# Questions tagged [landau-notation]

Questions about asymptotic notations such as Big-O, Omega, etc.

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### Asymptotic growth of a function containing a sum

How to compare the asymptotic growth of a function containing a sum with another function? I'm not sure how I'm supposed to dissolve the sum. Usually I just take the limis of f(x)/g(x). If that fails ...
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### Is this a correct way to thing about asymptotic notations?

I am reading a book on algorithms. It says that $2n^2+3n+1=2n+\Theta(n)$. For a person like me who has studied some set theory but not from axioms, this notation seems a bit insane. I was wondering ...
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### How do I simplify $O\left({n^2}/{\log{\frac{n(n+1)}{2}}}\right)$

I'm not very certain about how to deal with asymptotics when they are in the denominator. For $$O\left(\frac{n^2}{\log{\frac{n(n+1)}{2}}}\right)$$, my intuition tells me that it should be treated in a ...
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### Is O((n^2)*log(n)) greater than O(n^(2.5))?

I know that $O(n^2\times \log(n))$ is greater than $O(n^2)$, but is $O(n^2\times \log(n))$ greater than $O(n^{2.5})$?
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### Justification for neglecting constant factors in Big O

Many a times if the complexities are having constants such as 3n, we neglect this constant and say O(n) and not O(3n). I am unable to understand how can we neglect such three fold change? Some thing ...
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### What is wrong with this solution for $\mathcal{O}({\log({n \choose \frac{n}{2}})})$?

In this recitation on MIT OCW, the instructor uses Stirling's approximation to calculate that $\mathcal{O}({\log({n \choose \frac{n}{2}})}) = \mathcal{O}(n)$. However, I went through the following ...
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### How can a quadratic algorithm be faster than a linearithmic one?

I have to solve the following problem: Al and Bob are arguing about their algorithms. Al claims his $O(n\log n)$ time method is always faster than Bob’s $O(n^2)$ time method. To settle the issue, ...
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### Asymptotic relation between n! and (n+1)!

I am having difficulty writing this formally. I know that by L'Hospital's rule we can reduce it to $\lim_{n \to \infty} \frac{n+1}{n}$ which is a constant and hence $n = \theta (n+1)!$. But I am not ...