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Thread: [SOLVED] Chebyshev polynomial approximation

  1. #1
    Senior Member
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    [SOLVED] Chebyshev polynomial approximation

    The series

    $\displaystyle \Phi_n(x) = \frac{1}{2} c_0 T_0(x) + c_1 T_1(x) + \ldots+ c_n T_n(x)$

    is a least squares Chebyshev polynomial approximation of f(x) on [-1,1], if

    $\displaystyle c_j = \frac{2}{\pi} \int_{-1}^1\frac{f(x) T_j(x)}{\sqrt{1-x^2}} d x, \quad j=0,1,2,\ldots,n.$

    Find the quadratic least squares Chebyshev polynomial approximation of
    $\displaystyle g(z) = \pi \left(9+3 z^2\right) \sqrt{(2-z) z}$, on $\displaystyle z \in [0,2]$.
    -----------------------------
    Their answer (I don't know where I went wrong):
    $\displaystyle \Phi_2(z) = \frac{1}{15} \left( 42+ \left(1032 \right) z + \left(-456 \right) z^2\right)$

    ---------------

    My working:


    Let $\displaystyle x=z-1$ (to map to [-1,1] interval).
    $\displaystyle \therefore z = x+1$.

    $\displaystyle \therefore f(x)=\pi(9+3(x+1)^2)\sqrt{(2-(x+1))(x+1)}$
    $\displaystyle = \pi(9+3(x+1)^2)\sqrt{1-x^2}$.

    Then

    $\displaystyle c_j = 2\int_{-1}^1(9+3(x+1)^2)T_j(x)\,dx$
    So, (you can compute on Maple),
    $\displaystyle c_0 = 52,\; c_1=8,\;c_2=\frac{164}{5}$

    So $\displaystyle \Phi_2({\color{red}x})=\frac{1}{2}\cdot 52\cdot 1+8x+\frac{164}{5}(2x^2-1)$

    Then to map back to $\displaystyle z \in [0,2]$,
    $\displaystyle \Phi_2(z)=26+8(z-1)+\frac{164}{5}(2(z-1)^2-1)$.

    Now, simplifying is optional, but this answer is not correct (apparently).

    Any idea what I did wrong? Or is this the wrong forum for this stuff?
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  2. #2
    Senior Member
    Joined
    Jul 2006
    Posts
    364
    I found the error.

    Oops. Silly me. I entered the data incorrectly into maple. $\displaystyle c_2$ should be $\displaystyle \frac{-76}{5}$, not $\displaystyle \frac{164}{5}
    $.

    The rest is fine.
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