From a function in $\Theta(m + n^2)$ and $0 < m < n^2$, We conclude it is in $\Theta(n^2)$. Does a function in $\Theta(m\log n)$ so that $0 < m < n^2$, imply that it is in $\Theta(n^2\log n)$?
Does $\Theta(m \log n)$ and 0 < m < n^2 imply $\Theta(n^2 \log n)$?
-
1@Godot: I suppose that [should actually be $\Theta$, not $\theta$](http://en.wikipedia.org/wiki/Big_O_notation#Family_of_Bachmann.E2.80.93Landau_notations). – 2012-12-26
1 Answers
I assume that, by $\Theta$, you mean this notation.
From $f(m,n) \in \Theta(m \log n)$ and $0 < m < n^2$, we can conclude that $f(m,n) \in O(n^2 \log n)$. We cannot conclude $f(m,n) \in \Theta(n^2 \log n)$, since it's possible that $m \in o(n^2)$. For example, if $m = n$, then $f(m,n) \in \Theta(n \log n)$.
The reason we could conclude from $f(m,n) \in \Theta(m + n^2)$ and $0 < m < n^2$ that $f(m,n) \in \Theta(n^2)$ is that, even if $m \in o(n^2)$, the expression $m + n^2$ still contains an $n^2$ term which will dominate any smaller $m$. That is, we already know that $\Theta(m + n^2) \subset \Omega(n^2)$, and $m < n^2$ implies that $\Theta(m + n^2) \subset O(n^2)$, and thus $\Theta(m + n^2) \subset (\Omega(n^2) \cap O(n^2)) = \Theta(n^2)$.
However, if we know that $m \in \Theta(n^2)$, then we can indeed conclude from $f(m,n) \in \Theta(m \log n)$ that $f(m,n) \in \Theta(n^2 \log n)$.
-
0Yes as you guess it's big theta. Tnx for your clear and complete explanation. I will vote up as soon as i get 15 reputation. – 2012-12-26