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1

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.