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Attempting to compute the kslog numerically (i.e., Kneser's construction)
Okay, here is the convergence analysis. The blue lines represent (from top to bottom) the error (in bits) in the 105, 315, 945, and 2835 point systems, relative to the 2835 point 10-interval interpolated system.


A note here: The solutions are found by solving an integral using a system of equations, which basically means solving a matrix equation. Just as extra accuracy for an integral can be achieved by using weights to simulate polynomial interpolation, I've achieved extra accuracy in the matrix system by weighting each point appropriately. Note that this must be tuned by hand. For example, Fejer quadrature would seem an obvious choice here, but the singularities at each end of the interval cause Fejer quadrature to give horrible results, which I did not expect.

Anyway, to demonstrate the increased accuracy empirically, note the red lines. These correspond to the 105, 315, and 945 point systems, using 10-interval interpolation polynomials (11-points, with common endpoints). Note that in all three cases, they are about 2-3 bits more accurate than the "rectangle" quadratures. Indeed, they are almost as accurate as using a system three times as large!

Convergence is fairly linear, about 2.5 to 3 bits per system doubling. And the convergence near the singularities is even better, which is a good thing, as the accuracy is quite bad near the singularities (0 and 1, the endpoints of the interval).

My next test will be to use 14 point intervals. For a large enough system, this should improve accuracy, but I anticipate the 105-point system might actually get worse.
~ Jay Daniel Fox

Messages In This Thread
RE: Attempting to compute the kslog numerically (i.e., Kneser's construction) - by jaydfox - 09/24/2009, 12:17 AM

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