\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 solve for $\vec{v}$. 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*}

where $v_{i} = v(x_{i})$ and $f_{i} = f(x_{i})$. 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*}
    Since the boundary values are equal to $0$ the RHS can be renamed $g_{i} = h^{2} f_{i}$ for all $i$. 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 $n=m-2$.