5n) + O(n\log n)$

Asked Jun 19 '21 at 04:59

Active Jun 19 '21 at 07:50

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I think it will be $T(n)=O(n^2)$.

This is because...
Suppose $T(n) \leq cn^2$
Then $T(n) \leq c\cfrac{13}{25}n^2+O(n\log n) \leq cn^2$
($\because$) $there\ \ is\ \ a\ \ positive\ \ c\ \ such\ \ that\ \ O(n\log n) \leq c\frac{12}{15}n^2 $
Am I right? Thanks.

edited Jun 19 '21 at 07:50

asked Jun 19 '21 at 04:59

alsrbok

Do you mean $T(n)=O(n^2)$? – Kenta S Jun 19 '21 at 05:20
Yes. I made a mistake. I will fix it now. – alsrbok Jun 19 '21 at 07:50

Find a Big-O Notation for following recurrence formula : $T(n) = T(2/5n) + T(3/5n) + O(n\log n)$

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