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Repeating digits of rational numbers

Project description

Recent changes:

  • add some (La)TeX output (see bottom of this page)

  • change contFractionQ function output

This package provides a way to access repeating decimals in the decimal representation of rational numbers.

class object

>>> from dvtDecimal import *

Once package importation completed, you have to create a rational number using:

  • a fraction representation
>>> f = dvtDecimal(-604, 260)

for the fraction whose numerator is -604 and denominator is 260.

  • or a decimal representation, it could be an integer
>>> f = dvtDecimal(2.5)
  • or repeating decimals as a string
>>> f = dvtDecimal('00765')

thus creating a number (w/o irregular part) between 0 and 1. In the example, 0.007650076500765... and so on.

object methods and variables

Once you created the object, you can access to those variables:

>>> f.initValues
[-604, 260]
>>> f.simpValues
[-151, 65]
>>> f.intPart
-2
>>> f.repPart
[2, 3, 0, 7, 6, 9]
>>> f.sign
-1
>>> f.gcd
4

and to those methods:

>>> f.fraction()
-604/260
>>> f.irrPart()
0.3
>>> f.repPartC()
230769
>>> f.periodLen()
6
>>> f.mixedF()
[-2, 21, 65]
>>> f.isDecimal()
False
>>> f.dotWrite(20)
-2.32307692307692307692
>>> f.dispResults()
For fraction: -604/260
    integer   part : -2
    irregular part : 0.3
    periodic  part : [2, 3, 0, 7, 6, 9]
    mixed fraction : [-2, 21, 65]
    simp. fraction : [-151, 65]
               gcd : 4
    Python outputs : -2.3230769230769233

Entering via repeating decimals string allows:

>>> f = dvtDecimal('0123456789')
>>> f.simpValues
[13717421, 1111111111]

dvtDecimal also supports minimal operations (+,-,*,/) in between elements of the class but also with integers:

>>> f = dvtDecimal(1, 5)
>>> g = dvtDecimal(10, 3)
>>> h = f + g
>>> h.mixedF()
[3, 8, 15]
>>> i = f / g
>>> i.mixedF()
[0, 3, 50]
>>> f = dvtDecimal(1, 5)
>>> g = 5
>>> h = f * g
>>> h.isDecimal()
True
>>> f = dvtDecimal(1, 5)
>>> g = dvtDecimal(7, 5)
>>> h = f - g
>>> h.simpValues
[-6, 5]

egyptian fractions

IMPORTANT 1: Egyptian fractions features are quite slow (could be VERY slow) because it uses dvtDecimal representation of numbers (unlike previous versions). As it doesn't use algebraic computation, you have to untrust some results or at least to verify them with another method...

IMPORTANT 2: This part gives you results for the fractionnal part of the rational number you work with i.e. the last two numbers in the mixedF list, so that you only get results for <=1 fractions.

dvtDecimal provides the method egyptFractions to get all the egyptian fractions equal to your current fraction.

egyptFractions outputs a list of lists. Each of these lists are denominators (increasing) for unitary fractions whose sum equals to your fraction.

Two optional arguments:

  • eF number of fractions in the sums, default is 3
  • lim max. number of solutions in the results, default is 10
  • lim can be 0 for all fractions!
>>> f = dvtDecimal(18,5)
>>> f.mixedF()
[3, 3, 5]
>>> f.egyptFractions()
[[2, 11, 110], [2, 12, 60], [2, 14, 35], [2, 15, 30], [2, 20, 20], [3, 4, 60], [3, 5, 15], [3, 6, 10]]
>>> f.egyptFractions(lim=5)
[[2, 11, 110], [2, 12, 60], [2, 14, 35], [2, 15, 30], [3, 4, 60]]
>>> f.egyptFractions(eF=4, lim=1)
[[2, 11, 111, 12210]]
>>> f.egyptFractions(eF=4, lim=0)
[[2, 11, 111, 12210], [2, 11, 112, 6160], [2, 11, 114, 3135], [2, 11, 115, 2530], [2, 11, 120, 1320], [2, 11, 121, 1210], [2, 11, 130, 715], [2, 11, 132, 660], [2, 11, 135, 594], [2, 11, 154, 385], [2, 11, 160, 352], [2, 11, 165, 330], [2, 11, 210, 231], [2, 12, 61, 3660], [2, 12, 62, 1860], [2, 12, 63, 1260], [2, 12, 64, 960], [2, 12, 65, 780], [2, 12, 66, 660], [2, 12, 68, 510], [2, 12, 69, 460], [2, 12, 70, 420], [2, 12, 72, 360], [2, 12, 75, 300], [2, 12, 76, 285], [2, 12, 78, 260], [2, 12, 80, 240], [2, 12, 84, 210], [2, 12, 85, 204], [2, 12, 90, 180], [2, 12, 96, 160], [2, 12, 100, 150], [2, 12, 105, 140], [2, 12, 108, 135], [2, 12, 110, 132], [2, 13, 44, 2860], [2, 13, 45, 1170], [2, 13, 50, 325], [2, 13, 52, 260], [2, 13, 60, 156], [2, 13, 65, 130], [2, 14, 36, 1260], [2, 14, 40, 280], [2, 14, 42, 210], [2, 14, 60, 84], [2, 15, 31, 930], [2, 15, 32, 480], [2, 15, 33, 330], [2, 15, 34, 255], [2, 15, 35, 210], [2, 15, 36, 180], [2, 15, 39, 130], [2, 15, 40, 120], [2, 15, 42, 105], [2, 15, 45, 90], [2, 15, 48, 80], [2, 15, 50, 75], [2, 15, 55, 66], [2, 16, 27, 2160], [2, 16, 28, 560], [2, 16, 30, 240], [2, 16, 32, 160], [2, 16, 35, 112], [2, 16, 40, 80], [2, 16, 48, 60], [2, 17, 25, 850], [2, 17, 34, 85], [2, 18, 23, 1035], [2, 18, 24, 360], [2, 18, 25, 225], [2, 18, 27, 135], [2, 18, 30, 90], [2, 18, 35, 63], [2, 18, 36, 60], [2, 20, 21, 420], [2, 20, 22, 220], [2, 20, 24, 120], [2, 20, 25, 100], [2, 20, 28, 70], [2, 20, 30, 60], [2, 20, 36, 45], [2, 21, 28, 60], [2, 21, 35, 42], [2, 24, 30, 40], [3, 4, 61, 3660], [3, 4, 62, 1860], [3, 4, 63, 1260], [3, 4, 64, 960], [3, 4, 65, 780], [3, 4, 66, 660], [3, 4, 68, 510], [3, 4, 69, 460], [3, 4, 70, 420], [3, 4, 72, 360], [3, 4, 75, 300], [3, 4, 76, 285], [3, 4, 78, 260], [3, 4, 80, 240], [3, 4, 84, 210], [3, 4, 85, 204], [3, 4, 90, 180], [3, 4, 96, 160], [3, 4, 100, 150], [3, 4, 105, 140], [3, 4, 108, 135], [3, 4, 110, 132], [3, 5, 16, 240], [3, 5, 18, 90], [3, 5, 20, 60], [3, 5, 24, 40], [3, 6, 11, 110], [3, 6, 12, 60], [3, 6, 14, 35], [3, 6, 15, 30], [3, 7, 10, 42], [3, 8, 10, 24], [3, 9, 10, 18], [4, 5, 7, 140], [4, 5, 8, 40], [4, 5, 10, 20], [4, 5, 12, 15], [4, 6, 10, 12]]

In the above example, egyptian fractions are thus calculated with fraction 3/5.

First solution [2, 11, 111, 12210] (in the last command result) has to be interpreted as: 3/5 = 1/2 + 1/11 + 1/111 + 1/12210

dvtDecimal also provides egyptG2 method. This gives you a list of two denominators of unitary fractions whose sum equals your fraction.

This is an implementation of the greedy algorithm. It gives you naturally matching pairs for unitary fractions input.

The result may be an empty list since all numbers cannot be written as so.

>>> f = dvtDecimal(18,5)
>>> f.intPart
3
>>> f.egyptG2()
[2, 10]

So: 18/5 = 3 + 1/2 + 1/10

continued fractions

what is implemented and what is not

you can:

  • transform a dvtDecimal in a (finite) sequence which terms are integers in the continued fraction representation. This is done with the contFraction method;

  • get successive rational approximations from a list representing a continued fraction. This is done using ratApp function on a list. Output of this function is a list of dvtDecimal objects.

you cannot:

  • thus so far, no periodic continued fraction forms

  • transform irrationals...

some examples

  • Looking for continued fractions

Below, search for 634 / 75 and for 3.14159265

>>> f = dvtDecimal(634, 75)
>>> f.contFraction()
[8, 2, 4, 1, 6]
>>> f = dvtDecimal(3.14159265)
>>> f.contFraction()
[3, 7, 15, 1, 288, 1, 2, 1, 3, 1, 7, 4]

Every exact number gives a truthy continued fraction form.

In the latter example, the continued fraction do not represent π for the decimal 3.14159265 is not π

We can observe that numbers are quickly false: true continued fraction of π begins with [3, 7 15, 1, 292, 1, 1, 1, ...]

  • Looking for rational approximations
>>> f = dvtDecimal(634, 75)
>>> c = f.contFraction()
>>> print(ratApp(c))
[<dvtDecimal.dvtDecimal object at 0x7f40b8c9ec18>, <dvtDecimal.dvtDecimal object at 0x7f40b8c9ec50>, <dvtDecimal.dvtDecimal ob
ject at 0x7f40b8c9ec88>, <dvtDecimal.dvtDecimal object at 0x7f40b8c9ecc0>, <dvtDecimal.dvtDecimal object at 0x7f40b8c9ecf8>]
>>> for d in ratApp(c):
        print("{:<1}+{:>3}/{:<2}  {:<11}".format(*d.mixedF(), d.dotWrite(10)))
8+  0/1   8.0000000000
8+  1/2   8.5000000000
8+  4/9   8.4444444444
8+  5/11  8.4545454545
8+ 34/75  8.4533333333
>>> f = dvtDecimal(3.14159265)
>>> c = f.contFraction()
>>> for d in ratApp(c):
        print("{:<1}+{:>7}/{:<8}  {:<8}".format(*d.mixedF(), d.dotWrite(15)))
3+      0/1         3.000000000000000
3+      1/7         3.142857142857142
3+     15/106       3.141509433962264
3+     16/113       3.141592920353982
3+   4623/32650     3.141592649310872
3+   4639/32763     3.141592650245703
3+  13901/98176     3.141592649934810
3+  18540/130939    3.141592650012601
3+  69521/490993    3.141592649997046
3+  88061/621932    3.141592650000321
3+ 685948/4844517   3.141592649999989
3+2831853/20000000  3.141592650000000

The solar year is approximately 365 days, 5 hours, 48 minutes and 45.198 seconds leading to 365.2421898 days (see Wikipedia).

Let's play with this!

We are going to round this number with a precision from 1 to 7 digits then transform in continued fractions and find all the rational approx. given by these numbers using a mixed fraction. As some approx. will be equal, get unique representations and sort then using numerators (in mixed form).

>>> aS = 365.2421898
>>> mF = []
>>> for a in range(7):
        aaS = round(aS,7-a)
        f = dvtDecimal(aaS)
        c = f.contFraction()
        for d in ratApp(c):
            mF+=[d.mixedF()]

OK! I got all my mixed fractions from all roundings. I, now, create a list of uniques since a lot of them are equal, for this I've used this code from StackOverflow:

>>> mF = [list(x) for x in set(tuple(x) for x in mF)]
>>> mF.sort(key=lambda l:l[1])
>>> mF
[[365, 0, 1], [365, 1, 5], [365, 1, 4], [365, 6, 25], [365, 7, 29], [365, 8, 33], [365, 15, 62], [365, 31, 128], [365, 53, 219
], [365, 121, 500], [365, 132, 545], [365, 163, 673], [365, 295, 1218], [365, 458, 1891], [365, 752, 3105], [365, 814, 3361], 
[365, 1211, 5000], [365, 2473, 10211], [365, 3287, 13572], [365, 4543, 18758], [365, 5760, 23783], [365, 9838, 40621], [365, 1
4807, 61138], [365, 20567, 84921], [365, 24219, 100000], [365, 35374, 146059], [365, 126689, 523098], [365, 542130, 2238451], 
[365, 1210949, 5000000]]

This done, it's difficult to use for deciding leap years since denominators are quite maladjusted except for 5, 4, 25, 500, 5000, 100000, and 5000000. Now looking at numerators for these fractions, we recognize 1 year out of 4 and that's all! Where is 97 out of 400?

Actually, we do not use 365.2421898 but 365.2425, let's see with it:

>>> f = dvtDecimal(365.2425)
>>> c = f.contFraction()
>>> for d in ratApp(c):
        print("{:<1}+{:>3}/{:<3}  {:<4}".format(*d.mixedF(), d.dotWrite(4)))
365+  0/1    365.0000
365+  1/4    365.2500
365+  8/33   365.2424
365+ 97/400  365.2425
  • Continued fraction of a quadratic non rational number

contFractionQ function (it's not a method!) provides this:

>>> import math
>>> contFractionQ(math.sqrt(2019))
[44, [1, 13, 1, 88]]

The nested list is the periodic part of the continued fraction.

(La)TeX output

Since 1.3.0 version, dvtDecimal offers some (La)TeX output functions. Actullaly this is native TeX, so that it doesn't require any special package.

MY usage is to format continued fractions with \displaystyle, if you don't want them, just search and destroy with your (La)TeX editor!

  • toTeX method for dvtDecimal objects

Outputs fraction (w/o dps!), mixed fraction and decimal representation.

>>> f = dvtDecimal(1, 7)
>>> f.toTeX()
['{1\\over7}', '{0\\raise.21em\\hbox{$\\scriptscriptstyle\\frac{1}{7}$}}', '0.\\overline{142857}']
>>> print(f.toTex()[1])
{0\raise.21em\hbox{$\scriptscriptstyle\frac{1}{7}$}}
  • egToTeX function

Outputs egyptian fractions sum for list as argument. The list is typically output of the egyptG2 method.

>>> f = dvtDecimal(1, 7)
>>> f.egyptG2()
[8, 56]
>>> egToTeX(f.egyptG2())
'{1\\over8}+{1\\over56}'
>>> print(egToTeX(f.egyptG2()))
{1\over8}+{1\over56}
  • continued fractions functions: cfToTeX and cfQToTeX

First outputs full continued fraction of a finite (flat) list. Typically to use with contFraction method.

>>> f = dvtDecimal(123, 43)
>>> f.contFraction()
[2, 1, 6, 6]
>>> cfToTeX(f.contFraction())
'2+\\displaystyle{\\strut1\\over\\displaystyle1+{\\strut1\\over\\displaystyle6+{\\strut1\\over6}}}'
>>> print(cfToTeX(f.contFraction()))
2+\displaystyle{\strut1\over\displaystyle1+{\strut1\over\displaystyle6+{\strut1\over6}}}

Last outputs partial continued fraction of a periodic continued fraction. Typically to use with contFractionQ function.

So the list to provide contains numbers and a list (see contFractionQ) and needs an integer as a second argument: this is the index of the end, you'll get \dots in the last unit fraction.

>>> import math
>>> contFractionQ(math.sqrt(23))
[4, [1, 3, 1, 8]]
>>> cfQToTex(contFractionQ(math.sqrt(23)), 7)
[4] [1, 3, 1, 8]
'4+\\displaystyle{\\strut1\\over\\displaystyle1+{\\strut1\\over\\displaystyle3+{\\strut1\\over\\displaystyle1+{\\strut1\\over\\displaystyle8+{\\strut1\\over\\displaystyle1+{\\strut1\\over\\displaystyle3+{\\strut1\\over\\displaystyle\\dots}}}}}}}'
>>> print(cfQToTex(contFractionQ(math.sqrt(23)), 7))
[4] [1, 3, 1, 8]
4+\displaystyle{\strut1\over\displaystyle1+{\strut1\over\displaystyle3+{\strut1\over\displaystyle1+{\strut1\over\displaystyle8+{\strut1\over\displaystyle1+{\strut1\over\displaystyle3+{\strut1\over\displaystyle\dots}}}}}}}

further

I hope to:

  • improve contFractionQ function and provide a way to get succesive rational approximations ;

  • improve egyptian fractions methods to speed up calculations.

about

dvtDecimal is rather an attempt to publish on the PyPi packages index than a fully completed python project, I do not recommend dvtDecimal usage for professionnal use. You have to consider this package as an experiment.

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