<p>This platform is not a place to get your homework solved by the kind people on the internet. Just keep that in mind for your future posts.</p>
<p>Getting down to the question, I'll leave you with a small hint. Big-O of f(n) denotes a class of functions that have a higher rate of growth than f(n) after a certain point. Hence, O(n^2) includes n, nlogn, n^2 and many more! It should be easy for you to get to the solution now.</p>

CC BY-SA 4.0
2020-09-14T18:29:49.053Z
$O(n^2)$ does not contain $n^3$.
Christian Komusiewicz
CC BY-SA 4.0
2020-09-14T18:32:26.950Z
@ChristianKomusiewicz thanks for pointing it out. It was a stupid mistake on my part.
Kairav Shah
CC BY-SA 4.0
2020-09-13T21:26:43.443Z
2020-09-14T18:31:27.727Z
2020-09-14T18:31:27.727Z
Kairav Shah
<p>Why is O(f(n)) − O(f(n)) not equal to 0? Full disclosure this is a question from practice problems for my theory class.</p>

<time-complexity>
Big O question concerning time complexity
Erik
# Big O question concerning time complexity

(Asked by: username_0 on 2020-09-13)

Why is O(f(n)) − O(f(n)) not equal to 0? Full disclosure this is a question from practice problems for my theory class.

**Tags:** time-complexity

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## 1 answers

### Answer by username_1 on 2020-09-13. Score: 1

This platform is not a place to get your homework solved by the kind people on the internet. Just keep that in mind for your future posts.

Getting down to the question, I'll leave you with a small hint. Big-O of f(n) denotes a class of functions that have a higher rate of growth than f(n) after a certain point. Hence, O(n^2) includes n, nlogn, n^2 and many more! It should be easy for you to get to the solution now.

> **Comments:**
> **username_2**: $O(n^2)$ does not contain $n^3$.  
> **username_1**: @ChristianKomusiewicz thanks for pointing it out. It was a stupid mistake on my part.  

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