3 <= pi <= 4 ?

[m_f_h]

Since the original problem seems solved, I'd like to post here my favourite one: Show that pi is between 3 and 4.
Or a bit more explicitely (for the pleasure of pedants...) :

Show that the circumference of a circle in a 2-dim normed vector space is between 3 and 4 times its diameter, and that both of these limits may be reached.

(It is somehow trivial to show that the two limits are reached for particular (easy to guess) metrics, but less trivial to give a nice proof of the main statement.)
[You may use that any symmetric convex set defines a metric (and norm), for which it is the unit disc.]

[cheesehead]

Originally Posted by m_f_h

Show that pi is between 3 and 4.
< snip >

By considering a unit circle inscribed within a unit square, we see that the circumference of the circle, pi, is less than the perimeter of the square, 4.

By considering a regular hexagon inscribed within a unit circle, we see that the circumference of the circle, pi, is greater than the hexagon's perimeter, 3 (the hexagon can be divided into six equilateral triangles, each of side equal to the circle's radius, 1/2).

-- -- --

I composeded the above without consulting any reference, but long ago I did see Archimedes's method (http://physics.weber.edu/carroll/Archimedes/pi.htm or, more authentically, http://itech.fgcu.edu/faculty/clinds...rchimedes.html), and my subconscious memory probably informed me.

[davieddy]

That's neat but I think he was after something
rather less comprehensible!

BTW you could invoke Davar55's convex polygon
theorem to show that the perimeter of the
circle is less than that of the square and greater
than that of the hexagon.

David

[xilman]

Originally Posted by cheesehead

By considering a unit circle inscribed within a unit square, we see that the circumference of the circle, pi, is less than the perimeter of the square, 4.

By considering a regular hexagon inscribed within a unit circle, we see that the circumference of the circle, pi, is greater than the hexagon's perimeter, 3 (the hexagon can be divided into six equilateral triangles, each of side equal to the circle's radius, 1/2).

Your statement is true in a Euclidean space. In Euclidean space the ratio of the circumference of a circle to its radius is indeed \pi *2.

In non-Eclidean space this statement is not necessarily true and, indeed, the ratio can depend on the size of the circle, on the assumption that a circle is defined in such a space. An attractive definition is that a "circle" is the set of all points which lie at a constant distance, which we'll call the "radius", from a given point which we'll call the "centre".

For instance, on the surface of a sphere a circle will have a ratio larger than \pi (except in the limiting case of a circle of zero radius.) The equator is a circle (all its points are equidistant from a given point, the pole) but the ratio of circumference to radius is 4.

Somehow, I feel that this is what m_f_h was getting at.

Paul

[Wacky]

Originally Posted by xilman

For instance, on the surface of a sphere a circle will have a ratio larger than \pi (except in the limiting case of a circle of zero radius.) The equator is a circle (all its points are equidistant from a given point, the pole) but the ratio of circumference to radius is 4.

And the diameter is twice the radius. Therefore the ratio of the circumference to the diameter is 2.

Which is less than 3.

In fact, as the radius approaches the diameter of the sphere, the ratio tends to zero.

So I must conclude that this is not what m_f_h had in mind.

[m_f_h]

Originally Posted by xilman

(...) An attractive definition is that a "circle" is the set of all points which lie at a constant distance, which we'll call the "radius", from a given point which we'll call the "centre".
(...) Somehow, I feel that this is what m_f_h was getting at.
Paul

Indeed. (I should not have written the first "simplified" version of the question; at present I can't find the "edit" link to fix that... But the second part (of the explicit version) should have made this clear: the ratio CAN be equal to 3, and it CAN be equal to 4 (and any of the intermediate values), depending on the norm/metric.)
So the shape of the circle may vary (e.g. it may be non smooth, i.e. have angles...)
The (length of) the circumference is measured through the same norm which gives the shape of the circle.
To start with and get an idea, you may consider R² equipped with the infinity- (a.k.a. sup-) norm, | (x,y) | = max{ |x|, |y| }, or the 1-norm.

Originally Posted by xilman

(...)
For instance, on the surface of a sphere a circle will have a ratio larger than \pi (except in the limiting case of a circle of zero radius.) The equator is a circle (all its points are equidistant from a given point, the pole) but the ratio of circumference to radius is 4.

That's a bit confusing - let's stick to the 2-dimensional setting (and/or terminology), without "curved spacetime"... once again: R² equipped with an arbitrary norm (which defines resp. is defined by the unit disc : see the hidden hint in the original post - don't worry, it does not give the solution.

The given references suggest that cheesehead misinterpreted the question and worked in Euclidian space; else his solution would have qualified...

[cheesehead]

Originally Posted by m_f_h

(I should not have written the first "simplified" version of the question

But you did ... and by quoting only that version at the start of my posting, I signalled that my reply was to that "simplified" version rather than to the "explicit" version that followed it. So non-Euclidean objections to my intended-to-be-Euclidean-only reply are irrelevant.

The given references suggest that cheesehead misinterpreted the question and worked in Euclidian space; else his solution would have qualified...

I misinterpreted nothing; I chose to limit my reply to Euclidean space. I readily admit that I am rusty at algebras, non-Euclidean geometry, and topology, so deliberately avoided taking a chance of blundering by treading on unfamiliar ground, which I might have chanced had I tried being more pedantic. :-)

[xilman]

Originally Posted by m_f_h

Indeed. (I should not have written the first "simplified" version of the question; at present I can't find the "edit" link to fix that... But the second part (of the explicit version) should have made this clear: the ratio CAN be equal to 3, and it CAN be equal to 4 (and any of the intermediate values), depending on the norm/metric.])

Even in pathological spaces, such as those which are everywhere discontinuous?

I think you have to include certain non-singularity conditions.

Anyway, I claim that \pi is a constant equal to 3.14159..., independent of geometry, and therefore it's trivial that 3 <= \pi <= 4.

In a Euclidean space the ratio of the circumference of a circle to its radius is 2\pi, just as the volume of a sphere is 4\pi r^3 / 3. Neither are particularly good (though they are undoubtedly useful) definitions of \pi, which is a constant of much wider applicability than Euclidean geometry.


Paul

[xilman]

Originally Posted by Wacky

And the diameter is twice the radius. Therefore the ratio of the circumference to the diameter is 2.

Which is less than 3.

In fact, as the radius approaches the diameter of the sphere, the ratio tends to zero.

So I must conclude that this is not what m_f_h had in mind.

Correct, and my apologies for making the misleading statement.

The ratio circumference/radius in a spherical geometry is less than or equal to \pi. It is greater than or equal to \pi in a hyperbolic geometry.


Paul

[R.D. Silverman]

Originally Posted by xilman

Correct, and my apologies for making the misleading statement.

The ratio circumference/radius in a spherical geometry is less than or equal to \pi. It is greater than or equal to \pi in a hyperbolic geometry.


Paul

And it is no longer a constant. Its value depends on the diameter of the
circle!!!!!

[Wacky]

Originally Posted by xilman

Correct, and my apologies for making the misleading statement.

The ratio circumference/radius in a spherical geometry is less than or equal to \pi. It is greater than or equal to \pi in a hyperbolic geometry.


Paul

Paul, again you are mixing radius and diameter. As you noted earlier, the ratio is 4, which is greater than \pi in the case of the equator.

You need to use 2\pi when you refer to the radius.

Other than that, your observation is correct and relevant to the class of geometries referenced.