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The value of binomial(-1,-1) is?

asked 10 years ago

Peter Luschny gravatar image

Maple says 1, Mathematica says 1, Sage says 0 (at least in the version installed on SMC).

Peter

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The doc says "If `m < 0`, return `0`."

FrédéricC gravatar imageFrédéricC ( 10 years ago )

Yes. binomial(n,k) is often defined only for n>=0, k>=0, and otherwise set to 0. The result is in accordance with this definition. But is this restriction adequate for a system like Sage? I do not think so. In this sense I feel it is a bug.

Peter Luschny gravatar imagePeter Luschny ( 10 years ago )

An important example is binomial(n-1, k-1), an expression which arises in many combinatorial formulas. The Sage-definition prevents the right extension of all these formulas to (n,k) = (0,0). I could point at such cases in the OEIS where in consequence Maple and Mathematica compute different results compared to Sage, a very unsatisfactory situation.

Peter Luschny gravatar imagePeter Luschny ( 10 years ago )

Magma says 0.

Peter Luschny gravatar imagePeter Luschny ( 10 years ago )

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answered 10 years ago

I think that Mathematica says that binomial(-1, 0) is 1, as is binomial(-1, -1). (They may be taking their extension to negative entries from http://arxiv.org/pdf/1105.3689.pdf?) So if I'm right, then the identity

binomial(n,k) + binomial(n, k+1) = binomial(n+1, k+1)

fails when n=-1, k=-1. Sage seems to use a different convention so that this identity is preserved, or at least that's my guess as to what's going on:

sage: binomial(-1, -1)
0
sage: binomial(-1, 0)
1
sage: binomial(-1, -1) + binomial(-1, 0) == binomial(0, 0)
True

See http://en.wikipedia.org/wiki/Pascal%2....

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Set binomial(n,r) = limit(gamma(n+t+1)/(gamma(r+1)*gamma(n+t-r+1)), t=0). By symmetry binomial(-1,-1)= binomial(-1,0). Substituting n=-1, r=0 we get limit(gamma(t)/(gamma(1)*gamma(t)), t = 0) which is 1.

Peter Luschny gravatar imagePeter Luschny ( 10 years ago )

Yes, but my point was exactly that there are several choices about how to define binomial coefficients sensibly for negative numbers. This limit involving Gamma functions is one of them, but not the only one.

John Palmieri gravatar imageJohn Palmieri ( 10 years ago )

Whatever the convention is, a convention which violates the symmetry relation binomial (n, k) = binomial (n, n - k) (here binomial(-1,-1) = binomial(-1,0)) is not acceptable.

Peter Luschny gravatar imagePeter Luschny ( 10 years ago )

I would say the same about the relation binomial(n,k)+binomial(n,k+1)=binomial(n+1,k+1).

John Palmieri gravatar imageJohn Palmieri ( 10 years ago )

OK, at this point we have to stop arguing and start afresh giving first a definition of binomial(n,k) for negative integers n, k. After reading Kronenburg (thanks for the reference!) I for my part take his theorem 2.1 as the definition. And I am very happy that than binomial(-1,-1) turns out to be 1. (As a side note: Somehow the discussion is similar to what the value of 0^0 is. My experience shows me that things work better out when binomial(-1,-1) and 0^0 have the same value. And 0^0 is 1, isn't it?)

Peter Luschny gravatar imagePeter Luschny ( 10 years ago )

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Asked: 10 years ago

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Last updated: Oct 08 '14