Is -2n a big Oh of n as n goes to infinity?

Question

Based on the definition of the upper asymptotic bound, we have that :

$$|n| < K |-2n| $$

for $ K = 1$ for example and as $ n $ goes to $+\infty$, so the definition would lead us to say that $-2n$ is an asymptotic upper bound of $n$; Which doesn't make sense since $-2n$ is obviously smaller than $n$.

I would say $-2n$ is less than $n$, but not "smaller than" $n$. Anyway, big O only looks at the absolute values. — Robert Israel, Jan 03 '18 at 21:46
Please, if you are ok, you can accept the answer and set it as solved. Thanks! — user, Feb 02 '18 at 21:37

score 1 · Accepted Answer · answered Jan 03 '18 at 21:47

1

What counts is the $\color{red}{magnitude}$ of a general term, as $n\to \infty$, $$ O(-2n)=O(|-2n|)=O(2n)=O(n), $$ we are considering absolute values here.

answered Jan 03 '18 at 21:47

Olivier Oloa

120,989

user · Answer 2 · 2018-01-04T13:05:11.510

1

By definition

$$ \frac{f(n)}{n}\to k\in \mathbb{R}\implies f(n)=O(n)$$

thus $$O(n)=O(-2n) \quad n\to+\infty$$

edited Jan 04 '18 at 13:05

answered Jan 03 '18 at 21:56

user

154,566

Your first line is false. Only the $\Leftarrow$ direction is true. See, e.g., here. – Antonio Vargas Jan 04 '18 at 12:50

score 0 · Answer 3 · answered Jan 03 '18 at 21:46

0

What matters here is not the sign but the "size". That's why the definition uses the absolute value. In that sense $-2n$ is larger than $n$.

answered Jan 03 '18 at 21:46

Dan

382

Is -2n a big Oh of n as n goes to infinity?

3 Answers3