MHB Is the Limit of Regular Polygon Areas π? Using Infinitesimal Analysis to Prove

[lfdahl]

Let $A_n$ be the area of the regular polygon whose vertices are given by the $n$ roots of unity in the complex plane.

Prove: \[\lim_{n \rightarrow \infty }A_n = \pi \]

 

[kaliprasad]

Let $A_n$ be the area of the regular polygon whose vertices are given by the $n$ roots of unity in the complex plane.

Prove: \[\lim_{n \rightarrow \infty }A_n = \pi \]

Spoiler

Let n be number of sides of polygon. and O be the centre of polygon .let s be side length of polygon.

let AB be one side of polygon Join OA and OB

the radius of circumcircle is R=1 as they are n roots of unity in complex plane.
and r be radius of inscribed circle.
So are of each triangle( joining the centre with end vertices of each side of polygon) =
$\frac{1}{2}sr = R \sin(\frac{\pi}{n}) R \cos(\frac{\pi}{n})$
$=\frac{1}{2}R^2 (2\sin(\frac{\pi}{n}) \cos(\frac{\pi}{n})$
$=\frac{1}{2}R^2 (\sin{\frac{2\pi}{n}})$

so the area of polygon = $=\frac{1}{2}nR^2 (\sin{\frac{2\pi}{n}})= \frac{1}{2}nR^2 \frac{2\pi}{n}\frac{\sin{\frac{2\pi}{n}}}{\frac{2\pi}{n}} $
$= \pi R^2 \frac{\sin{\frac{2\pi}{n}}}{\frac{2\pi}{n}} $
as n goes to iniinite we get
$= \pi R^2$

As R is 1 result is $\pi$

 

[I like Serena]

Let $A_n$ be the area of the regular polygon whose vertices are given by the $n$ roots of unity in the complex plane.

Prove: \[\lim_{n \rightarrow \infty }A_n = \pi \]

My attempt:

Spoiler

The inner circle has radius $\cos\frac\pi n$ while the outer circle has radius 1.
So the following holds for the respective areas:
$$\pi \cos^2\frac\pi n < A_n < \pi$$
Taken to the limit it follows from the squeeze theorem that:
$$\lim_{n \rightarrow \infty }A_n = \pi$$

 

[HallsofIvy]

Let $A_n$ be the area of the regular polygon whose vertices are given by the $n$ roots of unity in the complex plane.

Prove: \[\lim_{n \rightarrow \infty }A_n = \pi \]

Spoiler

The nth roots of unity all lie on the unit circle, the circle with center at 0 and radius 1, in the complex plane. As n goes to infinity, that polygon goes to the unit circle which has area $$\pi(1)^2= \pi$$.

 

[lfdahl]

Spoiler

Let n be number of sides of polygon. and O be the centre of polygon .let s be side length of polygon.

let AB be one side of polygon Join OA and OB

the radius of circumcircle is R=1 as they are n roots of unity in complex plane.
and r be radius of inscribed circle.
So are of each triangle( joining the centre with end vertices of each side of polygon) =
$\frac{1}{2}sr = R \sin(\frac{\pi}{n}) R \cos(\frac{\pi}{n})$
$=\frac{1}{2}R^2 (2\sin(\frac{\pi}{n}) \cos(\frac{\pi}{n})$
$=\frac{1}{2}R^2 (\sin{\frac{2\pi}{n}})$

so the area of polygon = $=\frac{1}{2}nR^2 (\sin{\frac{2\pi}{n}})= \frac{1}{2}nR^2 \frac{2\pi}{n}\frac{\sin{\frac{2\pi}{n}}}{\frac{2\pi}{n}} $
$= \pi R^2 \frac{\sin{\frac{2\pi}{n}}}{\frac{2\pi}{n}} $
as n goes to iniinite we get
$= \pi R^2$

As R is 1 result is $\pi$

Great job, kaliprasad! Your quantitative approach was exactly, what I had in mind.

- - - Updated - - -

My attempt:

Spoiler

The inner circle has radius $\cos\frac\pi n$ while the outer circle has radius 1.
So the following holds for the respective areas:
$$\pi \cos^2\frac\pi n < A_n < \pi$$
Taken to the limit it follows from the squeeze theorem that:
$$\lim_{n \rightarrow \infty }A_n = \pi$$

Very short and elegant, I like Serena! (Cool) Thankyou for your participation!

 

[lfdahl]

Spoiler

The nth roots of unity all lie on the unit circle, the circle with center at 0 and radius 1, in the complex plane. As n goes to infinity, that polygon goes to the unit circle which has area $$\pi(1)^2= \pi$$.

Thankyou very much, HallsofIvy, for your clear and short qualitative solution. The limit (area of a unit circle) is obviously $\pi$, I was just thinking of a quantitative solution. Maybe, I should have written this right at the start.

 

[topsquark]

Spoiler

The nth roots of unity all lie on the unit circle, the circle with center at 0 and radius 1, in the complex plane. As n goes to infinity, that polygon goes to the unit circle which has area $$\pi(1)^2= \pi$$.

Way cool! Remind me to slip you a lolli-pop the next time I see you.

-Dan

 

[solakis1]

Let $A_n$ be the area of the regular polygon whose vertices are given by the $n$ roots of unity in the complex plane.

Prove: \[\lim_{n \rightarrow \infty }A_n = \pi \]

Using infinitesimal analysis we have ;

[sp]For a very large $$n$$ the angle $$d\theta$$ of each triangular sector of the regular polygon is very small and the area of each sector is :

$$\frac{d\theta}{2}$$ plus infinitesimal of higher order

And integrating over from 0 to $$2\pi$$ we get $$\pi$$

Note: $$sind\theta$$ is nearly equal to $$d\theta$$[/sp]

 

[lfdahl]

Using infinitesimal analysis we have ;

[sp]For a very large $$n$$ the angle $$d\theta$$ of each triangular sector of the regular polygon is very small and the area of each sector is :

$$\frac{d\theta}{2}$$ plus infinitesimal of higher order

And integrating over from 0 to $$2\pi$$ we get $$\pi$$

Note: $$sind\theta$$ is nearly equal to $$d\theta$$[/sp]

Thankyou, Solakis, for your nice solution on the infinitesimal scale!