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Questions tagged [asymptotics]

Questions about asymptotic notations and analysis

Asymptotic analysis is a method of describing the limiting behavior. It is used widely in computer science, particularly in analyzing the asymptotic behavior of algorithms and computational processes.

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1451 questions
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"For small values of n, O(n) can be treated as if it's O(1)"

I've heard several times that for sufficiently small values of n, O(n) can be thought about/treated as if it's O(1). Example: The motivation for doing so is based on the incorrect idea that O(1) is always better than O(lg n), is always better than…
rianjs
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What is an asymptotically tight upper bound?

From what I have learned asymptotically tight bound means that it is bound from above and below as in theta notation. But what does asymptotically tight upper bound mean for Big-O notation?
Vivek Kumar
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Difference between the tilde and big-O notations

Robert Sedgewick, at his Algorithms - Part 1 course in Coursera, states that people usually misunderstand the big-O notation when using it to show the order of growth of algorithms. Instead, he advocates for the tilde notation. I understand the…
thyago stall
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Big-O complexity of sqrt(n)

I'm trying to backfill missing CS knowledge and going through the MIT 6.006 course. It asks me to rank functions by asymptotic complexity and I want to understand how they should be reduced rather than just guessing. The question is to reduce this…
SimplGy
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Is an integer variable constant or logarithmic space?

My lecturer says that a variable takes up one memory position in RAM. This is the slide in question: But CLRS (Introduction to Algorithms by Cormen, end of page 23) says an integer is represented by $c\ lg\ n$. Are both statements true? How can…
Bee
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Useful functions between polylogarithmic and polynomial?

I'm wondering if there are any useful functions asymptotically greater than a polylogarithmic function and less than a polynomial function. That is, a function $f(n)$ such that $f(n) = \omega(\log(n)^k)$ for some constant $k > 0$ and $f(n) =…
ryan
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Confused about proof that $\log(n!) = \Theta(n \log n)$

So I was able to show that: $\log(n!) = O(n\log n)$ without any problems. My question is when trying to prove that $\log (n!) = \Omega(n\log n)$. I was able to show that: $$\begin{align*} \log n! &= \log(1 \cdot 2 \cdot 3 \cdots n )\\ &= \log 1 +…
David Velasquez
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3 answers

Show that $\log n = o(n^\epsilon)$

I am trying to understand how to prove that a polynomial will always grow faster than a logarithm. $\log n = o(n^\epsilon), \epsilon>0$ Intuitively, it is obvious, and plugging in a few numbers always yields true, but how can I prove this? Maybe…
Rodman Huey
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Is log n! = Θ(n log n)?

Why is $\log(n!)=\Theta(n\log n)$? I tried: $\log(n!) = \log1 + \dots + \log n \leq n \log n \Rightarrow \log(n!) = O(n \log n)$. But how can we prove $\log(n!) = \Omega(n \log n)$ without Sterling's approximation?
Z.S.CS
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Big Theta Proof on polynomial function

This is not homework. I have the solution but it's not what I'm getting. I know there are multiple solutions to the problem but I want to make sure that I'm not missing anything. The question is as follows: Prove that 2$n^2$ - 4n + 7 = Θ…
Harrison Nguyen
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Is $n^{1/\log \log n} = O(1)$?

Is $n^{1/\log \log n} = O(1)$ ? Suppose that $n^{1/\log \log n} = c$ where $c$ is constant. Taking logs of both sides, $$\frac{1}{\log \log n}\log n = \log c.$$ I am not able to spot an error. Please help
user110834
7
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1 answer

Asymptotics of a function that decreases as n increases

A homework assignment asks me to state the complexity in Big-O notation of the function $$f(n) = 7n – 3n \log n + 100000 $$ I graphed this function and decreases all the way down to zero nearly its entire lifespan. Therefore I concluded that the…
CodyBugstein
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How to determine $10^{\log n}$ and $3n^2$ which grows faster asymptotically?

My think is pretty easy that $10^{\log n} = n$, which is growing slower than $3n^2$. However, many tutorial shows that $3n^2$ ranks before $10^{\log n}$. I'm really confused.
user59001
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Do not understand why log n = O(n^c) (for any c>0)

Can anyone help me understand this equation? $\log (n) = O(n^c)$ (for any $c>0$) Does it mean that $O(\log (n)) < O(n^c)$ (for any $c>0$)? Added: Please also prove that $\log (n) = O(n^c)$ is true.
Xin
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Can subtracting o(1) from the parameter of a function change its Θ-class?

I would like to know if it is possible that two functions $f(n), g(n)$ can exist such that both of the following conditions are met: $g(n) = o(1)$ $f(n-g(n)) \neq \Theta (f(n))$ I though I found $1/n$ and $1/n^2$ but I think I got it all…
wannabe programmer
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