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diff --git a/Master/texmf-dist/doc/latex/exframe/exfsamp.tex b/Master/texmf-dist/doc/latex/exframe/exfsamp.tex new file mode 100644 index 00000000000..f1f59d5d908 --- /dev/null +++ b/Master/texmf-dist/doc/latex/exframe/exfsamp.tex @@ -0,0 +1,218 @@ +%% +%% This is file `exfsamp.tex', +%% generated with the docstrip utility. +%% +%% The original source files were: +%% +%% exframe.dtx (with options: `samplesingle') +%% +%% 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{exfsamp.tex}[2019/05/01 v3.2 standalone sample 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{composepointspairmargin}[2]{ + \ifdim#2pt=0pt#1p.% + \else\ifdim#1pt=0pt+#2p.% + \else#1+#2p.% + \fi\fi} + +\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},points=99+4] + +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+0.5,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+0.5} +\end{solution} + +You may give the final part a try: + +\begin{subproblem}[optional={optional}, + difficulty={requires inspiration},points={+3.5}] +Compute the series +\showpoints +\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 `exfsamp.tex'. |