user14969315
Reputation:
Is the following true: log(k!)=O(k)
Is it correct to claim that log(k!)=O(k)? I searched for a proof but didn't find any on the web.
Upvotes: 0
Views: 152
Answers (1)
Zvi Mints
Reputation: 1142
No, By Striling's approximation
k! ~ (k/e)^k * sqrt2 of (2*phi*k)
Therefore its Omega of (klogk) which is not O(k) See the following:
Left side:
log(n!) = ...
Note: Sorry for this type of writing, still learning Stackoverflow math writing...
Upvotes: 3
Related Questions
- Which one is bigger (loglogn)! or (log(n!))?
- Is log(n!) = Θ(n·log(n))?
- Why is O(log(n)) coming equal to O(log(n!))?
- What is the performance of log(n!) compared to n?
- is log(n!) equivalent to nlogn? (Big O notation)
- Is log(n!) = O((log(n))^2)?
- Is O((log n)!) polynomial?
- Comparing O((logn)!) and O(2^n)
- Is O(log(n*log n) can be considered as O(log n)
- why log(n)! isn't zero?