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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}[2019/01/21 v3.1 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'.