I'm having a hard time trying to understand how to solve this recurrence relation using the Master Method:

$$T(n) = 10T\Big(\frac{n}{2}\Big) + \frac{n^4}{\log(n)}$$

First, we have:

$a = 10,\ b = 2$ so we have $n^{\log_2^{10}} = n^{\lg10}$

Now we need to find how $f(n)$ compares to $n^{\lg10}$.

And here I'm stuck. What do I have to pay attention to in order to know whether I'm dealing with Case 1 (and look for $O(n^{{\lg10} - \varepsilon}))$ or Case 3 (and look for $\Omega(n^{{\lg10} + \varepsilon})) $ of the method?

My attempt:

From what I know $\log(n) < n^\varepsilon\quad \forall \varepsilon > 0$. So

$$\frac{1}{\log(n)} > \frac{1}{n^\varepsilon}$$ $$\frac{n^4}{\log(n)} > \frac{n^4}{n^\varepsilon}$$

$$\implies \frac{n^4}{\log(n)} = \Omega( n^{4-\varepsilon})$$

but I have no idea how to relate this (assuming no mistakes) with one of the two cases. In Case 1 we deal with $O(\cdot)$ and not $\Omega(\cdot)$ while in the 3rd Case the form of the exponents doesn't seem to match.


1 Answer 1


In order to apply Case 3, you have to show that $\frac{n^4}{\log n} = \Omega(n^c)$ for some $c > \log_2 10$, as well as the regularity condition (which holds for functions of the form $n^\alpha \log^\beta n$).

You take it from here. Hint: $4 > \log_2 10$.

  • $\begingroup$ Is it true that $\frac{n}{\log n} = \Theta(n)$? If so, can we say that $f(n) = \Theta(n^4)$? $\endgroup$
    – asd
    Commented Feb 10, 2019 at 16:20
  • $\begingroup$ No, that's actually false. You'll need to do something more subtle. Spend a few hours on it. It's the only way to learn and get better. $\endgroup$ Commented Feb 10, 2019 at 16:20
  • $\begingroup$ I graphed $\frac{n^4}{\log n}$. It looks like outgrows $n^k$ for any $k>1$. That right? Then it seems that all we need is, for sufficiently large $n$, $\Omega(n^4)$. Am I on the right track? $\endgroup$
    – asd
    Commented Feb 10, 2019 at 17:49
  • $\begingroup$ It is just not the case that $\frac{n^4}{\log n} = \Omega(n^4)$. Dividing both sides by $n^4$, you would get the nonsensical $\frac{1}{\log n} = \Omega(1)$. However, as you showed in the post, if $c < 4$ then $\frac{n^4}{\log n} = \Omega(n^c)$. $\endgroup$ Commented Feb 10, 2019 at 17:51

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