Talk:Factorial/Archive 1

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

and in C:

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 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:
 * 5? = 4!
 * 6? = 5!
 * 7? = 6!
 * 8? = 7!
 * 9? = 8!


 * 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 (talk • contribs) 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 (talk • contribs) 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 (talk • contribs) 04:31, 5 August 2004 (UTC)


 * Yes, a triangular number. --- User:Karl Palmen 5 August 2004


 * Adding an external link:
 * 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 (talk • contribs) 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 (talk • contribs) 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 (talk • contribs) 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
 * $$3\mathrm{S}\!\!\!\!\!\;\,{!}=6\uparrow\uparrow6={^6}6=6^{6^{6^{6^{6^6}}}}$$
 * 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)

Notationally one always prefers the larger number since the smaller number can be produced by a simpler formulation. Victor Kosko (talk) 01:07, 21 August 2018 (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 (talk • contribs) 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 &Gamma;(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! = &Gamma;(1.5) = 0.5 &Gamma;(0.5) = &radic;&pi; / 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 &Gamma;(.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 (talk • contribs) 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 (talk • contribs) 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 (talk • contribs) 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 (talk • contribs) 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:

$$(a+b)! = \frac{a!b!}{|b-a|!}$$

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.

--Hugo Dufort

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:


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

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)

Rational combination, Rational permutation. Should be in that article, not factorial. Victor Kosko (talk) 23:32, 18 August 2018 (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 (&Gamma;(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 &times; (n&minus;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 (talk • contribs) 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 ≤ i ≤ n. 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:


 * $$ n!=\prod_{k=1}^n k \qquad \mbox{for all } n \in \mathbb{N} \ge 0. \!$$


 * 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 a &le; k &le; b. 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))

additive analogue
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)

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 (talk • contribs) 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 (talk • contribs) 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! &mdash; 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). — Q uantling (talk &#124; 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 (talk • contribs) 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 

-

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: I looked at the ratio of these two expressions:
 * 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)
 * $$n!\approx \frac{2\sqrt {2 e n}}{3}\left(\frac{2n+1}{2 e}\right)^n.$$
 * An even more accurate approximation can be made using the following (provided by Brendon Phillips)
 * $$n!\approx \frac{2\sqrt {2 e n}}{3}\left(\frac{2n+1}{2 e}\right)^n.$$
 * $$n!\approx \frac{2\sqrt {2 e n}}{3}\left(\frac{2n+1}{2 e}\right)^n.$$
 * $$n!\approx \frac{2\sqrt {2 e n}}{3}\left(\frac{2n+1}{2 e}\right)^n.$$

\begin{align} & {} \qquad \frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{\frac{2\sqrt {2 e n}}{3}\left(\frac{2n+1}{2 e}\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{align} $$

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 is much better than Stirling's $$n!\approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n.$$
 * {|class=wikitable

!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)
 * 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)
 * 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)

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)

Addition Factorials
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..."
$$(\frac{n}{3})^n < n! < (\frac{n}{2})^n$$

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 $$ e\left(\frac ne\right)^n \leq n! \leq e\left(\frac{n+1}e\right)^{n+1} $$
 * 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 $$e\left(\frac ne\right)^n > \left(\frac ne\right)^n > \left(\frac{n}{3}\right)^n$$ 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)

&lt;nowiki&gt;!&lt;/nowiki&gt; ?
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 (talk • contribs) 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.-- JohnBlackburne wordsdeeds 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 (talk • contribs) 02:56, 24 October 2010‎ (UTC)