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Math Help - What straight line is closest to x^3 over the interval −1 ≤ x ≤ 1?

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    What straight line is closest to x^3 over the interval −1 ≤ x ≤ 1?

    I'm studying for a test, and this is one of the practice problems:

    What straight line is closest to x^3 over the interval −1 ≤ x ≤ 1?
    I have the answer, but can't make much sense of it:

    Solution: The answer is:

    ((x^3, 1)/(1, 1)) 1 + ((x^3, x)/(x, x)) x = ((1/2)/(2/3))x .

    Here, (f,g) = Integral from -1 to 1 of (fg dx).
    I figure this has something to do with least squares, since you're trying to fit a line to a curve, but other than that I'm lost.

    Could someone please provide a detailed explanation of the above solution? It would highly benefit my understanding of the material. Thank you!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by leesperare View Post
    I figure this has something to do with least squares, since you're trying to fit a line to a curve, but other than that I'm lost.

    We have to find the orthogonal projection of x^3 over the subspace of \mathbb{R}[x] :

    F=\{f:[-1,1]\to \mathbb{R},\;f(x)=a+bx\}

    A basis of F is B=\{1,x\} . As \int_{-1}^11\cdot x\;dx=0 then, B'=\{1/\sqrt{(1,1)},x/\sqrt{(x,x)}\} is an orthonormal basis of F . Now, apply a well known theorem.
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by leesperare View Post
    I'm studying for a test, and this is one of the practice problems:



    I have the answer, but can't make much sense of it:



    I figure this has something to do with least squares, since you're trying to fit a line to a curve, but other than that I'm lost.

    Could someone please provide a detailed explanation of the above solution? It would highly benefit my understanding of the material. Thank you!
    If 'closest' is in the 'least square sense' , then the coefficients of 'best polynomial' of order n approximating y(x) in [-1,1] minimize the quantity...

    \displaystyle I= \int_{-1}^{1} [y(x) - \sum_{k=0}^{n} a_{k}\ P_{k} (x)]^{2}\ dx (1)

    ... where P_{k}(x) is the Legendre Polynomial of order k. The a_{k} minimizing (1) are found with standard procedure to be...

    \displaystyle a_{k}= \frac{2k+1}{2}\ \int_{-1}^{1} y(x)\ P_{k}(x)\ dx (2)

    In Your case is y(x)=x^{3} , n=1, P_{0}=1 and P_{1}=x, so that...

    \displaystyle a_{0}= \frac{1}{2}\ \int_{-1}^{1} x^{3}\ dx

    \displaystyle a_{1}= \frac{3}{2}\ \int_{-1}^{1} x^{4}\ dx (3)

    Kind regards

    \chi \sigma
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