# Solve a recurrence using the master theorem

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.

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No. $(n\lg n)^{1/2} = o(n^{2/3})$. – JeffE Apr 16 '12 at 8:38
There is no question here. – Raphael Apr 16 '12 at 19:00

$f(n) = (n\cdot \lg n)^{1/2}$ 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 $T(n) = \Theta(n^{2/3})$.