Remove comment for problem 3

1 solve problem 1

.gitignore

@@ -30,3 +30,9 @@
 *.exe
 *.out
 *.app
+
+# Latex
+*.aux
+*.log
+*.out
+*.bib

latex/assignment_1.tex

@@ -74,8 +74,8 @@
 
 \begin{document}
 
-\title{Title of the document}      % self-explanatory
+\title{Project 1}      % self-explanatory
-\author{Cory Balaton \& Janita}          % self-explanatory
+\author{Cory Balaton \& Janita Willumsen}          % self-explanatory
 \date{\today}                             % self-explanatory
 \noaffiliation                            % ignore this, but keep it.
 
@@ -86,7 +86,38 @@
     
 \section*{Problem 1}
 
-% Do the double integral
+\begin{align*}
+    u(x) &= \int \int \frac{d^2 u}{dx^2} dx^2\\
+         &= \int \int -100 e^{-10x} dx^2 \\
+         &= \int \frac{-100 e^{-10x}}{-10} + c_1 dx \\
+         &= \int 10 e^{-10x} + c_1 dx \\
+         &= \frac{10 e^{-10x}}{-10} + c_1 x + c_2 \\
+         &= -e^{-10x} + c_1 x + c_2
+\end{align*}
+
+Using the boundary conditions, we can find $c_1$ and $c_2$ as shown below:
+
+\begin{align*}
+    u(0) &= 0 \\
+    -e^{-10 \cdot 0} + c_1 \cdot 0 + c_2 &= 0 \\
+    -1 + c_2 &= 0 \\
+    c_2 &= 1
+\end{align*}
+
+\begin{align*}
+    u(1) &= 0 \\
+    -e^{-10 \cdot 1} + c_1 \cdot 1 + c_2 &= 0 \\
+    -e^{-10} + c_1 + c_2 &= 0 \\
+    c_1 &= e^{-10} - c_2\\
+    c_1 &= e^{-10} - 1\\
+\end{align*}
+
+Using the values that we found for $c_1$ and $c_2$, we get
+
+\begin{align*}
+    u(x) &= -e^{-10x} + (e^{-10} - 1) x + 1 \\
+         &= 1 - (1 - e^{-10}) - e^{-10x}
+\end{align*}
 
 \section*{Problem 2}
 
@@ -94,7 +125,40 @@
 
 \section*{Problem 3}
 
-% Show how it's derived and where we found the derivation.
+To derive the discretized version of the Poisson equation, we first need
+the taylor expansion for $u(x)$ around $x + h$ and $x - h$.
+
+\begin{align*}
+    u(x+h) &= u(x) + u'(x) h + \frac{1}{2} u''(x) h^2 + \frac{1}{6} u'''(x) h^3 + \mathcal{O}(h^4)
+\end{align*}
+
+\begin{align*}
+    u(x-h) &= u(x) - u'(x) h + \frac{1}{2} u''(x) h^2 - \frac{1}{6} u'''(x) h^3 + \mathcal{O}(h^4)
+\end{align*}
+
+If we add the equations above, we get this new equation:
+
+\begin{align*}
+    u(x+h) + u(x-h) &= 2 u(x) + u''(x) h^2 + \mathcal{O}(h^4) \\
+    u(x+h) - 2 u(x) + u(x-h) + \mathcal{O}(h^4) &= u''(x) h^2 \\
+    u''(x) &= \frac{u(x+h) - 2 u(x) + u(x-h)}{h^2} + \mathcal{O}(h^2) \\
+    u_i''(x) &= \frac{u_{i+1} - 2 u_i + u_{i-1}}{h^2} + \mathcal{O}(h^2) \\
+\end{align*}
+
+We can then replace $\frac{d^2u}{dx^2}$ with the RHS (right-hand side) of the equation:
+
+\begin{align*}
+    - \frac{d^2u}{dx^2} &= 100 e^{-10x} \\
+    \frac{ - u_{i+1} + 2 u_i - u_{i-1}}{h^2} + \mathcal{O}(h^2) &= 100 e^{-10x} \\
+\end{align*}
+
+And lastly, we leave out $\mathcal{O}(h^2)$ and change $u_i$ to $v_i$ to 
+differentiate between the exact solution and the approximate solution, 
+and get the discretized version of the equation:
+
+\begin{align*}
+align*    \frac{ - u_{i+1} + 2 u_i - u_{i-1}}{h^2} &= 100 e^{-10x} \\
+\end{align*}
 
 \section*{Problem 4}
 
@@ -106,9 +170,9 @@
 
 \subsection*{b)}
 
-\section{Problem 6}
+\section*{Problem 6}
 
-\subsection{a)}
+\subsection*{a)}
 
 % Use Gaussian elimination, and then use backwards substitution to solve the equation