I have a question about the following recurrence relation:

$$T(n) = 27 \cdot T\left(\frac n 3\right ) + n^3 \log n$$

Using the master theorem, will this be

  • $T(n) = \Theta(n^3)$, or
  • $T(n) = \Theta(n^3 \log \log n)$?
  • $\begingroup$ Plug the possible answers in the recurrence, and you'll know. $\endgroup$
    – user16034
    Commented May 18, 2022 at 8:19

2 Answers 2


If a recurrence relation is of the form

$$ T(n)= aT \left( \frac{n}{b} \right) + {n^k}(\log n)^p $$

then, as per the Master Theorem, we have six conditions depending on value of $a,b,k$ and $p$

  • If $\log_b a>k$ : Answer is $\Theta(n^{\log_b a})$
  • If $\log_b a=k$ and $p>1$ : Answer is $\Theta({n^k}({\log n})^{p+1})$
  • If $\log_b a=k$ and $p=1$ : Answer is $\Theta({n^k}\log\log n)$
  • If $\log_b a=k$ and $p<1$ : Answer is $\Theta(n^k)$
  • If $\log_b a<k$ and $p \geq 0$ : Answer is $\Theta({n^k}({\log n})^p)$
  • If $\log_b a<k$ and $p < 0$ : Answer is $\Theta({n^k})$

In any problem, our main motive is to find $a,b,k$ and $p$.

For the problem with

  • $a=27$
  • $b=3$
  • $k=3$
  • $p=1$,

$\log_3 27 = 3$, so $\log_b a = k$, and since $p=1$, the answer is $\Theta({n^k}\log\log n)$

$$\Theta({n^3}\log\log n)$$

  • $\begingroup$ Please don't revert back changes that actually fix math notation. 1. It's $\Theta(\cdot)$ and not $\theta(\cdot)$. 2. It's $\log$ not $log$. 3. Don't use \bigg when \left will do. 4. $p = 1$, not $p = -1$. $\endgroup$
    – Pål GD
    Commented Sep 21, 2021 at 9:39
  • $\begingroup$ Thanks for the information @PålGD. Wasn't aware of these. $\endgroup$ Commented Sep 21, 2021 at 9:48

Quick check:

$$\Theta(n^3)=27\,\Theta\left(\frac{n^3}{27}\right)+n^3\log\log n$$

does not hold.


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