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latex/problems/problem6.tex

@@ -3,6 +3,46 @@
 \subsection*{a)}
 
 % Use Gaussian elimination, and then use backwards substitution to solve the equation
+Renaming the sub-, main-, and supdiagonal of matrix $\boldsymbol{A}$ 
+\begin{align*}
+    \vec{a} &= [a_{2}, a_{3}, ..., a_{n-1}, a_{n}] \\
+    \vec{b} &= [b_{1}, b_{2}, b_{3}, ..., b_{n-1}, b_{n}] \\
+    \vec{c} &= [c_{1}, c_{2}, c_{3}, ..., c_{n-1}] \\
+\end{align*}
+
+Following Thomas algorithm for gaussian elimination, we first perform a forward sweep
+\begin{algorithm}[H]
+    \caption{Foreward sweep}\label{algo:foreward}
+    \begin{algorithmic}
+        \State Create new vector \vec{\hat{b}} of length n.
+        \State \hat{b}[0] &= b[0] \Comment{Handle first element in main diagonal outside loop}
+        \For{$i = 1, ..., n-1$}
+        \State d = \frac{a[i-1]}{b[i-1]}
+        \State b -= d*(*sup_diag)(i-1);
+        (*g_vec)(i) -= d*(*g_vec)(i-1);
+        \EndFor
+        \While{Some condition}
+        \State Do something more here 
+        \EndWhile
+        \State Maybe even some more math here, e.g $\int_0^1 f(x) \dd x$
+    \end{algorithmic}
+\end{algorithm}
+Here the index i does not refer to the element in the vector, ie. $b_{i}$, but to the index in the vector where the element is found ie. $b[i] = b_{i+1}$.  
+
+\begin{align*}
+    \hat{b}_{1} &= b_{1} \\
+    \hat{b}_{2} &= b_{2} - \frac{a_{2}}{\hat{b}_{1}} \cdot c_{1} \\
+    \vdots & \\
+    \hat{b}_{i} &= b_{i} - \frac{a_{i}}{\hat{b}_{i-1}} \cdot c_{i-1} \\
+    \vdots & \\
+\end{align*}
+
+
+% as g_hat also make use of the ratio (ai/bi-1_hat) it only needs to be calculated once
+
+% for n steps we have n+1 points, that is values of x to evaluate u(x) at, when boundary values are known 
+% we have to perform (n+1)-2 = n-1 calculations (equal number of col in matrix). Dealing
+% with a quadratic matrix number of col = number of rows 
 
 \subsection*{b)}