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Math Help - sequence of polynomials

  1. #1
    mms
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    sequence of polynomials

    let <br />
\left\{ {p_n } \right\}<br /> <br />
be a sequence of polynomials of real variable defined by

    <br />
p_1  = 0<br /> <br />

    <br />
p_{n + 1} = \frac{1}{2}(x + 2p_n (x) - \left( {p_n (x)} \right)^2 )<br /> <br />
    show that for all <br />
x \in \left[ {0,1} \right]<br /> <br />

    <br />
0 \le \sqrt x - p_n (x) \le \frac{{2\sqrt x }}{{2 + n\sqrt x }}<br /> <br />

    and then show that <br />
\left\{ {p_n } \right\}<br />
converges uniformly to <br />
f(x) = \sqrt x <br /> <br />

    i need help, i tried to do the first part by induction but i dont seem to get to something useful
    Last edited by mms; May 3rd 2010 at 06:14 AM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mms View Post
    let <br />
\left\{ {p_n } \right\}<br /> <br />
be a sequence of polynomials of real variable defined by

    <br />
p_{n + 1} = \frac{1}{2}(x + 2p_n (x) - \left( {p_n (x)} \right)^2 )<br /> <br />
    show that for all <br />
x \in \left[ {0,1} \right]<br /> <br />

    <br />
0 \le \sqrt x - p_n (x) \le \frac{{2\sqrt x }}{{2 + n\sqrt x }}<br /> <br />

    and then show that <br />
\left\{ {p_n } \right\}<br />
converges uniformly to <br />
f(x) = \sqrt x <br /> <br />

    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?
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  3. #3
    mms
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    im sorry but what do you mean by LHS and RHS??
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  4. #4
    MHF Contributor chisigma's Avatar
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    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
    Last edited by chisigma; May 3rd 2010 at 09:09 AM.
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  5. #5
    mms
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    Quote Originally Posted by Drexel28 View Post
    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
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