0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.