2
$\begingroup$

Assume unit radius. The area of regular polygon with $n$ sides = $n\sin\left(\pi/n\right)$, how does $n\sin\left(\pi/n\right)$ approach $1$ as $n \to \infty$ ?.

$$ \mbox{I have tried}\quad \pi\,{\sin\left(\pi/n\right) \over \pi/n}\quad \mbox{so it looks like}\quad \pi\,{\sin(x) \over x} $$ L'Hopital's Rule and I still can not get $\quad\lim_{x \to \infty}\sin\left(x\right)/x = 1$.

$\endgroup$
  • $\begingroup$ If you're considering unit radius then the area of the regular polygon with n sides converges to the area of the circle of radius one, so it converges to $\pi$, which is what you have found. What is the problem? $\endgroup$ – Federico Oct 13 '16 at 18:31
  • $\begingroup$ $\lim_{n\to \color{red}\infty}\sin(\pi/n)] / (\pi/n)=\lim_{x\to\color{red}0+}\sin x / x = 1$ $\endgroup$ – georg Oct 13 '16 at 18:33
  • $\begingroup$ Of course, I don't know why i could not see that myself. Both x and sin x approaches 0 from positive. My bad...Thanks $\endgroup$ – W. Davis Oct 13 '16 at 20:37
4
$\begingroup$

As $\epsilon\to0$, we have $\sin\epsilon=\epsilon+o(\epsilon)$ (it's Taylor's formula around $0$, at order $1$), hence, as $n\to\infty$,

$$n\sin\frac\pi n=n\left(\frac\pi n+o(\frac1n)\right)=\pi+o(1)\to\pi$$

$\endgroup$
3
$\begingroup$

For small values of $x$, $\sin x\sim x$. So for large values of $x$, $\sin(1/x)\sim 1/x$. So as $n$ increases, $n\sin(\pi/n) \sim n\cdot\pi/n = \pi$.

$\endgroup$
  • 2
    $\begingroup$ For $n=1000$, my calculator says that $n\sin(\pi/n)\approx3.14159$. $\endgroup$ – Mark McClure Oct 13 '16 at 18:47
  • 1
    $\begingroup$ You forgot a π factor. $\endgroup$ – Bernard Oct 13 '16 at 19:01
2
$\begingroup$

As often, the shortest is to use equivalents:

$\sin x\sim_0 x$, hence $\sin\dfrac\pi n\sim_\infty\dfrac\pi n$ and $$n\sin\frac\pi n\sim_\infty n\dfrac\pi n=\pi.$$

$\endgroup$
  • $\begingroup$ I haven't seen this notation or terminology before -- can you provide a reference? It looks like an interesting way to think about these problems which I would like to learn about $\endgroup$ – Chill2Macht Oct 13 '16 at 19:32
  • 1
    $\begingroup$ It's a way to remove irrelevant details when calculating limits. $\sim_a$ reads as ‘equivalent in a neighbourhood of $a$’. You can take a look at the Wikipedia note on Asymptotic analysis. $\endgroup$ – Bernard Oct 13 '16 at 20:08
0
$\begingroup$

I just want to point out this can be derived from Archimedes' Method for calculating the area of a (unit) circle (here we apply it for a polygon with $2n$ sides)

enter image description here

$\endgroup$

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.