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| Title of object: |
generalized binomial coefficients |
| Canonical Name: |
GeneralizedBinomialCoefficients |
| Type: |
Definition |
| Created on: |
2004-10-06 12:41:27 |
| Modified on: |
2005-03-09 14:24:22 |
| Classification: |
msc:05A10, msc:11B65 |
| Defines: |
Pascal's formula, Vandermonde's formula |
Preamble:
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Content:
The binomial coefficients
\begin{align}
{n\choose r} = \frac{n!}{(n-r)!r!},
\end{align}
where $n$ is a non-negative integer and \,$r \in \{0,\,1,\,2,\,...,\,n\}$,
can be generalized for all integer and non-integer values of $n$ by using the reduced form
\begin{align}
{n\choose r} = \frac{n(n-1)(n-2)...(n-r+1)}{r!};
\end{align}
here $r$ may be any non-negative integer. \,Then the \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+...
\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)...(\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 the {\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. |
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