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February 5th, 2019, 12:53 PM   #1
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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.
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February 5th, 2019, 08:13 PM   #2
SDK
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This looks perfect to me.
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February 6th, 2019, 03:50 AM   #3
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Thank you, can be closed.
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