Why can the third case of master theorem not be applied here T(n)=2T(n/2)+nlgn

Question

Why is nlgn polynomially larger than n when: "Polynomially larger" means that the ratio of the functions falls between two polynomials, asymptotically

Here n^0.1 < log n < n^0.4 , so nlgn should be polynomially larger than n and hence we should be able to apply third case of Master Theorem in it?

Here is the graph plot of y=n^0.1,y=log n and y=n^0.4 https://www.desmos.com/calculator/vjq0j1ri3f

You really need to fix the formatting; this is unreadable as-is. — chepner, Apr 05 '18 at 17:26

score 1 · Answer 1 · answered Apr 05 '18 at 20:58

Here n^0.1 < log n < n^0.4 , so nlgn should be polynomially larger than n ...

Where did you get this idea?

See this Math SE post. Therefore, log n is less than any positive power polynomial; hence case 2 of the Master Theorem applies here, not 3. Proceed and you obtain T(n) = Θ(n (log n)^2).

Why can the third case of master theorem not be applied here T(n)=2T(n/2)+nlgn

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