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]
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