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I tried to use the master theorem to solve $T(n) = 2T(\frac n 2) +\frac n {\log n}$, and I got $\frac n {\log n} \overset{?} = O(n^{1-\epsilon})$, now it looks like it should work:

$\frac n {\log n} \le cn\Rightarrow 1 \le c \log n$ which is true for every $c$, so from the master thorem $T(n)=\Theta (n)$, but I see in my notes that's it's actually not, but why?

shinzou
  • 3,981
  • Let $G(n)=T(n)/n$. Then $G(n)=G(n/2)+\frac 1 {\log n}$ which yields $G(n)\sim H_{\log n}\sim\log\log n$, hence $T(n)\sim n\log\log n\ne \Theta(n)$. – A.S. Feb 17 '16 at 07:48
  • @A.S. Yes I know how to solve it in other ways, but why does the master theorem fails here? – shinzou Feb 17 '16 at 07:49
  • Because the recurrence doesn't satisfy any of its 3 cases: https://en.wikipedia.org/wiki/Master_theorem#Inadmissible_equations. You need to use a more general https://en.wikipedia.org/wiki/Akra%E2%80%93Bazzi_method instead. – A.S. Feb 17 '16 at 07:51
  • I see, thank you. @A.S. – shinzou Feb 17 '16 at 07:59
  • Several different approaches to this recurrence appeared at this MSE link. – Marko Riedel Feb 17 '16 at 21:41

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