This is the recursive formula for which I'm trying to find an asymptotic closed form by the master theorem: $$T(n)=9T(n/27)+(n \cdot \lg(n))^{1/2}$$

I started with $a=9,b=27$ and $f(n)=(n\cdot \lg n)^{1/2}$ for using the master theorem by $n^{\log_b(a)}$, and if so $n^{\log_{27}(9)}=n^{2/3}$ but I don't understand how to play with the $(n\cdot \lg n)^{1/2}$.

I think that the $(n\cdot \lg n)^{1/2}$ is bigger than $n^{2/3}$ but I'm sure I skip here on something.

I think it fits to the third case of the master theorem.

  • 2
    $\begingroup$ No. $(n\lg n)^{1/2} = o(n^{2/3})$. $\endgroup$
    – JeffE
    Commented Apr 16, 2012 at 8:38
  • 2
    $\begingroup$ There is no question here. $\endgroup$
    – Raphael
    Commented Apr 16, 2012 at 19:00

1 Answer 1


$f(n) = (n\cdot \lg n)^{1/2}$ and $n^{\log_b a}=n^{2/3}$, thus $f(n) = O(n^{\log_b a})$ and even $f(n) = O(n^{\log_b a - \epsilon})$ for $\epsilon < 1/6$.

Why? because $$\lim_{n\to\infty} \frac{f(n)}{n^{\log_b a - \epsilon}} = \lim_{n\to\infty}\frac{n^{1/2}\lg^{1/2}n}{n^{2/3-\epsilon}} = \lim_{n\to\infty} \frac{\lg^{1/2}n}{n^{1/6-\epsilon}} =0 \quad\text{for }\epsilon< 1/6$$

Thus case 1 of the Master theorem should apply, and $T(n) = \Theta(n^{2/3})$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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