08/10/2014, 12:30 AM

(08/09/2014, 11:10 PM)tommy1729 Wrote: Considering your lastest posts Jay , are you still convinced that J(x)/1.08... is a very good approximation to A(x) ?

I mean considering your plots , it does not seem like

lim n-> + oo J(n)/A(n) = 1.08... holds true.

The absolute error grows without bound, so in that sense, no, it's not a good approximation. However, the relative error decreases to 0 in the limit, so yes, I still consider it a good first-order approximation. If A(k) is approximated by J(k)/a_1, then the absolute error is (A(k) - J(k)/a_1), which I graphed previously. The relative error is the absolute error divided by A(k), or (A(k) - J(k)/a_1) / A(k), which is 1-J(k)/(a_1 A(k)).

The graphs below use an alternate relative error of (a_1 A(k))/J(k) - 1. For small errors, this alternate definition is essentially equivalent (e.g., 1-999/1000 ~= 1000/999 - 1), and I didn't want to redo the graphs, so they'll have to do. You can't actually see the difference in the graphs (i.e., comparing side by side), but since the equation is written at the top of the graphs, I figured I should disclose the discrepancy.

Skipping a doubling of the scale on the k-axis:

And skipping several more:

As you can see, the relative error continues to shrink. The sequence A(n) will cross the function J(x) an infinite number of times, and the absolute error will grow without bound, but the relative error will shrink to 0. I'm still working on proving it formally, but it seems intuitively clear, based on analysis up to 2^21 terms in the sequence, plus crude analysis of the 2^m terms up to m=150.

There's also the semi-relative

~ Jay Daniel Fox