# Talk:Factorial

## Moved from article

An imperative way of calculating factorial may be more understandable. In Python code:

def fact(x):
result = 1
while x > 1:
result *= x   # multiply result by x
x -= 1        # decrease x by 1
return result


// function factorial int fac(int fc)

  {
int i, ret = 1;
for (i = 1; i<=fc; i++)
{
ret*= i;
}
return (ret);
}


Moved from the article. Is this really worth mention? -- Taku 21:38 Apr 25, 2003 (UTC)

## Derivative of the factorial

Derivative of the Infoliofaktorial or infoliofactorial

• 5? = 4!
• 6? = 5!
• 7? = 6!
• 8? = 7!
• 9? = 8!

Does anyone know some values of the function that can be recieved by differentiating the factorial, using the ? symbol for this function?? A graphing calculator can be helpful. Try to see if you know the values of:

• 1? =
• 2? =
• 3? =
• 4? =
• 5? =
• 6? =
• 7? =
• 8? =
• 9? =
• 10? =

Dou you see a pattern for the values of n? for the integers?? (Look! We now have a mathematical meaning for just about every symbol on the computer; any counterexamples??) —Preceding unsigned comment added by 66.32.253.51 (talkcontribs) 21:17, 9 March 2004 (UTC)

How can you differentiate the factorial when it isn't a continuous function? You could differentiate the gamma function, though. Eric119 22:41, Jun 30, 2004 (UTC)
It's continuous over certain intervals. It just has pole as non-positive integers. Its not continuous, but its still differentiable. He Who Is 01:06, 6 June 2006 (UTC)
Of course you can differntiate the factorial function! It is not defined for integral values only but for all real values execept {-1,-2,-3,...}. The definition is x! = Gamma(x+1). Peter, 20 October 2006. —Preceding unsigned comment added by 84.136.150.222 (talkcontribs) 15:51, 20 October 2006 (UTC)
Wouldn't it be better to mention the integral definition
$z!=\int _{0}^{\infty }e^{-t}t^{z+1}\,dt$ in the discussion of the gamma function? 4pq1injbok 02:32, 16 Jul 2004 (UTC)
No, it has obvious relevance here also. Dysprosia 02:38, 16 Jul 2004 (UTC)
Sure, but isn't it still primarily about the gamma function? After all, the factorial is discrete, so defining it for nonintegral arguments is essentially defining the gamma function. It seems a bit scattered to me to talk about nonintegral z in two distinct places. 4pq1injbok 21:27, 16 Jul 2004 (UTC)
Is there a term for the sum of positive integers less than or equal to the given number? Bless me if I can't find reference to one. —Preceding unsigned comment added by 69.196.178.201 (talkcontribs) 04:31, 5 August 2004 (UTC)
Yes, a triangular number. --- User:Karl Palmen 5 August 2004
http://www.luschny.de/math/factorial/FastFactorialFunctions.htm
This site shows several interesting algorithms to compute the factorial. —Preceding unsigned comment added by 82.126.207.162 (talkcontribs) 22:26, 2 September 2004 (UTC)
This is easy. However, I prefer the systematic mathematical notation !' to the private invention ?.
• 0!' = -gamma
• 1!' = 1-gamma
• 2!' = 3-2*gamma
• 3!' = 11-6*gamma
• 4!' = 50-24*gamma
• 5!' = 274-120*gamma
• 6!' = 1764-720*gamma
• 7!' = 13068-5040*gamma
• 8!' = 109584-40320*gamma
• 9!' = 1026576-362880*gamma
The 'pattern' is:
n!' = (-1)^(n+1) * StirlingNumberFirstKind(n + 1, 2) - n! * gamma
and 'gamma' is the Euler-Mascheroni constant. See also sequence A000254 on the On-Line Encyclopedia of Integer Sequences! Peter, 20 October 2006. —Preceding unsigned comment added by 84.136.150.222 (talkcontribs) 15:51, 20 October 2006 (UTC)

## Incorrect reference to "the above recursive relation"

"Proper attention to the value of the empty product is important in this case, because <...> it makes the above recursive relation work for n = 1;"

But the recursive relation is given at the end of the next section (Applications), not above. —Preceding unsigned comment added by Mgedmin (talkcontribs) 18:01, 25 October 2004 (UTC)

## Superfactorials

Superfactorials get large very rapidly. Between what two consecutive superfactorials does Graham's number lie?? 1$= 1 and 2$ = 4, but even 3\$ is too big to write; it is 6^6^6^6^6^6. 66.32.244.149 21:43, 2 Nov 2004 (UTC)

In the Superfactorials (alternative definition) and above it isn't perfectly clear whether
$3\mathrm {S} \!\!\!\!\!\;\,{!}=6\uparrow \uparrow 6={^{6}}6=6^{6^{6^{6^{6^{6}}}}}$ represents
${\left({\left({\left({\left(6^{6}\right)}^{6}\right)}^{6}\right)}^{6}\right)}^{6}$ or
$6^{\left(6^{\left(6^{\left(6^{\left(6^{6}\right)}\right)}\right)}\right)}$ —DIV (128.250.80.15 (talk) 04:49, 21 August 2008 (UTC))

What is the reference to "superduperfactorials"? Can some please remove this if it is not legitimate. —Preceding unsigned comment added by 65.103.203.33 (talk) 03:35, 23 November 2009 (UTC)

## factorials for halves?

How does one calculate factorials that are in the form (n + .5)!, where n is a whole number? On windows calculator, it says 3.5! = 11.631728396567448929144224109426. how is this calculated? —Preceding unsigned comment added by 12.223.117.71 (talkcontribs) 02:57, 9 February 2005 (UTC)

With the gamma function. Fredrik | talk 02:59, 9 Feb 2005 (UTC)
Yup, 3.5! is the same as Γ(4.5). --MarkSweep 06:19, 9 Feb 2005 (UTC)

Strange... The Windows calculator returns about 0.88 as .5!, which is less than 1 and clearly not sqrt(pi)... He Who Is 00:51, 6 June 2006 (UTC)

Looks right to me. 0.5! = Γ(1.5) = 0.5 Γ(0.5) = √π / 2 = 0.88623... —Steven G. Johnson 00:59, 6 June 2006 (UTC)
Ahhh... I forgot about the phase change between factorials and gammas, and interpereted Γ(.5) as root pi. I've always wondered why the gamma function is more commonly used in evaluating factorials. The Pi function has no phase change and a slightly simpler formula, so why does everyone opt for the one that requires more work? He Who Is 01:02, 6 June 2006 (UTC)
Well, not everyone. Gauss did not! Sometimes the so called 'Mellin-transform' is said to be the reason: The Gamma function is the Mellin transform of the exponential function. See also the remarks at: http://www.luschny.de/math/factorial/factandgamma.html Peter, 20 October 2006. —Preceding unsigned comment added by 84.136.150.222 (talkcontribs) 19:33, 20 October 2006 (UTC)
The Pi function has offset coincide with the factorial, and thus may be considered a direct generalization of factorial (unlike the gamma). It should be mentioned in the section discussing generalizations, although more briefly than the gamma function, since the latter is more commonly used.--173.206.70.204 (talk) 06:44, 19 December 2009 (UTC)

## Fast computation

Luschny's "prime swing" algorithm is quite neat. I've translated his source code into Python/gmpy, and found it to be four times faster than Mathematica's n! and twice as fast as gmp's built-in factorial (testing with n up to 1000000). The algorithm is unfortunately poorly documented, so there doesn't seem to be much hope for writing about it here. Anyone here with more knowledge about this? Fredrik | tc 13:53, 25 November 2005 (UTC)

where can i get this program??? theres no1 at school 2day so decided 2 hav sum fun calculatin factorials but its to hard manualy an the calculater doesnt giv any nice answers(only upto 13!) —Preceding unsigned comment added by Msknathan (talkcontribs) 09:04, 16 January 2007

It'd be nice if this page mentioned the record largest factorial computed, maybe along with the number of digits of the result. —Preceding unsigned comment added by 66.235.40.197 (talkcontribs) 14:36, 16 August 2007

## Factorials for Fun

I've seen numerous videos and countless images using factorials utalizing their unique designs. Someone should make a section with a picture maybe showing this. 65.9.89.94 03:46, 2 March 2006 (UTC)

What "designs" are you referring to? Michael Hardy 20:04, 2 March 2006 (UTC)

I think he is confusing "factorials" and "fractals". Hugo Dufort

definitly —Preceding unsigned comment added by Msknathan (talkcontribs) 12:05, 19 January 2007

## Identities

Some identies for evaluating the factuarials of common binary functions would make a good addition to this article. For instance:

Or hopefully something without the recursivity in the denomitator. He Who Is 00:59, 6 June 2006 (UTC)

I don't think you mean that. For example if a=3 and b=4, you would get (a+b)!=5040 while the right hand side would be 6*24/1. --Audiovideo 22:22, 21 June 2006 (UTC)
Oh. I had a feeling I had that wrong, since I wrote it in a hurry. But either way, I think you got my point. It would be good to add mathematically correct formulas for factorials of sums, differences, products, quotients, etc. -- He Who Is[ Talk ] 22:49, 21 June 2006 (UTC)
We definitely need a section about identities and formulas (if any). That's what I was looking for when I wiki'ed this page. Of course, the example formula is wrong. Some identities for factorials are presented on the MathWorld site, for example the "raising factorials" and the "falling factorials" using the Pochhammer symbol.

## Superfactorial notation

By the magic of TeX, it's possible to wrangle the notation for superfactorials so it indeed does look like a superimposed S and !: $\mathrm {S} \!\!\!\!\!\;\,{!}$ . Should this go live in the article? Or are we happy with how this looks already? Dysprosia 13:17, 4 July 2006 (UTC)

i think it should be changed, althought the "code" would look pretty extensive. anyways, i vote change it. Cako 02:34, 30 November 2006 (UTC)

## Alternating Product

Does anyone know if there's a name for the function defined by:

$\phi (n)=\prod _{i=1}^{n}i^{(-1)^{i}}$ ?

It is essentially the factorial, but it alternates from multiplication to division. -- He Who Is[ Talk ] 18:35, 21 July 2006 (UTC)

So thats
$\phi (n)={\frac {2\cdot 4\cdot 6\ldots (n-1)}{1\cdot 3\cdot 5\ldots n}}={\frac {(n-1)!!}{n!!}}$ for n odd
$\phi (n)={\frac {2\cdot 4\cdot 6\ldots n}{1\cdot 3\cdot 5\ldots (n-1)}}={\frac {n!!}{(n-1)!!}}$ for n even
Just looking at n even, say s=n/2. Then using some identities for the double factorial which relate it to the Gamma function we get
$\phi (n)={\frac {{\sqrt {\pi }}\,\Gamma (s+1)}{\Gamma (s+1/2)}}$ which, if you write $\pi$ as $-\Gamma (-1/2)/2$ gives a Beta function:
$\phi (n)=-{\frac {1}{2}}B\left(-{\frac {1}{2}},s+1\right)$ I think you can do something similar for n odd. PAR 16:43, 20 October 2006 (UTC)

—Preceding unsigned comment added by 41.242.188.116 (talk) 14:29, 18 March 2009 (UTC)

## 0!

Can someone explain 0!=1 because intuitivle I would think 0!=0...

• This page is to discuss editing problems, not for discussing elementary learner problems. See newsgroups like sci.math for questions of this kind.
• Actually, I would say intuitively it is undefined. However, the Gamma function is the continuous version of the factorial (Γ(x+1)=x!) it effectively defines the factorial of any number, and it says 0!=1. Also, there's the binomial expansion. For example, raising (x+1) to the third power, we have
$(x+1)^{3}=\sum _{n=0}^{3}{\frac {x^{n}n!}{3!(3-n)!}}$ which works for all the terms, including the n=3 term, if 0!=1. Its like that over and over, every time you want to generalize to use 0!, it turns out to be 1. PAR 04:04, 5 November 2006 (UTC)
• If we define the factorial recursively by n! = n × (n−1)!, we must take 0! = 1 or we would get that 1! = 0, 2! = 0, 3! = 0, ..., which surely doesn't make sense. If we define n! as the product of the list {1, 2, ..., n}, 0! would be the product of a list of zero numbers, which is the empty product and equals 1. What goes on in both cases is that 0! must be 1 because 1 is neutral under multiplication. Your intuition is based on the behavior of functions defined through addition, where 0 is neutral. - Fredrik Johansson 09:41, 5 November 2006 (UTC)

I still don't quite get it. Given that the definition of a factorial is the value of a number, N, multiplied by every integer N through 1 inclusive, the factorial of 0 is the value of 0 multiplied by the integers 0 through 1 inclusive. The only integers that qualify are 0 and 1. Therefore, 0!=0*1=0. Please critique. —Preceding unsigned comment added by 2k6168 (talkcontribs) 22:01, 21 May 2008 (UTC)

If n denotes a natural number (including 0), then n! is, by definition, the product of the integers from 1 to n inclusive, that is, the set of integers i such that 1 ≤ in. If n = 3, for example, this is the set {1, 2, 3}, and indeed 3! = 1×2×3 = 6. If n = 0, you get the empty set: there are no integers i that satisfy 1 ≤ i ≤ 0. The product of the elements of the empty set is known as the empty product, and its value is 1.
The recurrence relation (n+1)! = (n+1) × n! shows that n is a factor of n! if n ≥ 1. You can revert the direction, giving
n! = (n+1)! / (n+1).
For example, 3! = 4! / 4 = 24 / 4 = 6. Then 0! = 1! / 1 = 1 / 1 = 1.  --Lambiam 16:27, 23 May 2008 (UTC)
So then because 0 doesn't fit that rule, we have an empty set, and although by definition factorializ(s)ing (correct term?) means multiplying that 0 by some number, that isn't what we do, because it's an empty set. Urgh. My mind's still stuck on that to factorializ(s)e, you must at least multiply n by something else, and when n is 0, the result is always 0. Oh well, that's my problem. Thanks. 2k6168 (talk) 12:17, 29 May 2008 (UTC)
You write "by definition factorializ(s)ing (correct term?) means multiplying that 0 by some number". In no way does the definition of the factorial mean that 0! is the result of multiplying 0 by some number. It is just like the powers of 10. Most are a multiple of 10, since 10n+1 = 10 × 10n. This does not mean that all powers of 10 are the result of multiplying 10 by some number. In particular, 100 = 1, which is not a multiple of 10.

## definition

Fixed definition to >0, =>0 can't work for the recursive part of the def, since then you would always get 0.

I take that back. It does work with =>. Although I did leave in the nonnegative integer bit. I'm getting rusty. . . .71.102.186.234 05:30, 11 November 2006 (UTC)

If I look at the definition:
$n!=\prod _{k=1}^{n}k\qquad {\mbox{for all }}n\in \mathbb {N} \geq 0.\!$ I get for n=0
$0!=\prod _{k=1}^{0}k=1\cdot 0\,$ which is zero. Shouldn't the proper definition be for n >= 1, with 0! not being defined by this particular definition? PAR 17:11, 11 November 2006 (UTC)

No, $\prod _{k=1}^{0}k$ is the empty product, which is 1. EdC 19:21, 11 November 2006 (UTC)

The empty product is a product over a null set. Here the set is not null, its {0,1}. If the definition were changed to n>0 rather than n>=0 it would read:
$n!=\prod _{k=1}^{n}k\qquad {\mbox{for all }}n\in \mathbb {N} >0.\!$ and then, for n=0, it would be a product over a null set. Does this sound right? PAR 19:43, 11 November 2006 (UTC)

The lower and upper bounds are ordered. $\prod _{k=a}^{b}$ means that the sum is taken over the values of k satisfying akb. Therefore b < a gives a null set. Fredrik Johansson 21:38, 11 November 2006 (UTC)

Could you give me a Wikipedia page that states this clearly, because I would like to clarify that point in this page by a link. I couldn't find such a page. I thought product might be the page, but it doesn't help. If there isn't such a page, could you edit the proper page and put this fact in it, so that I could reference it? Thanks - PAR 21:55, 11 November 2006 (UTC)

I've added a line to the product page. By the way, the article sum also says about big sigma notation (which is analogous to big pi notation) that "if m = n in the definition above, then there is only one term in the sum; if m = n + 1, then there is none.". Fredrik Johansson 17:14, 12 November 2006 (UTC)
I don't see an addition to either the product page or the product (mathematics) page. Probably the "product (mathematics)" page is where it should go, not to the "product" page, which is just a disambiguation page. PAR 17:34, 12 November 2006 (UTC)
Sorry, I meant multiplication. Fredrik Johansson 17:35, 12 November 2006 (UTC)

## Gamma Function

I have a couple minor complaints about the "Non-integer factorials" section:

1. Γ(z) is represented inconsistently. It is defined at the top of the section as Γ(z+1)=... for z > -1, but is later said to be "defined for all complex numbers z except for the nonpositive integers (z = 0, −1, −2, −3, ...)", which would be true of Γ(z), not Γ(z+1). The graph also represents Γ(z). I would suggest redoing the formula and definition to reflect Γ(z) as well, or adjust the rest of the section to increase clarity. I had to read it twice.

2. "The Gamma function is in fact defined for all complex numbers z except for the nonpositive integers (z = 0, −1, −2, −3, ...) where it goes to infinity." It doesn't really "go to infinity." In fact, it approaches positive or negative infinity near these values, depending upon the direction you come from. Barring a discussion of transfinite mathematics, it would be more accurate to write, "...) where it is undefined," or to simply omit it entirely, because the definition would be accurate without it (it states an exception to the domain).

3. "That is, it is the only function that could possibly be a generalization of the factorial function." This statement is a simplification of the previous statement that isn't accurate. You could, for example, construct a piecewise linear function that intersects the points of the factorial function and have a generalization. Maybe: "... it is the only holomorphic function that could ..."

I'm new here, so I'm not trying to step on anyone's toes, just trying to help with clarity.

Fishcorn 06:10, 26 November 2006 (UTC)

I agree (mostly). Problem 1 is not lack of rigor, but lack of clarity, right? Problem 2 is lack of rigor. Problem 3 - I think "generalization" was meant in terms of the recursion relationships, not that it happens to agree with the factorial function at certain integer values. I could fix it, but why don't you do it instead? PAR 16:53, 26 November 2006 (UTC)
Done. Fishcorn 20:55, 26 November 2006 (UTC)

4. the last equation (the infinite product) cannot be correct, as the k'th element goes to infinity, and so does the whole product. —Preceding unsigned comment added by 213.131.238.25 (talk) 13:53, 20 November 2007 (UTC)

### Application of the Gamma function

This section only mentions the use of the gamma function in calculating the volume of an n-dimensional hypersphere. However, if the number of dimensions is an integer, it seems that using the gamma function has no advantage over using the simple factorial. If there is some coherence to the notion of a space with a non-integer number of dimensions, it would be good to spell that out, and perhaps note who uses such a notion and why. (Yekwah 09:56, 20 August 2007 (UTC))

What is the operation called when you add a natural number to all of the natural numbers that are less than it? For example, 7? = 7+6+5+4+3+2+1 = 28, n? = n+(n-1)+(n-2)+...+1 This is like factorials, but with addition instead of multiplication. If there is a word for this, it seems important enough to reference in this article. Hermitage 18:16, 23 December 2006 (UTC) ɢ

Triangular number. And yes, a link is worthwhile. –EdC 02:27, 24 December 2006 (UTC)

## Why is (-1/2)! = Square Root of π?

TI-83 series calculators use this definition, but why is that? What's the origin of this definition and how is this useful? 74.112.121.40 04:53, 22 March 2007 (UTC)

Read the section "Non-integer factorials" in the article. Fredrik Johansson 08:09, 22 March 2007 (UTC)

I was wondering if there's a specific answer to just my question, not a general case for all non-integer factorials. This is because I assume the text book of my level of education would not ask us to learn what a Gamma Function is by ourselves. Any help would be appreciated, thanks in advance. 74.112.121.40 02:25, 23 March 2007 (UTC)

There's no difference between the general case and the specific case. Since $x!=x(x-1)!$ for all x, the value of (n + 1/2)! for some integer n is simply a rational multiple of the value for any other n. Why ${\sqrt {\pi }}$ in the first place? One explanation is that the gamma function integral turns into the Gaussian integral when x = 1/2. It is also a consequence of the reflection formula for the gamma function:
$\Gamma (z)\Gamma (1-z)={\frac {\pi }{\sin {\pi z}}}$ Inserting z = 1/2 gives the answer. Fredrik Johansson 13:47, 24 March 2007 (UTC)
You're right - your calculator is using more advanced mathematics than your textbook. Dcoetzee 00:32, 26 February 2008 (UTC)
Many calculators and mathematical software (such as Maple) use the more general definition of factorial, rather than just for positive (or non-negative) integers. The general definition is very commonly seen in more advanced mathematical tools, and should be mentioned in the introduction of the article, even if very brief (the details will be in its own sections, of course). --173.206.70.204 (talk)

## Computation of the factorial

Is it worth mention the basic commands for the factorial in some programming languages or math environments like matlab or maple? 9 April 2007 —Preceding unsigned comment added by 63.150.207.3 (talkcontribs) 19:38, 29 April 2007

No, I do not think so. In Maple, for example, you can write 5! —Preceding unsigned comment added by 82.149.175.194 (talkcontribs) 06:48, 30 June 2007

## Editing the references

The page says: "Peter Luschny. The Homepage of Factorial Algorithms (no longer existent)." This is no longer true. The page does exist again. However, I was not able to eliminate the "(no longer existent)" from the text. If you can change it please remove this misleading comment. Thanks. —Preceding unsigned comment added by 82.149.175.194 (talkcontribs) 06:48, 30 June 2007

## Double Factorial

just a heads up. those identities for the double factorial using the gamma function, are wrong. i suggest someone fixes them (i dont know how to). here is one identity: n!! == (2/Pi)^((1/4) (1 - Cos[Pi n])) 2^(n/2) Gamma[n/2 + 1] (source wolfram) —Preceding unsigned comment added by 85.166.237.71 (talkcontribs) 17:02, 10 July 2007

Is there any information available about the appearence of the notation "!!" and/or terminology of "double factorial"? (I know a paper from 1948 where it appears, but I think it should be much older.) Any hint is welcome -- Thanks! — MFH:Talk 22:50, 1 February 2009 (UTC)

I added an alternative definition of n!!, as shown in Wolfram Research Mathematica. Alex Vermeulen, Zoetermeer, The Netherlands, June 14, 2009. Jordaan12 (talk) 15:47, 14 June 2009 (UTC)jordaan12Jordaan12 (talk) 15:47, 14 June 2009 (UTC)

It seems that since we have 1!! = 1, 3!! = 3, 5!! = 15, 7!! = 105, and so on, this could be extended for negative odd integers as well by noting that

$(2n-1)!!={\frac {(2n+1)!!}{2n+1}};(-1)!!={\frac {1!!}{1}}=1.$ Then:
$(-3)!!={\frac {(-1)!!}{-1}}=-1={\frac {-1}{1!!}}$ $(-5)!!={\frac {(-3)!!}{-3}}={\frac {1}{3}}={\frac {1}{3!!}}$ $(-7)!!={\frac {(-5)!!}{-5}}={\frac {-1}{15}}={\frac {-1}{5!!}}\dots$ This leads to an interesting general double factorial formula for negative odd integers other than -1 in terms of positive odd integers:

$(-2n-1)!!={\frac {(-1)^{n}}{(2n-1)!!}}.$ Glenn L (talk) 07:45, 3 December 2010 (UTC)

I suspect that this is an interesting special case of Euler's reflection formula (also here). —Quantling (talk | contribs) 14:55, 3 December 2010 (UTC)

## Bang and anti-bangs

The term "bang" is sometimes used by journalists in reference to the ! symbol. It is also the sign for factorials. It then occurred to me that just as you can do something like 4!=1x2x3x4 to equal 24 (shorter is n!+1x...n), one can undo a factorial product by doing 24/4/3/2/1= 4. I could express is as 24?=4 or n?=n/.../1= n. I also found that you could produce n¿= 1/.../n and could also be expressed as n¿=1/n!. I came up with the idea of using ? and ¿ to express undoing factorials. True, it's not official but it's an idea worth sharing. It should also be noted that n! x n?= 1 and n! ÷ n?= n!^2 and n? ÷ n!=1/n!^2. Any comments or questions? R3hall 00:56, 14 October 2007 (UTC)R3hall

You might be interested in this equation: y=ln((x^x)/(x!))
It is essentially saying that for positive values, Euler's Gamma function has a direct relationship with x^x. The graph of that equation is funny in the same way that the value of e^(Pi*i) is. What you want is an antigamma function. See: http://mathoverflow.net/questions/12828/inverse-gamma-function 75.70.89.124 (talk) 07:07, 19 May 2013 (UTC)

## Isn't it an error: "Γ(n+1)=nΓ(n)" ?

Cos I think, it should be: Γ(n+1)=(n+1)Γ(n) if gamma function is to meet factorial's condition. And for consistency reasons there shouldn't be n and n-1 with factorial when there is n+1 and n with gamma. —Preceding unsigned comment added by BartekBl (talkcontribs) 18:40, 24 February 2008 (UTC)

Γ(n+1) = n! = n·(n−1)! = n·Γ(n).  --Lambiam 22:37, 24 February 2008 (UTC)

## Difference between n to the power of n upto nth position defined

The 4 steps for "Anupam's Formula" are as follows

Following are the steps for the "Anupam's Formula"

Step 1

Let a = xn - (x-1)n

b = (x-2)n - (x-1)n

c = (x-3)n - (x-2)n ...

p = (x-n)n - (x-n-1)n

Step 2

a1 = a - b - c - .. - z

a2 = b -c - ...- z

a3 = c- .. - z

...

p1 = a1 - a2 - a3 - ..

Step 3

Follow Step 2 repeatedly until there is only one amount left

Step 4

This amount is equal to n!

Example :

Take

n=2

122 - 112 = 144 - 121 = 23

112 - 102 = 121 - 100 = 21

So, 23 -21 = 2 = 2!

Again for n=3

Step 1

163 - 153 = 4096 - 3375 = 721

153 - 143 = 3375 - 2744 = 631

143 - 133 = 2744 - 2197 = 547

133 - 123 = 2197 - 1728 = 469

Step 2

721 - 631 = 90

631 - 547 = 84

Step 3

90 - 84 = 6

Step 4

6 = 3!

This is have tested upto 10 and have found to be correct.

122.161.30.232 (talk) 10:59, 13 March 2008 (UTC)Anupam Dutta <anupamdutta@rediffmail.com>

## Computation

I'm not comfortable with the section on computation leaving out the fact that most of the examples given use some small floating point representation for calculations, and therefore the value given for factorials will not be exact even for relatively small numbers. In some cases the returned (printed) value might not necessarily be using the maximum precision of the assumed representation. For example, google calculator reports a truncated value for 16!, even though this value can be represented exactly by an IEEE-754 64-bit variable. But regardless of the printout, most of the examples given use some underlying float representation and one must be careful. —Preceding unsigned comment added by 71.176.104.251 (talk) 02:38, 28 September 2008 (UTC)

Yeah, actually I went back, and the information is really misleading, claiming that the value returned by the calculators is indicative of the use of a 1024 bit integer. As it happens that is the maximum most likely because of an underlying IEEE-754 64-bit floating point value, not an integer, and therefore will only have 56 bits of precision. —Preceding unsigned comment added by 71.176.104.251 (talk) 02:42, 28 September 2008 (UTC)

The law of universal gravitation developes the concept that every particle attracts every other particle. This results in the gravitational force value being a factorial of the number of particles involved. But nobody believes that. So we change to the integral of lost free energy concept, to find the energy increment value. Is that logically and mathematically correct? WFPMWFPM (talk) 16:52, 16 October 2008 (UTC)

## 2^1024 and "10^100 in hexadecimal"

Don't know about you, despite i use hexadecimal numbers quite often, the number 10^100 were a bit confusing for me. Then i understood it means 16^256, but i was wondering then why is noted that in that form. I am a programmer so i would have used 0x10 ^ 0x100, but OK, Wikipedia readers are not programmers. If i would like to know the magnitude, i would rather write 10^308 (decimal), or the amount of memory used for representation, then 128 bytes, but definitely not 10^100. —Preceding unsigned comment added by 157.181.161.14 (talk) 13:53, 22 October 2008 (UTC)

## Allegedly more accurate estimate

The following was recently added to this article by an anonymous user:

When n is large, n! can be estimated quite accurately using Stirling's approximation:
$n!\approx {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}.$ An even more accurate approximation can be made using the following (provided by Brendon Phillips)
$n!\approx {\frac {2{\sqrt {2en}}}{3}}\left({\frac {2n+1}{2e}}\right)^{n}.$ I looked at the ratio of these two expressions:

{\begin{aligned}&{}\qquad {\frac {{\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}}{{\frac {2{\sqrt {2en}}}{3}}\left({\frac {2n+1}{2e}}\right)^{n}}}={\frac {3}{2}}{\sqrt {\frac {\pi }{e}}}\left({\frac {n}{n+1/2}}\right)^{n}\\\\&\longrightarrow {\frac {3}{2}}{\sqrt {\frac {\pi }{e}}}e^{-1/2}={\frac {3{\sqrt {\pi }}}{2e}}\cong 0.978\ldots \cong {\frac {1}{1.0224\ldots }}\quad {\text{ as }}n\to \infty .\end{aligned}} Since the ratio of Stirling's approximation to the factorial approaches 1 as n grows, the ratio of this new expression to the factorial must approach about 1.0224 times the factorial. I also tried it with some small numbers and it was not quite as close as Stirling's.

I didn't find any "Brendon Phillips" via Google Scholar, and Google Books shows a Brendon Phillips in the entertainment industry with no immediate suggestion of any material that could be cited here as a source.

As I said in the edit summary, I suspect WP:OR.

Hence I reverted to the earlier version. Michael Hardy (talk) 00:20, 4 February 2009 (UTC)

"A much better approximation for log n! was given by Srinivasa Ramanujan (Ramanujan 1988)" - I found this approximation to be worse than Stirlings, not better. Victor (talk) 21:06, 23 October 2010 (UTC)

My computations agree that the approximation by Ramanujan in Factorial#Rate of growth is much better than Stirling's $n!\approx {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}.$ n log(n!) log(Stirling's) log(Ramanujan's)
10 15.1044125730755 15.0960820096422 15.1044119983597
20 42.3356164607535 42.3314501410615 42.3356163817570
30 74.6582363488302 74.6554586739004 74.6582363246742
40 110.320639714757 110.318556424819 110.320639704405
50 148.477766951773 148.476100307326 148.477766946422
60 188.628173423672 188.626784547642 188.628173420556
70 230.439043565777 230.437853097684 230.439043563806
80 273.673124285694 273.672082624452 273.673124284369
90 318.152639620209 318.151713698094 318.152639619276
100 363.739375555563 363.738542225008 363.739375554882
PrimeHunter (talk) 22:07, 23 October 2010 (UTC)
Try putting this into Google: Brandon Phillips fredonia mathematics
Apparently, it was a typo? The thing about original research and mathematics, is that a formula like that is pretty simple to test with any of the millions of computers in the world with algebraic software. Still, without a source, it is OR and can lead to plagiarism arguments as well. 75.70.89.124 (talk) 07:14, 31 July 2013 (UTC)

## precedence of factorial

(This is my first edit of a new topic on a talk page, I hope I did it right). Why is there no mention of precedence of the factorial. For example, (and currently of interest to me) what is -5! ?~~ —Preceding unsigned comment added by Mortgagemeister (talkcontribs) 10:56, 31 July 2009 (UTC)

Close to right. Please put new comments at the bottom, and use four tildes (~~~~, or use the signature button just above the text window) to sign your messages. The generalization of factorials to numbers other than positive integers is given by the gamma function, but it diverges (essentially, has an infinite value) for negative integers. —David Eppstein (talk) 15:03, 31 July 2009 (UTC)

I asked why there is no mention of precedence, not about generalization of factorials to numbers other than positive. You are viewing -5! as the factorial of the number negative 5. That makes sense of course, but it also begs the question. Do you read -5! as the factorial of negative 5, or unary minus operating on 5 which has the factorial operator operating on it. That is when the issue of precedence comes up. While I (think I) agree that -5! should be read the way you do, other's read it as resulting in -120 (since they think the factorial should come first, followed by the unary minus - which looks just like the negative) Note that -5^2 is not read the way you read -5!. -5^2 is equal to -25 (the exponent operator comes first)Mortgagemeister (talk) 16:11, 31 July 2009 (UTC)

I would say that the factorial binds as tightly as exponentiation, so that -5! is -(5!), and ab! is a(b!). That's just my two cents; I don't have any references to back me up. Quantling (talk) 15:12, 12 March 2010 (UTC)

Hmm. So what about 2^n! ? Marc van Leeuwen (talk) 17:10, 12 March 2010 (UTC)

I'd write either of these:

$2^{n}!,\quad 2^{n!}.$ Both are clear. Michael Hardy (talk) 19:09, 12 March 2010 (UTC)

In a programming language one might have to deal with 2^n! without the visual aid of having some or all of it in a superscript. Both Google and Maple (version 7) say 2^3! is 64 implying that the factorial binds more tightly than the exponentiation. Quantling (talk) 13:56, 15 March 2010 (UTC)

There is two aspects to a question like this: 1) Order of operation and 2) Operator associativity. Precedence of operators may vary between interpretations/languages (e.g., Google, Maple, Python, etc...), but generally factorial is given (along with exponentiation) the highest precedence. When precedence is the same, it's only a question of associativity. Factorial (just like exponentiation) is usually "right associative". So "2^n!" would be interpreted "2^(n!)" and not "(2^n)!" (i.e., as if it were left-associative). Jwesley78

Likewise "2^3^4" should interpreted as "2^(3^4)". To interpret it otherwise (left-associative), one would obtain what is likely to be an undesirable interpretation: "2^3^4" = "(2^3)^4" = "2^(3*4)". Jwesley78 15:59, 15 March 2010 (UTC)

The factorial operator is a unary operator, not a binary operator like exponentiation, and I don't know what right-associative means for a unary operator. For instance, if I have 3!!, should the right-associative interpretation be 3(!!), whatever that means? I'm not saying you are wrong; only that I am confused. Quantling (talk) 20:33, 15 March 2010 (UTC)

Sorry. I 'went off on a tangent there. You have a good point. Typically, one only considers "associativity" when considering the order of evaluation for equal-precedent (k>1)-ary operations written in infix notation. Factorial is a unary operator written in the postfix position. Associativity appears to have no meaning in this context.
• For example, define fuctions "!(x)" and "^(x,y)" to be prefix expressions for the factorial and exponent operations.
• "3!^4" = "^(!(3),4)", and
• "3^4!" = "^(3,!(4))".
• In both cases the exponent operation is evaluated last.
(I remember this being a huge headache when I was writing the specification for a parser in a compilers class that I took. And now I'm fairly rusty in my knowledge of formal grammar parsing.)Jwesley78 21:06, 15 March 2010 (UTC) (Updated: --Jwesley78 22:23, 15 March 2010 (UTC))
• From this, it appears to me that "for all intents-and-purposes", the factorial operation has a higher precedence than exponentiation. I cannot see a situation for which exponentiation, or some other operation (excluding the use of parenthesis), would be evaluated first.Jwesley78 21:16, 15 March 2010 (UTC)
• Consider "-4!". If the precedence of "-" and "!" were the same, then an associativity rule might provide clarification as to which to evaluate first. The need for such a rule also extends to other even more ambiguous cases involving (prefix and postfix) unary operators, such as "ln -3!!". Should this be interpreted as "(ln -3)!!" or "ln -(3!!)", or something else? Jwesley78 23:31, 15 March 2010 (UTC)

## Display problem

I've removed this line from the table as it seemed to be causing weird problems with the width of the page. I have no idea why, unfortunately - any ideas? Shimgray | talk | 22:15, 9 September 2009 (UTC)

Something a lot of people go looking for but often cant find is a method of Addition factorials. IE, 1+2+3+4+5... It would be cool if someone would put in the method for that, either as a section of this page, or as a new page altogether.

That equation is 1+2+3+4+5...N(N+1)/2.

I dont have the wiki know-how to do this correctly, so I leave it up to you. —Preceding unsigned comment added by 208.126.145.232 (talk) 16:25, 15 September 2009 (UTC)

See triangular number. Robo37 (talk) 09:34, 23 November 2009 (UTC)
See also Gauss and the problem of adding all whole numbers from 1 to 100. 1+100+2+99+3+98+...+50+51=50(101)=5050 Not sure if the story is true but here it is: http://www.jimloy.com/algebra/gauss.htm Oh, and with addition, it wouldn't be a factorial, but a summation (Like the Sigma symbol in calculus does). 75.70.89.124 (talk) 07:58, 19 May 2013 (UTC)

## "A weak version that can easily be proved with..."

For the sake of completeness, has anyone an "easy" example of how to prove this? —Preceding unsigned comment added by 87.162.220.155 (talk) 19:18, 21 October 2009 (UTC)

I edited the section to include different, stronger bounds on n! for which the proof is really simple. As for proving these weaker bounds, the right one can be proved by pairing the elements of n! as (1*n) * (2*(n-1)) * ... and then noticing that the product in each pair is at most (n/2)*(n/2). I don't see an equally easy proof for the left inequality -- and anyway, the proof via integrals I just added is more practical. Misof (talk) 20:03, 13 December 2009 (UTC)
By removing the "easy" formula, you robbed me and my students of an easy check on the result of a factorial calculation in so-called retrospective fault detection. Where in the Factorial article is your replacement? Gpermant (talk) 12:12, 4 January 2010 (UTC)
I find the "robbing" part of your post insulting and completely unnecessary. I did not rob anyone, I just provided a stronger (i.e., more precise) formula, along with a reasonably simple explanation why it works. I have no idea what your "retrospective fault detection" is, but regardless of what it is I don't see a case where the more precise estimate cannot be applied.
The replacement is the explained formula $e\left({\frac {n}{e}}\right)^{n}\leq n!\leq e\left({\frac {n+1}{e}}\right)^{n+1}$ This formula works for all $n$ , whereas in the previous one the upper bound only works for $n\geq 6$ .
On the left side it is obvious that we have $e\left({\frac {n}{e}}\right)^{n}>\left({\frac {n}{e}}\right)^{n}>\left({\frac {n}{3}}\right)^{n}$ for all $n$ .
The right hand side obviously grows asymptotically slower than $\left({\frac {n}{2}}\right)^{n}$ , hence for almost all $n$ (which is precisely for all $n\geq 12$ ) it gives a smaller upper bound than $\left({\frac {n}{2}}\right)^{n}$ .
However, you gave me the idea that adding the simpler bounds back cannot hurt the article, I'll go and make the edit.
cheers, Misof (talk) 19:39, 23 August 2010 (UTC)

## <nowiki>!</nowiki> ?

The article often (always?) escapes the factorial shriek as indicated. Can anyone explain why? Nothing bad seems to happen without it, and I have not found any particular use of exclamation marks in wiki markup, except in wikitables. Certainly one can use ordinary punctuation in text without escaping ?!? Marc van Leeuwen (talk) 09:32, 14 March 2010 (UTC)

## Infoliofaktorial or infoliofactorial

Infoliofaktorial ("infoliofactorial "), as well as the Gamma function extends the factorial to x> 1, for example:

x! = x * (M! * m + (M-1) * (1-m))
where x = M + м
5.0133852! = 127 (or 5,0133852! = 126,9999803)

For integers x infoliofaktorial not differ from the Gamma function.
Derivative and inverse factorial is calculated taking into account the fact that M = int (x), м = x - M


М = х!/1/2/3/4... (ru.math.wikia.com/wiki/Обратный_факториал) — Preceding unsigned comment added by Infoliokrat (talkcontribs) 14:14, 16 January 2011 (UTC)

Wikia though is not a reliable source, and we'd need one to add this otherwise it looks like original research.--JohnBlackburnewordsdeeds 21:48, 16 June 2011 (UTC)

## infoliokratnaya-funkciya or infoliokratnaya function

infoliokratnaya-funkciya or infoliokratnaya function х!? = х (n!? = n)

From infoliofaktoriala, after dividing the left and right side in (M-1! have: Since x! / (M-1), = x (Mm + (1-m)), where x = M + m. Letting Z = X ! / (M-1)! have Z = (m + m) (Mm-m +1), m is for the usual quadratic equation m2 - m (M2-M +1) + M - Z = 0, where b = (M2-M +1) / (M-1), c = (M / (M-1) - (xi / (M-2)!) and determined part for ,+M=x. — Preceding unsigned comment added by 178.122.42.114 (talk) 21:41, 16 June 2011 (UTC)

## Gamma function

In the Wiki Article, that is quoted in part below the dashed line, on the binomial theorem and its extension to negative non integer, we find the Gamma Function. It uses capital Pi in two completely different ways that seems designed to confuse readers. It uses it as: (1.) a function Pi(z) = integral of t^Z exp(-t) dt, and as (2.) a PRODUCT operator, that multiples a series of terms index by a dummy variable k. The person who wrote this article was talking to themselves and made about 100 hidden assumptions. If he or she would share those hidden assumption, as they are made, and stop using the same symbol for multiple meaning, the article might actually help readers, instead of just confusing them.

Reading it gives one no clue of what (-1/2)! is or how to calculate it. If an illustration of the expansion of (-1/2)! were added, it would make the article 10 times easier to understand. The author blithely writes (-1/2)! and expects people to know what it means. Furthermore, the author asserts that (if the secret computation were revealed), (-1/2)! = square root of pi, with not one shred of evidence as to why. Not one reader concerned about the topic, in 10,000 will have any idea what is going on here.

The article also states Euler's original Gamma function as capital Pi function = to a Limit, as n goes to infinity, of a ratio with n^z n! as the numerator. Most readers will have no clue what infinity raised to a power is, or what infinity factorial is. Most will not not know if such terms are well defined or exist. Some explanation is obviously require to make this readable.

Will some one figure out what the author was trying to say, determine if it is correct, and then rewrite it so it is correct and can be understood.

The Gamma and Pi functions Main article: Gamma function The Gamma function, as plotted here along the real axis, extends the factorial to a smooth function defined for all non-integer values. The factorial function, generalized to all complex numbers except negative integers. For example, 0! = 1! = 1, (−0.5)! = √π, (0.5)! = √π/2.

Besides nonnegative integers, the factorial function can also be defined for non-integer values, but this requires more advanced tools from mathematical analysis. One function that "fills in" the values of the factorial (but with a shift of 1 in the argument) is called the Gamma function, denoted Γ(z), defined for all complex numbers z except the non-positive integers, and given when the real part of z is positive by

   \Gamma(z)=\int_0^\infty t^{z-1} e^{-t}\, \mathrm{d}t. \!


Its relation to the factorials is that for any natural number n

   n!=\Gamma(n+1).\,


Euler's original formula for the Gamma function was

   \Gamma(z)=\lim_{n\to\infty}\frac{n^zn!}{\prod_{k=0}^n (z+k)}. \!


It is worth mentioning that there is an alternative notation that was originally introduced by Gauss which is sometimes used. The Pi function, denoted Π(z) for real numbers z no less than 0, is defined by

   \Pi(z)=\int_0^\infty t^{z} e^{-t}\, \mathrm{d}t\,.


In terms of the Gamma function it is

   \Pi(z) = \Gamma(z+1) \,.


—Preceding unsigned comment added by Jaimster (talkcontribs)

## I read the extension to the complex plane bit of this article, but something seems to be missing...

...what is i!? Robo37 (talk) 09:33, 6 July 2011 (UTC)

Ah, I got it, it's 0.498015668 - 0.154949828i. Should this be inserted into this article do you think?
If there is a closed form version, in terms of the γ, π, and/or whatever, that would be even better. —Quantling (talk | contribs) 13:00, 8 July 2011 (UTC)

## Factorial root?

What is x when x! = n? Robo37 (talk) 09:58, 14 July 2011 (UTC)

Are you asking whether there is a name or whether there is a notation (or both)? I would say that, as with nearly any function, it is fair to speak of an inverse. So, "the factorial inverse of $n$ is $x$ ." As for notation, if you like working with $\Pi$ , you could write $\Pi ^{-1}(n)=x$ . I don't know that I've seen it written with an exclamation point. —Quantling (talk | contribs) 14:05, 14 July 2011 (UTC)
Thanks. I'm more looking for some kind of closed equation that would calculate an integar when a factoral number is entered? Robo37 (talk) 18:07, 15 July 2011 (UTC)
For small integers, just use an array. This would be an excellent application of a binary search. Think of the "lower/higher" guessing game. You could even use the logarithmic operator and one of the approximations listed in the article (invert the exponent into a ln function) to get an estimate on where to start your search in the array. However, the array would obviously only have a very small number of values to search through so it's probably not worth it. To fill the array is an exercise I'll leave to readers.  ;) An alternative (obviously poor, IMO) is to use a select...case structure.
See these: http://en.wikipedia.org/wiki/Stirling%27s_approximation http://mathforum.org/kb/thread.jspa?messageID=342551 75.70.89.124 (talk) 07:27, 31 July 2013 (UTC)

## Notation for Factorial

In the thirteenth edition of the "Handbook of Chemistry and Physics", Copywrite 1914,... (USA), published by the "Chemical Rubber Publishing Co.", Authors Hodgman, C. D. and Lange, N. A. use the following notation for factorial:

|n, where |n = n!

Possible reasons for this are that:

(i) Recalling that the factorial is related to the Gamma function, the notation used looks like the capital Greek letter, gamma (Γ), flipped about a horizontal line.
(ii) the used notation looks like the line(s) used for short/long divison.
(iii) it may have something to with Blaise Pascal's version of the well known triangle (to the right) that is of the right angled variety and thus forms L shapes.

Reasons aside, I thought that others might be interested in this notation. Mhallwiki (talk) 19:26, 9 August 2011 (UTC)

All three reasons proposed seem highly implausible to me; lacking any sources they would certainly be WP:OR. Also, unless any more recent uses can be found, it would seem like an attempt to introduce (yet) another notation for n!, which after almost a century can be safely considered to have failed. So I don't see much reason to mention it in the article (or maybe just briefly in the history section). Marc van Leeuwen (talk) 11:12, 10 August 2011 (UTC)

## Hmmm....

The sum of the reciprocals of the sum of the first n integers (the Triangular numbers) is 2. The sum of the reciprocals of the sum of the first n Triangular numbers (the Tetrahedral numbers) is 1.5.

The sum of the reciprocals of the product of the first n integers (the Factorials) is e, so I wonder, is the sum of the reciprocals of the product of the first n Factorials (the Superfactorials), which is 1.5868056, or the sum of the reciprocals of the sum of the first n Factorials which is 1.47608642, expressible in terms of e?

## Why is this?

It says at http://www.wolframalpha.com/input/?i=%28-1%29%5Ex that the series expansion of (-1)^x at x=0 is

... notice a bit of Factorialness in the demoninators there. Why is this? — Preceding unsigned comment added by Robo37 (talkcontribs)

You would be much better off asking such questions at the Mathematics Reference Desk, where people who answer such questions hang out. This talk page is for discussions related to improving this article only.--JohnBlackburnewordsdeeds 01:32, 19 August 2011 (UTC)
The answer is actually in the article. Factorial#Applications says:
"Factorials also turn up in calculus; for example they occur in the denominators of the terms of Taylor's formula, basically to compensate for the fact that the n-th derivative of xn is n!."
The first derivative of xn is n×xn-1. It follows from induction that he nth derivative of xn is n!. So if we start with xn/n! then the nth derivative becomes 1. PrimeHunter (talk) 02:00, 19 August 2011 (UTC)
-1 = ei pi and ex = Sum(xn/n!), ergo -1x = ei pi x = Sum[(i pi x)n/n!]79.113.230.39 (talk) 00:14, 9 March 2013 (UTC)

## philosophically

Besides the permutations of rearranging a given number: factorials may also be considered philosophically pertinent if all numbers are considered things-in-themselves. Wilst requiring all constituent numbers as an infrastructure upon which the following integer can be extant and therefore subsequently enumerated. Nagelfar (talk) 06:12, 12 September 2011 (UTC)

Thanks, now I can sleep properly tonight. McKay (talk) 07:34, 12 September 2011 (UTC)

## Last table entry

Why has the table at the beginning an entry for n = Template:Val? Does this number enjoy some particular interesting non-obvious property?  --Lambiam 19:30, 23 December 2011 (UTC)

To answer my own question: it appears to be (within the given precision) equal to 2210. But is that sufficiently interesting to list this here? And isn't that rather pointless without further explanation?  --Lambiam 19:44, 23 December 2011 (UTC)

It was once linked. I agree the current form doesn't make much sense after the link was removed in . PrimeHunter (talk) 23:06, 23 December 2011 (UTC)
So it is the maximum value representable in the IEEE floating-point standard's double-precision floating-point format. From a mathematical point of view, that number has no special significance. From a numeric-computational point of view: (A) Why stop with double precision? Why not quadruple precision? (B) It seems more relevant what the largest n is such that n! does not exceed the maximally representable value. If I did not make a mistake, the representability limits are reached with 170! for double precision and 1754! for quadruple precision, which underscores how fast the factorial grows. But the reference to IEEE floating-point formats seems unnecessarily abstruse for this article on such an elementary integer-valued function, and my preference is to avoid referring to floating-point formats and to delete this table entry.  --Lambiam 00:18, 24 December 2011 (UTC)
The table already shows 170! and 171! for this reason, mentioned in Factorial#Computation. But I wouldn't object to removing all table entries caused by IEEE limits. PrimeHunter (talk) 04:54, 24 December 2011 (UTC)

## Sample code?

Do we really need a code example? The current implementation is overly clumsy and does not even compile. But as noted at the start of the section, it's trivial to write, so instead of fixing it, I'd propose to remove it altogether. Any objections? --Clickingban (talk) 08:16, 26 June 2012 (UTC)

Agreed. The only point of showing such code would be to show how utterly trivial it is to compute factorials without recursion, thereby implicitly showing how silly it is to systematically use factorials as (first) examples of recursive functions. Marc van Leeuwen (talk) 11:13, 26 June 2012 (UTC)
Agree completely. -- Elphion (talk) 15:02, 26 June 2012 (UTC)
The current implementation also has the disadvantage that it is unnecessarily inefficient: it would be faster to precompute a lookup table for the first 20 factorials (anything larger overflows). To me, the sample code says more about the limitations of C++ than about computing factorials. I don't mind having sample code in some articles (as long as it doesn't start looking like a code farm instead of an article) and in particular I think pseudocode can often be helpful in pointing out some non-obvious implementation tricks (like the one mentioned above about multiplying bignums out of order). But there's nothing non-obvious in the present example, it needs even more cruft to be adequately engineered (it should not just overflow silently), and I think C++ was a bad choice (pseudocode or something closer to pseudocode like Python would have been better), so I agree that we're better off without it. —David Eppstein (talk) 16:55, 26 June 2012 (UTC)
It added nothing and is gone. McKay (talk) 07:48, 27 June 2012 (UTC)

## table at the introduction of the sample values far too wide to the right margin

Need a wiki-table expert to correct this -- it "glues" at the right browser-visible margin with its data in the right most column. :-( About 1/2 half inch free space needed (as else in the whole article). — Preceding unsigned comment added by 2001:638:504:C00E:214:22FF:FE49:D786 (talk) 14:49, 19 September 2012 (UTC)

## Articles to Factorial and Gamma function

These two articles overlap very much -- FAR too much, IMHO. Especially I wonder why the picture "File:Factorial05.jpg" is shown here, but not in the article Gamma function -- the truth is: this fact triggered my comment here. This "politic" is really for head-shaking.

I think either these two articles should be joined, or clearly separated for (integer argument and real/complex) consideration. Now it looks like there is an ongoing fighting of two author groups in wikipedia which can present the same topic better. Maybe this is indeed the case? Too many amateurs are proud to show their basic mathematical knowledge and especially press their oppinion how to present this. :-( Regards. — Preceding unsigned comment added by 2001:638:504:C00E:214:22FF:FE49:D786 (talk) 11:01, 21 September 2012 (UTC)

## Derivative Definition

How about we add the definition (d/dx)^n x^n = n! ? Okidan (talk) 11:38, 15 March 2013 (UTC)

It's already mentioned under Applications.--JohnBlackburnewordsdeeds 14:11, 15 March 2013 (UTC)
Do you think it warrants a mention in the definition section? Okidan (talk) 00:27, 17 March 2013 (UTC)
It seems an odd way to define the factorial. Have you seen reliable sources do that? PrimeHunter (talk) 00:49, 17 March 2013 (UTC)

I found it on the MIT Open Courseware Notes - it's near the bottom on the third page. Adding external link:

Okidan (talk) 00:03, 18 March 2013 (UTC)

Huh? It says 'The notation n! is called "n factorial" and defined by n!=n·(n−1)·2·1.' They then prove the property (d/dx)^n x^n = n! I don't see why that should make the property belong in the definition section. PrimeHunter (talk) 18:34, 18 March 2013 (UTC)

## Multifactorial

The recursive definition of the multifactorial in this article differs from that given in the article "Fakultät" in the german Wikipedia (for example, 2!!! is defined here as = 1, whereas on the german site is defined 2!!! = 2). Which definition is now the right one? I remember, on the german site the definition has previously been the same as here, but has then been changed with the purpose to correct it. Should the definition here be adapted to the one on the german site, or has the correction been wrong? --79.243.235.186 (talk) 21:53, 1 July 2013 (UTC)

I don't know what meaning the multifactorials exactly have, but I think the following definition is more consistent than the one in our article:
n!(k) = { 1 , for (1-k) ≤ n < 1 ; n*(n-k)!(k) , for n ≥ 1 .
Although written differently, this definition should have the same results than the one on the German Wikipedia site, except the expansion onto negative integers > -k. With that definition, 2!!! would be = 2. --79.243.224.89 (talk) 17:26, 27 July 2013 (UTC)

## Jargon; plus N, n and n!

Ok - so wikipedia is supposed to be readable as if general interest to the reasonably intelligent person without foreknowledge. I am sure this article is very readable to mathemticians, but I came here on a circuitous path trying to pin down the different uses of n for https://en.wiktionary.org/wiki/n. The WP article and the disambiguation page seems in some ways most helpful, although as the specialist pages please go a littlebit more slowly given the confusion and explain (gently, slowly, thoroughly) notational use/variations of n as a necessary introduction for your your general readership, please. Kathybramley (talk) 12:13, 18 November 2013 (UTC)

I'm very confused by your comment -- this article (which is about the factorial function, denoted "!") uses "n" only in the completely standard way that it is a variable on a certain domain (nonnegative integers, mostly). So, I don't see the relationship with your search, nor can I make sense of most of your other comments (the comments about jargon are vague; the comments about other pages, also). Can you clarify? --JBL (talk) 13:44, 18 November 2013 (UTC)
I see it is indeed listed at N (disambiguation) but that is debatable. It's really a meaning of "!" and not of "n". n is just a variable name. n is the most commonly used variable name for factorials but also for natural numbers in general (competing with x in some environments but mathematicians use x more for reals than integers). We could also have written x!, a! and so on. For a dictionary it's much better suited for wikt:! and is already there. It's also at our own article about ! (a redirect to Exclamation mark) as well as ! (disambiguation). I don't edit Wiktionary and don't know their practices but I wouldn't expect to see n! or other mathematical expressions like -n, n2, and so on at wiktionary:n, but there might be a mention that n is a common variable name for a natural number. A capital N (or sometimes $\mathbb {N}$ if you're fancy) denotes the set of natural numbers. PrimeHunter (talk) 15:52, 18 November 2013 (UTC)
Oy, that's terrible -- I've removed the link to this page from the disambiguation page for N. --JBL (talk) 17:22, 18 November 2013 (UTC)