yes, but $n \log^2 n = n \log n \log n,$ not $2n \log n$. That is for $n \log n^2.$

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Bikram
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in Algorithms
Oct 4, 2016

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2 votes

Order the following functions by growth rate :

- $\log n$
- $n/\log n$
- $(3/2)^n$
- $n\log^2 n$
- a. $\log n$ $\quad$ b. $n/\log n$ $\quad$ d. $n\log^2 n$ $\quad$ c. $(3/2)^n$
- d. $n\log^2 n$ $\quad$ c. (3/2)$^n$ $\quad$ a. $\log n$$\quad$ b. $n/log n$
- a. $\log n$ $\quad$ d.$n\log^2 n$ $\quad$ b. $n/\log n$ $\quad$ c. (3/2)$^n$
- a. $\log n$ $\quad$ c. (3/2)$^n$ $\quad$ b. $n/\log n$ $\quad$ d. $n\log^2 n$