Continued fractions¶

A continued fraction is a representation of a real number in terms of a sequence of integers denoted \([a_0; a_1, a_2, \ldots]\). The well known decimal expansion is another way of representing a real number by a sequence of integers. The value of a continued fraction is defined recursively as:

\[[a_0; a_1, a_2, \ldots] = a_0 + \frac{1}{[a_1; a_2, \ldots]} = a_0 + \frac{\displaystyle 1} {\displaystyle a_1 + \frac{\displaystyle 1} {\displaystyle a_2 + \frac{\displaystyle 1} {\ldots}}}\]

In this expansion, all coefficients \(a_n\) are integers and only the value \(a_0\) may be non positive. Note that \(a_0\) is nothing else but the floor (this remark provides a way to build the continued fraction expansion from a given real number). As examples

\[\frac{45}{38} = 1 + \frac{\displaystyle 1} {\displaystyle 5 + \frac{\displaystyle 1} {\displaystyle 2 + \frac{\displaystyle 1} {\displaystyle 3}}}\]

\[\pi = 3 + \frac{\displaystyle 1} {\displaystyle 7 + \frac{\displaystyle 1} {\displaystyle 15 + \frac{\displaystyle 1} {\displaystyle 1 + \frac{\displaystyle 1} {\displaystyle 292 + \frac{\displaystyle 1} {\ldots}}}}}\]

It is quite remarkable that

any real number admits a unique continued fraction expansion
finite expansions correspond to rationals
ultimately periodic expansions correspond to quadratic numbers (ie numbers of the form \(a + b \sqrt{D}\) with \(a\) and \(b\) rationals and \(D\) square free positive integer)
two real numbers \(x\) and \(y\) have the same tail (up to a shift) in their continued fraction expansion if and only if there are integers \(a,b,c,d\) with \(|ad - bc| = 1\) and such that \(y = (ax + b) / (cx + d)\).

Moreover, the rational numbers obtained by truncation of the expansion of a real number gives its so-called best approximations. For more informations on continued fractions, you may have a look at Wikipedia article Continued_fraction.

EXAMPLES:

If you want to create the continued fraction of some real number you may either use its method continued_fraction (if it exists) or call continued_fraction():

sage: (13/27).continued_fraction()
[0; 2, 13]
sage: 0 + 1/(2 + 1/13)
13/27

sage: continued_fraction(22/45)
[0; 2, 22]
sage: 0 + 1/(2 + 1/22)
22/45

sage: continued_fraction(pi)
[3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ...]
sage: continued_fraction_list(pi, nterms=5)
[3, 7, 15, 1, 292]

sage: K.<cbrt5> = NumberField(x^3 - 5, embedding=1.709)
sage: continued_fraction(cbrt5)
[1; 1, 2, 2, 4, 3, 3, 1, 5, 1, 1, 4, 10, 17, 1, 14, 1, 1, 3052, 1, ...]

It is also possible to create a continued fraction from a list of partial quotients:

sage: continued_fraction([-3,1,2,3,4,1,2])
[-3; 1, 2, 3, 4, 1, 2]

Even infinite:

sage: w = words.ThueMorseWord([1,2])
sage: w
word: 1221211221121221211212211221211221121221...
sage: continued_fraction(w)
[1; 2, 2, 1, 2, 1, 1, 2, 2, 1...]

To go back and forth between the value (as a real number) and the partial quotients (seen as a finite or infinite list) you can use the methods quotients and value:

sage: cf = (13/27).continued_fraction()
sage: cf.quotients()
[0, 2, 13]
sage: cf.value()
13/27

sage: cf = continued_fraction(pi)
sage: cf.quotients()
lazy list [3, 7, 15, ...]
sage: cf.value()
pi

sage: w = words.FibonacciWord([1,2])
sage: cf = continued_fraction(w)
sage: cf.quotients()
word: 1211212112112121121211211212112112121121...
sage: v = cf.value()
sage: v
1.387954587967143?
sage: v.n(digits=100)
1.387954587967142336919313859873185477878152452498532271894917289826418577622648932169885237034242967
sage: v.continued_fraction()
[1; 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2...]

Recall that quadratic numbers correspond to ultimately periodic continued fractions. For them special methods give access to preperiod and period:

sage: K.<sqrt2> = QuadraticField(2)
sage: cf = continued_fraction(sqrt2); cf
[1; (2)*]
sage: cf.value()
sqrt2
sage: cf.preperiod()
(1,)
sage: cf.period()
(2,)

sage: cf = (3*sqrt2 + 1/2).continued_fraction(); cf
[4; (1, 2, 1, 7)*]

sage: cf = continued_fraction([(1,2,3),(1,4)]); cf
[1; 2, 3, (1, 4)*]
sage: cf.value()
-2/23*sqrt2 + 36/23

On the following we can remark how the tail may change even in the same quadratic field:

sage: for i in range(20): print(continued_fraction(i*sqrt2))
[0]
[1; (2)*]
[2; (1, 4)*]
[4; (4, 8)*]
[5; (1, 1, 1, 10)*]
[7; (14)*]
...
[24; (24, 48)*]
[25; (2, 5, 6, 5, 2, 50)*]
[26; (1, 6, 1, 2, 3, 2, 26, 2, 3, 2, 1, 6, 1, 52)*]

Nevertheless, the tail is preserved under invertible integer homographies:

sage: apply_homography =  lambda m,z: (m[0,0]*z + m[0,1]) / (m[1,0]*z + m[1,1])
sage: m1 = SL2Z([60,13,83,18])
sage: m2 = SL2Z([27,80,28,83])
sage: a = sqrt2/3
sage: a.continued_fraction()
[0; 2, (8, 4)*]
sage: b = apply_homography(m1, a)
sage: b.continued_fraction()
[0; 1, 2, 1, 1, 1, 1, 6, (8, 4)*]
sage: c = apply_homography(m2, a)
sage: c.continued_fraction()
[0; 1, 26, 1, 2, 2, (8, 4)*]
sage: d = apply_homography(m1**2*m2**3, a)
sage: d.continued_fraction()
[0; 1, 2, 1, 1, 1, 1, 5, 2, 1, 1, 1, 1, 5, 26, 1, 2, 1, 26, 1, 2, 1, 26, 1, 2, 2, (8, 4)*]

Todo

Gosper’s algorithm to compute the continued fraction of (ax + b)/(cx + d) knowing the one of x (see Gosper (1972, http://www.inwap.com/pdp10/hbaker/hakmem/cf.html), Knuth (1998, TAOCP vol 2, Exercise 4.5.3.15), Fowler (1999). See also Liardet, P. and Stambul, P. “Algebraic Computation with Continued Fractions.” J. Number Th. 73, 92-121, 1998.
Improve numerical approximation (the method _mpfr_() is quite slow compared to the same method for an element of a number field)
Make a class for generalized continued fractions of the form \(a_0 + b_0/(a_1 + b_1/(...))\) (the standard continued fractions are when all \(b_n= 1\) while the Hirzebruch-Jung continued fractions are the one for which \(b_n = -1\) for all \(n\)). See Wikipedia article Generalized_continued_fraction.
look at the function ContinuedFractionApproximationOfRoot in GAP

AUTHORS:

Vincent Delecroix (2014): cleaning, refactorisation, documentation from the old implementation in contfrac (trac ticket #14567).

class sage.rings.continued_fraction.ContinuedFraction_base¶

Bases: sage.structure.sage_object.SageObject

Base class for (standard) continued fractions.

If you want to implement your own continued fraction, simply derived from this class and implement the following methods:

def quotient(self, n): return the n-th quotient of self as a Sage integer
def length(self): the number of partial quotients of self as a Sage integer or Infinity.

and optionally:

def value(self): return the value of self (an exact real number)

This base class will provide:

computation of convergents in convergent(), numerator() and denominator()
comparison with other continued fractions (see __richcmp__())
elementary arithmetic function floor(), ceil(), sign()
accurate numerical approximations _mpfr_()

All other methods, in particular the ones involving binary operations like sum or product, rely on the optional method value() (and not on convergents) and may fail at execution if it is not implemented.

additive_order()¶

Return the additive order of this continued fraction, which we defined to be the additive order of its value.

EXAMPLES:

sage: continued_fraction(-1).additive_order()
+Infinity
sage: continued_fraction(0).additive_order()
1

ceil()¶

Return the ceil of self.

EXAMPLES:

sage: cf = continued_fraction([2,1,3,4])
sage: cf.ceil()
3

convergent(n)¶

Return the n-th partial convergent to self.

EXAMPLES:

sage: a = continued_fraction(pi); a
[3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ...]
sage: a.convergent(3)
355/113
sage: a.convergent(15)
411557987/131002976

convergents()¶

Return the list of partial convergents of self.

If self is an infinite continued fraction, then the object returned is a lazy_list_generic which behave like an infinite list.

EXAMPLES:

sage: a = continued_fraction(23/157); a
[0; 6, 1, 4, 1, 3]
sage: a.convergents()
[0, 1/6, 1/7, 5/34, 6/41, 23/157]

Todo

Add an example with infinite list.

denominator(n)¶

Return the denominator of the n-th partial convergent of self.

EXAMPLES:

sage: c = continued_fraction(pi); c
[3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ...]
sage: c.denominator(0)
1
sage: c.denominator(12)
25510582
sage: c.denominator(152)
1255341492699841451528811722575401081588363886480089431843026103930863337221076748

floor()¶

Return the floor of self.

EXAMPLES:

sage: cf = continued_fraction([2,1,2,3])
sage: cf.floor()
2

is_minus_one()¶

Test whether self is minus one.

EXAMPLES:

sage: continued_fraction(-1).is_minus_one()
True
sage: continued_fraction(1).is_minus_one()
False
sage: continued_fraction(0).is_minus_one()
False
sage: continued_fraction(-2).is_minus_one()
False
sage: continued_fraction([-1,1]).is_minus_one()
False

is_one()¶

Test whether self is one.

EXAMPLES:

sage: continued_fraction(1).is_one()
True
sage: continued_fraction(5/4).is_one()
False
sage: continued_fraction(0).is_one()
False
sage: continued_fraction(pi).is_one()
False

is_zero()¶

Test whether self is zero.

EXAMPLES:

sage: continued_fraction(0).is_zero()
True
sage: continued_fraction((0,1)).is_zero()
False
sage: continued_fraction(-1/2).is_zero()
False
sage: continued_fraction(pi).is_zero()
False

multiplicative_order()¶

Return the multiplicative order of this continued fraction, which we defined to be the multiplicative order of its value.

EXAMPLES:

sage: continued_fraction(-1).multiplicative_order()
2
sage: continued_fraction(1).multiplicative_order()
1
sage: continued_fraction(pi).multiplicative_order()
+Infinity

n(prec=None, digits=None, algorithm=None)¶

Return a numerical approximation of this continued fraction with prec bits (or decimal digits) of precision.

INPUT:

prec – precision in bits
digits – precision in decimal digits (only used if prec is not given)
algorithm – ignored for continued fractions

If neither prec nor digits is given, the default precision is 53 bits (roughly 16 digits).

EXAMPLES:

sage: w = words.FibonacciWord([1,3])
sage: cf = continued_fraction(w)
sage: cf
[1; 3, 1, 1, 3, 1, 3, 1, 1, 3, 1, 1, 3, 1, 3, 1, 1, 3, 1, 3...]
sage: cf.numerical_approx(prec=53)
1.28102513329557

The method \(n\) is a shortcut to this one:

sage: cf.n(digits=25)
1.281025133295569815552930
sage: cf.n(digits=33)
1.28102513329556981555293038097590

numerator(n)¶

Return the numerator of the \(n\)-th partial convergent of self.

EXAMPLES:

sage: c = continued_fraction(pi); c
[3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ...]
sage: c.numerator(0)
3
sage: c.numerator(12)
80143857
sage: c.numerator(152)
3943771611212266962743738812600748213157266596588744951727393497446921245353005283

numerical_approx(prec=None, digits=None, algorithm=None)¶

Return a numerical approximation of this continued fraction with prec bits (or decimal digits) of precision.

INPUT:

prec – precision in bits
digits – precision in decimal digits (only used if prec is not given)
algorithm – ignored for continued fractions

If neither prec nor digits is given, the default precision is 53 bits (roughly 16 digits).

EXAMPLES:

sage: w = words.FibonacciWord([1,3])
sage: cf = continued_fraction(w)
sage: cf
[1; 3, 1, 1, 3, 1, 3, 1, 1, 3, 1, 1, 3, 1, 3, 1, 1, 3, 1, 3...]
sage: cf.numerical_approx(prec=53)
1.28102513329557

The method \(n\) is a shortcut to this one:

sage: cf.n(digits=25)
1.281025133295569815552930
sage: cf.n(digits=33)
1.28102513329556981555293038097590

p(n)¶

Return the numerator of the \(n\)-th partial convergent of self.

EXAMPLES:

sage: c = continued_fraction(pi); c
[3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ...]
sage: c.numerator(0)
3
sage: c.numerator(12)
80143857
sage: c.numerator(152)
3943771611212266962743738812600748213157266596588744951727393497446921245353005283

q(n)¶

Return the denominator of the n-th partial convergent of self.

EXAMPLES:

sage: c = continued_fraction(pi); c
[3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ...]
sage: c.denominator(0)
1
sage: c.denominator(12)
25510582
sage: c.denominator(152)
1255341492699841451528811722575401081588363886480089431843026103930863337221076748

quotients()¶

Return the list of partial quotients of self.

If self is an infinite continued fraction, then the object returned is a lazy_list_generic which behaves like an infinite list.

EXAMPLES:

sage: a = continued_fraction(23/157); a
[0; 6, 1, 4, 1, 3]
sage: a.quotients()
[0, 6, 1, 4, 1, 3]

Todo

Add an example with infinite list.

sign()¶

Returns the sign of self as an Integer.

The sign is defined to be 0 if self is 0, 1 if self is positive and -1 if self is negative.

EXAMPLES:

sage: continued_fraction(tan(pi/7)).sign()
1
sage: continued_fraction(-34/2115).sign()
-1
sage: continued_fraction([0]).sign()
0

str(nterms=10, unicode=False, join=True)¶

Return a string representing this continued fraction.

INPUT:

nterms – the maximum number of terms to use
unicode – (default False) whether to use unicode character
join – (default True) if False instead of returning a string return a list of string, each of them representing a line

EXAMPLES:

sage: print(continued_fraction(pi).str())
                             1
3 + ----------------------------------------------------
                                1
     7 + -----------------------------------------------
                                   1
          15 + -----------------------------------------
                                     1
                1 + ------------------------------------
                                         1
                     292 + -----------------------------
                                           1
                            1 + ------------------------
                                              1
                                 1 + -------------------
                                                1
                                      1 + --------------
                                                   1
                                           2 + ---------
                                                1 + ...
sage: print(continued_fraction(pi).str(nterms=1))
3 + ...
sage: print(continued_fraction(pi).str(nterms=2))
        1
3 + ---------
     7 + ...

sage: print(continued_fraction(243/354).str())
           1
-----------------------
             1
 1 + ------------------
                1
      2 + -------------
                  1
           5 + --------
                     1
                3 + ---
                     2
sage: continued_fraction(243/354).str(join=False)
['           1           ',
 '-----------------------',
 '             1         ',
 ' 1 + ------------------',
 '                1      ',
 '      2 + -------------',
 '                  1    ',
 '           5 + --------',
 '                     1 ',
 '                3 + ---',
 '                     2 ']

sage: print(continued_fraction(243/354).str(unicode=True))
           1
───────────────────────
             1
 1 + ──────────────────
                1
      2 + ─────────────
                  1
           5 + ────────
                     1
                3 + ───
                     2

class sage.rings.continued_fraction.ContinuedFraction_infinite(w, value=None, check=True)¶

Bases: sage.rings.continued_fraction.ContinuedFraction_base

A continued fraction defined by an infinite sequence of partial quotients.

EXAMPLES:

sage: t = continued_fraction(words.ThueMorseWord([1,2])); t
[1; 2, 2, 1, 2, 1, 1, 2, 2, 1...]
sage: t.n(digits=100)
1.422388736882785488341547116024565825306879108991711829311892452916456747272565883312455412962072042

We check that comparisons work well:

sage: t > continued_fraction(1) and t < continued_fraction(3/2)
True
sage: t < continued_fraction(1) or t > continued_fraction(2)
False

Can also be called with a value option:

sage: def f(n):
....:     if n % 3 == 2: return 2*(n+1)//3
....:     return 1
sage: w = Word(f, alphabet=NN)
sage: w
word: 1,1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,1,1,14,1,1,16,1,1,18,1,1,20,1,1,22,1,1,24,1,1,26,1,...
sage: cf = continued_fraction(w, value=e-1)
sage: cf
[1; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, 1, 1...]

In that case a small check is done on the input:

sage: cf = continued_fraction(w, value=pi)
Traceback (most recent call last):
...
ValueError: value evaluates to 3.141592653589794? while the continued fraction evaluates to 1.718281828459046? in Real Interval Field with 53 bits of precision.

length()¶

Returns infinity.

EXAMPLES:

sage: w = words.FibonacciWord([3,13])
sage: cf = continued_fraction(w)
sage: cf.length()
+Infinity

quotient(n)¶

The n-th partial quotient of self.

EXAMPLES:

sage: w = words.FibonacciWord([1,3])
sage: cf = continued_fraction(w)
sage: cf.quotient(0)
1
sage: cf.quotient(1)
3
sage: cf.quotient(2)
1

quotients()¶

Return the infinite list from which this continued fraction was built.

EXAMPLES:

sage: w = words.FibonacciWord([1,5])
sage: cf = continued_fraction(w)
sage: cf.quotients()
word: 1511515115115151151511511515115115151151...

value()¶

The value of self.

If this value was provided on initialization, just return this value otherwise return an element of the real lazy field.

EXAMPLES:

sage: def f(n):
....:     if n % 3 == 2: return 2*(n+1)//3
....:     return 1
sage: w = Word(f, alphabet=NN)
sage: w
word: 1,1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,1,1,14,1,1,16,1,1,18,1,1,20,1,1,22,1,1,24,1,1,26,1,...
sage: cf = continued_fraction(w, value=e-1)
sage: cf
[1; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, 1, 1...]
sage: cf.value()
e - 1

sage: w = words.FibonacciWord([2,5])
sage: cf = continued_fraction(w)
sage: cf
[2; 5, 2, 2, 5, 2, 5, 2, 2, 5, 2, 2, 5, 2, 5, 2, 2, 5, 2, 5...]
sage: cf.value()
2.184951302409338?

class sage.rings.continued_fraction.ContinuedFraction_periodic(x1, x2=None, check=True)¶

Bases: sage.rings.continued_fraction.ContinuedFraction_base

Continued fraction associated with rational or quadratic number.

A rational number has a finite continued fraction expansion (or ultimately 0). The one of a quadratic number, ie a number of the form \(a + b \sqrt{D}\) with \(a\) and \(b\) rational, is ultimately periodic.

Note

This class stores a tuple _x1 for the preperiod and a tuple _x2 for the period. In the purely periodic case _x1 is empty while in the rational case _x2 is the tuple (0,).

length()¶

Returns the number of partial quotients of self.

EXAMPLES:

sage: continued_fraction(2/5).length()
3
sage: cf = continued_fraction([(0,1),(2,)]); cf
[0; 1, (2)*]
sage: cf.length()
+Infinity

period()¶

Return the periodic part of self.

EXAMPLES:

sage: K.<sqrt3> = QuadraticField(3)
sage: cf = continued_fraction(sqrt3); cf
[1; (1, 2)*]
sage: cf.period()
(1, 2)

sage: for k in xsrange(2,40):
....:     if not k.is_square():
....:         s = QuadraticField(k).gen()
....:         cf = continued_fraction(s)
....:         print('%2d %d %s' % (k, len(cf.period()), cf))
 2 1 [1; (2)*]
 3 2 [1; (1, 2)*]
 5 1 [2; (4)*]
 6 2 [2; (2, 4)*]
 7 4 [2; (1, 1, 1, 4)*]
 8 2 [2; (1, 4)*]
10 1 [3; (6)*]
11 2 [3; (3, 6)*]
12 2 [3; (2, 6)*]
13 5 [3; (1, 1, 1, 1, 6)*]
14 4 [3; (1, 2, 1, 6)*]
...
35 2 [5; (1, 10)*]
37 1 [6; (12)*]
38 2 [6; (6, 12)*]
39 2 [6; (4, 12)*]

preperiod()¶

Return the preperiodic part of self.

EXAMPLES:

sage: K.<sqrt3> = QuadraticField(3)
sage: cf = continued_fraction(sqrt3); cf
[1; (1, 2)*]
sage: cf.preperiod()
(1,)

sage: cf = continued_fraction(sqrt3/7); cf
[0; 4, (24, 8)*]
sage: cf.preperiod()
(0, 4)

quotient(n)¶

Return the n-th partial quotient of self.

EXAMPLES:

sage: cf = continued_fraction([(12,5),(1,3)])
sage: [cf.quotient(i) for i in range(10)]
[12, 5, 1, 3, 1, 3, 1, 3, 1, 3]

value()¶

Return the value of self as a quadratic number (with square free discriminant).

EXAMPLES:

Some purely periodic examples:

sage: cf = continued_fraction([(),(2,)]); cf
[(2)*]
sage: v = cf.value(); v
sqrt2 + 1
sage: v.continued_fraction()
[(2)*]

sage: cf = continued_fraction([(),(1,2)]); cf
[(1, 2)*]
sage: v = cf.value(); v
1/2*sqrt3 + 1/2
sage: v.continued_fraction()
[(1, 2)*]

The number sqrt3 that appear above is actually internal to the continued fraction. In order to be access it from the console:

sage: cf.value().parent().inject_variables()
Defining sqrt3
sage: sqrt3
sqrt3
sage: ((sqrt3+1)/2).continued_fraction()
[(1, 2)*]

Some ultimately periodic but non periodic examples:

sage: cf = continued_fraction([(1,),(2,)]); cf
[1; (2)*]
sage: v = cf.value(); v
sqrt2
sage: v.continued_fraction()
[1; (2)*]

sage: cf = continued_fraction([(1,3),(1,2)]); cf
[1; 3, (1, 2)*]
sage: v = cf.value(); v
-sqrt3 + 3
sage: v.continued_fraction()
[1; 3, (1, 2)*]

sage: cf = continued_fraction([(-5,18), (1,3,1,5)])
sage: cf.value().continued_fraction() == cf
True
sage: cf = continued_fraction([(-1,),(1,)])
sage: cf.value().continued_fraction() == cf
True

class sage.rings.continued_fraction.ContinuedFraction_real(x)¶

Bases: sage.rings.continued_fraction.ContinuedFraction_base

Continued fraction of a real (exact) number.

This class simply wraps a real number into an attribute (that can be accessed through the method value()). The number is assumed to be irrational.

EXAMPLES:

sage: cf = continued_fraction(pi)
sage: cf
[3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ...]
sage: cf.value()
pi

sage: cf = continued_fraction(e)
sage: cf
[2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, 1, 1, ...]
sage: cf.value()
e

length()¶

Return infinity

EXAMPLES:

sage: continued_fraction(pi).length()
+Infinity

quotient(n)¶

Returns the n-th quotient of self.

EXAMPLES:

sage: cf = continued_fraction(pi)
sage: cf.quotient(27)
13
sage: cf.quotient(2552)
152
sage: cf.quotient(10000)   # long time
5

The algorithm is not efficient with element of the symbolic ring and, if possible, one can always prefer number fields elements. The reason is that, given a symbolic element x, there is no automatic way to evaluate in RIF an expression of the form (a*x+b)/(c*x+d) where both the numerator and the denominator are extremely small:

sage: a1 = pi
sage: c1 = continued_fraction(a1)
sage: p0 = c1.numerator(12); q0 = c1.denominator(12)
sage: p1 = c1.numerator(13); q1 = c1.denominator(13)
sage: num = (q0*a1 - p0); num.n()
1.49011611938477e-8
sage: den = (q1*a1 - p1); den.n()
-2.98023223876953e-8
sage: a1 = -num/den
sage: RIF(a1)
[-infinity .. +infinity]

The same computation with an element of a number field instead of pi gives a very satisfactory answer:

sage: K.<a2> = NumberField(x^3 - 2, embedding=1.25)
sage: c2 = continued_fraction(a2)
sage: p0 = c2.numerator(111); q0 = c2.denominator(111)
sage: p1 = c2.numerator(112); q1 = c2.denominator(112)
sage: num = (q0*a2 - p0); num.n()
-4.56719261665907e46
sage: den = (q1*a2 - p1); den.n()
-3.65375409332726e47
sage: a2 = -num/den
sage: b2 = RIF(a2); b2
1.002685823312715?
sage: b2.absolute_diameter()
8.88178419700125e-16

The consequence is that the precision needed with c1 grows when we compute larger and larger partial quotients:

sage: c1.quotient(100)
2
sage: c1._xa.parent()
Real Interval Field with 353 bits of precision
sage: c1.quotient(200)
3
sage: c1._xa.parent()
Real Interval Field with 753 bits of precision
sage: c1.quotient(300)
5
sage: c1._xa.parent()
Real Interval Field with 1053 bits of precision

sage: c2.quotient(200)
6
sage: c2._xa.parent()
Real Interval Field with 53 bits of precision
sage: c2.quotient(500)
1
sage: c2._xa.parent()
Real Interval Field with 53 bits of precision
sage: c2.quotient(1000)
1
sage: c2._xa.parent()
Real Interval Field with 53 bits of precision

value()¶

Return the value of self (the number from which it was built).

EXAMPLES:

sage: cf = continued_fraction(e)
sage: cf.value()
e

sage.rings.continued_fraction.check_and_reduce_pair(x1, x2=None)¶

There are often two ways to represent a given continued fraction. This function makes it canonical.

In the very special case of the number \(0\) we return the pair ((0,),(0,)).

sage.rings.continued_fraction.continued_fraction(x, value=None)¶

Return the continued fraction of x.

INPUT:

\(x\) – a number or a list of partial quotients (for finite development) or two list of partial quotients (preperiod and period for ultimately periodic development)

EXAMPLES:

A finite continued fraction may be initialized by a number or by its list of partial quotients:

sage: continued_fraction(12/571)
[0; 47, 1, 1, 2, 2]
sage: continued_fraction([3,2,1,4])
[3; 2, 1, 4]

It can be called with elements defined from symbolic values, in which case the partial quotients are evaluated in a lazy way:

sage: c = continued_fraction(golden_ratio); c
[1; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...]
sage: c.convergent(12)
377/233
sage: fibonacci(14)/fibonacci(13)
377/233

sage: continued_fraction(pi)
[3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ...]
sage: c = continued_fraction(pi); c
[3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ...]
sage: a = c.convergent(3); a
355/113
sage: a.n()
3.14159292035398
sage: pi.n()
3.14159265358979

sage: continued_fraction(sqrt(2))
[1; 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ...]
sage: continued_fraction(tan(1))
[1; 1, 1, 3, 1, 5, 1, 7, 1, 9, 1, 11, 1, 13, 1, 15, 1, 17, 1, 19, ...]
sage: continued_fraction(tanh(1))
[0; 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, ...]
sage: continued_fraction(e)
[2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, 1, 1, ...]

If you want to play with quadratic numbers (such as golden_ratio and sqrt(2) above), it is much more convenient to use number fields as follows since preperiods and periods are computed:

sage: K.<sqrt5> = NumberField(x^2-5, embedding=2.23)
sage: my_golden_ratio = (1 + sqrt5)/2
sage: cf = continued_fraction((1+sqrt5)/2); cf
[(1)*]
sage: cf.convergent(12)
377/233
sage: cf.period()
(1,)
sage: cf = continued_fraction(2/3+sqrt5/5); cf
[1; 8, (1, 3, 1, 1, 3, 9)*]
sage: cf.preperiod()
(1, 8)
sage: cf.period()
(1, 3, 1, 1, 3, 9)

sage: L.<sqrt2> = NumberField(x^2-2, embedding=1.41)
sage: cf = continued_fraction(sqrt2); cf
[1; (2)*]
sage: cf.period()
(2,)
sage: cf = continued_fraction(sqrt2/3); cf
[0; 2, (8, 4)*]
sage: cf.period()
(8, 4)

It is also possible to go the other way around, build a ultimately periodic continued fraction from its preperiod and its period and get its value back:

sage: cf = continued_fraction([(1,1),(2,8)]); cf
[1; 1, (2, 8)*]
sage: cf.value()
2/11*sqrt5 + 14/11

It is possible to deal with higher degree number fields but in that case the continued fraction expansion is known to be aperiodic:

sage: K.<a> = NumberField(x^3-2, embedding=1.25)
sage: cf = continued_fraction(a); cf
[1; 3, 1, 5, 1, 1, 4, 1, 1, 8, 1, 14, 1, 10, 2, 1, 4, 12, 2, 3, ...]

Note that initial rounding can result in incorrect trailing partial quotients:

sage: continued_fraction(RealField(39)(e))
[2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 2]

Note the value returned for floating point number is the continued fraction associated to the rational number you obtain with a conversion:

sage: for _ in range(10):
....:     x = RR.random_element()
....:     cff = continued_fraction(x)
....:     cfe = QQ(x).continued_fraction()
....:     assert cff == cfe, "%s %s %s"%(x,cff,cfe)

sage.rings.continued_fraction.continued_fraction_list(x, type='std', partial_convergents=False, bits=None, nterms=None)¶

Returns the (finite) continued fraction of x as a list.

The continued fraction expansion of x are the coefficients \(a_i\) in

\[x = a_0 + 1/(a_1 + 1/(...))\]

with \(a_0\) integer and \(a_1\), \(...\) positive integers. The Hirzebruch-Jung continued fraction is the one for which the \(+\) signs are replaced with \(-\) signs

\[x = a_0 - 1/(a_1 - 1/(...))\]

Continued fractions¶

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