Fixed the explanation for renaming f

latex/problems/problem4.tex

@@ -2,7 +2,7 @@
 
 % 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   
+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} \\
@@ -10,7 +10,7 @@ The value of $u(x_{0})$ and $u(x_{1})$ is known, using the discretized equation
     - 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
+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} \\
@@ -40,5 +40,5 @@ We now have a number of linear eqations, corresponding to the number of unknown
       \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 $n=m-2$.
+    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 $n=m-2$.