Questions tagged [asymptotics]

For questions involving asymptotic analysis, including function growth, Big-$O$, Big-$\Omega$ and Big-$\Theta$ notations.

Questions involving asymptotic analysis, including function growth, Big-$O$, Big-$\Omega$ and Big-$\Theta$ notations.

  • $f(x) = O(g(x))$ as $x \to \infty$ is used to mean that for sufficiently large values of $x$, we have $|f(x)| \leq A g(x)$ for some constant $A$.

  • $f(x)=\Omega(g(x))$ is equivalent to saying that $g(x)=O(f(x))$.

  • $f(x)=\Theta(g(x))$ is used to mean that $f(x)=O(g(x))$ and that $f(x)=\Omega(g(x))$.

9469 questions
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Hello. I need to show that $\sqrt n$ grows faster than $(\log n)^{100}$

Is there an easy way to show that $$\lim_{n\to \infty}\frac {(\log n)^{100}}{\sqrt n}=0 $$
Jack
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3 answers

Proving $\log^3n=O(n)$

I am a little confused as in how to prove this: $$\log^3n=O(n)$$ Any tips/hints on how to proceed? I have tried taking logarithms or exponentiating both sides but nothing seems to help. EDIT: By $\log^3n$, I mean $(\log n)^3$.
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Asymptotic for $(1-\frac{1}{\sqrt{n}})^n$

I'm not sure but, I guess (from my book) an asymptotic approximation of $(1-\frac{1}{\sqrt{n}})^n$ would be $$(1-\frac{1}{\sqrt{n}})^n \space\space \to \space\space e^{-\sqrt{n}-\frac{1}{2} + O(n^{-1/2})}$$ as n increases. Can you explain to me…
David Lee
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1 answer

How is $(\log n)!$ a member of $\theta (\log n \log\log n)$?

Can you please explain how $(\log n)! = \theta((\log n)*(\log \log n))$? I myself used the equation $n! = \theta(n^n)$ and replaced $\log n$ for $n$ as below $(\log n)! = \theta((\log(n))^{\log(n)})$. But it is different. Why shouldn't we use this…
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1 answer

Asymptotic notation question

Arrange the following functions in increasing asymptotic order:
sittian
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