Periodic continued fraction explained

In mathematics, an infinite periodic continued fraction is a simple continued fraction that can be placed in the form

x=a₀+\cfrac{1}{a₁+\cfrac{1}{a₂+\cfrac{1}{ \ddots a_k+\cfrac{1}{a_k+1+\cfrac{\ddots}{ \ddots a_k+m-1+\cfrac{1}{a_k+m+\cfrac{1}{a_k+1+\cfrac{1}{a_k+2+{\ddots}}}}}}}}}

where the initial block

[a_0;a_1,...,a_k]

of k+1 partial denominators is followed by a block

[a_k+1,a_k+2,...,a_k+m]

of m partial denominators that repeats ad infinitum. For example,

\sqrt2

can be expanded to the periodic continued fraction

[1;2,2,2,...]

This article considers only the case of periodic regular continued fractions. In other words, the remainder of this article assumes that all the partial denominators a_i (i ≥ 1) are positive integers. The general case, where the partial denominators a_i are arbitrary real or complex numbers, is treated in the article convergence problem.

Purely periodic and periodic fractions

Since all the partial numerators in a regular continued fraction are equal to unity we can adopt a shorthand notation in which the continued fraction shown above is written as

\begin{align} x&=[a_0;a_1,a_2,...,a_k,a_k+1,a_k+2,...,a_k+m,a_k+1,a_k+2,...,a_k+m,...]\\ &=[a_0;a_1,a_2,...,a_{k,\overline{a}_k+1,a_k+2,...,a_k+m

}]\end

where, in the second line, a vinculum marks the repeating block. Some textbooks use the notation

\begin{align} x&=[a_0;a_1,a_2,...,a

•

k,a

_k+1,a_k+2,...,

•
a

_k+m] \end{align}

where the repeating block is indicated by dots over its first and last terms.

If the initial non-repeating block is not present - that is, if k = -1, a₀ = a_m and

x=[\overline{a_0;a_1,a_2,...,a_m-1

}],
the regular continued fraction x is said to be purely periodic. For example, the regular continued fraction

[1;1,1,1,...]

of the golden ratio φ is purely periodic, while the regular continued fraction
[1;2,2,2,...]

of
\sqrt2

is periodic, but not purely periodic.
As unimodular matrices

Periodic continued fractions are in one-to-one correspondence with the real quadratic irrationals. The correspondence is explicitly provided by Minkowski's question-mark function. That article also reviews tools that make it easy to work with such continued fractions. Consider first the purely periodic part

x=[0;\overline{a_1,a_2,...,a_m}],

This can, in fact, be written as
x=

\alphax+\beta
\gammax+\delta

with the
\alpha,\beta,\gamma,\delta

being integers, and satisfying
\alpha\delta-\beta\gamma=1.

Explicit values can be obtained by writing
S=\begin{pmatrix}1&0\ 1&1\end{pmatrix}

which is termed a "shift", so that
Sⁿ=\begin{pmatrix}1&0\ n&1\end{pmatrix}

and similarly a reflection, given by
T\mapsto\begin{pmatrix}-1&1\ 0&1\end{pmatrix}

so that
T^2=I

. Both of these matrices are unimodular, arbitrary products remain unimodular. Then, given
x

as above, the corresponding matrix is of the form
a₁
S
a₂
TS

T …

a_m
TS

=\begin{pmatrix}\alpha&\beta\ \gamma&\delta\end{pmatrix}

and one has
x=[0;\overline{a_1,a_2,...,a_m}]=

\alphax+\beta
\gammax+\delta

as the explicit form. As all of the matrix entries are integers, this matrix belongs to the modular group
SL(2,Z).

Relation to quadratic irrationals

A quadratic irrational number is an irrational real root of the quadratic equation

ax²+bx+c=0

where the coefficients a, b, and c are integers, and the discriminant,

b²-4ac

, is greater than zero. By the quadratic formula, every quadratic irrational can be written in the form
\zeta=

P+\sqrt{D

}
where P, D, and Q are integers, D > 0 is not a perfect square (but not necessarily square-free), and Q divides the quantity

P²-D

(for example
(6+\sqrt{8})/4

). Such a quadratic irrational may also be written in another form with a square-root of a square-free number (for example
(3+\sqrt{2})/2

) as explained for quadratic irrationals.
By considering the complete quotients of periodic continued fractions, Euler was able to prove that if x is a regular periodic continued fraction, then x is a quadratic irrational number. The proof is straightforward. From the fraction itself, one can construct the quadratic equation with integral coefficients that x must satisfy.

Lagrange proved the converse of Euler's theorem: if x is a quadratic irrational, then the regular continued fraction expansion of x is periodic. Given a quadratic irrational x one can construct m different quadratic equations, each with the same discriminant, that relate the successive complete quotients of the regular continued fraction expansion of x to one another. Since there are only finitely many of these equations (the coefficients are bounded), the complete quotients (and also the partial denominators) in the regular continued fraction that represents x must eventually repeat.

Reduced surds

\zeta=

P+\sqrt{D

} is said to be reduced if
\zeta>1

and its conjugate
η=

P-\sqrt{D

}satisfies the inequalities
-1<η<0

. For instance, the golden ratio
\phi=(1+\sqrt{5})/2=1.618033...

is a reduced surd because it is greater than one and its conjugate
(1-\sqrt{5})/2=-0.618033...

is greater than -1 and less than zero. On the other hand, the square root of two
\sqrt{2}=(0+\sqrt{8})/2

is greater than one but is not a reduced surd because its conjugate
-\sqrt{2}=(0-\sqrt{8})/2

is less than -1.
Galois proved that the regular continued fraction which represents a quadratic surd ζ is purely periodic if and only if ζ is a reduced surd. In fact, Galois showed more than this. He also proved that if ζ is a reduced quadratic surd and η is its conjugate, then the continued fractions for ζ and for (-1/η) are both purely periodic, and the repeating block in one of those continued fractions is the mirror image of the repeating block in the other. In symbols we have

\begin{align} \zeta&=[\overline{a_0;a_1,a_2,...,a_m-1

}]\\[3pt]\frac& = [\overline{a_{m-1};a_{m-2},a_{m-3},\dots,a_0}]\,\end
where ζ is any reduced quadratic surd, and η is its conjugate.

From these two theorems of Galois a result already known to Lagrange can be deduced. If r > 1 is a rational number that is not a perfect square, then

\sqrt{r}=[a_{0;\overline{a}_1,a_2,...,a_2,a_1,2a_0}].

In particular, if n is any non-square positive integer, the regular continued fraction expansion of contains a repeating block of length m, in which the first m - 1 partial denominators form a palindromic string.

Length of the repeating block

By analyzing the sequence of combinations

P_n+\sqrt{D

}
that can possibly arise when

\zeta=

P+\sqrt{D

} is expanded as a regular continued fraction, Lagrange showed that the largest partial denominator a_i in the expansion is less than
2\sqrt{D}

, and that the length of the repeating block is less than 2D.
l{O}(\sqrt{D}ln{D}).

Canonical form and repetend

The following iterative algorithm can be used to obtain the continued fraction expansion in canonical form (S is any natural number that is not a perfect square):

m₀₌₀

d₀₌₁

a_{0=\left\lfloor\sqrt{S}\right\rfloor}

m_n+1=d_na_n-m_n

d_n+1=

2
S-m
n+1
d_n

a_n+1=\left\lfloor

\sqrt{S
+m

_n+1

}\right\rfloor =\left\lfloor\frac\right\rfloor\!.
Notice that m_n, d_n, and a_n are always integers.The algorithm terminates when this triplet is the same as one encountered before.The algorithm can also terminate on a_i when a_i = 2 a₀, which is easier to implement.

The expansion will repeat from then on. The sequence

[a_0;a_1,a_2,a_3,...]

is the continued fraction expansion:
\sqrt{S}=a₀+\cfrac{1}{a₁+\cfrac{1}{a₂+\cfrac{1}{a_3+\ddots}}}

Example

To obtain as a continued fraction, begin with m₀ = 0; d₀ = 1; and a₀ = 10 (10² = 100 and 11² = 121 > 114 so 10 chosen).

\begin{align} \sqrt{114}&=

\sqrt{114
+0}{1}

=10+

\sqrt{114
-10}{1}

=10+

(\sqrt{114
-10)(\sqrt{114}+10)}{\sqrt{114}+10}

\\ &=10+

114-100
\sqrt{114

+10}=10+

1
\sqrt{114
+10

{14}}. \end{align}

m₁=d₀ ⋅ a₀-m₀=1 ⋅ 10-0=10.

d₁=

2
S-m
1
d₀

=

114-10²
1

=14.

a₁=\left\lfloor

a_0+m₁
d₁

\right\rfloor=\left\lfloor

10+10
14

\right\rfloor=\left\lfloor

20
14

\right\rfloor=1.

So, m₁ = 10; d₁ = 14; and a₁ = 1.

\sqrt{114
+10}{14}

=1+

\sqrt{114
-4}{14}

=1+

114-16
14(\sqrt{114

+4)}=1+

1
\sqrt{114
+4

{7}}.

Next, m₂ = 4; d₂ = 7; and a₂ = 2.

\sqrt{114
+4}{7}

=2+

\sqrt{114
-10}{7}

=2+

14
7(\sqrt{114

+10)}=2+

1
\sqrt{114
+10

{2}}.

\sqrt{114
+10}{2}=10+ \sqrt{114
-10}{2}=10+ 14
2(\sqrt{114

+10)}=10+

1
\sqrt{114
+10

{7}}.

\sqrt{114
+10}{7}=2+ \sqrt{114
-4}{7}=2+ 98
7(\sqrt{114

+4)}=2+

1
\sqrt{114
+4

{14}}.

\sqrt{114
+4}{14}=1+ \sqrt{114
-10}{14}=1+ 14
14(\sqrt{114

+10)}=1+

1
\sqrt{114
+10

{1}}.

\sqrt{114
+10}{1}=20+ \sqrt{114
-10}{1}=20+ 14
\sqrt{114

+10}=20+

1
\sqrt{114
+10

{14}}.

Now, loop back to the second equation above.

Consequently, the simple continued fraction for the square root of 114 is

\sqrt{114}=[10;\overline{1,2,10,2,1,20}].

is approximately 10.67707 82520. After one expansion of the repetend, the continued fraction yields the rational fraction

21194
1985

whose decimal value is approx. 10.67707 80856, a relative error of0.0000016% or 1.6 parts in 100,000,000.
Generalized continued fraction

A more rapid method is to evaluate its generalized continued fraction. From the formula derived there:

\begin{align} \sqrt{z}=\sqrt{x^2+y}&=x+\cfrac{y}{2x+\cfrac{y}{2x+\cfrac{y}{2x+\ddots}}}\\ &=x+\cfrac{2x ⋅ y}{2(2z-y)-y-\cfrac{y^2}{2(2z-y)-\cfrac{y^2}{2(2z-y)-\ddots}}} \end{align}

and the fact that 114 is 2/3 of the way between 10²=100 and 11²=121 results in

\begin{align} \sqrt{114}=\cfrac{\sqrt{1026}}{3}=\cfrac{\sqrt{32^2+2}}{3}&=\cfrac{32}{3}+\cfrac{2/3}{64+\cfrac{2}{64+\cfrac{2}{64+\cfrac{2}{64+\ddots}}}}\\ &=\cfrac{32}{3}+\cfrac{2}{192+\cfrac{18}{192+\cfrac{18}{192+\ddots}}}, \end{align}

which is simply the aforementioned

[10;1,2,10,2,1,20,1,2]

evaluated at every third term. Combining pairs of fractions produces
\begin{align} \sqrt{114}=\cfrac{\sqrt{32^2+2}}{3}&=\cfrac{32}{3}+\cfrac{64/3}{2050-1-\cfrac{1}{2050-\cfrac{1}{2050-\ddots}}}\\ &=\cfrac{32}{3}+\cfrac{64}{6150-3-\cfrac{9}{6150-\cfrac{9}{6150-\ddots}}}, \end{align}

which is now

[10;1,2,\overline{10,2,1,20,1,2}]

evaluated at the third term and every six terms thereafter.
References

Web site: Beceanu. Marius. 5 February 2003. Period of the Continued Fraction of sqrt(n). Theorem 2.3. 3 May 2022. https://web.archive.org/web/20151221205104/http://web.math.princeton.edu/mathlab/jr02fall/Periodicity/mariusjp.pdf. 21 December 2015.

Cohn. J. H. E.. 1977. The length of the period of the simple continued fraction expansion of d^1/2. Pacific J. Math.. 71. 21–32. 10.2140/pjm.1977.71.21. free.

Book: Davenport , H. . December 1982. The Higher Arithmetic. Cambridge University Press. 0-521-28678-6.

Book: Gliga , Alexandra Ioana . 17 March 2006. On continued fractions of the square root of prime numbers. Corollary 3.3. 3 May 2022.

Hickerson. Dean R.. 1973. Length of period of simple continued fraction expansion of vd. Pacific J. Math.. 46. 429–432. 10.2140/pjm.1973.46.429. free.

Book: Khinchin , A. Ya. . Aleksandr Khinchin. 1964. Continued Fractions. registration. Originally published in Russian, 1935. University of Chicago Press. 0-486-69630-8. (This is now available as a reprint from Dover Publications.)

Book: Long , Calvin T. . 1972. Elementary Introduction to Number Theory. 3 Sub. Waveland Pr Inc. 77-171950.

Book: Pettofrezzo. Anthony Joseph. Byrkit. Donald R.. 1970. Elements of Number Theory. Prentice Hall. Englewood Cliffs. 11. 9780132683005. 77-81766.

Podsypanin. E.V.. 1982. Length of the period of a quadratic irrational. Journal of Soviet Mathematics. 18. 6. 919–923. 10.1007/BF01763963. 119567810.

Book: Rockett. Andrew M.. Szüsz. Peter. Peter Szüsz. 1992. CONTINUED FRACTIONS. registration. World Scientific Publishing Company. 9789810210526.

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