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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 3 - Likningsløsning}\label{innlevering-3---likningsluxf8sning} De fleste likninger som detter ut av praktiske anvendelser er umulige å løse for hånd. Under ser du et bilde av en torsjonsfjær. Dette er en slik fjær du finner i musefeller og i gamle dører og den leverer et moment som er proporsjonalt med utslagsvinkelen \(\theta\). \begin{figure} \centering \includegraphics{L1030352.jpg} \caption{alt text} \end{figure} Vi fester en horisontal stang med lengde \(l\) i en vegg med en torsjonsfjær med stivhet \(k\) og legger en vekt \(m\) ytterst på stangen. Gravitasjonskraften fra vekten gir dreiemomentet \(ml\cos \theta\), så hvis vi har lyst til å regne ut hvor langt ned vekten drar stangen, må vi finne punktet der momentet fra fjæren balanserer momentet fra vekten: \begin{align*} k\theta = ml\cos \theta \end{align*} \begin{figure} \centering \includegraphics{torsjonsfjær.png} \caption{alt text} \end{figure} Vi setter \(ml/k=1\). Her er et plot av funksjonene \(f(\theta)=\theta\) og \(g(\theta)=\cos \theta\). Disse krysser hverandre, så løsningen til likningen eksisterer, men den er ikke helt god å finne med penn og papir. Løsningen kalles dottietallet og er oppkalt etter en professor i fransk. Professor Dottie oppdaget at hvis hun skrev inn et tilfeldig tall på en kalkulator og trykket gjentatte ganger på cosinusknappen, endte hun alltid opp på omtrent \(0.739085\). \begin{figure} \centering \includegraphics{dottie.png} \caption{alt text} \end{figure} \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item Løs likningen \(\theta=\cos \theta\) med vanlig fikspunktiterasjon. Spiller \(x_0\) noen rolle? Hvor mange interasjoner trengs for å nå maskinpresisjon? \end{enumerate} På skolen lærte du Newtons metode: \begin{align*} x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \end{align*} Denne leter etter nullpunkter til \(f\). \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \setcounter{enumi}{1} \tightlist \item Løs likningen \(\theta = \cos \theta\) med Newtons metode. Spiller \(x_0\) noen rolle? Hvor mange interasjoner trengs for å nå maskinpresisjon? \end{enumerate} Som du ser av oppgavene over, oppfører disse to metodene seg helt forskjellig. Dersom vi skal forstå hvorfor, trenger vi Taylors teorem, som sier at dersom \(f\) er \(n+1\) ganger kontinuerlig deriverbar på et åpent intervall som inneholder \(x\) og \(x\), finnes det en \(s\) mellom \(x\) og \(a\) slik at \begin{align*} f(x)&= f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2+\cdots+ \frac{f^{(n)}(a)}{n!}(x-a)^n + \frac{1}{n!}\int^x_a (x-s)^nf^{(n+1)}(s)\; ds \end{align*} Dersom \(n=0\) kalles denne sekantsetningen. \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \setcounter{enumi}{2} \tightlist \item Bruk Taylors teorem \(n=0\) til å vise at for fikspunktiterasjonen finnes det, dersom \(g\) er deriverbar og alle brouwers teoremers mor og så finnes \(s\) slik at \begin{align*} x_{n+1}-r=g'(s)(x_n-r) \end{align*} der \(r=g(r)\) og \(x_{n+1} = g(x_n)\). \end{enumerate} Dersom man innser at Newtons metode også er en fikspunktiterasjon \begin{align*} x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = g(x_n) \end{align*} kan vi analysere den på liknende vis, men vi må sette \(n=1\) i Taylors teorem. \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \setcounter{enumi}{3} \tightlist \item Vis at under noen betingelser på fandens oldemor finnes \(s\) slik at \begin{align*} x_{n+1}-r= \frac{f''(s)}{2f'(x_n)}(r-x_n)^2. \end{align*} \end{enumerate} Jacobis metode \begin{align*} x_{n+1} = D^{-1}\left(b - (L+U)x_n\right) \end{align*} for lineære systemer er forresten også en fikspunktiterasjon. At den konvergerer når \(A\) er strengt diagonaldominant er lett å se dersom du kan Banachs fikspunktteorem og vet hva en gershgorinsirkel er. \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \setcounter{enumi}{4} \tightlist \item Vis dette og test Jacobis metode på likningssystemene \begin{align*} 2x_1 + 3x_2 + 4x_3 = 4 \\ 3x_1 + 4x_2 + 5x_3 = 5 \\ 4x_1 + 5x_2 + 7x_3 = 3 \end{align*} og \begin{align*} -\frac{1}{\sqrt{2}}x_1 + \frac{1}{2}x_2 = 0 \\ \frac{1}{\sqrt{2}}x_1 + \frac{\sqrt{3}}{2}x_2 = 1 \end{align*} Prøv å spå om det konvergerer før du kjører. \end{enumerate} Og her er litt kode for cobwebplot forresten. \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder] \prompt{In}{incolor}{ }{\boxspacing} \begin{Verbatim}[commandchars=\\\{\}] \PY{c+c1}{\PYZsh{} pakkene vi trenger} \PY{o}{\PYZpc{}}\PY{k}{matplotlib} inline \PY{k+kn}{import} \PY{n+nn}{math} \PY{c+c1}{\PYZsh{} vi trenger tilgang til sinus\PYZhy{}funksjonen math.sin() } \PY{k+kn}{import} \PY{n+nn}{numpy} \PY{k}{as} \PY{n+nn}{np} \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{} fikser litt på pyplot} \PY{n}{newparams} \PY{o}{=} \PY{p}{\PYZob{}}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{figure.figsize}\PY{l+s+s1}{\PYZsq{}}\PY{p}{:} \PY{p}{(}\PY{l+m+mf}{8.0}\PY{p}{,} \PY{l+m+mf}{4.0}\PY{p}{)}\PY{p}{,} \PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{axes.grid}\PY{l+s+s1}{\PYZsq{}}\PY{p}{:} \PY{k+kc}{True}\PY{p}{,} \PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{lines.markersize}\PY{l+s+s1}{\PYZsq{}}\PY{p}{:} \PY{l+m+mi}{8}\PY{p}{,} \PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{lines.linewidth}\PY{l+s+s1}{\PYZsq{}}\PY{p}{:} \PY{l+m+mi}{2}\PY{p}{,} \PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{font.size}\PY{l+s+s1}{\PYZsq{}}\PY{p}{:} \PY{l+m+mi}{14}\PY{p}{\PYZcb{}} \PY{n}{plt}\PY{o}{.}\PY{n}{rcParams}\PY{o}{.}\PY{n}{update}\PY{p}{(}\PY{n}{newparams}\PY{p}{)} \PY{c+c1}{\PYZsh{} høyresiden i den logistiske ligninga} \PY{k}{def} \PY{n+nf}{f}\PY{p}{(}\PY{n}{x}\PY{p}{,}\PY{n}{r}\PY{p}{,}\PY{n}{K}\PY{p}{)}\PY{p}{:} \PY{k}{return} \PY{n}{x}\PY{o}{+}\PY{n}{r}\PY{o}{*}\PY{n}{x}\PY{o}{*}\PY{p}{(}\PY{n}{K}\PY{o}{\PYZhy{}}\PY{n}{x}\PY{p}{)} \PY{c+c1}{\PYZsh{}parametre} \PY{n}{r}\PY{o}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{2} \PY{c+c1}{\PYZsh{}vekstfaktor } \PY{n}{K}\PY{o}{=}\PY{l+m+mi}{1} \PY{c+c1}{\PYZsh{}bærekapasitet} \PY{n}{u0}\PY{o}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{4} \PY{c+c1}{\PYZsh{}startverdi} \PY{n}{nmax}\PY{o}{=}\PY{l+m+mi}{10} \PY{c+c1}{\PYZsh{} vi itererer nmax ganger} \PY{n}{n} \PY{o}{=} \PY{n}{np}\PY{o}{.}\PY{n}{arange}\PY{p}{(}\PY{l+m+mf}{0.0}\PY{p}{,} \PY{n}{nmax}\PY{o}{+}\PY{l+m+mi}{1}\PY{p}{,} \PY{l+m+mi}{1}\PY{p}{)} \PY{c+c1}{\PYZsh{} itererer u\PYZus{}\PYZob{}n+1\PYZcb{} = f(u\PYZus{}n,r,K) nmax ganger; 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