Thursday, February 3, 2022

Proving the Perimeter of a Circle


What is the perimeter of a circle? This is a very easy question right? $2 \pi r$ of course. Well why? This was the simple question that bothered my mind last weekend. I thought of some proofs myself and made many mistakes along the way.

It seemed like a simple integral: $ \int_0^{2 \pi} \! r \, \mathrm{d} \theta $. Since $d \theta \times r$ gives us the arc length, taking the integral from $0$ to $2 \pi$ (basically making a full turn) gave us the perimeter. The problem with this solution was that it felt like cheating! How did I even know the arc length? 

My next idea was pretty straight forward as well. Firstly I defined a simple sequence: let $x_n$ be the perimeter of an n-sided regular convex shape circumscribed in a circle of radius $r$. So for example:
Then clearly as n goes to infinity, we would get the perimeter of the circle. So all that was left to do was to find the formula and take the limit. 

Since the sum of all outer angles in a convex shape equals $2 \pi$ and since this shape is regular, $2 \theta= \pi - \frac{2 \pi}{n}$. Then one side of the polygon is $2 \times r \times cos(\theta)$.
Thus, the perimeter of $x_n$ circumscribed in a circle is: 
\[x_n = n \times 2 \times r \times cos(\theta)\]
\[ x_n = n \times 2 \times r \times cos(\frac{ \pi - \frac{2 \pi}{n}}{2}) \]
\[\lim_{n \to  \infty} x_n = \lim_{n \to \infty} \left[ n  \times 2 \times r \times cos(\frac{ \pi - \frac{2 \pi}{n}}{2}) \right] \]
\[ \lim_{n \to  \infty} x_n= 2 \times r \times  \lim_{n \to \infty} \left[ n  \times cos(\frac{ \pi - \frac{2 \pi}{n}}{2}) \right] \]
Let's solve the limit:
\[\lim_{n \to \infty} \left[ n \times cos(\frac{ \pi - \frac{2 \pi}{n}}{2}) \right] = \lim_{n \to \infty} \left[ n \times cos(\frac{\pi}{2} - \frac{\pi}{n}) \right]  \]
\[= \lim_{n \to \infty} \left[ n \times sin(\frac{\pi}{n}) \right]= \lim_{n \to \infty} \left[ \frac{sin(\frac{\pi}{n})}{\frac{1}{n}} \right]  \]
\[\stackrel{L'H}{=}\lim_{n \to \infty} \left[ \frac{cos(\frac{\pi}{n})\times (-\frac{\pi}{n^2})}{-\frac{1}{n^2}}  \right] = \pi \times \lim_{x \to \infty} cos(\frac{\pi}{n})= \pi\]
Then finally we find the perimeter of a circle: $2 \pi r$. We could have proved the perimeter of a circle in a similar manner, this time defining $x_n$ as the perimeter of an n-sided regular convex shape inscribed in a circle of radius $r$.
This proof didn't satisfy me as well but I didn't exactly know why. Using trigonometric functions felt like cheating, since, their intuition came to me from a circle. These trigonometric functions probably already used the perimeter formula in some way. I was using the thing that I wanted to prove as an hypothesis. 
I got annoyed.  The perimeter formula had $\pi$ in it! How was I supposed to find $\pi$ without using trigonometric functions or radians (which were already based on the perimeter of a circle)? Then it suddenly hit me! I had forgotten one of the sacred rules of mathematics, I had ignored the definitions.My stupid mistake hit me in the face:
\[  \pi = \frac{\textrm{perimeter}}{\textrm{diameter}}\]
OF COURSE THE PERIMETER WAS $2 \pi r$! It was in the definition. There was nothing to prove. 
But I didn't want to come out with my hands empty. Since $\pi$ was just a ratio, I wanted to prove that this ratio was constant and didn't depend on the radius. This is where my  sequence came in handy! If we have two different circles, their $x_n$ ratios will equal to their radius ratios. In other words, the ratio of their perimeters will equal the ratio of their radiuses. Thus there really is a fixed ratio between the perimeter and the diameter, independent of the length of the radius. 
And we call this ratio $\pi$!
PS. I'm guessing that cosine function kind of depends on this ratio being fixed so I am probably just repeating my mistake. This problem will stay in my mind and you can expect to see a new blog posts when I have some answers.



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