Question

Arrange the following functions in increasing growth order : (i)0 (n2) (ii)0 (log n) (iii)0 (2") (iv)0 (n log n)​

Answer 1

so simple you can do it by yourself

Explanation:

0

of 0 vote

a) 22n = O(n)

b) 2n2 = O(n) (or if it's 2^n * 2, then O(2^n), or if it's 2 * n^2, then O(n^2))

c) n2log(n) = O(n log(n)) (or if it's n^2 log (n), then O(n^2 log (n))

d) n = O(n)

e) n2n = O(n^2) (or if it's n^2 * n, then O(n^3)

log(n) = O(n^a) for any a > 0, so n * log (n) grows faster than n, but slower than n^2.

Any sort of exponential function grows faster than any power of n, so 2^n grows the fastest.

To encompass all possibilities, here is the order of growth in increasing order

n, 22n

n * log (n)

n^2, 2n^2

n^2 * log (n)

n^3

Answer 2

omg the original broooooooooooooo