How would one go about computing a Chebyshev-type quadrature problem with general integration limits of [a,b] instead of [-1,1]?

280 Views Asked by At

More specifically, a problem of the form

$\int_a^b\frac{f(x)}{\sqrt{1-x^2}}dx = \sum_{i=1}^{N}w_if(x_i)$,

where $a,b \in [-1,1]$, $w_i$ are the weights, and $x_i$ are the abscissa. A quick search online yields information about standard Gauss-Chebyshev quadrature (where the weight function $w(x) = \frac{1}{\sqrt{1-x^2}}$) and where a = -1 and b = +1. But when one performs an affine coordinate transform to change the limits of the integral in question to [-1,1] the weight function changes, rendering Gauss-Chebyshev quadrature seemingly useless.

Is there a way of transforming the weights ($w_i$), much like one would transform the abscissae ($x_i$) (using the same transformation described in the previous paragraph) for Gauss-Chebyshev quadrature? Or must one define a new quadrature scheme by searching for a set of polynomials which are orthogonal to one another (with the new weight function), every time the values of a and b are changed?

1

There are 1 best solutions below

0
On

I think you have the right idea: the nodes and weights for a Gaussian quadrature formula are completely specific to the weight function and the interval. Polynomials which were orthogonal over one interval with one weight function won't in general be orthogonal over another. This is a big limitation when the weight function is other than the Gauss-Legendre $w(x)=1$.

That having been said, if your interval includes one of the endpoints a Gauss-Jacobi quadrature formula may give better results because it can take into account the singularity at that endpoint. Otherwise the normal thing to do would be to use Gauss-Legendre quadrature or Kronrod or Lobatto, or whatever.

There is a technique called Clenshaw-Curtis quadrature which effectively expands a function in Chebyshev polynomials using Gauss-Chebyshev quadrature and produces the integral of the Chebyshev polynomials. Don't know why I mentioned that, though... In that case the effective interval of integration is still $[-1,1]$ so it's not related to your question... just blathering...