%% %% 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'.