>>>>> "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)
I think you have not really understood from help(choose)
that we have
choose(n,k) := n(n-1)...(n-k+1) / k!
defined for non-negative integer k and arbitrary real n
This definition has the very important property that it is used
in the 'binomial theorem' and everything depending on that
theorem,
(1 + x) ^ \alpha = \sum_{k=0}^\infty choose(x,k) x^k
which holds for all |x| < 1 for all \alpha \in \R
-------
Your "fundamental" property does only hold when n is
non-negative integer.
Definitely no bug in R here!
Martin Maechler, ETH Zurich
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