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](https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_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!



How to solve for $a$ in $\sum_{j=i}^n (a -j) \binom{n}{j} y^j (1-y)^{n-j} = 0$
$$\sum\_{j=i}^n (a -j) \binom{n}{j} y^j (1-y)^{n-j}
=0$$
$$\Rightarrow a=i-\frac{y \binom{n}{i+1} \, \_2F\_1\left(2,i-n+1;i+2;\frac{y}{y-1}\right)}{(y-1) \binom{n}{i} \, \_2F\_1\left(1,i-n;i+1;\frac{y}{y-1}\right)}.$$
Check that for $i=1$, this simplifies to $a=n y \left(\frac{1}{(1-y)^{-n}-1}+1\right)$, which is the correct solution.


