# Master's Theorem?

When the ratio is $1$, why does the efficiency of the algorithm evaluate to $\mathcal{O} \left( n^d \log n \right)$?

The total work done would be:

$$T(n) = \mathcal{O}(n^d) (1+1+\cdots+1^k)$$

$$= \mathcal{O}(n^d) (1)(\log_b n)$$

where $n$ is the size of the problem, $d$ the efficiency exponent, $k$ the height of the tree and $b$ the factor the size of the problem is reduced by at each iteration

Therefore wouldn't the efficiency be $\mathcal{O}(n^d) \log_b n$ as opposed to $\mathcal{O}(n^d \log n)$?

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 The title is misleading, it seems the OP is concerned with (some misconceptions concerning) the big-O notation. – Did May 10 '12 at 5:54 The Master theorem is a theory regarding big-O notation... My question is regarding a conclusion of the Master's theorem making the title valid. – user26649 May 10 '12 at 11:45 No: The MT is a list of recipes to deduce the asymptotic behaviour of a sequence from a recursion that the sequence satisfies. Your question assumes this step is done and is concerned with the result produced by the MT. As I said, the question is internal to the big-O notation. (But this is no big deal.) – Did May 10 '12 at 12:41

$$\log_b(n) = \log_b(e) \times \log_e(n)$$ Hence, it doesn't matter in the big-$\mathcal{O}$ notation.
This might be a stupid question, but what's $e$? Is it some constant in the efficiency equation or is it Euler's number $e$? – user26649 May 10 '12 at 3:27
@FarhadYusufali $e$ is the Euler's constant. I should have mentioned it in the post i.e. $\log_e(n)$ is the natural logarithm of $n$. – user17762 May 10 '12 at 3:28
You can replace e above with any constant to get any kind of log you like. $\log_b(n)=\log_b(7)\times\log_7(n),$ for example. – Charles May 10 '12 at 13:27