Arselection the adhering to functions in enhancing order of development rate (through g(n) following f(n) in your list if and just if f(n)=O(g(n))).

You are watching: Arrange the following functions in increasing order of growth rate

a)2^log(n) b)2^2log(n) c)n^5/2 d)2^n^2 e)n^2 log(n) So i think answer is in boosting order isCEDABis it correct? i have confusion in choice A and also B. i think choice A must be at first location.. less one i expect so please assist just how to fix this.This question I confronted in algorithm course component 1 assignment (Coursera) .

Firstly, any positive power of n is always better than log n, so E comes before C, not after.

Also, D comes after eincredibly various other feature, as either interpretation of 2^n^2 (could be 2^(n^2) or (2^n)^2 = 2^(2n); I could be wrong in ignoring BIDMAS though...) are exponentials of n itself.

Taking log to be base a, some arbitrary constant:

a)

b)

Thus, unfortunately, the actual order depends of the worth of a, e.g. if the value of

is higher than 2, then A comes after E, otherwise before. Curiously the base of the log term in E is irpertinent (it still maintains its place).

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answered Jul 17 "16 at 16:52
user3235832user3235832
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The answer is aecbd

The most basic method to check out why is to develop a table with different worths of n and also compare amongst them. But some intuition:

a grows lesser than any kind of others, specially c because of the log term in the power as opposed to the term itself

e is a via a n**2 term multiplied in, which is better than it being in an exponent

b is a dual exponent, but still much better than a quadratic power

d is the noticeable worst bereason it grows tremendously with a quadratic power!

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answered May 16 "18 at 10:43

WboyWboy
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