What value does the area approach?

Algebra Level 4

$x^n = 1$ When the solutions to the equation above are plotted in the complex plane for values of $n$ approaching infinity, the area of the shape obtained by mapping consecutive solutions of $x$ to one another approaches which of the following values?

2\pi

\pi

i

-\pi

-1

2

Indeterminate

1

2 solutions

Akeel Howell
Jun 16, 2016

Let $A_\circ$ be the area of the shape. The shape obtained by mapping the consecutive solutions of $x$ in the equation $x^n = 1$ , is a regular polygon inscribed in a circle of radius $1$ . Thus, the shape described is a regular $n-$ gon inscribed in the unit circle. As $n$ approaches infinity, the shape has an infinite number of sides, each of which are at a constant distance from it's center. All points $1$ unit from the center of the shape are on the polygon. This shape, by definition, approaches a circle as $n$ approaches infinity. $\therefore \mbox{A}_\circ = \pi \times 1 \times 1$ $= \pi$ Therefore the area of the shape approaches $\boxed{\pi}$ as $n$ approaches infinity.

Alternatively, since $x^n = 1$ , we know that the shape described is a regular polygon with $n$ sides, inscribed in a circle of radius 1. Therefore, area of the shape is: $A_\circ = r^2 \times n \times \sin\bigg(90^\circ - \frac{180^\circ}{n}\bigg) \times \cos\bigg(90^\circ - \frac{180^\circ}{n}\bigg)$ However, we soon realize that this does not work since plugging in $\infty$ for $n$ yields an indeterminate form: $A_\circ = 1^2 \times \infty \times \sin\bigg(90^\circ - \frac{180^\circ}{\infty}\bigg) \times \cos\bigg(90^\circ - \frac{180^\circ}{\infty}\bigg)$ $\frac{180}{\infty}$ $\longrightarrow$ $0$ . So $A_\circ = 1^2 \times \infty \times \sin\bigg(90^\circ - \frac{180^\circ}{\infty}\bigg) \times \cos\bigg(90^\circ - \frac{180^\circ}{\infty}\bigg)$ $= 1 \times \infty \times \sin\big(90^\circ\big) \times \cos\big(90^\circ\big)$ $= \infty \times 1 \times 0$ $= \infty \times 0$ This value is indeterminate. We now resort to tedious case study. Let $n$ be equal to $10$ . Therefore, the area of the shape is: $A_\circ = 1^2 \times 10 \times \sin\bigg(90^\circ - \frac{180^\circ}{10}\bigg) \times \cos\bigg(90^\circ - \frac{180^\circ}{10}\bigg)$ $= 10 \times \sin(72^\circ) \times \cos(72^\circ)$ $\ = 2.93892626164$ $\therefore \mbox{for } n = 10, \mbox{A}_\circ = 2.93892626164$ $n = 100, \mbox{A}_\circ = 3.13952597647$ , $n = 1000, \mbox{A}_\circ = 3.14157198278$ , $n = 10000, \mbox{A}_\circ = 3.14159244688$ , $n = 1000000, \mbox{A}_\circ = 3.14159265362 \ldots$ . etc. This value is clearly approaching $\pi$ . Therefore, the area of the shape approaches $\boxed\pi$ as $n$ approaches infinity. NOTE: This method is clearly very tedious, but it could be intuitive. It raises the question of whether a circle could be a regular polygon with infinite sides. If the statement, "a circle is by no means a regular polygon" is correct, then why does the area of a regular polygon approach that of it's circumcircle as the number of sides bounding the polygon approach infinity? Is there such a thing as an infinite-sided polygon? If so, why does it's area appear indeterminate when evaluated through conventional methods? I'd love to see some thoughts on these.

Akeel Howell - 4 years, 12 months ago

Indeed write $A_0$ as:-

$A_0=\dfrac{n}2\times\sin\left(\dfrac{2\pi}n\right)=\pi \left(\dfrac{ \sin\left(\frac{2\pi}n\right)}{\frac{2\pi}n}\right)$

Clearly as $n\to\infty$ , $A_0$ approaches $\pi$ since $\small{\displaystyle\lim_{x\to 0}\dfrac{\sin x}x=1}$ .

Rishabh Jain - 4 years, 12 months ago

As you go on increasing the number of sides of the regular polygons keeping the circumradius constant, the figure quickly begins to look like a circle. If you look at a figure of a circle, and a figure of a regular $100$ -gon, it is very difficult to distinguish one from the other. Instead of saying "a circle is an infinite-sided polygon", it is better to say that the circle is a regular $n$ -gon as $\lim _{n \to \infty}$ .

The limit of the polygon's area can be an indeterminate form, depending on the expression of area. However, as Rishabh has shown, a fair bit of manipulation allows us to find the exact value of the limit of the polygon's area.

Pranshu Gaba - 4 years, 11 months ago

Edwin Gray
Mar 30, 2019

The nth roots of unity lie on the circumference of a circle of radius 1. The chord distance between consecutive points is 2*pi/n. As n approaches infinity, the figure approches a unit circle with area of pi.

What value does the area approach?

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