0

So I think this falls under case 1...but I am not sure how to formally prove that $n^{lg_{10} 6}$ is asymptotically greater than $\lg^2 n$. I tried using L'hopitals but kinda reached a dead end. Here's what i have so far

$$\lim_{n \rightarrow \infty} \frac{d(\lg^2 n)}{d(n^{\lg_{10}6})} = \frac{2 \lg n (\frac{1}{n})}{(\lg_{10}6) n ^{(\lg_{10}6) - 1}}$$

Is there some better way I can think about proving case 1?

Raphael
  • 72,336
  • 29
  • 179
  • 389
wieiooof
  • 71
  • 1
  • 6

1 Answers1

-1

Yuval Filmus pretty much showed the way in the comment above. I will just elaborate on his approach.

First, we raise both to the power lg610.

(nlg106)lg610 = n(lg106 x lg610) = n

and,

(lg2n)lg610 = (lg n)2 x lg610 = (lg n)2.57

Now, taking log of both of these, we get lg(n) and 2.57xlg(lg(n)) respectively.

lg(n) is asymptotically greater than lg(lg(n)), and thus, our case is proved.

Sajid Hasan Apon
  • 14
  • 1
  • 1
    Anyone who's unwilling to just accept that polynomial dominates polylog should also be unwilling to just accept that log dominates log-log. Essentially, you've just done some algebraic manipulations and asserted a fact that's essentially equivalent to the thing you're trying to prove. – David Richerby Feb 11 '18 at 12:36
  • @David Richerby, it's just as you said. But maybe the questioner has just started learning about time complexities? And, maybe s/he found my algebraic manipulations a bit easier to understand and that's why accepted the answer? Also, "log > log-log" is easier to grasp I think. – Sajid Hasan Apon Feb 11 '18 at 19:07