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Hello all I am kinda confused on how to solve the above recurrence since the masters theorem would not work with the above equation. Could anyone please help me solving this and showing me the steps? Requirement is to find asymptotically tightest bound.

$$T(n) = 4T(\frac{n}{2})+n!$$

Raphael
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user81271
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  • @OmG Note that "big" LaTeX in titles is not always better than plain text, since it'll have to be rendered in all question lists, and fail to render in emails and RSS readers. – Raphael Dec 04 '17 at 17:34

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"... the masters theorem would not work with the above equation" is not right. From the master theorem, as we know $n! = \Omega(n^3)$, and $4\left(\frac{n}{2}\right)! < 0.9 n!$ ($k = 0.9$) for large enough $n$, so $T(n) = \Theta(n!)$.

To know more about this case, follow this link and find third case of the master theorem.

OmG
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  • An even better bound is $(1+o(1))n!$ (and so $T(n) \sim n!$). We can even improve on this, for example $n! + (1+o(1))4(n/2)!$. And so on. – Yuval Filmus Dec 04 '17 at 18:00