multiplicityMultiplicity2004-06-09 17:42:512009-11-14 12:27:36Definitionfundamental theorem of algebra resultzero of ordermultiple zerosimple zerosimple% this is the default PlanetMath preamble. as your knowledge
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% define commands hereIf a polynomial $f(x)$ in $\mathbb{C}[x]$ is divisible by $(x-a)^m$ but not by $(x-a)^{m+1}$ ($a$ is some complex number, \,$m \in \mathbb{Z}_+$), we say that \, $x = a$ \, is a\, {\em zero of the polynomial with multiplicity $m$} or alternatively a {\em zero of order $m$}.
Generalization of the multiplicity to \PMlinkname{real}{RealFunction} and complex functions (by rspuzio): \,If the function $f$ is continuous on some open set $D$ and \,$f(a) = 0$\, for some \,$a \in D$, then the zero of $f$ at $a$ is said to be of multiplicity $m$ if $\frac{f(z)}{(z\!-\!a)^m}$ is continuous in $D$ but $\frac{f(z)}{(z-a)^{m+1}}$ is not.
If\, $m \geqq 2$, we speak of a {\em multiple zero}; if\, $m = 1$, we speak of a {\em simple zero}.\, If\, $m = 0$, then actually the number $a$ is not a zero of $f(x)$, i.e.\, $f(a) \neq 0$.
Some properties (from which 2, 3 and 4 concern only polynomials):
\begin{enumerate}
\item The zero $a$ of a polynomial $f(x)$ with multiplicity $m$ is a zero of the \PMlinkescapetext{derivative} $f'(x)$ with multiplicity $m\!-\!1$.
\item The zeros of the polynomial $\gcd(f(x), f'(x))$ are same as the multiple zeros of $f(x)$.
\item The quotient $\displaystyle\frac{f(x)}{\gcd(f(x), f'(x))}$ has the same zeros as $f(x)$ but they all are \PMlinkescapetext{simple}.
\item The zeros of any irreducible polynomial are \PMlinkescapetext{simple}.
\end{enumerate}