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Math Help - Polynomial Interpolation

  1. #1
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    Polynomial Interpolation

    Let h_i = x_i-x_{i-1}>0. Suppose f is three times differentiable and let P_i be the linear interpolant through the points (x_{i-1}, f(x_{i-1})) and (x_i, f(x_i)).
    I want to show, for h_i \rightarrow 0
    \\
    f'(x)-P'_i(x) = \frac{1}{2}(2x-x_{i-1}-x_i)f''(x)+\mathcal{O}(h_i^2)= \left(\begin{array}{cc}\mathcal{O}(h_i^2),&\mbox{ if } x= \frac{1}{2}(x_{i-1}+x_i)\\\mathcal{O}(h_i), & \mbox{ otherwise} \end{array}\right)
    Last edited by bram kierkels; October 2nd 2009 at 02:36 AM.
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  2. #2
    Junior Member
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    Johannesburg, South Africa
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    I almost have the solution. I just made a mistake somewhere with some minussign but i can't find a mistake;

    P_i(x) is of degree 1. So  P_i'(x) is a constant and P_i'(x) = \frac{f(x_i)-f(x_{i-1})}{h_i}

    f(x_i)=f(x)+f'(x)(x-x_i)+\frac{1}{2}f"(x)(x-x_i)^2+\mathcal{O}(h_i^3)
    f(x_{i-1})=f(x)+f'(x)(x-x_{i-1})+\frac{1}{2}f"(x)(x-x_{i-1})^2+\mathcal{O}(h_i^3)
    So f(x_i)-f(x_{i-1}) = f'(x)(x_{i-1}-x_i)+\frac{1}{2}f"(x)(x^2-2xx_i+x_i^2-x^2+2xx_{i-1}-x_{i-1}^2)+\mathcal{O}(h_i^3)
    =f'(x)(x_i-x_{i-1})+\frac{1}{2}f"(x)(-2x(x_i-x_{i-1})+(x_i-x_{i-1})(x_i+x_{i-1}))+\mathcal{O}(h_i^3)
    While  h_i=x_{i-1}-x_i the equation becomes
    =-f'(x)h_i+\frac{1}{2}f"(x)(-2xh_i+ (x_i+x_{i-1})h_i)+\mathcal{O}(h_i^3)
    And so P_i'(x) = \frac{f(x_i)-f(x_{i-1})}{h_i}=-f'(x)+\frac{1}{2}f"(x)(-2x+ x_i+x_{i-1})+\mathcal{O}(h_i^2)
    Thus f'(x)-P_i'(x)=f'(x)-(-f'(x)+\frac{1}{2}f"(x)(-2x+ x_i+x_{i-1})+\mathcal{O}(h_i^2)

    So somewhere there is a small mistake, can you find it anywhere?
    thanks
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