According to various sources e.g.


and Wolfram themselves

http://functions.wolfram.com/GammaBetaErf/Binomial/02/ , the binomial coefficient ${n\choose k}$ is is defined as 0 whenever $k$ and $n$ are negative integers and $k\le n$. But when I type


Mathematica returns


I looked up the documentation for the definition of Binomial and it says

In general, ${n\choose m}$ is defined by $\Gamma(n+1)/\Gamma(m+1)\Gamma(n-m+1)$ or suitable limits of this.

Apparently, when $n=-1$ or $m=-1$ since $\Gamma(0)$ is not defined the suitable limit case is applied.

So, why does Mathematica return 1 for ${-1\choose -1}$? What precisely is the "suitable limits"?


Looked up the rules on this. This is how it works. If $n$ and $r$ are negative integers, there is a symmetry relation $\binom{n}{r}=\binom{n}{n-r}$ and now the limit is used.

Mathematica graphics

Mathematica graphics

But now $\binom{n}{r}=\binom{-1}{0}$ from above. Hence the above limit is, where $n=-1$ and $r=0$ is

    n = -1; r = 0;
    Limit[ Gamma[n + t + 1]/(Gamma[r + 1] Gamma[n + t - r + 1]), t -> 0]
    (* 1 *)

Maple also agrees with Mathematica

Mathematica graphics

Reference: Maple help on Binomial

| improve this answer | |
  • $\begingroup$ Thanks. I've never used Maple. Do you mean Mathematica uses the same definition as Maple? $\endgroup$ – Y. Pei Apr 10 '14 at 20:51
  • $\begingroup$ Also, the documentation of Maple is not clear either. It says "In the case that n is a negative integer, binomial(n,r) is defined by this limit." Note here's a full stop at the end of the sentence. So binomial(-1,-1) should be using this limit $\lim_{t\to0}{\Gamma(-1+t+1)\over \Gamma(-1+1)\Gamma(-1+t-(-1)+1)}$, which is ill-defined because of the $\Gamma(0)$ in the denominator. Even if when k is a negative integer we use the symmetry property, then the definition does not cover binomial(-3,-2) properly because then binomial(-3,-2) turns into binomial(-3,-1), which is not well-defined. $\endgroup$ – Y. Pei Apr 10 '14 at 21:02
  • $\begingroup$ Result from PARI (version 2.7.2): binomial(-1,-1) = 0 $\endgroup$ – Vaclav Kotesovec Jun 13 '15 at 21:25

Without expanding Binomial into Gamma functions, you can also see that the result is correct based on the following true statement:

SeriesCoefficient[(1 + x)^n, {x, 0, k}, Assumptions -> k >= 0]

(* ==> Binomial[n, k] *)

This is the binomial expansion, valid in particular for n = -1. But that case leads to the alternative expression

SeriesCoefficient[(1 + x)^-1, {x, 0, k}, 
 Assumptions -> k >= 0]

(* ==> (-1)^k *)

which by comparison with the previous line implies that the following is the correct result:

Binomial[-1, 0]

(* ==> 1 *)

Now we have the symmetry relation

FullSimplify[Binomial[n, k] == Binomial[n, n - k]]

(* ==> True *)

Using this with n = -1 and k = -1 then confirms what Mathematica says:

Binomial[-1, 0] == Binomial[-1, -1]

(* ==> True *)

Combined with the previous equation the result Binomial[-1, -1] == 1 follows.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.