How can I prove that $T(n)=\sqrt{n}T(\sqrt{n})+n$ grows as $\Theta(n\log \log{n})$?

Question

How can I prove that if $T(n) = \sqrt{n}T(\sqrt{n}) + n$ then $T(n) = \Theta(n\log\log n)$?

I tried to define $T(n)$ by $G(n)$, prove about $G(n)$, and then to return to $T(n)$, but it's not working..

Start by dividing both sides of the recurrence by $n$ to get $\frac{T(n)}{n}=\frac{T(\sqrt{n})}{\sqrt{n}}+1$. Write $\frac{T(n)}{n}=F(n)$. Now you have a much easier recurrence to work with: $F(n)=F(\sqrt{n})+1$ — mursalin, Apr 02 '18 at 13:09

score 2 · Accepted Answer · answered Apr 02 '18 at 12:55

You can use repeated substitution: $$ \begin{align*} T(n) &= n + \sqrt{n}T(\sqrt{n}) \\ &= n + n + n^{3/4} T(n^{1/4}) \\ &= n + n + n + n^{7/8} T(n^{1/8}) \\ &= \cdots \end{align*} $$ You reach the base case after after $\ell$ iterations when $n^{1/2^\ell} = 2$ (say), whose solution is $\ell = \log_2\log_2 n$. Assuming for simplicity that $T(2) = 0$, we get $T(n) = n\log_2 \log_2 n$.

How can I prove that $T(n)=\sqrt{n}T(\sqrt{n})+n$ grows as $\Theta(n\log \log{n})$?

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