three binomial(n,k) definitions
Henry Gould
gould at math.wvu.edu
Fri Dec 13 20:45:01 CET 2002
Yes, there has never been agreement aboiuyt what to do with the
binomial coefficient C(n,k) when k is negative.
Please see my paper
Note on a Paper of Klamkin concerning Stirling numbers, Amer. Math.
Monthly, 68(1961), 477-479.
Reviews in: MR 23(1962), #A1548; Zentralblatt 146(1968), pp.50-51; Ref
Zhur. 1962, #5A163.
I showed there that Klamkin (1957)made a mistake of this sort when he
used manipulations
with binomial identities and thought to have found a nice way to express
a certain series
using Stirling numbers of the f i r s t kind, but when I examined his
work I found that
nothing new could be gotten and the work was a mere tautology, due to
the ambiguity of
C(-n,-k) when n and k are positive.
Moreover, if you examine John Riordan's famous book "An Introduction to
Combinatorial
Analysis," published by Wiley, N.Y., 1958, in particular page 5, he
chose one way to give a
meaning for C(-n,-k) and yet it is easily and equally argueable to have
the negative of this.
My conclusion has been that since the only number X having the property
that X = -X is
zero, then a truly suitable and satisfying continuation or
generalization is not really
possible. I have never seen any definition that breaches the gap.
This subject is dear to my heart, and I would enjoy hearing from anyone
about it.
Sincerely,
Henry W. Gould
Michael Somos wrote:
> seqfan,
> I was afraid it would come to this. I just came
> to realize that the three natural variations of the
> binomial table are actually in use without any real
> acknowledgment. Here is the three binomial tables :
>
> binomial(n,k) according to Maple
>
> n\k -4 -3 -2 -1 0 1 2 3
> +--------------------------------------
> -4 | 1 0 0 0 1 -4 10 -20
> -3 | -3 1 0 0 1 -3 6 -10
> -2 | 3 -2 1 0 1 -2 3 -4
> -1 | -1 1 -1 1 1 -1 1 -1
> 0 | 0 0 0 0 1 0 0 0
> 1 | 0 0 0 0 1 1 0 0
> 2 | 0 0 0 0 1 2 1 0
> 3 | 0 0 0 0 1 3 3 1
>
> binomial(n,k) according to PARI and Mathematica
>
> n\k -4 -3 -2 -1 0 1 2 3
> +--------------------------------------
> -4 | 0 0 0 0 1 -4 10 -20
> -3 | 0 0 0 0 1 -3 6 -10
> -2 | 0 0 0 0 1 -2 3 -4
> -1 | 0 0 0 0 1 -1 1 -1
> 0 | 0 0 0 0 1 0 0 0
> 1 | 0 0 0 0 1 1 0 0
> 2 | 0 0 0 0 1 2 1 0
> 3 | 0 0 0 0 1 3 3 1
>
> binomial(n,k) = if(k<0|k>n,0,n!/(k!*(n-k)!))
>
> n\k -4 -3 -2 -1 0 1 2 3
> +--------------------------------------
> -4 | 0 0 0 0 0 0 0 0
> -3 | 0 0 0 0 0 0 0 0
> -2 | 0 0 0 0 0 0 0 0
> -1 | 0 0 0 0 0 0 0 0
> 0 | 0 0 0 0 1 0 0 0
> 1 | 0 0 0 0 1 1 0 0
> 2 | 0 0 0 0 1 2 1 0
> 3 | 0 0 0 0 1 3 3 1
>
> Notice that for n and k nonnegative, they all agree as
> expected. They first two agree when n<0 and k>=0. Now I
> think it is obvious that these are all natural and valid
> sequence tables. Also obvious that they do not agree for
> all integer n and k. Now which should deserve the name
> of "binomial coefficients"? More to the point, when the
> binomial(n,k) is used in a formula which involves n<0 or
> k<0, how are we to guess which of the three interpretations
> is to apply? This also relates to my earlier attempt to
> raise the issue of two-way infinite and one-way infinite
> sequences. Shalom, Michael
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