Prove that if the formula
is exact for all polynomials of degree and if the nodes are symmetrically placed about the origin, then the formula is exact for all polynomials of degree
So, clearly if then
But, obviously I am suppose to show that if then
[strike]So, I was thinking that I had to show that has roots and therefore . However, I cannot see how to do this.[/strike]
EDIT: I figured out a lot more. If where is even then Thus we see that . On the other hand, the are evenly distributed about zero and is an odd function. Thus, as long as the cancel out. Now I just have to ensure that these will cancel.
EDIT2: So I think there is a problem here. Since 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 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 , call it , as the linear combination of lower degree monomials and the monomial . Thus where . Now if we look at the error we see that this can be simplified to
where . But, we know that the quadrature for polynomials of degree n or less must be exact so
By symmetry and the fact that is an odd function the error further reduces to
Therefore, the quadrature is exact for polynomials of degree or less.