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: $$ a = \frac{\sum_{j=i}^n j \binom{n}{j} (1-y)^{n-j} y^j} {\sum_{j=i}^n \binom{n}{j} (1-y)^{n-j} y^j} \\ = \frac{\sum_{j=i}^n j \binom{n}{j} (1-y)^{n-j} y^j} {B_{i, j}(y)} \\ $$

Any advice on how to simplify this further? Thanks!

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!

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.

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: $$ a = \frac{\sum_{j=i}^n j \binom{n}{j} (1-y)^{n-j} y^j} {\sum_{j=i}^n \binom{n}{j} (1-y)^{n-j} y^j} \\ = \frac{\sum_{j=i}^n j \binom{n}{j} (1-y)^{n-j} y^j} {B_{i, j}(y)} \\ $$

Any advice on how to simplify this further? Thanks!