If I have a function $$f = \exp(\sqrt{n} \cdot \frac{\sqrt{\log{n}}}{\sqrt{n}-\sqrt{\log n}}),$$ I can notice, that $$\lim_{n \to \infty} f = \infty,$$ but also I can notice that it goes very slowly to infinity, for example for $$n = 1000, f(n) \approx exp(2.87),$$ $$n = 10000, f(n) \approx exp(3.13).$$
How can I write that the rate of convergence is very slow? I mean I don't want to just write that $f \to \infty$, but I want use more accurate asymptotic notation which will help me to express that?

  • $\begingroup$ Have you heard of the big-O notation? $\endgroup$ – user122283 May 31 '14 at 17:20
  • $\begingroup$ it goes to infinity linearly. $$f(n)\sim n$$ $\endgroup$ – Jika May 31 '14 at 17:23
  • 2
    $\begingroup$ And even, $f(n)\geqslant n$ for every $n$. $\endgroup$ – Did May 31 '14 at 17:26
  • $\begingroup$ @QiaochuYuan Yes, you're right. I made two mistakes writing my functions here. $\endgroup$ – Rop May 31 '14 at 18:45
  • $\begingroup$ @Did I made a mistake, while writting my formula here. Now it is what I meant. $\endgroup$ – Rop May 31 '14 at 18:45

I guess one way of seeing it rewriting the function $\frac{\sqrt{n} \log n }{\sqrt{n}-\log n} = \frac{\log n }{1 -\frac{\log n}{\sqrt{n}}}$. Hence your function becomes $n^\frac{1}{1-\frac{\log n }{\sqrt{n}}} \sim n^{1 +\frac{\log n}{\sqrt{n}} + O\big(\frac{1}{n^2}\big)}$, where each term $n^{a_n} \to_n 1$. Hence your expression grows $\sim n$ for large enough $n$.

| cite | improve this answer | |
  • $\begingroup$ Thanks for your help. I made a mistake in my formula, but I think I know now how to estimate the rate of convergence $\endgroup$ – Rop May 31 '14 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.