I need help to find asymptotic solution of this recurrence equation $T(n)=\sqrt{n}T(\sqrt{n})+cn$ where $c$ is constant.
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Here are two thoughts, either one may help. $$\begin{split} T(n) &= n^{1/2} T\left(n^{1/2}\right) + cn \\ &= n^{1/2} \left[ n^{1/4} T\left(n^{1/4}\right)+cn^{1/2} \right] + cn\\ &= n^{1/2+1/4} T\left(n^{1/4}\right) + cn^{1/2+1/2} + cn\\ &= n^{1/2+1/4} \left[ n^{1/8} T\left(n^{1/8}\right)+cn^{1/4} \right] + 2cn\\ &= n^{1/2+1/4+1/8} T\left(n^{1/8}\right) + 3cn \\ &= \ldots \\ &= n^{1-1/2^k} T\left(n^{1/2^k}\right) + kcn \\ &= \ldots \text{ let } k = \log \log n \text{ so } 2^k = \log n \text{ and } n^{2^{-k}} = n^{1/\log n} = 2 \ldots\\ &= \frac{n}{2} T\left(2\right) + cn\log \log n \\ &= \Theta (n \log \log n) \end{split}$$
A different approach is to divide by $n$: $$ \frac{T(n)}{n} = \frac{T \left( \sqrt{n} \right)}{\sqrt{n}}+c $$ and changing variables to $S(n) = T(n)/n$ yields $S(n) = S\left(n^{1/2}\right)+c$, which should be easier to solve that the one above directly.
gt6989b
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I do not understand your step from $ n^{1/2+1/4+1/8} \dots$ to $n^{2-1/2^k},;$ I would expect $n^{1-1/2^k}$. And then IMO the rest would break down. – gammatester Feb 25 '14 at 15:00
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@gammatester thank you, found and fixed the typo. – gt6989b Feb 25 '14 at 18:39
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The change of variable $$S(k)=2^{-2^k}T(2^{2^k})\implies S(k)=S(k-1)+c\implies S(k)=\Theta(k)$$ suggests (but cannot prove) that $$ T(n)=\Theta(n\log\log n). $$
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