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November 5

Pi is a constant

Hello,

Where can I find a rigorous proof that the ratio of a circle's circumference to its diameter is a constant, independent of the diameter? Thanks, 09:09, 5 November 2015 (UTC) — Preceding unsigned comment added by 212.179.21.194 (talk)

Here is your rigorous proof. The diameter of a circle is a constant. The circumference of a circle is a constant. The ratio of two constant number is another constant number. Thus the ratio of circumference of a circle to the diameter of a circle is a constant. Thus Pi is a constant. QED. 175.45.116.66 (talk) 00:27, 6 November 2015 (UTC)[reply]

I think you asked the question incorrectly. You mean to ask the following. Pi for circle A is a constant. Pi for circle B is a constant. Circle B is "bigger" than Circle A. Show that Pi for circle A has the same value as Pi for circle B. 175.45.116.66 (talk) 00:32, 6 November 2015 (UTC)[reply]

Let the diameter for Circle_A be 1 unitlength. Thus the circumference for Circle_A would be " Pi_A * 1 unitlength ". Let Circle_B be equal to Circle_A scaled up linearly by a factor F. Thus diameter for Circle_B is " F * 1 unitlength " because this is what linear scaling means.

All you have to do is show that circumference for Circle_B be " F * Pi_A * 1 unitlength ".

I will do it like this. First split the circumference into 4 equal points, Join the points up to look like a square. The circumference of the square is 4 times the length of the side. If the square is scaled up Linearly by a factor of F, then the new circumference is " F * 4 * length of the old side"

Next do it for 8 equal points.

Next do it for 16 equal points.

The limit as n->infinity, the splitting the circumference into n equal points approaches the circumference of a proper circle, hence circumference for Circle_B is " F * circumference for Circle_A " 175.45.116.66 (talk) 00:50, 6 November 2015 (UTC)[reply]

As for a rigorous proof, the circumference of a circle of radius r is given by the integral

2\int _{-r}^{r}{\sqrt {1+{\frac {x^{2}}{r^{2}-x^{2}}}}}\,dx.

After the change of variables $u=x/r$ , this becomes

2\int _{-1}^{1}{\sqrt {1+{\frac {u^{2}}{1-u^{2}}}}}\,\,r\,du=2r\int _{-1}^{1}{\sqrt {1+{\frac {u^{2}}{1-u^{2}}}}}\,du.

So the circumference of a circle of radius r is equal to r times the circumference of a circle of radius 1. Sławomir
Biały 01:46, 6 November 2015 (UTC)[reply]

As you can see the proof can be difficult depending on which techniques you are "allowed" to use. Using calculus may be unsatisfying if you want a proof in the style of the ancient geometry school, which certainly was aware of the fact. The first proof seems to go back to Archimedes- see this nice article for his proof and some more history: http://arxiv.org/abs/1303.0904. Staecker (talk) 01:50, 6 November 2015 (UTC)[reply]

The circumference requires analysis to define rigorously (e.g., the Hausdorff measure of a plane set). Archimedes' argument would probably not be considered "rigorous" by modern standards. It relies on an implicit assumption that the circumference of a circle inscribed in a polygon is less than the perimeter of the polygon. This is something we now know to be true, as a theorem in analysis. The same techniques that rigorously lead to this conclusion also lead rather directly to the desired fact, because with analysis it becomes very straightforward to calculate the circumference directly, without introducing any auxiliary assumptions. Sławomir
Biały 02:06, 6 November 2015 (UTC)[reply]

Another rigorous proof is first to define π in some other way, independently of circles, and then show that a circle of radius r has circumference 2πi. One such approach is to let π be the smallest positive root of $e^{i\pi }=-1$ . Then Cauchy's integral formula gives

\oint {\frac {dz}{z}}=2\pi i

around any simple plane curve containing 0. The circle can be parameterized by $z=re^{i\theta }$ . So $dz/z=id\theta$ . The circumference of a circle of radius r is therefore

\oint |dz|=\oint rd\theta =-ir\oint {\frac {dz}{z}}=2\pi r.

-- Sławomir
Biały 02:18, 6 November 2015 (UTC)[reply]

The ratio between circumference and diameter of circles depend of the diameter in non-euclidean geometry, where the fifth axiom of Euclids elements is denied. So the proof that "pi is constant" must depend on the fifth axiom. Bo Jacoby (talk) 05:03, 6 November 2015 (UTC).[reply]

It can't be derived from the Euclidean postulates anyway. There is no postulate that allows you to unroll a circle onto a line to obtain its circumference. Sławomir
Biały 11:14, 6 November 2015 (UTC)[reply]

Well, Euclid's Elements does have theorems about lengths of arcs; it's just that as far as I know he only compares them to lengths of other arcs, not to straight lines. He could have tried to prove that C₁/C₂ = d₁/d₂ for any two circles 1 and 2, even if he didn't have the concepts to allow that to be rearranged as C₁/d₁ = C₂/d₂. --70.49.170.168 (talk) 05:39, 7 November 2015 (UTC)[reply]

I couldn't find anything in the Elements about arc lengths, not to say it isn't there since there is a lot of material to scan through. But Book XII prop. 2 says the area of a circle is proportional to the square of the diameter and is proved using the method of exhaustion. I don't think the method of exhaustion can be applied to arc length though. --RDBury (talk) 19:43, 8 November 2015 (UTC)[reply]