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As I was studying some limit problems, I came across

$$\sqrt[n]{n}$$

and astoundingly found out that the graph of this has a maximum when $n = e$.

I thought there is no way that this is not a famous fact and I am very interested in it. I looked up some words such as "nth roots" or "rational exponents" but I haven't found this fact right away.

Can someone guide me to a link or tell me at least what this expression goes by in order to do a little bit of researching ?

Antonio Vargas
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hyg17
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    $\sqrt[n]{n}=n^{(1/n)}=e^{\ln(n)/n}$. Show that ${\ln(n)/n}$ maxes out at $n=e$, and then $e^{\ln(n)/n}$ maxes out at $e$ too. – 2'5 9'2 Jun 01 '13 at 22:19
  • If you were studying limits problems couldn't it be that $,n\in\Bbb N,$ and thus it can not be $,n=e,$ ? – DonAntonio Jun 01 '13 at 22:25
  • @DonAntonio: Derivatives are part of limits, right ? – hyg17 Jun 01 '13 at 22:28
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    Well, derivatives are limits, @hyg17 , yet when we use the letter $,n,$ we usually refer to discrete limits: sequences indexed by the natural numbers. Of course, it may be you decided to use that letter as continuous variable...] – DonAntonio Jun 01 '13 at 22:35
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    As a general rule, mathematicians use $n$ to represent natural numbers and integers. Not a law or anything, but it is confusing if you write $\sqrt[n]{n}$ and not explicitly say what your domain is for $n$ - almost all mathematicians will read it as the natural numbers. (If you had written $\sqrt[x]{x}$, most mathematicians will read it as positive real numbers...) – Thomas Andrews Jun 01 '13 at 22:35
  • By the way, can anyone tell me where I can find some facts about $\sqrt[x]{x}$ ? – hyg17 Jun 01 '13 at 22:52
  • Awfully snotty for someone asking people to help you out for free. – Thomas Andrews Jun 02 '13 at 00:03
  • To better understand the phenomenon you are observing, pretend you did not know about the number $e$. Try to work out the derivative without using $e$ or the base $e$ logarithm. How have you defined the number $e$? One common way is by the limit $e = \lim_{x\rightarrow \infty}(1 +x^{-1})^x$. I suspect that the limit defining $f'(x) = 0$ where $f(x) = x^{1/x}$ is equivalent to the above definition of $e$ by some algebraic manipulation. – tghyde Jun 02 '13 at 01:29
  • @ThomasAndrews: You know what, I noticed how I sounded like when you commented about what I said. I deeply apologize. My ignorance lead to a horrible comment and I'm sorry for what I did. Thank you for teaching me something I did not know, and I will not forget this thanks to you. Again, I'm truly sorry for how I reacted. I will delete my comment as it is an eyesore. – hyg17 Jun 03 '13 at 09:48

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Simply derive to find the maximum, but first note that $$\sqrt[n]{n} =n^{1/n}=\exp(\log(n^{1/n}))$$ This should explain the connection with $e$. In more detail: $$\frac{dn^{1/n}}{dn}=n^{1/n}\frac{1-\log n}{n^2}$$ So the derivative is only zero when $\log n = 1 \to \ n=e$.

Nathaniel Bubis
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  • Where would I be able to find more facts about this expression ? – hyg17 Jun 01 '13 at 23:18
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    @hyg17 There aren't a lot of facts to be found. What specifically do you want to know? – Potato Jun 02 '13 at 03:25
  • @hyg17 - you'll have to specify what else you want to know.. – Nathaniel Bubis Jun 02 '13 at 03:56
  • It would have been best if I could know what this expression is known for. Because I am more interested in what application to mankind it would be used as and its practicality. Mathematically its also fascinating, but I would only be able to find out a very limited amount since my knowledge is not as deep as most people who uses this site. – hyg17 Jun 03 '13 at 09:50
  • @hyg17 - In short, AWAIK this expression is not known for anything practical and is not otherwise useful in mathematics. It's just another function. Hope this answers your question. – Nathaniel Bubis Jun 03 '13 at 09:56