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I've been trying to find analytical solutions for the zeros of the following expression Zeros in terms of variables n, m and f (here I represented two forms of the expression R, I've worked with both but I couldn't find a general solution, the code is intended to be used on Mathematica):

R[x_, n_, m_] := 
  x^(m - 1)*(Sum[
      Binomial[n - 1, k]*x^(k - m + 1)*(1 - x)^(n - 1 - k), {k, m, 
       n - 1}] + 
     n*(Sum[Binomial[n - 1, k]*x^(k - m + 1)*(1 - x)^(n - 1 - k), {k, 
         0, m - 1}]) + m*Binomial[n - 1, m - 1]*(1 - x)^(n - m))= 
1 - (1 - N)* (Sum[
    Binomial[n - 1, k]*x^(k)*(1 - x)^(n - 1 - k), {k, 0, 
     m - 1}]) + m*Binomial[n - 1, m - 1]*x^(m - 1)*(1 - x)^(n - m);

Zeros[x_, n_, m_, f_] := R[x, n, m] - n/f;

I also tried solving the equation by using Reduce[R[x, n, m] - n/F == 0, x] but he couldn't evaluate it and get to a conclusion. I'd really appreciate some suggestions. Thanks for your time.

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  • $\begingroup$ Your two expressions for $R$ don't give the same answer. For example, R[x,5,2] gives $-20 x^4+56 x^3-48 x^2+8 x+5$ using the first definition and $-8 x^4+12 x^3+8 x^2-16 x+\frac{4}{x}+1$ using the second. $\endgroup$ – Misha Lavrov Apr 12 '17 at 4:02
  • $\begingroup$ Sorry i wrote the second expression wrong. I've corrected it now. Thanks for alerting me. $\endgroup$ – Matt Apr 13 '17 at 0:19

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