Merge branch 'Exercise-7' into 'master'

Add exercise 7 and 8

See merge request oysteikt/ma0001-matematikk!1

.gitignore

@@ -8,3 +8,4 @@
 *.out
 *.synctex.gz
 *.toc
+*.listing

Exercise 2/main.tex

@@ -1,7 +1,7 @@
 \documentclass{article}
 
-\author{Øystein Tveit}
+\newcommand{\theTitle}{MA0001 Øving 2}
-\title{MA0001 Øving 2}
+\newcommand{\theAuthor}{Øystein Tveit}
 
 \input{../lib/lib.tex}
 

Exercise 3/main.tex

@@ -1,5 +1,8 @@
 \documentclass{article}
 
+\newcommand{\theTitle}{MA0001 Øving 3}
+\newcommand{\theAuthor}{Øystein Tveit}
+
 \input{../lib/lib.tex}
 
 \begin{document}

Exercise 4/main.tex

@@ -1,5 +1,8 @@
 \documentclass{article}
 
+\newcommand{\theTitle}{MA0001 Øving 4}
+\newcommand{\theAuthor}{Øystein Tveit}
+
 \input{../lib/lib.tex}
 
 \usepackage{amssymb}

Exercise 5/main.tex

@@ -1,5 +1,8 @@
 \documentclass{article}
 
+\newcommand{\theTitle}{MA0001 Øving 5}
+\newcommand{\theAuthor}{Øystein Tveit}
+
 \input{../lib/lib.tex}
 
 \usepackage{amssymb}

Exercise 6/main.tex

@@ -1,5 +1,8 @@
 \documentclass{article}
 
+\newcommand{\theTitle}{MA0001 Øving 6}
+\newcommand{\theAuthor}{Øystein Tveit}
+
 \input{../lib/lib.tex}
 
 \begin{document}

Exercise 7/main.tex

@@ -0,0 +1,46 @@
+\documentclass{article}
+
+\newcommand{\theTitle}{MA0001 Øving 7}
+\newcommand{\theAuthor}{Øystein Tveit}
+
+\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 7/tasks/1.tex

@@ -0,0 +1,11 @@
+\begin{deloppgaver}
+  \delo
+    \[f'(x) = 3\]
+
+  \delo
+    \[f^{-1}(x) = \frac{x}{3} - 1\]
+  
+  \delo
+    \[\frac{d}{dx} f^{-1}(x) = \frac{1}{3} \]
+  
+\end{deloppgaver}

Exercise 7/tasks/2.tex

@@ -0,0 +1,12 @@
+\[ f(x) = x^2 \]
+
+\begin{align*}
+  \frac{d}{dx} f(x) &= \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x} \\[2ex]
+  &= \lim_{\Delta x \to 0} \frac{(x+\Delta x)^2 - x^2}{\Delta x} \\[2ex]
+  &= \lim_{\Delta x \to 0} \frac{x^2 + 2x \Delta x + {\Delta x}^2 - x^2}{\Delta x} \\[2ex]
+  &= \lim_{\Delta x \to 0} \frac{2x \Delta x + {\Delta x}^2}{\Delta x} \\[2ex]
+  &= \lim_{\Delta x \to 0} \frac{2x + \Delta x}{1} \\[2ex]
+  &= \lim_{\Delta x \to 0} \frac{2x + \Delta x}{1} \\[2ex]
+  &= 2x + 0 \\[2ex]
+  &= 2x \\[2ex]
+\end{align*}

Exercise 7/tasks/3.tex

@@ -0,0 +1,6 @@
+\[ h(x) = -\frac{1}{4}x^4 + \frac{2}{3}x^3 + 2x^2 - \sqrt{2}x + \pi^e \]
+
+\begin{align*}
+  h'(x) &= 4 \cdot -\frac{1}{4}x^{4-1} + 3 \cdot \frac{2}{3}x^{3-1} + 2\cdot2x^{2-1} - 1 \cdot \sqrt{2}x^{1-1} + 0 \cdot \pi^e x^{0-1} \\
+  &= -x^{3} + 2x^{2} + 4x - \sqrt{2} \\
+\end{align*}

Exercise 7/tasks/4.tex

@@ -0,0 +1,12 @@
+\begin{align*}
+  \left(x^{1/3}-x\right) \cdot \left(x^{2/3}+x^2\right)
+  &= x^{1/3} \cdot x^{2/3} + x^{1/3} \cdot x^2 - x\cdot x^{2/3} - x \cdot x^{2} \\
+  &= x + x^{7/3} - x^{5/3} - x^3 \\
+\end{align*}
+
+
+\begin{align*}
+  \frac{d}{dx} \left( x + x^{7/3} - x^{5/3} - x^3 \right)
+  &= 1 + \frac{7}{3}x^{4/3} - \frac{5}{3}x^{2/3} - 3x^2 \\
+  &=  - 3x^2 - \frac{5}{3}x^{2/3} + \frac{7}{3}x^{4/3} + 1 
+\end{align*}

Exercise 7/tasks/5.tex

@@ -0,0 +1,14 @@
+\[ f(x): \mathbb{R} \to \mathbb{R} = \begin{cases}
+  \sqrt{x},\qquad &x \ge 0 \\
+  x, &x < 0
+\end{cases} \]
+
+Ettersom \[\sqrt{0} = 0\] og \[\lim_{x\to 0}x = 0\] er funksjonen kontinuerlig.
+
+Den deriverte av den første delen av funksjonen blir
+\begin{align*}
+  \frac{d}{dx} \sqrt{x} &= \frac{1}{2} x^{-\frac{1}{2}} \\[2ex]
+  &= \frac{1}{2\sqrt{x}}
+\end{align*}
+
+Ved $x=0$ blir \[\frac{1}{2\sqrt{x}} = \frac{1}{0}\] som er udefinert, og dermed er ikke den funksjonen deriverbar for denne $x$-verdien

Exercise 8/main.tex

@@ -0,0 +1,47 @@
+\documentclass{article}
+
+\newcommand{\theTitle}{Øystein Tveit}
+\newcommand{\theAuthor}{MA0001 Øving 8}
+
+\input{../lib/lib.tex}
+\input{../lib/lst.tex}
+
+\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 8/scripts/4.py

@@ -0,0 +1,19 @@
+from math import e
+
+def f(x): return e ** x - 2
+def df(x): return e ** x
+
+def Newton(a, f, df):
+  return a - f(a)/df(a)
+
+def main():
+  i = 0
+  a = [5]
+  while True:
+    a.append(Newton(a[i], f, df))
+    if abs(a[i] - a[i-1]) < 0.001:
+      print(f'Iterasjoner: {i-1}, a: {a[i-1]}')
+      break
+    i += 1
+    
+main()

Exercise 8/tasks/1.tex

@@ -0,0 +1,10 @@
+
+\[f(x) = \frac{1}{2}x^3 -7x +22\]
+
+\[f'(x) = \frac{3}{2}x^2 -7\]
+
+\[f''(x) = 3x \]
+
+\[f'''(x) = 3 \]
+
+\[f^{(100)} = 0\]

Exercise 8/tasks/2.tex

@@ -0,0 +1,53 @@
+\begin{enumerate}[label=\arabic*.]
+  \item \[ f(x) = ln\left(\frac{1}{x^2}\right) \]
+
+    La $u = \frac{1}{x^2}$
+
+    Jeg bruker kjerneregelen:
+
+    \begin{align*}
+      \frac{dy}{dx} &= \frac{dy}{du} \cdot \frac{du}{dx} \\
+      &= \frac{d}{du} ln(u) \cdot \frac{d}{dx} \frac{1}{x^2} \\
+      &= \frac{1}{u} -2 \frac{1}{x^3} \\
+      &= \frac{1}{\frac{1}{x^2}} -2 \frac{1}{x^3} \\
+      &= x^2 -2 \frac{1}{x^3}
+    \end{align*}
+
+
+
+  \item \[g(x) = \frac{1 + \sin x}{1 + e^x + x^2}\]
+
+  Jeg bruker kvotientregelen:
+  
+  $u = 1 + \sin x$
+
+  $v = 1 + e^x + x^2$
+
+  \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{ \frac{d}{dx}(u) \cdot v - u \cdot \frac{d}{dx}(v)}{v^2} \]
+
+  \begin{align*}
+    \frac{d}{dx} u &= \frac{d}{dx} 1 + \sin x \\
+    &= \cos x
+  \end{align*}
+
+  \begin{align*}
+    \frac{d}{dx} v &= \frac{d}{dx} 1 + e^x + x^2 \\
+    &= e^x + 2x
+  \end{align*}
+
+  \begin{align*}
+    \frac{dg}{dx} &= \frac{(\cos x)(1+e^x+x^2) - (1+\sin x)(e^x + 2x)}{\left(1 + e^x + x^2\right)^2}
+  \end{align*}
+  
+  \item \[h(x) = \sqrt{1 + \sqrt{x}}\]
+
+    $u = 1 + \sqrt{x}$
+
+    \begin{align*}
+      \frac{dy}{dx} &= \frac{dy}{dv} \cdot \frac{du}{dx} \\
+      &= \frac{d}{du} \sqrt{u} \cdot \frac{d}{dx} \left( 1 + \sqrt{x} \right) \\
+      &= \frac{1}{2\sqrt{u}} \cdot \frac{1}{2\sqrt{x}} \\
+      &= \frac{1}{2\sqrt{1+\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} \\
+      &= \frac{1}{4\sqrt{1+\sqrt{x}} \sqrt{x}} 
+    \end{align*}
+\end{enumerate}

Exercise 8/tasks/3.tex

@@ -0,0 +1,41 @@
+\[f(t) = \frac{t^2 -1}{t+1} + 6t^{1/3} + \sqrt{\sin t} + 4^t\]
+
+Jeg deriverer funksjonen ledd for ledd
+
+Ledd 1:
+
+\begin{align*}
+  \frac{d}{dt} \frac{t^2-1}{t+1} &= \frac{d}{dt}\frac{(t+1)(t-1)}{t+1} \\
+  &= \frac{d}{dt} t-1 \\
+  &= 1
+\end{align*}
+
+Ledd 2:
+
+\begin{align*}
+  \frac{d}{dt} 6t^{1/3} &= \frac{1}{3} \cdot 6t^{(1/3 - 1)} \\
+  &= 2t^{-2/3}\\
+  &= \frac{2}{\sqrt[3]{t^2}}
+\end{align*}
+
+Ledd 3:
+  
+\[ \frac{d}{dt} \sqrt{\sin t} \]
+
+$u = \sin t$
+
+\begin{align*}
+  \frac{dy}{dt} &= \frac{dy}{du} \cdot \frac{du}{dt} \\
+  &= \frac{d}{du} \sqrt{u} \cdot \frac{d}{dt} \sin t \\
+  &= \frac{1}{2\sqrt{u}} \cdot \cos t \\
+  &= \frac{\cos t}{2\sqrt{\sin t}}
+\end{align*}
+
+Ledd 4:
+
+\begin{align*}
+  \frac{d}{dt} 4^t = 4^t ln(t) \\
+\end{align*}
+
+\[\frac{df}{dx} = 1 + \frac{2}{\sqrt[3]{t^2}} + \frac{\cos t}{2\sqrt{\sin t}} + 4^t ln(t)\]
+  

Exercise 8/tasks/4.tex

@@ -0,0 +1,7 @@
+
+Her har jeg brukt python for å løse oppgaven.
+
+\pythonBlock{./scripts/4.py}
+
+Output:
+\fbox{Iterasjoner: 7, a: 0.6932882713164431}

Exercise 8/tasks/5.tex

@@ -0,0 +1,18 @@
+\[f(y) = a^y \]
+\[f'(y) = a^y \ln a\]
+\[f^{-1}(y) = \log_{a} y\]
+
+\begin{align*}
+  \frac{d}{dy}f^{-1}(y) &= \frac{1}{f' \circ f^{-1}(y)} \\[2ex]
+  &= \frac{1}{a^{\log_{a} y} \ln a} \\[2ex]
+  &= \frac{1}{y \ln a} \\
+\end{align*}
+
+Gitt at vi vet at $\frac{d}{dx} ln(x) = \frac{1}{x}$, så kan vi også løse det på følgende måte
+
+\begin{align*}
+  \frac{d}{dy} \log_{a} y &= \frac{d}{dy} \frac{\ln y}{\ln a} \\[2ex]
+  &= \frac{1}{\ln a} \cdot \frac{d}{dy} \ln y \\[2ex]
+  &= \frac{1}{\ln a} \cdot \frac{1}{y} \\[2ex]
+  &= \frac{1}{y \ln a} 
+\end{align*}

lib/lib.tex

@@ -11,22 +11,22 @@
 
 \definecolor{ntnublue}{RGB}{0,80,158}
 
-\input{../lib/geometry.tex}
 \input{../lib/header.tex}
+\input{../lib/geometry.tex}
 \input{../lib/math.tex}
 \usetikzlibrary{angles, quotes}
 
 \hypersetup{
-    colorlinks=true,
+  colorlinks=true,
-    linkcolor=blue,
+  linkcolor=blue,
-    filecolor=magenta,
+  filecolor=magenta,
-    urlcolor=blue,
+  urlcolor=blue,
-}
+  }
   
-\pgfplotsset{compat=newest}
+  \pgfplotsset{compat=newest}
   
-\author{Øystein Tveit}
+  \author{\theAuthor}
-\title{MA0001 Øving 6}
+  \title{\theTitle}
 
 \input{../lib/titling.tex}
 

lib/lst.tex

@@ -0,0 +1,41 @@
+\usepackage{listings}
+
+\definecolor{background}{RGB}{39, 40, 34}
+\definecolor{string}{RGB}{230, 219, 116}
+\definecolor{comment}{RGB}{117, 113, 94}
+\definecolor{normal}{RGB}{248, 248, 242}
+\definecolor{identifier}{RGB}{166, 226, 46}
+
+\lstdefinestyle{monokaiPython}{
+  frame=none,
+  language=python,                		% choose the language of the code
+  numbers=left,                   		% where to put the line-numbers
+  stepnumber=1,                   		% the step between two line-numbers.        
+  numbersep=5pt,                  		% how far the line-numbers are from the code
+  numberstyle=\color{white}\ttfamily,
+  backgroundcolor=\color{background},  		% choose the background color. You must add \usepackage{color}
+  showspaces=false,               		% show spaces adding particular underscores
+  showstringspaces=false,         		% underline spaces within strings
+  showtabs=false,                 		% show tabs within strings adding particular underscores
+  tabsize=2,                      		% sets default tabsize to 2 spaces
+  captionpos=b,                   		% sets the caption-position to bottom
+  breaklines=true,                		% sets automatic line breaking
+  breakatwhitespace=true,         		% sets if automatic breaks should only happen at whitespace
+  title=\color{white}\lstname,                 		% show the filename of files included with \lstinputlisting;
+  basicstyle=\color{normal}\ttfamily,					% sets font style for the code
+  keywordstyle=\color{magenta}\ttfamily,	% sets color for keywords
+  stringstyle=\color{string}\ttfamily,		% sets color for strings
+  commentstyle=\color{comment}\ttfamily,	% sets color for comments
+  emph={format_string, eff_ana_bf, permute, eff_ana_btr},
+  emphstyle=\color{identifier}\ttfamily
+}
+
+\usepackage{framed}
+\definecolor{shadecolor}{named}{background} 
+
+\newcommand{\pythonBlock}[1]
+{
+  \begin{shaded}
+    \lstinputlisting[style=monokaiPython]{#1}
+  \end{shaded}
+}