17. Benford law

Benford's law (binary numbering):

for any number greater than zero,

"1" appears as the first digit in

 100% of cases.

The problem was this: consider the first digit in the decimal expansion of 2ⁿ for n ≥ 0: 1, 2, 4, 8, 1, 3, 6, 1, 2, 5, 1, 2, 4, ...

Does the digit 7 appear in this sequence?

Does 7 or 8 appear more frequently? 

How much more frequently?

At the time I did not know which the correct way was to solve the problem, but I approached it like this: on a logarithmic scale the products are transformed into sums, for example multiplying by 2 is equivalent to adding log (2) and we obtain in sequence: 1, 2, 4, 8, 16, 32, 64, 128, ...

The first digits of the sequence are: 1, 2, 4, 8, 1, 3, 6, 1, 2, 5, 1, 2, 4, 8, 1, 3, 6, 1, 2, 5, 1, 2, 4, 8, 1, 3, 6, 1, 2, 5, 1, 2, 4, 8, 1, 3, 6, 1, 2, 5, 1, 2, 4, 8, 1, 3, 7, 1, 2, 5, 1, 2, 4, 9, 1, 3, 7, 1, 2, 5, 1, 2, 4, 9, 1, 3, 7, 1, 2, 5, 1, 2, 4, 9, 1, 3, 7, 1, 3, 6, 1, 2, 4, 9, 1, 3, 7, 1, 3, 6, 1, 2, 4, 9, 1, 3, ...

Sloane's On-Line Encyclopedia of Integer Sequences A008952

By definition, position 1 is set to log₁₀ 1 = 0, while position 10 to 1 (log₁₀ 10 = 1) and 100 to 2 (log₁₀ 100 = 2), the intermediate positions are at:

log₁₀ 2 = 0.3010, log₁₀ 3 = 0.4771, log₁₀ 4 = 0.6020,…, log₁₀ 9 = 0.9542

Each interval I (m) of numbers starting with the digit m is between log (m) and log (m + 1), so the various intervals hold:

log (m + 1) - log (m) = log ((m + 1) / m) = log (1 + 1/m)

Returning to the problem, the digit 7 appears log (1 + 1/7) = 0.05799 = 5.8% and to see it appear we have to wait 2⁴⁶ = 70368744177664.

The second question can be answered that it appears more 7, and

for n > 209, the frequency f (7) of the digit 7 is greater than that of 8:

f (7) / f (8) tends to log₁₀ (1 + 1/7) / log₁₀ (1 + 1/8) = 1.133706496

Note: I(1) = I(2) + I(3) = I(4) + I(5) + I(6) + I(7)

Benford's law arises from the observation that in many collections of numbers, eg. mathematical tables, real-life data, or combinations thereof in various units of measurement, the initial significant digits are not evenly distributed, as you might expect, but are larger for the smaller digits.

It claims that the significant figures in many data sets follow a logarithmic distribution. In its most common formulation, it is also known as the "law of the first digit" and is named after the American physicist Frank Benford (1883-1948) who formulated it in 1938, although it had already been highlighted by the American astronomer Simon Newcomb (1835 -1909) in 1881.

Benford observed that the logarithmic tables, used at the time for calculations, had very crumpled first pages and therefore deduced that the numbers beginning with 1 occurred more often than those beginning for the other digits.

This distribution has a characteristic property known as "scale invariance" and is often used to discover "forged" data.

If you were to falsify numbers, make sure that the number 1 appears in about 30% of the cases, 2 in 17% and so on.

For a given number of digits, it is possible to calculate the probability of encountering a number starting with the digit string n of that length. Below what is reported in Wikipedia under "Benford's law".

The distribution of the n-th digit, as n increases, quickly approaches a uniform distribution with 10% for each of the ten digits, as shown below:

Benford Online Bibliography

https://mathworld.wolfram.com/BenfordsLaw.html

Benford's Law (mathpages.com)

Index to OEIS: Section Be - OeisWiki

https://oeis.org/A008952

16. Binary code

If you divide a square of unit area into 2 equal parts, each will have a value of 0.5.

By dividing one of these into 2 equal parts and continuing to dissect in the same way, the values will be obtained: 0.5; 0.25; 0.125; 0.0625; 0.03125; ...

In other words, a geometric series is obtained whose sum converges to 1:

A scale with available weights: 0.5 lb; 0.25 lb; etc., could measure any object less than 1 pound. What we are doing is using a numbering system based on 2.

Probably everyone has tried their hand at binary numbers, but few people will have used this base to deal with decimals, here we propose an example: 

3,14159265358979323846…  = 11,00100100001111110110101...

The whole part will probably be clear to everyone, while the other part will require a minimum of reasoning.

These tables can perhaps help:

 

Here are some examples of pi in various numbering systems.

As an exercise, you can see how simple the transition from base 2 to base 4 and from base 4 to base 16 is.

base-2  (binary code) :  11.0010010000111111011010101000100010000101101000110000100011010011000100110001100110001010001011100000 …

base-4 : 3.0210033312222020201122030020310301030121202202320003130013031010221000210320020202212133030131000020 …

base-16  (exadecimal code): 3.243F6A8885A308D313198A2E03707344A4093822299F31D0082EFA98EC4E6C89452821E638D01377BE5466CF34E90C6CC0AC …

 

base-3 : 10.0102110122220102110021111102212222201112012121212001211001001012220222120120121112101210112002201202 ...

base-5 : 3.0323221430334324112412240414023142111430203100220034441322110104033213440043244401441042334133011323 ...

base-6 : 3.0503300514151241052344140531253211023012144420041152525533142033313113553513123345533410015154344401 ...

base-7 : 3.0663651432036134110263402244652226643520650240155443215426431025161154565220002622436103301443233631 ...

base-8 : 3.1103755242102643021514230630505600670163211220111602105147630720020273724616611633104505120207461615 ...

base-9 : 3.12418812407442788645177761731035828516545353462652301126321450283864034354163303086781327871588 ...

 

Binary number - Wikipedia

Zibaldone Scientifico: pi (zibalsc.blogspot.com)

https://robertlovespi.net/2014/06/09/the-beginning-of-the-number-pi-in-binary-through-hexadecimal-etc/

15. Oloid

Discovered by Paul Schatz in 1929, the Oloid is defined as the convex hull of 2 circumferences of radius R equal to each other, arranged on two orthogonal planes and such that each of the 2 steps to the center of the other.

This grooved surface has many interesting properties (for example, all of its generating lines have the same length).

The Oloid has a shape particularly suitable for the mixing of fluids.

The Oloid surface S is equal to the area of the sphere of radius R;

while by means of numerical calculation, it can estimate the volume V:

The Oloid is the only three-dimensional shape that can rotate on its entire surface.

When is made to roll on a flat horizontal surface, the Oloid moves uniformly because the distance from its center of mass at the surface is almost constant.

Prof. Dr. Peter Berger:    http://www.prof-dr-berger.de/_figurint-tork.htm

 The dell'Oloide equation corresponds to an algebraic surface of order 8 :

http://dictionnaire.sensagent.leparisien.fr/Oloide/es-es/

 https://demonstrations.wolfram.com/MovementOfAnOloid/

https://en.wikipedia.org/wiki/Oloid

https://www.youtube.com/watch?v=GM3_JuFgJ2E

http://www.dimnp.unipi.it/gabiccini-m/robotica.html

http://www.ucke.de/christian/physik/ftp/lectures/udine2003.pdf

http://www.heldermann-verlag.de/jgg/jgg01_05/jgg0113.pdf

http://www.mathcurve.com/surfaces.gb/orthobicycle/orthobicycle.shtml

http://www.mathcurve.com/surfaces/orthobicycle/orthobicycle.shtml

14. Geodesics on a Polyhedron

In mathematics, a geodesic is the shortest curve that connects two points. The geodesics in the plane are straight lines, on a sphere they are the arcs of maximum circle. At each point the main normal to it coincides with the normal to the surface at that point; that is, the osculating plane line is normal to the surface at that point.

What are the shortest paths on polyhedron?

A polyhedron is a solid bounded by a finite number of polygonal plane faces and, if 2 points lie on the same face, the shortest path is the straight line segment. If instead the 2 points lie on adjacent faces, then the shortest path belongs to a straight line on the developed surface (open on the plane).

Prolonging this segment, beyond its extremes, we can join all the pairs of points on the surface of the polyhedron.

A cubic-edged planet with an edge equal to the diameter of the Earth (12,746 km) would have a diagonal of 22,077 km. The difference between the highest and the deepest point would be 4,665 km (4,665,000 meters). On Earth between Mount Everest (8,844 m) and the Mariana Trench (-10,911 m) there is a difference in height of 19,755 m. Assuming 10,000 m deep oceans, the 8 vertices would be mountains 500 times the height of Mount Everest. To go from one of the 6 faces to the other you should overcome passes with an altitude equal to 300 times Mount Everest. The oceans could be at most 6.

 

On the Carlos Furuti website, you can find many Cartographic Projections:

http://www.progonos.com/furuti/MapProj/Normal/ProjPoly/Foldout/Cube/cube.html

and you can even print a cubic Earth.

The great British mathematician Henry Ernest Dudeney (1857 - 1930) invented hundreds of mathematical games a couple will be reported dealing with geodesics on polyhedron.

THE JOURNEY OF THE MOSCOW

A fly, starting from point A, can travel the 4 sides of the base of a cubic block in 4 minutes. How long will it take to get from A to the opposite corner B?

The fly would select the path shown by the line in the illustration, which will require 2.236 minutes. It will not go in the direction indicated by the dotted line that might seem the suggested one. This path is longer and takes more time (2.414 minutes).

THE SPIDER AND THE FLY

A rectangular room has the dimensions shown in the figure (in English measurements). A spider, indicated with the yellow star, is located in the center of one of the two bottom walls one foot from the ceiling. A fly, indicated with the dark star, is instead found on the opposite wall, one foot from the floor. What is the shortest distance the spider has to travel to reach the fly?

The easiest way to solve the problem is to develop the room on the floor. In this way the shortest path is the hypotenuse of a right triangle, equal to 40 feet.

The paradox, in this case, consists in the fact that the horizontal path might seem shorter, but it is not difficult to verify that it is equal to 42 feet (2 more).

http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Dudeney.html

http://mathworld.wolfram.com/SpiderandFlyProblem.html
https://pbbmath.weebly.com/blog/the-logic-puzzles-of-henry-dudeney
http://www.ilovephilosophy.com/viewtopic.php?f=4&t=181293
https://math.stackexchange.com/questions/292495/gentle-introduction-to-quasi-geodesics
http://www.progonos.com/furuti/MapProj/Normal/ProjPoly/projPoly3.html

13. Squaring the Circle in n-Dimensions

In the previous post we saw how, starting from the Wallis sieve, we can arrive at a 3D-cubic fractal-like and by assembling 8 of these cubes we have the same volume as a sphere of unitary radius. Below we will explain in an analytical way how to build a sieve in 2D, 3D, etc.

For 2D we have a square divided into 9 equal parts:

The number of elements (or cardinality) of the Cartesian product of 2 sets is the product of the number of elements of the 2 sets. Generalizing, the number of elements of the Cartesian product of n sets is the product of the number of elements of each set.

We can give a tabular representation, which consists in writing all the possible pairs enclosed in braces:

{(1, 1); (1, 2); (1, 3); (2, 1); (2, 2); (2, 3); (3, 1); (3, 2); (3, 3)}

or by means of a double entry table, obtained with the elements of the first set placed in the column and those of the second in the first row; the boxes contain the various pairs that are obtained:

As can be seen, the only square removed is (2, 2).

In the case of 3D fractal-like we can do a similar reasoning and the 7 cubes removed will be:

(1, 2, 2); (3, 2, 2); (2, 1, 2); (2, 2, 2); (2, 3, 2); (2, 2, 1); (2, 2, 3)

The following table summarizes (for each dimension) how many groups of numbers 2 we can find; eg in 3D we have 8 triples with zero 2, 12 with only 2, 6 with 2 elements equal to 2 and 1 with all the numbers equal to 2 (which corresponds to the central cube):

the combinations with at least 2 elements equal to 2 are 6 + 1 = 7, on a total of 27 cubes (shown in the last column) and for a cube of side 3.

We can do the same reasoning for a cube of side 5:

last column shows the number of cubes remaining.

Note: the coefficients that appear in the table can be easily calculated using the formula of the Newton binomial - in this case (4 + 1) n -

Now, in the previous post we saw that in 2D for an initial value of 4, multiplied by 8/9, 24/25, 48/49, 80/81, ..., (n² - 1) / n²

we have the formula derived from the Wallis product:

which can be summarized as follows:

While in 3D we have this result:

where the numbers in gray represent the various denominators.

I omit the passages and report what I calculated with Wolfram|Alpha:

which can be summarized:

In the formulas the Euler Gamma function appears which extends the factorial concept (even to complex numbers). At the end of the post I report the chart and some notable formulas concerning it.

Here I am only interested in noting that the right part of the formulas containing the index n tends to 1 when tending n to infinity.

So what remains of the 3 formulas is pi and the number is in the denominator. Remembering that each formula must be multiplied by the number of "quadrants", the first (2D) 2² = 4, the second (3D) for 2³ = 8 and the third (4D) for 16.

Thus obtaining the "Volumes" of the n-dimensional spheres:

This is not yet a real demonstration, but it contains all the elements and could also be used as an alternative method to calculate the volume of a hyper-sphere.

https://community.wolfram.com/groups/-/m/t/822984

https://www.wolframalpha.com/input/?i=product+(2k%2F(2k%2B1))%5E4+((2k%2B5)%2F(2k%2B1)),+k%3D1+to+infinity

https://math.stackexchange.com/questions/1783732/is-a-hypersphere-of-non-integer-dimension-a-fractal

https://math.stackexchange.com/search?q=fractal+sphere+gamma

http://zeta137.blogspot.com/2019/06/12-wallis-and-squaring-of-circle.html?_sm_au_=iQW2QM3DR7VWvD1j

http://zeta137.blogspot.com/2015/06/6-wallis-and-pippenger-infinite-products.html?_sm_au_=iQW2QM3DR7VWvD1j

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

https://blogs.scientificamerican.com/roots-of-unity/a-few-of-my-favorite-spaces-the-wallis-sieve/?redirect=1

https://demonstrations.wolfram.com/WallisSievePiApproximation/

https://community.wolfram.com/groups/-/m/t/822984

https://mathsedideas.blogspot.com/2018/02/on-pi-day.html?_sm_au_=iQWkRvfFNHZnNJj6