Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $0<w_1, \ldots, w_n$ 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.

share|improve this question
1  
An easier route would be to replace $P(x)$ with monomials $x^k$, $k=0\dots n-1$. You now have a Vandermonde system of equations for the corresponding $w_k$... –  J. M. Feb 8 '12 at 23:28
    
I don't know what a Vandermonde system is. The wikipedia entry makes me think I would have to prove too many preliminary results before that approach could be useful. –  nullUser Feb 9 '12 at 0:36
    
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 linear-algebra, so proving the conditions for the invertibility of a Vandermonde matrix should be routine...) –  J. M. Feb 9 '12 at 0:46
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.