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Convergence of matrix solution for base e
#4
Hmm, I'm not quite sure if we mean the same thing about the first graph. The first graph has, for the x-axis, the coefficient index "n" of the power series, and for the y-axis, the log of the cumulative absolute error for all coefficients from 1 to n. In other words, assuming the x-axis is linear in n, then the y-axis is:



Here, is the k-th coefficient of a particular solution of a truncated matrix, and is the k-th coefficient of the "theoretical" infinite solution. (In fact, in these graphs, it's the 1200-term truncated solution.)

As you can see, for any particular power series of a finite truncation, the cumulative error rises rapidly for the first dozen or so coefficients, but tapers off quickly, approaching some value asymptotically. This is because the terms decrease exponentially (per the radius of convergence), so the absolute error in the later terms is negligible. This is assuming we evaluated at some point on the unit circle, and took the absolute value of each term in the power series, so this is only a useful bound within the unit disc.


BTW, if I were to do the same calculation, but assuming a radius of 4/3 (1.333...), then the total error would rise higher and continue to rise much further into the series. It would be interesting to see, as it would dictate a very conservative upper bound on error within the disc with radius 4/3. I could extend this logic up to, but not including, the radius of convergence, at which point, the total error for any finite truncation would in theory be infinite, making comparison...difficult Tongue.

As for the error relationship, I think yes, we can say we have found a very conservative upper bound for error, but only within the unit disc. I need to try this for various radii and see if we can find an upper bound as a function of n and radius. That would be quite nice!
~ Jay Daniel Fox
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RE: Convergence of matrix solution for base e - by jaydfox - 12/14/2007, 05:55 PM

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