-6
$\begingroup$

The number $\pi$ can be expressed as $\pi=\lim_{n\to\infty} \frac{n\sqrt[n]{-1}-n}{\sqrt{-1}}$ or more poetically $\pi=\frac{\infty\sqrt[\infty]{-1}-\infty}{\sqrt{-1}}$. Here we choose the principal branch of the root. Is there a proof using only roots, without resorting to sines and cosines? (Note: Use of the hyperreals is not required.) Does this formula have historical sources? Feel free to use l'Hopital's rule and Taylor formula, too.

$\endgroup$
8
  • 3
    $\begingroup$ What is this expression even supposed to mean? $\endgroup$
    – Wojowu
    Mar 29, 2016 at 11:17
  • 5
    $\begingroup$ I am not familiar with this fomula. Could you please give a reference, preferably to a source that would explain what it means? $\endgroup$
    – Ben McKay
    Mar 29, 2016 at 11:18
  • $\begingroup$ @Wojowu, if $\infty$ denotes an infinite number, then $\pi$ is the unique real number infinitely close to the righthandside. $\endgroup$ Mar 29, 2016 at 11:51
  • 1
    $\begingroup$ You should specify what arithmetic system of infinite numbers you are dealing with. Also, $\sqrt[n]{-1}$ is ambiguous due to multivaluedness of complex exponentiation. $\endgroup$
    – Wojowu
    Mar 29, 2016 at 11:53
  • 2
    $\begingroup$ I think all of this should be said in the question body so that one doesn't have to ask about this. Either way, I feel like this question doesn't fit on MO. $\endgroup$
    – Wojowu
    Mar 29, 2016 at 11:55

1 Answer 1

11
$\begingroup$

$$\frac{\infty\sqrt[\infty]{-1}-\infty}{\sqrt{-1}}=\frac{1}{i}\lim_{x\rightarrow \infty}\left(x e^{i \pi/ x}-x\right)=\pi$$

$\endgroup$
5
  • 4
    $\begingroup$ I'm perfectly happy with this explanation. I think I would have made it even more freshman-like by interpolating $= \frac1{i} \lim_{h \to 0} \frac{e^{i\pi h} - 1}{h}$ before the final answer $\pi$. $\endgroup$
    – Todd Trimble
    Mar 29, 2016 at 12:00
  • $\begingroup$ At the same time, I imagine the OP has no plans to accept this answer; by mentioning "transfer principle", it seems clear he wants some sort of short nonstandard analysis proof. (Why, I don't know.) $\endgroup$
    – Todd Trimble
    Mar 29, 2016 at 12:08
  • $\begingroup$ @ToddTrimble, I added the comment about transfer because user wojowu asked me to. Since a moderator objects I will redelete it. $\endgroup$ Mar 29, 2016 at 12:09
  • 1
    $\begingroup$ I'm not objecting, @katz -- indeed, I think you ought to amplify even more what you're really after in this question. $\endgroup$
    – Todd Trimble
    Mar 29, 2016 at 12:10
  • 2
    $\begingroup$ @ToddTrimble, I like Carlo's answer but I was hoping for something more elementary that does not involve sines and cosines and taylor expansions. I am thinking about how to reformulate the question but it may soon be too late :-) I seem to have a knack for asking lousy questions that generate popular answers :-) The one about the intended interpretation has been downvoted by a whopping 12 people, while the leading answer has been upvoted by 13 :-) $\endgroup$ Mar 29, 2016 at 12:14

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