# Bernstein's polynomial

1. Oct 25, 2012

### Artusartos

Find the sequence $(B_nf)$ of Bernstein's polynomials in

a) f(x)=x and

b) $f(x)=x^2$

a) $B_nf(x) = x$ for all n.

b) $B_nf(x) = x^2 + \frac{1}{n} x (1-x)$

I know that the bernstein's polynomial is:

$B_nf(x) = \sum_{k=0}^n f (\frac{k}{n}) \binom{n}{k} x^k (1-x)^{n-k}$

...but I don't know how they got the answer from this...

2. Oct 25, 2012

### HallsofIvy

Staff Emeritus
Have you used that formula to calculate, say, B0 through B5 for f(x)= x and f(x)= x2? That should give you an idea.

3. Dec 18, 2012

### Artusartos

But how can I calculate $$B_0$$? If I say n=0, then

$B_nf(x) = \sum_{k=0}^n f (\frac{k}{0}) \binom{0}{k} x^k (1-x)^{0-k}$

So f(k/0) is undefined?

4. Dec 18, 2012

### HallsofIvy

Staff Emeritus
Sorry. Clearly "B0" is not defined so calculate B1, B2, etc.

For example, with f(x)= x,
$$B_1(x)= f(0)\begin{pmatrix}1 \\ 0\end{pmatrix}x^0(1- x)^1+ f(1)\begin{pmatrix}1 \\ 1\end{pmatrix}x^1(1- x)^0= 0(1- x)+ 1x= x$$