most recent 30 from http://mathoverflow.net 2013-05-25T04:17:11Z http://mathoverflow.net/feeds/question/10124 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10124/the-factorial-of-1-2-3 Bruce Arnold 2009-12-30T11:32:10Z 2012-01-08T23:00:35Z

<p>Well, n! is for integer n &lt; 0 not defined -- as yet.</p> <p>So the question is: How could a sensible generalization of the factorial for negative integers look like?</p> <p>Clearly a good generalization should have a clear combinatorial meaning which combines well with the nonnegative case.</p>

http://mathoverflow.net/questions/10124/the-factorial-of-1-2-3/10129#10129 Gjergji Zaimi 2009-12-30T12:00:47Z 2009-12-30T12:00:47Z

<p>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 <a href="http://en.wikipedia.org/wiki/Gamma_function" rel="nofollow">Gamma function</a>. 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.</p>

http://mathoverflow.net/questions/10124/the-factorial-of-1-2-3/10145#10145 John D. Cook 2009-12-30T15:06:44Z 2009-12-30T15:06:44Z

<p>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 <a href="http://www.johndcook.com/computing%5Fbinomial%5Fcoefficients.html" rel="nofollow">chart</a> for determining how to compute binomial coefficients for general arguments.</p>

http://mathoverflow.net/questions/10124/the-factorial-of-1-2-3/10150#10150 Michael Lugo 2009-12-30T16:10:52Z 2009-12-30T16:10:52Z

<p>I think it's worth pointing out here that near z = -a, we have</p> <p>$$ \Gamma(z) = (-1)^a {1 \over a!} {1 \over {z-a}} + O(1) $$</p> <p>and so it might be tempting to say that, in some sense,</p> <p>$$ \Gamma(-a) = (-1)^a {1 \over a!} \infty $$</p> <p>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.</p> <p>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.</p>

http://mathoverflow.net/questions/10124/the-factorial-of-1-2-3/10166#10166 Richard Stanley 2009-12-30T18:56:21Z 2009-12-30T18:56:21Z

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

http://mathoverflow.net/questions/10124/the-factorial-of-1-2-3/10195#10195 Bruce Arnold 2009-12-30T23:46:51Z 2009-12-30T23:46:51Z

<p>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.</p> <p>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 &lt; 0 we get the factorial numbers, if we sum the rows for k > 0 we get the Bell numbers.</p> <p>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? </p> <p>See what <a href="http://arxiv.org/abs/math/9205211" rel="nofollow">Knuth</a> says about the origin of this duality (table on page 11). </p> <p><br/> {120} <br/> ......{24} <br/> .1,.......{6} <br/> 10, .1,......{2} <br/> 35, .6, .1,.....{1} <br/> 50, 11, 3, 1, <br/> 24, .6, 2, 1, 1,....{1} <br/> .0, .0, 0, 0, 0, 1, <br/> .0, .0, 0, 0, 0, 0, 1,...{1} <br/> .0, .0, 0, 0, 0, 0, 1, .1,....{2} <br/> .0, .0, 0, 0, 0, 0, 1, .3, .1,....{5} <br/> .0, .0, 0, 0, 0, 0, 1, .7, .6, .1,....{15} <br/> .0, .0, 0, 0, 0, 0, 1, 15, 25, 10, 1,....{52} <br/></p>

http://mathoverflow.net/questions/10124/the-factorial-of-1-2-3/32302#32302 Gottfried Helms 2010-07-17T18:39:47Z 2010-07-17T18:39:47Z

<p>I think, a better 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 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 if the factorials at negative parameters. For instance, we get -1! = 1 + 1/2 + 1/3 + 1/4 + ... -2! = (1) + (1+1/4) + (1+1/4+1/9) + ... -3! = (1) + (2+1/8) + (3+2/8+1/27) + ... and so on. The terms can be computed by the direct definition of Eulerian-numbers (see formula in wikipedia,for instance). I have discussed this in a hobby-treatize of the Eulerian-triangle in <a href="http://go.helms-net.de/math/binomial_new/01_12_Eulermatrix.pdf" rel="nofollow">http://go.helms-net.de/math/binomial_new/01_12_Eulermatrix.pdf</a></p> <p>Gottfried Helms</p>

http://mathoverflow.net/questions/10124/the-factorial-of-1-2-3/85197#85197 John O 2012-01-08T15:37:52Z 2012-01-08T23:00:35Z

<p>As 0! and 1! = 1 ,<br> 2! = 2,<br> 3! = 6 and so on</p> <p>Can we not have</p> <p>-1! = -1<br> -2! = 2 = -1 X -2<br> -3! = -6 = -1 X -2 X -3<br> -4! = 24 = -1 X -2 X -3 X -4<br> -5! = -120 = -1 X -2 X -3 X -4 X -5 </p> <p>...and so on?</p> <p>It would produce some sort of series but would they be of any use anywhere?</p>