From: rusin@math.niu.edu (Dave Rusin) Subject: Re: generalization of binomial coefficient Date: 10 May 1999 21:26:52 GMT Newsgroups: sci.math.research Keywords: extending, integrating binomial coefficients with Beta function, q-hypergeometric functions In article <7gg7of$ab2@gap.cco.caltech.edu>, Toby Bartels wrote: >In this post, 0 is a natural number. >n choose i is most basically defined when n and i are both natural. >Then n choose i is an ith degree polynomial in n with rational coefficients, >so n choose i makes sense for n an element of any >commutative associative Qalgebra with identity. [...] >You can also use the Gamma function to get finite/infinity = 0 >(but that's no good when n is a negative integer, >because it gets you infinity/infinity -- hard to interpret). [...] >Does anyone know of a situation where n choose i >is useful or even just interesting for i not a natural number? >Even if it doesn't generalize to all possibilities for n, >I would be interested. Seems to me the right answer is, the binomial coefficients can be defined in every case except when n, i, and n-i are all negative integers, although in the cases {n=negative integer, i nonintegral} the right value is "oo" in the Riemann sphere. You're looking for a function f(n,i) which agrees with the binomial-coefficient function binom(n,i) in the cases you know about (e.g. i = natural number, n arbitrary). As you remarked, this can also be written g(n,i)=GAMMA(n+1)/GAMMA(n-i+1)/GAMMA(i+1) in the cases you're familiar with, and moreover, _this_ g is defined for almost every combination of n and i. It even satisfies the usual functional relationships wherever defined, since GAMMA does (whether the arguments are integral or not). So my suggestion would be to take f to be this g itself, with the domain extended as much as possible by continuity. Since GAMMA is defined and nonzero for all inputs except non-positive integers, we need only be concerned with the cases that one or more of n, i, or n-i is a negative integer. If i is a negative integer but n is not, then we write g(n,z)=(z-i)/[GAMMA(n+(-z)+1)/GAMMA(n+1)]/[(z-i)GAMMA(z) * z] for z near i. The term [GAMMA(n+(-z)+1)/GAMMA(n+1)] in the denominator will approach GAMMA(n+(-i)+1)/GAMMA(n+1) as z->i ; this is the polynomial (n+(-i))(n+(-i-1))...(n+1), which has a finite, nonzero value. The term (z-i)GAMMA(z) in the denominator approaches the residue of GAMMA at i, which is (-1)^i/ (-i)!. So the behaviour of g is that of (z-i)/z, up to a constant. In particular, we will have g(n,i) = 0 by continuity. We may handle the case in which n-i is a negative integer in the same way, or simply note g(n,i) = g(n, n-i) = 0 by the previous paragraph. If n is a negative integer but i is not integral, then in the same way we see g behaves as a constant multiple of 1/(z-n) near n, and so g(n,i) is undefined (or "=infinity" if we accept values in the Riemann sphere). If n is a negative integer and i is a positive integer, g is given by a polynomial in n, which then has a finite nonzero value. Likewise if n is a negative integer and n-i is a positive integer. There remains only the case that n, i, and n-i are all negative integers. In this case we have g(w,z)=GAMMA(w+1)/GAMMA(w-z+1)/GAMMA(z+1) for all (w,z) near (n,i) with w<>n, z<>i, w-z<>n-i. We also may write this g(w,z)=[(w/(w-n))(w-n)GAMMA(w)]/ ([((w-z)/(w-z-n+i))(w-z-n+i)GAMMA(w-z)]*[(z/(z-i))(z-i)GAMMA(z)]) As in an earlier paragraph, each of the products with GAMMA approaches one of the residues of GAMMA at a negative integer, which is finite and nonzero; thus g behaves as the rational function (z-i)*(w-z-n+i)/(w-n) of two variables w and z. Now, how would you like to define g right at (n,i) if it has this kind of singularity at this point? This rational function has no limit at (n,i) in the usual sense, and limits along curves which approach (n,i) can take on any finite value, the value "oo", or be undefined. So I don't think e.g. g(-7, -3) can be given a reasonable value any more than h(w,z) = (z+3)*(w+z+4)/(w+7) can. dave ============================================================================== From: bruck@pacificnet.net (Ronald Bruck) Subject: Re: generalization of binomial coefficient Date: Mon, 10 May 1999 23:24:25 -0700 Newsgroups: sci.math.research In article <7h7ius$gl9$1@gannett.math.niu.edu>, rusin@math.niu.edu (Dave Rusin) wrote: > In article <7gg7of$ab2@gap.cco.caltech.edu>, > Toby Bartels wrote: > >In this post, 0 is a natural number. > >n choose i is most basically defined when n and i are both natural. > >Then n choose i is an ith degree polynomial in n with rational coefficients, > >so n choose i makes sense for n an element of any > >commutative associative Qalgebra with identity. > [...] > >You can also use the Gamma function to get finite/infinity = 0 > >(but that's no good when n is a negative integer, > >because it gets you infinity/infinity -- hard to interpret). > [...] > >Does anyone know of a situation where n choose i > >is useful or even just interesting for i not a natural number? > >Even if it doesn't generalize to all possibilities for n, > >I would be interested. > > Seems to me the right answer is, the binomial coefficients can be defined > in every case except when n, i, and n-i are all negative integers, > although in the cases {n=negative integer, i nonintegral} the right > value is "oo" in the Riemann sphere. > > > You're looking for a function f(n,i) which agrees with the > binomial-coefficient function binom(n,i) in the cases you know about > (e.g. i = natural number, n arbitrary). As you remarked, this can also > be written > g(n,i)=GAMMA(n+1)/GAMMA(n-i+1)/GAMMA(i+1) > in the cases you're familiar with, and moreover, _this_ g is defined for > almost every combination of n and i. It even satisfies the usual > functional relationships wherever defined, since GAMMA does (whether the > arguments are integral or not). Here's an interesting question: so let's extend the binomial coefficients by using the gamma function; is it true, then, that \int_0^n \binomial{n}{x} dx = 2^n ? Numerically this seems to be true (well, OK, I checked for a few positive values of n). But Mathematica doesn't seem to be able to do the symbolic integral. I think this should be an easy consequence of the reflection principle, \Gamma(x)\Gamma(1-x) = \pi \csc \pi x for 0 < x < 1, but I haven't worked out the details. (It gives you a specific rational function of x times sine of pi x to integrate, which shouldn't be hard.) Probably this can be found among well-known formulas for the Beta function, but I don't have such at hand right now... --Ron Bruck ============================================================================== From: Dave Rusin Subject: Re: generalization of binomial coefficient Date: Tue, 11 May 1999 09:42:52 -0500 (CDT) Newsgroups: sci.math.research To: bruck@pacificnet.net In article you write: >> g(n,i)=GAMMA(n+1)/GAMMA(n-i+1)/GAMMA(i+1) >Here's an interesting question: so let's extend the binomial coefficients >by using the gamma function; is it true, then, that > > \int_0^n \binomial{n}{x} dx = 2^n ? > >Numerically this seems to be true (well, OK, I checked for a few positive >values of n). But Mathematica doesn't seem to be able to do the symbolic >integral. I think this should be an easy consequence of the reflection >principle, > > \Gamma(x)\Gamma(1-x) = \pi \csc \pi x for 0 < x < 1, > >but I haven't worked out the details. (It gives you a specific rational >function of x times sine of pi x to integrate, which shouldn't be hard.) >Probably this can be found among well-known formulas for the Beta >function, but I don't have such at hand right now... Yes. Maple succeeds with n is a given integer, but I think you want something else, e.g. \int_0^\infty. I suppose I could carry out the calculations myself to prove _this_ integral is 2^n for any given positive integer n; maybe I could even do it for all positive integer n at once; I doubt I could do it for all n, but it's probablly still true. Here's n=6 and n=2\pi : |\^/| Maple V Release 4 (Northern Illinois University) ._|\| |/|_. Copyright (c) 1981-1996 by Waterloo Maple Inc. All rights \ MAPLE / reserved. Maple and Maple V are registered trademarks of <____ ____> Waterloo Maple Inc. | Type ? for help. > G:=GAMMA(n+1)/GAMMA(n-x+1)/GAMMA(x+1); GAMMA(n + 1) G := ----------------------------- GAMMA(n - x + 1) GAMMA(x + 1) > H:=simplify(subs(n=6,G)); sin(Pi (x + 1)) H := -720 ---------------------------------------------------------- x (-1 + x) (-2 + x) (-3 + x) (-4 + x) (-5 + x) (-6 + x) Pi > antider:=int(H,x); antider := -720 (- 1/720 Si(Pi x) - 1/120 Si(-5 Pi + Pi x) - 1/36 Si(-3 Pi + Pi x) - 1/48 Si(-2 Pi + Pi x) - 1/120 Si(-Pi + Pi x) - 1/720 Si(-6 Pi + Pi x) - 1/48 Si(-4 Pi + Pi x))/Pi > defint:=evalf(int(H,x=0..6)); defint := 63.39674448 > defint:=evalf(int(H,x=-1..7)); defint := 64.01801320 > defint:=evalf(int(H,x=-100..106)); defint := 63.99999998 > evalf(int(subs(n=6,G),x=-infinity..infinity)); 64. > evalf(int(subs(n=2*Pi,G),x=-100..100)); 77.88023365 > evalf(2^(2*Pi)); 77.88023369 ============================================================================== From: Dale Smith Subject: Re: generalization of binomial coefficient Date: Sun, 09 May 1999 00:46:32 +0000 Newsgroups: [missing] To: Dave Rusin [deletia --djr] If the gamma function thing won't work, what about the symbol (a)_n = a(a+1)(a+2)...(a+n-1), n<> 0, = 1 if n = 0 Maybe writing things like this will help - it gives another parameter to play with, I dunno if it will help or not. The two references for q-hypergeometric functions, etc are Richard Askey and James Wilson, "Some basic hypergeometric orthogonals that generalize Jacobi polynomials", Memoirs of the AMS, volume 54, number 319 Richard Askey and Mourad Ismail, "Recurrence relations, continued fractions, and orthogonal polynomials", Memoirs of the AMS, volume 49, number 300 There are more recent refs (papers) but I really can't supply a full list. I suggest a search through Math Reviews. I would provide such a list as I am trying to get back into the area of research, even though I've been working in industry but I don't have access to the online search for MR. Dave Rusin wrote: > > You wrote > >Write the binomial coefficient as a ratio of gamma functions. Wherever > >the gamma functions are defined, the ratio can be extended to the > >complex plane. > > I had argued the same thing to the original poster. His concern was > that this wouldn't apply in the cases he was interested in. In response > to my (=your) suggestion, he inserted the last long sentence in this > paragraph: > >My question is: what if i is not a natural number? > >If n is natural, n choose i = n choose n-i > >implies n choose i is 0 for i a negative integer. > >You can also use the Gamma function to get finite/infinity = 0 > >(but that's no good when n is a negative integer, > >because it gets you infinity/infinity -- hard to interpret). > > So I don't know what exactly the original poster wants, but somehow it > seems our suggestion isn't what he wants. Thus I'm inclined not to approve > your post. On the other hand, you added > >You can also write down the q-binomial coefficients for 0 < q < 1, but > >that's another story. > > Perhaps you can expand this thought? [deletia --djr] ERROR: REALITY.SYS Corrupted! Reboot universe? (Y/n)