Re: (1 Combination 2) = 0 -- Better explanation?

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Sujet : Re: (1 Combination 2) = 0 -- Better explanation?
De : jbb (at) *nospam* notatt.com (Jeff Barnett)
Groupes : comp.lang.python sci.math sci.lang
Date : 16. Jul 2024, 00:20:54
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v747c8$t5ch$1@dont-email.me>
References : 1 2 3
User-Agent : Mozilla Thunderbird
On 7/15/2024 12:49 PM, HenHanna wrote:

On 7/14/2024 8:44 PM, Jeff Barnett wrote:
On 7/14/2024 2:57 PM, HenHanna wrote:

Python says:  (1 Combination 2) = 0

         Ok... It's Impossible (to do).

              ------- is there a Better explanation?



(5 Combination 0) = 1  <---- This is explained by  Comb(5,0)=Comb(5,5)

                                      in general:  
Comb(N,r)=Comb(N,N-r)

_______________________________________

from math import comb

for i in range(6):      print( comb(5,i) )

print( comb(1,2)  )



Let combination of n things taken m at a time be represented by [n,m].
Then [n,m] = [n,n-m] as you correctly note above. Further, we have the
computational formula [n,m] = n!/(m!(n-m)!) where x!  is simply x
factorial. So [1,2] = 1!/(2!((-1)!)), or 1/2 divided by (-1)!. However
factorial of a negative integer is, by convention, an infinite value
so [1.2] = 0.


THank you...


  Bard.Google.com   says that

                     Comb(1,2) is not defined

                 factorial(-1) is not defined
                 factorial(-2) is not defined

             GammaFunction(-1) is not defined
             GammaFunction(-2) is not defined


They are partially correct. However, one can say that 1/0 = oo is
defined (where oo is infinity). In particular. 1/oo = 0 certainly makes
sense and that's all we  need to accept for the above deductions.

Something I forgot to say in my original above is why x! = oo when x is
a negative integer. I cure that omission now. We want to define the
factorial as 1! = 1 and n! = n*(n-1)! when n > 1. We would also like to
be able to pedal backwards, i.e.. derive (n-1)! from n! and n. This is
certainly straightforward when n > 0. However the cases for other
integer n is trickier. For example, from our recursion formula we have
0! = 0*(-1)! and we know that (by definition) 0! =1. Thus, 1 = 0*(-1)!
which only has the possible symbolic solutions (-1)! = oo or -oo. Now a
trivial induction argument will draw the same conclusion for all
negative integers.

The importance of this bit of sophistry is our desire to do symbolic
manipulations with various classes of formulas without having to
numerically separate out a lot of special cases. Look at the above
definition of combination. With the established conventions, we have 1)
[n,n] = 1 and this can't work unless 0! = 1; 2) [n.0] = 1 which says
that the only o element subset of n elements is the empty set; etc.

It is a goal of mathematicians to make their definitions work not only
for the obvious cases but where there is darkness in our knowledge that
might be trivially illuminated by formulas that allow straightforward
consistent treatment throughout.
--
Jeff Barnett


Date Sujet#  Auteur
14 Jul 24 * (1 Combination 2) = 0 -- Better explanation?5HenHanna
15 Jul 24 +* Re: (1 Combination 2) = 0 -- Better explanation?3Jeff Barnett
15 Jul 24 i`* Re: (1 Combination 2) = 0 -- Better explanation?2HenHanna
16 Jul 24 i `- Re: (1 Combination 2) = 0 -- Better explanation?1Jeff Barnett
16 Jul 24 `- Re: (1 Combination 2) = 0 -- Better explanation?1Ben Bacarisse

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