The Ratio of a Circle's Circumference to Diameter

[mite]

can we prove the ratio of circumference to diameter is same for all circles & is equal to pi?

 

[CompuChip]

Well, the circumference of a circle is [itex]L = 2\pi r[/itex] source and by definition the diameter is twice the radius (d = 2r). So [itex]L/d = 2 \pi r / (2 r) = \pi[/itex].

 

[tiny-tim]

can we prove the ratio of circumference to diameter is same for all circles & is equal to pi?

Hi mite!

It depends what axioms (basic definitions) you start with.

Euclid regarded the similarity of two circles as an axiom, so there was nothing to prove!

And π is defined as the ratio.

 

[HallsofIvy]

Historically, the fact that the ratio of circumference to diameter is a constant was a numerical observation. The Greeks proved it by calculating the ratio of the perimeter of a regular n-gon to its "diameter" and then seeing what happened as n got larger and larger (a limit process). As for the fact that that ratio is equal to pi- that's essentially the definition of pi.

A modern proof would be something like this: Since sin²(t)+ cos²(t)= 1 for all t, x= Rcos(t) and y= Rsin(t) are parametric equations for a circle of radius R. The circumference, then, is given by
[tex]\int_0^{2\pi}\sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2}dt[/tex]
[tex]= \int_0^{2\pi}\sqrt{R^2sin^2(t)+ R^2cos^2(t)}dt= \int_0^{2\pi}Rdt= 2\pi R[/tex]
Since the circumference is [itex]2\pi R[/itex] and the diameter is 2R, the ratio of circumference to diameter is [itex]2\pi R/(2R)= \pi[/itex].

(The fact that sin(t) and cos(t) have period [itex]2\pi[/itex], which is critical to this proof, can be shown by using the fact that the second derivative of sin(t) is -sin(t) and the second derivative of cos(t) is -cos(t).)

 

[tiny-tim]

The Greeks proved it by calculating the ratio of the perimeter of a regular n-gon to its "diameter" and then seeing what happened as n got larger and larger (a limit process).

Hi HallsofIvy!

No, surely that's how they calculated π …

they were already convinced that it was the same for all circles?

 

[HallsofIvy]

I didn't say that was how they calculated it. I said that was how they proved it was the same ratio for all circles. It was Archimedes who did that. I am sure that Greeks before that just assumed it was a constant.

 

[tiny-tim]

I am sure that Greeks before that just assumed it was a constant.

And they were right to do so!

From their point of view, because it was axiomatic and/or obvious …

from our point of view, because of the scalar symmetry of Euclidean space.

(Of course, circumference/diameter isn't a constant in non-Euclidean space. )

(If you'd said to them "you've proved that it's a constant", they'd have replied "no we haven't, we've only calculated the constant … we implicitly used a symmetry theorem on polygons in the course of that calculation, and that applies to circles anyway" )

 

[dynamicsolo]

The arclength integration doesn't really constitute a proof. The choice of integration limit for the "angle parameter" of [tex]2 \pi[/tex] is based on the definition of angle as the ratio of arclength to radius for a circle: [tex]\theta = \frac{s}{R}[/tex]. So it is completely unsurprising that the result of the integration for the circumference of a circle is [tex]2\pi R[/tex].

I haven't explored the history of the number thoroughly (although there are at least two or three histories of pi out there now), but I believe that pi is simply defined as the ratio of circumference to diameter for a circle (the specific letter was chosen somewhere around the 18th Century), it having already been understood in antiquity that the ratio is a constant. So there is no proof involved for this. (One of the continuing mysteries is why [tex]\pi[/tex] is so deeply imbedded in the structure of mathematics and turns up in other relations which has little to do with circles...)

 

[granpa]

can we prove the ratio of circumference to diameter is same for all circles & is equal to pi?

real world circles or mathematical circles? the circumference of the latter depends on your metric.