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Is this statement true? $$f(n) \notin o(g(n))~~~\wedge~~~f(n) \notin \omega(g(n))~~\implies f(n)\in\Theta(g(n))$$

I'm thinking that because $f(n)$ is never greater than or less than $cg(n)$, then it must be growing at the same rate? However I don't know how to derive this solution from first principles.

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  • $\begingroup$ what's your definition of those 3 simbols? (If it's what I have in mind, then it's false, since $f(n)=sin (n)$ and $g(n)=1$ are not comparable) $\endgroup$
    – Exodd
    Commented Sep 28, 2015 at 22:25
  • $\begingroup$ what is this $\omega$ thing? $\endgroup$
    – zhw.
    Commented Sep 28, 2015 at 22:47

1 Answer 1

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In general, not every two functions are comparable asymptotically. Consider $f(x) = e^{x\cos(x)}$ and $g(x) = x$.

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