On Nov 10, 10:56 am, "Edward Jensen" <edw...@jensen.invalid> wrote:
> Hi.
> I'm working on a problem where I got some coefficients of the form
> (p1 + q1) * (p2 + q2) * ... * ... (pn * qn)
> where pi and qi are different for i = 1 to n
conditions:
pi=/=qi for each i, but for i=/=j, you could have any of pi=pj, qi=qj,
or pi=qj?
pi=/=qi and for i=/=j, we also have pi=/=pj and qi=/=qj, but it is
possible to have pi=qj?
The pi and qj are pairwise distinct?
Which?
> I want to write them up in a "nice" form term by term. In the case where the
> pi's and qi's are equal, we got the binomial thereom. Is there any clean way
> to write this?
i and j, and qi=qj for all i and j", so that adds many possibilities
to the meaning of the phrase you wrote above. It now seems you could
also mean:
If i=/=j, then pi=/=pj and qi=/=qj; however, we may have pi=pj, qi=qj,
pi=qi, or pi=qj ?
and many other combinations. You are going to have to be more
precise.
> I suspect that we generelly got n+1 terms.
For pairwise distinct, with two terms, you get
(p+q)(r+s) = pr + ps + qr + qs
with is 4 terms, not 3.
> I suspect the following pattern:
> for i = 1 and i = n we got the product of all pi's and qi's respectively.
> for 1 < j < n we got Choose(n, j) sums of all Choose(n,j) combinations of
> pi's multiplied by all Choose(n,j-1) combinations of qk's such that all i !=
> k in each term.
That wouldn't be n+1 terms, because each of those "terms" is a sum of
terms itself.
> Is there a clean way to write this term by term as a sum from 1 to n?
you get 2^n binomials in general. The binomial theorem lets you
combine two of the four because the choice (p,q) [p from the first
binomial, q from the second] yields a term that is equal to (q,p) [q
from the first binomial, p from the second].
So rather than combinations, you are better off with tuples: you are
dealing with all functions from {1,...,n} to {1,2}.
--
Arturo Magidin
