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).
Zulfi.
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1Does this answer your question? Sorting functions by asymptotic growth – Luke Mathieson Jan 29 '20 at 12:23
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Sorry I can't understand that stuff. It has very little discussion about lower bounds i.e. Omega notations. – user2994783 Jan 29 '20 at 15:15
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Being a quadratic function, it is both lower and upper bounded by n^2. – Ṃųỻịgǻňạcểơửṩ Jan 29 '20 at 21:52
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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 $n^2$.
Ṃųỻịgǻňạcểơửṩ
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"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. – user2994783 Jan 30 '20 at 03:04
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Often when referring to a function f(n) where n isnt declared we simply write f as the function. Namely f=f(n) – Ṃųỻịgǻňạcểơửṩ Jan 30 '20 at 03:25
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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. – user2994783 Jan 30 '20 at 04:48
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