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!

Edit I wanted to put it in a form which would make the big O of $\frac{n}{\log(n)}$ easier to see

  • $\begingroup$ More understandable? In what sense? And the inverse of the logarithm is the exponential. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 4 '14 at 15:59
  • $\begingroup$ The functional inverse of the logarithm is the exponential. The multiplicative inverse of log(n) is log(n)^-1. $\endgroup$ – Tyler Mar 4 '14 at 16:01
  • $\begingroup$ as for simplification, i wouldn't even say you simplified it at all. it's two ways for writing the same thing, since log(n)^-1 = 1/log(n) by definition. it doesn't look to me like there is any way to simplify it. $\endgroup$ – Tyler Mar 4 '14 at 16:02
  • $\begingroup$ $\log n=e^{\ln(\log n)}$ $\endgroup$ – Semsem Mar 4 '14 at 16:28

$\frac n{\log(n)}$ 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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