Talk:Incomplete gamma function

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Relation between types?

Question: What is the relation between two definitions of incomplete Gamma functions ? Are they the same ? It seems not. —The preceding unsigned comment was added by 169.229.120.24 (talk • contribs) .

They're not the same, but the relationship between them is

\gamma (a,x)+\Gamma (a,x)=\Gamma (a)\,

. Evil saltine 21:12, 6 May 2006 (UTC)[reply]

It would seem that the definition of the two gammas here is different from that on the Poisson distribution page. There the cumulative distribution is said to be related to capital Gamma; but by this definition, it would be lower case Gamma. —Preceding unsigned comment added by YouRang? (talk • contribs) 16:17, 8 August 2008 (UTC)[reply]

Sum or programming solution

Does anyone know if theres are sums that are exactly equal to the incomplete gamma functions? I'm looking for a programming solution for this, and I'd rather not make-shift an approximation. Fresheneesz 19:06, 13 September 2006 (UTC)[reply]

Check out the Abramowitz and Stegun link in the references, Equation 6.5.13. Here's a quick link. See also 6.5.29.

PS - why not include your findings in the article? PAR 02:21, 14 September 2006 (UTC)[reply]

I always do - it makes homework take forever. haha. Fresheneesz 03:31, 14 September 2006 (UTC)[reply]

Alternative name for upper incomplete gamma function

It may be useful also to note that, in older literature in particular, the upper incomplete gamma function is often referred to as "Prym's function". (Reference(s) available, if required.)

Hair Commodore 17:43, 5 November 2006 (UTC)[reply]

lower incomplete gamma function for small x'es

The current article states

 $\gamma (a,x)=\int _{0}^{x}t^{a-1}\,e^{-t}\,dt.\,\!$ 
and
 ${\frac {\gamma (a,x)}{x^{a}}}\rightarrow 1\quad \mathrm {as\ } x\rightarrow 0,\,$ 
shouldn't it maybe be something like, - say - 
 ${\frac {\gamma (a,x)}{x^{a}}}\rightarrow a^{-1}\quad \mathrm {as\ } x\rightarrow 0,\,$

I have just created my wikipedia-account today, so Im not really sure how this stuff works, but wouldn't you agree that an extra a in the above formula makes sense? (b.t.w. I didn't place that retangular dashed boxed in my post on purpose)

Robert (talk) 21:59, 3 December 2008 (UTC)[reply]

There is no $a$ in the formula. As mentioned in the section on derivatives, $\partial _{x}\gamma (a,x)=x^{a-1}e^{-x}$ , and so as x goes to zero, ${\frac {\gamma (a,x)}{x^{a}}}\approx e^{-x}\rightarrow 1.$

Petrelharp (talk) 20:33, 12 November 2009 (UTC)[reply]

derivatives

I have just added some material concerning derivatives of the (upper) incomplete gamma function with respect to its argument(s). I decided to be bold and also added some material indicating how and why this is useful. This material has been in the literature for more than 20 years. I thought it was high time it was added. —Preceding unsigned comment added by 171.71.55.135 (talk) 22:38, 12 January 2009 (UTC)[reply]

great! very useful. Petrelharp (talk) 20:33, 12 November 2009 (UTC)[reply]

Thank you! -- Tony 69.106.255.20 (talk) 06:33, 25 January 2010 (UTC)[reply]

asymptotic formula?

I wonder what the restriction |arg s| < 1.5*pi in the asymptotic formula of Gamma(s,z) really means and I suspect the constraints of the formula are not correctly cited here. If one takes arg running from -pi to pi (the principle value, as defined here http://en.wikipedia.org/wiki/Argument_%28complex_analysis%29), then the restriction always holds, and can simply be omitted. To my knowledge, formulas related to the incomplete gamma function often hold in a (complex) half plane only, so some kind of constraint might be necessary here, but it is certainly not the one mentioned here. I got some books (Temme, Cruyt, Abramowitz/Stegun and others) that cover the incomplete gamma function, and I'm going to look that up these days.

Wolf Lammen —Preceding unsigned comment added by 85.183.23.210 (talk) 09:23, 15 February 2011 (UTC)[reply]