Subject: simplification of sums with binomial coefficients
From: Andrej Vodopivec
Date: Sun, 9 Apr 2006 12:27:04 -0700
Hi,
the default routines cant handle this case. Here's how you do it with
Zeilberger. You have
sm[k] = sum(r/k*binom(n,r)*binom(m,k-r)/binom(n+m,k),r,-inf,inf)
so you can find the recurrence for sm[k]:
(%i1) load(zeilberger)$
(%i2) load(solve_rec)$
(%i3) r/k*binom(n,r)*binom(m,k-r)/binom(n+m,k);
(%o3) (binom(m,k-r)*binom(n,r)*r)/(k*binom(n+m,k))
(%i4) summand_to_rec(%, r, k);
(%o4) sm[k+1]-sm[k]
which tells you that the sum is constant. Now
(%i5) sum(''%o3, r, 0, k), k=1;
(%o5) n/(n+m)
tells you that it is equal to n/(n+m).
Note that summand_to_rec is a wrapper for Zeilberger, which gives the
recurrence and some more data:
(%i6) Zeilberger(%o3, r, k);
(%o6) [[((r-1)*(r+m-k))/((n+m-k)*(r-k-1)),[-1,1]]]
If you get a more complicated recurrence you can try solving it with solve_rec.
Andrej
On 4/9/06, az <zoav1 at uic.nnov.ru> wrote:
> Hello,
> How do I make Maxima simplify
> sum(r/k*binom(n,r)*binom(m,k-r)/binom(n+m,k),r,0,k) ?
>
> simpsum doesn't seem to help. Maybe somebody can give an example of the
> Zeilberger
> share package usage in this situation?
>
> --
> Andrei Zorine
>
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