=
Note: Conversion is based on the latest values and formulas.
Why is $\\log(n!)$ $O(n\\log n)$? - Mathematics Stack Exchange I thought that $\\log(n!)$ would be $\\Omega(n \\log n )$, but I read somewhere that $\\log(n!) = O(n\\log n)$. Why?
logarithms - why $n^ {\log {\log {n}}}=\log {n}^ {\log {n ... 1 Mar 2021 · I was reading the solution of one exercise on my book on algorithms. What I need to do in short is order some function from fast to slow. This is the link of the solution: site. At some point I get...
Is log n! = Θ(n log n)? - Computer Science Stack Exchange 17 Oct 2015 · 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 Sterli...
How to solve T (n)=2T (√n)+log n with the master theorem? OK but we already have two answers saying "change variables to cm c m, solve that recurrence and substitute to get T(n) = Θ(log n log log n) T (n) = Θ (log n log log n). So what does your answer add? (And note that your form of the answer confusingly uses different bases in the logarithms: the suppressed constant factors in the big-Theta notation let you change all the …
n*log n and n/log n against polynomial running time A general rule is that multiplying (or dividing) by log n will eventually be negligible compared to multiplying (or dividing) by n^f for any f > 0. To show this more clearly, let us consider what happens as n increases.
algorithms - How is $O (\log (\log (n)))$ also $O ( \log n ... 30 May 2015 · How is O(log(log(n))) O (log (log (n))) also O(log n) O (log n)? I have seen this result somewhere with this but I still didn't quite understand how this is true. This would also help me compute Big Omega of the same function.
How to prove $\\log n < n$? - n$? - Mathematics Stack Exchange 17 Sep 2011 · You would say that a O(log N) function grows asymptotically slower than a O(N) function. Note in many cases like comparing f(N) = N and g(N) = log N it will be true over the entire domain of N (equivalent to N> 0).
Is there a property for log (n)/n? - Mathematics Stack Exchange Then it just skipped and say that the answer was n = 43 n = 43. I was wondering if there is some kind of property for log(n)/n log (n) / n I don't know about. Or otherwise, how was this solved? EDIT: This is the exercise Suppose we are comparing implementations of insertion sort and merge sort on the same machine. For inputs of size n, insertion sort runs in 8n2 steps, while …
logarithms - Difference between $\log n$ and $\log^2 n I'm researching the different execution time of various sorting algorithms and I've come across two with similar times, but I'm not sure if they are the same. Is there a difference between log n log n and log2 n log 2 n? EDIT: Follow up question: in terms of complexity , which would be faster, O(log n) O (log n) or O(log2 n) O (log 2 n)?
notation - What is the difference between $\log^2 (n)$, $\log (n)^2 ... 8 Jan 2016 · Log^2 (n) means that it's proportional to the log of the log for a problem of size n. Log (n)^2 means that it's proportional to the square of the log.