Thread Subject: Quadgk and Legendre-Gauss

I am rather curious as to why quadgk uses a 7 point
Gauss-Legendre basis for its Kronrod extension.

Does the accuracy of the calculation decrease if instead of
doing Integrate[f[x],{x,-1,1}] I change variables to
sin(x*pi)=y so that Integrate[f[y]D[x,y],{y,-1,1}] becomes a
Chebyshev integral? (pardon the heretical Mathematica notation)

If not, why can't I just compute a regular 2N+1 point
Chebyshev quadrature which is automatically the Kronrod
extension of the regular N point formula? Unless I am
missing something subtle, the 7-15 quadrature is accurate up
to 22nd order, while a straight Chebyshev 15 point
quadrature is accurate up to 30th order with all the good
exponential suppression intrinsic to full Gauss quadratures.

I am sorry, I realize this is not correct for regular
Chebyshev, but is certainly correct for Chebyshev-Lobatto,
which uses abcissas located at x_j=cos(j*pi/N) rather than
at x_j=cos((2*j-1)*pi/2/N) Basic question still holds.



"Rodrigo " <guerra.remove.this@physics.harvard.edu> wrote in
message <ft9foi$hn0$1@fred.mathworks.com>...
> I am rather curious as to why quadgk uses a 7 point
> Gauss-Legendre basis for its Kronrod extension.
>
> Does the accuracy of the calculation decrease if instead of
> doing Integrate[f[x],{x,-1,1}] I change variables to
> sin(x*pi)=y so that Integrate[f[y]D[x,y],{y,-1,1}] becomes a
> Chebyshev integral? (pardon the heretical Mathematica
notation)
>
> If not, why can't I just compute a regular 2N+1 point
> Chebyshev quadrature which is automatically the Kronrod
> extension of the regular N point formula? Unless I am
> missing something subtle, the 7-15 quadrature is accurate up
> to 22nd order, while a straight Chebyshev 15 point
> quadrature is accurate up to 30th order with all the good
> exponential suppression intrinsic to full Gauss quadratures.

Never mind. That transformation produces equally spaced
abscissas, with equal weights and can't possibly be accurate.