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Binomial coefficient

  1. Jul 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate
    [tex]{-3 \choose 0}[/tex], [tex]{-3 \choose 1}[/tex], [tex]{-3 \choose 2}[/tex]

    2. Relevant equations
    In case of integer ##n## and ##k##
    [tex]{ n \choose k}=\frac{n!}{k!(n-k)!}=\frac{n(n-1)(n-2)...(n-k+1)}{k!} [/tex]

    3. The attempt at a solution
    I am not sure how to calculate this. Any idea?
     
  2. jcsd
  3. Jul 8, 2015 #2

    haruspex

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    I've never worked with such things, but I believe binomial coefficients can be extended to noninteger and negative arguments by using the gamma function instead of factorials. Google it.
     
  4. Jul 8, 2015 #3
    Yes but gamma function diverge for integer negative values. So ##\Gamma(-3)=(-4)!=\infty##
     
  5. Jul 8, 2015 #4

    haruspex

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    Can you avoid that by some cancellation? You will have negative arguments in gamma functions both in the numerator and the denominator.
     
  6. Jul 8, 2015 #5

    hilbert2

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    The limit ##\lim_{x\to n}\frac{\Gamma(x+1)} {\Gamma(k+1)\Gamma(x-k+1)}## can exist even if ##n## is negative. Try it with Wolfram Alpha.
     
  7. Jul 8, 2015 #6
    Not sure. For example ##{-3 \choose 0}=\frac{(-3)!}{0!(-3-0)!}=1##. This is OK. But in case ##{-3 \choose 1}=\frac{(-3)!}{(-4)!}=\frac{\Gamma(-2)}{\Gamma(-3)}=\frac{-3\Gamma(-3)}{\Gamma(-3)}=-3##
    I think this is OK. Tnx :)
     
  8. Jul 16, 2015 #7

    Ray Vickson

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    The standard _definition_ of ##{n \choose k}## for non-negative integer ##k## and general real ##n## is given by the second formula you wrote in Post #1 (i.e., the formula that does not involve ##n!## or ##(n-k)!##). Sometimes that formula can be expressed in terms of Gamma functions, and sometimes not.

    Using that definition we obtain
    [tex] {-m \choose k} = (-1)^k {m+k-1 \choose k} [/tex]
    for integers ##m,k \geq 0##.
     
    Last edited: Jul 18, 2015
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