Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

$$T (n) = 3T (n/5) + \log^2 n$$

i am trying to solve this recurrence i have been told that it is master methods case 1 but how do i compare $\log^2n$ with $n^ {\log_5 3}$

share|cite|improve this question
i mean how can i decide weather $log^2n$ = $O(n^{log_5 3})$ or not – Bunny Rabbit Dec 28 '10 at 14:50
up vote 4 down vote accepted

A rule of thumb is that polynomial functions always grow faster than logarithmic functions. E.g. this lecture. So your polynomial function asymptotically dominates your log function, giving you case 1.

If you want to know why powers grow faster than logs, you can put the functions in a ratio and then evaluate $\lim_{n\to \infty} \frac{f(n)}{g(n)}$.

share|cite|improve this answer
polynomials grow faster than logs agreed, but in this case it's the square of the log function how to go about that ? – Bunny Rabbit Dec 28 '10 at 15:46
am i doing something incorrect here ? – Bunny Rabbit Dec 28 '10 at 15:52
Polys grow faster than logs, no matter what power you raise the logs to. You can take the limit of the ratio to prove this, or look at Trevor's answer below where he works through the math. – dsolimano Dec 28 '10 at 16:00
@Bunny Rabbit: Consider $n=e^x$ and the fact that $e^{\varepsilon x}$ grows faster than $x^m$, for any $\varepsilon>0$ (small as you wish) and $m>0$ (large as you wish). – Shai Covo Dec 28 '10 at 16:04
@dsolimano Thanks a lot for your comment it helped a lot. – Bunny Rabbit Dec 28 '10 at 19:55

This comment below makes explicit what the first answer already said, and hopefully removes the confusion regarding polylogs that seems to persist in the OP's mind.

To see why, consider $$\lim_{n \to \infty} \frac{(\log n)^k}{n^\epsilon}$$ for any positive $\epsilon$ and any positive integer $k$ (integrality is not really necessary, but suffices for our purposes).

We have $$\lim_{n \to \infty} \frac{(\log n)^k}{n^\epsilon} = \lim_{n \to \infty} \underbrace{\frac{\log n}{n^{\epsilon/k}}\cdots \frac{\log n}{n^{\epsilon/k}}}_{k-\mathrm{times}}.$$

Thus, it all boils down to proving that $$\lim_{n \to \infty} \frac{\log n}{n^\gamma} = 0,$$ where $\gamma > 0$, and we already know how to prove this.

So, I hope this provides the clarification the OP sought.

share|cite|improve this answer

So we want to show that $T(n) = \Theta(n^{\log_{b}a})$ and $f(n) = O(n^{\log_{b}(a)-\epsilon})$. So $a=3, b=5, f(n) = \log^{2} n$, and $\log_{5} 3 = 0.6826$. So we check the following:

  • $f(n) = \log^{2} n = O(n^{0.6826- \epsilon})$. So choose any $\epsilon >0$. In other words, $\log^{2} n \in o(n^{0.6826- \epsilon})$. Note the derivative of $(\log n)^2 = \frac{2 \log n}{n}$ and the derivative of $n^{0.6826}$ is $\frac{0.6826}{n^{0.3174}}$.

The result follows.

share|cite|improve this answer
but this is what my question was how do we arrive at the result that $n^{0.6826- \epsilon}$ asymptotically larger than $\log^{2} n$ – Bunny Rabbit Dec 28 '10 at 15:50

There is another closely related recurrence that admits an exact solution. Suppose we have $T(0)=0$ and $T(1)=T(2)=T(3)=T(4)=1$ and for $n\ge 5$ $$T(n) = 3 T(\lfloor n/5 \rfloor) + \lfloor \log_5 n \rfloor^2.$$

Then we can unroll the recurrence to obtain the following exact formula for $n\ge 5$ $$T(n) = 3^{\lfloor \log_5 n \rfloor} + \sum_{j=0}^{\lfloor \log_5 n \rfloor -1} 3^j \times (\lfloor \log_5 n \rfloor - j)^2.$$ Note that this is independent of the actual digits of $n.$

This simplifies to $$\frac{5}{2} 3^{\lfloor \log_5 n \rfloor} - \frac{1}{2} \lfloor \log_5 n \rfloor^2 - \frac{3}{2} \lfloor \log_5 n \rfloor - \frac{3}{2}.$$

Now the first term is obviously the dominant one asymptotically, so that we get the following answer for the complexity of the recurrence:

$$T(n)\in\Theta\left(3^{\lfloor \log_5 n \rfloor}\right) = \Theta\left(3^{\log_5 n}\right) = \Theta\left(5^{\log_5 3 \times \log_5 n}\right) = \Theta\left(n^{\log_5 3}\right).$$

This agrees with what the Master Theorem would say.

This MSE link has a series of similar calculations.

share|cite|improve this answer

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.