I have a recurrence relation which is like the following:

$T(n) = 2T(\frac{n}{2}) + \log_{2}n$

I am using recursion tree method to solve this. And at the end, i came up with the following equation:

$T(n)=(2\log_{2}n)(n-1)-(1\times 2 + 2\times 2^{2} + \ldots + k\times2^{k})$ where $k=\log_{2}n$

I am trying to find a theta notation for this equation. But i cannot find a closed formula for the sum $(1\times 2 + 2\times 2^{2} + \ldots + k\times2^{k})$. How can I find a big theta notation for $T(n)$?


2 Answers 2


This looks like the running time of a divide-and-conquer algorithm. You can apply the Akra-Bazzi method here. The method works for recurrences of the form:

$$T(n)=\sum_i^k a_iT(b_i n+h_i(n))+g(n), \forall n\ge n_0$$

In your case:


$g(n)=\log n$



For this to work, we have to find a number $p$ satisfying $ab^p=1$. That's easy,let $p=1$.

The method tells us that $T(n)=\Theta(n^p(1+\int_1^n\frac{g(x)}{x^{p+1}}dx))$

I computed the integral on WolframAlpha. The answer is $1-\frac{\log n + 1}{n}$.

Therefore, $$T(n)=\Theta(2n-\log n)=\Theta(n)$$

Here is another way of solving this. Let $n=2^m$


So $T(n)=\Theta(n)$

  • 1
    $\begingroup$ Master Theorem applies here and is easier than Akra-Bazzi. Plus if you're willing to use Wolfram Alpha you can get the recurrence solution directly. $\endgroup$
    – SamM
    Mar 7, 2013 at 20:15
  • 1
    $\begingroup$ The first half makes it harder than it has to be, but +1 for the second half. $\endgroup$
    – Joe
    Mar 8, 2013 at 3:45
  • $\begingroup$ @SamM The point is to expose the Akra-Bazzi method. $\endgroup$
    – mrk
    Mar 8, 2013 at 11:37
  • $\begingroup$ @Joe Thanks! Change the order if that makes more sense to you :D $\endgroup$
    – mrk
    Mar 8, 2013 at 12:23

For $\sum_{0 \le r \le k} r \cdot 2^r$, start with: $$ \frac{1 - z^{k + 1}}{1 - z} = \sum_{0 \le r \le k} z^r $$ As this is a polynomial, the following mangling doesn't need any justification: $$ z \frac{d}{dz} \frac{1 - z^{k + 1}}{1 - z} = \sum_{0 \le r \le k} r z^r $$ Doing the operations indicated: $$ \sum_{0 \le r \le k} r z^r = \frac{z^k (k z^2 - (k + 1) z) + z}{(1 - z)^2} $$ Finally: $$ \sum_{0 \le r \le k} r 2^r = (2 k - 2) 2^k + 2 = \Theta(k 2^k) = \Theta(n \log n) $$

  • $\begingroup$ This answer is incorrect; it is $\Theta(n)$. $\endgroup$
    – SamM
    Mar 7, 2013 at 20:17
  • 2
    $\begingroup$ @SamM, Impossible, just the last term is $\Theta(k 2^k) = \Theta(n \log n)$. $\endgroup$
    – vonbrand
    Mar 7, 2013 at 20:26
  • $\begingroup$ Ah, I take it back. You were solving for the term he was subtracting from the original equation, not the recurrence itself. If you substitute back in you get $T(n) = \Theta(n)$. $\endgroup$
    – SamM
    Mar 7, 2013 at 20:31

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