}(x) + \]}[L_{1}(x)=1-x,L_{0}(x)=1,x=0.5] +\end{verbatim} +$\Longrightarrow$ \recur[do=13,see1=4,see2=2,vvd={,}\\(vv)\\,*]{\[ + L_{n+1}(x)=\frac{2n+1-x}{n+1}L_{n}(x)- + \frac{n}{n+1}L_{n-1}(x) + \]}[L_{1}(x)=1-x,L_{0}(x)=1,x=0.5] + +\noindent and for $x=5$: +\begin{verbatim} + \recur[p=.,do=13,see1=4,see2=2,vvd={,}\\(vv)\\,*]{\[ + L_{n+1}(x)=\frac{2n+1-x}{n+1}L_{n}(x)- + \frac{n}{n+1}L_{n-1}(x) + \]}[L_{1}(x)=1-x,L_{0}(x)=1,x=5] +\end{verbatim} +$\Longrightarrow$ \recur[p=.,do=13,see1=4,see2=2,vvd={,}\\(vv)\\,*]{\[ + L_{n+1}(x)=\frac{2n+1-x}{n+1}L_{n}(x)- + \frac{n}{n+1}L_{n-1}(x) + \]}[L_{1}(x)=1-x,L_{0}(x)=1,x=5] + +\noindent The results (reassuringly) coincide with those provided +in \emph{HMF }Table 22.11. + +\subsection{Nesting} + +It is possible to use the \verb`\recur` command (with star in v.1 +of \verb`numerica`, with or without star in v.2 of \verb`numerica-plus`) +within an \verb`\eval`, \verb`\iter`, or \verb`\solve` command, +and indeed in \verb`\recur` itself, but with this caveat: if \verb`\recur` +is nested within another command, the initial terms of the recurrence +\textendash{} e.g., $f_{1}=1,f_{0}=1$, for the Fibonacci series, +or $L_{1}(x)=1-x,L_{0}(x)=1$ for the Laguerre polynomials \textendash{} +\emph{must be located in the vv-list of that inner }\verb`\recur`\emph{ +command}. Other shared variables can often be shifted to the vv-list +of the outer command, but not these initial terms. + +In the following example I multiply together (rather futilely) the +third and fourth members of the sequence of Laguerre polynomials for +$x=5$ (the answer expected is \verb`$ \eval{3.5\times2.666667} $` +$\Longrightarrow$ $ \eval{3.5\times2.666667} $). Note that although +it is tempting to shift the shared vv-lists of the inner \verb`\recur*` +commands to the vv-list of the outer \verb`\eval` command, in fact +only the \verb`x=5` entry has been transferred: +\begin{verbatim} + \eval[p=.]{$ + \recur*[do=3] + { L_{n+1}(x)=\frac{2n+1-x}{n+1}L_{n}(x)- + \frac{n}{n+1}L_{n-1}(x)} + [L_{1}(x)=1-x,L_{0}(x)=1] + \times + \recur*[do=4] + { L_{n+1}(x)=\frac{2n+1-x}{n+1}L_{n}(x)- + \frac{n}{n+1}L_{n-1}(x)} + [L_{1}(x)=1-x,L_{0}(x)=1] + $}[x=5] +\end{verbatim} +$\Longrightarrow$ \eval[p=.]{$ + \recur*[do=3] + { L_{n+1}(x)=\frac{2n+1-x}{n+1}L_{n}(x)- + \frac{n}{n+1}L_{n-1}(x)} + [L_{1}(x)=1-x,L_{0}(x)=1,x=5] + \times + \recur*[do=4] + { L_{n+1}(x)=\frac{2n+1-x}{n+1}L_{n}(x)- + \frac{n}{n+1}L_{n-1}(x)} + [L_{1}(x)=1-x,L_{0}(x)=1,x=5] + $} + +\noindent\begin{minipage}[t]{1\columnwidth}% +\begin{shaded}% +The terms of a recurrence relation are multi-token variables but \texttt{numerica} +requires single tokens for its calculations. The problem for \verb`\recur` +is that the terms in the recurrence relation in the main (mandatory) +argument differ from the terms in the vv-list: for instance $f_{n}$ +in the main argument, $f_{0}$ in the vv-list. If left like that, +when \texttt{numerica} does its conversion from multi-token to single +token variables, $f_{n}$ would not be found since it differs from +$f_{0}$. Hence a crucial first step for \verb`\recur` is to reconcile +the different forms, which it does by converting the forms in the +vv-list to the forms in the recurrence in the main argument. To be +available for this form change, they must reside in the \emph{inner} +vv-list. In the outer vv-list they would be inaccessible to the inner +command. + +{*}{*}{*} + +This suggests an alternative way of proceeding: write the inital values +of the recurrence terms in the \emph{same} form in which they occur +in the recurrence relation, together with an initial value for the +recurrence variable, e.g., $f_{n+1}=1,f_{n}=1,n=0$. This is not how +mathematicians write the initial values in recurrence relations, which +is why I did not pursue it, but it neatly sidesteps what is otherwise +an initial awkwardness. \end{shaded}% +\end{minipage} + +\chapter{Reference summary} + +\section{Commands defined in \texttt{numerica-plus}} +\begin{enumerate} +\item \texttt{\textbackslash nmcIterate, \textbackslash iter} +\item \texttt{\textbackslash nmcSolve, \textbackslash solve} +\item \textbackslash\texttt{nmcRecur, \textbackslash recur} +\end{enumerate} + +\section{Settings for the three commands} + +\subsection{Settings for \texttt{\textbackslash nmcIterate}} + +Settings option of \verb`\nmcIterate`: + +\begin{center} +\begin{tabular}{llll} +\toprule +{\small key} & {\small type} & {\small meaning} & {\small default}\tabularnewline +\midrule +{\small\texttt{var}} & {\small token(s)} & {\small iteration variable} & \tabularnewline +{\small\texttt{+}} & {\small int} & {\small fixed point extra rounding} & {\small\texttt{0}}\tabularnewline +{\small\texttt{max}} & {\small int > 0} & {\small max. iteration count (fixed points)} & {\small\texttt{100}}\tabularnewline +{\small\texttt{do}} & {\small int > 0} & {\small number of iterations to perform} & {\small\texttt{5}}\tabularnewline +{\small\texttt{see}} & {\small int > 0} & {\small number of final iterations to view} & {\small\texttt{4}}\tabularnewline +{\small\texttt{reuse}} & {\small int ($\mathtt{0}/\mathtt{1}/\mathtt{2}$)} & {\small form of result saved with }{\small{\small\verb`\reuse`}} & {\small\texttt{0}}\tabularnewline +\bottomrule +\end{tabular} +\par\end{center} + +Configuration settings for \verb`\nmcIterate`: + +\begin{center} +\begin{tabular}{ll} +\toprule +{\small key} & {\small default}\tabularnewline +\midrule +{\small iter-extra-rounding} & {\small\texttt{0}}\tabularnewline +{\small iter-max-iterations} & {\small\texttt{100}}\tabularnewline +{\small iter-do} & {\small\texttt{5}}\tabularnewline +{\small iter-see-last} & {\small\texttt{4}}\tabularnewline +{\small iter-reuse} & {\small\texttt{0}}\tabularnewline +\bottomrule +\end{tabular} +\par\end{center} + +\subsection{Settings for \texttt{\textbackslash nmcSolve}} + +Settings option of \verb`\nmcSolve`: + +\begin{center} +\begin{tabular}{llll} +\toprule +{\small key} & {\small type} & {\small meaning} & {\small default}\tabularnewline +\midrule +{\small\texttt{var}} & {\small token(s)} & {\small equation variable} & \tabularnewline +{\small\texttt{dvar}} & {\small real $\ne0$} & {\small initial step size} & {\small\texttt{1}}\tabularnewline +{\small\texttt{+}} & {\small int} & {\small extra rounding} & {\small\texttt{0}}\tabularnewline +{\small\texttt{max}} & {\small int > 0} & {\small max. number of steps before cut off} & {\small\texttt{100}}\tabularnewline +{\small\texttt{reuse}} & {\small int ($\mathtt{0}/\mathtt{1}$)} & {\small form of result saved with }{\small{\small\verb`\reuse`}} & {\small\texttt{0}}\tabularnewline +\bottomrule +\end{tabular} +\par\end{center} + +Configuration settings for \verb`\nmcSolve`: + +\begin{center} +\begin{tabular}{ll} +\toprule +{\small key} & {\small default}\tabularnewline +\midrule +{\small solve-first-step} & 1\tabularnewline +{\small solve-extra-rounding} & {\small\texttt{0}}\tabularnewline +{\small solve-max-steps} & {\small\texttt{100}}\tabularnewline +{\small solve-reuse} & {\small\texttt{0}}\tabularnewline +\bottomrule +\end{tabular} +\par\end{center} + +\subsection{Settings for \texttt{\textbackslash nmcRecur}} + +Settings option of \verb`\nmcRecur`: + +\begin{center} +\begin{tabular}{llll} +\toprule +{\small key} & {\small type} & {\small meaning} & {\small default}\tabularnewline +\midrule +{\small\texttt{do}} & {\small int$\ge0$} & {\small number of terms to calculate} & {\small\texttt{7}}\tabularnewline +{\small\texttt{see1}} & {\small int$\ge0$} & {\small number of initial terms to display} & {\small\texttt{3}}\tabularnewline +{\small\texttt{see2}} & {\small int$\ge0$} & {\small number of final terms to display} & {\small\texttt{2}}\tabularnewline +{\small\texttt{...}} & {\small chars} & {\small follow display of values with an ellipsis} & \tabularnewline +{\small\texttt{reuse}} & {\small int ($\mathtt{0}/\mathtt{1}/\mathtt{2}$)} & {\small form of result saved with }{\small{\small\verb`\reuse`}} & {\small\texttt{0}}\tabularnewline +\bottomrule +\end{tabular} +\par\end{center} + +Configuration settings for \verb`\nmcRecur`: + +\begin{center} +\begin{tabular}{ll} +\toprule +{\small key} & {\small default}\tabularnewline +\midrule +{\small recur-do} & {\small\texttt{7}}\tabularnewline +{\small recur-see-first} & {\small\texttt{3}}\tabularnewline +{\small recur-see-last} & {\small\texttt{2}}\tabularnewline +{\small recur-reuse} & {\small\texttt{0}}\tabularnewline +\bottomrule +\end{tabular} +\par\end{center} +\end{document} diff --git a/Master/texmf-dist/tex/latex/numerica-plus/numerica-plus.sty b/Master/texmf-dist/tex/latex/numerica-plus/numerica-plus.sty new file mode 100644 index 00000000000..445c2b9ec6a --- /dev/null +++ b/Master/texmf-dist/tex/latex/numerica-plus/numerica-plus.sty @@ -0,0 +1,916 @@ +% This is file `numerica-plus.sty'. +% +% This work may be distributed and/or modified under the conditions +% of the LaTeX Project Public License, either version 1.3c of this +% license or any later version; see +% http://www.latex-project.org/lppl.txt +% +% Andrew Parsloe (ajparsloe@gmail.com) +% +\RequirePackage{numerica}[2021/12/07] +\@ifpackagelater{numerica}{2021/12/07} + {} + {% + \PackageError{numerica-plus}{Package numerica too old.} + {Please update the package.} + \endinput + } +\ProvidesExplFile + {numerica-plus.sty} + {2021/12/10} + {2.0.0} + {Iterate functions, find zeros/extrema of functions, evaluate recurrences } +%---------------------------------------------------------- +\cs_generate_variant:Nn \tl_if_novalue:nTF { o } + +\clist_map_inline:nn { iter,solve } + { + \tl_new:c { g__nmc_info_#1_tl } + \tl_gset:cn { g__nmc_info_#1_tl } { 0 } + } +\clist_gput_right:Nn \g__nmc_info_proc_clist { iter, solve } + +\cs_new_protected:Npn \__nmc_plus_get_var:NNn #1#2#3 + { + \tl_if_empty:NT #1 + { + \tl_set:Nx \l_tmpa_tl + { \seq_item:Nn \l__nmc_vv_all_seq { \g__nmc_consts_vv_int } } + \quark_if_no_value:NTF \l_tmpa_tl + { + \__nmc_error_where:n { \__nmc_verb:n { #2 } } + \__nmc_error_what:n { No~#3~variable~specified~in } } + { + \__nmc_vv_split_item:V \l_tmpa_tl + \tl_set_eq:NN #1 \l__nmc_eq_var_tl + } + } + } +%#1 vv-list reversed #2 var_tl #3 \nmc<cmd> #4 {adjective} +\cs_new_protected:Npn \__nmc_plus_vv_digest:NNNn #1#2#3#4 + { + \clist_push:NV \l_nmc_formula_tl #2 + \__nmc_vv_digest:N #1 + \tl_if_empty:NTF #2 + { \__nmc_plus_get_var:NNn #2 #3 { #4 } } + { \clist_pop:NN \l_nmc_formula_tl #2 } + } +\cs_new_protected:Npn \__nmc_plus_reuse:nNN #1#2#3 + { + \int_case:nnF { #1 } + { + { 1 } + { + \seq_pop:NN #2 \l_tmpa_tl + \seq_pop:NN #3 \l_tmpb_tl + \tl_gset:Nx \g__nmc_reuse_tl + { { \l_tmpa_tl , \l_tmpb_tl } } + \seq_mapthread_function:NNN #2 #3 \__nmc_plus_reuse_display:nn + } + { 2 } + { + \tl_gset:Nx \g__nmc_reuse_tl + { \seq_use:Nn #3 { , } } + } + } + { \tl_gset_eq:NN \g__nmc_reuse_tl \l_nmc_result_tl } + } +\cs_new_protected:Npn \__nmc_plus_reuse_display:nn #1#2 + { \clist_gput_right:Nx \g__nmc_reuse_tl { { #1 , #2 } } } + +% \nmcIterate %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\int_new:N \l__nmc_iter_total_int +\int_new:N \l__nmc_iter_view_int +\int_new:N \l__nmc_iter_round_int +\tl_new:N \l__nmc_iter_var_tl +\fp_new:N \l__nmc_iter_first_fp +\seq_new:N \l__nmc_iter_result_seq +\seq_new:N \l__nmc_iter_index_seq + +\fp_new:N \l__nmc_iter_fixedpti_fp +\fp_new:N \l__nmc_iter_fixedptii_fp + +\tl_new:N \l__nmc_iter_begin_tl +\tl_new:N \l__nmc_iter_end_tl + +\nmc_define:NnN \nmcIterate { iter } \iter + +\cs_gset_protected:Npn \__nmc_iter_initialize: + { + \tl_set:Nn \l__nmc_dbg_idiii_tl { function } + \tl_set:Nn \l__nmc_dbg_idv_tl { stored } + } +\cs_gset_protected:Npn \__nmc_iter_settings_digest: + { + \bool_set_false:N \l__nmc_allow_TF_out_bool + \int_set:Nn \l__nmc_iter_total_int + { \int_max:nn { \l__nmc_iter_total_int } { 1 } } + \int_set:Nn \l__nmc_iter_view_int + { \int_min:nn { \int_max:nn { \l__nmc_iter_view_int } + { 1 } } { \l__nmc_iter_total_int } } + } +\cs_gset_protected:Npn \__nmc_iter_vv_digest:N #1 + { + \__nmc_plus_vv_digest:NNNn #1 + \l__nmc_iter_var_tl \nmcIterate {iteration} + } +\cs_gset_protected:Npn \__nmc_iter_process: + { + \int_set:Nn \l__nmc_iter_round_int + { \l__nmc_round_int + \l__nmc_iter_extra_int } + % first iterate + \__nmc_error_where:n { formula } + \tl_clear:N \l__nmc_fp_expr_tl + \__nmc_fpify:VN \l_nmc_formula_tl \l__nmc_fp_expr_tl + \fp_set:Nn \l__nmc_result_fp { \l__nmc_fp_expr_tl } + \__nmc_error_fpflag: + \bool_if:NF \g__nmc_error_bool + { \fp_set_eq:NN \l__nmc_iter_first_fp \l__nmc_result_fp } + \bool_if:NF \g__nmc_error_bool + { + \__nmc_iter_do: + \tl_set_eq:NN \l__nmc_fp_expr_tl \l__nmc_fp_exprn_tl + } + \bool_if:nF { \g__nmc_error_bool || \l__nmc_num_only_bool } + { + \__nmc_iter_environ: + \__nmc_iter_write: + } + \int_if_zero:nF \l__nmc_dbg_int + { \__nmc_dbg_get_data: } + } +%-------------------------------------- +\cs_new_protected:Npn \__nmc_iter_do: + { + \bool_if:NTF \l__nmc_num_only_bool + { + \__nmc_iter_fixed_pt: + \bool_if:NF \g__nmc_error_bool + { + \__nmc_num_format:nNnN { \l__nmc_iter_fixedpti_fp } + \l_nmc_result_tl { \l__nmc_round_int } + \l__nmc_sci_num_out_bool + \int_decr:N \l__nmc_iter_total_int + \tl_gset:Nx \g__nmc_info_iter_tl + { \int_use:N \l__nmc_iter_total_int } + } + } + { % don't print initial iterations + \int_step_function:nnnN { 1 } { 1 } + { \l__nmc_iter_total_int - \l__nmc_iter_view_int - 1 } + \__nmc_iter_current:n + \bool_if:NF \g__nmc_error_bool + { % store then print these ones + \seq_clear:N \l__nmc_iter_result_seq + \int_step_function:nnnN { 1 } { 1 } + { \int_min:nn { \l__nmc_iter_view_int } + { \l__nmc_iter_total_int - 1 } } \__nmc_iter_current_store:n + } + } + } +\cs_new_protected:Npn \__nmc_iter_current:n #1 + { % stepping function + \bool_if:NF \g__nmc_error_bool + { + \__nmc_calc_fn_val:VNnN \l__nmc_iter_var_tl \l_nmc_formula_tl + { \l__nmc_result_fp } \l__nmc_result_fp + } + } +\cs_new_protected:Npn \__nmc_iter_current_store:n #1 + { % stepping function + \__nmc_calc_fn_val:VNnN \l__nmc_iter_var_tl \l_nmc_formula_tl + { \l__nmc_result_fp } \l__nmc_result_fp + \bool_if:NF \g__nmc_error_bool + { + \__nmc_num_format:nNnN { \l__nmc_result_fp } \l_nmc_result_tl + { \l__nmc_round_int } \l__nmc_sci_num_out_bool + \seq_put_right:NV \l__nmc_iter_result_seq \l_nmc_result_tl + \int_set:Nn \l_tmpa_int + { \l__nmc_iter_total_int - \l__nmc_iter_view_int + #1 } + \seq_put_right:Nx \l__nmc_iter_index_seq { \int_use:N \l_tmpa_int } + } + } +\cs_new_protected:Npn \__nmc_iter_fixed_pt: + { % already 1 iteration + \int_set:Nn \l__nmc_iter_total_int { 1 } + \fp_set_eq:NN \l__nmc_iter_fixedpti_fp \l__nmc_iter_first_fp + \fp_set:Nn \l__nmc_iter_fixedptii_fp { \l__nmc_iter_fixedpti_fp + 1 } + \bool_until_do:nn + { + \fp_compare_p:nNn { 0 } = + { round( \l__nmc_iter_fixedpti_fp - \l__nmc_iter_fixedptii_fp, + \l__nmc_iter_round_int ) } + || \g__nmc_error_bool + } + { + \int_incr:N \l__nmc_iter_total_int + \fp_set_eq:NN \l__nmc_iter_fixedptii_fp \l__nmc_iter_fixedpti_fp + \__nmc_calc_fn_val:VNnN \l__nmc_iter_var_tl \l_nmc_formula_tl + { \l__nmc_result_fp } \l__nmc_result_fp + \fp_set_eq:NN \l__nmc_iter_fixedpti_fp \l__nmc_result_fp + \int_compare:nNnF { \l__nmc_iter_total_int } < + { \l__nmc_iter_max_int } + { + \bool_gset_true:N \g__nmc_error_bool + \__nmc_error_what:n + { + No~fixed~point~attained~after~$\int_use:N + \l__nmc_iter_total_int$~iterations~of + } + } + } + } +\cs_new_protected:Npn \__nmc_iter_environ: + { + \tl_set:Nx \l__nmc_iter_begin_tl + { + \mode_if_math:F { \exp_not:o \l__nmc_math_delimi_tl } + \exp_not:N \begin{array}{r@{}l} + \bool_if:NTF \l__nmc_wraps_math_bool + { \exp_not:o \l__nmc_formula_dup_tl &{}= } + { &{} } + } + \tl_set:Nx \l__nmc_iter_end_tl + { + \exp_not:N \end{array} + \mode_if_math:F { \exp_not:o \l__nmc_math_delimii_tl } + } + } +\cs_new_protected:Npn \__nmc_iter_write: + { + \__nmc_num_format:nNnN { \l__nmc_iter_first_fp } \l_tmpa_tl + { \l__nmc_round_int } \l__nmc_sci_num_out_bool + \tl_gset:Nx \g__nmc_info_iter_tl { \int_use:N \l__nmc_iter_total_int } + \__nmc_if_mod_zero:nnTF { \l__nmc_dbg_int } { 7 } + { \seq_get_right:NN \l__nmc_iter_result_seq \l_nmc_result_tl } + { + \tl_set:Nx \l_nmc_result_tl + { + \exp_not:o \l__nmc_iter_begin_tl + \l_tmpa_tl + \exp_not:o \l__nmc_vv_display_tl + \int_compare:nNnT + { \l__nmc_iter_total_int } > { \l__nmc_iter_view_int + 1 } + { + \\ & \exp_not:N \ldots + \ \exp_not:N\mbox{final\ + \int_use:N \l__nmc_iter_view_int\ of\ \g__nmc_info_iter_tl :} + } + \int_compare:nNnT { \l__nmc_iter_total_int } > { 1 } + { \\ & \exp_not:N \hookrightarrow } + \seq_use:Nn \l__nmc_iter_result_seq { \\ & \hookrightarrow } + \exp_not:o \l__nmc_iter_end_tl + } + } + } +\cs_gset_protected:Npn \__nmc_iter_display: + { + \bool_if:NTF \l__nmc_num_only_bool + { \tl_gset_eq:NN \g__nmc_reuse_tl \l_nmc_result_tl } + { + \__nmc_plus_reuse:nNN { \l__nmc_iter_reuse_int } + \l__nmc_iter_index_seq \l__nmc_iter_result_seq + } + \l_nmc_result_tl + } +%-------------------------------------- +%\nmcSolve +\int_new:N \l__nmc_solve_round_int +\int_new:N \l__nmc_solve_steps_int +\int_new:N \l__nmc_solve_slope_int +\int_new:N \l__nmc_solve_slopei_int +\int_new:N \l__nmc_solve_signs_int +\fp_new:N \l__nmc_solvea_fp +\fp_new:N \l__nmc_solveb_fp +\fp_new:N \l__nmc_solvec_fp +\fp_new:N \l__nmc_solved_fp + +\fp_new:N \l__nmc_solvefa_fp +\fp_new:N \l__nmc_solvefb_fp +\fp_new:N \l__nmc_solvefc_fp +\bool_new:N \l__nmc_solve_stop_bool + +\nmc_define:NnN \nmcSolve { solve } \solve + +\cs_gset_protected:Npn \__nmc_solve_initialize: + { + \tl_set:Nn \l__nmc_dbg_idiii_tl { function } + \tl_set:Nn \l__nmc_dbg_idv_tl { stored } + } +\cs_gset_protected:Npn \__nmc_solve_settings_digest: + { + \bool_set_false:N \l__nmc_allow_TF_out_bool + \tl_if_empty:NF \l__nmc_solve_step_tl + { + \__nmc_fpify_set:NV \l__nmc_solve_step_fp \l__nmc_solve_step_tl + \fp_compare:nNnT \l__nmc_solve_step_fp = { 0 } + { \__nmc_error_what:n { Non-zero~initial~step~required~in } } + } + } +\cs_gset_protected:Npn \__nmc_solve_vv_digest:N #1 + { + \__nmc_plus_vv_digest:NNNn #1 + \l__nmc_solve_var_tl \nmcSolve {equation} + } +\cs_gset_protected:Npn \__nmc_solve_process: + { + \int_set:Nn \l__nmc_solve_round_int + { \l__nmc_round_int + \l__nmc_solve_extra_int } + \bool_if:NF \g__nmc_error_bool + { \__nmc_solve_get_trial_vals: } + \bool_if:NF \g__nmc_error_bool + { + \__nmc_error_where:n { function } + \__nmc_solve_do: + } + \bool_if:NF \g__nmc_error_bool + { + \__nmc_num_format:nNnN { \l__nmc_solvea_fp } \l_nmc_result_tl + { \l__nmc_round_int } \l__nmc_sci_num_out_bool + \tl_gset:Nx \g__nmc_info_solve_tl + { \clist_use:Nn \g__nmc_info_solve_tl { + } } + \tl_set_eq:NN \l__nmc_fp_expr_tl \l__nmc_fp_exprn_tl + } + \int_if_zero:nF \l__nmc_dbg_int + { \__nmc_dbg_get_data: } + } +\cs_new_protected:Npn \__nmc_solve_get_trial_vals: + { + \prop_get:NVN \l__nmc_subst_var_prop + \l__nmc_solve_var_tl \l__nmc_subst_tl + % ensure a < b + \int_case:nn { \fp_sign:n { \l__nmc_solve_step_fp } } + { + { 1 } + { + \fp_set:Nn \l__nmc_solvea_fp { \l__nmc_subst_tl } + \fp_set:Nn \l__nmc_solveb_fp + { \l__nmc_solvea_fp + \l__nmc_solve_step_fp } + } + { -1 } + { + \fp_set:Nn \l__nmc_solveb_fp { \l__nmc_subst_tl } + \fp_set:Nn \l__nmc_solvea_fp + { \l__nmc_solveb_fp + \l__nmc_solve_step_fp } + } + } + } +% find opp. signs, zero, or fn min. +% a b = var vals; fa fb = fn vals +\cs_new_protected:Npn \__nmc_solve_do: + { + \__nmc_solve_calc_values: + \int_zero:N \l__nmc_solve_steps_int + \tl_gclear:N \g__nmc_info_solve_tl + \bool_do_until:nn + { + \g__nmc_error_bool || \l__nmc_solve_stop_bool + || \int_compare_p:nNn + { \l__nmc_solve_steps_int } > { \l__nmc_solve_max_int } + || \fp_compare_p:nNn { 0 } = + { round(\l__nmc_solveb_fp - \l__nmc_solvea_fp, + \l__nmc_solve_round_int) } + } + { + \int_incr:N \l__nmc_solve_steps_int + \int_case:nn { \l__nmc_solve_signs_int } + { + { 0 } { \__nmc_solve_do_bingo: } + { -1 } { \__nmc_solve_do_bisect: } + { 1 } { \__nmc_solve_do_slope: } + } + } + \bool_if:nF { \g__nmc_error_bool || \l__nmc_solve_stop_bool } + { + \__nmc_error_where:n { $\l_nmc_formula_tl$ } + \__nmc_error_what:n + { + No~zero/extremum~found~after~$\int_use:N + \l__nmc_solve_max_int$~steps~for~function + } + } + } +\cs_new_protected:Npn \__nmc_solve_do_bingo: + { % fn = 0 to 16 figures + \fp_compare:nNnTF { \l__nmc_solvefb_fp } = { 0 } + { + \fp_set_eq:NN \l__nmc_solvea_fp \l__nmc_solveb_fp + \fp_set_eq:NN \l__nmc_solvefa_fp \l__nmc_solvefb_fp + } + { + \fp_set_eq:NN \l__nmc_solveb_fp \l__nmc_solvea_fp + \fp_set_eq:NN \l__nmc_solvefb_fp \l__nmc_solvefa_fp + } + \bool_set_true:N \l__nmc_solve_stop_bool + } +\cs_new_protected:Npn \__nmc_solve_do_bisect: + { + \tl_gset:Nx \g__nmc_info_solve_tl { \int_use:N \l__nmc_solve_steps_int } + \int_zero:N \l__nmc_solve_steps_int + \fp_set:Nn \l__nmc_solvec_fp { ( \l__nmc_solvea_fp + \l__nmc_solveb_fp ) / 2 } + \__nmc_calc_fn_val:VNnN \l__nmc_solve_var_tl \l_nmc_formula_tl + { \l__nmc_solvec_fp } \l__nmc_solvefc_fp + \fp_set:Nn \l__nmc_solved_fp { \l__nmc_solvec_fp + 1 } + \bool_until_do:nn + { + \g__nmc_error_bool || + \fp_compare_p:nNn { 0 } = + { round( \l__nmc_solvec_fp - \l__nmc_solved_fp , + \l__nmc_solve_round_int ) } + } + { + \int_incr:N \l__nmc_solve_steps_int + \fp_set_eq:NN \l__nmc_solved_fp \l__nmc_solvec_fp + \fp_compare:nNnTF { 0 } = { \l__nmc_solvefc_fp } + { + \fp_set_eq:NN \l__nmc_solvea_fp \l__nmc_solvec_fp + \fp_set_eq:NN \l__nmc_solvefa_fp \l__nmc_solvefc_fp + } + { + \fp_compare:nNnTF + { sign(\l__nmc_solvefa_fp)sign(\l__nmc_solvefc_fp) } = { 1 } + { + \fp_set_eq:NN \l__nmc_solvea_fp \l__nmc_solvec_fp + \fp_set_eq:NN \l__nmc_solvefa_fp \l__nmc_solvefc_fp + } + { + \fp_set_eq:NN \l__nmc_solveb_fp \l__nmc_solvec_fp + \fp_set_eq:NN \l__nmc_solvefb_fp \l__nmc_solvefc_fp + } + \fp_set:Nn \l__nmc_solvec_fp + { ( \l__nmc_solvea_fp + \l__nmc_solveb_fp ) / 2 } + \__nmc_calc_fn_val:VNnN \l__nmc_solve_var_tl + \l_nmc_formula_tl { \l__nmc_solvec_fp } \l__nmc_solvefc_fp + } + } + \bool_set_true:N \l__nmc_solve_stop_bool + \clist_gput_right:Nx \g__nmc_info_solve_tl { \int_use:N \l__nmc_solve_steps_int } + } +\cs_new_protected:Npn \__nmc_solve_do_slope: + { + \bool_if:NF \g__nmc_error_bool + { + \int_if_zero:nTF { \l__nmc_solve_slope_int } + { % contract + \fp_add:Nn \l__nmc_solvea_fp + { ( \l__nmc_solveb_fp - \l__nmc_solvea_fp ) / 4 } + \fp_sub:Nn \l__nmc_solveb_fp + { ( \l__nmc_solveb_fp - \l__nmc_solvea_fp ) / 4 } + } + { % always towards x-axis + \fp_compare:nNnTF { 0 } < + { \l__nmc_solvefa_fp * \l__nmc_solve_slope_int } + { \__nmc_solve_do_slope_left: } + { \__nmc_solve_do_slope_right: } + } + \fp_set_eq:NN \l__nmc_solved_fp \l__nmc_solvea_fp + \int_set_eq:NN \l__nmc_solve_slopei_int \l__nmc_solve_slope_int + \__nmc_solve_calc_values: + } + \bool_if:NF \g__nmc_error_bool + { + \int_compare:nNnF { \l__nmc_solve_slope_int } = + { \l__nmc_solve_slopei_int } + { + \fp_set:Nn \l__nmc_solvec_fp { ( \l__nmc_solvea_fp + \l__nmc_solveb_fp ) / 2 } + \int_case:nn { \l__nmc_solve_slopei_int } + { + { 1 } { \fp_set_eq:NN \l__nmc_solvea_fp \l__nmc_solvec_fp } + { -1 } { \fp_set_eq:NN \l__nmc_solveb_fp \l__nmc_solvec_fp } + } + \__nmc_solve_calc_values: + } + } + \fp_compare:nNnT { 0 } = + { round(\l__nmc_solveb_fp - \l__nmc_solvea_fp, + \l__nmc_solve_round_int) } + { \bool_set_true:N \l__nmc_solve_stop_bool } + } +\cs_new_protected:Npn \__nmc_solve_do_slope_left: + { + \fp_set:Nn \l__nmc_solvec_fp { 2 \l__nmc_solvea_fp - \l__nmc_solveb_fp } + \fp_set_eq:NN \l__nmc_solveb_fp \l__nmc_solvea_fp + \fp_set_eq:NN \l__nmc_solvea_fp \l__nmc_solvec_fp + } +\cs_new_protected:Npn \__nmc_solve_do_slope_right: + { + \fp_set:Nn \l__nmc_solvec_fp { 2 \l__nmc_solveb_fp - \l__nmc_solvea_fp } + \fp_set_eq:NN \l__nmc_solvea_fp \l__nmc_solveb_fp + \fp_set_eq:NN \l__nmc_solveb_fp \l__nmc_solvec_fp + } +\cs_new_protected:Npn \__nmc_solve_calc_values: + { + \__nmc_calc_fn_val:VNnN \l__nmc_solve_var_tl \l_nmc_formula_tl + { \l__nmc_solvea_fp } \l__nmc_solvefa_fp + \bool_if:NF \g__nmc_error_bool + { + \__nmc_calc_fn_val:VNnN \l__nmc_solve_var_tl \l_nmc_formula_tl + { \l__nmc_solveb_fp } \l__nmc_solvefb_fp + } + \bool_if:NF \g__nmc_error_bool + { + \int_set:Nn \l__nmc_solve_slope_int + { \fp_eval:n { sign(\l__nmc_solvefb_fp - \l__nmc_solvefa_fp) } } + \int_set:Nn \l__nmc_solve_signs_int + { \fp_eval:n { sign(\l__nmc_solvefa_fp) sign(\l__nmc_solvefb_fp) } } + } + } +\cs_gset_protected:Npn \__nmc_solve_display: + { + \tl_gset_eq:NN \g__nmc_reuse_tl \l_nmc_result_tl + \bool_if:NF \l__nmc_num_only_bool + { + \__nmc_num_format:nNnN { \l__nmc_solvefa_fp } \l_tmpa_tl + { \l__nmc_round_int } \l__nmc_sci_num_out_bool + \tl_set:Nx \l_nmc_result_tl + { + \exp_not:o \l__nmc_math_delimi_tl + \bool_if:NTF \l__nmc_wraps_math_bool + { + \exp_not:o \l_nmc_formula_tl = \l_tmpa_tl + \bool_if:NTF \l__nmc_vv_multline_bool + { \exp_not:o \l__nmc_vv_display_tl } + { + \mathchoice{ \exp_not:o \l__nmc_vv_display_tl } + { \exp_not:o \l__nmc_vv_inline_tl }{}{} + } + \rightarrow \exp_not:N \quad + \l__nmc_solve_var_tl = \l_nmc_result_tl + } + { + \l__nmc_solve_var_tl = \l_nmc_result_tl + \mathchoice{ \exp_not:o \l__nmc_vv_display_tl } + { \exp_not:o \l__nmc_vv_inline_tl }{}{} + } + \l__nmc_punc_tl + \exp_not:o \l__nmc_math_delimii_tl + } + \int_if_zero:nT { \l__nmc_solve_reuse_int } + { \tl_gset_eq:NN \g__nmc_reuse_tl \l_nmc_result_tl } + } + \l_nmc_result_tl + } +%-------------------------------------- +% \nmcRecur +\bool_new:N \l__nmc_recur_ellipsis_bool +\int_new:N \l__nmc_recur_last_int +\fp_new:N \l__nmc_recur_result_fp + +\int_new:N \l__nmc_recur_subscr_ini_int +\int_new:N \l__nmc_recur_subscr_val_int +\int_new:N \l__nmc_recur_order_int +\int_new:N \l__nmc_recur_var_int + +\tl_new:N \l__nmc_recurrence_tl +\tl_new:N \l__nmc_recur_base_var_tl +\tl_new:N \l__nmc_recur_subscr_var_tl + +\seq_new:N \l__nmc_recur_result_seq +\seq_new:N \l__nmc_recur_vars_seq + +\nmc_define:NnN \nmcRecur { recur } \recur + +\cs_gset_protected:Npn \__nmc_recur_initialize: + { + \tl_set:Nn \l__nmc_dbg_idiii_tl { relation } + \tl_set:Nn \l__nmc_dbg_idv_tl { stored } + } +\cs_gset_protected:Npn \__nmc_recur_settings_digest: + { + \bool_set_false:N \l__nmc_allow_TF_out_bool + \int_set:Nn \l__nmc_recur_total_int + { \int_max:nn { \l__nmc_recur_total_int } { 1 } } + \int_set:Nn \l__nmc_recur_last_int + { \int_max:nn { 0 } { \int_min:nn + { \l__nmc_recur_last_int } { \l__nmc_recur_total_int } } } + \int_set:Nn \l__nmc_recur_first_int + { \int_max:nn { 0 } { \int_min:nn { \l__nmc_recur_first_int } + { \l__nmc_recur_total_int - \l__nmc_recur_last_int } } } + \int_if_zero:nT { \l__nmc_recur_first_int } + { \int_decr:N \l__nmc_recur_first_int } + } +\cs_gset_protected:Npn \__nmc_recur_vv_digest:N #1 + { % #1 = reversed vv clist + \bool_set_true:N \l__nmc_multitok_bool + \__nmc_recur_elements: + \__nmc_recur_vars_change:N #1 + \__nmc_vv_digest:N #1 + \__nmc_recur_vv_post: + \tl_set_eq:NN \l_nmc_formula_tl \l__nmc_recurrence_tl + \int_if_zero:nF \l__nmc_dbg_int + { \__nmc_dbg_get_data: } + } +% \l__nmc_recurrence_tl, \l__nmc_recur_base_var_tl, +% \l__nmc_recur_subscr_var_tl, \l__nmc_recur_subscr_val_int +\cs_new_protected:Npn \__nmc_recur_elements: + { + \tl_clear:N \l_tmpa_tl + \bool_set_false:N \l_tmpa_bool + \tl_map_inline:Nn \l_nmc_formula_tl + { + \bool_if:NTF \l_tmpa_bool + { + \tl_set:Nn \l_tmpa_tl { ##1 } + \tl_map_break: + } + { + \token_if_math_subscript:NTF ##1 + { + \tl_set:NV \l__nmc_recur_base_var_tl \l_tmpa_tl + \bool_set_true:N \l_tmpa_bool + } + { \tl_put_right:Nn \l_tmpa_tl { ##1 } } + } + } + \__nmc_recur_parse_subscr:N \l_tmpa_tl + \exp_last_unbraced:NV\__nmc_split_eq:w \l_nmc_formula_tl \q_stop + \tl_set:NV \l__nmc_recurrence_tl \l__nmc_eq_val_tl + \tl_set_rescan:Nno \l__nmc_recurrence_tl { \ExplSyntaxOn } \l__nmc_recurrence_tl + } +\cs_new_protected:Npn \__nmc_recur_parse_subscr:N #1 + { + \tl_clear:N \l__nmc_recur_subscr_var_tl + \tl_set:Nn \l_tmpb_tl { 0 } + \int_zero:N \l__nmc_recur_subscr_val_int + \bool_set_false:N \l_tmpa_bool + \tl_map_inline:Nn #1 + { + \bool_if:NTF \l_tmpa_bool + { \tl_put_right:Nn \l_tmpb_tl { ##1 } } + { + \tl_if_in:nnTF { +- } { ##1 } + { + \tl_put_right:Nn \l_tmpb_tl { ##1 } + \bool_set_true:N \l_tmpa_bool + } + { \tl_put_right:Nn \l__nmc_recur_subscr_var_tl { ##1 } } + } + } + \int_set:Nn \l__nmc_recur_subscr_val_int { \l_tmpb_tl } + } +\cs_new_protected:Npn \__nmc_recur_vars_change:N #1 + { % f_{1} etc ==> f_{n-1} etc in #1 (reverse order vv-list) + \clist_reverse:N #1 + \int_zero:N \l__nmc_recur_order_int + \clist_clear:N \l_tmpa_clist % --> \l__nmc_recur_vars_seq + \clist_clear:N \l_tmpb_clist % --> #1 + \int_set:Nn \l_tmpb_int { \l__nmc_recur_subscr_val_int - 1 } + \tl_set_rescan:Nno #1 { \ExplSyntaxOn } #1 + \clist_map_inline:Nn #1 + { + \seq_set_split:Nnn \l_tmpa_seq {_} { ##1 } + \seq_pop:NN \l_tmpa_seq \l_tmpa_tl + \seq_if_empty:NTF \l_tmpa_seq + { \clist_put_left:NV \l_tmpb_clist \l_tmpa_tl } + { + \tl_if_eq:NNTF \l_tmpa_tl \l__nmc_recur_base_var_tl + { % change e.g. f_{1}(x) to f_{n-1}(x) + \int_incr:N \l__nmc_recur_order_int + \tl_put_right:Nn \l_tmpa_tl { _ } + \tl_set_eq:NN \l_tmpb_tl \l__nmc_recur_subscr_var_tl + \int_case:nn { \int_sign:n { \l_tmpb_int } } + { + { -1 } + { + \tl_put_right:Nx \l_tmpb_tl + { \int_use:N \l_tmpb_int } + } + { 0 } { \prg_do_nothing: } + { 1 } + { + \tl_put_right:Nn \l_tmpb_tl { + } + \tl_put_right:Nx \l_tmpb_tl + { \int_use:N \l_tmpb_int } + } + } + \tl_put_right:Nx \l_tmpa_tl { { \l_tmpb_tl } } + \int_decr:N \l_tmpb_int + \seq_pop:NN \l_tmpa_seq \l_tmpb_tl + \int_set:Nn \l__nmc_recur_subscr_ini_int + { \tl_head:N \l_tmpb_tl } + \tl_put_right:Nx \l_tmpa_tl + { \tl_range:Nnn \l_tmpb_tl { 2 } { -1 } } + \clist_put_left:NV \l_tmpb_clist \l_tmpa_tl + \seq_set_split:NnV \l_tmpb_seq { = } \l_tmpa_tl + \seq_pop:NN \l_tmpb_seq \l_tmpa_tl + \clist_put_left:NV \l_tmpa_clist \l_tmpa_tl + } + { \clist_put_left:Nn \l_tmpb_clist { ##1 } } + } + } + \int_set:Nn \l__nmc_recur_var_int + { \l__nmc_recur_subscr_ini_int + \l__nmc_recur_order_int + - \l__nmc_recur_subscr_val_int } + \clist_set_eq:NN #1 \l_tmpb_clist + \clist_put_left:NV \l_tmpa_clist \l__nmc_recur_subscr_var_tl + \clist_concat:NNN \l_nmc_formula_tl \l__nmc_recurrence_tl \l_tmpa_clist + } +\cs_new_protected:Npn \__nmc_recur_vv_post: + { + \clist_pop:NN \l_nmc_formula_tl \l__nmc_recurrence_tl + \clist_pop:NN \l_nmc_formula_tl \l__nmc_recur_subscr_var_tl + \tl_set:Nx \l_tmpa_tl { \int_use:N \l__nmc_recur_var_int } + \__nmc_vv_record:NVN \l__nmc_recur_subscr_var_tl \l_tmpa_tl \c_empty_prop + \tl_set_eq:NN \l_tmpa_tl \l__nmc_recur_subscr_var_tl + \tl_put_right:Nn \l_tmpa_tl { =0 } % formal value + \seq_put_left:NV \l__nmc_calc_fn_seq \l_tmpa_tl + \seq_set_from_clist:NN \l__nmc_recur_vars_seq \l_nmc_formula_tl + } +%%%%%%%%%%%%%%%%%%% +\cs_gset_protected:Npn \__nmc_recur_process: + { % store initial vals; generate later vals + \__nmc_recur_store_ini: + \__nmc_error_where:n { recurrence~formula } + \__nmc_recur_generate: + \seq_get_right:NN \l__nmc_recur_result_seq \l_nmc_result_tl + \tl_set_eq:NN \l__nmc_fp_expr_tl \l__nmc_fp_exprn_tl + } +\cs_new_protected:Npn \__nmc_recur_store_ini: + { + \seq_set_eq:NN \l_tmpa_seq \l__nmc_recur_vars_seq + \int_step_inline:nnnn { 1 } { 1 } { \l__nmc_recur_order_int } + { + \seq_pop:NN \l_tmpa_seq \l_tmpa_tl + \prop_get:NVN \l__nmc_subst_var_prop \l_tmpa_tl \l__nmc_subst_tl + \__nmc_num_format:nNnN { \l__nmc_subst_tl } \l_tmpa_tl + { \l__nmc_round_int } \l__nmc_sci_num_out_bool + \seq_put_right:NV \l__nmc_recur_result_seq \l_tmpa_tl + \tl_set:Nx \l_tmpb_tl { \int_eval:n + { \l__nmc_recur_subscr_ini_int + ##1 -1 } } + } + } +\cs_new_protected:Npn \__nmc_recur_generate: + { + \prop_get:NVN \l__nmc_subst_var_prop + \l__nmc_recur_subscr_var_tl \l__nmc_subst_tl + \int_set:Nn \l__nmc_recur_var_int { \l__nmc_subst_tl } + \int_set:Nn \l_tmpa_int { \l__nmc_recur_var_int + + \l__nmc_recur_total_int - \l__nmc_recur_order_int - 1 } + \__nmc_error_where:n { recurrence~relation } + \__nmc_if_mod_zero:nnT { \l__nmc_dbg_int } { 7 } + { \__nmc_fpify:VN \l__nmc_recurrence_tl \l__nmc_fp_expr_tl } + \int_step_function:nnnN { \l__nmc_recur_var_int } { 1 } + {\l_tmpa_int } \__nmc_recur_generate_loop:n + } +\cs_new_protected:Npn \__nmc_recur_generate_loop:n #1 + { + \bool_if:NF \g__nmc_error_bool + { % calc. the next term + \fp_set:Nn \l_tmpa_fp { #1 } + \__nmc_calc_fn_val:VNnN \l__nmc_recur_subscr_var_tl + \l__nmc_recurrence_tl { \l_tmpa_fp } \l__nmc_recur_result_fp + } + \bool_if:NF \g__nmc_error_bool + { % store the result + \__nmc_num_format:nNnN { \l__nmc_recur_result_fp } + \l_tmpa_tl { \l__nmc_round_int } \l__nmc_sci_num_out_bool + \seq_put_right:NV \l__nmc_recur_result_seq \l_tmpa_tl + % shift vals "down variable"; tmpa above, tmpb below + \seq_set_eq:NN \l_tmpa_seq \l__nmc_recur_vars_seq + \seq_pop:NN \l_tmpa_seq \l_tmpb_tl % low var + \int_step_inline:nnnn {2} { 1 } { \l__nmc_recur_order_int } + { + \seq_pop:NN \l_tmpa_seq \l_tmpa_tl % hi var + \prop_get:NVN \l__nmc_subst_var_prop \l_tmpa_tl \l__nmc_subst_tl + \prop_put:NVV \l__nmc_subst_var_prop \l_tmpb_tl \l__nmc_subst_tl + \prop_put:NVV \l__nmc_vv_change_prop \l_tmpb_tl \l__nmc_subst_tl + \tl_set_eq:NN \l_tmpb_tl \l_tmpa_tl + } + % use tmpb, not tmpa, in case order = 1 + \prop_put:NVx \l__nmc_subst_var_prop \l_tmpb_tl + { \fp_use:N \l__nmc_recur_result_fp } + \prop_put:NVx \l__nmc_vv_change_prop \l_tmpb_tl + { \fp_use:N \l__nmc_recur_result_fp } + } + } +\cs_gset_protected:Npn \__nmc_recur_display: + { + \bool_if:NTF \l__nmc_num_only_bool + { \tl_gset_eq:NN \g__nmc_reuse_tl \l_nmc_result_tl } + { + \seq_clear:N \l_tmpa_seq + \seq_clear:N \l_tmpb_seq + \seq_clear:N \l_tmpc_seq + \__nmc_recur_result:NN \l_tmpa_seq \l_tmpb_seq + \tl_set:Nx \l_nmc_result_tl + { + \exp_not:o \l__nmc_math_delimi_tl + \bool_if:NT \l__nmc_wraps_math_bool + { + \exp_not:o \l__nmc_formula_dup_tl + \bool_if:NTF \l__nmc_vv_multline_bool + { \exp_not:o \l__nmc_vv_display_tl } + { + \mathchoice{ \exp_not:o \l__nmc_vv_display_tl } + { \exp_not:o \l__nmc_vv_inline_tl }{}{} + } + \rightarrow \exp_not:N \quad + } + \seq_use:Nn \l_tmpa_seq { ,\ } + \l__nmc_punc_tl + \exp_not:o \l__nmc_math_delimii_tl + } + \__nmc_plus_reuse:nNN { \l__nmc_recur_reuse_int } \l_tmpc_seq \l_tmpb_seq + } + \l_nmc_result_tl + } +\cs_new_protected:Npn \__nmc_recur_result:NN #1#2 + { + \int_zero:N \l_tmpa_int + \seq_map_inline:Nn \l__nmc_recur_result_seq + { + \int_compare:nNnTF { \l_tmpa_int } < { \l__nmc_recur_first_int } + { \seq_put_right:Nn #1 { ##1 } } + { + \int_compare:nTF { \l_tmpa_int = \l__nmc_recur_first_int + < \l__nmc_recur_total_int - \l__nmc_recur_last_int } + { \seq_put_right:Nn #1 { \ldots } } + { + \int_compare:nNnT { 1 + \l_tmpa_int } > + { \l__nmc_recur_total_int - \l__nmc_recur_last_int } + { + \seq_put_right:Nn #1 { ##1 } + \seq_put_right:Nn #2 { ##1 } + \seq_put_right:Nx \l_tmpc_seq { \int_eval:n { \l_tmpa_int + \l__nmc_recur_subscr_ini_int } } + } + } + } + \int_incr:N \l_tmpa_int + } + \bool_if:NT \l__nmc_recur_ellipsis_bool + { \seq_put_right:Nn #1{ \ldots } } + } +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\keys_define:nn { numerica-plus } + { + iter-do .int_set:N = \l__nmc_iter_total_int, + iter-see-last .int_set:N = \l__nmc_iter_view_int, + iter-max-iterations .int_set:N = \l__nmc_iter_max_int, + iter-extra-rounding .int_set:N = \l__nmc_iter_extra_int, + iter-reuse .int_set:N = \l__nmc_iter_reuse_int, + + iter-do .default:n = 5, + iter-see-last .default:n = 4, + iter-max-iterations .default:n = 100, + iter-extra-rounding .default:n = 0, + iter-reuse .default:n = 0, +% + solve-max-steps .int_set:N = \l__nmc_solve_max_int, + solve-extra-rounding .int_set:N = \l__nmc_solve_extra_int, + solve-first-step .tl_set:N = \l__nmc_solve_step_tl, + solve-reuse .int_set:N = \l__nmc_solve_reuse_int, + + solve-max-steps .default:n = 100, + solve-extra-rounding .default:n = 0, + solve-first-step .default:n = 1, + solve-reuse .default:n = 0, +% + recur-do .int_set:N = \l__nmc_recur_total_int, + recur-see-first .int_set:N = \l__nmc_recur_first_int, + recur-see-last .int_set:N = \l__nmc_recur_last_int, + recur-reuse .int_set:N = \l__nmc_recur_reuse_int, + + recur-do .default:n = 7, + recur-see-first .default:n = 3, + recur-see-last .default:n = 2, + recur-reuse .default:n = 0 + } +\keys_set_known:nn { numerica-plus } + { + iter-do,iter-see-last,iter-max-iterations, + iter-extra-rounding,iter-reuse, + solve-variable,solve-first-step,solve-max-steps, + solve-extra-rounding,solve-reuse, + recur-do,recur-see-first,recur-see-last,recur-reuse + } +\keys_define:nn { numerica/iter } + { + var .tl_set:N = \l__nmc_iter_var_tl, + + .int_set:N = \l__nmc_iter_extra_int, + max .int_set:N = \l__nmc_iter_max_int, + do .int_set:N = \l__nmc_iter_total_int, + see .int_set:N = \l__nmc_iter_view_int, + reuse .int_set:N = \l__nmc_iter_reuse_int + } +\keys_define:nn { numerica/solve } + { + var .tl_set:N = \l__nmc_solve_var_tl, + + .int_set:N = \l__nmc_solve_extra_int, + max .int_set:N = \l__nmc_solve_max_int, + dvar .tl_set:N = \l__nmc_solve_step_tl, + reuse .int_set:N = \l__nmc_solve_reuse_int + } +\keys_define:nn { numerica/recur } + { + do .int_set:N = \l__nmc_recur_total_int, + see1 .int_set:N = \l__nmc_recur_first_int, + see2 .int_set:N = \l__nmc_recur_last_int, + ... .code:n = \bool_set_true:N \l__nmc_recur_ellipsis_bool, + reuse .int_set:N = \l__nmc_recur_reuse_int + } +\file_if_exist:nT { numerica-plus.cfg } + { + \file_get:nnN { numerica-plus.cfg } {} \l_tmpa_tl + \exp_args:NnV \keys_set_known:nn { numerica-plus } \l_tmpa_tl + } +% end of `numerica-plus.sty'. diff --git a/Master/tlpkg/tlpsrc/numerica-plus.tlpsrc b/Master/tlpkg/tlpsrc/numerica-plus.tlpsrc new file mode 100644 index 00000000000..e69de29bb2d --- /dev/null +++ b/Master/tlpkg/tlpsrc/numerica-plus.tlpsrc |