Zeta 137: 12. Wallis and Squaring of the Circle

The squaring of the circle, together with the problem of angle trisection and that of cube duplication, is a classical problem in Greek mathematics, whose purpose is to construct a square that has the same area as a given circle, with exclusive use of straightedge and compass.

In 1882, Ferdinand Lindemann, of the University of Munich, showed that pi is transcendent, thus forever drawing a line on the problem of squaring the circle; from the proof provided by Lindemann, it turns out that it is impossible to construct, only with straightedge and compass, a square of area equal to that of a given circle, a problem that has tormented entire generations of mathematicians since before Euclid. Lindemann showed that pi is not an algebraic number.

Any geometry problem that can be solved only with the ruler and the compass, when placed under its equivalent algebraic form, leads to one or more algebraic equations with integer rational coefficients that can be solved by successive extractions of square roots. Since π does not satisfy any of these equations, we cannot arrive at the quadrature of the circle with the instruments in question.

Note: with row and compass it is obviously possible to "construct" the positive integers as whole distances from the given origin. Every rational number is constructible, and this thanks to Thales' beautiful theorem on similar triangles; furthermore it can be shown that the square root of any constructible number is itself constructible.

A fractal is a geometric object that is repeated in its form in the same way on different scales, and by enlarging any part of it you get a figure similar to the original.

In the previous post we saw the Sierpinski carpet, a fractal obtained from a square, described by the Polish mathematician Wacław Sierpiński in 1916. At each step the squares that make up the figure are divided into 9 smaller squares and the central square is removed. In this way, for each step the area is reduced by 8/9, so the fractal dimension of the carpet is log 8 / log 3, equal to 1.892789 ...

Now let's see another object that "might" look similar to the previous one.

The sieve of Wallis is thus constructed:

it starts with a 2 x 2 square and is divided into 4 squares;

divide each new sub-panel by 9 and remove the central square (8/9);

each new sub-panel is divided by 25 and the central square (24/25) is removed;

each new sub-panel is divided by 49 and the central square (48/49) is removed;

each new sub-panel is divided by 81 and the central square (80/81) is removed

and so on.

Here's how it looks after a few steps:

I said "could" because the Wallis sieve is a fractal-like, at every step the rule changes and therefore self-similarity is not maintained. However, while in the case of the carpet the area (Lebesgue measure) is zero, for the sieve the area has a well-known value.

The initial value is by definition 4, then in the order it is multiplied by 8/9, 24/25, 48/49, 80/81, ..., (n² - 1) / n²

This is the famous Wallis product:

which can also be rewritten in this way:

The wonder lies in the fact that the areas of the Wallis sieve and the Circle are the same.

Let's see 2 more views:

in the first figure we start with only 1 square and relate to the relative inscribed circle, while the same object is then decomposed and recomposed to better highlight the equivalence of circle and sieve.

With straightedge and compass you can therefore build a "square" equivalent to the circle, just have patience and the necessary time ... if it is not exactly a square, it somehow looks like it.

Another way could be the following: take a square of side 2, remove a concentric square of area 4/3, add one of area 4/5 and continue like this to infinity ... you will have thus obtained the series of Gregory – Leibniz multiplied by 4, that is pi.

Size of yellow area is equal to pi

Let us ask ourselves this question:

is there a Menger sponge equivalent to the Wallis sieve?

It starts from a cube of side 1 and divides the side into 3 parts (27 cubes as for the Rubik's cube), the central cube and the 6 central cubes are removed with each face (so 20 cubes remain), in the next step divide the side of each new cube by 5 removing the central cube and the cubes that extend from the center, then repeat the same procedure with 7, 9, 11 and the subsequent odd numbers.

Wonder of wonders, 8 of these cubes have the same volume as a sphere of unit radius

or equivalent

cube of side 1 has the same volume as sphere of unitary diameter