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discrete mathematics - (logn)^(logn) = n^(log10+logn). WHY ... 15 Jan 2018 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …
Comparison between $n\\log n$ and $n^2$ sorting algorithms 16 May 2016 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …
logarithms - why $n^ {\log {\log {n}}}=\log {n}^ {\log {n ... 1 Mar 2021 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …
algorithms - How is $O (\log (\log (n)))$ also $O ( \log n ... 30 May 2015 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …
n*log n and n/log n against polynomial running time I understand that $\Theta(n)$ is faster than $\Theta(n\log n)$ and slower than $\Theta(n/\log n)$. What is difficult for me to understand is how to actually compare $\Theta(n \log n)$ and …
calculus - How to prove $\log n - Mathematics Stack Exchange 17 Sep 2011 · In your comments you seem ask about this in the context of the big O notation -- e.g., the concept frequently used in computer science that when analyzing an algorithm for …
asymptotics - Why is $\log (n!)$ $O (n\log n)$? - Mathematics … One idea to prove the claim : $$\log n! =\sum_{k=1}^n \log k \leq \sum_{k=1}^n \log n=n\log n$$ The other approach would be :
Is log n! = Θ(n log n)? - Computer Science Stack Exchange 17 Oct 2015 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …
notation - What is the difference between $\log^2(n)$, $\log(n)^2 ... 8 Jan 2016 · |f(n)| ≤ k*|log(n)|, for n>N <=> |f(n)| ≤ k/2*|2log(n)|, for n>N (++) Hence, we can just choose a constant k2=k/2 and it follows from (++) that f(n) is in O(log(n) . Key point: The …
logarithms - Difference between $\log n$ and $\log^2 n $\begingroup$ It means as input size, N grows, time complexity, log^2 N grows much faster than time complexity, log^2 N. So for large input size, N, the algorithm which has time complexity, …