I am trying to simplify $f(n) = \frac{n}{\log(n)}$ into a more easily understandable function. Up until now, I got as far as $n\cdot(\left(\log(n)\right)^{-1})$. Is there any way I can further simplify this function? I am thinking it will have something to do with directly representing the inverse-logarithm as something else, but I am blanking out on what that something is. Please remind me! Thank you!
EditI wanted to put it in a form which would make the big O of $\frac{n}{\log(n)}$ easier to see
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$\begingroup$$\frac n{\log(n)}$ seems quite simple to me. Your alternate, $n(\log(n))^{-1}$ is equivalent, but I don't find it simpler. The proper form is determined by what is useful in following calculation, or in the eye of the beholder if it is the final answer. I would stay with the first.
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