How should this be proved? I know that $f(n)$ is $Θ(n)$ if and only if $f(n)$ is $O(g(n))$ and $f(n)$ is $Ω(g(n))$ and that If $f(n)=\theta(g(n))$ then $\lim_{n \to \infty}\frac{f(n)}{g(n)}$ will be a constant. Whats the better way to solve this? I have to directly find the constants k, C1, and C2


closed as off-topic by kingW3, Alexander Gruber Apr 29 at 1:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ limit of $f(n)/n^4$ as $n $ to infinity is a constant and that's pretty easy to find. $\endgroup$ – kingW3 Apr 21 at 22:26
  • $\begingroup$ I dont think I'm allowed to answer this in terms of limits. I need to use the theorems above in terms of discrete math. $\endgroup$ – ionics Apr 21 at 22:31
  • $\begingroup$ Do you mean that you need to explicitly find constants $c_1, c_2$ and $n_0$ such that $c_1 g(n)\le f(n)\le c_2 g(n)$ for all integers $n > n_0$? $\endgroup$ – Minus One-Twelfth Apr 21 at 22:43
  • $\begingroup$ Yes! Thank you. $\endgroup$ – ionics Apr 21 at 22:49
  • $\begingroup$ You should include the definition you use of the theta function. $\endgroup$ – kingW3 Apr 21 at 23:09

Note that for $n>1$, $$8n^4>5n^4+3n^3>5n^4+3n^3-5>5n^4-2>3n^4$$


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