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Gaussian quadrature allows us to integrate polynomials up to order $2 n-1$ using only $n$ function values.

$\int_{x_0}^{x_1} ( \sum_{i=0}^{2 n-1} a_i x^i ) dx = f(a_0, \dots , a_{2 n-1}) $

thus, the function $f(a_0, \dots , a_{2 n-1})$, which naively has $2 n$ parameters, can actually be calculated using only $n$ parameters. Or, there exists a coordinate transform $a_i\rightarrow a_i^{\star}$

$f(a_0, \dots , a_{2 n-1}) \rightarrow f^{\star}(a_0^{\star}, \dots , a_n^{\star} , \dots , a_{2 n-1}^{\star}) $

such that $\frac{\partial f^{\star} }{a_i^{\star}} \equiv 0$ if $i>n$

I understand that this is true, and that the construction of gaussian quadrature has $2n$ free parameters, i.e. sampling positions and weights, which seemingly explains why it can integrate polynomials up to order $2 n-1$.

Nevertheless, I wonder if there is a "geometric" argument that makes it clear that the function $f(a_0, \dots , a_{2 n-1})$, in $2n$-dimensional space, can be calculated using only $n$ parameters.

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This may be a bit too elementary for this site. The observation you need is that the integral of an odd function (over an interval symmetric around the origin) is zero. –  Jitse Niesen Jan 4 '11 at 9:33
    
I see, thanks. So the $n$ function values are in fact sufficient because one only needs the even-degree terms in the polynomial to calculate the integral. I guess I should have thought of that ... –  someone Jan 4 '11 at 10:00
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1 Answer 1

Look at it this way: what you are computing is a PROJECTION. In this sense, you are not surprised that in order to compute the value of the function $f(x_1, x_2)=x_1$ you need only 1 parameter. Something similar happens here. Instead of taking as parameters the coefficients $a_j$ of the expansion of the given polynomial in the basis of monomials, take the $a_j$ to be the expansion of this polynomial in terms of the orthogonal polynomials with respect to the integration weight. Then the quadrature formula is given by the linear combination of the coefficients $a_n, \dots, a_{2n-1}$ only.

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