I am an electrical engineer and I am interested in maximizing an objective function that contains a binomial coefficient $\binom{n}{k} = \frac{{n!}}{{\left( {n - k} \right)!k!}}$.
This is a follow up question from a previous question.
The problem can be stated as:
$\begin{array}{*{20}{c}} {\mathop {{\rm{Max}}}\limits_{} }&{m \times \sum\limits_{k = 2}^{k = n} {\left[ {\left( {1 - {{\left( {1 - kq{p^{n - 1}}} \right)}^{m - 1}}} \right)\binom{n}{k}{q^k}{p^{n - k}}} \right]} }\\ {{\rm{s.t:}}}&{p = 1 - q}\\ {}&{0 \le p \le 1}\\ {}&{m,n \in Z} \end{array}$
From this post, it seems that the binomial coefficient can be thought of as some kind of gamma function but this gamma function is so strange and I am not sure if its derivative even exists:
$\binom{n}{k}:=\frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}$
Case 1: $n$ is a known parameter, $m$ and $p$ are decision variables.
Case 2: $m$ is a known parameter, $n$ and $p$ are decision variables. I think case 2 is very difficult, it looks like some sort of online optimization problem.
Case 3: Both $m$ and $n$ are known parameters, $p$ or $q$ is a decision variable. I think this case might be easier.
Case 4: Both $m$ and $p$ are known parameters, $n$ is a decision variable.
Effort so far for case 3: this case is quite complicated than it looks, I have set $m=n$ for simplicity and optimize the variable $q$. Since this problem has only one variable $q$, I have taken the derivative of the objective function and recive the following
$f'\left( q \right) = n\sum\limits_{k = 2}^n {\left( {{q^k}\left( {\begin{array}{*{20}{c}} n\\ k \end{array}} \right)\left( { - k{n^2}q{{(1 - q)}^{n - 2}}{{\left( {1 - kq{{(1 - q)}^{n - 1}}} \right)}^{n - 2}} - kq{{(1 - q)}^{n - 2}}{{\left( {1 - kq{{(1 - q)}^{n - 1}}} \right)}^{n - 2}} - k{{(1 - q)}^{n - 1}}{{\left( {1 - kq{{(1 - q)}^{n - 1}}} \right)}^{n - 2}} + kn{{(1 - q)}^{n - 1}}{{\left( {1 - kq{{(1 - q)}^{n - 1}}} \right)}^{n - 2}} + 2knq{{(1 - q)}^{n - 2}}{{\left( {1 - kq{{(1 - q)}^{n - 1}}} \right)}^{n - 2}}} \right){{(1 - q)}^{n - k}} + k{q^{k - 1}}\left( {\begin{array}{*{20}{c}} n\\ k \end{array}} \right)\left( {1 - {{\left( {1 - kq{{(1 - q)}^{n - 1}}} \right)}^{n - 1}}} \right){{(1 - q)}^{n - k}} + {q^k}\left( {\begin{array}{*{20}{c}} n\\ k \end{array}} \right)\left( {1 - {{\left( {1 - kq{{(1 - q)}^{n - 1}}} \right)}^{n - 1}}} \right)\left( {k{{(1 - q)}^{ - k + n - 1}} - n{{(1 - q)}^{ - k + n - 1}}} \right)} \right)}$
Now, I cannot solve $f'\left( q \right) = 0$
Would you kindly help me with this ?
Thank you for your enthusiasm !