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Oystein Kristoffer Tveit 2020-09-27 11:29:20 +02:00
parent c22cf7206e
commit b2e3d7ff2f
11 changed files with 204 additions and 1 deletions
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1
.gitignore vendored
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# Latex build files
*.aux
*.auxlock
*.fdb_latexmk
*.fls
*.log

20
Exercise 5/figures/1a.tex Normal file
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\begin{tikzpicture}
\begin{axis}[
xlabel = $x$, ylabel = $y$,
width = 10cm, height=6cm,
axis lines = middle,
grid,
xmin=-1.5, xmax=5.5,
ymin=-1.5, ymax=1.5,
legend pos=outer north east,
samples=10000
]
\addplot+ [
no marks,
domain = 0:5.5,
]{sin(deg(sqrt(x) * (e^(x))))};
\addlegendentry{$y=sin(\sqrt{x}e^x)$}
\end{axis}
\end{tikzpicture}

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Exercise 5/figures/1b.tex Normal file
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\begin{tikzpicture} [
% declare function={
% func(\x)= (\x >= -3) * (\x^2) +
% and(\x < -3) * (-\x);
% }
]
\begin{axis}[
xlabel = $x$, ylabel = $y$,
width = 10cm, height=10cm,
axis lines = middle,
grid,
xmin=-9, xmax=5,
ymin=-7, ymax=7,
legend pos=outer north east,
legend style={cells={align=left}}
]
\addplot [
blue,
no marks,
domain= -3:100
]{2*x};
\addplot [
blue,
no marks,
domain= -100:-3
]{-x};
\addplot+[blue,mark=*,mark options={fill=white}] (-3,3);
\addplot+[blue,mark=*] (-3,-6);
\addlegendentry{$g(x) = \begin{cases}
2x\text{, hvis } x \geq -3 \\
-x \text{, hvis } x < -3
\end{cases}$ \\}
\end{axis}
\end{tikzpicture}

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Exercise 5/main.pdf Normal file
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Exercise 5/main.tex Normal file
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\documentclass{article}
\input{../lib/lib.tex}
\usepackage{amssymb}
\begin{document}
\thispagestyle{plain}
\tittel
\tableofcontents
\newpage
\section{Forberedende oppgaver}
\begin{oppgaver}
\oppg
\input{tasks/1.tex}
\end{oppgaver}
\newpage
\section{Innleveringsoppgaver}
\begin{oppgaver}
\setoppg{1}
\oppg
\input{tasks/2.tex}
\oppg
\input{tasks/3.tex}
\oppg
\input{tasks/4.tex}
\oppg
\input{tasks/5.tex}
\end{oppgaver}
\end{document}

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Exercise 5/tasks/1.tex Normal file
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\begin{deloppgaver}
\delo
\[f(x) = sin(\sqrt{x}e^x)\]
er kontinuerlig for alle punkt hvor den er definert
\begin{graphbox}
\input{figures/1a.tex}
\end{graphbox}
\delo
\[g(x) = \begin{cases}
2x\text{, hvis } x \geq -3 \\
-x \text{, hvis } x < -3
\end{cases}\]
er ikke kontinuerlig ved $x = -3$, hvor $y$-verdien gjør et hopp fra $3$ til $-6$
\begin{graphbox}
\input{figures/1b.tex}
\end{graphbox}
\end{deloppgaver}

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Exercise 5/tasks/2.tex Normal file
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\begin{align*}
\lim_{n\to\infty} a_n &= \lim_{n\to\infty} \frac{\sqrt{n^2+6n+9}}{\sqrt{n^2+1}} \\[1ex]
\lim_{n\to\infty} a_n &= \lim_{n\to\infty} \sqrt{\frac{n^2+6n+9}{n^2+1}} \\
\end{align*}
Ettersom $\lim\limits_{n\to\infty} (a_n b_n) = (\lim\limits_{n\to\infty} a_n)(\lim\limits_{n\to\infty}b_n)$
\[\lim_{n\to\infty} a_n = \sqrt{ \lim_{n\to\infty} \frac{n^2+6n+9}{n^2+1}} \]
Ettersom $\lim\limits_{n\to\infty} \frac{a_n}{b_n}= \frac{\lim\limits_{n\to\infty} a_n}{\lim\limits_{n\to\infty}b_n}$ \\
I tillegg deler vi både teller og nevner på $n^2$
\[ \lim_{n\to\infty} a_n = \sqrt{ \frac{ \lim_{n\to\infty} 1+\frac{6}{n}+\frac{9}{n^2}}{ \lim_{n\to\infty} 1+\frac{1}{n^2}}} \]
Både $\frac{9}{n^2}$, $\frac{6}{n}$ og $\frac{1}{n^2}$ går mot $0$ når $n \to \infty$
\begin{align*}
\lim_{n\to\infty} a_n &= \sqrt{ \frac{ 1+0+0}{ 0+1}} \\
&= \sqrt{1} \\
&= 1 \\
\end{align*}

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Exercise 5/tasks/3.tex Normal file
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\begin{align*}
\lim_{n\to\infty} a_n &= \lim_{n\to\infty} \left(\sqrt{n^2-n+9} - \sqrt{n^2+9}\right) \\[1ex]
&= \lim_{n\to\infty} \left(\sqrt{n^2-n+9} - \sqrt{n^2+9}\right)
\frac{\sqrt{n^2-n+9} + \sqrt{n^2+9}}{\sqrt{n^2-n+9} + \sqrt{n^2+9}} \\[1ex]
&= \lim_{n\to\infty} \frac{(n^2-n+9) - (n^2+9)}{\sqrt{n^2-n+9} + \sqrt{n^2+9}} \\[1ex]
&= \lim_{n\to\infty} \frac{-n}{\sqrt{n^2-n+9} + \sqrt{n^2+9}} \\[1ex]
&= -\lim_{n\to\infty} \frac{n}{\sqrt{n^2-n+9} + \sqrt{n^2+9}} \\[1ex]
&= -\lim_{n\to\infty} \frac{1}{n^{-1}\sqrt{n^2-n+9} + n^{-1}\sqrt{n^2+9}} \\[1ex]
&= -\lim_{n\to\infty} \frac{1}{\sqrt{n^{-2}(n^2-n+9)} + \sqrt{n^{-2}(n^2+9)}} \\[1ex]
&= -\lim_{n\to\infty} \frac{1}{\sqrt{1-\frac{1}{n}+\frac{9}{n^2}} + \sqrt{1+\frac{9}{n^2}}} \\[1ex]
\end{align*}
Dermed blir
\begin{align*}
-\lim_{n\to\infty} \frac{1}{\sqrt{1-\frac{1}{n}+\frac{9}{n^2}} + \sqrt{1+\frac{9}{n^2}}} &= -\frac{1}{\sqrt{1-0+0} + \sqrt{1+0}} \\
&= -\frac{1}{1 + 1} \\
&= -\frac{1}{2} \\
\end{align*}

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Exercise 5/tasks/4.tex Normal file
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\[ a_{k+1} = 2-\frac{1}{a_k} \]
Fikspunktet vil være punktet hvor $a_{k+1} = a_k$
\begin{align*}
a &= 2 - \frac{1}{a} \\
a - 2 &= - \frac{1}{a} \\
a^2 -2a &= - 1 \\
a^2 -2a + 1 &= 0 \\
(a-1)^2 &= 0 \quad \Leftrightarrow \quad a = 1
\end{align*}
$a$ har ett fikspunkt ved $a = 1$

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Exercise 5/tasks/5.tex Normal file
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\[
\begin{cases}
b_0 = 1 \\
b_1 = 2 \\
b_{n+1} = b_n + 2 \cdot b_{n-1}
\end{cases}
\]
\begin{align*}
b_2 &= b_1 + 2 \cdot b_{0} = 2 + 2 \cdot 1 = 4 \\
b_3 &= b_2 + 2 \cdot b_{1} = 4 + 2 \cdot 2 = 8 \\
b_4 &= b_3 + 2 \cdot b_{2} = 8 + 2 \cdot 4 = 16 \\
b_5 &= b_4 + 2 \cdot b_{3} = 16 + 2 \cdot 8 = 32 \\
b_6 &= b_5 + 2 \cdot b_{4} = 32 + 2 \cdot 16 = 64 \\
b_7 &= b_6 + 2 \cdot b_{5} = 64 + 2 \cdot 32 = 128
\end{align*}
Jeg gjetter at $f(n) = 2^n$.
\begin{align*}
b_n + 2 \cdot b_{n-1} &= 2^n + 2 \cdot 2^{n-1} \\
&= 2^n + 2^n \\
&= 2 \cdot 2^{n} \\
&= 2^{n+1} \\
&= b_{n+1}
\end{align*}

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lib/lib.tex
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\pgfplotsset{compat=newest}
\author{Øystein Tveit}
\title{MA0001 Øving 4}
\title{MA0001 Øving 5}
\input{../lib/titling.tex}