I have a probably simple big $\mathcal{O}$ question.
Is the following statement correct?
$$\mathcal{O}(x \log x)=\mathcal{O}(\sqrt x \log x)$$
why?
2026-04-06 22:40:21.1775515221
A probably simple big $\mathcal{O}$ question
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No. $x\log x\in O(x\log x)$ (that shold be obvious), but assume that $x\log x\in O(\sqrt{x}\log x)$, then there exist $N$ and $y$ such that for $x>y$: $$x\log x < N\sqrt{x}\log x$$ but that means that for $x>y$: $$x < N\sqrt{x}$$ but then choose $x>N^2$.