I was just wondering if anyone can help me understand this problem and its solution:
Note that: lg $n$ = $\log_2 n$
The recurrence is: $T(n) = 7T(n/2)+ n^3\text{lg } n$
Then the solution given to us by our professor is: a = 7, b = 2, and $f(n) = n^3 \text{lg } n.$ Now, $\log_b a = \log_2 7 < 2.81. $
Thus $f(n) = Ω(n^3 ) = Ω(n^{\log_27+ε} )$ applies with ε = 0.1.
We also have $af(n/b) = 7(n^3/8)(\text{lg } n−1) < (7/8)f(n)$. Hence case 3 of the master theorem applies. Thus $T(n) = Θ(n^3 \text{lg } n)$.
Where did the $ε = 0.1. $ came from? and what does $\log_b a = \log_2 7 < 2.81$ mean? Isn't it supposed to be $\log_b a = \log_2 7 = 2.81? $