Talk:Double factorial

Extension

The current extension to complex numbers is misleading. It correctly reproduces the double factorial for odd integers, but does not reproduce the double factorial for even integers. Interestingly, Mathematica can evaluate fractional arguments for Factorial2, which reproduces both, but I can't find the definition they use to compute it.

--Kaba3 (talk) 10:53, 15 June 2015 (UTC)[reply]

The relevant formula for Mathematica's Factorial2 is:

:z!!=2^{([1+2z-\cos(\pi z)]/4)}\pi ^{([\cos(\pi z)-1]/4)}\Gamma (1+{\frac {1}{2}}z).:

Perhaps what you mention should be stressed on the page a little more. 163.1.18.120 (talk) 10:24, 30 September 2024 (UTC)[reply]

Uhm...

I think this article is not bad, but it seems to me, it risks to be a bit misleading as it misses the main point of the double factorial, which is, first of all, just giving a useful notation for a common expression, rather than defining some new function to be extended to negative or complex values. For a non-negative integer n we give a special notation to the product of all positive integers not larger than n and with the same parity of n, namely $n!!:=\textstyle \prod _{{1\leq k\leq n} \atop {k=n\mod 2}}k$ because it arises frequently in mathematical expressions --such as coefficients of certain series expansions, enumerative formulas, repeated integrals etc, disregarding to the parity of the number n (even if within formulas that may depend on the parity of n). For this reason, I don't see any reason for treating differently odd and even numbers, and I do not see much interest in making emphasis on x!! for negative or complex numbers, nor I see other reason of dropping the alternative definition $0!!={\sqrt {2/\pi }}$ than indulging oneself in mathematical mysticism or amazing naive readers. --pm a 10:27, 29 November 2013 (UTC)[reply]

Agreed. Zaslav (talk) 16:52, 18 July 2014 (UTC)[reply]

Name

The traditional name of this function is semifactorial. I am moving the page. It is not "double", it is half of a factorial as it omits half the factors of a factorial. Zaslav (talk) 02:59, 17 July 2014 (UTC)[reply]

Do you have a reference for this? I have never seen semifactorial used in anything and double factorial is used all over the place (just look at the references). By WP:COMMONNAME I suspect that the name of this article should not have been changed. Bill Cherowitzo (talk) 03:59, 17 July 2014 (UTC)[reply]

I would also like to see a reference for the new name, *before* your bulk edits to articles imposing your opinion on the many other articles that use this term. The weight of current sourcing is strongly in favor of the double factorial name. It's a little hard to tell from searches because of the alternative meaning of both names in statistical design, but MathSciNet has 22 hits in which "double factorial" is related to this function (and two others where it means something unrelated) and absolutely zero hits for this sense of "semifactorial" (with nine others where it is used to mean something unrelated); that looks pretty definitive to me. As for which name makes sense: that's not the usual basis for naming articles here (see WP:SOAPBOX. —David Eppstein (talk) 04:37, 17 July 2014 (UTC)[reply]

I see no evidence that "semifactorial" is the most commonly name used, assuming it is even used at all. Revert this move. --Kinu ^t/_c 04:48, 17 July 2014 (UTC)[reply]

Unfortunately it is not easy (as far as I know) to revert all the link edits though it would be a good idea at this point. Also to revert the move. I agree there is widespread use of "double factorial". I use "semifactorial" myself, which I learned in college.

Please see my remarks at Wikipedia_talk:WikiProject_Mathematics/Archive/2014/Jul#Double_factorial_vs_semifactorial for a presentation of the three incompatible definitions of the "double factorial" of an even number given in the one article. Something should be done about that. For instance, in combinatorics I think it's safe to say there is only one definition and it's the one I called "semifactorial" (the product n(n-2)(n-4)...).

I disagree with David Eppstein about what is more logical. Since the "semifactorial" is the product of about half the integers that are multiplied in the factorial for both even and odd integers, "semi" seems more logical than "double". However, logic is not the deciding factor for WP and I didn't intend it to be. Please see my notes linked above. Note my apology, for what it's worth, for jumping into a complicated matter (even aside from the terminology).

Anyone know an easy reversion method for all this? Zaslav (talk) 05:12, 17 July 2014 (UTC)[reply]

As an admin, I could rollback all your edits, but any rollbacker should be able to revert the edits in question, if you can tell me when you started. — Arthur Rubin (talk) 06:27, 17 July 2014 (UTC)[reply]

Both odd and even numbers

The double factorial in the sense of semifactorial, $n(n-2)(n-4)\cdots ,$ is defined for all nonnegative integers. That is the core meaning, as far as I can see from other WP articles and my own experience. I've revised the article to make this "the" definition. I kept the parts about extending odd double factorials to negatives or complexes. Is this acceptable?

I also removed the term "odd factorial" as, in reviewing various WP articles, I never found an "odd factorial" of an even number (or hardly ever and then I forgot it). I have no problem with "odd factorial" = 1⋅3⋅5⋅... up to n if odd or n−1 if even, but I don't know that it is existing terminology. I did, however, find this incorrect definition of double factorial (possibly of an odd number only, but not so stated) in a few places: n!! = 1⋅3⋅5⋅..., which needs correction.

I hope the interested persons will review this and either improve it or decide it's another mistake. Zaslav (talk) 17:48, 18 July 2014 (UTC)[reply]

A search of Math Rev (MathSciNet) revealed no instances of "odd factorial" and two instances of "odd double factorial", both meaning the double factorial of an odd number. Zaslav (talk) 00:25, 19 July 2014 (UTC)[reply]

The phrase "odd factorial" with this meaning does appear in a few places found by Google scholar, e.g. Henderson, Daniel J.; Parmeter, Christopher F. (2012), "Canonical higher-order kernels for density derivative estimation", Statistics & Probability Letters, 82 (7): 1383–1387, doi:10.1016/j.spl.2012.03.013, MR 2929790 and Nielsen, B. (1999), "The likelihood-ratio test for rank in bivariate canonical correlation analysis", Biometrika, 86 (2): 279–288, doi:10.1093/biomet/86.2.279, MR 1705359. But there aren't very many of them so I don't think it's very important to include in the article. —David Eppstein (talk) 01:39, 19 July 2014 (UTC)[reply]

Thanks. Both mean the double factorial of an odd number. I'll add the term (with a redirect), as they seem to think it's standard. Zaslav (talk) 06:32, 19 July 2014 (UTC)[reply]

Relation to factorial

In the Relation to section, I'm wondering why it isn't stated that n! = n!! x (n-1)!! for all n>0. It is clearly true.??98.21.70.161 (talk) 19:54, 25 October 2017 (UTC)[reply]

Well, it's actually given in the generalizations section of the factorial function page in more generality as

{\begin{aligned}n!&=n!!\cdot (n-1)!!,\ n\geq 1\\&=n!!!\cdot (n-1)!!!\cdot (n-2)!!!,\ n\geq 2\\&=\prod _{i=0}^{k-1}(n-i)!^{(k)},\ {\text{ for }}k\in \mathbb {Z} ^{+},n\geq k-1.\end{aligned}}

Maxie (talk) 22:01, 25 October 2017 (UTC)[reply]

Log convex edits

@Quantling: One result of this edit is that in the sentence beginning "From this", the referent of "this" is now the fact that something is log convex. I suspect that this is not right. Can you please look over the section and check for global coherence? (This version before you moved things around perhaps does not have the same problem.) Thanks, --JBL (talk) 20:38, 1 February 2023 (UTC)[reply]

Thank you for the good suggestion. I have made a try. Your further suggestions or direct edits are welcome. —

Q

uantling (talk | contribs) 21:25, 1 February 2023 (UTC)[reply]

Thanks! --JBL (talk) 18:41, 5 February 2023 (UTC)[reply]

Trouble in "Relation to the factorial" last equation

The last equation in this section reads: $(2k-1)!!={\frac {_{2k}P_{k}}{2^{k}}}={\frac {(2k)^{\underline {k}}}{2^{k}}}\,.$ I don't understand the exponent ${(2k)^{\underline {k}}}$ which may indicate a lack of my math knowledge. ${\underline {k}}$ means ... what? I would write the last term $...={\frac {(2k)!}{k!2^{k}}}$ , which works for small $k$ , at least. Sorry if I'm missing something basic here. JohnH~enwiki (talk) 23:00, 9 August 2023 (UTC)[reply]

See falling and rising factorials for this notation. —David Eppstein (talk) 00:08, 10 August 2023 (UTC)[reply]