08/19/2014, 01:36 AM

(08/11/2014, 05:19 PM)jaydfox Wrote:(08/11/2014, 04:51 PM)sheldonison Wrote: Nice post. What is the most accurate version of the scaling constant that you know of? Earlier, you posted, "1.083063".

I'm currently working on a version that would allow me to calculate all the terms of the form [(2^(n-1)+1)...(2^n)]*(2^m), for a given small n and arbitrary m, which should run in about O(2^n m^5) time.

For example, for n=3, I could calculate 5..8 * 2^m, so let's say:

5, 6, 7, 8,

10, 12, 14, 16,

20, 24, 28, 32,

40, 48, 56, 64,

...

5*2^m, 6*2^m, 7*2^m, 8*2^m

This would allow me to more accurately estimate a_1, because I would be able to monitor the peaks and troughs of the oscillations in the relative error.

So hopefully, within a week or two, I should have an estimate of a_1 to at least 1-2 more digits than the current 1.083063. Who knows, I might soon have the most accurate estimate? Short of a closed form to calculate it, the only way I know to determine its value accurately is calculate very large terms in the sequence, isolate a local minimum and local maximum, and estimate the midpoint.

I'm getting closer to finding a better approximation of a_1. So far, I think it's slightly less than 1.083063, perhaps in the neighborhood of 1.0830628-1.0830629.

Time for more pictures. I finished the code I mentioned above. I calculated A(k) for k = m*2^n, where 513 <= m <= 1024, and n up to 150 (so the largest value was 1024 * 2^150, or 2^160). I have a calculation running in the background to get me up to 2^260, so I'll have new pictures tomorrow.

Note that earlier, I defined a_1 as the limit of f(k)/A(k). However, to me it is more useful to graph A(k)/f(k), because this helps me remember that A(k) over estimates for even k, and underestimates for odd k. So the graph will be a graph of 1/a_1 in the limit as k goes to infinity. So I made a secondary axis on the right side of the graphs, to show the value of a_1(k).

I'll start by graphing A(k)/f(k) versus k:

Then I have a few graphs versus log_2(k).

Finally, I have a graph versus log_2(log_2(k)), which I think helps show how quickly (or not) the series converges on f(x).

Stay tuned for more exciting pictures tomorrow (or sometime this week). Also, I'll try to post an updated version of the code later this week, which has a minor bug fix, as well as some shortcuts which cut the running time by about 40% for calculating the polynomials.

~ Jay Daniel Fox