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| Title of object: |
gamma function |
| Canonical Name: |
GammaFunction |
| Type: |
Definition |
| Created on: |
2001-11-17 15:28:54 |
| Modified on: |
2006-10-02 03:03:22 |
| Classification: |
msc:33B15, msc:30D30 |
Revision comment (for changes between this and next version):
Preamble:
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{graphicx}
\usepackage{xypic}
\def\C{\mathbb{C}}
\def\Re{\operatorname{Re}} |
Content:
\PMlinkescapeword{entire}
\PMlinkescapeword{generated by}
\PMlinkescapeword{satisfies}
The \emph{gamma function} is defined by
\[
\Gamma(z) = \int_0^\infty e^{-t} t^{z-1} dt
\]
for $z\in\C$ with $\Re(z)>0$,
and by analytic continuation for the rest of the complex plane,
except for the non-positive integers (where it has simple poles).
The gamma function can equivalently be defined by the formula
\[
\Gamma(z) = \frac{e^{-\gamma z}}{z}
\prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n},
\]
where $\gamma$ is Euler's constant.
Another formula is
\[
\Gamma(z) = \lim_{n\to\infty}\frac{n^z n!}{\prod_{k=0}^n(z+k)}.
\]
The gamma function satisfies the functional equation
\[
\Gamma(z+1) = z \Gamma(z).
\]
As $\Gamma(1)=1$, it follows by induction that
\[
\Gamma(n) = (n-1)!
\]
for positive integer values of $n$.
Another functional equation satisfied by the gamma function is
\[
\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin \pi z}.
\]
The gamma function for real $z$ looks like this:
\begin{center}
\begin{tabular}{c}
\includegraphics[scale=.9]{gammafunc.eps} \\
{\tiny (generated by GNU Octave and gnuplot) }
\end{tabular}
\end{center}
Some approximate values of the gamma function for small arguments are:
\[
\begin{array}{cc}
\Gamma(1/5) \approx 4.5909 & \Gamma(1/4) \approx 3.6256 \\
\Gamma(1/3) \approx 2.6789 & \Gamma(2/5) \approx 2.2182 \\
\Gamma(3/5) \approx 1.4892 & \Gamma(2/3) \approx 1.3541 \\
\Gamma(3/4) \approx 1.2254 & \Gamma(4/5) \approx 1.1642
\end{array}
\]
and the ever-useful $\Gamma(1/2)=\sqrt{\pi}$. These values allow a quick calculation of $\Gamma(n+f)$
where $n$ is a natural number and $f$ is any fractional value for which the gamma function's value is known. Since $\Gamma(z+1)=z\Gamma(z)$, we have
\begin{eqnarray*}
\Gamma(n+f) & = & (n+f-1)\Gamma(n+f-1) \\
& = & (n+f-1)(n+f-2)\Gamma(n+f-2) \\
& \vdots & \\
& = & (n+f-1)(n+f-2)\cdots(f)\Gamma(f)
\end{eqnarray*}
which is easy to calculate if we know $\Gamma(f)$.
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