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Re: [Axiom-developer] Issue 336

From: Martin Rubey
Subject: Re: [Axiom-developer] Issue 336
Date: 17 Mar 2007 20:16:02 +0100
User-agent: Gnus/5.09 (Gnus v5.9.0) Emacs/21.4

Der Waldek,

Waldek Hebisch <address@hidden> writes:

> Martin Rubey wrote:
> > If the second argument k is negative, we could return zero. However, I 
> > vaguely
> > remember that the definition in terms of Gamma functions does not yield 
> > this as
> > a limit, at least not for any n. Do you happen to remember?
> > 
> > In fact, if n=k, the definition in terms of Gamma functions always gives 
> > one,
> > no matter whether n is negative of not...

> I would tend to say that binomial(n, k) with k beeing a negative integer is
> an error.  Definition in term of Gamma functions has discontinuity there: if
> n is non-integer and k tends to negative integer then we get 0 as a limit.
> Now, if we keep k fixed and let n go to any value we get 0.  But if we allow
> n and k to change simultaneously we may get nonzero limit (so limit as a
> function of two variables does not exist).  We have discontinuity only for
> integer n and k such that k <= n < 0 (otherwise limit is 0).

I should know, but I don't: is there a sensible definition of the binomial
coefficient which disagrees with the definition in terms of Gamma functions at
certain values?

(I don't think so...)

If the definition in terms of the Gamma function is "canonical", I'd suggest to
simplify only when the limit is unique. (Contrary to Mathematica) So, as you
and Mathworld say

* for 0 <= k \in Z we return the product
* if we can prove that n<k, we return 0

I wonder about the case n=k. Why isn't that equal to 1? We obtain

binomial(m,m)=limit(k->m, n->m) Gamma(n+1)/(Gamma(k+1)Gamma(n-k+1))

isn't that limit equal to one?

In ibinom (that is, for FunctionSpaces over arbitrary R we currently have


If the above limit does not exist, they are all wrong, I guess...

BTW, does binomial(n,k)=binomial(n,n-k) hold always?


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