explicit variable from equation

[terminal]

hi all!!

I'm facing this problem and I can't get a solution, I want to explicit variable m from the following equation, I tried with binomial theorem to break up binomial coefficient but it was useless...
can someone help me please?

\sum_{i=n}^{n+m}\binom{n+m} {i}p^i(1-p)^{n+m-i} = k

all variables are known except m, and k is a known constant.
virtually I'd like something in the form:
m = g(n,p,k)

thanks!!

[HallsofIvy]

Look at a couple of simple examples. Suppose m= n= 1. Then that says
\sum_{i=1}^{2}\binom{2}{i}p^i(1-p)^{2-i} = k
which is
\binom{2}{1}p(1- p)+ \binom{2}{2}p^2= 2p- 2p^2+ p^2= 2p- p^2= k

In general, that is an m+n degree polynomial in p and you are not going to find any explicit formula for the solution of such a polynomial.

[terminal]

Look at a couple of simple examples. Suppose m= n= 1. Then that says
\sum_{i=1}^{2}\binom{2}{i}p^i(1-p)^{2-i} = k
which is
\binom{2}{1}p(1- p)+ \binom{2}{2}p^2= 2p- 2p^2+ p^2= 2p- p^2= k

In general, that is an m+n degree polynomial in p and you are not going to find any explicit formula for the solution of such a polynomial.

mm let me understand better...you are talking about a polynomial in p, but I told that p is known! my incognita is m... did I miss something?

[HallsofIvy]

Sorry, for some reason I thought you had said "p". But because the m occurs both in the binomial coefficient and in the exponent, I would say that there is much less likely to be an explicit formula for m.

[terminal]

Sorry, for some reason I thought you had said "p". But because the m occurs both in the binomial coefficient and in the exponent, I would say that there is much less likely to be an explicit formula for m.

mm ok! :)
do you see another way to calculate m (or an approximated value)??
It's ok for me to get the minimum value of m that satisfies:

g(n,m,p)>=k