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I ‎need ‎to ‎simplify‎ Division of two gamma functions whose ‎inputs ‎differ ‎by a‎ ‎positive‎ ‎real number ‎‎‎as ‎follow:‎

‎$‎\dfrac{‎‎\Gamma‎(‎x+y‎)}{‎‎\Gamma‎(‎y‎)‎},‎‎\qquad‎‎ x,y\in (0,+‎\infty‎)‎‎‎$‎

If $‎x‎$ and $‎y‎$ were positive integers ‎and greater than one‎, I would use the ‎equality ‎‎‎$‎‎\Gamma‎(‎y‎‎)=(‎y‎-1)‎‎\Gamma‎(‎y-1)‎$ ‎and ‎simplify ‎the ‎func‎tin. ‎But ‎they ‎can ‎be ‎every ‎positive ‎real ‎number.‎ ‎Also, I‎ ‎tried ‎to ‎simplify ‎the‎ ‎expression ‎by ‎using ‎of‎ the q-gamma function, or basic gamma function as follow:‎‎ ‎

‎$\mathop {\lim }\limits_{q \to {1^ + }}\dfrac{‎‎\Gamma‎(‎x+y‎‎)}{‎‎\Gamma‎(‎y‎)‎}= \mathop {\lim }\limits_{q \to {1^ + }} \frac{{{\Gamma _q}(x + y)}}{{{\Gamma _q}(y)}}$

$= \mathop {\lim }\limits_{q \to {1^ + }} \frac{{{{(1 - q)}^{1 - (x + y)}}}}{{{{(1 - q)}^{1 - y}}}} \times \frac{{\prod\limits_{i = 0}^{ + \infty } {\frac{{1 - {q^{i + 1}}}}{{1 - {q^{i + x + y}}}}} }}{{\prod\limits_{i = 0}^{ + \infty } {\frac{{1 - {q^{i + 1}}}}{{1 - {q^{i + y}}}}} }}$

$= \mathop {\lim }\limits_{q \to {1^ + }} {(1 - q)^{ - x}}\prod\limits_{i = 0}^{ + \infty } {\frac{{1 - {q^{i + y}}}}{{1 - {q^{i + x + y}}}}} .‎‎$‎‎

But I could not be successful. Please, somebody help me!!!

E. T
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1 Answers1

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You could call it $\text{pochhammer}(y,x)$.

Robert Israel
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  • So, I should write: – E. T Oct 20 '16 at 15:31
  • Thanks a lot. So, I should write $‎\dfrac{‎‎\Gamma‎(‎x+y‎)}{‎‎\Gamma‎(‎y‎)‎}=y^{(x)}$. But it is just a new form of the function. – E. T Oct 20 '16 at 15:34
  • Anything you write that isn't wrong is going to be "just a new form of the function". – Robert Israel Oct 20 '16 at 19:11
  • You're right. I need to find $x$ in terms of $y$ and $z$ when ‎$‎\dfrac{‎‎\Gamma‎(‎x+y‎)}{‎‎\Gamma‎(‎y‎)‎}=z$. I thought that I should ‎simplify the left side of the equation and then find x. Sorry, I shoulddn't say "it is just a new form of the function"... Thank you and excuse me. – E. T Oct 21 '16 at 12:58