Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I solved the recurrence $T(n) = T(\sqrt[4]{n}) + 1$, $T(2) = 1$ through thinking it through. $T(n) = O(\log{\log{n}})$ since the number of times we add 1 is the number of times we can take the 4th root of $n$. Assume $\log$ means base 2.

Apparently, I can systematically solve this recurrence by setting $n = \log{m}$ and substituting. Why? Can someone clarify that for me?

This isn't homework. I'm trying to develop a systematic way of solving similar recurrences.

share|cite|improve this question

If I substitute $n =$log $m$ then that implies that $ m= a^n $ where $a$ is the base of your logarithm. Thus the recurrence will transform to

$$T(a^n) = T(a^{\frac{n}{4}}) +1 $$

and if we then use the transformation that $S(n) = T(a^n)$ we will have a much simpler recurrence solvable by Master theorem, or unrolling etc.

share|cite|improve this answer
Thanks, what do you mean by use the transformation that $S(n) = T(a^n)$? I know the master theorem, just not sure how it applies here. – John Hoffman Feb 23 '14 at 20:20
What I mean is that, the transformation $S(n)=T(a^n)$ will allow you to simplify the recurrence greatly and allow you to use master theorem. You'll end up with $S(m) = S(\frac{m}{4}) +1$ – Millardo Peacecraft Feb 23 '14 at 21:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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