\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,