# Rate of convergence from asymptotic limit

I am working on an algorithm in which one parameter $$a(t)$$ verifies the following asymptotic limit: $$f(a(t))=\mathcal{O}(1/t^{2})$$. I am using it to assess an empirical rate of convergence of $$f(a(t))$$ as $$\frac{f(a(t))}{f(a(10t))}\approx100$$. It works more or less well, but I know that mathematically it is not formal. Do you have any suggestion on how to compute $$\frac{f(a(t))}{f(a(10t))}$$ from the asymptotic limit?

$$f(a(t))=O(1/t^2)$$ means that $$f(a(t))=\frac{g(t)}{t^2}$$, $$g$$ being a bounded function. So $$\frac{f(a(t))}{f(a(10t))} = \frac{\frac{g(t)}{t^2}}{\frac{g(10t)}{100t^2}} = 100\frac{g(t)}{g(10t)}$$ Problem is, you have to know both bounds for $$g$$ to make this accurate. If you know $$(\forall t)\ 0 which is quite common in algorithms theory, then $$\frac{f(a(t))}{f(a(10t))} \le \frac{100B}{A}$$ But in general, you can not find any good bound (and this bound may not exist).
• Thanks @Nicolas FRANCOIS, in this particular case, $f(a(t))\leq \frac{C}{t^{2}}$. Could we use 100 as an upper bound for $\frac{f(a(t))}{f(a(10t))}$? Dec 17, 2020 at 19:08