Finished exercise 4

latex/problems/problem4.tex

@@ -1,3 +1,44 @@
 \section*{Problem 4}
 
 % Show that each iteration of the discretized version naturally creates a matrix equation.
+
+The value of $u(x_{0})$ and $u(x_{1})$ is known, using the discretized equation we can approximate the value of $f(x_{i}) = f_{i}$. This will result in a set of equations   
+\begin{align*}
+    - v_{0} + 2 v_{1} - v_{2} &= h^{2} \cdot f_{1} \\
+    - v_{1} + 2 v_{2} - v_{3} &= h^{2} \cdot f_{2} \\
+    \vdots & \\
+    - v_{m-2} + 2 v_{m-1} - v_{m} &= h^{2} \cdot f_{m-1} \\
+\end{align*}
+
+Rearranging the first and last equation, moving terms of known boundary values to the RHS
+\begin{align*}
+    2 v_{1} - v_{2} &= h^{2} \cdot f_{1} + v_{0} \\
+    - v_{1} + 2 v_{2} - v_{3} &= h^{2} \cdot f_{2} \\
+    \vdots & \\
+    - v_{m-2} + 2 v_{m-1} &= h^{2} \cdot f_{m-1} + v_{m} \\
+\end{align*}
+
+We now have a number of linear eqations, corresponding to the number of unknown values, which can be represented as an augmented matrix 
+\begin{align*}
+    \left[
+      \begin{matrix}
+        2v_{1} & -v_{2} & 0 & \dots & 0 \\
+        -v_{1} & 2v_{2} & -v_{3} & 0 & \\
+        0 & -v_{2} & 2v_{3} & -v_{4} & \\
+        \vdots  & & & \ddots & \vdots \\
+        0 & & & -v_{m-2} & 2v_{m-1} \\
+      \end{matrix}
+      \left|
+        \,
+        \begin{matrix}
+          g_{1}  \\
+          g_{2}  \\
+          g_{2}  \\
+          \vdots  \\
+          g_{m-1}  \\
+        \end{matrix}
+      \right.
+    \right]
+    \end{align*}
+    where $g_{i} = h^{2} f_{i}$. An augmented matrix can be represented as $\boldsymbol{A} \vec{x} = \vec{b}$. In this case $\boldsymbol{A}$ is the coefficient matrix with a tridiagonal signature $(-1, 2, -1)$ and dimension $n \cross n$, where $n=m-2$.
+