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## Re: [Axiom-developer] Issue 336

**From**: |
Waldek Hebisch |

**Subject**: |
Re: [Axiom-developer] Issue 336 |

**Date**: |
Sat, 17 Mar 2007 19:41:39 +0100 (CET) |

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

**[Axiom-developer] Issue 336**, *Waldek Hebisch*, `2007/03/13`
**Re: [Axiom-developer] Issue 336**, *Martin Rubey*, `2007/03/13`
**Re: [Axiom-developer] Issue 336**,
*Waldek Hebisch* **<=**
**Re: [Axiom-developer] Issue 336**, *Martin Rubey*, `2007/03/17`
**Re: [Axiom-developer] Issue 336**, *Waldek Hebisch*, `2007/03/17`
**Re: [Axiom-developer] Issue 336**, *Martin Rubey*, `2007/03/17`
*Message not available***Re: [Axiom-developer] Issue 336**, *Martin Rubey*, `2007/03/17`
**Re: [Axiom-developer] Issue 336**, *Waldek Hebisch*, `2007/03/19`