I need to prove $$ 2n^2+5n+20log(n)=O(n^2) $$
I know that the right needs change term to the greatest degree I got: $$ 2n^2+5n+20log(n)≤2n^2+5n^2+... $$
I want to how to deal with the log
Asked
Active
Viewed 274 times
0
-
2simply log(n) < n – Anatoliy R Dec 03 '19 at 20:55
-
Note that $\ln(x)\le x-1$ – Maximilian Janisch Dec 03 '19 at 20:56
1 Answers
0
Perhaps a (logarithmic) graph will help:
David G. Stork
- 29,774
-
That means I can replace $20logn$ by $n^2$ or directly replace $20logn$ to $n^2$ – Antonymous Dec 03 '19 at 21:04
-
-
No. Given that only the largest asymptotic term is relevant, my graph reveals that the logarithmic term will always be smaller than the $n^2$ term. Hence the logarithm can be ignored when calculating the order. – David G. Stork Dec 03 '19 at 21:13