The Problem
Given $i, n, y$, I am trying to find a closed form solution for $a$ in the equation $$ \sum_{j=i}^n (a -j) \binom{n}{j} y^j (1-y)^{n-j} = 0 $$ Do you have any recommendations on how to solve this problem?
Background/Motivation
Let me explain. I am trying to compute the derivative of a Bernstein polynomial with respect to y. A Bernstein polynomial is defined as $$ B_{i:n}(y) = \sum_{j=i}^n \binom{n}{j} y^j (1-y)^{n-j} $$ and I've worked out the derivative to be $$ \frac{d}{dy} \bigg[ B_{i:n}(y) \bigg] = \sum_{j=i}^n \binom{n}{j} y^{j-1} (1-y)^{n-j} j \\ + \sum_{j=i}^n \binom{n}{j} y^j (1-y)^{n-j - 1} (n-j) $$ However, I want to simplify the derivative so it contains the expression $B_{i:n}$. Using the first sum of the derivative as an example, I want to do something like this: $$ \sum_{j=i}^n \binom{n}{j} y^{j-1} (1-y)^{n-j} j = \frac{1}{y} \sum_{j=i}^n \binom{n}{j} y^j (1-y)^{n-j} j \\ = \frac{a}{y} \sum_{j=i}^n \binom{n}{j} y^j (1-y)^{n-j} = \frac{a}{y} B_{i:n}(y) $$ for some value $a$ that allows us go get rid of the $j$ that we are multiplying each term of the summation by. However, I am unsure how to compute $a$. To compute $a$, I simply have to solve $$ a \sum_{j=i}^n \binom{n}{j} y^j (1-y)^{n-j} = \sum_{j=i}^n \binom{n}{j} y^j (1-y)^{n-j} j \\ \sum_{j=i}^n (a -j) \binom{n}{j} y^j (1-y)^{n-j} = 0 $$ Yet I do not know how to solve this problem for $a$. I know there are various numerical methods for root solvers but I am looking for a closed form solution. Any suggestions would be greatly appreciated.
One more thing: this problem can be generalized to solving: $$ \sum_{j=1}^n f(j) j = a \sum_{j=1}^n f(j) \\ \sum_{j=1}^n (a - j) f(j) = 0 $$ where $a \neq i$. Any recommendations on techniques/methods I can use to solve problems of this form? To me, this generalized problem is trying to remove the $j$ by which we multiply every term in a summation when that summation is defined over $j$. (Perhaps this problem formulation is a more well known problem than my specific problem)
This problem is especially difficult because according to the the Abel–Ruffini theorem, obtaining a closed form solution for the root of a polynomial such as the one we have is incredibly difficult for polynomials of degree five and higher and our polynomial seems to be of degree five or higher.
Update 1
Qiaochu Yuan, thanks your right. We can just write the solution as: $$ a = \frac{\sum_{j=i}^n j \binom{n}{j} (1-y)^{n-j} y^j} {\sum_{j=i}^n \binom{n}{j} (1-y)^{n-j} y^j} \\ = \frac{\sum_{j=i}^n j \binom{n}{j} (1-y)^{n-j} y^j} {B_{i, n}(y)} \\ $$
Any advice on how to simplify this further? Thanks!
