Talk:Gamma function/Archive 1

Archive 1

Archive 2

Opera 5.11

Ted: Using Opera 5.11 on Win95 I get those pretty little boxes for &infin, &Gamma and &int. -- Stephen Gilbert —Preceding undated comment added 10:47, 29 June 2001

Complain to the opera people. Those entities are valid HTML 4.0.— Preceding unsigned comment added by AxelBoldt (talk • contribs) 14:44, 9 August 2001 (UTC)

Integral sign

How did you do the integral sign?— Preceding unsigned comment added by Conversion script (talk • contribs) 15:51, 25 February 2002 (UTC)

like this: &int; you can also do it like this: ∫— Preceding unsigned comment added by JeffBobFrank (talk • contribs) 00:36, 28 February 2004 (UTC)

Graph

I think we should have a graph of the gamma function between say -3 and 3 here. It's a really beautiful graph and illustrates why the gamma function is such a fascinating topic. Barnaby dawson 12:47, 18 September 2004 (UTC)

---

Any ideas how to compute the gamma function quickly? (mainly interested in real values) Fredrik | talk 20:59, 15 February 2005 (UTC)

Nevermind, I found what I was looking for ([1]). Maybe something for this article? Fredrik | talk 22:06, 16 February 2005 (UTC)

Contrasting with special relativity

I think that this page should have a reference to the γ (v) which is used in special relativity. Many physics students will not have heard of "the" gamma function and may be confused.— Preceding unsigned comment added by TomViza (talk • contribs) 22:01, 3 May 2005 (UTC)

Lorentz factor

γ (v) is sometimes written as a constant and called the Lorentz_factor.

Where should it go and what form should it take?

Hmm, hope these same physics students don't think that gamma is the same thing as the gyromagnetic ratio γ, or we're really in trouble. linas 23:52, 4 May 2005 (UTC)

Lanczos approximation

I have created a rough writeup about the Lanczos approximation. It would be helpful if someone with greater expertise could check the article for accuracy and add information about the approximation's derivation, known improvements, and error estimates. Fredrik | talk 12:09, 3 June 2005 (UTC)

Particular values table

If anyone objects to the "particular values" table formatting, please realize that the argument involves the formatting choice in the individual's "Rendering math" preferences. The first three choices are most relevant:

1. Always render PNG
2. HTML if very simple or else PNG
3. HTML if possible or else PNG

I think we can agree that the table should be consistently rendered as much as possible. Since the entries with square roots will always be rendered in PNG, we want the table to be rendered in PNG for option number 2. This is the way I have reformatted it. PAR 13:53, 9 June 2005 (UTC)

The new image

I wonder if the new image is an improvement

• It has pixellation problems
• I don't think the scale 0-2-4 is better than 0-1-2-3-4.
• I'ts smaller - thats not an issue since it has to be reduced for the page anyway, but if anyone wants to use it for a presentation or something it wont be very good.
• It has an English title - its less usable for other language wikis

69.143.43.101 19:40, 9 October 2005 (UTC)

Vertical asymptote

I'm interested in an explaination of why there is a vertical asymptote at Gamma(0), when 0! = 1. Thanks! Turidoth 16:38, 27 March 2006 (UTC)

Note that the argument must be shifted; Γ(0) = (-1)!, Γ(1) = 0!, etc. As for the vertical asymptote, consider the property Γ(z+1) = z Γ(z) and think about what happens when z approaches 0. Then Γ(z+1) approaches 1, and so must the right-hand side. In order for z Γ(z) to approach 1 when z approaches 0, Γ(z) must approach infinity. Fredrik Johansson 17:06, 27 March 2006 (UTC)

Colorful plots

The plots in the article are all aesthethically pleasing and whatnot, but they're all lacking a colorbar. Not being graphically inclined, I'm not in a place to upgrade the plots, but feel the plots would be much more illustrative/understandable (especially to the non-mathhead) if they gave a colorbar. They sure are pretty, though. Just a thought if anyone knows how to easily do this. 07:45, 4 September 2006 (UTC)

Improving those plots (with proper color bars, etc) has been on my to-do list for a while. In the mean time, you can check out the surface plots on this page. Fredrik Johansson 10:45, 4 September 2006 (UTC)

Gamma versus factorial

I have often heard the gamma function being sighted as an example of 0!=1 If Gamma(1) = 1, does it mean 0! is one? note that gamma (1/2) exists but 1/2! may not, as per the defintion of factorial: The factorial is defined for a positive integer as n! =n.(n-1)..3.2.1 Just because gamma shows the recursive property of (n-1).(n-2).. is it right to extrapolate?

What is going on here, is that the Gamma function is defined for all complex numbers z except z = 0,  −1, −2, −3, ... , and it agrees with the factorial function for the positive integers. There are good mathematical reasons for saying that the Gamma function is the most natural way to extend the Gamma function to the complex numbers, so it makes sense to define 0! to be 1. Madmath789 08:25, 16 June 2006 (UTC)

In combinatorics it also turns out that 0!=1 is useful and convenient in many formulas. E.g., Choose(n,k) = n! / k!(n-k)! works for n=k or k=0 if 0!=1.

Another way you can get to 0! is by looking at the factorial definition backwards. Solving for n!,

$(n+1)!=(n+1)(n!)$

becomes

$n!=(n+1)!/(n+1)$

so if you are ok with 1! = 1, then 0! = 1! / 1 = 1. Following this through with gamma forces gamma to give the same value... --Jake 16:11, 24 October 2006 (UTC)

Binet's integrals

Might it be a good idea to add Binet's first and second integrals (for the logarithm of the gamma function) to this page? They certainly deserve a mention - somewhere.

Hair Commodore 22:08, 30 October 2006 (UTC)

Sure, but they would fit even better on the Stirling's approximation page than here. Fredrik Johansson 22:17, 30 October 2006 (UTC)

I agree with this. Hair Commodore 22:04, 6 November 2006 (UTC)

A question

Should the Pi function section of this article have it's own? Sr13 05:03, 15 November 2006 (UTC)

No. It's just a notation; hardly a function in its own right. Fredrik Johansson 05:58, 15 November 2006 (UTC)

PNG forcing

I reverted the \! forcing of PNG to the \, forcing. The standard way of doing math equations is to have in-line equations not forced to PNG, and in-line forced to PNG with "\," but not "\!".

You have three basic choices in your Wikipedia "preferences"

1. Always render PNG
2. HTML if very simple or else PNG
3. HTML if possible or else PNG

	1	2	3
forced with \,	PNG	PNG	HTML
forced with \!	PNG	HTML	HTML

Editors should force with \,. As a reader, then, if you want to see PNG all the time, pick option 1. If you want to see HTML where the editor wanted it and PNG where the editor wanted it, pick option 2. If you want HTML wherever possible, pick option 3.

You don't have this freedom when forcing with \!, which is why \, is chosen to force. I think the person who changed the \, to \! has option 3 set. Just change it to option 2 and you will be good to go on almost all math pages. PAR 04:42, 21 November 2006 (UTC)

Connection to String Theory

does anyone have an opinion on adding a section regarding the connection that Gabriele Veneziano found between the Gamma function and the strong force? this observation allowed the field of String Theory to first take shape in the 1970s.

I don't know the technicalites of how the strong force relates to the gamma function, but NOVA's Elegant Universe does mention this relationship and should be added to this article. Sr13 08:43, 22 November 2006 (UTC) Actually, the beta function (which involves the gamma function in one of its forms) is related to Veneziano, not the gamma function. Sr13 08:47, 22 November 2006 (UTC)

So NOVA was WRONG when it showed a shot of the paper with "Gamma function" on it, then? mike4ty4 02:17, 19 April 2007 (UTC)

Plots are mathematically strange

Huh?

A few complaints about some of the plots:

• Image:Log gamma imag.png should really cycle colors every 2pi, like Image:Complex gamma.jpg naturally does. Actually I'm not sure why we need the former when we already have the latter.
• I'm not sure what Image:LogGammaAbsPlot.png and Image:Log gamma absolute.png are really showing. Is that quantity even meaningful?
• If Image:LogGammaImPlot.png is going to look like that, at least the discontinuities on the right-hand corners should be removed.
• The 3D images should give the real and imaginary axes equal scales, so that the poles look circular. Contrast Image:GammaAbsSmallPlot.png.

Melchoir (talk) 06:08, 5 April 2008 (UTC)

...I'm just going to remove the section. There will always be the Commons gallery linked at the bottom of the article. Melchoir (talk) 00:11, 7 April 2008 (UTC)

Convergence for Re(z) < 1

Can someone explain how the convergence of the integral comes about if Re(z) < 1 ? Ok, it has a pole at zero, but for instance for z=i we seem to have a finite value for gamma(z), whereas the integrand is oscillating wildly and diverging as it approaches t=0. Thanks, Mauritsmaartendejong (talk) 07:14, 23 April 2009 (UTC)

There is no problem with the (absolute) convergence of the integral, as long as $\mathrm {Re} z>0.$ If you want to look at $z=i,$ then clearly there is a problem with the integral, that is certainly not absolutely convergent, and I don't even know for sure how to give it a meaning by some more clever procedure like Cauchy principal value. What people usually do is to use the functional equation $\Gamma (z+1)=z\Gamma (z)$ in order to extend the function. For example, $\Gamma (i)=\Gamma (i+1)/i.$ --Bdmy (talk) 09:31, 23 April 2009 (UTC)

Typo

I think there is a typo in Gamma_function#Alternative_definitions when substituting for z+1. There is a n+1 instaed of n. Can someone please check? --kupirijo (talk) 08:51, 26 October 2009 (UTC)

You are quite right. I have corrected the error. JamesBWatson (talk) 11:37, 28 October 2009 (UTC)

Relations to differential equations/Laplace transforms

Should there be a section on the relationship to Laplace Transforms, i.e. ${\mathcal {L}}\left\{t^{a}\right\}={\frac {\Gamma (a+1)}{S^{a+1}}}\$ I don't think it's mentioned in the article. 76.93.191.58 (talk) 06:29, 30 March 2010 (UTC)

Definition

What's the t for in the definition, can somebody please post an explanation? It just appears without mention. Even if it's an arbitrary constant (which it isn't), this should be explained. AdamSebWolf 01:19, 13 December 2006 (UTC)

It doesn't need any explanation if the reader knows what an integral is. The t is the dummy variable of integration since it is in the "dt" term. Explaining this in the article would only gum it up. Baccyak4H (Yak!) 03:08, 13 December 2006 (UTC)

Is this the mathematical equivalent of giving someone the fingers? Explaining something necessary to the understanding of the article will spoil the article, is this what you're saying? Koro Neil (talk) 10:01, 29 February 2008 (UTC)

Perhaps a reference to the term "integral" or "integration" would be appropriate., but anyone who doesn't understand integrals won't understand most of the rest of the article. — Arthur Rubin | (talk) 10:04, 29 February 2008 (UTC)

In the Definition paragraph it says "For a comlex vale of z with a positive real part, the Gammafunction is defined by the integral" (the integral according to Legendre, that is). But in the plot, values are shown for Re(z)<0, as well as in the Particular Values paragraph. Would it not be better to write something like "The definition is valid for any complex number z, provided Re(z)is not a negative integer"? Alex Vermeulen, Zoetermeer, The NetherlandsJordaan12 (talk) 13:18, 31 March 2010 (UTC)

Perhaps it is not explained well enough in the article, but there is a subtle distinction being made here. The integral is defined only for Re(z)>0. The function defined by that integral can be extended by analytic continuation to the entire complex plane except the non-positive integers. The function defined by the integral has no name, but its analytic continuation is called the Gamma function. (Well, I guess, we could "name" the integral as "the Gamma function restricted to Re(z)>0.") I'll see what I can do to make all this clearer in the article. Quantling (talk) 14:28, 31 March 2010 (UTC)

Baffling rendering of definition

In the image of the definition of the function, it's displaying t^≈-1 instead of t^z-1. This appeared in this revision though I'm not sure why. The math markup wasn't changed, and is still correct to this day. I'm not sure how to regenerate the image or otherwise fix this. —Preceding unsigned comment added by Hyuga (talk • contribs) 15:19, 31 October 2008 (UTC)

Quite. I have no idea, either. Perhaps you could try Wikipedia talk:WikiProject Mathematics. I'm only a subject expert, not a MathML expert. — Arthur Rubin (talk) 18:21, 31 October 2008 (UTC)

Is it so baffling? The letter z as rendered in TeX and the "≈" symbol simply look similar to each other, so maybe on some browsers the differences are not conspicuous. Michael Hardy (talk) 18:49, 31 October 2008 (UTC)

Perhaps I and the original poster have a font or rendering problem. I see no pixel diferences between them on my machine, though. — Arthur Rubin (talk) 21:14, 31 October 2008 (UTC)

It seems to be the addition of a full stop at the end of the formula that changed the rendering. I've changed the space before the full stop to "\;" (TeX for a very slightly thinner space). For some reason this alters the rendering on my setup so that the central stroke of the z is visible again. I don't really understand the cause of the problem though. Some sort of aliasing issue perhaps? Qwfp (talk) 22:34, 31 October 2008 (UTC)

I've done a tweak on the second statement of the same formula, which had a similar problem on my computer. Very strange. Septentrionalis PMAnderson 03:44, 2 November 2008 (UTC)

This is a bug in dvipng, which is being used to generate the math images. See bugzilla:15777 for the details. — Carl (CBM · talk) 02:40, 3 November 2008 (UTC)

At any rate, the image is fixed now. Looks fine to me. Thanks! And in response to Mr. Hardy, I did consider that the two characters could justr be rendered very similarly. The baffling part was that when I looked a previous version of the page, the function was clearly rendered correctly. It was after a revision that seemed to make little difference (though apparently it did) that the z became a "≈". — Hyuga (talk) 16:04, 17 November 2008 (UTC)

The rendering bug is still present. It appears in all previous revisions that I've viewed (except the pre-<math> versions). —Deadcode (talk) 18:55, 31 March 2010 (UTC)

History

In his book "Riemann's Zeta Function", H. M. Edwards claims that Gauss introduced the Pi function and writes in a footnote on page 8: "Unfortunately, Legendre subsequently introduced the notation $\Gamma (s)$ for $\Pi (s-1)$. Legendre's reason for considering $(n-1)!$ instead of $n!$ are obscure (perhaps he felt it was more natural to have the first pole occur at $s=0$ rather than $s=-1$) but, whatever the reason, this notation prevailed in France and, by the end of the nineteenth century, in the rest of the world as well." -- Tobias Bergemann 13:52, 2005 Apr 10 (UTC)

Thats interesting - why not condense it somewhat and put it in the introduction? Include the book as a reference, take out the smaller reference to Legendre in the definition section. PAR 17:16, 10 Apr 2005 (UTC)

I'd like to do some more research first. Edward references a Gauss publication from 1813 for the Pi function notation. The entry about the Gamma function in the german wikipedia (de:Gammafunktion) reports Leonhard Euler as the inventor of the first interpolation formula for faculties. (In 1730! Apparently it is really true that in mathematics theorems are usually named after the first mathematician who rediscovers them after Euler.) -- Tobias Bergemann 15:16, 2005 Apr 11 (UTC)

The French Wikipedia tells another story. It has been the year 1729 and it was Daniel Bernoulli who found the Gamma function. This has been recorded by

Paul Heinrich Fuss, Correspondance mathématique et physique de quelques célèbres géomètres du XVIIIeme siècle .. , St. Pétersbourg, 1843, Tome II, page 324-325. Paul Heinrich Fuss was Euler's great-grandson. 78.55.75.14 (talk) —Preceding undated comment added 20:52, 5 July 2010 (UTC).

Is Gamma Unique

That is, is the Gamma function the only generalization of factorial that has the properties:

Gamma(1) = 1

Gamma(z+1) = z Gamma(z)

For all complex numbers z? If so (or even if not), this would be nice to mention on the page. But really I'm just personally interested. Luqui 07:41, 14 September 2005 (UTC)

That depends on what you mean by generalization. For example, you could consider a function that agrees with the factorial for all natural numbers but is zero everywhere else a generlization of the factorial.

So you need additional requirements on the generalization for uniqueness. The article mentions the Bohr-Mollerup theorem which states that Gamma is the only logarithmically convex generalization of the factorial. -- Tobias Bergemann 12:56, 14 September 2005 (UTC)

It's unique if you pose the further condition of log-convexity. See Bohr-Mollerup theorem. Michael Hardy 01:03, 15 September 2005 (UTC)

In other words: Every holomorphic function f:C to C that is limited on the "stripe" of the complex plane (1<Re(z)<2) and fulfils Gamma(z+1)=z*Gamma(z) is necessarily of the form f(z)=f(1).Gamma(z). (This statement is known as Wielandt's Theorem, albeit is is rarely taught in complex calculus classes.) Feel free to implement it in the article, I don't have time right now. (Simple proof via Liouville's Theorem, after constructing a periodic function from f over full C) 85.212.32.132

In the section on uniqueness, it might help to mention a uniqueness property that is NOT true (also because the casual reader might come away with the impression that it is true), namely "the Gamma function is the only analytic function on the right half plane which equals (n-1)! at z=n." This is not true because we can add any analytic function which is zero on the integers. This is of course not anything special about the gamma function, but it wouldn't hurt to mention it here, due to the popularity/importance of the gamma function and the natural motivation for wanting to know the various senses in which it is a "unique" extension of n!. Perhaps include a link to the theorem stating that the zero set of a holomorphic function is discrete and mention that this theorem can be used to prove many uniqueness results, although it does not apply here? 72.51.124.202 (talk) 04:08, 19 April 2010 (UTC) Brian Maurizi

Good point. I have added a note about this. Previously the article contained wording which certainly gave the impression that this uniqueness property was valid. JamesBWatson (talk) 11:47, 19 April 2010 (UTC)

I updated the Motivation section to mention how one can obtain the gamma function from Bohr-Mollerup and analytic continuation. I felt the section left me hanging otherwise, ending with "but this can be done in infinitely many ways, so... well, we haven't actually motivated it. Sorry." Feel free to reword, etc. 67.158.43.41 (talk) 10:49, 26 October 2010 (UTC)

I changed your wording a little bit. Please check that I haven't thwarted your intentions. However, I'm not 100% sure I support this text in the motivation section. It doesn't address why I should care about logarithmic convexity. Quantling (talk) 19:48, 26 October 2010 (UTC)

What makes Γ(z) unique to me is that it is the only function defined on all the complex numbers z (other than the non-positive integers) that satisfies all of

${\begin{aligned}\Gamma (1)&=1\\\Gamma (z+1)&=z\Gamma (z)\\\lim _{n\rightarrow \infty }{\frac {\Gamma (n+z)}{(n-1)!\,n^{z}}}&=1\end{aligned}}$

where n is a positive integer in the last expression. I'd add it to the article if I had a citation to back me up. Quantling (talk) 20:59, 26 October 2010 (UTC)

I've updated the motivation again. The recurrence relation needs to be satisfied for positive real inputs, as mentioned in Bohr-Mollerup theorem. If I wrote complex last night (not worth looking in the history), oops, my bad; I've changed the variable to x since z tends to imply complex. I'd be happy to hear a motivation for logarithmic convexity, or for an entirely different true uniqueness condition to be used. The one you wrote above, if it's true, would be fine, so long as you motivated the final condition. Of the two conditions under consideration, I prefer logarithmic convexity, but I almost always prefer mental pictures to algebra.

The main point of my edit was that the motivation section didn't motivate the gamma function, it only motivated one of infinitely many alternatives--where the alternatives are nontrivial. Any text which accomplishes the goal of giving a brief uniqueness argument is exactly what I had in mind. 67.158.43.41 (talk) 08:23, 27 October 2010 (UTC)

I rearranged the paragraphs above to put them in chronological order, without so much indenting; hope that's okay. The motivation for the limit above is the observation that

${\frac {(n+k)!}{n!}}=(n+k)\cdot \ldots \cdot (n+1)\sim n^{k}\,$

for fixed integer k, as integer $n\rightarrow \infty$. Personally, I am motivated to have this be true for all complex values of k, not just the positive integers. And, lucky for me, such a requirement is sufficient to define the Gamma function uniquely! That is, the limit above is merely the requirement that this asymptotic result for positive integer k holds for any complex value of z, offset by 1 for the usual discrepancy between the Gamma function and the factorial function. Quantling (talk) 13:10, 27 October 2010 (UTC)

I strongly prefer the logarithmically convexity condition. The other one is just saying the Gamma function is the only function such that $\Gamma (z+1)=z\Gamma (z)$ and that is equal to the Gamma function! In fact, using the first condition, the third one is clearly equivalent to Euler definition of the Gamma function (and the proof of the equivalence between Euler definition and the classical one is quite easy); also, the condition $\Gamma (1)=1$ is not really needed, it is implied by the other 2.--Sandrobt (talk) 19:53, 27 October 2010 (UTC)

Sandrobt, thank you for your observation that the second two conditions imply the first! Also, I agree that, together, the second and third expressions do lead very naturally to Euler's original definition of the Gamma function.

However, could you clarify the rest of your statement a bit? For instance, the first two conditions (Γ(1) = 1 and Γ(z+1) = zΓ(z)) are not sufficient to uniquely determine the Gamma function; the third expression or logarithmic convexity (or another alternative) is required, yes? I think the question we are discussing is, which is the most motivating among these "third conditions"? I am having trouble understanding whether you are addressing this question … or another question that I am being too dense to pick up. If the former, would you say why you consider logarithmic convexity to be more motivating than the asymptotic property? Clearly, I am not understanding your intent; please help. Thanks —Quantling (talk) 20:38, 27 October 2010 (UTC)

Some third condition is most definitely required. For instance, the relatively trivial solution of f = (n-1)! for positive integers n, f = 0 elsewhere satisfies the first two conditions. After taking a closer look at your proposal, $\lim _{n\rightarrow \infty }{\frac {\Gamma (n+z)}{(n-1)!\,n^{z}}}=1$ (which, incidentally, I believe has an off-by-one error) is, as you both say, just a rearrangement of Euler's definition, which itself isn't terribly motivating, to me at least. Stating the result in terms of asymptotics might be more intuitive. Something like

${\frac {(n+k)!}{n!}}\sim n^{k}\,$

so we require

${\frac {\Gamma ((n+k)!)}{n!}}\sim n^{k}$

which is equivalent to Euler's definition with the recursion formula (where any off-by-one errors have been fixed).

Of the two alternatives so far I continue to prefer logarithmic convexity. The other gets detail-heavy quite quickly, while on the other hand it doesn't seem horrible to gloss over why we require logarithmic convexity above anything else. I've tried thinking of motivations for logarithmic convexity, but haven't been terribly successful beyond "the factorial increases so fast that a continuous extension of it has its logarithm's tangent lines continually increase".

Also, to be clear, f(1) = 1 is a consequence of the other two definitions using Quantling's suggestion, while f(1) = 1 is very much required for my suggestion. I think that's what you both meant anyway. 67.158.43.41 (talk) 00:17, 28 October 2010 (UTC)

I am having trouble understanding why the ease of the third condition in leading to Euler's original formulation is considered a drawback. If anything, usually ease is considered a positive! In this case, I see the ease as neither a drawback nor an advantage. Rather, for me, it all rests on whether the stated "third condition" is appealing to the not-so-expert reader; contrastingly, how that "axiom" is then used to come up with this or that formulation of the Gamma function is more-or-less insignificant. So, I ask myself, when teaching students about the Gamma function for the first time, would I rather say that the chosen function is motivated by

${\frac {\Gamma (n+k+1)}{n!}}\sim n^{k}\qquad (1)$

for k any complex number, in the asymptotic limit as positive integer n goes to infinity, or would I rather say it is motivated by

${\frac {d^{2}}{dx^{2}}}\ln \Gamma (x)>0\qquad (2)$

for all positive real x? The choice is easy for me: it is expression (1). Quantling (talk) 12:46, 28 October 2010 (UTC)

To answer your question above, of course the recurrence relation together with being 1 at 1 is not enough to determine the Gamma function. Also, the Γ(1)=1 condition is surely needed in the Bohr–Mollerup theorem. Anyway, I think we all agree with those two facts. I prefer the log-convexity condition for two reasons: firstly because that one is the classic way to do it and secondly because Bohr–Mollerup theorem is not a trivial theorem at all (and it's a condition that could even be useful in some cases. for example one of the proofs of Stirling's formula uses that theorem), while the other one (together with the recurrence relation) can be quite easily shown to be equivalent to the classic definition, so it's more like a definition of Γ than a characterization of it. Also, the log-convexity doesn't sound so strange to me: if you draw the "simplest" smooth line passing through (1,1!), (2,2!), (3,3!), ..., you probably would draw a log-convex function.--Sandrobt (talk) 15:37, 28 October 2010 (UTC)

It seems we all understand each other and the alternatives are clear. So far it's 2 votes for log convexity, 1 for the asymptotic condition. If anyone else has an opinion or alternative, please chime in. Otherwise, it seems to me log convexity happens to win for now. 67.158.43.41 (talk) 23:29, 28 October 2010 (UTC)

For what it is worth, instead of having Equation 1 hold for any complex number $k$, it would be sufficient to have it hold for any positive real $k$. That is, this would uniquely determine the Gamma function on the positive reals. Then the function definition could be extended to the complex plan exactly as it is done after use of the Bohr–Mollerup theorem, with analytic continuation. —Quantling (talk | contribs) 01:22, 7 November 2010 (UTC)

Also, for what it is worth, it is sufficient for Equation 2 to hold only in a limiting case. That is, if

$\liminf _{x\rightarrow \infty }{\frac {d^{2}}{dx^{2}}}\ln \Gamma (x)\geq 0\qquad (2,{\text{limiting}})$

then that, together with $\Gamma (1)=1$ and $\Gamma (x+1)=x\Gamma (x)$, is sufficient to prove logarithmic convexity over all real values of $x$, and thus is sufficient to uniquely define the Gamma function (via analytic continuation). —Quantling (talk | contribs) 14:48, 10 November 2010 (UTC)

Well done!

This extremely well-written article did a great job explaining the concept to me and, as a historian, I found the history section fascinating. Well done all! -- Michael Scott Cuthbert (talk) 03:57, 24 August 2011 (UTC)

Plot of reciprocal gamma function

Somewhere near the mention of the reciprocal gamma function, I'd like to see a plot of it. I think there is value in seeing how (much better) it behaves for non-positive real values. Can someone add that? Perhaps plot from −5 to +5 as is done for several of the other plots. Thanks —Quantling (talk | contribs) 16:10, 23 September 2011 (UTC)

Limiting property

I do not have a citation for the fact that for any $k>0$

$\lim _{z\to 0^{+}}{\frac {\Gamma (kz)}{\Gamma (z)}}={\frac {1}{k}}$

This is something I proved as part of my own work and know it to be true. Using the

$\Gamma (z)={\frac {1}{z}}\prod _{n=1}^{\infty }{\frac {\left(1+{\frac {1}{n}}\right)^{z}}{1+{\frac {z}{n}}}}$

characterization of the gamma function for a non-negative argument, this property is proven trivially. I included this because I felt this could be of some use to someone, somewhere. I'm not sure what the criteria for inclusion is, but the note in the edit history indicated to bring it up here. —Preceding unsigned comment added by 141.211.222.131 (talk) 02:54, 18 March 2010 (UTC)

I don't think this is particularly interesting (btw, if you want you can drop the + in the limit), it's just a trivial consequence of the fact that Gamma has a simple pole in zero.--Sandrobt (talk) 03:02, 18 March 2010 (UTC)

It's pretty easy to prove from the recurrence relation, yes?

$\lim _{z\to 0}{\frac {\Gamma (kz)}{\Gamma (z)}}=\lim _{z\to 0}{\frac {\Gamma (1+kz)/kz}{\Gamma (1+z)/z}}=\lim _{z\to 0}{\frac {\Gamma (1+kz)}{\Gamma (1+z)}}{\frac {z}{kz}}={\frac {1}{k}}$

—Quantling (talk | contribs) 16:17, 23 September 2011 (UTC)

Gamma or gamma

There is really no reason to capitalize the name of the gamma function—it's a function like the sine or the logarithm, none of which are capitalized. The fact that TeX \Gamma and HTML &Gamma; need a capital G is not relevant.
—Herbee 15:02, 2004 Mar 3 (UTC)

The needs of TeX and html are not the reason why people capitalize it. The reason is that the capital Greek letter Γ is used. Are there really people who learned TeX and html before learning the Greek alphabet? I suppose nowadays there probably are, but it seems bizarre. Michael Hardy 19:48, 3 Mar 2004 (UTC)

Agree with Michael. Put Gamma instead of gamma. Oleg Alexandrov 07:01, 21 Feb 2005 (UTC)

Every source I believe to be authoritatative says "Gamma" except Erdelyi (Higher Transcendental Functions). I think "Gamma" is right. Paul Reiser 14:34, 21 Feb 2005 (UTC)

It also avoids confusion with the Euler-Mascheroni constant, which is conventionally written as a lower-case "gamma".

Hair Commodore 13:51, 31 October 2006 (UTC)

I concur with Gamma. JJL 16:36, 31 October 2006 (UTC)

2011

I read where the article said there is no accepted standard for whether the function name should be written "gamma function" or "Gamma function" before I noticed this old talk section. I already switched it to lower case, per that statement and MOS:CAPS. Most sources do use lower case, and WP style is to do the same. Dicklyon (talk) 14:33, 19 October 2011 (UTC)

Plot in Motivation section does not agree with text

The text beginning the section on Motivation says "Find a smooth curve that connects the points (x, y) given by y = (x − 1)! at the positive integer values for x." However, the y-axis for the figure is n!, not (n - 1)! IapetusWave (talk) 18:11, 20 February 2012 (UTC)IapetusWave

Value table

I can't see any discussion on why the table of values that used to be here was removed. I can't think of a reasonable reason to have removed it - the half integer values were interesting, and although not wp pc, also useful for reference. Could we get some debate about this? —Preceding unsigned comment added by 64.206.141.60 (talk) 17:36, 19 February 2011 (UTC)

totally agree, I want to know Gamma(n/2)=?, and I don't want to spend time to read the shit, and then do the calculation. Jackzhp (talk) 15:50, 16 March 2011 (UTC)

Totally agree - its done. PAR (talk) 03:45, 27 July 2011 (UTC)

I don't see these -- were they removed again? — Preceding unsigned comment added by 65.123.216.4 (talk) 23:01, 4 May 2012 (UTC)

Articles to Factorial and Gamma function

Hi

I just made this comment [2] on the talk-page of the article factorial. I disliked to repeat it word by word again, but I think the very informative picture "File:Factorial05.jpg" should be (also/only?) here. It remains me, what nice identities are missing. Look e.g. on the central line Im(Gamma(z)) = 0. ;-) Or the simple closed formulae for Gamma(iy) and Gamma(1/2+iy). Regards. 2001:638:504:C00E:214:22FF:FE49:D786 (talk) —Preceding undated comment added 11:13, 21 September 2012 (UTC)

Alternative definition related to Laguerre polynomials

In the third of the alternative definitions (related to Laguerre polynomials) there is a t on the right but not on the left. — Preceding unsigned comment added by 74.15.52.33 (talk) 23:13, 16 August 2012 (UTC)

I'don't know and I'm too lazy to check it now. But here is a good start to recover the correct formula:

$\Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}L_{n}^{z}(t)\,dt$

This t in the wrong article formula is probable a remainder from integration by parts! :-) 2001:638:504:C00E:214:22FF:FE49:D786 (talk) 14:56, 21 September 2012 (UTC)

I have corrected it just. I think it should better be deleted. Far too high math for wikipedia. ;-) I didn't found the incomplete Gamma function here. Regards 178.203.233.57 (talk) —Preceding undated comment added 12:30, 22 September 2012 (UTC)

Positive Infinity

Right now the definition:

$\Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}\,{\rm {d}}t.$

uses $\infty$ for $+\infty$. I think this is a bad idea, as unsigned infinity is something completely different than a limit to positive infinity, i.e. if the domain were to be considered as $\mathbb {C} \cup \{\infty \}$. When dealing with sequences, it's fine, because negative infinity isn't used. However, here there is a danger of ambiguity. GFauxPas (talk) 21:49, 25 January 2013 (UTC)

Maybe it's not a good idea but it's very standard to use infty in writing an improper integral. I think it's simpler for us to just stick to a standard writing. -- Taku (talk) 03:04, 26 January 2013 (UTC)

Gamma and Euler's constant (e)

We know that e is defined as the (discrete) sum of the multiplicative inverses of the factorials (of natural numbers): $e=\sum _{n=0}^{\infty }{\frac {1}{n!}}$ . Now, since the integral is the continuous equivalent of a discrete sum, and the gamma function the generalization of the factorial function, it would follow that $\int _{0}^{\infty }{\frac {dx}{\Gamma (x)}}$ would be the continuous equivalent of the mathematical definition of the number e. When I tried to compute this value with Mathematica, it yielded a value close, though not identical, to that of the number e: 2.807770... instead of 2.71828..., the error being about 0.089... I guess my question would be two-fold: (1) is this correct ?, and (2) does it have any meaning ? (Like, for instance, the constant $\gamma$, which is defined as the difference between the [discrete] sum of the inverses of natural numbers, a.k.a the harmonic series, and the [continuous] integral of the same: $\gamma =\sum _{n=1}^{\infty }{\frac {1}{n}}-\int _{1}^{\infty }{\frac {dx}{x}}$). — 79.113.223.19 (talk) 09:51, 27 February 2013 (UTC)

Maybe you can get more answers at Wikipedia:Reference desk/Mathematics. -- Taku (talk) 12:44, 27 February 2013 (UTC)

Thanks! Apparently it's called the Fransén–Robinson constant. :-) — 79.113.225.144 (talk) 17:41, 2 March 2013 (UTC)

Splitting some complex math-tags into simple parts

Due to reports of slow reformatting, I have modified several equations, to use many simple <math> tags, rather display with just a single complex math tag. When using separate tags for each symbol, or short expression ("xy" or "x+y"), then the page seems to reuse rapid cache images for many of the symbols, at the rate of 600-1,000 simple math tags per second. While many complex equations format slowly (0.5~0.8 second), others are extremely rapid, as if some whole complex equations were pre-stored, and so the speed of reformatting has been hard to predict, from equation-to-equation. In some cases, I have put a clearer variable 'z' in an equation, while changing to the faster display format. The result is as follows, using numerous small math tags:

• <math>\Gamma(z)</math> <math>=</math> <math>\int_0^\infty</math> <math>x</math>^{''z'' -1} <math>e</math>^-''x'' <math>{\rm d}x</math>.
• $\Gamma (z)$ $=$ $\int _{0}^{\infty }$ $x$^{z -1} $e$ ^-x ${\rm {d}}x$.

Over the past few days, some complex <math> tags had become very slow, during edit-preview, due to a lack of image-cache activity, while other complex equations were running quickly. -Wikid77 (talk) 02:33, 10 March 2013 (UTC)

According to the discussion at WP:VPT, breaking up the < math> tags makes any problem worse. I've reverted all your edits (even to those within {{math}} tags, which may have been improvements. Please do not do that without, at least, verifying that all the equations appear identical. — Arthur Rubin (talk) 17:29, 10 March 2013 (UTC)

The fact that < math> tags no longer render on preview may make more difficult to edit, as well. But you need to verify that the equations actually resemble one that can be understood by a mathematician. — Arthur Rubin (talk) 17:31, 10 March 2013 (UTC)

Questions about recent edit

I was wondering why the recent edit (16:50, 27 December 2011‎ RogierBrussee) is necessary. This probably reveals my lack of math skills, but what is the advantage of using the new notation? Also, I a find the "z" in the superscript rather hard to read. Is there a better way to typeset this equation? Sirsparksalot (talk) 20:12, 3 January 2012 (UTC)

• Equations re-typeset with clearer 'z': Although it has been more than a year since requested, I have put a clearer variable 'z' in the equation, while changing to a faster display format. The result is as follows:

$\Gamma (z)$ $=$ $\int _{0}^{\infty }$ $x$^{z -1} $e$ ^-x ${\rm {d}}x$.

Some complex [math] tags had become very slow, during edit-preview, due to a lack of image-cache activity, while other complex equations were running quickly. -79.113.221.97 (talk) 20:57, 19 March 2013 (UTC)

Why is the Gamma function defined in unnatural way, off by one

That is why it is defined in such a way that Γ(n + 1) = n! and not just Γ(n) = n! (for integer n)? This is clearly unnatural because in every single formula where the Gamma function takes part, its argument is shifted by one in order to compensate for this. Its very definition is artificially shifted by one: $\Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}\,\mathrm {d} t.\!$

Example of a formula which utilizes the Gamma function is the volume of n-dimentional hypersphere: $V_{n}={\pi ^{n/2}R^{n} \over \Gamma ((n/2)+1)}.$

Generally this definition has the effect of adding a bit of extra (unnecessary) complexity in all formulas that use the function. —The preceding unsigned comment was added by 213.240.234.31 (talk) 12:51, August 23, 2007 (UTC)

It is natural when you come to the riemann zeta function, and several other functions. anyways the gamma function you speak of is actually the Pi function which also appears in this article. —Preceding unsigned comment added by T.Stokke (talk • contribs) 21:39, August 26, 2007 (UTC)

Do you refer to $\zeta (s)=2^{s}\pi ^{s-1}\sin \left({\frac {\pi s}{2}}\right)\Gamma (1-s)\zeta (1-s)$ ?

If you substitute Gamma with Pi in this equation, you get $\zeta (s)=2^{s}\pi ^{s-1}\sin \left({\frac {\pi s}{2}}\right)\Pi (-s)\zeta (1-s)$

which is hardly more complex than the first one.

In that sense the Gamma function's definition (as shifted by one) doesn't appear to be more natural in this case.

Yes I saw the paragraph about the that Pi function. But why there is the Gamma function and why is it more "standard" and more widely used than this Pi function although it makes the formulae more complex?

Also we could have some more redundant functions, like say "Lambda" function defined so that L(x)=(x-2)! and Mu function that M(x)=(x+3)!, etc. You get the idea.

Anyway i just hoped that someone knew the reason behind this strange definition. It always makes me wonder when I come across this function. I don't know of another similar case in the entire mathematics—Preceding unsigned comment added by 213.240.234.31 (talk) 14:29, 24 September 2007 (UTC)

Euler defined the Gamma function in a letter to Goldbach like $\Gamma (s):=\int _{0}^{1}(-\log(u))^{s-1}\mathrm {d} u$, and I personally think he made it so that it would be convergent for $\Re s>0$ instead of 1. A year earlier he also had a letter to Goldbach which defined $\Gamma _{p}(x):={\frac {p!p^{x}}{x(x+1)...(x+p)}}={\frac {p^{x}}{x(1+x/1)...(1+x/p)}}$ (which may also lead to the "off by one" integral definition?). Substituting $p^{x}=e^{x(\log(p)-1-1/2-...-1/p}e^{x+x/2+...+x/p}$. And letting $\lim _{p\to \infty }$ You get the Weierstrass definition of the Gamma function. ${\frac {1}{\Gamma (x)}}=xe^{\gamma x}\prod _{n=1}^{\infty }\left(1+{\frac {x}{n}}\right)e^{-x/n}$ Both of these two latter are mentioned in the article, the definite integral is not, still i belive these latter two to be the main reason for the offset definition. T.Stokke (talk) 11:14, 15 March 2008 (UTC)

Hi T.Stokke. I'm interested in the $\Gamma _{p}(x)$ formula you have included. I wonder if an upper part say $\Gamma _{u}(x)$ has also been defined such that $\Gamma _{p}(x)\Gamma _{u}(x)=\Gamma (x)$? Do you know of any sources for $\Gamma _{p}(x)$ and indeed the hypothetical $\Gamma _{u}(x)$? Mhallwiki (talk) 13:41, 27 July 2011 (UTC)

This discussion http://mathoverflow.net/questions/20960/why-is-the-gamma-function-shifted-from-the-factorial-by-1/25973] at Mathoverflow may be of interest to you. pma 06:45, 20 September 2012 (UTC).

It seems to me appropriate to include some of the arguments/quotes from that discussion on the Wikipedia article. Right now we write The gamma function can be seen as a solution to the following interpolation problem: "Find a smooth curve that connects the points (x, y) given by y = (x − 1)! at the positive integer values for x." which doesn't sound exactly right. --Thomasda (talk) 23:24, 28 May 2013 (UTC)

Another expression for Gamma ?

$\Gamma (x+1)=\int _{0}^{\infty }{e^{-{\sqrt[{x}]{t}}}}dt$ — 79.113.237.34 (talk) 07:19, 6 March 2013 (UTC)

Yes, using the change of variables $u={\sqrt[{x}]{t}}$.--Jasper Deng (talk) 06:33, 18 August 2014 (UTC)

Don't understand the motivation section

Hello, math is not my ability, but I found this confusing:

A more restrictive property than satisfying the above interpolation is to satisfy the recurrence relation defining a translated version of the factorial function,

   f(1) = 1 and f(x+1) = x f(x)

Is it
f(x + 1) = x f(x)
or
f(x + 1) = (x + 1) f(x) ???

Please add a clarification to the "motivation" section if the current formula is correct. Thanks. 200.28.224.36 (talk) 02:44, 13 May 2015 (UTC)

There is no error. See the functional equation discussed the "definition" section. It might help to notice that the gamma function and the factorial are related by: $\Gamma (n+1)=n!$. I get you (and everyone else) wonders why there is the +1 shift. There are at least two reasons: the one for the historical reason, the other has to do with the fact that dx/x is the invariant measure (cf. Haar measure) for the multiplicative action. I do agree that it makes sense to add some clarification (but I'm lazy busy and also don't know the good way to explain this confusing point.) -- Taku (talk) 23:34, 13 May 2015 (UTC)

An Entire Gamma Function Formula

a.k.a. How Euler was wrong.

https://sites.google.com/site/thebohrmagneton/the-entire-gamma-function 188.29.164.25 (talk) 11:49, 27 January 2015 (UTC)

I thought, by the uniqueness of analytic continuation, you can't really just change the behavior of the gamma function at certain areas. -- Taku (talk) 14:37, 27 January 2015 (UTC)

Although I'm sure, as is TakuyaMurata, the paper is almost certainly wrong, the problem with respect to use in Wikipedia is that it is not published. — Arthur Rubin (talk) 01:21, 29 January 2015 (UTC)

"Almost certainly wrong" - A logically inconsistent statement. Either it is or it isn't. The mathematics is supplied. 188.29.165.4 (talk) 22:57, 18 September 2015 (UTC)

The uniqueness of analytical continuation (and the non uniqueness of the gamma function (hence the log convex constraint)), is the problem. The entire gamma function is defined by relation to the Hurwitz zeta, and a differential relation. There is only one definition of it if Gamma(1) = 1. The functional equation does not hold, but then such an equation which does not in itself infer just one analytical continuation (there must be just one), leads to lots of further issues of the underspecified (under constrained) type. — Preceding unsigned comment added by 188.29.165.4 (talk) 22:46, 18 September 2015 (UTC)

Formal maths equations versus social sciences? Published as in made public for view versus some other definition of publishing? Polotics in maths versus the actual maths? 188.29.165.4 (talk) 23:07, 18 September 2015 (UTC)

Also of note is the noticing that it is ONLY the negative values of x for which the function appears to differ. As far as I've worked out the product of numbers down to 1 (or up to 1 for negatives), is replaced. The poles of digamma, half come from zeros in the entire gamma, and half come from poles in a limit of the differential of the entire gamma. That is to say it only has half the poles of the Euler gamma. This is likely linked to the Bernoulli numbers having a odd and even indexed, zero to value pattern. 188.29.164.91 (talk) 16:03, 19 September 2015 (UTC)

No reliable source -- can be ignored. If it ever gets written up in a reputable journal, we can reconsider. --Macrakis (talk) 19:23, 19 September 2015 (UTC)

No my early point was that the "uniqueness" of analytic continuation prohibits the functions to differ only at negative numbers. You can only uniquely extend the gamma function on the positive half-plane to the negative side. If your function differ at some points, it may or may not be an interesting function to study but it's just not the gamma function; there cannot be multiple gamma functions. (Analytic continuation is something everyone learns in an elementary school but maybe this needs to be clarified in the article?) -- Taku (talk) 21:16, 19 September 2015 (UTC)

The functional equation counts down in ones for positives, and counts up in ones for negatives. onto a known strip (to be analytically continued from) of width one. In this sense the functional equation is a constraint assumed to hold for all values both positive and negative. If this functional equation is used as part of the process of continuation, you end up with the Euler gamma. If however the functional equation is assumed to not hold for all values (this does not prevent it holding for some values), and other methods are used for analytic continuation, by definition a different function is defined for all values. This does not prevent parts of those functions being identical where the functional equation holds, just as they would be expected to differ where it does not hold. Considering that the need for the functional equation to hold, so to be a factorial extension falls on the positive x axis (for all or almost all uses), the need for it to hold for negative values (quite abstract kind of need), may prevent a more useful analytic continuation. In that sense the functional equations role in the Euler continuation is what causes the lack of utility. The negative side just becomes a kind of reflection of all that holds true for positive values, instead of being a more useful continuation.

In fact it maybe that many of the only half plane defined issues of modular form functions comes from this pre-constraint to not bother to find the actual analytic continuation, but to assume a fully valid functional equation range to make a bastardised "continuation" to have something to play with. — Preceding unsigned comment added by 188.29.164.91 (talk) 17:07, 20 September 2015 (UTC)

To put it bluntly. How can the full range validity of the functional equation be assumed before the continuation is complete? The functional equation then becomes definition and can in no way ever be proved without circular reasoning. 188.29.164.91 (talk) 17:17, 20 September 2015 (UTC)

To put it bluntly, that no sense makes. The functional equation holds for Re z >= 1; so it must hold everywhere. It's not immediately obvious, but, as there can be only one solution to the functional equation, there can be only one analytic continuation. — Arthur Rubin (talk) 06:40, 21 September 2015 (UTC)

"The functional equation holds for Re z >= 1; so it must hold everywhere" - This is incorrect logic, just as I can breathe above water (without apparatus), so I must be able to breathe below water" - The functional equation (assumed valid everywhere) is used in the Euler continuation, therefore there is no surprise it is valid by definition. This does not prevent a different continuation existing where no assumption on range of validity is made for the functional equation, and continuation is made via a differential equation from a different source. After this continuation is made, it is then found that the functional equation does not haven to hold for the entire range of z. 188.29.164.63 (talk) 12:02, 21 September 2015 (UTC)

Analytic continuations are (locally) unique. In other words, if f is analytic on an open set U, and g and h are analytic on a larger open set V, and f, g, and h are identical on U, then g and h are identical on V. In this case, let $f(x)=x\Gamma (x)$ on $U=\{z\mid \operatorname {Re} z>1\}$, let $g(x)=\Gamma (x+1)$, and let w be your proposed extension , and let $h(x)=xw(x)$, and let $k(x)=w(x+1)$. Then h and k are identical on their common domain. — Arthur Rubin (talk) 13:57, 21 September 2015 (UTC)

Sounds like a proof by negation which might have not taken into account Godel's work, and the alternate negation possibility of system (or assumption) inconsistency. On V (the whole real set), it is not assumed by me that h(x) = x w(x). In fact on V the larger open set, I know this could not be true, as the functional equation does not hold there. But as the digamma is a ratio of the differential of gamma over gamma, the function described by me is to a multiplicative constant equal to gamma for all z >= 1. Otherwise the particular relation of digamma to the Hurwitz zeta would not be true. This is what then leads to a differential equation with the one constant of integration (the multiplicative constant), and the log poles at the negative integers and zero, lead to alternate signed infinities raised as a power of e. That is to say alternate zeros and poles. Thus it would not be possible for the function equation to hold, as an infinity times a finite no zero number, would never equal zero, in the limits. 188.29.164.63 (talk) 19:09, 21 September 2015 (UTC)

Sounds like nonsense. In the case of a merimorphic (sp?) function on all of C, the whole is the analytic continuation of any "patch" (its restriction to any open subset of C). If there are more "complex" singularities, it might only be true in the sense of multi-valued analytic functions (see sheaf theory). — Arthur Rubin (talk) 22:59, 21 September 2015 (UTC)

Yes the negative half plane is multivalued on C. As the 2 k pi i from a log is multiplied by a non integer sometimes, before being the raising power of e. 188.29.164.220 (talk) 12:17, 22 September 2015 (UTC)

I'm absolutely sure that is nonsense. log Gamma is multivalued in the complex plain, but Gamma cannot be. — Arthur Rubin (talk) 14:57, 22 September 2015 (UTC)

I didn't carefully read the posts but a functional equation and a differential equation don't matter in this issue. Let me put my early point in this way:

1. On some open set, a meromohic function has poles or is holomorphic otherwise (this is known but requires a proof.)
2. If there is an analytic function f that agrees with the gamma function Γ(t) on some open set, then the two functions must agree on where they are both defined.

1. and 2. now imply that if your function is entire, then it cannot be the gamma function Γ(t), which has poles. The key point is that the process of how analytic continuation is done is irrelevant. -- Taku (talk) 20:49, 22 September 2015 (UTC)

Definitely it is meromorphic, and I'd agree with 1. As for 2. I would not agree with. Maybe it's not entire, but I have seen definitions of entire which take into account ignoring the poles, as having a pole does not necessarily prevent a differential of the function existing at the pole. Which would make such a function entirely analytic. 188.29.164.146 (talk) 22:43, 22 September 2015 (UTC)

Maybe you mean smooth, with no discontinuities and resultant infinities. So on point 2, as to the method of continuation having a possible effect on the continued function. (not in a confined annulus or half plane convergence sense), but meromorphic identity. OK? — Preceding unsigned comment added by 188.29.164.146 (talk) 22:55, 22 September 2015 (UTC)

In fact, we don't even need to cite analytic continuation: Let $U\subset V\subset \mathbb {C}$ be nonempty open subsets. If f and g are holomorphic (i.e., analytic) functions on V and if f and g coincide on U, then f - g has a nonisolated zero on V; only the zero among holomorphic functions on V can have a nonisolated zero; so f = g on V. (This proves the uniqueness of analytic continuation by the way). Apply this with g = the gamma function, U = { Re z > 1 }, V the complement of the poles of g in C. If f is entire meaning holomorphic everywhere and if f = g on U, then g is a meromorphic function that is holomorphic on C and by 1. g cannot have poles, contradiction.

(Here it is important to remember two meromorphic functions are considered the same if they differ on some finite set; that is, as a meromorphic function, z^2/z is considered holomorphic at z = 0, since it doesn't have a pole at z = 0.)

The point is that things like convergence of integral or series is irrelevant to analytic continuation. The use of functional equation is perhaps most elegant but need not be the only way. -- Taku (talk) 02:31, 23 September 2015 (UTC).

But given f(z) = g'(z)/g(z) then g(z) = Const.exp(integral(f(z), z)). For a particular f(z), the integral has to be unique, to have the same differential, and on the real line g(z) hence has to be unique for unique f(z) to within multiplication by a constant. So for the digamma Hurwitz zeta relation it uniquely defines a unique g(z) for real z (it may be multivalued and complex). L'hopitals rule in the limit does give such a function.

Proof by negation is not mathematically sound. NOT(red) can not be assumed to be GREEN. The implicit binary assumption of truth of a postulate is not valid. — Preceding unsigned comment added by 188.29.164.146 (talk) 13:41, 23 September 2015 (UTC)

What you call "proof by negation" is generally considered mathematically sound; and, even in intuitionistic logic, one can show that "your" entire function is not identical to the Gamma function at some specific complex number z with Re z > 1. I cannot name it because you haven't identified the function.

In any case, your material is unpublished, so should not be included in the article. If published in a normally reliable journal, we could then discuss the fact that it is considered impossible by subject-matter experts here should then determine that the publication is not reliable, and cast doubt on the journal. — Arthur Rubin (talk) 19:19, 23 September 2015 (UTC)

Complex infinity a singular object? And how does Zeta(z) = -1/12? I heard a half explination on a WKB course. 188.29.164.146 (talk) 19:37, 23 September 2015 (UTC)

If complex numbers are a combination of two numbers, what prevents both x and y being complex? What are imaginary solutions? Simon Jackson, BEng. (Electrical and Electronic) 188.29.164.146 (talk) 19:42, 23 September 2015 (UTC)

Most complex analysis works in the complex projective line, with one point at infinity, rather than the real projective plane, with a line at infinity. And ζ(1) = -1/12 is explained in a number of places on Wikipedia, including Particular values of Riemann zeta function and 1 + 2 + 3 + 4 + ⋯.

See Bicomplex number for what happens with x + i y where x and y are complex.

In any case, none of this is relevant here. — Arthur Rubin (talk) 21:19, 23 September 2015 (UTC)

Talk pages are "not for general discussion of the topic". Without a reliable source for an alternate definition, there is no point in this discussion. --Macrakis (talk) 22:18, 23 September 2015 (UTC)

$\Gamma (z)=e^{\left[\sum _{n=0}^{\infty }{\frac {1}{n+1}}\sum _{k=0}^{n}(-1)^{k+1}\left({\begin{matrix}n\\k\end{matrix}}\right)\ln \left({\frac {k+z}{k+1}}\right)\right]}$ 188.29.165.184 (talk) 17:32, 3 October 2015 (UTC)

Undefined Variables

What is the x in the improper integral at the beginning of the article? It doesn't say.

This article would be a lot more useful and informative if it said what "x" was representing in the equation.75.128.143.229 (talk) 01:54, 12 March 2015 (UTC)

If you mean the integral in the sixth line from the top of the article, the "x" is defined. It disappears when the integration is carried out. It always appears in integrals of this type. — Preceding unsigned comment added by 212.159.119.123 (talk) 13:24, 9 March 2016 (UTC)

The "x" is the positive numbers and 0 from 0 to infinity. — Preceding unsigned comment added by 212.159.119.123 (talk) 13:29, 9 March 2016 (UTC)

arcGamma

It would be good to include also some formulas for the evaluation of the arcGamma function. dima (talk) 11:39, 14 January 2009 (UTC)

You have just volunteered to do a lot of work. — Preceding unsigned comment added by 212.159.119.123 (talk) 13:50, 11 March 2016 (UTC)

Which is Euler's definition?

Both the Euler integral of the second kind and the infinite product formulation are attributed to Euler. Is that right? In the article only the latter is referred to as "Euler's definition"; that seems confusing to me. 𝕃eegrc (talk) 12:10, 21 July 2016 (UTC)

Non-positive integer arguments

The (sub)section on particular values lists entries for three non-positive integer arguments as infinity: $\Gamma (-2)=\Gamma (-1)=\Gamma (0)=\infty$. However, both the lede and the separate article Particular values of the gamma function state that $\Gamma$ is undefined for non-positive integers. Moreover, the picture graphing $\Gamma$ suggests that for any non-positive integer argument $\Gamma$ goes to $+\infty$ when approaching from one side and to $-\infty$ when approaching from the other, furthering my suspicion that $\Gamma (-2)$, $\Gamma (-1)$ and $\Gamma (0)$ really are undefined and therefore incorrectly stated in said table to be of value $\infty$. Any thoughts? -- Skysmurf  (Talk) 00:47, 29 November 2016 (UTC)

I agree that the complex-valued gamma function is not defined at these values. However, the reciprocal of the gamma function is defined and analytic everywhere on the complex plane. I have edited the section accordingly. 𝕃eegrc (talk) 13:13, 29 November 2016 (UTC)

That looks good to me, thanks. -- Skysmurf  (Talk) 14:05, 29 November 2016 (UTC)

I edited it further. I hope you like it. 𝕃eegrc (talk) 14:51, 29 November 2016 (UTC)

Incorrect branch cut for log Gamma(z)

The function log Gamma(z) was computed using incorrectly chosen branch cuts, and therefore incorrectly shows zeros that do not exist. In Python, the incorrect call is np.log(scipy.special.gamma(x+y*1j)). The correct call is scipy.special.loggamma(x+y*1j)). I will load a correctly computed function of log Gamma(z). A quick way to verify the incorrectness of the function is a call to minimize the incorrect function, which yields a nonexistent zero:

xf = scipy.optimize.minimize(lambda x: np.abs(np.log(sps.gamma(x[0]+x[1]*1j))), np.array([4.,4.]), method='Nelder-Mead', tol=1.e-12)
print(np.log(sps.gamma(xf.x[0]+xf.x[1]*1j)))
print(sps.loggamma(xf.x[0]+xf.x[1]*1j))

The correct function has a value of 2πi at this value, not 0 as incorrectly shown in the original image. Stsmith (talk) 14:17, 26 April 2017 (UTC)

You might be right, but I think it would be better to include the same type of plot (i.e. a color 2D plot) instead. Or even better, include both so one can see the difference between both versions. --Deacon Vorbis (talk) 14:31, 26 April 2017 (UTC)

The color 2-D plots used in the page do not intuitively illustrate the 4-dimensionality of analytic functions. Rather, the 2-D plots should be replaced with traditional 3-D plots like Jahnke's and complex surfaces found in places like Charles and Ray Eames's Mathematica exhibit and elsewhere. Stsmith (talk) 14:54, 26 April 2017 (UTC)

Jahnke's 3-D gamma function illustration

Parametrization

In the section 'In terms of generalized Laguerre polynomials' an unusual parametrization is mentioned. I wonder what the role of the variable t is? Madyno (talk) 12:21, 16 August 2017 (UTC)

Return period

I forgot to add to the edit summary, but the text "time between occurrences of ..." shouldn't really be changed to "return period of ..." per MOS:JARGON. As a side note, that article is in pretty poor shape, but that's not really too important here I guess. --Deacon Vorbis (talk) 04:08, 21 September 2017 (UTC)

Caption

The caption of this figure was as shown here.

It is easy graphically to interpolate the factorial function to non-integer values, but is there a formula that describes the resulting curve?

A caption is supposed to give information, not just ask a question to which the answer is known without answering it, so I extended the caption by one sentence to say

It is easy graphically to interpolate the factorial function to non-integer values, but is there a formula that describes the resulting curve? The gamma function is the most useful such formula.

This was reverted by @Sapphorain: with the edit summary The gamma function is not a formula.

First, the lead of the article Formula says

In mathematics, a formula is an entity constructed using the symbols and formation rules of a given logical language.

So the gamma function is a formula according to this, although I would have no problem changing to a different word if "formula" is also replaced in the preceding sentence of the caption.

But second, the main point is that we cannot justify having a caption, like the one that has been restored, (a) that asks a question without answering it, even though answering it is the whole point of the article, and (b) that doesn't even mention the subject matter of the article.

Please suggest something better than what you've restored. Thanks. Loraof (talk) 20:48, 15 October 2017 (UTC)

No. The gamma function is not a formula, the gamma function is a function. It can be defined by a formula. Sapphorain (talk) 06:40, 16 October 2017 (UTC)

I replaced the comment under the caption by « The gamma function interpolates the factorial function to non-integer values ». Everything else was verbiage. There are many ways to graphically perform this interpolation, and thus many functions that describe the resulting curve. The statement that the gamma function is « the most useful such function » is very vague and doesn’t provide any valuable information. Sapphorain (talk) 21:34, 21 October 2017 (UTC)

Polynomial Expansion

Recognizing that for all x>0, I can find Gamma(x) simply by repeated multiplication x*(x-1)*(x-2)... until x-n is less than 1. We really only need a table or a means of calculating Gamma(x-n) for 0<x-n<1. I attempted to find a convergent polynomial expansion of Gamma(x). I chose to center the expansion around alpha=0.5, being in the middle of the range I required convergence. What I found was quite odd though. All coefficients follow an interesting pattern. The constant term is naturally sqrt(pi)=1.77..., just under 2. And for all subsequent terms, the coefficient is "just under" the next power of 2. The powers of two is fascinating and I cant explain it, but there doesnt seem to be any pattern for how far under the power of two the coefficient is. But I can say that coefficients get closer and closer; the coefficients themselves converge to powers of two. So Gamma(x+0.5) is roughly an alternating series with coefficients of powers of two (just under). What Im trying to find is the variation in the coefficient. I think this polynomial expansion is worth discussing as part of the many representation of Gamma(x). I dont see any polynomial representations in the article, and this one Ive come across by accident has an odd connection to powers of two. If anyone has legitimate insights to contribute, lets talk, instead of deleting my contribution to a TALK PAGE. 50.125.86.70 (talk) 16:57, 27 January 2019 (UTC)

Okay, now you talk about adding it to the article (before, it was not clear what you wanted to do regarding the article). I don't know what you mean by "polynomial expansion" (the gamma function is most certainly not a polynomial as it has no roots and isn't constant; do you mean a Taylor series? (even then, the Taylor series isn't particularly interesting since its coefficients can't be computed that easily; the gamma function is a transcendentally transcendental function and therefore not even a generalized hypergeometric function). To insert it into the article would require a rigorous definition of that, and supporting reliable sources.--Jasper Deng (talk) 20:30, 27 January 2019 (UTC)

Presumably he means the Taylor series. Wolfram Alpha seems to confirm this (link). That is, $\Gamma ^{(n)}(1/2)\sim (-1)^{n}\cdot 2^{n+1}\cdot n!$ I'd be surprised if no one has pointed this out before, but I don't know why this happens, and I don't know where you'd be able to cite anything about it. –Deacon Vorbis (carbon • videos) 00:42, 28 January 2019 (UTC)

If we can find where someone has pointed it out before we may also find an explanation in the same place. But this sort of thing seems difficult to search for. —David Eppstein (talk) 01:00, 28 January 2019 (UTC)

I think this is how it goes. The nth derivative at $x=0.5$ is $\int _{0}^{\infty }\ln(x)^{n}x^{-1/2}e^{-x}dx$. Change variables with $u=\ln(x),du={\frac {dx}{x}},dx=xdu=e^{u}du$ to obtain $\int _{-\infty }^{\infty }u^{n}e^{-u/2}e^{u}e^{-e^{u}}du$. Since ${\frac {d}{du}}e^{-e^{u}}=-e^{u}e^{-e^{u}}$, I believe we can repeatedly integrate by parts to obtain an inductive formula in n, with the power of 2 and factorial resulting from integrating $u^{n}e^{-u/2}$. This is only a rough idea. I'll need to return to it later. Note that this is for my own curiosity only, absent reliable sources which would be needed to insert it into the article.--Jasper Deng (talk) 04:37, 28 January 2019 (UTC)

Fractional?

It says that Gamma extends the factorial to "fractional and complex" values. Does fractional not tend to imply only rational numbers? And in fact, would not just "complex" be sufficient? I could be very wrong here. I'm not sure whether there's a technical meaning of "fractional." Hence my asking. — Preceding unsigned comment added by 68.185.202.7 (talk) 08:28, 28 February 2006 (UTC)

Interesting point. Neither is fully accurate though. "Complex" and "rational" also include the negative integers, doesnt it? But those are clearly excluded from consideration. Thus neither set is an appropriate term to use. Though gamma is extended to all real numbers (excluding negative integers), one probably would not compute such a value but on a computer. And computers only deal in rationals, as all floating point numbers are of finite-digit precision. The term "fractional" is used colloquially rather than technically; the implication is that values between the integers are allowed. In other words, a non-zero "fractional part". I think by "fractional numbers" they simply mean numbers other than integers. 50.125.86.70 (talk) 07:32, 17 February 2019 (UTC)

Double factorial

At one point in the article, it gives a formula for the value of Γ(x) at x = n/2 + 1 for n odd:

                       Γ(n/2 + 1) = sqrt(π)n!!·2(n+1)/2 for n odd,

and explains that "n!!" means a double factorial, with this same link provided (to the article "Factorial" in which the double factorial (for n odd) is defined as [what's equivalent to]

            n!! := n·(n-2)· . . . ·3·1.

But this double factorial notation terminology and notation is very much not universally understood. In many venues both the terminology and notation each refer to the factorial of the factorial (so, e.g., 3!! = 6! = 720 (and not 3).

Because there is a compact way to express Γ(n/2 + 1) (for n odd) in a more readily and universally-understood way -- namely as:

            Γ(n/2 + 1)  = sqrt(π)n!/(2n((n-1)/2)!)  n odd,

or what's even clearer,

            Γ(n/2 + 1)  =  sqrt(π)n!/(2nk!)   for all odd n = 2k + 1

-- I suggest that this is preferable to using by-no-means-universally-accepted notation and terminology.

Perhaps it's also worth mentioning that Γ(n/2 + 1) may be thought of as a way to define (n/2)! for odd n.Daqu 02:48, 16 May 2007 (UTC)

I would disagree. I have been around the block a few times in the mathematical field. Perhaps subjective and anecdotal, but I have NEVER seen 3!! = 720. If they meant that they'd write (3!)! or perhaps, if you want to be lazy, 3!,! Every source and reference I have ever seen clearly make 3!! = 6 as the "double factorial". I would like to see the sources you claim say otherwise.50.125.86.70 (talk) 07:38, 17 February 2019 (UTC)

\operatorname

${\begin{aligned}&\left.{\begin{array}{l}n\Gamma (n)\\[4pt]n\operatorname {\Gamma } (n)\end{array}}\right\}\longleftarrow {\text{These conspicuously differ from each other typographically.}}\\[6pt]&n\cos(n)\end{aligned}}$

It does appear that the spacing is the same as the standard spacing with \cos, \log, \exp, etc. only when \operatorname{} is used. That does not surprise me. Michael Hardy (talk) 16:20, 25 February 2019 (UTC)

See User talk:A1E6#\operatorname for some more detailed explanation. –Deacon Vorbis (carbon • videos) 16:23, 25 February 2019 (UTC)

Semi-protected edit request on 20 July 2019

Grammar - The portion of the opening statement included, should have a comma between the word numbers and the clause except the non-positive integers. Shown below. The way it is stated, when first read, makes it say that it is defined for all complex numbers except non-pos integers and positive integers, which would be all integers, and it's not until you continue reading that you realize the second part is setting you up to say, and for any positive integer n, so forth and so on. Reads much cleaner with comma. The gamma function is defined for all complex numbers, except the non-positive integers, and for any positive integer n {\displaystyle n} n, 2601:445:4480:490:14FC:344A:D522:1F9B (talk) 16:28, 20 July 2019 (UTC)

 Not done: the comma after the phrase "the non-positive integers" makes it unambiguous. Adding your comma would actually induce ambiguity. DVdm (talk) 16:40, 20 July 2019 (UTC)

Wrong integral representation of the Riemann zeta

The exponent of the integrand of the third equation of the section "Relation to other functions" is wrong. http://mathworld.wolfram.com/RiemannZetaFunction.html. The integral representation of the Riemann $z$ appears to be wrong all over Wikipedia, and it carries with it other mistakes (i.e. log-gamma integral expression). — Preceding unsigned comment added by 160.78.67.137 (talk) 12:14, 24 January 2020 (UTC)

Please put new talk page messages at the bottom of talk pages and sign your messages with four tildes (~~~~) — See Help:Using talk pages. Thanks.

Look at the integrand: it is divided by u. That's the factor -1 you are looking for:

$\zeta (z)\Gamma (z)=\int _{0}^{\infty }{\frac {u^{z}}{e^{u}-1}}\,{\frac {du}{u}}=\int _{0}^{\infty }{\frac {u^{z-1}}{e^{u}-1}}\,du$

- DVdm (talk) 12:31, 24 January 2020 (UTC)

Right, I rewrote the three occurances the standard way. I believe that the log-gamma function was indeed wrong, now it should match Gradshteyn and Ryzhik 8.341.3 — Preceding unsigned comment added by 160.78.67.137 (talk • contribs) 13:38, 24 January 2020 (UTC)

Please indent your talk page messages as outlined in wp:THREAD and wp:INDENT, and sign them with four tildes (~~~~) — See Help:Using talk pages. Thanks. - DVdm (talk) 15:26, 24 January 2020 (UTC)

23 April 2020 additions

@Leo.Mathematics: Your additions amount to WP:OR (as you yourself stated in the first edit summary) and thus can't be added. Less important (but still somewhat important) is that it was formatted extremely poorly (things like an asterisk for multiplication) to the point where it's impossible to unambiguously parse the intended meaning of the formula. The tone was also highly inappropriate (the "great care has been taken" bit, etc). Please take some time and read the links that are in the welcome message that I dropped on your talk page, as well as some related pages, like the math MOS page. –Deacon Vorbis (carbon • videos) 01:11, 24 April 2020 (UTC)

• unfortunately they keep deleting this code from the "Gamma function" page;
• however: the file https://github.com/discoleo/R/blob/master/Math/Integrals.Gamma.R on Github represents the original material: any other derivative will need to reference this original - even if it is personal research;

[add in the section with the Multiplication Theorem]

= = Real uses of the Multiplication Formula: = =

Relations derived using the multiplication theorem (Ref ):

References

$\Gamma (1/6)=\Gamma (1/3)^{2}/{\sqrt {\pi }}*2^{2/3}*\sin({\pi /3}).$

$\Gamma (5/6)=1/\Gamma (1/3)^{2}*{\sqrt {\pi }}^{3}*2^{4/3}/{\sqrt {3}}.$

$\Gamma (1/10)=\Gamma (1/5)*\Gamma (2/5)/{\sqrt {\pi }}*2^{4/5}*\sin({2*\pi /5}).$

$\Gamma (3/10)=\Gamma (1/5)/\Gamma (2/5)*{\sqrt {\pi }}/2^{3/5}/\sin({3*\pi /10}).$

$\Gamma (7/10)=\Gamma (2/5)/\Gamma (1/5)*{\sqrt {\pi }}*2^{3/5}.$

$\Gamma (9/10)=1/(\Gamma (1/5)*\Gamma (2/5))*{\sqrt {\pi }}^{3}/2^{4/5}/(\sin(\pi /10)*\sin({2*\pi /5})).$

See the reference for additional formulas.

[Comment]

(great care has been taken to copy correctly the formulas; the above formulas should be checked and corrected against the formulas in the reference; - this is a comment to help find errors;). -- Leo.Mathematics (talk) 05:18, 24 April 2020 (UTC)

Please put new talk page messages at the bottom of talk pages and indent as outlined in wp:THREAD and wp:INDENT — See Help:Using talk pages. Thanks.

@Leo.Mathematics: As was said by user Deacon Vorbis, Wikipedia does not accept orinal research. See wp:No original research.

Also, you cannot ask editors to help find errors. Here we discuss the article, based on wp:reliable sources, not the subject — see wp:Talk page guidelines. - DVdm (talk) 09:13, 24 April 2020 (UTC)

Proofreading: Errors may happen when copying the formula to Wikipedia. It makes sense to verify by another person that copying such complicated formulas to Wikipedia was correct - and the formulas have a certain level of complexity.

By the way, books may contain errors in formulas as well. And the errors may propagate from one book to the next. The R code on the Github page enables one at least to check numerically that the formulas (on the Github page) are correct. But it is impossible to copy a formula from Wikipedia and run it inside a (free) tool to check it for corectness. --Leo.Mathematics (talk) 15:24, 24 April 2020 (UTC)

Wikipedia does not allow you to put your work in article talk pages to be verified by another person. Third level warning on your user talk page. - DVdm (talk) 15:31, 24 April 2020 (UTC)

Also, note that changing your own talk page messages after someone has replied to them, as you just did here, is bad practice and discouraged. See wp:REDACT. - DVdm (talk) 15:47, 24 April 2020 (UTC)

Missing Integrals

The article is missing generalizations of important integrals (like the Gaussian integral), which involve the Gamma function.

1.) Generalization of the Gaussian integral to higher orders:

${\frac {\Gamma (1/n)}{n}}=\int _{0}^{\infty }e^{-x^{n}}\,dx,\ \qquad n>=0\ .$

Note: one obtains the simple Gaussian integral for n = 2.

2.) Generalization of the Gaussian integral to multiple dimensions:

$({\frac {\Gamma (1/n)}{n}})^{k}=\int _{0}^{\infty }...\int _{0}^{\infty }e^{-(x_{1}^{n}+...+x_{k}^{n})}\,dx_{1}...dx_{k},\ \qquad n>=0\ .$

The 2 formulas are easy to derive and therefore references might be available. In the case that no references are available, see: https://github.com/discoleo/R/blob/master/Math/Integrals.2D.R

Note: the R code tests the formulas numerically (up to 2 dimensions for the "bi-dimensional" variant).

Note: there is likely no reference for the 3rd integral on Github, which is personal research. That one is a separate topic anyway. --Leo.Mathematics (talk) 21:40, 24 April 2020 (UTC)

No proper source, no place in Wikipedia. Our own work doesn't count. Reliable sources are not only necessary to verify correctness, they also establish whether content is sufficiently interesting and notable to be mentioned in this encyclopedia. We can calculate anything, and it can be 100% correct, but only when our research is sufficiently mentioned in the literature, it can be mentioned here. Now unless you first find such a good source, please stop insisting. Final warning on your user talk page. - DVdm (talk) 09:36, 25 April 2020 (UTC)

There is a brief mention on the Gaussian integral page on Wikipedia: https://en.wikipedia.org/wiki/Gaussian_integral. Unfortunately, it lacks any reference as well. Do you plan to delete that section as well? Most of that page does not include any references. Do you consider that references from Stack-Exchange are sufficient? --Leo.Mathematics (talk) 10:58, 25 April 2020 (UTC)

The generalization of the Gaussian integral puts Gamma(1/n) into the right perspective and explains why it is useful and how it is useful. A comprehensive encyclopedia should explain also the higher level concepts, like this functional relation. See also the section [Relation to the Gamma Function|https://en.wikipedia.org/wiki/Gaussian_integral#Relation_to_the_gamma_function] - although it has only a very brief mention to this. It also jumps directly to a formula that doesn't look like the Gaussian integral (see the integral posted in this section for a comparison), without explaining it. --Leo.Mathematics (talk) 11:17, 25 April 2020 (UTC)

It's not because unsourced content exists, that we can add even more. See the essay wp:otherstuffexists. That other stuff might have survived by wp:consensus. There seems to be no consensus to add yours. - DVdm (talk) 11:44, 25 April 2020 (UTC)

@Leo.Mathematics: Please note proper notation:

right: $\left({\frac {\Gamma (1/n)}{n}}\right)^{k}$

wrong: $({\frac {\Gamma (1/n)}{n}})^{k}$

right: $n\geq 0$

wrong: $n>=0$

Also it is vulgar to use an asterisk for ordinary multiplication in this context. You can write $3\cdot 5$ or $3\times 5$. Michael Hardy (talk) 20:00, 28 April 2020 (UTC)

Never mind, I have reported the user for talk page abuse. Indef blocked now. - DVdm (talk) 21:44, 28 April 2020 (UTC)

Android Calculator

Removed statement about Android calculator returning Gamma function values when a non-integer value is passed as argument to the factorial function, as it's false (it returns "Domain error" instead.) — Preceding unsigned comment added by 148.87.23.4 (talk) 03:27, 24 November 2021 (UTC)