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I'm trying to solve a recurrence using the master method:

$T(n) = 10*T(n/3) + n^2\log^5n$

I set up the comparison

$n^2\log^5n$ and $n^{\log_3 10}$

and reduced them to:

$\log^5 n$ and $n^{\log_3(10/9)}$

I'm struggling to understand how these functions compare asymptotically.

From one side I understand that in general polynomials grow slower than logarithms, as written in a similar question here: Sorting functions by asymptotic growth.

However, I graphed both of the functions using a graphing calculator and it appeared that the logarithmic function grew faster than the polynomial, even at large n.

Basically, is it really true that $(\log n)^{\alpha} = o(n^\beta)$ for all $\alpha,\beta>0$? If it is, how would you go about rigorously proving it one way or the other?

Raphael
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Alexander Christenson
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    Where was that written in the linked question? ​ ​ –  Jun 06 '17 at 21:58
  • It's like the 9th bullet on Raphaels answer – Alexander Christenson Jun 06 '17 at 22:00
  • I only see 8 bullets in that answer. ​ ​ –  Jun 06 '17 at 22:02
  • Hmmm I'm on my phone so that might be affecting how I view it. It's right after "It is also useful to keep a pool of basic relations you prove once and use often, such as:" – Alexander Christenson Jun 06 '17 at 22:03
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    The thing just after that is the opposite of what you wrote. ​ ​ –  Jun 06 '17 at 22:04
  • That answer is the best you will do; that's why we call it reference question. Plots are not a suitable tool here; neither do they prove anything, nor should they make you uncertain (for the same reasons). See also here. – Raphael Jun 06 '17 at 22:06
  • How so? Maybe I don't fully understand the notation, but what I'm trying to say is that $n^\beta$ is a tight asymptotic upper bound for the logarithmic function – Alexander Christenson Jun 06 '17 at 22:07
  • you wrote "in general polynomials grow slower than logarithms" but the referenced answer states the opposite: "logarithms grow slower than polynomials". So it is really true that $(\log n)^{\alpha} = o(n^\beta)$ and your graphic calculator is misleading you. – miracle173 Jun 07 '17 at 02:29
  • and your question is "How to proof that $(\log n)^{\alpha} = o(n^\beta)$"? – miracle173 Jun 07 '17 at 02:33
  • Look at CLRS 3ed page 57 for a proof. – ryan Jun 07 '17 at 03:07

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