1. ## sequence of polynomials

let $
\left\{ {p_n } \right\}

$
be a sequence of polynomials of real variable defined by

$
p_1 = 0

$

$
p_{n + 1} = \frac{1}{2}(x + 2p_n (x) - \left( {p_n (x)} \right)^2 )

$

show that for all $
x \in \left[ {0,1} \right]

$

$
0 \le \sqrt x - p_n (x) \le \frac{{2\sqrt x }}{{2 + n\sqrt x }}

$

and then show that $
\left\{ {p_n } \right\}
$
converges uniformly to $
f(x) = \sqrt x

$

i need help, i tried to do the first part by induction but i dont seem to get to something useful

2. Originally Posted by mms
let $
\left\{ {p_n } \right\}

$
be a sequence of polynomials of real variable defined by

$
p_{n + 1} = \frac{1}{2}(x + 2p_n (x) - \left( {p_n (x)} \right)^2 )

$

show that for all $
x \in \left[ {0,1} \right]

$

$
0 \le \sqrt x - p_n (x) \le \frac{{2\sqrt x }}{{2 + n\sqrt x }}

$

and then show that $
\left\{ {p_n } \right\}
$
converges uniformly to $
f(x) = \sqrt x

$

i need help, i tried to do the first part by induction but i dont seem to get to something useful
What's wrong with induction? I'll let you handle the LHS of the inequality.

Have you tried showing that the sequence of polynomials is decreasing, that it converges and that it's limit is the suggested RHS?

3. im sorry but what do you mean by LHS and RHS??

4. The 'attack procedure' used in...

http://www.mathhelpforum.com/math-he...nvergence.html

... works succesfully also in this case...

The equation generating the sequence of functions can be written as...

$\Delta_{n} (x) = p_{n+1} (x) - p_{n} (x) = \frac{x - p_{n}^{2} (x)}{2} = \varphi \{p_{n} (x)\}$ (1)

The $\varphi (\xi)$ is represented here...

The only 'attractive fixed point' is the positive zero of the equation $\varphi (\xi) =0$ that is...

$\xi_{+} = \sqrt{x}$ (2)

... so that if $p_{n} (x)$ converges, it does converge to $f(x)= \sqrt {x}$. The condition for convergence is $|\varphi (\xi)| \le |r_{2, +}|$, where $r_{2,+}$ is the 'red line' crossing the $\xi$ axis in $\xi=\xi_{+}$ with slope equal to $-2$ and in this case is...

$|x-\xi^{2}|\le 4\cdot |\sqrt{x} - \xi|$ (3)

... which is satisfied for...

$-\sqrt{x} < p_{0} (x) < 4 - \sqrt{x}$ (4)

Kind regards

$\chi$ $\sigma$

5. Originally Posted by Drexel28
What's wrong with induction? I'll let you handle the LHS of the inequality.

Have you tried showing that the sequence of polynomials is decreasing, that it converges and that it's limit is the suggested RHS?

hmm ive been trying to show that the sequence is decreasing but i cant