>>>>> "MM" == Martin Maechler <[EMAIL PROTECTED]>
>>>>> on Wed, 26 Mar 2008 11:02:37 +0100 (CET) writes:
>>>>> "JL" == Jerry Lewis <[EMAIL PROTECTED]>
>>>>> on Wed, 26 Mar 2008 05:10:04 +0100 (CET) writes:
JL> Full_Name: Jerry W. Lewis Version: 2.7.0 (2008-03-23
JL> r44847) OS: Windows XP Professional Submission from:
JL> (NULL) (71.184.230.48)
JL> choose(n,k) = choose(n,n-k) is not satisfied if either
JL> 1. n is a negative integer with k a positive integer
JL> (due to automatically returning 0 for n-k<0)
JL> 2. n is not an integer with k a positive integer (due to
JL> rounding n-k to an integer, compounded by automatically
JL> returning 0 if n<0 which implies n-k<0)
MM> I think you have not really understood from help(choose)
MM> that we have
MM> choose(n,k) := n(n-1)...(n-k+1) / k!
MM> defined for non-negative integer k and arbitrary real n
MM> This definition has the very important property that it is used
MM> in the 'binomial theorem' and everything depending on that
MM> theorem,
MM> (1 + x) ^ \alpha = \sum_{k=0}^\infty choose(x,k) x^k
Oops! "Of course", I meant 'choose(\alpha,k)' , not 'choose(x,k)'
MM> which holds for all |x| < 1 for all \alpha \in \R
MM> -------
MM> Your "fundamental" property does only hold when n is
MM> non-negative integer.
MM> Definitely no bug in R here!
Martin Maechler, ETH Zurich
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