`continuedfractions.continuedfraction`#

class continuedfractions.continuedfraction.ContinuedFraction(numerator=0, denominator=None)[source]#

An object-oriented representation of a (finite) simple continued fraction.

An implementation of simple continued fractions as Python objects and instances of the standard library fractions.Fraction class, with various properties for the continued fraction, including its coefficients, the order, convergents, and remainders.

The term “simple continued fraction” denotes a specific type of continued fraction where the fractional terms only have numerators of \(1\).

Arguments can be any which are valid for creating objects of the fractions.Fraction superclass.

For clarification, valid arguments can be one of the following:

a single instance of numbers.Rational, including int, fractions.Fraction or ContinuedFraction, named or unnamed
a pair of numbers.Rational instances, including int, fractions.Fraction and ContinuedFraction, named or unnamed
a single float or decimal.Decimal value that is not a special value such as math.nan, float('inf'), or Decimal('infinity')
a single numeric valid string (str) - validity is determined in the superclass by the fractions._RATIONAL_FORMAT test

Attributes

`coefficients`	A generator of the (ordered) sequence of coefficients of the continued fraction.
`convergents`	Generates an enumerated sequence of all convergents of the continued fraction.
`counter`	A counter for the coefficients.
`even_order_convergents`	Generates an enumerated sequence of all even-order convergents of the continued fraction.
`height`	The height of this rational number.
`khinchin_mean`	The Khinchin mean of the continued fraction, which is defined as the geometric mean of all the tail coefficients.
`log_height`	The (natural) logarithm of the height of this rational number as defined above.
`odd_order_convergents`	Generates an enumerated sequence of all odd-order convergents of the continued fraction.
`order`	The order of the continued fraction, which is the number of its coefficients minus \(1\).
`remainders`	Generates an enumerated sequence of all remainders of the continued fraction in descending order of index.

Methods

`as_decimal`()	Returns the `decimal.Decimal` value of the continued fraction.
`convergent`(k, /)	Returns the \(k\)-th (simple) convergent of the continued fraction.
`extend`(*new_coeffs)	Performs an in-place tail extension of the current sequence of coefficients.
`from_coefficients`(*coeffs)	Returns a `ContinuedFraction` instance from a sequence of (integer) coefficients of a (finite) simple continued fraction.
`left_mediant`(other, /, *[, k])	Returns the \(k\)-th left-mediant of the continued fraction with another rational number.
`mediant`(other, /)	Returns the simple mediant of the continued fraction with another continued fraction.
`remainder`(k, /)	Returns the \(k\)-th remainder of the continued fraction.
`right_mediant`(other, /, *[, k])	Returns the \(k\)-th right-mediant of the continued fraction with another rational number.
`semiconvergent`(k, m, /)	Returns the \(m\)-th semiconvergent of two consecutive convergents \(p_{k - 1}\) and \(p_k\) of the continued fraction.
`truncate`(*tail_coeffs)	Performs an in-place truncation of a contiguous trailing segment of the coefficients in the tail.

property coefficients: Generator[int, None, None]#

A generator of the (ordered) sequence of coefficients of the continued fraction.

Examples

>>> cf = ContinuedFraction('.12345')
>>> cf
ContinuedFraction(2469, 20000)
>>> tuple(cf.coefficients)
(0, 8, 9, 1, 21, 1, 1, 5)

Type:: typing.Generator

property order: int#

The order of the continued fraction, which is the number of its coefficients minus \(1\).

Examples

>>> cf = ContinuedFraction('.12345')
>>> cf
ContinuedFraction(2469, 20000)
>>> tuple(cf.coefficients)
(0, 8, 9, 1, 21, 1, 1, 5)
>>> cf.order
7

Type:: int

property counter: Counter#

A counter for the coefficients.

Examples

>>> cf = ContinuedFraction(928374923, 8249234)
>>> cf.counter
Counter({1: 6, 2: 3, 24: 2, 112: 1, 5: 1, 3: 1})

Type:: collections.Counter

property khinchin_mean: Decimal | None#

The Khinchin mean of the continued fraction, which is defined as the geometric mean of all the tail coefficients.

We define the Khinchin mean \(K_n\) of a (simple) continued fraction \([a_0; a_1, a_2, \ldots, a_n]\) as:

\[K_n := \sqrt[n]{a_1a_2 \cdots a_n} = \left( a_1a_2 \cdots a_n \right)^{\frac{1}{n}}, \hskip{3em} n \geq 1\]

This property is intended to make it easier to study the limit of \(K_n\) as \(n \to \infty\).

In the special case of integers or fractions representing integers, whose continued fraction representations consist of only a single element, a null value is returned.

Examples

Note that the default decimal context precision of \(28\) is used in these examples.

>>> tuple(ContinuedFraction(649, 200).coefficients)
(3, 4, 12, 4)
>>> ContinuedFraction(649, 200).khinchin_mean
Decimal('5.76899828122963409526846589869819581508636474609375')
>>> tuple(ContinuedFraction(415, 93).coefficients)
(4, 2, 6, 7)
>>> ContinuedFraction(415, 93).khinchin_mean
Decimal('4.37951913988788898990378584130667150020599365234375')
>>> tuple((ContinuedFraction(649, 200) + ContinuedFraction(415, 93)).coefficients)
(7, 1, 2, 2, 2, 1, 1, 11, 1, 2, 12)
>>> (ContinuedFraction(649, 200) + ContinuedFraction(415, 93)).khinchin_mean
Decimal('2.15015313349074244086978069390170276165008544921875')
>>> ContinuedFraction(5000).khinchin_mean

Type:: decimal.Decimal or None

classmethod from_coefficients(*coeffs: int) → ContinuedFraction[source]#

Returns a ContinuedFraction instance from a sequence of (integer) coefficients of a (finite) simple continued fraction.

Invalid coefficients will trigger a ValueError.

Parameters:

*coeffsint: An ordered sequence of integer coefficients of a (finite) simple continued fraction.

Returns:

ContinuedFraction: A new and fully initialised instance of ContinuedFraction with the given sequence of coefficients.

Raises:

ValueError: If the given coefficients include non-integers, or where the tail segment contains non-positive integers.

Examples

Constructing a continued fraction for the rational \(\frac{649}{200}\) using the sequence of coefficients \(3, 4, 12, 4\).

>>> c1 = ContinuedFraction.from_coefficients(3, 4, 12, 4)
>>> c1
ContinuedFraction(649, 200)

Constructing the continued fraction of the (multiplicative) inverse \(\frac{200}{649}\) using the sequence of coefficients \(0, 3, 4, 12, 4\).

>>> c2 = ContinuedFraction.from_coefficients(0, 3, 4, 12, 4)
>>> c2
ContinuedFraction(200, 649)

Validation of coefficients.

>>> ContinuedFraction.from_coefficients('0', 1)
Traceback (most recent call last):
...
ValueError: Continued fraction coefficients must be integers, and all coefficients from the 1st onwards must be positive.
>>> ContinuedFraction.from_coefficients(0, 1, 2.5)
Traceback (most recent call last):
...
ValueError: Continued fraction coefficients must be integers, and all coefficients from the 1st onwards must be positive.
>>> ContinuedFraction.from_coefficients(1, 0)
Traceback (most recent call last):
...
ValueError: Continued fraction coefficients must be integers, and all coefficients from the 1st onwards must be positive.
>>> ContinuedFraction.from_coefficients(1, -1)
Traceback (most recent call last):
...
ValueError: Continued fraction coefficients must be integers, and all coefficients from the 1st onwards must be positive.

extend(*new_coeffs: int) → None[source]#

Performs an in-place tail extension of the current sequence of coefficients.

Raises a ValueError if the given sequence of values is empty, or includes non-integers or non-positive integers.

Note

As this method performs an in-place modification of the existing/ current instance the object ID remains the same.

Parameters:

new_coeffsint: An (ordered) sequence of new (positive) integer coefficients by which the tail of the existing sequence of coefficients is to be extended.

Raises:

ValueError: If the given sequence is empty or includes non-integers or non- positive integers.

Examples

>>> cf = ContinuedFraction('3.245')
>>> cf
ContinuedFraction(649, 200)
>>> tuple(cf.coefficients)
(3, 4, 12, 4)
>>> cf.order
3
>>> tuple(cf.convergents)
((0, ContinuedFraction(3, 1)), (1, ContinuedFraction(13, 4)), (2, ContinuedFraction(159, 49)), (3, ContinuedFraction(649, 200)))
>>> tuple(cf.remainders)
((3, ContinuedFraction(4, 1)), (2, ContinuedFraction(49, 4)), (1, ContinuedFraction(200, 49)), (0, ContinuedFraction(649, 200)))
>>> cf.extend(5, 2)
>>> cf
ContinuedFraction(7457, 2298)
>>> tuple(cf.coefficients)
(3, 4, 12, 4, 5, 2)
>>> cf.order
5
>>> tuple(cf.convergents)
((0, ContinuedFraction(3, 1)), (1, ContinuedFraction(13, 4)), (2, ContinuedFraction(159, 49)), (3, ContinuedFraction(649, 200)), (4, ContinuedFraction(3404, 1049)), (5, ContinuedFraction(7457, 2298)))
>>> tuple(cf.remainders)
((5, ContinuedFraction(2, 1)), (4, ContinuedFraction(11, 2)), (3, ContinuedFraction(46, 11)), (2, ContinuedFraction(563, 46)), (1, ContinuedFraction(2298, 563)), (0, ContinuedFraction(7457, 2298)))
>>> cf = ContinuedFraction(649, 200)
>>> cf.extend(0, 1)
Traceback (most recent call last):
...
ValueError: The coefficients to be added to the tail must be positive integers.
>>> cf.extend(1, -1)
Traceback (most recent call last):
...
ValueError: The coefficients to be added to the tail must be positive integers.

truncate(*tail_coeffs: int) → None[source]#

Performs an in-place truncation of a contiguous trailing segment of the coefficients in the tail.

Note

As this method performs an in-place modification of the existing/ current instance the object ID remains the same.

Parameters:

tail_coeffsint: An (ordered) sequence of (positive) integers which form a contiguous trailing segment of the current sequence of coefficients and is to be truncated.

Raises:

ValueError: If the tail coefficients provided are not positive integers, or do not form a contiguous trailing segment, or their number exceeds the length of the existing tail, i.e. the order of the continued fraction represented by the instance.

Examples

>>> cf = ContinuedFraction('3.245')
>>> cf
ContinuedFraction(649, 200)
>>> tuple(cf.coefficients)
(3, 4, 12, 4)
>>> cf.order
3
>>> cf.counter
Counter({4: 2, 3: 1, 12: 1})
>>> tuple(cf.convergents)
((0, ContinuedFraction(3, 1)), (1, ContinuedFraction(13, 4)), (2, ContinuedFraction(159, 49)), (3, ContinuedFraction(649, 200)))
>>> tuple(cf.remainders)
((3, ContinuedFraction(4, 1)), (2, ContinuedFraction(49, 4)), (1, ContinuedFraction(200, 49)), (0, ContinuedFraction(649, 200)))
>>> cf.truncate(12, 4)
>>> cf
ContinuedFraction(13, 4)
>>> tuple(cf.coefficients)
(3, 4)
>>> cf.order
1
>>> tuple(cf.convergents)
((0, ContinuedFraction(3, 1)), (1, ContinuedFraction(13, 4)))
>>> tuple(cf.remainders)
((1, ContinuedFraction(4, 1)), (0, ContinuedFraction(13, 4)))
>>> cf = ContinuedFraction(649, 200)
>>> cf.truncate(4, 12)
Traceback (most recent call last):
...
ValueError: The coefficients to be truncated from the tail must consist of positive integers and form a contiguous trailing segment of the tail.
>>> cf.truncate(3, 4, 12, 4)
Traceback (most recent call last):
...
ValueError: The coefficients to be truncated from the tail must consist of positive integers and form a contiguous trailing segment of the tail.

as_decimal() → Decimal[source]#

Returns the decimal.Decimal value of the continued fraction.

Returns:

decimal.Decimal: The decimal.Decimal representation of the continued fraction.

Examples

Note that the default decimal context precision of \(28\) is used in these examples.

>>> import math
>>> math.pi
3.141592653589793

Now construct a ContinuedFraction instance from it, and check the float value.

>>> cf = ContinuedFraction(math.pi)
>>> cf
ContinuedFraction(884279719003555, 281474976710656)
>>> cf.as_decimal()
Decimal('3.141592653589793115997963469')

convergent(k: int, /) → ContinuedFraction[source]#

Returns the \(k\)-th (simple) convergent of the continued fraction.

Given a (simple) continued fraction \([a_0;a_1,a_2,\ldots]\) the \(k\)-th convergent is defined as:

\[C_k = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 \ddots \cfrac{1}{a_{k-1} + \cfrac{1}{a_k}}}}\]

The result is a ContinuedFraction instance.

If the continued fraction is of order \(n\) then it has exactly \(n + 1\) convergents \(C_0,C_1,C_2,\ldots,C_n\) where the \(k\)-th convergent \(C_k = \frac{p_k}{q_k}\) is given by the recurrence relation:

\[\begin{split}\begin{align} p_k &= a_kp_{k - 1} + p_{k - 2} \\ q_k &= a_kq_{k - 1} + q_{k - 2}, \hskip{3em} k \geq 3 \end{align}\end{split}\]

where \(p_0 = a_0\), \(q_0 = 1\), \(p_1 = p_1p_0 + 1\), and \(q_1 = p_1\).

Parameters:

kint: The index of the convergent, as described above.

Returns:

ContinuedFraction: A new ContinuedFraction instance representing the \(k\)-th (simple) convergent of the original continued fraction, as described above.

Examples

>>> cf = ContinuedFraction('.12345')
>>> cf
ContinuedFraction(2469, 20000)
>>> cf.convergent(0)
ContinuedFraction(0, 1)
>>> cf.convergent(2)
ContinuedFraction(9, 73)
>>> cf.convergent(6)
ContinuedFraction(448, 3629)
>>> cf.convergent(7)
ContinuedFraction(2469, 20000)

property convergents: Generator[tuple[int, ContinuedFraction], None, None]#

Generates an enumerated sequence of all convergents of the continued fraction.

The convergents are generated as tuples of int and ContinuedFraction instances, where the integers represent the indices of the convergents.

If \(n\) is the order of the continued fraction then \(n + 1\) convergents \(C_0, C_1, \ldots, C_n\) are generated in that order.

Yields:

tuple: A tuple of convergent index (int) and convergents (ContinuedFraction) of the continued fraction.

Examples

>>> cf = ContinuedFraction('3.245')
>>> tuple(cf.convergents)
((0, ContinuedFraction(3, 1)), (1, ContinuedFraction(13, 4)), (2, ContinuedFraction(159, 49)), (3, ContinuedFraction(649, 200)))

property even_order_convergents: Generator[tuple[int, ContinuedFraction], None, None]#

Generates an enumerated sequence of all even-order convergents of the continued fraction.

The convergents are generated as tuples of int and ContinuedFraction instances, where the integers represent the indices of the convergents.

If \(n\) is the order of the continued fraction then only the even- indexed convergents \(C_0, C_2, C_4, \ldots\) are generated.

Yields:

tuple: A tuple of convergent index (int) and convergents (ContinuedFraction) of the continued fraction.

Examples

>>> tuple(ContinuedFraction('3.245').even_order_convergents)
((0, ContinuedFraction(3, 1)), (2, ContinuedFraction(159, 49)))

property odd_order_convergents: Generator[tuple[int, ContinuedFraction], None, None]#

Generates an enumerated sequence of all odd-order convergents of the continued fraction.

The convergents are generated as tuples of int and ContinuedFraction instances, where the integers represent the indices of the convergents.

If \(n\) is the order of the continued fraction then only the odd- indexed convergents \(C_1, C_3, C_5, \ldots\) are generated.

Yields:

tuple: A tuple of convergent index (int) and convergents (ContinuedFraction) of the continued fraction.

Examples

>>> tuple(ContinuedFraction('3.245').odd_order_convergents)
((1, ContinuedFraction(13, 4)), (3, ContinuedFraction(649, 200)))

semiconvergent(k: int, m: int, /) → ContinuedFraction[source]#

Returns the \(m\)-th semiconvergent of two consecutive convergents \(p_{k - 1}\) and \(p_k\) of the continued fraction.

The integer \(k\) must be positive and determine two consecutive convergents \(p_{k - 1}\) and \(p_k\) of a (finite, simple) continued fraction.

The integer \(m\) can be any positive integer.

Parameters:

kint: The integer \(k\) determining two consecutive convergents \(p_{k - 1}\) and \(p_k\) of a (finite, simple) continued fraction \([a_0; a_1, \ldots, a_{k}, a_{k + 1}, \ldots, a_n]\).
mint: The index of the semiconvergent of the convergents \(p_{k - 1}\) and \(p_k\).

Returns:

ContinuedFraction: The \(m\)-th semiconvergent of the convergents \(p_{k - 1}\) and \(p_k\).

Raises:

ValueError: If \(k\) or \(m\) are not positive integers, or \(k\) is an integer that does not satisfy \(1 \leq k \leq n\) where \(n\) is the order of the (finite, simple) continued fraction of which \(p_{k - 1}\) and \(p_k\) are convergents.

Examples

>>> cf = ContinuedFraction(-415, 93)
>>> tuple(cf.coefficients)
(-5, 1, 1, 6, 7)
>>> tuple(cf.convergents)
((0, ContinuedFraction(-5, 1)), (1, ContinuedFraction(-4, 1)), (2, ContinuedFraction(-9, 2)), (3, ContinuedFraction(-58, 13)), (4, ContinuedFraction(-415, 93)))
>>> cf.semiconvergent(3, 1)
ContinuedFraction(-67, 15)
>>> cf.semiconvergent(3, 2)
ContinuedFraction(-125, 28)
>>> cf.semiconvergent(3, 3)
ContinuedFraction(-183, 41)
>>> cf.semiconvergent(3, 4)
ContinuedFraction(-241, 54)
>>> cf.semiconvergent(3, 5)
ContinuedFraction(-299, 67)
>>> cf.semiconvergent(3, 6)
ContinuedFraction(-357, 80)
>>> cf.semiconvergent(3, 7)
ContinuedFraction(-415, 93)

remainder(k: int, /) → ContinuedFraction[source]#

Returns the \(k\)-th remainder of the continued fraction.

Given a (simple) continued fraction \([a_0;a_1,a_2,\ldots]\) the \(k\)-th remainder \(R_k\) is the (simple) continued fraction \([a_k; a_{k + 1}, a_{k + 2}, \ldots]\):

\[R_k = a_k + \cfrac{1}{a_{k + 1} + \cfrac{1}{a_{k + 2} \ddots }}\]

where \(R_0\) is just the original continued fraction, i.e. \(R_0 = [a_0; a_1, a_2, \ldots]\).

The result is a ContinuedFraction instance.

Parameters:

kint: The index of the remainder, as described above.

Returns:

ContinuedFraction: A new ContinuedFraction instance representing the \(k\)-th remainder of the original continued fraction, as described above.

Examples

>>> cf = ContinuedFraction('.12345')
>>> cf
ContinuedFraction(2469, 20000)
>>> cf.remainder(0)
ContinuedFraction(2469, 20000)
>>> cf.remainder(2)
ContinuedFraction(2469, 248)
>>> cf.remainder(6)
ContinuedFraction(6, 5)
>>> cf.remainder(7)
ContinuedFraction(5, 1)

property remainders: Generator[tuple[int, ContinuedFraction], None, None]#

Generates an enumerated sequence of all remainders of the continued fraction in descending order of index.

If \(n\) is the order of the continued fraction then there are \(n + 1\) remainders \(R_0, R_1, \ldots, R_n\), and the method generates these in reverse order \(R_0, R_1, \ldots, R_n\).

The remainders are generated as tuples of int and ContinuedFraction instances, where the integers represent the indexes of the remainders.

Yields:

tuple: A tuple of remainder indices (int) and remainders (ContinuedFraction) of the continued fraction.

Examples

>>> tuple(ContinuedFraction('3.245').remainders)
((3, ContinuedFraction(4, 1)), (2, ContinuedFraction(49, 4)), (1, ContinuedFraction(200, 49)), (0, ContinuedFraction(649, 200)))
>>> tuple(ContinuedFraction(-415, 93).remainders)
((4, ContinuedFraction(7, 1)), (3, ContinuedFraction(43, 7)), (2, ContinuedFraction(50, 43)), (1, ContinuedFraction(93, 50)), (0, ContinuedFraction(-415, 93)))

left_mediant(other: Fraction, /, *, k: int = 1) → ContinuedFraction[source]#

Returns the \(k\)-th left-mediant of the continued fraction with another rational number.

For a positive integer \(k\), the \(k\)-th left-mediant of two rational numbers \(r = \frac{a}{b}\) and \(s = \frac{c}{d}\), where \(b, d, b + d \neq 0\), is defined as:

\[\frac{ka + c}{kb + d}, \hskip{3em} k \geq 1\]

Parameters:

otherfractions.Fraction, ContinuedFraction: The second fraction to use to calculate the \(k\)-th mediant with the first.
kint, default=1: The order of the mediant, as defined above.

Returns:

ContinuedFraction: The \(k\)-th left-mediant of the original fraction and the second fraction, as a ContinuedFraction instance.

Examples

>>> c1 = ContinuedFraction('1/2')
>>> c2 = ContinuedFraction(3, 5)
>>> c1, c2
(ContinuedFraction(1, 2), ContinuedFraction(3, 5))
>>> c1.left_mediant(c2)
ContinuedFraction(4, 7)
>>> c1.left_mediant(c2, k=2)
ContinuedFraction(5, 9)
>>> c1.left_mediant(c2, k=3)
ContinuedFraction(6, 11)
>>> c1.left_mediant(c2, k=100)
ContinuedFraction(103, 205)
>>> assert c1.left_mediant(c2, k=2) < c1.right_mediant(c2, k=2)
>>> assert c1.left_mediant(c2, k=3) < c1.right_mediant(c2, k=3)
>>> assert c1.left_mediant(c2, k=100) < c1.right_mediant(c2, k=100)

right_mediant(other: Fraction, /, *, k: int = 1) → ContinuedFraction[source]#

Returns the \(k\)-th right-mediant of the continued fraction with another rational number.

For a positive integer \(k\), the \(k\)-th right-mediant of two rational numbers \(r = \frac{a}{b}\) and \(s = \frac{c}{d}\), where \(b, d, b + d \neq 0\), is defined as:

\[\frac{a + kc}{b + kd}, \hskip{3em} k \geq 1\]

Parameters:

otherfractions.Fraction, ContinuedFraction: The second fraction to use to calculate the \(k\)-th mediant with the first.
kint, default=1: The order of the mediant, as defined above.

Returns:

ContinuedFraction: The \(k\)-th right-mediant of the original fraction and the second fraction, as a ContinuedFraction instance.

Examples

>>> c1 = ContinuedFraction('1/2')
>>> c2 = ContinuedFraction(3, 5)
>>> c1, c2
(ContinuedFraction(1, 2), ContinuedFraction(3, 5))
>>> c1.right_mediant(c2)
ContinuedFraction(4, 7)
>>> c1.right_mediant(c2, k=2)
ContinuedFraction(7, 12)
>>> c1.right_mediant(c2, k=3)
ContinuedFraction(10, 17)
>>> c1.right_mediant(c2, k=100)
ContinuedFraction(301, 502)
>>> assert c1.left_mediant(c2, k=2) < c1.right_mediant(c2, k=2)
>>> assert c1.left_mediant(c2, k=3) < c1.right_mediant(c2, k=3)
>>> assert c1.left_mediant(c2, k=100) < c1.right_mediant(c2, k=100)

mediant(other: Fraction, /) → ContinuedFraction[source]#

Returns the simple mediant of the continued fraction with another continued fraction.

The simple mediant of two rational numbers \(r = \frac{a}{b}\) and \(s = \frac{c}{d}\), where \(b, d, b + d \neq 0\), is defined as:

\[\frac{a + c}{b + d}\]

The resulting value \(\frac{a + c}{b + d}\) is the same as the 1st order left- or right-mediant of \(r = \frac{a}{b}\) and \(s = \frac{c}{d}\). So this method would produce the same result as the left_mediant() or right_mediant() methods where the order \(k\) is set to \(1\).

Parameters:

otherfractions.Fraction, ContinuedFraction: The other continued fraction.

Returns:

ContinuedFraction: The simple mediant of the original fraction and the other continued fraction.

Examples

>>> ContinuedFraction(1, 2).mediant(ContinuedFraction(3, 5))
ContinuedFraction(4, 7)
>>> assert ContinuedFraction(1, 2).mediant(ContinuedFraction(3, 5)) == ContinuedFraction(1, 2).left_mediant(ContinuedFraction(3, 5), k=1)
>>> assert ContinuedFraction(1, 2).mediant(ContinuedFraction(3, 5)) == ContinuedFraction(1, 2).right_mediant(ContinuedFraction(3, 5), k=1)

property height: int#

The height of this rational number.

Returns:

int: The height of the fraction as a rational number represented by homogeneous coordinates in projective space \(\mathbb{P}^1(\mathbb{Q})\).

Examples

>>> ContinuedFraction(0, 1).height
1
>>> ContinuedFraction(-1, 2).height
2
>>> ContinuedFraction(3, -2).height
3

Type:: int

property log_height: Decimal#

The (natural) logarithm of the height of this rational number as defined above.

Returns:

decimal.Decimal: The (natural) logarithm of the height of this fraction as a rational number as defined above.

Examples

>>> ContinuedFraction(0, 1).log_height
Decimal('0')
>>> ContinuedFraction(-1, 2).log_height
Decimal('0.69314718055994528622676398299518041312694549560546875')
>>> ContinuedFraction(3, -2).log_height
Decimal('1.0986122886681097821082175869378261268138885498046875')

Type:: decimal.Decimal

__neg__() → ContinuedFraction[source]#

The negation algorithm for a finite simple continued fraction.

The basic algorithm can be described as follows: if \([a_0; a_1,\ldots, a_n]\) is the simple continued fraction of a positive rational number \(\frac{a}{b}\) of finite order \(n\) then \(-\frac{a}{b}\) has the simple continued fraction:

\[\begin{split}\begin{cases} [-a_0;] \hskip{3em} & n = 0 \\ [-(a_0 + 1); 2] \hskip{3em} & n = 1 \text{ and } a_1 = 2 \\ [-(a_0 + 1); a_2 + 1, a_3,\ldots, a_n] \hskip{3em} & n \geq 2 \text{ and } a_1 = 1 \\ [-(a_0 + 1); 1, a_1 - 1, a_2, \ldots,a_n] \hskip{3em} & n \geq 2 \text{ and } a_1 \geq 2 \end{cases}\end{split}\]

In applying this algorithm there is an assumption that the last coefficient \(a_n > 1\), as any simple continued fraction of type \([a_0; a_1,\ldots, a_{n} = 1]\) can be reduced to the simple continued fraction \([a_0; a_1,\ldots, a_{n - 1} + 1]\).

continuedfractions.continuedfraction#

`continuedfractions.continuedfraction`#