Simplification of an expression.

Question

I've been working on analysing one system lately, I arrived to an expression which I'd like to simplify more, still couldn't manage to.

I arrived to the following expression : $\sum_{k = n+2}^{N+1}\gamma^k\prod_{t=n+2}^{k}(1-(t-n-1)\alpha); n \in N , \alpha \in R$, and $\gamma<1$.

I just included the sum operator for illustration.

What I'd like to know, if there's a way to write :

$\prod_{t=n+2}^{k}(1-(t-n-1)\alpha)$

in a more simplified way. I was thinking about the Gamma function, since this expression looks like a combination of factorials somehow to me, but didn't manage to figure out how can I arrive there.

Thank you

Answer 1

We can employ the Gamma function to get a convenient representation.

We obtain for $\alpha\in\mathbb{C}\setminus\mathbb{Z}$: \begin{align*} \color{blue}{\prod_{t=n+2}^{k}}&\color{blue}{\left(1-\left(t-n-1\right)\alpha\right)}\\ &=\prod_{t=1}^{k-n-1}\left(1-t\alpha\right)\tag{1}\\ &=\alpha^{k-n-1}\prod_{t=1}^{k-n-1}\left(\frac{1}{\alpha}-t\right)\tag{2}\\ &=\alpha^{k-n-1}\binom{\frac{1}{\alpha}-1}{k-n-1}(k-n-1)!\tag{3}\\ &=\alpha^{k-n-1}\frac{\Gamma\left(\frac{1}{\alpha}\right)}{\Gamma\left(\frac{1}{\alpha}-k+n+1\right)\Gamma(k-n)}(k-n-1)!\tag{4}\\ &\,\,\color{blue}{=\frac{\alpha^{k-n-1}\Gamma\left(\frac{1}{\alpha}\right)}{\Gamma\left(\frac{1}{\alpha}-k+n+1\right)}} \end{align*}

Comment:

• In (1) we shift the index by $n+1$ to start with $t=1$.

• In (2) we factor out $\alpha^{k-n-1}$.

• In (3) we use the definition of binomial coefficients for $\beta \in \mathbb{C}, q\in \mathbb{N}$ \begin{align*} \binom{\beta}{q}=\frac{\beta(\beta-1)\cdots(\beta-q+1)}{q!} \end{align*}

• In (4) we use the identity $\binom{\beta}{q}=\frac{\Gamma\left(\beta+1\right)}{\Gamma\left(\beta - q + 1\right)\Gamma\left(q+1\right)}$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\left. \sum_{k\ =\ n + 2}^{N + 1}\gamma^{k} \prod_{t\ =\ n + 2}^{k}\bracks{1 -\pars{t - n - 1}\alpha} \,\right\vert_{{\Large\ n\ \in\ \mathbb{N}} \atop {{\Large\ \alpha\ \in\ \mathbb{R}} \atop{\Large\ \gamma\ <\ 1}}}} \\[5mm] = &\ \sum_{k\ =\ n + 2}^{N + 1}\gamma^{k}\,\pars{-1}^{k - n - 1}\, \alpha^{k - n - 1} \prod_{t\ =\ n + 2}^{k}\pars{t - n - 1 - {1 \over \alpha}} \\[5mm] = &\ \pars{-\alpha}^{-n - 1}\sum_{k\ =\ n + 2}^{N + 1}\pars{-\alpha\gamma}^{k} \pars{1 - {1 \over \alpha}}^{\overline{k - n - 1}} \\[5mm] = &\ \pars{-\alpha}^{-n - 1}\sum_{k\ =\ n + 2}^{N + 1}\pars{-\alpha\gamma}^{k}\, {\Gamma\pars{k - n - 1/\alpha} \over \Gamma\pars{1 - 1/\alpha}} \\[5mm] = &\ \bbx{{1 \over \pars{-\alpha}^{n + 1}\,\Gamma\pars{1 - 1/\alpha}}\,\, \sum_{k\ =\ n + 2}^{N + 1}\pars{-\alpha\gamma}^{k}\, \pars{k - n - 1 - {1 \over \alpha}}!} \\ & \end{align}