f(n) = 32n^2 + 17n + 1.

The lecture slide says that lower bound can be Omega(n^2) or Omega(n).

Some body please guide me why the lower bound can be Omega (n). i know the upper bound which is O(n^2).


  • 1
    $\begingroup$ Does this answer your question? Sorting functions by asymptotic growth $\endgroup$ Jan 29, 2020 at 12:23
  • $\begingroup$ Sorry I can't understand that stuff. It has very little discussion about lower bounds i.e. Omega notations. $\endgroup$ Jan 29, 2020 at 15:15
  • $\begingroup$ Being a quadratic function, it is both lower and upper bounded by n^2. $\endgroup$ Jan 29, 2020 at 21:52

1 Answer 1


This is a quadratic function; therefore $f$ is big-Theta of $n^2$. Therefore, it is both Big-O and Big-Omega of $n^2$.

However, when a function $f$ is big-Omega of another function $g$, then $f$'s growth is of greater or equal order than $g$. However, since a quadratic grows strictly faster than a linear in the long run, $n^2$ is big-Omega of $n$. Think of as if big-Omega means "greater or equal than".

Also note when $f$ is big-Omega of $g$, then $g$ is big-O of $f$. Obviously, $n$ is big-O of both $n$ and $n^2$.

  • $\begingroup$ "Obviously, n is big-O of both n and n2. " . This does not seem correct. Big-O notation is used for functions not for variables. $\endgroup$ Jan 30, 2020 at 3:04
  • $\begingroup$ Often when referring to a function f(n) where n isnt declared we simply write f as the function. Namely f=f(n) $\endgroup$ Jan 30, 2020 at 3:25
  • $\begingroup$ You are not saying 'f', you are saying 'n' is big-O of both 'n' and 'n^2'. I think its a typo. Please correct it, your last line. $\endgroup$ Jan 30, 2020 at 4:48
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    $\begingroup$ f(n)=n, g(n)=n^2, n in O(f(n)) and O(g(n)) $\endgroup$ Jan 30, 2020 at 4:58

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