# Questions tagged [landau-notation]

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

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### Finding the constants in Landau notation

I am trying to find the constants $n_0$ and $c$ to show that some given functions belong to the $O(\cdot)$ equivalence class. But, while it seems easy, I am not sure whether I am allowed to do what I ...
• 173
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### Trying to understand the basic about recurrence trees

I have little background on recurrence trees, and I am working on the following exercise: Exercise. Take $T(n) = 2T(n/2) + 3\log(n)$. Draw the recurrence trees for $n=2$ and $n=4$. What can we ...
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1 vote
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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 ...
• 113
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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 $O(n^{2.5})$?
1 vote
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### Summation of asymptotic notation

How can we solve summation of asymptotic notations like given below: $$\sum_{k=1}^{n-1} O(n).$$
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### When are log complexities considered equivalent?

Would we consider $O(\log_2(n))$ to be the same complexity as $O(\log_2(n-1))$? Why or why not? I'm specifically wondering about how the number we take the log of affects the time complexity.
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### Why is $\sum_{i=1}^n O(i)$ not the same as $O(1)+O(2)+\dots+O(n)$?

The well-known textbook Introduction to Algorithms ("CLRS", 3rd edition, chapter 3.1) claims the following: $$\sum_{i=1}^n O(i)$$ is not the same as (I'm not using DNE because the book explicitly ...
1 vote
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### Substitution for Landau's O notation formula

I found the following description when I was reading a paper on computational complexity theory. This can be done ... in time 2n･poly(logs,n)+2O(logs)c. For s≤2no(1), the runtime is 2n･poly(n). I ...
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### which rule can conduct this formula $\log n = O(n^{0.000001})$? [duplicate]

i am learning this post about Big O, which gives this formula $$\log n = O(n^{0.000001})$$ why is that?
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### Run time of pseudo code in big theta notation [duplicate]

I am looking for the run time of the following pseudo code. ...
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### How can i prove this asymptotic comparison? [duplicate]

This is an exercise that's part of my assignment, but it is optional and flagged as a "challenge". I would like to discuss its solution: Prove that: $$27\log{n} + \sqrt{n} = \theta(\sqrt{n})$$ ...
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### If my algorithm has complexity O(n!*n), can I just write O(n!), or do I have to keep it like O(n!*n)?

Just as I asked in the title: if my algorithm has complexity $O(n!\times n)$, can I just write $O(n!)$, or I have to keep it like $O(n!\times n)$?
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### O(·) is not a function, so how can a function be equal to it?

I totally understand what big $O$ notation means. My issue is when we say $T(n)=O(f(n))$ , where $T(n)$ is running time of an algorithm on input of size $n$. I understand semantics of it. But $T(n)$ ...
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### How come O(n) + O(logn) = O(logn)

How come O(n) + O(logn) = O(logn)? When talking for example about an algorithm that has two operations. One of them takes O(n) and the other O(logn) and in the end we say that the total complexity is ...
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### Is this a valid use of big-O notation?

Suppose that $m=O(n^{c+1/2})$ for some real $c>0$ and $x=O(\sqrt{\log m})$. Are the following two computations valid? I understand that I'm abusing notations a bit to get at the desired results. ...
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1 vote
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### Introductory explanation of the Big-Oh properties

I've noticed that Big-Oh notation actually has some properties such as summation, product but i couldn't find an introductory explanation for their use or how they can help to solve asymptotic ...
• 173
1 vote
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### How to prove $\Theta(g(n))\cup o(g(n))\ne O(g(n))$

How to prove $\Theta(g(n))\cup o(g(n))\ne O(g(n))$ ? Is there a simple example for understanding? Seems there's a gap between $O(g(n))- \Theta(g(n))$ and $o(g(n))$ just from the definition. But I ...
1 vote
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### Is this correct in term of big-oh notation: given $g = O(f)$ and $h = O(f)$ can we say $g = O(h)$?

We have two equations $g = O(f)$ and $h = O(f)$ , then can we derive that $g = O(h)$. I came up with following proof but i dont know it's correct or not. $$g = O(f)$$ $$g \le c_1*f$$ $$h \le c_2*f$$ ...
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### Asymptotic Notation Analysis

2^n=O(3^n) : This is true or it is false if n>=0 or if n>=1 since 2^n may or not be element of O(3^n) I need a hint to figure the problem