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Math Help - Polynomial interpolation

  1. #1
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    Polynomial interpolation

    Let x_{0}, x_{1}, \cdots , x_{n} be distinct points in the interval [a,b] and f \in C^{1}[a,b].


    We show that for any given \epsilon >0 there exists a polynomial p such that

    \left\| f-p \right\|_{\infty} < \epsilon and p(x_{i}) = f(x_{i}) for all i=1,2, \cdots , n


    I know \left\| f\right\| _{\infty}= max_{x \in [a,b]}|f(x)| and I wonder if the polynomial they are asking for is the Lagrangian polynomial interpolating f at the nodes x_{0}, x_{1}, \cdots , x_{n}. If yes, I am not sure how to prove the problem.
    Last edited by math8; April 11th 2010 at 10:09 AM.
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  2. #2
    Super Member Deadstar's Avatar
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    Lol you might wanna check out your post again... Whatever you copied of physics forum has been saved as an image so won't show up I guess.
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  3. #3
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    Its probably cached by their browser so they can't tell. I'll try to repair it...

    Quote Originally Posted by math8 View Post
    Let x_0,x_1,\cdots x_n be distinct points in the interval [a,b] and f\in C^1([a,b]).

    We show that for any given \epsilon > 0 there exists a polynomial p such that ||f-p||_\infty < \epsilon and p(x_i)=f(x_i) for all i=1,\cdots, n.

    I know ||f||_\infty = \max_{x\in[a,b]}|f(x)| and I wonder if the polynomial they are asking for is the Lagrangian polynomial interpolating f at the nodes x_1,\cdots,x_n. If yes, I am not sure how to prove the problem.
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  4. #4
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    I'll give it a shot I guess...

    Assume WLOG that both a and b are in the x_i, and replace f with f-g, where g is the polynomial which interpolates the nodes x_i. Clearly if f-g can be approximated by a polynomial, f can be approximated by that polynomial plus g. So we have to find a polynomial which is within \epsilon of f and vanishes at x_i.

    Chuse as g a polynomial which is everywhere within \epsilon\over 2n of f by the Stone–Weierstrass theorem. Let d_i = f(x_i) - g(x_i) which is no more than \epsilon\over 2n in magnitude. We will chuse polynomials h_i to add to g to correct these errors so that the final approximating polynomial is g+\sum h_i. Then ||f-g-\sum h_i||_\infty \le ||f-g||_\infty + \sum ||h_i|| \le {\epsilon\over 2n} + n{\epsilon \over 2n} \le \epsilon if all the ||h_i||_\infty \le {\epsilon\over 2n}.

    Therefore we have only to prove the following claim: for every \epsilon>0 and x,y\in[a,b] ( y<\epsilon), there is a polynomial h which is everywhere less than \epsilon in magnitude on [a,b] and h(x)=y.

    To do this, chuse a cosine wave centered at x and scale it down to amplitude y. We can take its Taylor polynomial to enough terms that inside [a,b] it is closer to zero at every point than the cosine wave [perhaps someone should check this, as I only have a rough sketch], so that this Taylor polynomial is y at x and it is \le \epsilon on [a,b], as desired.



    ...god, what a mess. Its probably wrong too. Be warned!
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