Get the Asymptotic Notation of The Run time

Question

I have the runtime below:

  T(n) = (1- (1/2^n)) ((n+1)/2)

I know its upperbound can be something like : O(2^n)

But I couldn't find its lowerbound or omega. Is it possible that an algorithm has no lower bound? What about theta ? little oh and little omega?

@OmG: technically speaking, you are right. But it is clear that the OP misintepreted the expression. — , Aug 14 '18 at 13:02

score 2 · Answer 1 · answered Aug 14 '18 at 08:00

2

The inverse exponential is negligible in front of 1, so Θ(n), which implies all other bounds.

answered Aug 14 '18 at 08:00

OmG · Answer 2 · 2018-08-14T14:02:28.557

1

As 1- 1/2^n >= 0.5 for all n >= 1, we can say T(n) >= 0.5(n+1)/2 = (n+1) /4. Hence, T(n) = \Omega(n). Also, as 1- 1/2^n < 2 for all n >= 0, T(n) <= n + 1 we can say T(n) = O(n). Therefore, we can say T(n) = \Theta(n).

edited Aug 14 '18 at 14:02

answered Aug 14 '18 at 11:59

OmG

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How did you came up with 0.25 in the equation 0.25(n+1)/2 ? Please englighten me :) – alyssaeliyah Aug 14 '18 at 12:45
Why are you separating (1- (1/2^n)) and ((n+1)/2). They are on the same equation. – alyssaeliyah Aug 14 '18 at 12:47
1

@Eliyah I've split them to simplify the part of `1-1/2^n`. As, this part of the equation can be bounded from its upper and lower, in contrast with `(n+1)/2` which cannot be bounded from its upper. – OmG Aug 14 '18 at 12:59
Okay, I get it but how do you came up with 0.25(n+1)/2? Sorry but please let me understand this :) – alyssaeliyah Aug 14 '18 at 13:05
@Eliyah just replace the minimum value of the `1 - 1/2^n` part. – OmG Aug 14 '18 at 13:10
What about T(n) <= n + 1 ? How did it happen? – alyssaeliyah Aug 14 '18 at 14:12
@Eliyah The same. Instead of the lower bound, upper bound of the specified term is replaced. – OmG Aug 14 '18 at 14:18
So what do you think the upper bound is? – alyssaeliyah Aug 14 '18 at 21:51

Get the Asymptotic Notation of The Run time

2 Answers2