My Math Forum Sorting functions asymptotically, solution correct?

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 February 5th, 2019, 12:53 PM #1 Newbie   Joined: Feb 2019 From: Earth Posts: 3 Thanks: 0 Sorting functions asymptotically, solution correct? The task is to sort functions $\displaystyle f_i$ for $\displaystyle i = 1, \dots, 4$ defined as follows: $\displaystyle f_1 := log(n^{3n+1})$ $\displaystyle f_2 := \sum_{i=0}^{2n} 3i$ $\displaystyle f_3 := 4 \cdot {n \choose 2}$ $\displaystyle f_4 := 2n^3$ Ideas: $\displaystyle f_1 = log(n^{3n+1}) = (3n + 1) \cdot log(n) = 3n \cdot log(n) + log(n) \in \Theta(n \cdot log(n))$ $\displaystyle f_2 = \sum_{i=0}^{2n} 3i = 3 \cdot \sum_{i=1}^{2n} i = 3 \cdot ((2n)^2 + 2n) = 6n^2 + 6n \in \Theta(n^2)$ $\displaystyle f_3 = 4 \cdot {n \choose 2} \in \Theta(4 \cdot n^2) = \Theta(n^2)$ [we are allowed to use the fact that $\displaystyle {n \choose k} \in \Theta (n^k)$] $\displaystyle f_4 = 2n^3 \in \Theta(n^3)$ It is true that $\displaystyle n \cdot log(n) \in o(n^2)$ and $\displaystyle n^2 \in o(n^3)$. It follows $\displaystyle o(f_1) \subset o(f_2) = o(f_3) \subset o(f_4)$. I just need to know if I've made any mistakes and if yes a hint Last edited by electron; February 5th, 2019 at 01:07 PM.
 February 5th, 2019, 08:13 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics This looks perfect to me. Thanks from electron
 February 6th, 2019, 03:50 AM #3 Newbie   Joined: Feb 2019 From: Earth Posts: 3 Thanks: 0 Thank you, can be closed.

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