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I have studied the following recurrence. The ratio between f(n) and n^log_b(a) is log n so there is non polynomial difference but I have studied from book that it can be solved by master method.

$T (n) = 2T (n/2) + n log n$

On the other hand ratio in $T (n) = 2T (n/2) + n /log n$ is $1/log n$ so this also has non polynomial difference but it can not be solved by master method. Because of this, I am having confusion in recognizing recurrences which can be solved by master method. I know what Master method is and how it works. I just want to understand why both of above recurrences cannot be solved by master theorem.

user4129542
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  • Check out our reference question; this recurrence can clearly be solved with the Master theorem. Hint: "polynomial difference" means "polynomially bounded difference". here. – Raphael Mar 06 '15 at 17:56
  • The reference answer does not answer which recurrences can b solved by master theorem neither does it explain the polynomial difference. It only explains the method which I already know. Kindly unmark it as a duplicate. – user4129542 Mar 07 '15 at 03:00
  • The formulation of the Master theorem itself specified the set of recurrences it is applicable to. I don't know what you mean by "(non-)polynomial difference". The Master theorem applies to your recurrences, so your last sentence is not a meaningful question. Maybe you need help with determining the relative asymptotic growth of two functions? – Raphael Mar 07 '15 at 19:13
  • I've confirmed from an authenticated source that my second recurrence cannot be solved by master theorem. If you're sure that my second recurrence can be solved by master theorem then show me how else unmark it as duplicate. – user4129542 Mar 08 '15 at 05:12
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    I'm sorry; I forgot the condition that $k \geq 0$ in case 2. So the second can indeed not be "solved" with the Master theorem. But there is no insight here: a recurrence for which no case applies can not be "solved". Period. Since the question I marked this a duplicate of lists all conditions, the question has already been answered. That's the meaning of duplicate. (And I also gave you a reference helping you with checking the conditions.) – Raphael Mar 08 '15 at 07:54

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