Dissecting Andrew's slog solution
#1
I've been looking at the terms to Anrew's slog function, as computed by various matrix sizes. Andrew notes in his paper that there is a periodicity of 7 in the terms. As a first rough guess, he's correct, but a closer look reveals a more interesting pattern.

Perhaps it's best if I walk you down the path I followed to get where I'm at now. I started, of course, with the graph of the root test for power series. Here I show values for the 400-term solution, with root test values between 0.4 and 0.8, to expose a little more detail:

   

So far, nothing terribly interesting stands out. Sure, there's a periodicity, though it's not exactly at 7 terms. As we get about halfway through the series, the periodicity breaks down, but that would appear to be an artifact of the finite matrix we solved. For a 200x200 matrix, the periodicity is pretty consistent out to about 100 terms. For a 500x500 matrix, it's pretty consistent out to about 250 terms, give or take.

My first "aha!" moment came when I forgot to set the plotjoined flag when I saved the graph to disk. I was zooming in, which also helped. This fortuitous accident made something very striking stand out:

   

Notice the overlapping parabolic-looking arcs? The overlapping part is key, because it exposes the pattern that's nearly indiscernible in the connected plot. I knew at once that I needed to look at evens and odds. Here's the next graph I looked at, with the odd terms of the series graphed in red, the evens in blue:

   

I decided to take a closer look at the unaltered terms. The first thing that struck me was that the sign of the even terms seemed to alternate. If I graph the even terms, they alternate, so if I reverse the sign of the even evens, I should hope to see a smooth (relatively speaking) curve. Of course, the terms diminish quite rapidly. In the following graph, I added a scaling factor in addition to the alternately reversing the sign, by using (-2)^(k/2) as a scaling factor. For evens such as 0, 4, 8, 12, etc., this would be positive, while for 2, 6, 10, etc., it would be negative. Here's the resulting graph, again for the 400 term solution:

   
~ Jay Daniel Fox
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Dissecting Andrew's slog solution - by jaydfox - 08/28/2007, 07:23 AM

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