[texhax] \binom
John Wheeler
jwheeler at ucsd.edu
Mon Mar 2 06:59:16 CET 2009
Actually, there's not a problem here, since 1/Gamma(0) = 0 and
1/Gamma(n) = 0 for any negative
integer n. As a consequence, \binom{3}{4} = 3!/(4!(-1)!) =
Gamma(4)/[Gamma(5)Gamma(0)] = 0, which is the correct answer for the
coefficient of x^4y^{-1} in (x+y)^3.
Alternatively, the binomial coefficient represented by \binom{n}{m} is
defined as zero whenever
the integer m is greater than the integer n.
Admittedly, the example \binom{4}{3} would have been less likely to lead
to confusion.
John Wheeler
Rolf Turner wrote:
> On 2/03/2009, at 10:47 AM, P. R. Stanley wrote:
>
>
>> An excellent answer! That's exactly the sort of info I'm after!
>> Cheers,
>> Paul
>>
>>
>>> Hi
>>>
>>>> What does it do? What do you get when you compile $\binom{3}{4}$
>>>> in a
>>>> document? Thanks
>>>> Paul
>>>>
>>> It sets a 3 above a 4 and encloses them both in a single pair of
>>> sufficiently
>>> tall parentheses. The 3 and the 4 are the same size as each other.
>>> You
>>> probably already know what it denotes, but just in case you don't,
>>> here's one
>>> way to describe it: $\binom{n}{r}$ denotes the coefficient of $x^r
>>> y^{n-r}$ in
>>> the expansion of $(x + y)^n$.
>>>
>
> There's a wee problem in that \binom{3}{4} would denote the
> coefficient of
> x^4 y^{-1} in the expansion of (x + y)^3 --- and there isn't one.
>
> As far as I can discern there is no way to make \binom{3}{4}
> meaningful, since
> Gamma(0) is undefined.
>
> Now \binom{4}{3} --- that's a different story! :-)
>
> cheers,
>
> Rolf Turner
>
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