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Concerning the Master Theorem. I have found the following equation as the base of analysis:

$\quad T(n) = aT(n/b) + \Theta(n^k)$

but I also found the following:

$\quad T(n) = aT(n/b) + \Theta(n^k\cdot\log_p n)$

where the base $p$ is a real number.

Can anyone explain the second equation? I understand the proof with the first equation but can not understand the second formula.

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    $\begingroup$ Your question is not clear? Where do you have "found" the formulas? What proof? $\endgroup$
    – A.Schulz
    Jan 20 '13 at 11:03
  • $\begingroup$ They are from text-books.By proof I mean how to extract the 3 branches that specify the complexities in Master Theorem $\endgroup$
    – Cratylus
    Jan 20 '13 at 11:06
  • $\begingroup$ Could be related to another similar question. $\endgroup$
    – Dmitri Chubarov
    Jan 20 '13 at 11:28
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    $\begingroup$ If you look up the wikipedia article you could see that the generic form of the second branch is $f(n) = \Theta \left(n^{\mathop{\text{log}}{}_b a} \mathop{\text{log}}{}^k n \right)$ that matches both formulas in your question. $\endgroup$
    – Dmitri Chubarov
    Jan 20 '13 at 13:10
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    $\begingroup$ Without the algorithm being analysed we won't be able to tell you where the $\log$ comes from. $\endgroup$
    – Raphael ♦
    Jan 20 '13 at 15:27
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Sometimes the master theorem is only given for recursions of the form $T(n) = aT(n/b) + \Theta(n^k)$, but the Wikipedia article includes a more general version which can handle functions of the form $T(n) = aT(n/b) + \Theta(n^k(\log n)^l)$, and even more general ones when $k \neq \log_b a$. The similar Akra-Bazzi theorem handles more general situations.

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