Re: [Axiom-developer] Issue 336

[Waldek Hebisch]

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...
> 
> Axiom gives 0 currently, due to the definition of ibinom. This looks all quite
> inconsistent...
>

Hmm, really inconsistent:

(4) ->  binomial(n, n)
   (4)  1
                                                     Type: Expression Integer
(5) -> binomial(-3, -3)
   (5)  0
                                                     Type: NonNegativeInteger

I would tend to say that binomial(n, k) with k beeing a negative
integeriis 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).

-- 
                              Waldek Hebisch
address@hidden