\documentclass[11pt]{article} \usepackage[breakable]{tcolorbox} \usepackage{parskip} % Stop auto-indenting (to mimic markdown behaviour) % Basic figure setup, for now with no caption control since it's done % automatically by Pandoc (which extracts ![](path) syntax from Markdown). \usepackage{graphicx} % Maintain compatibility with old templates. Remove in nbconvert 6.0 \let\Oldincludegraphics\includegraphics % Ensure that by default, figures have no caption (until we provide a % proper Figure object with a Caption API and a way to capture that % in the conversion process - todo). \usepackage{caption} \DeclareCaptionFormat{nocaption}{} \captionsetup{format=nocaption,aboveskip=0pt,belowskip=0pt} \usepackage{float} \floatplacement{figure}{H} % forces figures to be placed at the correct location \usepackage{xcolor} % Allow colors to be defined \usepackage{enumerate} % Needed for markdown enumerations to work \usepackage{geometry} % Used to adjust the document margins \usepackage{amsmath} % Equations \usepackage{amssymb} % Equations \usepackage{textcomp} % defines textquotesingle % Hack from http://tex.stackexchange.com/a/47451/13684: \AtBeginDocument{% \def\PYZsq{\textquotesingle}% Upright quotes in Pygmentized code } \usepackage{upquote} % Upright quotes for verbatim code \usepackage{eurosym} % defines \euro \usepackage{iftex} \ifPDFTeX \usepackage[T1]{fontenc} \IfFileExists{alphabeta.sty}{ \usepackage{alphabeta} }{ \usepackage[mathletters]{ucs} \usepackage[utf8x]{inputenc} } \else \usepackage{fontspec} \usepackage{unicode-math} \fi \usepackage{fancyvrb} % verbatim replacement that allows latex \usepackage{grffile} % extends the file name processing of package graphics % to support a larger range \makeatletter % fix for old versions of grffile with XeLaTeX \@ifpackagelater{grffile}{2019/11/01} { % Do nothing on new versions } { \def\Gread@@xetex#1{% \IfFileExists{"\Gin@base".bb}% {\Gread@eps{\Gin@base.bb}}% {\Gread@@xetex@aux#1}% } } \makeatother \usepackage[Export]{adjustbox} % Used to constrain images to a maximum size \adjustboxset{max size={0.9\linewidth}{0.9\paperheight}} % The hyperref package gives us a pdf with properly built % internal navigation ('pdf bookmarks' for the table of contents, % internal cross-reference links, web links for URLs, etc.) \usepackage{hyperref} % The default LaTeX title has an obnoxious amount of whitespace. 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\lccode`\~`\,\lowercase{\def~}{\discretionary{\hbox{\char`\,}}{\Wrappedafterbreak}{\hbox{\char`\,}}}% \lccode`\~`\;\lowercase{\def~}{\discretionary{\hbox{\char`\;}}{\Wrappedafterbreak}{\hbox{\char`\;}}}% \lccode`\~`\:\lowercase{\def~}{\discretionary{\hbox{\char`\:}}{\Wrappedafterbreak}{\hbox{\char`\:}}}% \lccode`\~`\?\lowercase{\def~}{\discretionary{\hbox{\char`\?}}{\Wrappedafterbreak}{\hbox{\char`\?}}}% \lccode`\~`\!\lowercase{\def~}{\discretionary{\hbox{\char`\!}}{\Wrappedafterbreak}{\hbox{\char`\!}}}% \lccode`\~`\/\lowercase{\def~}{\discretionary{\hbox{\char`\/}}{\Wrappedafterbreak}{\hbox{\char`\/}}}% \catcode`\.\active \catcode`\,\active \catcode`\;\active \catcode`\:\active \catcode`\?\active \catcode`\!\active \catcode`\/\active \lccode`\~`\~ } \makeatother \let\OriginalVerbatim=\Verbatim \makeatletter \renewcommand{\Verbatim}[1][1]{% %\parskip\z@skip \sbox\Wrappedcontinuationbox {\Wrappedcontinuationsymbol}% \sbox\Wrappedvisiblespacebox {\FV@SetupFont\Wrappedvisiblespace}% \def\FancyVerbFormatLine ##1{\hsize\linewidth \vtop{\raggedright\hyphenpenalty\z@\exhyphenpenalty\z@ \doublehyphendemerits\z@\finalhyphendemerits\z@ \strut ##1\strut}% }% % If the linebreak is at a space, the latter will be displayed as visible % space at end of first line, and a continuation symbol starts next line. % Stretch/shrink are however usually zero for typewriter font. \def\FV@Space {% \nobreak\hskip\z@ plus\fontdimen3\font minus\fontdimen4\font \discretionary{\copy\Wrappedvisiblespacebox}{\Wrappedafterbreak} {\kern\fontdimen2\font}% }% % Allow breaks at special characters using \PYG... macros. \Wrappedbreaksatspecials % Breaks at punctuation characters . , ; ? ! and / need catcode=\active \OriginalVerbatim[#1,codes*=\Wrappedbreaksatpunct]% } \makeatother % Exact colors from NB \definecolor{incolor}{HTML}{303F9F} \definecolor{outcolor}{HTML}{D84315} \definecolor{cellborder}{HTML}{CFCFCF} \definecolor{cellbackground}{HTML}{F7F7F7} % prompt \makeatletter \newcommand{\boxspacing}{\kern\kvtcb@left@rule\kern\kvtcb@boxsep} \makeatother \newcommand{\prompt}[4]{ {\ttfamily\llap{{\color{#2}[#3]:\hspace{3pt}#4}}\vspace{-\baselineskip}} } % Prevent overflowing lines due to hard-to-break entities \sloppy % Setup hyperref package \hypersetup{ breaklinks=true, % so long urls are correctly broken across lines colorlinks=true, urlcolor=urlcolor, linkcolor=linkcolor, citecolor=citecolor, } % Slightly bigger margins than the latex defaults \geometry{verbose,tmargin=1in,bmargin=1in,lmargin=1in,rmargin=1in} \begin{document} \section{Innlevering 4 - Numerisk integrasjon}\label{innlevering-4---numerisk-integrasjon} Standardnormalfordelingen er gitt ved \[ f(x) = \frac{1}{\sqrt{2\pi}} e^{-x ^2/2}. \] Standardnormalfordelingen er et eksempel på en sannsynlighetstetthetsfunksjon, og for TMA4400 holder det å vite at sannsynligheten for å havne mellom \(a\) og \(b\) er gitt ved \[ \frac{1}{\sqrt{2\pi}} \int_a^b e^{-x ^2/2} \; dx. \] Det er bare ett problem - standardnormalfordelingen kan ikke antideriveres. Under er kode for å beregne arealet ved riemannsummer. \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder] \prompt{In}{incolor}{ }{\boxspacing} \begin{Verbatim}[commandchars=\\\{\}] \PY{k+kn}{import} \PY{n+nn}{numpy} \PY{k}{as} \PY{n+nn}{np} \PY{c+c1}{\PYZsh{}integrasjonsgrenser} \PY{n}{a}\PY{o}{=}\PY{l+m+mi}{0} \PY{n}{b}\PY{o}{=}\PY{l+m+mi}{1} \PY{c+c1}{\PYZsh{}gitter} \PY{n}{n}\PY{o}{=}\PY{l+m+mi}{1000} \PY{n}{x}\PY{o}{=}\PY{n}{np}\PY{o}{.}\PY{n}{linspace}\PY{p}{(}\PY{n}{a}\PY{p}{,}\PY{n}{b}\PY{p}{,}\PY{n}{n}\PY{o}{+}\PY{l+m+mi}{1}\PY{p}{)} \PY{n}{h}\PY{o}{=}\PY{n}{b}\PY{o}{/}\PY{n}{n} \PY{c+c1}{\PYZsh{}venstre riemannsum} \PY{n}{f}\PY{o}{=}\PY{n}{h}\PY{o}{*}\PY{p}{(}\PY{n}{np}\PY{o}{.}\PY{n}{sum}\PY{p}{(}\PY{n}{np}\PY{o}{.}\PY{n}{exp}\PY{p}{(}\PY{o}{\PYZhy{}}\PY{n}{x}\PY{o}{*}\PY{o}{*}\PY{l+m+mi}{2}\PY{o}{/}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{np}\PY{o}{.}\PY{n}{exp}\PY{p}{(}\PY{o}{\PYZhy{}}\PY{n}{b}\PY{o}{*}\PY{o}{*}\PY{l+m+mi}{2}\PY{o}{/}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{o}{/}\PY{n}{np}\PY{o}{.}\PY{n}{sqrt}\PY{p}{(}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{np}\PY{o}{.}\PY{n}{pi}\PY{p}{)} \PY{n+nb}{print}\PY{p}{(}\PY{n}{f}\PY{p}{)} \end{Verbatim} \end{tcolorbox} \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item Modifiser koden til å beregne arealet ved trapesmetoden, kjør alt med \(a=0\), \(b=1\), sammenlikne begge metoder med den korrekte sannsynligheten \[ \frac{1}{\sqrt{2\pi}} \int_0^1 e^{-x ^2/2} \; dx \approx 0.3413447460685429485852. \] Kan du forklare forskjellen? (Fun fact: Det er denne beregningen som forteller deg at om du har IQ på 115 er du smartere enn om lag 84 prosent av befolkningen. Mer om dette i TMA4240/4245.) \end{enumerate} \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \setcounter{enumi}{1} \tightlist \item Man kan bruke polynominterpolasjon til å finne bedre teknikker for numerisk integrasjon. Utled Simpsons metode ved å finne lagrangepolynomene til gitteret \[ \left\{ a, \frac{a+b}{2}, b \right\} \] og integrere dem. Test Simpsons metode på arealet over og forklar forskjellen. \end{enumerate} De numeriske integrasjonsrutinene gir oss et sett med korresponderende teknikker for å løse ordinære differensiallikninger numerisk. La oss si vi ønsker å løse initialverdiproblemet \[ \dot x = f\left( x \right) \hspace{10mm} x(0) = 0 \] på intervallet \([0,T]\). Vi deler tidsintervallet inn i en partisjon med \(n+1\) punkter \[ t_k = kh \hspace{10mm} 0\leq k \leq n \] der \(h=T/n\), og integrerer differensiallikningen fra ett gitterpunkt til neste: \[ x\left( t_{k+1} \right) - x\left( t_k \right) = \int_{t_k}^{t_{k+1}} f\left( x\left( t \right) \right) \; dt \] Hvis vi setter inn en enkelt rektangulær riemannboks (venstre, høyre eller sentrert) eller trapesmetoden for integralet på høyresiden, og så setter inn approksimasjonen \(x_k \approx x\left( t_k \right)\) overalt, får vi nå fire metoder på rappen: \begin{align*} \text{Fremlengs Euler:} \hspace{10mm} &x_{k+1} = x_k + hf\left( x_k \right) \\[3mm] \text{Baklengs Euler:} \hspace{10mm} &x_{k+1} = x_k + hf\left( x_{k+1} \right) \\[3mm] \text{Midtpunktmetoden:} \hspace{10mm} &x_{k+1} = x_k + hf\left( \frac{x_k + x_{k+1}}{2} \right) \\[3mm] \text{Trapesmetoden:} \hspace{10mm} &x_{k+1} = x_k + \frac{h}{2}\left( f\left( x_k \right) + f\left( x_{k+1} \right) \right) \end{align*} Under finner du kode for fremlengs Euler på det logistiske problemet \[ \dot{x}=a\left( 1-x/k \right)x \hspace{10mm} x(0) = x_0 \hspace{10mm} a >0,\; k > 0. \] \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder] \prompt{In}{incolor}{ }{\boxspacing} \begin{Verbatim}[commandchars=\\\{\}] \PY{k+kn}{import} \PY{n+nn}{matplotlib}\PY{n+nn}{.}\PY{n+nn}{pyplot} \PY{k}{as} \PY{n+nn}{plt} \PY{c+c1}{\PYZsh{}parametre} \PY{n}{a}\PY{o}{=}\PY{l+m+mi}{1} \PY{c+c1}{\PYZsh{}vekstfaktor} \PY{n}{k}\PY{o}{=}\PY{l+m+mi}{1} \PY{c+c1}{\PYZsh{}bærekapasitet} \PY{c+c1}{\PYZsh{}tidsakse} \PY{n}{T}\PY{o}{=}\PY{l+m+mi}{10} \PY{n}{N}\PY{o}{=}\PY{l+m+mi}{10000} \PY{n}{h}\PY{o}{=}\PY{n}{T}\PY{o}{/}\PY{n}{N} \PY{n}{t}\PY{o}{=}\PY{n}{np}\PY{o}{.}\PY{n}{linspace}\PY{p}{(}\PY{l+m+mi}{0}\PY{p}{,}\PY{n}{T}\PY{p}{,}\PY{n}{N}\PY{o}{+}\PY{l+m+mi}{1}\PY{p}{)} \PY{c+c1}{\PYZsh{}løsning} \PY{n}{x}\PY{o}{=}\PY{n}{np}\PY{o}{.}\PY{n}{zeros}\PY{p}{(}\PY{n}{N}\PY{o}{+}\PY{l+m+mi}{1}\PY{p}{)} \PY{n}{x}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{]}\PY{o}{=}\PY{l+m+mf}{.1} \PY{k}{for} \PY{n}{i} \PY{o+ow}{in} \PY{n+nb}{range}\PY{p}{(}\PY{n}{N}\PY{p}{)}\PY{p}{:} \PY{n}{x}\PY{p}{[}\PY{n}{i}\PY{o}{+}\PY{l+m+mi}{1}\PY{p}{]}\PY{o}{=}\PY{n}{x}\PY{p}{[}\PY{n}{i}\PY{p}{]}\PY{o}{+}\PY{n}{h}\PY{o}{*}\PY{n}{x}\PY{p}{[}\PY{n}{i}\PY{p}{]}\PY{o}{*}\PY{p}{(}\PY{l+m+mi}{1}\PY{o}{\PYZhy{}}\PY{n}{x}\PY{p}{[}\PY{n}{i}\PY{p}{]}\PY{p}{)} \PY{n}{plt}\PY{o}{.}\PY{n}{plot}\PY{p}{(}\PY{n}{t}\PY{p}{,}\PY{n}{x}\PY{p}{)} \PY{n}{plt}\PY{o}{.}\PY{n}{savefig}\PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{logistisk.png}\PY{l+s+s1}{\PYZsq{}}\PY{p}{)} \end{Verbatim} \end{tcolorbox} \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \setcounter{enumi}{2} \tightlist \item Implementér de forskjellige metodene for det logistiske problemet og sammenlikne med den analytiske løsningen \[ x(t)=\frac{kx_0e^{at}}{k-x_0+x_0e^{at}}. \] Kan du forklare forskjellene? (Velg vekstfaktoren \(a\) og bærekapasiteten \(k\) selv. For midtpunkt- og trapesmetoden kan en numerisk likningsløser komme til nytte.) \end{enumerate} % Add a bibliography block to the postdoc \end{document}