# How do you do the multiplication of an infinite sided regular polygon by an infinitesimal side to get the perimeter of it?

Using the law of cosines to find each side length and multiplying by the number of sides, I see a pattern, which should approach $$~2\pi~$$ as the number of sides grows to infinity.

How can $$~2\pi~$$ be calculated, when it's infinity multiplied by an infintesimal?

Using the law of cosines, the missing side of each perimeter section can be found, where $$~x~$$ is the number of sides the polygon has. In this, the distance from each vertex on the polygon to the center $$~= 1~$$ unit.

The formula for each polygon's perimeter based on its number of sides $$(x)$$ is as follows:

perimeter $$= x * \sqrt{2 - 2 ~\cos\left(\frac{360}{x}\right)}~.$$

So, what is $$~2\pi~$$ , using this infinite by infinitesimal product? Calculus anyone?

Using Taylor series for large $$x$$
$$\cos \left(\frac{2 \pi }{x}\right)=1-\frac{2 \pi ^2}{x^2}+\frac{2 \pi ^4}{3 x^4}+O\left(\frac{1}{x^6}\right)$$ $$2-2\cos \left(\frac{2 \pi }{x}\right)=\frac{4 \pi ^2}{x^2}-\frac{4 \pi ^4}{3 x^4}+O\left(\frac{1}{x^6}\right)$$ $$\sqrt{2-2\cos \left(\frac{2 \pi }{x}\right)}=\frac{2 \pi }{x}-\frac{\pi ^3}{3 x^3}+O\left(\frac{1}{x^5}\right)$$ $$x\sqrt{2-2\cos \left(\frac{2 \pi }{x}\right)}=2 \pi -\frac{\pi ^3}{3 x^2}+O\left(\frac{1}{x^4}\right)$$
Use it for $$x=6$$; the exact perimeter is $$6$$ for sure and the above formula gives $$2 \pi -\frac{\pi ^3}{108}\approx 5.99609$$.
If you want a still better approximation, using Padé approximants, $$x\sqrt{2-2\cos \left(\frac{2 \pi }{x}\right)}=2 \pi -\frac{20 \pi ^3}{3 \left(20 x^2+\pi ^2\right)}$$ which, for $$x=6$$, would give $$5.99997$$.
\begin{align*} \text{If}\;\,& f(x) = x \sqrt{ 2- 2\cos \bigl( {\small{\frac{2\pi}{x}}} \bigr) } \\[12pt] \text{then}\;&\lim_{x\to\infty}(f(x))^2\\[8pt] &= \lim_{x\to\infty} x^2 \left( 2- 2\cos \bigl( {\small{\frac{2\pi}{x}}} \bigr) \right) \\[8pt] &= 4\pi^2\lim_{x\to\infty} \frac { 2- 2\cos \bigl( \frac{2\pi}{x} \bigr) } { \left(\frac{2\pi}{x}\right)^2 } \\[8pt] &= 4\pi^2\lim_{t\to 0^+} \frac{2-2\cos(t)}{t^2} \\[8pt] &= 4\pi^2\lim_{t\to 0^+} \frac{\sin(t)}{t} \qquad\text{[by L'Hopital's Rule]} \\[8pt] &= (4\pi^2)\,(1) \\[8pt] &= 4\pi^2 \\[12pt] \text{hence}\;&\lim_{x\to\infty}f(x)=2\pi \\[4pt] \end{align*}