1
$\begingroup$

Given recurrence relation $T(n)=8T(n/6)+n \log n$, I get that the running time of the leaves should be $n^{\log_6 8}$ and $f(n)$ should be $n \log n$, but how can I know which one is bigger ?

$\endgroup$
1
$\begingroup$

$n^{\log_6 8} = n^c$ for some $c>1$ (you can tell that $\log_6 8 > 1$ from the fact that $8>6$). Then $n^c = \omega(n \log n)$.

Indeed: $$\lim_{n \to +\infty} \frac{n^c}{n \log n} = \lim_{n \to +\infty} \frac{n^{c-1}}{\log n} = \lim_{n \to +\infty} \frac{(c-1)n^{c-2}}{1/n} = \lim_{n \to +\infty} (c-1)n^{c-1} = +\infty.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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