Finished question 3.

simmod/ass1/report.tex

@@ -104,6 +104,10 @@ Type                 & Operator & Million ops/sec \\
 \section{Summation} % {{{
 \label{sec:Summation}
 
+We've calculated $\sum_{i=1}^{N}\frac{1}{i}$ for $N = 10^8$ and $N = 2 \cdot
+10^8$ using a forward and backward summation approach, with data types
+\texttt{float} and \texttt{double}. The results of this are in the table below.
+
 \begin{table}[H]
 \begin{tabular}{l|lll}
 Type            & N              & Forward    & Backward \\
@@ -116,6 +120,34 @@ Type            & N              & Forward    & Backward \\
 \caption{Results of various summation approaches on floats and doubles.}
 \end{table}
 
+\noindent \textbf{Observations}
+\begin{itemize}
+    \item Since the results for the \texttt{double} datatype are eaqual for both
+    the forward and backward summation approach, we can say that these are the
+    correct results.
+    \item For the \texttt{float} data type, we observe that the backward approach
+    yields a higher result than the forward approach. This can be explained as
+    follows. When using the forward approach, we start with a small $i$, thus
+    with a large $1/i$. This means that the initial value of \texttt{sum} is
+    large. The value will keep growing untill the significance of $1/i$ is too
+    small to add to the result. From this point, no more $1/i$ will be added to
+    the result because the significane of the individual numbers is too small.
+    However, the sum of the ignored numbers would be a large enough number to
+    add to the result. This is why the backward approach yields a higher number:
+    the sum of the ignored numbers is computed and later the larger numbers
+    are added. The remaining inprecision is probably due to rounding problems and
+    the fact that $1/10^8$ and a range of larger numbers are represented as
+    zeroes in \texttt{float} representation and therefore not added to the
+    result. This problem does not occur when using the \texttt{double} data type
+    which has a higher precision, therefore yielding the (correct) higher
+    result.
+    \item We can see that both approaches yield the same result for $N = 10^8$
+    and $N = 2 \cdot 10^8$ when using the \texttt{float} data type. This is due
+    to the same problem as described above: all numbers in $[\frac{1}{10^8},
+    \frac{1}{2 \cdot 10^8}]$ are also represented as zero and therefore not added
+    to the result.
+\end{itemize}
+
 % }}}
 
 \end{document}