Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Well, n! is for integer n < 0 not defined -- as yet.

So the question is: How could a sensible generalization of the factorial for negative integers look like?

Clearly a good generalization should have a clear combinatorial meaning which combines well with the nonnegative case.

share|improve this question
3  
Meta (scnr): This question was asked in the first weeks of mathoverflow and has received a lot of attention. When such an elementary question would be asked nowadays, it would get closed almost immediately. –  Martin Brandenburg Aug 19 '13 at 13:03

8 Answers 8

It's not that it's not defined... Actually it has been defined more than it should have. There are plenty of functions that interpolate the factorials, some of them extend to the negative integers as well. Hadamard's Gamma function is entire, logarithmic single inflected factorial function is another example. But on the other hand, for some mysterious reason, the nice property that we want an extension of the factorial to enjoy is log-convexity. The Bohr-Mollerup-Artin Theorem tells us that the only function which is logarithmically convex on the positive real line and satisfies $f(z)=zf(z-1)$ there (also $f(1)=1$ and $f(z)>0$) is the Gamma function. Unfortunately the gamma function doesn't extend to negative integers, and that is why I guess people don't really care that much for defining them as they know that no "good" answer can be found.

share|improve this answer
14  
I would just add that the definition $n!=\infty$ when n<0 makes good sense; it extends the usual recursion of the factorial function and is consistent with the poles of the Gamma function. It also allows combinatorial formulas to work in degenerate cases. That is, when the factorial of a negative number appears in the denominator of a fraction, the usual convention is that the fraction is 0. –  Jonas Meyer Dec 30 '09 at 12:29
    
Am I mistaken? Hadamars Gamma function is not logarithmic single inflected. The 'mysterious reason of log-convexity' /does not/ carry over to the general gamma function, i. e., it does not characterise the complex gamma function. This might be regarded as a hint, that there /is/ room for other sensible definitions and that log-convexity is not the last word in this matter, it is just a substitute for some analytical condition. –  Bruce Arnold Dec 31 '09 at 0:22
3  
Hadamards gamma function and the logarithmic single inflected factorial function are different functions. –  Gjergji Zaimi Dec 31 '09 at 0:34
2  
I am just referring to various functions that one can interpolate the factorials with, mostly based on playing around with the diGamma function. There is no citeable reference but here's a "pictoresque" account :) luschny.de/math/factorial/hadamard/HadamardsGammaFunction.html –  Gjergji Zaimi Dec 31 '09 at 5:40
21  
Don't blame log-convexity for making $(-1)!$ undefined. The functional equation $n!=(n-1)!n$ does that job all by itself. First (with $n=1$) it forces the standard convention that $0!=1$, and then (with $n=0$) is requires $(-1)!$ to be the reciprocal of 0. –  Andreas Blass Jan 9 '12 at 1:15

I think it's worth pointing out here that, if $a\ge0$, then, near z = -a, we have

$$ \Gamma(z) = (-1)^a {1 \over a!} {1 \over {z+a}} + O(1) $$

and so it might be tempting to say that, in some sense,

$$ \Gamma(-a) = (-1)^a {1 \over a!} \infty $$

where the symbol $\infty$ represents the rate at which $\Gamma$ blows up near the pole at $a = 0$. That is, $\Gamma(0) = \infty, \Gamma(-1) = -\infty, \Gamma(-2) = \infty/2, \Gamma(-3) = -\infty/6$, and so on.

In particular, this interpretation might work in some formula in which $\Gamma$ evaluated at nonpositive integers appears in both the numerator and the denominator, and the symbol $\infty$ can be canceled to yield a real number.

share|improve this answer

For a related paper see D. Loeb, Sets with a negative number of elements, Adv. Math. 91 (1992), 64–74.

share|improve this answer

If you're wanting to compute factorials as an intermediate step to computing binomial coefficients, you may find a more satisfactory answer to your question. See this chart for determining how to compute binomial coefficients for general arguments.

share|improve this answer
    
See also "Getting results with negative thinking", D. E. Loeb et alia, especially the table on page 10 and the proposition on the six regions. arxiv.org/abs/math/9502214 –  Bruce Arnold Dec 30 '09 at 23:25

My question was intended somewhat along the line: Assume the Gamma function is not yet invented and Goldbach asks you the question: "What is (-n)! ?" You know that Goldbach expects a combinatorial answer in the domain of integer or rational numbers. What would you answer? I will give my answer in this sens.

Looking at GKP's ConMath, Table 253, the combined Stirling triangles in their dual form, we see: If we sum the columns in this triangle for k < 0 we get the factorial numbers, if we sum the rows for k > 0 we get the Bell numbers.

What about saying the Bell numbers are the factorial numbers at negative integers? Is the answer encoded in one of the most important triangles in combinatorics?

See what Knuth says about the origin of this duality (table on page 11).


{120}
......{24}
.1,.......{6}
10, .1,......{2}
35, .6, .1,.....{1}
50, 11, 3, 1,
24, .6, 2, 1, 1,....{1}
.0, .0, 0, 0, 0, 1,
.0, .0, 0, 0, 0, 0, 1,...{1}
.0, .0, 0, 0, 0, 0, 1, .1,....{2}
.0, .0, 0, 0, 0, 0, 1, .3, .1,....{5}
.0, .0, 0, 0, 0, 0, 1, .7, .6, .1,....{15}
.0, .0, 0, 0, 0, 0, 1, 15, 25, 10, 1,....{52}

share|improve this answer
4  
Then you should have definitely mentioned this in the question :). To convince us that your correspondence is appropriate one has to give a combinatorial theory where the Stirling number duality makes sense. This is precisely what this thread mathoverflow.net/questions/9721/… is about. –  Gjergji Zaimi Dec 31 '09 at 0:11
    
Thanks for the link. I just ask here another question to exemplify my main question. Looking at this triangle which was been considered by Kramp, Stanley, B. F. Logan, I. Gessel and D. Knuth, among others, the question is: Is there a simple and uniform mathematical explanation why summing rows and columns lines up the double infinite sequence ...203, 52, ,15, 5, 2, 1, 1, 1, 2, 6, 24, 120, 720,... and does this make the definition (-n)! = Bell(n) meaningful? Or do I have to conclude from your answer that there is no 'combinatorial theory where the Stirling number duality makes sense'? –  Bruce Arnold Jan 1 '10 at 20:29

I think, a sensical definition stems from the generalization of the triangle of eulerian numbers.

For positive integer indexes the rows sum to factorials, and even if the rows are interpolated to fractional indexes (based on the closed-form-formula for the direct computation) the rowsums are fractional factorials or gamma values.

Thus I assume the extension of the eulerian triangle to negative indexes gives the answer to a sensical definition of the factorials at negative parameters.

For instance, we get $$ \small \begin{array}{r|lllll|lll} \text{index k} & \\ \hline\\ &&&& \text{ extension to negative indexes} \\ \cdots &\cdots \\ -4& 1 & 3+\frac 1{16} & 6+\frac 3{16}+\frac 1{81}& 10+\frac 6{16}+\frac 3{81}+\frac 1{256}&\ldots &\tiny \sum \overset ?= & -4! \\ -3& 1 & 2+\frac 18 & 3+\frac 28+\frac 1{27}& 4+\frac 38+\frac 2{27}+\frac 1{64}&\ldots &\tiny \sum \overset ?= & -3! \\ -2& 1 & 1+\frac 14& 1+\frac 14+\frac 19& 1+\frac 14+\frac 19 + \frac 1{16}& \ldots &\tiny \sum \overset ?= & -2! \\ -1& 1 & 0+\frac 12 & 0+0+\frac 13 & 0+0+0+\frac 14 & \ldots &\tiny \sum \overset ?= & -1! \\ \hline \\ &&&& \text{ triangle of Eulerian numbers} \\ 0& 1 & . & . & . &. &\tiny\sum = & 0! \\ 1& 1 & . & . & . &. &\tiny\sum = & 1! \\ 2& 1 & 1 & . & . &. &\tiny\sum = & 2! \\ 3& 1 & 4 & 1 & . &. &\tiny\sum = & 3! \\ 4& 1 & 11 & 11 & 1 &. &\tiny\sum = & 4! \\ \cdots & \cdots \\ \end{array} $$ and so on.


I have a more involved discussion in a hobby-treatize about the Eulerian-triangle here

share|improve this answer

According do the definition of factorial, $1 = 0! $ and $ 0! = -1! * 0$. So, first negative integer factorial is $$-1! = 1/0 = \infty$$. I am not sure why it should be a negative infinity. Possibly because zero can be very small negative number as well as positive. I cannot derive the sign. But, I can prove that other integer negatives are also infinities.

Take -2! * -1 = -1!. It follows that $-2! = -1!/-1 = -\infty$.

Next, -3! * -2 = -2! whereupon, $-3! = -2!/-2 = {-1!\over (-1)(-2)} = +\infty$.

Generally, we see that all factorials are infinities with alternating sign, $$-n! = {-1! \over (-1)(-2)\cdots(1-n)} = \infty/(-1)(-2)\cdots = (-1)^n\infty$$

share|improve this answer
1  
Please see Michael Lugo's answer mathoverflow.net/a/10150/78 for a more precise version of this manipulation. (Also, note that if you believe that $-\infty \neq +\infty$, you might as well also believe in $2\infty \neq \infty$. Lugo's answer also makes this precise.) –  Theo Johnson-Freyd Jul 24 '13 at 15:56
    
He starts with "gamma" function but we, with the author of this post, show how to derive the answer without being aware of that. Also, Lugo makes the difference between + and - $\infty$. But, thanks for pointing out that they are equal. I think that my example shows that $+\infty$ must be equal to $-\infty$, as there is a simmetry. That is, we cannot guess that zero is positive infinitecimal rather than negative. –  Val Jul 24 '13 at 16:08
1  
Hi Val: Sorry, the point I was trying to make is that games with $\pm 0$ are fun, but mathematicians have much more advanced ways to make them precise, including residues of analytic functions. The Gamma function is the unique analytic function satisfying the functional equation $\Gamma(a+1) = a\Gamma(a)$. I'm sure that OP knows that the naive generalization to negative integers is $(-n)! = (-1)^n\ n!\ \infty$. –  Theo Johnson-Freyd Jul 24 '13 at 16:12

As 0! and 1! = 1 ,
2! = 2,
3! = 6 and so on

Can we not have

-1! = -1
-2! = 2 = -1 X -2
-3! = -6 = -1 X -2 X -3
-4! = 24 = -1 X -2 X -3 X -4
-5! = -120 = -1 X -2 X -3 X -4 X -5

...and so on?

It would produce some sort of series but would they be of any use anywhere?

share|improve this answer
1  
This is known as the Roman Factorial - see Qiaochu's comment to Michael Lugo's answer. –  S. Carnahan Jan 9 '12 at 5:35

protected by François G. Dorais Aug 19 '13 at 22:07

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.