Can Mathematica find the asymptotics of a function in the following sense?

I have

Log[1/n^2]/Log[4^(-Sqrt[Log[n]]) + (2^(-Sqrt[Log[n]]) - 1)^2]

and I would like to know an asymptotic approximation when $n$ is large. That is a simple function that is within a constant multiplicative value in the limit when $n \rightarrow \infty$. If instead it was

Log[1/n^2]/Log[4^(-Sqrt[Log[n]]) + (2^(-Sqrt[Log[n]]))^2]

then I know that

Log[1/n^2]/Log[4^(-Sqrt[Log[n]]) + (2^(-Sqrt[Log[n]]))^2]/Sqrt[Log[n]] 

tends to a constant value. My question is how could you use Mathematica to discover that $\sqrt{\log{n}}$ is the right answer in the second case and to find whatever the right solution is in the first case?

By trial and error I happen to know that the right answer in the first case is somewhere between $2^{\log^{1/2}{n}}$ and $2^{\log^{1/2+\epsilon}{n}}$.

  • $\begingroup$ Could you please put parentheses around 2^-... and 4^-... ? $\endgroup$ – b.gates.you.know.what Dec 12 '12 at 20:52
  • $\begingroup$ Done. I hope that is clearer. $\endgroup$ – lip1 Dec 12 '12 at 20:55
  • $\begingroup$ What is connection between 2nd and 3rd formulas? $\endgroup$ – Vitaliy Kaurov Dec 12 '12 at 21:00
  • $\begingroup$ You ought to help Mathematica ought with some simple analysis first. The solution in the first case looks like $O(2^{\sqrt{\text{Log}[n]}}\text{Log}\left[n\right])$; recognizing this, you can ask MMA to take the limit of the ratio for you (demonstrating it is correct and obtaining the constant in the process). $\endgroup$ – whuber Dec 12 '12 at 21:02
  • $\begingroup$ @VitaliyKaurov, The third is just the second divided by $\sqrt{\log{n}}$. This is a constant in the limit so $\sqrt{\log{n}}$ is asymptotically equal to the second formula under my definition. $\endgroup$ – lip1 Dec 12 '12 at 21:04

For the first equation, the substitution $z= 2^{\sqrt{\log n}}$ with the inverse relation $n = \exp[ (\log_2 z)^2]$ seems to be worth to try. Note that with $n\to\infty$ also $z\to\infty$. So we try

sub=Simplify[Log[1/n^2]/ Log[4^(-Sqrt[Log[n]]) + (2^(-Sqrt[Log[n]]) - 1)^2] /. n -> Exp[Log[2, z]^2], z > 1]

with the result

-((2 Log[z]^2)/(Log[2]^2 Log[(2 - 2 z + z^2)/z^2])).

And next

PowerExpand[Series[sub, {z, \[Infinity], 4}] /. {z -> 2^Sqrt[Log[n]]}]

which yields the asymptotic expansion

$$\log (n) 2^{\sqrt{\log (n)}}+\frac{2 \log (n)}{3\ 2^{\sqrt{\log (n)}}}+\frac{\log (n)}{\left(2^{\sqrt{\log (n)}}\right)^2}+\frac{56 \log (n)}{45 \left(2^{\sqrt{\log (n)}}\right)^3}+\frac{4 \log (n)}{3 \left(2^{\sqrt{\log (n)}}\right)^4}+O\left(\left(\frac{1}{2^{\sqrt{\log (n)}}}\right)^5\right).$$

I believe for the second problem another substitution might do the trick.

| improve this answer | |
  • $\begingroup$ That's very nice, thanks. $\endgroup$ – lip1 Dec 12 '12 at 21:24

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.