Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given $H(x) = lg(f(n))$, where $f(n)$ is an asymptotically positive function, is it always true that if $f(n) = \Theta(g(n))$, then

$H(x) = lg(\Theta(g(n)))$

$\Rightarrow H(x) = \Theta(lg(g(n)))$

To illustrate, Is $lg(n + c) = \Theta(lg(n))$ provided that $c > 0$ and $c$ is a big number?

share|cite|improve this question
It is not always true that if $f(n) = \Theta(g(n))$ then $\log f(n) = \Theta(\log g(n))$. For example, take $f(n) = 2$ and $g(n) = 1 + 1/n$. Then $f(n) = \Theta(g(n))$ but $\log f(n) \neq O(\log g(n)) = O(1/n)$. As for your example, $\log(n+c) = \log n + O(1/n)$, so that $\log(n+c) \sim \log n$. – Antonio Vargas Mar 18 '12 at 8:14
Please use \log or \ln. – Did Mar 18 '12 at 12:08

What is true is that if $f(n) = \Theta(g(n))$ with $f, g > 0$ then $\log f(n) = \log g(n) + O(1)$. If in addition either $g(n) \to \infty$ or $g(n) \to 0$ as $n \to \infty$, then $\log f(n) = \Theta(\log g(n))$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.