Solve Recurrence Equation Problem

Asked Aug 19 '14 at 08:39

Active Aug 19 '14 at 11:19

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How we calculate the answer of following recurrence?

$$T(n)=4T\left(\frac{\sqrt{n}}{3}\right)+ \log^2n\,.$$

Any nice solution would be highly appreciated.

My solution is to substitute $n=3^m$, giving $$T(3^m)=4T\left(\frac{3^{m/2}}{3}\right)+\log^2 3^m = F(m)=4F((m/2)-1)+m^2=O(m^2logm)= O(\log^2 n \log n \log n)\,.$$

edited Aug 19 '14 at 11:19

asked Aug 19 '14 at 08:39

Mina Simin

1

Give base case ? – hanugm Aug 19 '14 at 09:37
And is this discrete? Should it be, say, $\lfloor \tfrac13\sqrt{n}\rfloor,$? – David Richerby Aug 19 '14 at 10:13
Dear @hanu, we have no base case. – Mina Simin Aug 19 '14 at 10:13
Dear @DavidRicherby, just i know the answer is $O(log^2 n)$ – Mina Simin Aug 19 '14 at 10:14
2

@DavidRicherby The reference question does not explain how to deal with the $\sqrt{n}$. This question may be a better target. – FrankW Aug 19 '14 at 10:37
Dear @FrankW, i see this link, but i dont underestand how this one is solved. in my answer sheet just mentioned the final answer, not detail answer. – Mina Simin Aug 19 '14 at 10:50
@MinaSimin Note that $\log^2 n \log n \log n = \log^4 n$. – David Richerby Aug 19 '14 at 11:04
@MinaSimin How do you arrive at that $O$-term? – FrankW Aug 19 '14 at 11:10
have you taken a look here Akra Bazzi method? – Nikos M. Aug 19 '14 at 12:38
yes, but it's not related in my thoutgh. – Mina Simin Aug 19 '14 at 13:00
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The last line in your post should be $=O(\log^2n\log\log n)$. – Rick Decker Aug 19 '14 at 14:58
@DavidRicherby Careful, some people read $\log^2 n = \log(\log n)$. Splitting the terms hints in this direction, imho. – Raphael Aug 19 '14 at 17:27
@hanu The base case is often dropped if the implicit assumption is that we are only after $\Theta$ of the solution; then, the anchors are just "some constant" since they don't impact the result. For similar reasons, the recurrence is given as sloppily as it is (not really over naturals). – Raphael Aug 19 '14 at 17:31

Solve Recurrence Equation Problem

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