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.

-

2
No. $(n\lg n)^{1/2} = o(n^{2/3})$. – JeffE Apr 16 '12 at 8:38

2
There is no question here. – Raphael♦ Apr 16 '12 at 19:00

1 Answer

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

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9714753031730652, "perplexity_flag": "head"}