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| Title of object: |
signum function |
| Canonical Name: |
SignumFunction |
| Type: |
Definition |
| Created on: |
2003-05-05 17:13:42 |
| Modified on: |
2009-03-27 10:33:09 |
| Classification: |
msc:26A06, msc:30-00 |
| Defines: |
Heavyside step function, step function |
Revision comment (for changes between this and next version):
Preamble:
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\newcommand{\signum}[0]{\mathop{\mathrm{sgn}}}
\newcommand{\R}[0]{\mathbb{R}}
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Content:
\PMlinkescapeword{analysis}
\PMlinkescapeword{applications}
\PMlinkescapeword{arguments}
\PMlinkescapeword{calculate}
\PMlinkescapeword{clear}
\PMlinkescapeword{equation}
\PMlinkescapeword{onto}
\PMlinkescapeword{point}
\PMlinkescapeword{properties}
\PMlinkescapeword{relations}
The \emph{signum function} is the function $\signum\colon\R\to \R$
\begin{eqnarray*}
\signum (x) &=& \left\{ \begin {array}{ll}
-1 & \mbox{when}\,\, x<0, \\
0 & \mbox{when}\,\, x=0,\\
1 & \mbox{when}\,\, x>0. \\ \end{array} \right.
\end{eqnarray*}
The following properties hold:
\begin{enumerate}
\item For all $x\in \R$, $\signum(-x) = -\signum(x).$
\item For all $x\in \R$, $|x|=\signum(x) x.$
\item For all $x\neq 0$, $\frac{d}{dx}|x|=\signum(x)$.
\end{enumerate}
Here, we should point out that the signum function
is often defined simply as $1$ for $x>0$ and $-1$ for $x<0$.
Thus, at $x=0$, it is left undefined. See for example \cite{kreyszig93}.
In applications such as the Laplace transform this definition is adequate,
since the value of a function at a single point does not change the 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 $\signum(-\infty)=-1$.
A related function is the \emph{Heaviside step function}
defined as
\begin{eqnarray*}
H(x) &=& \left\{ \begin {array}{ll} 0 & \mbox{when}\,\, x< 0, \\
1/2 & \mbox{when}\,\, x= 0,\\
1 & \mbox{when}\,\, x> 0.\\
\end{array} \right.
\end{eqnarray*}
Again, this function is sometimes left undefined at $x=0$.
The motivation for setting $H(0)=1/2$ is that
for all $x\in\R$, we then have the relations
\begin{eqnarray*}
H (x) &=& \frac{1}{2}(\signum(x)+1), \\
H(-x) &=& 1-H(x).
\end{eqnarray*}
This first relation is clear. For the second, we have
\begin{eqnarray*}
1-H(x) &=& 1-\frac{1}{2}(\signum(x)+1) \\
&=& \frac{1}{2}(1-\signum(x)) \\
&=& \frac{1}{2}(1+\signum(- x)) \\
&=& H(-x).
\end{eqnarray*}
{\bf Example} Let $a<b$ be real numbers, and let $f:\R\to\R$ be the
piecewise defined function
\begin{eqnarray*}
f (x) &=& \left\{ \begin {array}{ll}
4 & \mbox{when}\,\, x\in(a,b), \\
0 & \mbox{otherwise.} \\
\end{array} \right.
\end{eqnarray*}
Using the Heaviside step function, we can write
\begin{eqnarray}
\label{almost}
f(x) &=& 4\big(H(x-a) - H(x-b)\big)
\end{eqnarray}
almost everywhere.
Indeed, if we calculate $f$ using equation \ref{almost} we obtain
$f(x)=4$ for $x\in(a,b)$, $f(x)=0$ for $x\notin[a,b]$,
and $f(a)=f(b)=2$. Therefore, equation \ref{almost}
holds at all points except $a$ and $b$.
$\Box$
\section{Signum function for complex arguments}
For a complex number $z$, the signum function is defined as \cite{bachman}
\begin{eqnarray*}
\signum (z) &=& \left\{ \begin {array}{ll}
0 & \mbox{when}\,\, z=0,\\
{z}/{|z|} & \mbox{when}\,\, z\neq 0. \\ \end{array} \right.
\end{eqnarray*}
In other words, if $z$ is non-zero, then $\signum z$ is the projection
of $z$ onto the unit circle $\{z\in \mathbb{C} \mid |z| = 1\}$.
Clearly, the complex signum function reduces to the real signum function
for real arguments.
For all $z\in \mathbb{C}$, we have
$$ z \signum \overline{z} = |z|,$$
where $\overline{z}$ is the complex conjugate of $z$.
\begin{thebibliography}{9}
\bibitem {kreyszig93} E. Kreyszig,
\emph{Advanced Engineering Mathematics},
John Wiley \& Sons, 1993, 7th ed.
\bibitem{bachman} G. Bachman, L. Narici,
\emph{Functional analysis}, Academic Press, 1966.
\end{thebibliography} |
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