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Given Recurrence Relation $$\mathcal{T}(n) = \sqrt{n}\text{ }\mathcal{T}(\sqrt{n}) + \mathcal{O}(n^{2}) $$

Master Theorem doesn't apply here.

Tried using $n= 2^{k}$, but got stuck at $$\mathcal{T}(2^{k}) = 2^{k/2}\mathcal{T}(2^{k/2}) + \mathcal{O}(2^{2k})$$

Help me in proceeding.

Rohit Singh
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  • Please. Have a look at https://math.stackexchange.com/questions/4395729/guess-on-big-o-of-recurrence-relation-sqrtn-cdot-t-sqrtn-sqrtn/4395846#4395846 – Cesareo May 04 '22 at 12:55
  • Let $T(n)=O(n^a)$, then $T(n)=O(n^\frac{a+1}2)+O(n^2)$, then $a=max(\frac{a+1}2,2)$, then $a=2$. – Ivan Kaznacheyeu May 04 '22 at 13:02

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