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Math Help - [SOLVED] Quadrature Degree of Precision on [-1,1] with n Even

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    [SOLVED] Quadrature Degree of Precision on [-1,1] with n Even

    Prove that if the formula
    <br />
        \int_{-1}^1\! f(x)\, dx \approx \sum_{i=0}^n A_i f(x_i)\; (n\text{ is even})<br />
    is exact for all polynomials of degree n, and if the nodes are symmetrically placed about the origin, then the formula is exact for all polynomials of degree n+1.

    So, clearly if f_n \in P_n then
    <br />
p_n = \int_{-1}^1\! f_n(x)\, dx - \sum_{i=0}^n A_i f_n(x_i) = 0.<br />
    But, obviously I am suppose to show that if f_{n+1} \in P_{n+1} then
    <br />
p_{n+1} = \int_{-1}^1\! f_{n+1}(x)\, dx - \sum_{i=0}^n A_i f_{n+1}(x_i) = 0.<br />
    [strike]So, I was thinking that I had to show that   p_{n+1} has n + 2 roots and therefore p_{n+1} \equiv 0. However, I cannot see how to do this.[/strike]

    EDIT: I figured out a lot more. If f_{n+1} \in P_{n+1} where n is even then \int_{-1}^1\! f_{n+1}(x)\, dx = 0 Thus we see that p_{n+1} = -\sum_{i=0}^n A_i f(x_i). On the other hand, the x_i are evenly distributed about zero and f_{n+1}(x) is an odd function. Thus, p_{n+1} = 0 as long as the A_i cancel out. Now I just have to ensure that these will cancel.

    EDIT2: So I think there is a problem here. Since n is even we have an odd number of points. Thus, one point is centered at zero, while the other points may result in a cancellation. However, there is no guarantee that the point at zero results in a zero function evaluation. hmmm.

    EDIT3: By symmetry we know that the A_i cancel. However, I am still stuck with the term at 0.

    EDIT4: Wow! This is much simpler than I thought. We can think of a polynomial of degree n+1, call it f(x), as the linear combination of lower degree monomials and the monomial x^{n+1}. Thus \int_{-1}^1\! f(x)\, dx = \sum_{i=0}^{n+1} c_i x^i\, dx where c_i \in \mathbb{R}/\{0\}. Now if we look at the error e = \int_{-1}^1\! f(x)\, dx - \sum_{i=0}^n A_i f(x_i) we see that this can be simplified to
    e = \int_{-1}^1\! c_{n+1} x^{n+1}\, dx + \int_{-1}^1\! p_n(x)\,dx - \sum_{i=0}^n A_i c_{n+1} x_i^{n+1} - \sum_{i=0}^n A_i p_n(x_i)
    where p_n(x) \in P_n. But, we know that the quadrature for polynomials of degree n or less must be exact so
    e = c_{n+1}\left[\int_{-1}^1\! x^{n+1}\, dx - \sum_{i=0}^n A_i x_i^{n+1}\right].
    By symmetry and the fact that x^{n+1} is an odd function the error further reduces to
    e = c_{n+1} A_{n/2 + 1} 0^{n+1} = 0.
    Therefore, the quadrature is exact for polynomials of degree n+1 or less.
    Last edited by lvleph; April 13th 2010 at 11:26 AM.
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