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

I am currently trying to learn the meaning of the Landau Symbols a bit better by solving exercices, namely the following three:

  1. $f_1(x) + f_2(x) = O(g_1(x) + g_2(x))$ given that for each f(x) the property already holds

  2. $n!n^{s} = o(n^n)$ for $n \rightarrow \infty, s \geq 0$

  3. For all $\epsilon > 0: 2^{n+\epsilon} = O(2^n)$ but $ 2^{n(n+\epsilon)} \neq O(2^n)$

where are the Landau-Symbol's Definitions are as follows: For $x \rightarrow a \ f(x) = O(g(x))$ when $\limsup_{x \rightarrow a} f(x)/g(x) < \infty$ and for $x \rightarrow a \ f(x) = o(g(x))$ if $\lim_{x \rightarrow a} f(x)/g(x) = 0$

I have no experience at all with the Landau Notations yet and would be quite glad for hints and help

share|cite|improve this question
Work with this definition of the big O notation, it will make it slightly easier. And it will be true even if $g$ vanishes in every punctured neighborhood of $a$. – 1015 Mar 11 '13 at 19:55
Just apply the definitions you gave, hint for 2. you can use the Stirling's approximation's_approximation – user63181 Mar 11 '13 at 19:58
Thank you for the hint, but I would prefer to work with the definition we were given in class.. – TestGuest Mar 11 '13 at 19:58
Presumably, you are restricted to positive functions with these definitions. The first part uses that if $c,d>0$ then $\frac{a+b}{c+d}$ is in the closed interval between $\frac{a}{c}$ and $\frac{b}{d}$. – Thomas Andrews Mar 11 '13 at 19:58
I actually was thinking about Stirling, but then thought maybe it is rather Jensen..still I got quite stuck at some point – TestGuest Mar 11 '13 at 20:03

Your Answer


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

Browse other questions tagged or ask your own question.