generalized binomial coefficientsGeneralizedBinomialCoefficients2004-10-06 12:41:272006-10-06 13:40:40Definitionbinomial coefficientPascal's formulaVandermonde's formula% this is the default PlanetMath preamble. as your knowledge
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% define commands hereThe binomial coefficients
\begin{align}
{n\choose r} = \frac{n!}{(n\!-\!r)!r!},
\end{align}
where $n$ is a non-negative integer and\, $r \in \{0,\,1,\,2,\,\ldots,\,n\}$,\,
can be generalized for all integer and non-integer values of $n$ by using the \PMlinkname{reduced}{Division} form
\begin{align}
{n\choose r} = \frac{n(n\!-\!1)(n\!-\!2)\ldots(n\!-\!r\!+\!1)}{r!};
\end{align}
here $r$ may be any non-negative integer.\, Then \PMlinkname{Newton's binomial series}{BinomialFormula} gets the \PMlinkescapetext{simple} form
\begin{align}
(1\!+\!z)^{\alpha} = \sum_{r = 0}^{\infty}{\alpha\choose r}z^r
= 1\!+\!{\alpha\choose1}z\!+\!{\alpha\choose 2}z^2\!+\cdots
\end{align}
It is not hard to show that the radius of convergence of this series is 1.\, This series expansion is valid for every complex number $\alpha$ when\, $|z| < 1$,\, and it presents such a \PMlinkname{branch}{GeneralPower} of the \PMlinkname{power}{GeneralPower} $(1\!+\!z)^{\alpha}$ which gets the value 1 in the point\, $z = 0$.
In the case that $\alpha$ is a non-negative integer and $r$ is great enough, one factor in the numerator of
\begin{align}
{\alpha\choose r} =
\frac{\alpha(\alpha\!-\!1)(\alpha\!-\!2)\ldots(\alpha\!-\!r\!+\!1)}{r!}
\end{align}
vanishes, and hence the corresponding binomial coefficient ${\alpha\choose r}$ equals to zero; accordingly also all following binomial coefficients with a greater $r$ are equal to zero.\, It means that the series is left to being a finite sum, which gives the binomial theorem.
For all complex values of $\alpha$, $\beta$ and non-negative integer values of $r$, $s$, the {\em Pascal's formula}
\begin{align}
{\alpha\choose r}\!+\!{\alpha\choose r\!+\!1} = {{\alpha\!+\!1}\choose{r\!+\!1}}
\end{align}
and {\em Vandermonde's convolution}
\begin{align}
\sum_{r = 0}^s{\alpha\choose r}\!{\beta\choose{s\!-\!r}} = {{\alpha\!+\!\beta}\choose s}
\end{align}
hold (the latter is proved by expanding the power $(1\!+\!z)^{\alpha+\beta}$ to series).\, Cf. Pascal's rule and Vandermonde identity.