Add exercise

.gitignore

@@ -1,6 +1,7 @@
 
 # Latex build files
 *.aux
+*.auxlock
 *.fdb_latexmk
 *.fls
 *.log

Exercise 5/figures/1a.tex

@@ -0,0 +1,20 @@
+\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}

Exercise 5/figures/1b.tex

@@ -0,0 +1,38 @@
+\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}

Exercise 5/main.tex

@@ -0,0 +1,43 @@
+\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}

Exercise 5/tasks/1.tex

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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}

Exercise 5/tasks/2.tex

@@ -0,0 +1,22 @@
+\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*}

Exercise 5/tasks/3.tex

@@ -0,0 +1,19 @@
+\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*}

Exercise 5/tasks/4.tex

@@ -0,0 +1,13 @@
+\[ a_{k+1} = 2-\frac{1}{a_k} \]
+
+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 $a = 1$

Exercise 5/tasks/5.tex

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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*}

lib/lib.tex

@@ -26,7 +26,7 @@
 \pgfplotsset{compat=newest}
 
 \author{Øystein Tveit}
-\title{MA0001 Øving 4}
+\title{MA0001 Øving 5}
 
 \input{../lib/titling.tex}