For example if $$ f(x)= \Theta (g(x)) $$

from the definition of the theta notation, there exist c1 and c2 constants such that

$$c_1 g(x) \le f(x) \le c_2 g(x)$$

then if only we took the constants $1/c_1$ and $1/c_2$ we could say from the definition that

$$ g(x)= \Theta (f(x)) $$


  • $\begingroup$ A different way to see it would be that f=O(g) is equivalent to g=Omega(f), and f=Theta(g) is equivalent to both f=O(g) and f=Omega(g). $\endgroup$
    – chirlu
    Sep 15, 2013 at 19:18

1 Answer 1


Right, except that the constants are actually $1/c_2$ and $1/c_1$. That is, $$c_1 g(x) \leq f(x) \leq c_2g(x) \Rightarrow \frac{1}{c_2}f(x) \leq g(x) \leq \frac{1}{c_1}f(x)\,.$$ Also, remember that the inequalities only apply for large enough $x$.

  • 4
    $\begingroup$ This assumes that both functions attain only positive values (in the limit), i.e. both constants are positive (or negative). If you have mixed signs, the calculations do not work. $\endgroup$
    – Raphael
    Sep 16, 2013 at 8:40
  • $\begingroup$ @Raphael, can you explain why the calculations do not work when signs are mixed? $\endgroup$
    – user127304
    Dec 16, 2020 at 18:59
  • $\begingroup$ @Raphael, In CLRS, it's said at the beginning that all the functions are asymptotically positive. At first, I didn't understand why the functions required to be asymptotically positive, now I kind of have thoughts about it. can you explain the reason for it? $\endgroup$
    – user127304
    Dec 16, 2020 at 19:02
  • $\begingroup$ @Imral For the first comment, what have you tried? Do you remember what happens with _in_equalities if you multiply both sides by a negative factor? $\endgroup$
    – Raphael
    Dec 17, 2020 at 22:13
  • 1
    $\begingroup$ @Imral Convenience. It's an introductory textbook; handling only non-negative functions (plus more assumptions, usually) makes many things easier, mathematically speaking. All the terms can be defined in more general cases -- in fact, mathematicians routinely define $O$ on the reals for $x \to x_0 \in \mathbb{R} \cup \{-\infty, \infty\}$. Not only do things get a little more complicated, but all kinds of explicit and implicit "lemmas" used for abusing Landau notation break down in more general settings. $\endgroup$
    – Raphael
    Dec 17, 2020 at 22:17

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