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Math Help - Find a polynomial approaching |x| with error 0.1

  1. #1
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    Find a polynomial approaching |x| with error 0.1

    How can I find a polynomial approaching |x| such that the maximum error is 0.1?
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  2. #2
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    I'm pretty sure that the answer will depend on what interval you are considering. I say this because most of the approximation theorems I am aware of (that would be useful in this case) rely on you working on some closed interval [a,b].
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    yes. From [-1,1]
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    Well the Stone-Weirestrass theorem states that this is possible. However, in most proofs I have seen you never actually see how to construct such polynomials.

    Though not all hope is lost. Look at Bernstein Polynomials. Here is a link to the Wikipedia page on them. Also a while back I wrote a blog entry on just this topic, here is a link to the post. I don't think that I mentioned this in the post say using the method described in the post you end up with a polynomial of degree n, there might be other polynomials of lower degree that also give you the desired error. This is because of some very conservative estimations that are made.
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    Well. I try to use the Therem, but no metter what degree I put, the expansion will just cancell evey term excerpt for |x|, this is not the one that I can find
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    Obviously, you only need even powers. If you tried, say ax^4+ bx^2, then the error, for given x is |ax^4+ bx^2- |x||. Because of symmetry, it is sufficient to look at x> 0 so that is |ax^4+ bx^2- x|. Is it possible to choose a and b so that is never larger than .1 for all x between -1 and 1?
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  7. #7
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    Quote Originally Posted by happybear View Post
    Well. I try to use the Therem, but no metter what degree I put, the expansion will just cancell evey term excerpt for |x|, this is not the one that I can find
    The theorems regarding approximation with polynomials often assume the domain of the function is [0,1]. If you want an approximation on [-1,1] you will need to make an adjustment.
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  8. #8
    MHF Contributor chisigma's Avatar
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    The function to be round up in the interval [-1,1] is y(x) = |x| and it is an even function, so that we search an even polinomial of degree 2n written as p(x)= \sum_{k=0}^{n} p_{k}\cdot x^{2k} that approximates y(x) in 'min-max' sense. If we indicate the 'error function' with e(x)= y(x)-p(x), the 'min-max condition' force that to be...

    e(\frac{k}{n+1}) = (-1)^{k}\cdot e , k=0,1,...,n+1 (1)

    The (1) is a system of linear equations in the unknown variables p_{0},p_{1}, ... , p_{n}, e that can be solved in standard way. In the case n=1 it becomes...

    p_{0} + e = 0

    p_{0} + \frac{p_{1}}{4} - e = \frac{1}{2}

    p_{0} + p_{1} + e = 1 (2)

    ... the solution of which is p_{0}= \frac{1}{8}, p_{1}= 1, e=-\frac{1}{8} , so that the 'min-max' polynomial is p(x)= \frac{1}{8} + x^{2} and the error is alternatively +.125 and  -.125. You require an error not grater than .1, so that the min-max polynomial is of order 4 , i.e. n=2 and (1) becomes...

    p_{0} + e =0

    p_{0} + \frac{p_{1}}{9} + \frac {p_{2}}{81} - e = \frac{1}{3}

    p_{0} + \frac{4}{9}\cdot p_{1} + \frac{16}{81}\cdot p_{2} + e = \frac{2}{3}

    p_{0} + p_{1} + p_{2} - e = 1 (3)

    ... the solution of which is left as exercise ...

    Kind regards

    \chi \sigma
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  9. #9
    MHF Contributor chisigma's Avatar
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    The 'solution' of the 'exercise' I have proposed in the last post is  p_{0}=\frac{1}{16}, p_{1}= 2, p_{2}= -\frac{9}{8}, e= -\frac{1}{16} , so that the min-max polynomial of degree 4 gives an approximation with maximum error e= .0625...




    If greater precision is required it is necessary to increase n and the maximum error I suppose is e= \frac{1}{2^{n+2}}...

    Kind regards

    \chi \sigma
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