fractionizer: Allows to approximate the fractional numbers between 0 and 1 with sum of two unit fractions of special types.

[ #AncientEgypt, language, library, math, mit, music, sound ] [ Propose Tags ]
This version is deprecated.

Inspired by the ancient Egyptian mathematics. See the ideas description and formulation in the related draft paper.


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Versions [RSS] 0.1.0.0, 0.1.0.1, 0.2.0.0, 0.3.0.0, 0.4.0.0, 0.5.0.0, 0.6.0.0, 0.6.1.0, 0.6.1.1, 0.6.1.2, 0.7.0.0, 0.7.0.1, 0.8.0.0, 0.9.0.0, 0.10.0.0, 0.11.0.0, 0.12.0.0, 0.13.0.0, 0.14.0.0, 0.14.1.0, 0.15.0.0, 0.16.0.0, 0.17.0.0 (info)
Change log CHANGELOG.md
Dependencies base (>=4.13 && <5) [details]
License MIT
Copyright Oleksandr Zhabenko
Author Oleksandr-Zhabenko
Maintainer oleksandr.zhabenko@yahoo.com
Category Math, Music, Sound, Language
Uploaded by OleksandrZhabenko at 2023-04-19T09:07:14Z
Distributions NixOS:0.17.0.0
Downloads 480 total (52 in the last 30 days)
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Status Docs available [build log]
Last success reported on 2023-04-19 [all 1 reports]

Readme for fractionizer-0.1.0.0

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Mathematical problem solution {#mathematical-problem-solution .unnumbered}

Let's consider the problem of decomposition (approximation) of fractions that lie in the range from 0 to 1 by the sum of two fractions with 1 in the numerators (so called unit fractions) and, possibly, natural, unequal numbers in the denominators.

We will consider one of the two denominators to be a natural number, smaller than the other number in the denominator, and we will look for such a pair of numbers that, when rounded to natural numbers, will give the minimum absolute error of the resulting approximation. Mathematically, this leads to a problem and its further solution below.

\[\dfrac{1}{a} + \dfrac{1}{b} = k,\quad k,\,b \in \mathrm{R},\; a \in \mathrm{N},\; 0 < k < 1,\; b>0,\, a < b\]

\[\dfrac{1}{a} + \dfrac{1}{b} = k = \dfrac{a+b}{ab}>\dfrac{2}{a+b}=\dfrac{1}{\dfrac{a+b}{2}}\]

\[b=\dfrac{a}{ka-1}\]

Logically, there are possible two situations, but the first one (the two equations below) leads to the contradiction with the idea of approximation, so we use the second option.

\[1<a<\dfrac{1}{k}<b\] \[1 < b < 0,\quad \emptyset\]

The second option is below:

\[1 < \dfrac{1}{k}<a<b\]

Let's consider the following option and prove that it leads to no solutions.

\[a \geq \dfrac{2}{k}\] \[\dfrac{1}{a} \leq \dfrac{k}{2}\] \[\dfrac{1}{b} \geq k - \dfrac{k}{2} = \dfrac{k}{2}\] \[a < b \leq \dfrac{2}{k} \leq a,\quad \emptyset\]

This contradiction above proves that we should seek for solution using the following:

\[a \in \left[\left[\dfrac{1}{k}\right]+1,\ldots ,\; \left[\dfrac{2}{k}\right]\right], \; b = \dfrac{a}{ka - 1}\] For the solutions above there is a criterion of their usability below:

\[b = na, n \geq 1\] \[na = \dfrac{a}{ka-1}\] \[a=\dfrac{n+1}{kn}<\left[\dfrac{2}{k}\right]\] \[n>\dfrac{1}{\left[\dfrac{2}{k}\right]\cdot k - 1}\]

Relation to music rhythm and meter {#relation-to-music-rhythm-and-meter .unnumbered}

It can be used to produce music by approximation of the meter. This leaeds to interesting structures, that have some commonality with the structures, which are written about in the work [@link1994long].

It also is directly connected to the irrational time signatures in music (see: [@wheatley2019use]).

Acknowledgements {#acknowledgements .unnumbered}

Author would like to support the foundation Gastrostars and its founder Emma Kok. The founder inspired him to conduct such a research. Besides, the author is grateful to the Hackage website for publishing the Haskell code related to the research. On the 2023-04-19 there is the founder's namesday, the memory of St. Emma of Lesum or Emma of Stiepel
(also known as Hemma and Imma). If you would like to share some financial support, please, contact the foundation using the URL:

Contact Foundation GASTROSTARS