Added example to the documentation for the NormTypes enum (L1, L2, INF)

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Jan S. (Milania1) 2015-10-15 19:16:07 +02:00
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2 changed files with 19 additions and 1 deletions

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@ -164,7 +164,25 @@ enum DecompTypes {
\f[norm = \forkthree{\frac{\|\texttt{src1}-\texttt{src2}\|_{L_{\infty}} }{\|\texttt{src2}\|_{L_{\infty}} }}{if \(\texttt{normType} = \texttt{NORM_RELATIVE_INF}\) }
{ \frac{\|\texttt{src1}-\texttt{src2}\|_{L_1} }{\|\texttt{src2}\|_{L_1}} }{if \(\texttt{normType} = \texttt{NORM_RELATIVE_L1}\) }
{ \frac{\|\texttt{src1}-\texttt{src2}\|_{L_2} }{\|\texttt{src2}\|_{L_2}} }{if \(\texttt{normType} = \texttt{NORM_RELATIVE_L2}\) }\f]
*/
As example for one array consider the function \f$r(x)= \begin{pmatrix} x \\ 1-x \end{pmatrix}, x \in [-1;1]\f$.
The \f$ L_{1}, L_{2} \f$ and \f$ L_{\infty} \f$ norm for the sample value \f$r(-1) = \begin{pmatrix} -1 \\ 2 \end{pmatrix}\f$
is calculated as follows
\f{align*}
\| r(-1) \|_{L_1} &= |-1| + |2| = 3 \\
\| r(-1) \|_{L_2} &= \sqrt{(-1)^{2} + (2)^{2}} = \sqrt{5} \\
\| r(-1) \|_{L_\infty} &= \max(|-1|,|2|) = 2
\f}
and for \f$r(0.5) = \begin{pmatrix} 0.5 \\ 0.5 \end{pmatrix}\f$ the calculation is
\f{align*}
\| r(0.5) \|_{L_1} &= |0.5| + |0.5| = 1 \\
\| r(0.5) \|_{L_2} &= \sqrt{(0.5)^{2} + (0.5)^{2}} = \sqrt{0.5} \\
\| r(0.5) \|_{L_\infty} &= \max(|0.5|,|0.5|) = 0.5.
\f}
The following graphic shows all values for the three norm functions \f$\| r(x) \|_{L_1}, \| r(x) \|_{L_2}\f$ and \f$\| r(x) \|_{L_\infty}\f$.
It is notable that the \f$ L_{1} \f$ norm forms the upper and the \f$ L_{\infty} \f$ norm forms the lower border for the example function \f$ r(x) \f$.
![Graphs for the different norm functions from the above example](pics/NormTypes_OneArray_1-2-INF.png)
*/
enum NormTypes { NORM_INF = 1,
NORM_L1 = 2,
NORM_L2 = 4,