Fixed typo's

parent 7a7b83d6
...@@ -21,7 +21,7 @@ speed: speed.c ...@@ -21,7 +21,7 @@ speed: speed.c
touch $@ touch $@
pr: extra_precision.o pr: extra_precision.o
$(CC) $(FLAGS) -o $@ $^ $(CC) $(FLAGS) -mfpmath=387 -O2 -o $@ $^
fp: floating_point.o fp: floating_point.o
$(CC) $(FLAGS) -o $@ $^ $(CC) $(FLAGS) -o $@ $^
......
...@@ -122,20 +122,20 @@ Type & N & Forward & Backward \\ ...@@ -122,20 +122,20 @@ Type & N & Forward & Backward \\
\noindent \textbf{Observations} \noindent \textbf{Observations}
\begin{itemize} \begin{itemize}
\item Since the results for the \texttt{double} datatype are eaqual for both \item Since the results for the \texttt{double} data type are equal for both
the forward and backward summation approach, we can say that these are the the forward and backward summation approach, we can say that these are the
correct results. correct results.
\item For the \texttt{float} data type, we observe that the backward approach \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 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 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 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 large. The value will keep growing until the significance of $1/i$ is too
small to add to the result. From this point, no more $1/i$ will be added to 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. the result because the significance of the individual numbers is too small.
However, the sum of the ignored numbers would be a large enough number to 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: 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 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 are added. The remaining imprecision is probably due to rounding problems and
the fact that $1/10^8$ and a range of larger numbers are represented as 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 zeroes in \texttt{float} representation and therefore not added to the
result. This problem does not occur when using the \texttt{double} data type result. This problem does not occur when using the \texttt{double} data type
......
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