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

  1. #1
    Senior Member
    Mar 2009

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

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

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

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

    EDIT2: So I think there is a problem here. Since $\displaystyle n$ 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 $\displaystyle A_i$ 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 $\displaystyle n+1$, call it $\displaystyle f(x)$, as the linear combination of lower degree monomials and the monomial $\displaystyle x^{n+1}$. Thus $\displaystyle \int_{-1}^1\! f(x)\, dx = \sum_{i=0}^{n+1} c_i x^i\, dx$ where $\displaystyle c_i \in \mathbb{R}/\{0\}$. Now if we look at the error $\displaystyle e = \int_{-1}^1\! f(x)\, dx - \sum_{i=0}^n A_i f(x_i)$ we see that this can be simplified to
    $\displaystyle 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 $\displaystyle p_n(x) \in P_n$. But, we know that the quadrature for polynomials of degree n or less must be exact so
    $\displaystyle e = c_{n+1}\left[\int_{-1}^1\! x^{n+1}\, dx - \sum_{i=0}^n A_i x_i^{n+1}\right]$.
    By symmetry and the fact that $\displaystyle x^{n+1}$ is an odd function the error further reduces to
    $\displaystyle e = c_{n+1} A_{n/2 + 1} 0^{n+1} = 0$.
    Therefore, the quadrature is exact for polynomials of degree $\displaystyle n+1$ or less.
    Last edited by lvleph; Apr 13th 2010 at 11:26 AM.
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