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This is a homework problem that I have been stuck on for a while.

Suppose $w:[a,b] \to \mathbb{R}^+$ is continuous and not identically zero. Show that that there is a unique quadrature formula with sample points $a=x_1 < x_2 < \ldots < x_{n-1} < x_n=b$ and weights $0 so that $ \int_a^bP(x)w(x)dx = \sum_{i=1}^nw_iP(x_i) $ for all polynomials $P$ of degree at most $2n-3$.

What I have done so far:

Consider $\phi(x) = \prod_{i=1}^n(x-x_i).$

Any polynomial $P$ of degree at most $2n-3$ can be written as $ \phi(x)p(x) + r(x) $ with $\deg p\leq n-3$ and $\deg r < n$. So $ \int_a^bP(x)w(x)dx = \int_a^b\phi(x)p(x)w(x)dx + \int_a^br(x)w(x)dx. $ We want that the first integral on the right side to be zero. Hence we need $\phi$ to be orthogonal to $\mathcal{P}_{n-3}$ (using the weighted integral inner product).

If that were the case then $ \phi = p_n(x) + c_1p_{n-1}(x)+ c_2p_{n-2}(x) $ for some $p_n,p_{n-1},p_{n-2}$ with degrees at most $n,n-1,n-2$ respectively.

Then we note $\phi(a) = \phi(b) = 0$ which gives us the system $ c_1p_{n-1}(a) + c_2p_{n-2}(a) = -p_n(a)\\ c_1p_{n-1}(b) + c_2p_{n-2}(b) = -p_n(b) $ which we need to be invertible. I am having trouble showing that it is.

Can anyone give me some help please? Pointers on how to proceed after showing it is invertible would be appreciated too.

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    Not really. You can easily demonstrate the conditions when the matrix \begin{pmatrix}1&1\\x_1&x_2\end{pmatrix} is singular right? How about \begin{pmatrix}1&1&1\\x_1&x_2&x_3\\x_1^2&x_2^2&x_3^2\end{pmatrix}? Generalize accordingly. (You tagged your question [tag:linear-algebra], so proving the conditions for the invertibility of a Vandermonde matrix should be routine...)2012-02-09

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