Difference between revisions of "Gamma function"
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| − | + | Let $\mathbb{T}$ be a [[time scale]] and define | |
$$p_f(t,s)=e_{\frac{f}{\mathrm{id}}}(t,s),$$ | $$p_f(t,s)=e_{\frac{f}{\mathrm{id}}}(t,s),$$ | ||
| − | where $\mathrm{id}$ denotes the identity map $\mathrm{id} \colon \mathbb{T} \rightarrow \mathbb{T}$ and $e_{\frac{f}{\mathrm{id}}}$ denotes the [[delta exponential]]. With these definitions, we define the gamma operator | + | where $\mathrm{id}$ denotes the identity map $\mathrm{id} \colon \mathbb{T} \rightarrow \mathbb{T}$ and $e_{\frac{f}{\mathrm{id}}}$ denotes the [[delta exponential]]. With these definitions, we define the gamma operator |
$$\Gamma_{\mathbb{T}}(f;s)=\mathscr{L}_{\mathbb{T}}\{p_{f \boxminus_{\mu} 1}(\cdot,s)\}(1)=\displaystyle\int_0^{\infty} p_{f \boxminus_{\mu}1}(\eta,s) e_{\ominus_{\mu}1}^{\sigma}(\eta,0) \Delta \eta.$$ | $$\Gamma_{\mathbb{T}}(f;s)=\mathscr{L}_{\mathbb{T}}\{p_{f \boxminus_{\mu} 1}(\cdot,s)\}(1)=\displaystyle\int_0^{\infty} p_{f \boxminus_{\mu}1}(\eta,s) e_{\ominus_{\mu}1}^{\sigma}(\eta,0) \Delta \eta.$$ | ||
Revision as of 17:40, 15 January 2023
Let $\mathbb{T}$ be a time scale and define $$p_f(t,s)=e_{\frac{f}{\mathrm{id}}}(t,s),$$ where $\mathrm{id}$ denotes the identity map $\mathrm{id} \colon \mathbb{T} \rightarrow \mathbb{T}$ and $e_{\frac{f}{\mathrm{id}}}$ denotes the delta exponential. With these definitions, we define the gamma operator $$\Gamma_{\mathbb{T}}(f;s)=\mathscr{L}_{\mathbb{T}}\{p_{f \boxminus_{\mu} 1}(\cdot,s)\}(1)=\displaystyle\int_0^{\infty} p_{f \boxminus_{\mu}1}(\eta,s) e_{\ominus_{\mu}1}^{\sigma}(\eta,0) \Delta \eta.$$
Properties of gamma functions
Theorem: If $s \in \mathbb{T}^+$, then $\Gamma_{\mathbb{T}}(x;s)$ converges for any $x \in \mathbb{R}^+$.
Proof: █
Theorem: If $s \in \mathbb{T}^+$, then $$\displaystyle\lim_{x \rightarrow 0^+} \Gamma_{\mathbb{T}}(x;s) = \infty.$$
Proof: █
Theorem: If $s \in \mathbb{T}^+$, then $$\displaystyle\lim_{x \rightarrow \infty} \Gamma_{\mathbb{T}}(x;s) = \infty.$$
Proof: █
Theorem: If $s \in \mathbb{T}^+$, then $\Gamma_{\mathbb{T}}(1;s)=1$.
Proof: █
Theorem: If $s \in \mathbb{T}^+$, then for all $x \in \mathbb{R}^+$, $$\Gamma_{\mathbb{T}}(x \boxplus_{\mu} 1;s) = \dfrac{x}{s} \Gamma_{\mathbb{T}}(x;s).$$
Proof: █
Define the bracket number operators (they are actually functions) $$[n]_{\mathbb{T}} = \left\{ \begin{array}{ll} 0 &; n=0 \\ [n-1]_{\mathbb{T}} \boxplus_{\mu} 1 &; n=1,2,\ldots \end{array} \right.$$ and the bracket factorial $$[n]_{\mathbb{T}}! = \left\{ \begin{array}{ll} 1&; n=0 \\ \displaystyle\prod_{j=1}^n [j]_{\mathbb{T}} &; n=1,2,\ldots \end{array} \right.$$
Theorem: Let $n \in \mathbb{Z}^+$ and assume that $[k]_{\mathbb{T}}$ is a constant function on $\mathbb{T}^+$ for all $k\in[1,n]\bigcap \mathbb{Z}^+$. Then $$\Gamma_{\mathbb{T}}\left( [n]_{\mathbb{T}};s \right) = \dfrac{[n-1]_{\mathbb{T}}!}{s^{n-1}}.$$
Proof: █
Examples of gamma functions
We write formulas for gamma functions defined for $x \in \mathbb{R}^+$ and $s \in \mathbb{T}^+$.
| $\mathbb{T}=$ | $\Gamma_{\mathbb{T}}(x;s)=$ |
| $\mathbb{R}$ | $\displaystyle\int_0^{\infty} \left( \dfrac{\tau}{s} \right)^{x-1}e^{-\tau} d\tau$ |
| $h\mathbb{Z};h>0$ | $h \displaystyle\sum_{k=0}^{\infty} \left( \displaystyle\prod_{j=s}^{k-1} \dfrac{j+x}{j+1} \right) \dfrac{1}{(1+h)^{k+1}}$ |
| $\overline{q^{\mathbb{Z}}}; q>1$ | $\dfrac{(q-1)s}{(1+(q-1)x)^{\log_q(s)}} \displaystyle\sum_{k=-\infty}^{\infty} \dfrac{(1+(q-1)x)^k}{\prod_{j=-\infty}^{k} (1+(q-1)q^k)}$ |
References
<bibtex>@inproceedings{
title="The Gamma Function on Time Scales", author="Bohner, Martin and Karpuz, Başak", booktitle="Dynamics of Continuous, Discrete & Impulsive Systems. Series A. Mathematical Analysis", volume="20", year="2013", pages="pp.507--522", url="http://online.watsci.org/abstract_pdf/2013v20/v20n4a-pdf/7.pdf"
} </bibtex>
