Bernstein's Polynomials for f(x)=x and f(x)=x^2: Sequence and Formula

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

a) f(x)=x and

b) f(x)=x^2

Answers (from my textbook):

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...
 
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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.
 
HallsofIvy said:
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.

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?
 
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
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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