# 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)                                                        # needs sage.symbolic
[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)                                         # needs sage.symbolic
[3, 7, 15, 1, 292]

sage: x = polygen(ZZ, 'x')
sage: K.<cbrt5> = NumberField(x^3 - 5, embedding=1.709)                             # needs sage.rings.number_field
sage: continued_fraction(cbrt5)                                                     # needs sage.rings.number_field
[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]); w                                             # needs sage.combinat
word: 1221211221121221211212211221211221121221...
sage: continued_fraction(w)                                                         # needs sage.combinat
[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)                                                   # needs sage.symbolic
sage: cf.quotients()                                                                # needs sage.symbolic
lazy list [3, 7, 15, ...]
sage: cf.value()                                                                    # needs sage.symbolic
pi

sage: # needs sage.combinat
sage: w = words.FibonacciWord([1,2])
sage: cf = continued_fraction(w)
sage: cf.quotients()
word: 1211212112112121121211211212112112121121...
sage: v = cf.value(); 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: # needs sage.rings.number_field
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                                 # needs sage.rings.number_field
[4; (1, 2, 1, 7)*]

sage: cf = continued_fraction([(1,2,3),(1,4)]); cf
[1; 2, 3, (1, 4)*]
sage: cf.value()                                                                    # needs sage.rings.number_field
-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))                        # needs sage.rings.number_field
[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: # needs sage.modular sage.rings.number_field
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

• 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 (github issue #14567).

class sage.rings.continued_fraction.ContinuedFraction_base#

Bases: 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:

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.

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
1

apply_homography(a, b, c, d, forward_value=False)#

Return the continued fraction of $$(ax + b)/(cx + d)$$.

This is computed using Gosper’s algorithm, see continued_fraction_gosper.

INPUT:

• a, b, c, d – integers

• forward_value – boolean (default: False) whether the returned continued fraction is given the symbolic value of $$(a x + b)/(cx + d)$$ and not only the list of partial quotients obtained from Gosper’s algorithm.

EXAMPLES:

sage: (5 * 13/6 - 2) / (3 * 13/6 - 4)
53/15
sage: continued_fraction(13/6).apply_homography(5, -2,  3, -4).value()
53/15


We demonstrate now the effect of the optional argument forward_value:

sage: cf = continued_fraction(pi)                                           # needs sage.symbolic
sage: h1 = cf.apply_homography(35, -27, 12, -5); h1                         # needs sage.symbolic
[2; 1, 1, 6, 3, 1, 2, 1, 5, 3, 1, 1, 1, 1, 9, 12, 1, 1, 1, 3...]
sage: h1.value()                                                            # needs sage.symbolic
2.536941776086946?

sage: h2 = cf.apply_homography(35, -27, 12, -5, forward_value=True); h2     # needs sage.symbolic
[2; 1, 1, 6, 3, 1, 2, 1, 5, 3, 1, 1, 1, 1, 9, 12, 1, 1, 1, 3...]
sage: h2.value()                                                            # needs sage.symbolic
(35*pi - 27)/(12*pi - 5)


REFERENCES:

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                                         # needs sage.symbolic
[3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ...]
sage: a.convergent(3)                                                       # needs sage.symbolic
355/113
sage: a.convergent(15)                                                      # needs sage.symbolic
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: # needs sage.symbolic
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()                                       # needs sage.symbolic
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()                                      # needs sage.symbolic
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()                         # needs sage.symbolic
+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])                                        # needs sage.combinat
sage: cf = continued_fraction(w); cf                                        # needs sage.combinat
[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)                                          # needs sage.combinat
1.28102513329557


The method $$n$$ is a shortcut to this one:

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

numerator(n)#

Return the numerator of the $$n$$-th partial convergent of self.

EXAMPLES:

sage: # needs sage.symbolic
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])                                        # needs sage.combinat
sage: cf = continued_fraction(w); cf                                        # needs sage.combinat
[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)                                          # needs sage.combinat
1.28102513329557


The method $$n$$ is a shortcut to this one:

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

p(n)#

Return the numerator of the $$n$$-th partial convergent of self.

EXAMPLES:

sage: # needs sage.symbolic
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: # needs sage.symbolic
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()#

Return 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()                                  # needs sage.symbolic
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())                                   # needs sage.symbolic
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))                           # needs sage.symbolic
3 + ...
sage: print(continued_fraction(pi).str(nterms=2))                           # needs sage.symbolic
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)#

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

EXAMPLES:

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


We check that comparisons work well:

sage: t > continued_fraction(1) and t < continued_fraction(3/2)                 # needs sage.combinat
True
sage: t < continued_fraction(1) or t > continued_fraction(2)                    # needs sage.combinat
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); w                                               # needs sage.combinat
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); cf                                 # needs sage.combinat sage.symbolic
[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)                                      # needs sage.combinat sage.symbolic
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()#

Return infinity.

EXAMPLES:

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

quotient(n)#

Return the n-th partial quotient of self.

INPUT:

• n – an integer

EXAMPLES:

sage: # needs sage.combinat
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])                                        # needs sage.combinat
sage: cf = continued_fraction(w)                                            # needs sage.combinat
sage: cf.quotients()                                                        # needs sage.combinat
word: 1511515115115151151511511515115115151151...

value()#

Return 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); w                                           # needs sage.combinat
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); cf                             # needs sage.combinat sage.symbolic
[1; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, 1, 1...]
sage: cf.value()                                                            # needs sage.combinat sage.symbolic
e - 1

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

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

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()#

Return 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: # needs sage.rings.number_field
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():
....:         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)*]

period_length()#

Return the number of partial quotients of the preperiodic part of self.

EXAMPLES:

sage: continued_fraction(2/5).period_length()
1
sage: cf = continued_fraction([(0,1),(2,)]); cf
[0; 1, (2)*]
sage: cf.period_length()
1

preperiod()#

Return the preperiodic part of self.

EXAMPLES:

sage: # needs sage.rings.number_field
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)

preperiod_length()#

Return the number of partial quotients of the preperiodic part of self.

EXAMPLES:

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

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                                                     # needs sage.rings.number_field
sqrt2 + 1
sage: v.continued_fraction()                                                # needs sage.rings.number_field
[(2)*]

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


Some ultimately periodic but non periodic examples:

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

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

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

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

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); cf                                           # needs sage.symbolic
[3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ...]
sage: cf.value()                                                                # needs sage.symbolic
pi

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

length()#

Return infinity

EXAMPLES:

sage: continued_fraction(pi).length()                                       # needs sage.symbolic
+Infinity

quotient(n)#

Return the n-th quotient of self.

EXAMPLES:

sage: # needs sage.symbolic
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: # needs sage.symbolic
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: # needs sage.rings.number_field
sage: x = polygen(ZZ, 'x')
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: # needs sage.symbolic
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: # needs sage.rings.number_field
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)                                            # needs sage.symbolic
sage: cf.value()                                                            # needs sage.symbolic
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                                   # needs sage.symbolic
[1; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...]
sage: c.convergent(12)                                                          # needs sage.symbolic
377/233
sage: fibonacci(14)/fibonacci(13)                                               # needs sage.libs.pari
377/233

sage: # needs sage.symbolic
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: # needs sage.symbolic
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: # needs sage.rings.number_field
sage: x = polygen(ZZ, 'x')
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: # needs sage.rings.number_field
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()                                                                # needs sage.rings.number_field
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)                              # needs sage.rings.number_field
sage: cf = continued_fraction(a); cf                                            # needs sage.rings.number_field
[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))                                      # needs sage.symbolic
[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)#

Return 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

$x = a_0 - 1/(a_1 - 1/(...))$

INPUT:

• x – exact rational or floating-point number. The number to compute the continued fraction of.

• type – either “std” (default) for standard continued fractions or “hj” for Hirzebruch-Jung ones.

• partial_convergents – boolean. Whether to return the partial convergents.

• bits – an optional integer that specify a precision for the real interval field that is used internally.

• nterms – integer. The upper bound on the number of terms in the continued fraction expansion to return.

OUTPUT:

A lits of integers, the coefficients in the continued fraction expansion of x. If partial_convergents is set to True, then return a pair containing the coefficient list and the partial convergents list is returned.

EXAMPLES:

sage: continued_fraction_list(45/19)
[2, 2, 1, 2, 2]
sage: 2 + 1/(2 + 1/(1 + 1/(2 + 1/2)))
45/19

sage: continued_fraction_list(45/19, type="hj")
[3, 2, 3, 2, 3]
sage: 3 - 1/(2 - 1/(3 - 1/(2 - 1/3)))
45/19


Specifying bits or nterms modify the length of the output:

sage: # needs sage.symbolic
sage: continued_fraction_list(e, bits=20)
[2, 1, 2, 1, 1, 4, 2]
sage: continued_fraction_list(sqrt(2) + sqrt(3), bits=30)
[3, 6, 1, 5, 7, 2]
sage: continued_fraction_list(pi, bits=53)
[3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14]
sage: continued_fraction_list(log(3/2), nterms=15)
[0, 2, 2, 6, 1, 11, 2, 1, 2, 2, 1, 4, 3, 1, 1]
sage: continued_fraction_list(tan(sqrt(pi)), nterms=20)
[-5, 9, 4, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 2, 4, 3, 1, 63]


When the continued fraction is infinite (ie x is an irrational number) and the parameters bits and nterms are not specified then a warning is raised:

sage: continued_fraction_list(sqrt(2))                                          # needs sage.symbolic
doctest:...: UserWarning: the continued fraction of sqrt(2) seems infinite,
return only the first 20 terms
[1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]
sage: continued_fraction_list(sqrt(4/19))                                       # needs sage.symbolic
doctest:...: UserWarning: the continued fraction of 2*sqrt(1/19) seems infinite,
return only the first 20 terms
[0, 2, 5, 1, 1, 2, 1, 16, 1, 2, 1, 1, 5, 4, 5, 1, 1, 2, 1, 16]


An examples with the list of partial convergents:

sage: continued_fraction_list(RR(pi), partial_convergents=True)                 # needs sage.symbolic
([3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 3],
[(3, 1),
(22, 7),
(333, 106),
(355, 113),
(103993, 33102),
(104348, 33215),
(208341, 66317),
(312689, 99532),
(833719, 265381),
(1146408, 364913),
(4272943, 1360120),
(5419351, 1725033),
(80143857, 25510582),
(245850922, 78256779)])

sage.rings.continued_fraction.convergents(x)#

Return the (partial) convergents of the number x.

EXAMPLES:

sage: from sage.rings.continued_fraction import convergents
sage: convergents(143/255)
[0, 1, 1/2, 4/7, 5/9, 9/16, 14/25, 23/41, 60/107, 143/255]

sage.rings.continued_fraction.last_two_convergents(x)#

Given the list x that consists of numbers, return the two last convergents $$p_{n-1}, q_{n-1}, p_n, q_n$$.

This function is principally used to compute the value of a ultimately periodic continued fraction.

OUTPUT: a 4-tuple of Sage integers

EXAMPLES:

sage: from sage.rings.continued_fraction import last_two_convergents
sage: last_two_convergents([])
(0, 1, 1, 0)
sage: last_two_convergents([0])
(1, 0, 0, 1)
sage: last_two_convergents([-1,1,3,2])
(-1, 4, -2, 9)

sage.rings.continued_fraction.rat_interval_cf_list(r1, r2)#

Return the common prefix of the rationals r1 and r2 seen as continued fractions.

OUTPUT: a list of Sage integers.

EXAMPLES:

sage: from sage.rings.continued_fraction import rat_interval_cf_list
sage: rat_interval_cf_list(257/113, 5224/2297)
[2, 3, 1, 1, 1, 4]
sage: for prec in range(10,54):                                                 # needs sage.rings.real_interval_field
....:     R = RealIntervalField(prec)
....:     for _ in range(100):
....:         x = R.random_element() * R.random_element() + R.random_element() / 100
....:         l = x.lower().exact_rational()
....:         u = x.upper().exact_rational()
....:         if l.floor() != u.floor():
....:             continue
....:         cf = rat_interval_cf_list(l,u)
....:         a = continued_fraction(cf).value()
....:         b = continued_fraction(cf+[1]).value()
....:         if a > b:
....:             a,b = b,a
....:         assert a <= l
....:         assert b >= u