I'd like to prove this identity:

$$\sum_{k=0}^n\left(x-\frac{k}{n}\right)^2 \binom{n}{k}x^k(1-x)^{n-k}=\frac{1}{n}x(1-x)$$ for $x\in[0,1]$ and $n\in\mathbb{N}$.

I've worked on this problem for the last few hours without success. I have tried induction and I have tried re-writing the expression as many ways as I could think of. Any help would be greatly appreciated. Thanks.

  • $\begingroup$ You can find this identity proved in Ross Elementary calculus book. I think this identity is used for proving Weirstrass Approximation theorem, via Bernstein Polynomials $\endgroup$ – crskhr Dec 8 '15 at 6:21
  • $\begingroup$ duplicate of math.stackexchange.com/q/463996 $\endgroup$ – Jean Marie Mar 14 '20 at 5:51

Brute force approach. There might be a more elegant approach.

Multiply by $n^2$, and you want:

$$\sum_{k=0}^n\left(nx-k\right)^2 \binom{n}{k}x^k(1-x)^{n-k}=nx(1-x)$$

First part: $$\sum_{k=0}^n n^2x^2\binom{n}{k}x^k(1-x)^{n-k} = n^2x^2(x+(1-x))^n = n^2x^2.\tag{1}$$

Second part: $$\begin{align} \sum_{k=0}^n 2nkx \binom{n}{k}x^ky^{n-k} &= 2nx^2\sum_{k=0}^n k\binom{n}{k}x^{k-1}y^{n-k}\\ &=2nx^2\frac{d}{dx}(x+y)^n \\ &= 2nx^2n(x+y)^{n-1}\end{align}$$

Letting $y=1-x$, then we get:

$$\sum_{k=0}^n 2nkx \binom{n}{k}x^k(1-x)^{n-k} = 2n^2x^2\tag{2}$$

Part 3:

$$\begin{align} \sum k^2\binom{n}{k}x^ky^{n-k} &= x\frac{d}{dx}\sum_{k=0}^nk\binom{n}{k}x^ky^{n-k}\\ &=\left(x\frac{d}{dx}\right)^2(x+y)^n\\ &=x\frac{d}{dx}\left(xn(x+y)^{n-1}\right)\\ &=x\left(n(x+y)^{n-1} + xn(n-1)(x+y)^{n-2}\right)\\ \end{align}$$

Letting $y=1-x$ we have:

$$\sum k^2\binom{n}{k}x^ky^{n-k}=xn + n(n-1)x^2\tag{3}$$

Computing $(1)-(2)+(3)$ we get:

$$\sum_{k=0}^n\left(nx-k\right)^2 \binom{n}{k}x^k(1-x)^k=n^2x^2-2n^2x^2+n(n-1)x^2 + nx=nx-nx^2=nx(1-x)$$

A more algebraic way might amount to noting that:

$$\sum_{k=0}^n ((n-k)x -ky)^2\binom{n}{k}x^ky^{n-k}$$

is a symmetric polynomial in $x,y$, and thus can be written in terms of $x+y,xy$. Not sure where to go from there, however.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.