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We defined the Bernstein polynomials as following $$ p_{nk} \ = \ \frac{n!}{k!(n-k)!} x^k(1-x)^{n-k}$$ I have to show this: $$ \sum_{k=0}^n \left(x - \frac kn\right)^2 p_{nk}(x) \ = \ \frac{x(1-x)}{n} $$

My own work

It is easy to show that $$ (1-x+tx)^n \ = \ \sum_{k=0}^n p_{nk}(x)t^k $$ From this I have deduced that $$ \sum_{k=0}^n p_{nk}(x) = 1 \qquad \text{and} \qquad \sum_{k=0}^n kp_{nk}(x) = nx \qquad \text{and} \qquad \sum_{k=0}^n k(k-1)p_{nk}(x) = n(n-1)x^2 $$

Now by dividing and plugging in I got that:

$$ \frac{1}{n}(x-x^2) \ = \ \frac{1}{n^2(n-1)} \left((n-1)\sum_{k=0}^n kp_{nk}(x) + \sum_{k=0}^n k(k-1)p_{nk}(x)\right) $$ Which equals $$ \frac{1}{n^2(n-1)} \sum_{k=0}^n k p_{nk}(x)\left( (n-1) - (k-1)\right) \ = \ \frac{1}{n^2(n-1)} \sum_{k=0}^n k p_{nk}(x)(n-k) $$ The left hand side is equal to this though: $$ \sum_{k=0}^n \left(x^2 - \frac {2k}n+ \frac{k^2}{n^2}\right) p_{nk}(x) $$ And these things don't look equal. Please help me!

Martin Sleziak
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Koenraad van Duin
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1 Answers1

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It is simpler to start from $\sum_{k=0}^n\left(x-\frac kn\right)^2p_{n,k}(x)$, using the fact that $\sum_{k=0}^nk^2p_{n,k}(x)=n(n-1)x^2+nx$. This gives \begin{align} \sum_{k=0}^n\left(x-\frac kn\right)^2p_{n,k}(x)&=x^2-\frac{2x}n\sum_{k=0}^nkp_{n,k}(x)+\frac 1{n^2}(n(n-1)x^2+nx)\\ &=x^2-\frac {2x}n\cdot nx+\frac 1{n^2}(n(n-1)x^2+nx)\\ &=x^2-2x^2+\frac{(n-1)x^2+x}n \\ &=-x^2+x^2+\frac{x-x^2}n\\ &=\frac{x(1-x)}n. \end{align}

Davide Giraudo
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