exframe (16jan19)

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diff --git a/Master/texmf-dist/doc/latex/exframe/README.txt b/Master/texmf-dist/doc/latex/exframe/README.txt
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+exframe v3.0
+Copyright 2011-2019 Niklas Beisert
+
+exframe is a LaTeX2e package which provides a general purpose
+framework to describe and typeset exercises and exam questions
+along with their solutions. The package features mechanisms to
+hide or postpone solutions, to assign and handle points, to
+collect problems on exercise sheets, to store and use metadata
+and to implement a consistent numbering. It also provides a
+very flexible interface for configuring and customising the
+formatting, layout and representation of the exercise content.
+
+The package consists of the files
+
+    README.txt   readme file
+    exframe.ins  installation file
+    exframe.dtx  source file
+    exframe.sty  package file
+    exfsmp1.tex  sample latex file
+    exframe.pdf  manual
+
+The distribution consists of the files
+README.txt, exframe.ins and exframe.dtx.
+
+Run "(pdf)latex exframe.dtx" to compile the manual exframe.pdf.
+
+Run "latex exframe.ins" to create the package exframe.sty
+and the sample exfsmp1.tex.
+Copy the file exframe.sty to an appropriate directory of your LaTeX
+distribution, e.g. texmf-root/tex/latex/exframe.
+
+This work may be distributed and/or modified under the
+conditions of the LaTeX Project Public License, either version 1.3
+of this license or (at your option) any later version.
+The latest version of this license is in
+  http://www.latex-project.org/lppl.txt
+and version 1.3 or later is part of all distributions of LaTeX
+version 2005/12/01 or later.
diff --git a/Master/texmf-dist/doc/latex/exframe/exframe.pdf b/Master/texmf-dist/doc/latex/exframe/exframe.pdf
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diff --git a/Master/texmf-dist/doc/latex/exframe/exfsmp1.tex b/Master/texmf-dist/doc/latex/exframe/exfsmp1.tex
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+%%
+%% This is file `exfsmp1.tex',
+%% generated with the docstrip utility.
+%%
+%% The original source files were:
+%%
+%% exframe.dtx  (with options: `sample1')
+%% 
+%% Copyright (C) 2011-2019 Niklas Beisert
+%% 
+%% This work may be distributed and/or modified under the
+%% conditions of the LaTeX Project Public License, either version 1.3
+%% of this license or (at your option) any later version.
+%% The latest version of this license is in
+%%   http://www.latex-project.org/lppl.txt
+%% and version 1.3 or later is part of all distributions of LaTeX
+%% version 2005/12/01 or later.
+%% 
+\NeedsTeXFormat{LaTeX2e}[1996/12/01]
+\ProvidesFile{exfsmp1.tex}[2019/01/16 v3.0 sample 1 for exframe]
+\documentclass[12pt]{article}
+
+\usepackage{geometry}
+\geometry{layout=a4paper}
+\geometry{paper=a4paper}
+\geometry{margin=2.5cm}
+\parindent0pt
+\parskip0.5ex
+
+\usepackage{amsmath}
+\usepackage{hyperref}
+\usepackage[extstyle]{exframe}
+
+\exercisesetup{solutions=true}
+%%\exercisesetup{solutions=false}
+
+\exercisestyle{plainheader}
+\exerciseconfig{composeheaderbelowright}{\getsheetdata{points}}%
+
+\exerciseconfig{countersheet}{\Roman{sheet}}
+\exerciseconfig{countersubproblem}{\roman{subproblem})}
+\exerciseconfig{countersubproblemmax}{vii)}
+
+\exerciseconfig{termsheet}{\"Ubungsblatt}
+\exerciseconfig{termsheets}{\"Ubungsbl\"atter}
+
+\exercisestyle{pointsat=margin}
+\reversemarginpar
+\exerciseconfig{composepointsmargin}[1]{#1p.}
+
+\exerciseconfig{styletitle}{\sffamily\bfseries}
+
+\exerciseconfig{skipproblembelow}{1.5cm}
+
+\exercisestyle{fracpoints}
+\exercisestyle{solutionbelow=problem}
+\exercisestyle{solutionsep}
+
+\exerciseconfig{composemetasheet}[2]{\getexercisedata{course},
+  \exerciseifempty{#2}{\getexerciseconfig{termsheet} #1}{#2}}
+
+\exercisesetup{pdfdata=sheet}
+\exercisedata{title=%
+  {\getexercisedata{course}, \getexercisedata{material}}}
+\exercisedata{author=%
+  {\getexercisedata{instructor}, \getexercisedata{institution}}}
+
+\exercisedata{institution={Katharinen-Volksschule}}
+\exercisedata{course={Mathematik}}
+\exercisedata{instructor={J.\ G.\ B\"uttner}}
+\exercisedata{period={ca.\ 1786}}
+\exercisedata{material={\"Ubungsaufgaben}}
+
+\begin{document}
+
+\begin{sheet}[number=5]
+
+\begin{problem}[title={Sums}]
+
+This problem deals with sums and series.
+
+\begin{subproblem}[points=2,difficulty=simple,label={\problemtag-simplesum}]
+Compute the sum
+\showpoints
+\begin{equation}
+1+2+3.
+\end{equation}
+
+\begin{solution}
+The result is
+\begin{equation}
+1+2+3=6.
+\end{equation}
+\end{solution}
+
+\end{subproblem}
+
+\begin{subproblem}[points=97,difficulty=lengthy]
+Compute the sum
+\begin{equation}
+1+2+3+\ldots+98+99+100.
+\end{equation}
+Keep calm and calculate!
+%%That ought to keep him occupied for a while
+\end{subproblem}
+
+\begin{solution}[author={C.\ F.\ Gau\ss}]
+We use the result $1+2+3=6$ from part \ref{\problemtag-simplesum}
+to jumpstart the calculation. The remaining sums yield
+\awardpoints[1 for each remaining sum]{97}
+\begin{equation}
+6+4+5+\ldots+99+100=5050.
+\end{equation}
+Alternatively the summands can be grouped into pairs as follows:
+\begin{align}
+1+100&=101,\\
+2+99&=101,\\
+3+98&=101,\\
+\ldots &\nonumber\\
+50+51&=101.
+\end{align}
+These amount to 50 times the same number 101.
+Therefore the sum equals
+\begin{equation}
+1+2+\ldots+99+100=50\cdot 101=5050.
+\end{equation}
+\textit{Ligget se!} \awardpoints*{97}
+\end{solution}
+
+You may give the final part a try:
+
+\begin{subproblem}[optional={optional}, difficulty={requires inspiration}]
+Compute the series
+\begin{equation}
+1+2+3+\ldots
+\end{equation}
+
+\begin{solution}
+The series is divergent, so the result is $\infty$ \awardpoints*{1}.
+\par
+However, after subtracting the divergent part,
+the result clearly is
+\begin{equation}
+\zeta(-1)=-\frac{1}{12}\,,
+\end{equation}
+where the zeta-function $\zeta(s)$ is defined by
+\begin{equation}
+\zeta(s):=\sum_{k=1}^\infty \frac{1}{k^s}\,.
+\end{equation}
+This definition holds only for $s>1$ where the sum is convergent,
+but one can continue the complex analytic function to $s<0$
+\awardpoints*{1.5}.
+\par
+Another way of understanding the result
+is to use the indefinite summation formula
+for arbitrary exponent $s$ in the summand
+(which also follows from the Euler--MacLaurin formula)
+\begin{equation}
+\sum_n n^s
+= \frac{n^{s+1}}{s+1}
+ -\sum_{j=0}^s  \frac{\zeta(j-s)\,s!}{(s-j)!\,j!}\,n^j
+= \ldots - \zeta(-s)\,n^0.
+\end{equation}
+Curiously, the constant term with $j=0$ is just the desired result
+but with the wrong sign
+(in fact, the constant term of an indefinite sum is ambiguous;
+for the claim we merely set $j=0$
+in the expression which holds for others values of $j$)
+\awardpoints*{0.5}.
+In order to understand the sign,
+we propose that the above formula describes the regularised result
+for the sum with limits $+\infty$ and $n$
+\begin{equation}
+\sum_{k=+\infty}^n k^s
+\simeq \frac{n^{s+1}}{s+1}
+ -\sum_{j=0}^s  \frac{\zeta(j-s)\,s!}{(s-j)!\,j!}\,n^j.
+\end{equation}
+Then we flip the summation limits of the desired sum
+to bring it into the above form
+\awardpoints*{0.5}
+\begin{equation}
+\sum_{k=1}^\infty k^s
+= -\sum_{k=\infty}^0 k^s
+\simeq \zeta(-s).
+\end{equation}
+\end{solution}
+
+\end{subproblem}
+
+\end{problem}
+
+\begin{problem}[points=1, difficulty=insane]
+Show that the equation
+\begin{equation}
+a^3+b^3=c^3
+\end{equation}
+has no positive integer solutions.
+\end{problem}
+
+\begin{solution}
+\normalmarginpar
+This is beyond the scope of this example.
+\marginpar{\footnotesize\raggedright does not fit here.\par}
+\end{solution}
+
+\end{sheet}
+
+\end{document}
+\endinput
+%%
+%% End of file `exfsmp1.tex'.