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Inmathematics, an infiniteperiodic continued fraction is asimple continued fraction that can be placed in the form

${\displaystyle x=a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{\ddots a_k+\frac{1}{a_{k+1}+\frac{\ddots }{\ddots a_{k+m-1}+\frac{1}{a_{k+m}+\frac{1}{a_{k+1}+\frac{1}{a_{k+2}+\ddots }}}}}}}}}$

where the initial block${\displaystyle [a_0;a_1,\dots ,a_k]}$ ofk+1 partial denominators is followed by a block${\displaystyle [a_{k+1},a_{k+2},\dots ,a_{k+m}]}$ ofm partial denominators that repeatsad infinitum. For example,${\displaystyle \sqrt{2}}$ can be expanded to the periodic continued fraction${\displaystyle [1;2,2,2,...]}$.

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

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

where, in the second line, avinculum marks the repeating block.^[1] Some textbooks use the notation

where the repeating block is indicated by dots over its first and last terms.^[2]

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

${\displaystyle x=[\overline{a_0;a_1,a_2,\dots ,a_{m-1}}],}$

the regular continued fractionx is said to bepurely periodic. For example, the regular continued fraction${\displaystyle [1;1,1,1,\dots ]}$ of thegolden ratio φ is purely periodic, while the regular continued fraction${\displaystyle [1;2,2,2,\dots ]}$ of${\displaystyle \sqrt{2}}$ is periodic, but not purely periodic. However, the regular continued fraction${\displaystyle [2;2,2,2,\dots ]}$ of thesilver ratio ⁠${\displaystyle \sigma =\sqrt{2}+1}$ is purely periodic.

Periodic continued fractions are in one-to-one correspondence with the realquadratic irrationals. The correspondence is explicitly provided byMinkowski'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

${\displaystyle x=[0;\overline{a_1,a_2,\dots ,a_m}],}$

This can, in fact, be written as

${\displaystyle x=\frac{\alpha x+\beta }{\gamma x+\delta }}$

with the${\displaystyle \alpha ,\beta ,\gamma ,\delta }$ being integers, and satisfying${\displaystyle \alpha \delta -\beta \gamma =1.}$ Explicit values can be obtained by writing

which is termed a "shift", so that

and similarly a reflection, given by

so that${\displaystyle T^2=I}$. Both of these matrices areunimodular, arbitrary products remain unimodular. Then, given${\displaystyle x}$ as above, the corresponding matrix is of the form^[3]

and one has

${\displaystyle x=[0;\overline{a_1,a_2,\dots ,a_m}]=\frac{\alpha x+\beta }{\gamma x+\delta }}$

as the explicit form. As all of the matrix entries are integers, this matrix belongs to themodular group${\displaystyle SL(2,\mathbb{Z}).}$

Aquadratic irrational number is anirrational real root of the quadratic equation

${\displaystyle a{x}^2+bx+c=0}$

where the coefficientsa,b, andc are integers, and thediscriminant,${\displaystyle b^2-4ac}$, is greater than zero. By thequadratic formula, every quadratic irrational can be written in the form

${\displaystyle \zeta =\frac{P+\sqrt{D}}{Q}}$

whereP,D, andQ are integers,D > 0 is not aperfect square (but not necessarily square-free), andQ divides the quantity${\displaystyle P^2-D}$ (for example${\displaystyle (6+\sqrt{8})/4}$). Such a quadratic irrational may also be written in another form with a square-root of a square-free number (for example${\displaystyle (3+\sqrt{2})/2}$) as explained forquadratic irrationals.

By considering thecomplete quotients of periodic continued fractions,Euler was able to prove that ifx is a regular periodic continued fraction, thenx is a quadratic irrational number. The proof is straightforward. From the fraction itself, one can construct the quadratic equation with integral coefficients thatx must satisfy.

Lagrange proved the converse of Euler's theorem: ifx is a quadratic irrational, then the regular continued fraction expansion ofx is periodic.^[4] Given a quadratic irrationalx one can constructm different quadratic equations, each with the same discriminant, that relate the successive complete quotients of the regular continued fraction expansion ofx 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 representsx must eventually repeat.

The quadraticsurd${\displaystyle \zeta =\frac{P+\sqrt{D}}{Q}}$ is said to bereduced if${\displaystyle \zeta >1}$ and itsconjugate${\displaystyle \eta =\frac{P-\sqrt{D}}{Q}}$satisfies the inequalities. For instance, the golden ratio${\displaystyle \varphi =(1+\sqrt{5})/2=\mathrm{1.618033...}}$ is a reduced surd because it is greater than one and its conjugate${\displaystyle (1-\sqrt{5})/2=-\mathrm{0.618033...}}$ is greater than −1 and less than zero. On the other hand, the square root of two${\displaystyle \sqrt{2}=(0+\sqrt{8})/2}$ is greater than one but is not a reduced surd because its conjugate${\displaystyle -\sqrt{2}=(0-\sqrt{8})/2}$ 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

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. Ifr > 1 is a rational number that is not a perfect square, then

${\displaystyle \sqrt{r}=[a_0;\overline{a_1,a_2,\dots ,a_2,a_1,2a_0}].}$

In particular, ifn is any non-square positive integer, the regular continued fraction expansion of√n contains a repeating block of lengthm, in which the firstm − 1 partial denominators form apalindromic string.

By analyzing the sequence of combinations

${\displaystyle \frac{P_n+\sqrt{D}}{Q_n}}$

that can possibly arise when${\displaystyle \zeta =\frac{P+\sqrt{D}}{Q}}$ is expanded as a regular continued fraction,Lagrange showed that the largest partial denominatora_i in the expansion is less than${\displaystyle 2\sqrt{D}}$, and that the length of the repeating block is less than 2D.

More recently, sharper arguments^[5][6][7] based on thedivisor function have shown that the length of the repeating block for a quadratic surd of discriminantD ison the order of${\displaystyle \mathcal{O}(\sqrt{D}\ln D).}$

The following iterative algorithm^[8] can be used to obtain the continued fraction expansion in canonical form (S is anynatural number that is not aperfect square):

${\displaystyle m_0=0}$

${\displaystyle d_0=1}$

${\displaystyle a_0=\lfloor \sqrt{S}\rfloor }$

${\displaystyle m_{n+1}=d_n{a}_n-m_n}$

${\displaystyle d_{n+1}=\frac{S-m_{n+1}^2}{d_n}}$

${\displaystyle a_{n+1}=\lfloor \frac{\sqrt{S}+m_{n+1}}{d_{n+1}}\rfloor =\lfloor \frac{a_0+m_{n+1}}{d_{n+1}}\rfloor .}$

Notice thatm_n,d_n, anda_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,^[9] which is easier to implement.

The expansion will repeat from then on. The sequence${\displaystyle [a_0;a_1,a_2,a_3,\dots ]}$ is the continued fraction expansion:

${\displaystyle \sqrt{S}=a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\ddots }}}}$

To obtain√114 as a continued fraction, begin withm₀ = 0;d₀ = 1; anda₀ = 10 (10² = 100 and 11² = 121 > 114 so 10 chosen).

${\displaystyle m_1=d_0\cdot a_0-m_0=1\cdot 10-0=10.}$

${\displaystyle d_1=\frac{S-m_1^2}{d_0}=\frac{114-{10}^2}{1}=14.}$

${\displaystyle a_1=\lfloor \frac{a_0+m_1}{d_1}\rfloor =\lfloor \frac{10+10}{14}\rfloor =\lfloor \frac{20}{14}\rfloor =1.}$

So,m₁ = 10;d₁ = 14; anda₁ = 1.

${\displaystyle \frac{\sqrt{114}+10}{14}=1+\frac{\sqrt{114}-4}{14}=1+\frac{114-16}{14(\sqrt{114}+4)}=1+\frac{1}{\frac{\sqrt{114}+4}{7}}.}$

Next,m₂ = 4;d₂ = 7; anda₂ = 2.

${\displaystyle \frac{\sqrt{114}+4}{7}=2+\frac{\sqrt{114}-10}{7}=2+\frac{14}{7(\sqrt{114}+10)}=2+\frac{1}{\frac{\sqrt{114}+10}{2}}.}$

${\displaystyle \frac{\sqrt{114}+10}{2}=10+\frac{\sqrt{114}-10}{2}=10+\frac{14}{2(\sqrt{114}+10)}=10+\frac{1}{\frac{\sqrt{114}+10}{7}}.}$

${\displaystyle \frac{\sqrt{114}+10}{7}=2+\frac{\sqrt{114}-4}{7}=2+\frac{98}{7(\sqrt{114}+4)}=2+\frac{1}{\frac{\sqrt{114}+4}{14}}.}$

${\displaystyle \frac{\sqrt{114}+4}{14}=1+\frac{\sqrt{114}-10}{14}=1+\frac{14}{14(\sqrt{114}+10)}=1+\frac{1}{\frac{\sqrt{114}+10}{1}}.}$

${\displaystyle \frac{\sqrt{114}+10}{1}=20+\frac{\sqrt{114}-10}{1}=20+\frac{14}{\sqrt{114}+10}=20+\frac{1}{\frac{\sqrt{114}+10}{14}}.}$

Now, loop back to the second equation above.

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

${\displaystyle \sqrt{114}=[10;\overline{1,2,10,2,1,20}].}$ (sequenceA010179 in theOEIS)

√114 is approximately 10.67707 82520. After one expansion of the repetend, the continued fraction yields the rational fraction${\displaystyle \frac{21194}{1985}}$ whose decimal value is approx. 10.67707 80856, a relative error of0.0000016% or 1.6 parts in 100,000,000.

A more rapid method is to evaluate itsgeneralized continued fraction. From the formula derivedthere:

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

which is simply the aforementioned${\displaystyle [10;1,2,10,2,1,20,1,2]}$ evaluated at every third term. Combining pairs of fractions produces

which is now${\displaystyle [10;1,2,\overline{10,2,1,20,1,2}]}$ evaluated at the third term and every six terms thereafter.

• Continued fraction – Mathematical expression
• Generalized continued fraction – Mathematical expressionPages displaying short descriptions of redirect targets
• Hermite's problem
• Continued fraction method of computing square roots – Algorithms for calculating square roots
• Restricted partial quotients – Analytic series
• Continued fraction factorization – an integer factorization algorithmPages displaying wikidata descriptions as a fallback

• Continued fractions
• Mathematical analysis

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