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%% exframe.dtx  (with options: `sample1')
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%% Copyright (C) 2011-2019 Niklas Beisert
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%% This work may be distributed and/or modified under the
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\documentclass[12pt]{article}

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\parindent0pt
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\usepackage{amsmath}
\usepackage{hyperref}
\usepackage[extstyle]{exframe}

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\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
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%% End of file `exfsmp1.tex'.
