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I searched here and I got the definition that the circumference of a curve would be the smallest upper bound of the sequence of the sum of lenghts of polygonal paths along the curve, but how can I proof that a circle's circumference is actual value? That it is well defined?

And after proofing that, how could I find it without using integrals? (Because I think using them is ciclical, since we need the circumference of the circle to get the relation between radians and a full turn on the circle if we use trigonometric functions or its inverses)

I tried to proof that pi is a constant to get it, but then I got stuck because I need to proof that $$\lim_{n\to \infty} nL = C$$, with n being the number of sides of a regular n-agon, L being its side and C being the circle's circumference, but I don't know how to do this formally (I can see it informally that the sum would get to it, but I wanted to be really formal) and I'd very well thank anyone who can help me with this.

@edit: I'm sorry for not mentioning that what I wanted was for a circle, I'm still getting used to English terminology of some things

@edit: my doubt is gone thanks to Paramanand ^^

Thanks in advance, everyone!

Geovane Coelho
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1 Answers1

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The length of a curve is defined as the supremum of lengths of polygonal arcs obtained by joining a finite number of points on the curve.

A necessary and sufficient condition for a curve to have a well defined length according to this definition is that it's components are of bounded variation. More formally if $f, g$ are continuous functions on interval $[a, b] $ then the curve given by parametric equation $x=f(t), y=g(t) $ has a well defined length (or is rectifiable) if and only if both $f$ and $g$ are of bounded variation.

The case of a circle is easy. Just consider the first quadrant of the unit circle $x^2+y^2=1$ which is parameterized by $x=t, y=\sqrt{1-t^2}$ for $t\in[0,1]$ and clearly these functions $t, \sqrt{1-t^2}$ are monotone and hence of bounded variation on $[0,1]$. Hence the first quadrant of the circle has a well defined length. The same can be done for other three quadrants and thus the circle turns out to be a rectifiable curve.

The evaluation of the length of circle depends on integrals and we have the circumference of unit circle given by $$4\int_{0}^{1}\frac{dx}{\sqrt{1-x^2}}=2\int_{0}^{1}\sqrt{1-x^2}\,dx$$ By the traditional definition of $\pi$ we see that we have the analytical result $$2\int_{0}^{1}\frac{dx}{\sqrt{1-x^2}} =\int_{0}^{1}\sqrt{1-x^2}\,dx=\pi$$

And there is no circular argument over here while using integrals. If we accept the definition of $\pi$ as ratio of circumference of a circle to its diameter and then this leads us to the analytical definition of $\pi$ in terms of the integrals given above.

Paramanand Singh
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  • Could you please explain how did you get the first equality? I can see that the first integral is equivalent to $\int_0^1 \sqrt{(1+(y')^2)}dx$ but I couldn't get past that... – Geovane Coelho Sep 25 '18 at 03:27
  • @GeovaneCoelho: for circle $y=\sqrt{1-x^2}$ so that $y'=-x/\sqrt{1-x^2}$ and $1+y'^2=1/(1-x^2)$. Taking roots we get desired integral. The second integral is due to integration by parts. – Paramanand Singh Sep 25 '18 at 04:19
  • Oh my God, it worked out beyond my expectation!! You even gave me a way to calculate the area of the circle without using the trigonometric functions or other things, thank you very much!! – Geovane Coelho Sep 25 '18 at 20:10