Question: Consider the quadrature rule $$\int_{-1}^1 f(x)dx \approx w_{-1}f(-1)+w_0f(0)+w_1f(1)+w'_0f'(0)$$

Compute the weights such that the polynomial is exact up to degree 3.

Answer: When I computed the coefficients I got that $$w_0'=0, w_1=1/3,w_0=4/3,w_{-1}=1/3$$

I was wondering if the fact that one of these weights is $0$ means that I don't get the degree 3 exactness?

I was thinking this since I think I get $n-1$ accuracy when the degree is n.


Whether or not one (or more than one) of the coefficients is zero is not related to the degree of precision (or exactness if you prefer) of the given quadrature rule. Instead, it depends on how you found these coefficients.

Assuming you found those coefficients by substituting $f(x) = 1, x , x^2, x^3$ into the quadrature rule and solving the $4\times 4$ linear system with variables $w_{-1}, w_0, w_1, w_0'$, then the answer to your question is no; the quadrature rule with coefficients found above is, by construction, exact for polynomials of degree at most 3.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.