# Recurrence relation

The recurrence relation $T(n)=4T\left(\frac{n}{2}\right)+n^2$ describes the running time of an algorithm $A$. Competing algorithm $A'$ has the running time of $T'=aT'\left(\frac{n}{4}\right)+n^2$. What is the largest value of $a$ such that $A'$ is asymptotically faster than $A$.

Please provide the solution steps also.

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What did you try? Which similar problems can you solve? –  Did Feb 6 '13 at 17:15
Can you get an asymptotic estimate of $T(n)$? If so, how did you do it? That will help us to know how much and how little detail the answer needs to have to be helpful to you. –  Rick Decker Feb 6 '13 at 18:01