Convergence of matrix solution for base e
#6
I threw together some graphs of the cumulative absolute error of the accelerated solution, at the radius of convergence:

The first graph is the accumulated error by coefficient index:

   

The bolder section of each line represents the coefficients actually calculated by a matrix solution, while the pale section represents the coefficients of the logarithms used to accelerate the solution. Notice that the error rises quite rapidly, then seems to get basically flat once we reach the end of the matrix solution. The flatness is due to the fact that the error has grown so large by the final term of the truncated matrix, that the remaining terms (from the power series for the logarithms) add negligible error, especially on this logarithmic scale. Thus, with larger systems, the initial error is very small, allowing for a slow and steady climb, which flattens off when we've reached peak error. I'll post more graphs later showing the term by term error.

The second graph is the plot of log(error) vs. log(size).

   

Once again, the plot appears linear, though we only get 3.6 additional bits of accuracy for each additional doubling of matrix size. With 36 bits for a 512x512 system, we're looking at about 40 bits for my 1200x1200 solution, perhaps as much as 42-45 bits in practice, because the errors in the coefficients will not all be oriented in the same direction (in the complex plane). So I don't have enough for double precision over the entire domain yet. But excluding values very close to the singularities, the accuracy should be enough for double precision math. (I'll need to investigate what regions can expect double precision, based on solution size.)
~ Jay Daniel Fox
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RE: Convergence of matrix solution for base e - by jaydfox - 12/17/2007, 01:58 PM

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