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Revision difference : signum function
Version 5 Version 4
\newcommand{\signum}[0]{\mathop{\mathrm{sign}}} \newcommand{\signum}[0]{\mathop{\mathrm{sign}}}
\newcommand{\R}[0]{\mathbb{R}} \newcommand{\R}[0]{\mathbb{R}}
The \emph{signum function} is the function $\signum :\R\to \R$ The \emph{signum function} is the function $\signum :\R\to \R$
\begin{eqnarray*} \begin{eqnarray*}
\signum (x) &=& \left\{ \begin {array}{ll} \signum (x) &=& \left\{ \begin {array}{ll}
-1 & \mbox{when}\,\, x<0, \\ -1 & \mbox{when}\,\, x<0, \\
0 & \mbox{when}\,\, x=0,\\ 0 & \mbox{when}\,\, x=0,\\
1 & \mbox{when}\,\, x>0. \\ \end{array} \right. 1 & \mbox{when}\,\, x>0. \\ \end{array} \right.
\end{eqnarray*} \end{eqnarray*}
The following properties hold: The following properties hold:
\begin{enumerate} \begin{enumerate}
\item For all $x\in \R$, $\signum(-x) = -\signum(x).$ \item For all $x\in \R$, $\signum(-x) = -\signum(x).$
\item For all $x\in \R$, $|x|=\signum(x) x.$ \item For all $x\in \R$, $|x|=\signum(x) x.$
\item For all $x\neq 0$, $\frac{d}{dx}|x|=\signum(x)$. \item For all $x\neq 0$, $\frac{d}{dx}|x|=\signum(x)$.
\end{enumerate} \end{enumerate}
Here, we should point out that the signum function is often Here, we should point out that the signum function is often
defined simply as $1$ for $x>0$ and $-1$ for $x<0$. defined simply as $1$ for $x>0$ and $-1$ for $x<0$.
Thus, at $x=0$, it is left undefined. See e.g. \cite{kreyszig93}. Thus, at $x=0$, it is left undefined. See e.g. \cite{kreyszig93}.
In applications, In applications,
such as the Laplace transform, this definition is adequate since such as the Laplace transform, this definition is adequate since
the value of a function at a single point does not change the the value of a function at a single point does not change the
analysis. One could then, in fact, set $\signum(0)$ to any analysis. One could then, in fact, set $\signum(0)$ to any
value. However, setting $\signum(0)=0$ is motivated by the above relations. On a related note, we can extend the definition to the extended real numbers $\overline{\mathbb{R}}=\mathbb{R}\cup\{\infty,-\infty\}$ by defining $\signum(\infty)=1$ and $\signum(-\infty)=-1$. value. However, setting $\signum(0)=0$ is motivated by the above relations..
A related function is the \emph{Heaviside step function} A related function is the \emph{Heaviside step function}
defined as defined as
\begin{eqnarray*} \begin{eqnarray*}
H(x) &=& \left\{ \begin {array}{ll} 0 & \mbox{when}\,\, x< 0, \\ H(x) &=& \left\{ \begin {array}{ll} 0 & \mbox{when}\,\, x< 0, \\
1/2 & \mbox{when}\,\, x= 0,\\ 1/2 & \mbox{when}\,\, x= 0,\\
1 & \mbox{when}\,\, x> 0.\\ 1 & \mbox{when}\,\, x> 0.\\
\end{array} \right. \end{array} \right.
\end{eqnarray*} \end{eqnarray*}
Again, this function is sometimes left undefined at $x=0$. Again, this function is sometimes left undefined at $x=0$.
The motivation for setting $H(0)=1/2$ is that The motivation for setting $H(0)=1/2$ is that
for all $x\in\R$, we then have the relations for all $x\in\R$, we then have the relations
\begin{eqnarray*} \begin{eqnarray*}
H (x) &=& \frac{1}{2}(\signum(x)+1), \\ H (x) &=& \frac{1}{2}(\signum(x)+1), \\
H(-x) &=& 1-H(x). H(-x) &=& 1-H(x).
\end{eqnarray*} \end{eqnarray*}
This first relation is clear. For the second, we have This first relation is clear. For the second, we have
\begin{eqnarray*} \begin{eqnarray*}
1-H(x) &=& 1-\frac{1}{2}(\signum(x)+1) \\ 1-H(x) &=& 1-\frac{1}{2}(\signum(x)+1) \\
&=& \frac{1}{2}(1-\signum(x)) \\ &=& \frac{1}{2}(1-\signum(x)) \\
&=& \frac{1}{2}(1+\signum(- x)) \\ &=& \frac{1}{2}(1+\signum(- x)) \\
&=& H(-x). &=& H(-x).
\end{eqnarray*} \end{eqnarray*}
{\bf Example} Let $a<b$ be real numbers, and let $f:\R\to\R$ be the {\bf Example} Let $a<b$ be real numbers, and let $f:\R\to\R$ be the
piecewise defined function piecewise defined function
\begin{eqnarray*} \begin{eqnarray*}
f (x) &=& \left\{ \begin {array}{ll} f (x) &=& \left\{ \begin {array}{ll}
4 & \mbox{when}\,\, x\in(a,b), \\ 4 & \mbox{when}\,\, x\in(a,b), \\
0 & \mbox{otherwise.} \\ 0 & \mbox{otherwise.} \\
\end{array} \right. \end{array} \right.
\end{eqnarray*} \end{eqnarray*}
Using the Heaviside step function, we can write Using the Heaviside step function, we can write
\begin{eqnarray} \begin{eqnarray}
\label{almost} \label{almost}
f(x) &=& 4\big(H(x-a) - H(x-b)\big) f(x) &=& 4\big(H(x-a) - H(x-b)\big)
\end{eqnarray} \end{eqnarray}
almost everywhere. almost everywhere.
Indeed, if we calculate $f$ using equation \ref{almost} we obtain Indeed, if we calculate $f$ using equation \ref{almost} we obtain
$f(x)=4$ for $x\in(a,b)$, $f(x)=0$ for $x\notin[a,b]$, $f(x)=4$ for $x\in(a,b)$, $f(x)=0$ for $x\notin[a,b]$,
and $f(a)=f(b)=2$. Therefore, equation \ref{almost} and $f(a)=f(b)=2$. Therefore, equation \ref{almost}
holds at all points except $a$ and $b$. holds at all points except $a$ and $b$.
$\Box$ $\Box$
\section{Signum function for complex arguments} \section{Signum function for complex arguments}
For a complex number $z$, the signum function is defined as \cite{bachman} For a complex number $z$, the signum function is defined as \cite{bachman}
\begin{eqnarray*} \begin{eqnarray*}
\signum (z) &=& \left\{ \begin {array}{ll} \signum (z) &=& \left\{ \begin {array}{ll}
0 & \mbox{when}\,\, z=0,\\ 0 & \mbox{when}\,\, z=0,\\
{z}/{|z|} & \mbox{when}\,\, z\neq 0. \\ \end{array} \right. {z}/{|z|} & \mbox{when}\,\, z\neq 0. \\ \end{array} \right.
\end{eqnarray*} \end{eqnarray*}
In other words, if $z$ is non-zero, then $\signum z$ is the projection In other words, if $z$ is non-zero, then $\signum z$ is the projection
of $z$ onto the unit circle $\{z\in \mathbb{C} \mid |z| = 1\}$. of $z$ onto the unit circle $\{z\in \mathbb{C} \mid |z| = 1\}$.
Clearly, the complex signum function reduces to the real signum function Clearly, the complex signum function reduces to the real signum function
for real arguments. for real arguments.
For all $z\in \mathbb{C}$, we have For all $z\in \mathbb{C}$, we have
$$ z \signum \overline{z} = |z|,$$ $$ z \signum \overline{z} = |z|,$$
where $\overline{z}$ is the complex conjugate of $z$. where $\overline{z}$ is the complex conjugate of $z$.
\begin{thebibliography}{9} \begin{thebibliography}{9}
\bibitem {kreyszig93} E. Kreyszig, \bibitem {kreyszig93} E. Kreyszig,
\emph{Advanced Engineering Mathematics}, \emph{Advanced Engineering Mathematics},
John Wiley \& Sons, 1993, 7th ed. John Wiley \& Sons, 1993, 7th ed.
\bibitem{bachman} G. Bachman, L. Narici, \bibitem{bachman} G. Bachman, L. Narici,
\emph{Functional analysis}, Academic Press, 1966. \emph{Functional analysis}, Academic Press, 1966.
\end{thebibliography} \end{thebibliography}