In mathematics, a continued fraction is an expression obtained through an iterative process of representing a number as the sum of its integer part and the reciprocal of another number, then writing this other number as the sum of its integer part and another reciprocal, and so on.[1] In a finite continued fraction (or terminated continued fraction), the iteration/recursion is terminated after finitely many steps by using an integer in lieu of another continued fraction. In contrast, an infinite continued fraction is an infinite expression. In either case, all integers in the sequence, other than the first, must be positive. The integers ai are called the coefficients or terms of the continued fraction.[2] Continued fractions have a number of remarkable properties related to the Euclidean algorithm for integers or real numbers. Every rational number p/q has two closely related expressions as a finite continued fraction, whose coefficients ai can be determined by applying the Euclidean algorithm to (p, q). The numerical value of an infinite continued fraction will be irrational; it is defined from its infinite sequence of integers as the limit of a sequence of values for finite continued fractions. Each finite continued fraction of the sequence is obtained by using a finite prefix of the infinite continued fraction's defining sequence of integers. Moreover, every irrational number α is the value of a unique infinite continued fraction, whose coefficients can be found using the nonterminating version of the Euclidean algorithm applied to the incommensurable values α and 1. This way of expressing real numbers (rational and irrational) is called their continued fraction representation. If arbitrary values and/or functions are used in place of one or more of the numerators or the integers in the denominators, the resulting expression is a generalized continued fraction. When it is necessary to distinguish the first form from generalized continued fractions, the former may be called a simple or regular continued fraction, or said to be in canonical form. The term continued fraction may also refer to representations of rational functions, arising in their analytic theory. For this use of the term see Padé approximation and Chebyshev rational functions. Consider a typical rational number 415/93, which is around 4.4624. As a first approximation, start with 4, which is the integer part; 415/93 = 4 + 43/93. Note that the fractional part is the reciprocal of 93/43 which is about 2.1628. Use the integer part, 2, as an approximation for the reciprocal, to get a second approximation of 4 + 1/2 = 4.5; 93/43 = 2 + 7/43. The fractional part of 93/43 is the reciprocal of 43/7 which is about 6.1429. Use 6 as an approximation for this to get 2 + 1/6 as an approximation for 93/43 and 4 + 1/(2 + 1/6), about 4.4615, as the third approximation; 43/7 = 6 + 1/7. Finally, the fractional part of 43/7 is the reciprocal of 7, so its approximation in this scheme, 7, is exact (7/1 = 7 + 0/1) and produces the exact expression 4 + 1/(2 + 1/(6 + 1/7)) for 415/93. This expression is called the continued fraction representation of the number. Dropping some of the less essential parts of the expression 4 + 1 / (2 + 1 / (6 + 1 / 7)) gives the abbreviated notation 415/93=[4;2,6,7]. Note that it is customary to replace only the first comma by a semicolon. Some older textbooks use all commas in the (n+1)tuple, e.g. [4,2,6,7].[3][4] If the starting number is rational then this process exactly parallels the Euclidean algorithm. In particular, it must terminate and produce a finite continued fraction representation of the number. If the starting number is irrational then the process continues indefinitely. This produces a sequence of approximations, all of which are rational numbers, and these converge to the starting number as a limit. This is the (infinite) continued fraction representation of the number. Examples of continued fraction representations of irrational numbers are: √19 = [4;2,1,3,1,2,8,2,1,3,1,2,8,…]. The pattern repeats indefinitely with a period of 6. Continued fractions are, in some ways, more "mathematically natural" representations of real number than other representations such as decimal representations, and they have several desirable properties: The continued fraction representation for a rational number is finite and only rational numbers have finite representations. In contrast, the decimal representation of a rational number may be finite, for example 137/1600 = 0.085625, or infinite with a repeating cycle, for example 4/27 = 0.148148148148…. Basic formulae A finite continued fraction is an expression of the form \[ a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{ \ddots + \cfrac{1}{a_n} }}}, \] where a0 is an integer, all other ai are positive integers, and n is a nonnegative integer. Thus, all of the following illustrate valid finite continued fractions:
An infinite continued fraction can be written as \[ a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{a_4 + \ddots}}}}, \] with the same constraints on the ai as in the finite case. Consider a real number r. Let i be the integer part and f the fractional part of r. Then the continued fraction representation of r is [i;a1,a2,…], where [a1;a2,…] is the continued fraction representation of 1/f. To calculate a continued fraction representation of a number r, write down the integer part (technically the floor) of r. Subtract this integer part from r. If the difference is 0, stop; otherwise find the reciprocal of the difference and repeat. The procedure will halt if and only if r is rational.
The number 3.245 can also be represented by the continued fraction expansion [3;4,12,3,1]; refer to Finite continued fractions below. The integers a0, a1 etc., are called the quotients of the continued fraction. One can abbreviate a continued fraction as \[ x = [a_0; a_1, a_2, a_3] \; \] or, in the notation of Pringsheim, as \[ x = a_0 + \frac{1 \mid}{\mid a_1} + \frac{1 \mid}{\mid a_2} + \frac{1 \mid}{\mid a_3}. \] Here is another related notation: \[ x = a_0 + {1 \over a_1 + {}} {1 \over a_2 + {}} {1 \over a_3 + {}}. \] Sometimes angle brackets are used, like this: \[ x = \left \langle a_0; a_1, a_2, a_3 \right \rangle.\;\] The semicolon in the square and angle bracket notations is sometimes replaced by a comma. One may also define infinite simple continued fractions as limits: [a_0; a_1, a_2, a_3, \,\ldots ] = \lim_{n \to \infty} [a_0; a_1, a_2, \,\ldots, a_n]. \] This limit exists for any choice of a0 and positive integers a1, a2, ... . Every finite continued fraction represents a rational number, and every rational number can be represented in precisely two different ways as a finite continued fraction. These two representations agree except in their final terms. In the longer representation the final term in the continued fraction is 1; the shorter representation drops the final 1, but increases the new final term by 1. The final element in the short representation is therefore always greater than 1, if present. In symbols: \[ [a_{0}; a_{1}, a_{2}, \,\ldots, a_{n1}, a_{n}, 1]=[a_{0}; a_{1}, a_{2}, \,\ldots, a_{n1}, a_{n} + 1]. \;\] For example, \[ 2.25 = 2 + 1/4 = [2; 4] = 2 + 1/(3 + 1/1) = [2; 3, 1], \;\] Continued fractions of reciprocals The continued fraction representations of a positive rational number and its reciprocal are identical except for a shift one place left or right depending on whether the number is less than or greater than one respectively. In other words, the numbers represented by [a_0;a_1,a_2,a_3,\ldots,a_n] and [0;a_0,a_1,a_2,\ldots,a_n] are reciprocals. This is because if a\ is an integer then if x<1\ then x = 0+1/(a+1/b)\ and 1/x = a+1/b\ and if x>1\ then x = a+1/b\ and 1/x = 0+1/(a+1/b)\ with the last number that generates the remainder of the continued fraction being the same for both x\ and its reciprocal. For example, \[ 2.25 = \frac{9}{4} = [2; 4], \;\] \[ \frac{1}{2.25} = \frac{4}{9} = [0; 2, 4]. \;\] Infinite continued fractions Every infinite continued fraction is irrational, and every irrational number can be represented in precisely one way as an infinite continued fraction. An infinite continued fraction representation for an irrational number is mainly useful because its initial segments provide excellent rational approximations to the number. These rational numbers are called the convergents of the continued fraction. Evennumbered convergents are smaller than the original number, while oddnumbered ones are bigger. For a continued fraction [a0;a1,a2,…], the first four convergents (numbered 0 through 3) are \[ \frac{a_0}{1},\qquad \frac{a_1a_0 + 1}{a_1},\qquad \frac{ a_2(a_1a_0+1)+a_0}{a_2a_1+1},\qquad \frac{a_3(a_2(a_1a_0+1)+a_0)+(a_1a_0+1)}{a_3(a_2a_1+1)+a_1}. \] In words, the numerator of the third convergent is formed by multiplying the numerator of the second convergent by the third quotient, and adding the numerator of the first convergent. The denominators are formed similarly. Therefore, each convergent can be expressed explicitly in terms of the continued fraction as the ratio of certain multivariate polynomials called continuants. If successive convergents are found, with numerators h1, h2, ... and denominators k1, k2, ... then the relevant recursive relation is: \[ h_n=a_nh_{n1}+h_{n2},\qquad k_n=a_nk_{n1}+k_{n2}.\] The successive convergents are given by the formula \[ \frac{h_n}{k_n}= \frac{a_nh_{n1}+h_{n2}}{a_nk_{n1}+k_{n2}}. \] Thus to incorporate a new term into a rational approximation, only the two previous convergents are necessary. The initial "convergents" (required for the first two terms) are 0⁄1 and 1⁄0. For example, here are the convergents for [0;1,5,2,2].
When using the Babylonian method to generate successive approximations to the square root of an integer, if one starts with the lowest integer as first approximant, the rationals generated all appear in the list of convergents for the continued fraction. Specifically, the approximants will appear on the convergents list in positions 0, 1, 3, 7, 15, … , 2^k  1… For example, the continued fraction expansion for √3 is [1;1,2,1,2,1,2,1,2,…]. Comparing the convergents with the approximants derived from the Babylonian method:
\[ x_0 = 1 = \frac{1}{1}\] Some useful theorems If a0, a1, a2, … is an infinite sequence of positive integers, define the sequences hn and kn recursively: For any positive \[ x\in\mathbb{R} \] \[ \left[a_0; a_1, \,\dots, a_{n1}, x \right]= \frac{x h_{n1}+h_{n2}} {x k_{n1}+k_{n2}}.\] Theorem 2 The convergents of [a0;a1,a2,…] are given by \[ \left[a_0; a_1, \,\dots, a_n\right]= \frac{h_n} {k_n}.\] Theorem 3 If the nth convergent to a continued fraction is \[ h_n/k_n, \] then \[ k_nh_{n1}k_{n1}h_n=(1)^n.\, \] Corollary 1: Each convergent is in its lowest terms (for if h_n and k_n had a nontrivial common divisor it would divide k_nh_{n1}k_{n1}h_n, which is impossible). Corollary 2: The difference between successive convergents is a fraction whose numerator is unity: \[ \frac{h_n}{k_n}\frac{h_{n1}}{k_{n1}} = \frac{h_nk_{n1}k_nh_{n1}}{k_nk_{n1}}= \frac{(1)^n}{k_nk_{n1}}. \] Corollary 3: The continued fraction is equivalent to a series of alternating terms: \[ a_0 + \sum_{n=0}^\infty \frac{(1)^{n}}{k_{n+1}k_{n}}. \] Corollary 4: The matrix \[ \begin{bmatrix} h_n & h_{n1} \\ k_n & k_{n1} \end{bmatrix} \] has determinant plus or minus one, and thus belongs to the group of 2×2 unimodular matrices SL^*(2,\mathbb{Z}). Each (sth) convergent is nearer to a subsequent (nth) convergent than any preceding (rth) convergent is. In symbols, if the nth convergent is taken to be [a_0;a_1,a_2,\ldots a_n]=x_n, then \[ \left x_r  x_n \right > \left x_s  x_n \right \] for all r < s < n. Corollary 1: the even convergents (before the nth) continually increase, but are always less than xn. Corollary 2: the odd convergents (before the nth) continually decrease, but are always greater than xn. \[ \frac{1}{k_n(k_{n+1}+k_n)}< \leftx\frac{h_n}{k_n}\right< \frac{1}{k_nk_{n+1}}. \] Corollary 1: any convergent is nearer to the continued fraction than any other fraction whose denominator is less than that of the convergent Corollary 2: any convergent which immediately precedes a large quotient is a near approximation to the continued fraction. If \[ \frac{h_{n1}}{k_{n1}}\text{ and }\frac{h_n}{k_n}\] are successive convergents, then any fraction of the form \[ \frac{h_{n1} + ah_n}{k_{n1}+ak_n} \] where a is a nonnegative integer and the numerators and denominators are between the n and n + 1 terms inclusive are called semiconvergents, secondary convergents, or intermediate fractions. Often the term is taken to mean that being a semiconvergent excludes the possibility of being a convergent, rather than that a convergent is a kind of semiconvergent. The semiconvergents to the continued fraction expansion of a real number x include all the rational approximations which are better than any approximation with a smaller denominator. Another useful property is that consecutive semiconvergents a/b and c/d are such that ad − bc = ±1. A best rational approximation to a real number x is a rational number n/d, d > 0, that is closer to x than any approximation with a smaller denominator. The simple continued fraction for x generates all of the best rational approximations for x according to three rules: Truncate the continued fraction, and possibly decrement its last term. For example, 0.84375 has continued fraction [0;1,5,2,2]. Here are all of its best rational approximations.
The strictly monotonic increase in the denominators as additional terms are included permits an algorithm to impose a limit, either on size of denominator or closeness of approximation. The "half rule" mentioned above is that when ak is even, the halved term ak/2 is admissible if and only if \[  x  [a_0 ; a_1, \dots, a_{k1}]  >  x  [a_0 ; a_1, \dots, a_{k1}, a_k/2].[6] \] This is equivalent[6] to: \[ \left[a_k;a_{k1},\ldots,a_1\right] > \left[a_k;a_{k+1},\ldots\right].[7] \] The convergents to x are best approximations in an even stronger sense: n/d is a convergent for x if and only if dx − n is the least relative error among all approximations m/c with c ≤ d; that is, we have dx − n < cx − m so long as c < d. (Note also that dkx − nk → 0 as k → ∞.) A rational that falls within the interval ((x, y)), for 0 < x < y, can be found with the continued fractions for x and y. When both x and y are irrational and \[ \begin{align} x &= [a_0; a_1, a_2, \ldots, a_{k1}, a_k, a_{k+1}, \ldots ]\\ y &= [a_0; a_1, a_2, \ldots, a_{k1}, b_k, b_{k+1}, \ldots ] \end{align} \] where x and y have identical continued fraction expansions up through \[ a_{k1} \], a rational that falls within the interval (x,y) is given by the finite continued fraction, \[ z(x,y) = [a_0; a_1, a_2, \ldots, a_{k1}, \min(a_k,b_k)+1]\,.\] This rational will be best in that no other rational in ((x, y)) will have a smaller numerator or a smaller denominator. If x is rational, it will have two continued fraction representations that are finite, \[ x_1 .\] and \[ x_2, .\] and similarly a rational y will have two representations, y_1 and y_2. The coefficients beyond the last in any of these representations should be interpreted as +∞; and the best rational will be one of z(x_1,y_1), z(x_1,y_2), z(x_2,y_1), or z(x_2,y_2). For example, the decimal representation 3.1416 could be rounded from any number in the interval [3.14155, 3.14165]. The continued fraction representations of 3.14155 and 3.14165 are \[ \begin{align} 3.14155 &= [3; 7, 15, 2, 7, 1, 4, 1, 1] = [3; 7, 15, 2, 7, 1, 4, 2]\\ 3.14165 &= [3; 7, 16, 1, 3, 4, 2, 3, 1] = [3; 7, 16, 1, 3, 4, 2, 4] \,, \end{align}\] and the best rational between these two is \[ [3; 7, 16] = \frac{355}{113} = 3.1415929\ldots\,. \] Thus, in some sense, 355/113 is the best rational number corresponding to the rounded decimal number 3.1416. A rational number, which can be expressed as finite continued fraction in two ways, \[ z = [a_0; a_1, \ldots, a_{k1}, a_{k}, 1] = [a_0; a_1, \ldots, a_{k1}, a_{k}+1]\] will be one of the convergents for the continued fraction expansion of a number, if and only if the number is strictly between \[ \begin{align} x & = [a_0; a_1, \ldots, a_{k1}, a_{k}, 2]\mathrm{~and}\\ y & = [a_0; a_1, \ldots, a_{k1}, a_{k} + 2]\,. \end{align}\] Note that the numbers x and y are formed by incrementing the last coefficient in the two representations for z, and that x < y when k is even, and x > y when k is odd. For example, the number 355/113 has the continued fraction representations \[ 355/113 = [3; 7, 15, 1] = [3; 7, 16]\,, \] and thus 355/113 is a convergent of any number strictly between \[ \begin{align} \,[3; 7, 15, 2] &= \frac{688}{219} \approx 3.1415525\mathrm{~and}\\ \,[3; 7, 17] &= \frac{377}{120} \approx 3.1416667\,. \end{align}\] Comparison of continued fractions Consider x = [a0; a1, ...] and y = [b0; b1, ...]. If k is the smallest index for which ak is unequal to bk then x < y if (−1)k(ak − bk) < 0 and y < x otherwise. If there is no such k, but one expansion is shorter than the other, say x = [a0; a1, ..., an] and y = [b0; b1, ..., bn, bn+1, ...] with ai = bi for 0 ≤ i ≤ n, then x < y if n is even and y < x if n is odd. To calculate the convergents of pi we may set a_0 = \lfloor \pi \rfloor = 3 , define u_1 = \frac {1}{\pi  3} \approx 7.0625 and a_1 = \lfloor u_1 \rfloor = 7 , u_2 = \frac {1}{u_1  7} \approx 15.9965 and a_2 = \lfloor u_2 \rfloor = 15 , u_3 = \frac {1}{u_2  15} \approx 1.003 . Continuing like this, one can determine the infinite continued fraction of π as \[ [3;7,15,1,292,1,1,…] \](sequence A001203 in OEIS). The third convergent of π is [3;7,15,1] = 355/113 = 3.14159292035..., sometimes called Milü, which is fairly close to the true value of π. Let us suppose that the quotients found are, as above, [3;7,15,1]. The following is a rule by which we can write down at once the convergent fractions which result from these quotients without developing the continued fraction. The first quotient, supposed divided by unity, will give the first fraction, which will be too small, namely, 3/1. Then, multiplying the numerator and denominator of this fraction by the second quotient and adding unity to the numerator, we shall have the second fraction, 22/7, which will be too large. Multiplying in like manner the numerator and denominator of this fraction by the third quotient, and adding to the numerator the numerator of the preceding fraction, and to the denominator the denominator of the preceding fraction, we shall have the third fraction, which will be too small. Thus, the third quotient being 15, we have for our numerator (22 × 15 = 330) + 3 = 333, and for our denominator, (7 × 15 = 105) + 1 = 106. The third convergent, therefore, is 333/106. We proceed in the same manner for the fourth convergent. The fourth quotient being 1, we say 333 times 1 is 333, and this plus 22, the numerator of the fraction preceding, is 355; similarly, 106 times 1 is 106, and this plus 7 is 113. In this manner, by employing the four quotients [3;7,15,1], we obtain the four fractions: \[ \frac{3}{1}, \frac{22}{7}, \frac{333}{106}, \frac{355}{113}, \,\ldots.\] These convergents are alternately smaller and larger than the true value of π, and approach nearer and nearer to π. The difference between a given convergent and π is less than the reciprocal of the product of the denominators of that convergent and the next convergent. For example, the fraction 22/7 is greater than π, but 22/7 − π is less than 1/(7 × 106), that is 1/742 (in fact, 22/7 − π is just less than 1/790). The demonstration of the foregoing properties is deduced from the fact that if we seek the difference between one of the convergent fractions and the next adjacent to it we shall obtain a fraction of which the numerator is always unity and the denominator the product of the two denominators. Thus the difference between 22/7 and 3/1 is 1/7, in excess; between 333/106 and 22/7, 1/742, in deficit; between 355/113 and 333/106, 1/11978, in excess; and so on. The result being, that by employing this series of differences we can express in another and very simple manner the fractions with which we are here concerned, by means of a second series of fractions of which the numerators are all unity and the denominators successively be the product of every two adjacent denominators. Instead of the fractions written above, we have thus the series: \[ \frac{3}{1}+\frac{1}{1 \times 7}\frac{1}{7 \times 106}+\frac{1}{106 \times 113}  \cdots. \] The first term, as we see, is the first fraction; the first and second together give the second fraction, 22/7; the first, the second and the third give the third fraction 333/106, and so on with the rest; the result being that the series entire is equivalent to the original value. A generalized continued fraction is an expression of the form \[ x = b_0 + \cfrac{a_1}{b_1 + \cfrac{a_2}{b_2 + \cfrac{a_3}{b_3 + \cfrac{a_4}{b_4 + \ddots\,}}}}\] where the an (n > 0) are the partial numerators, the bn are the partial denominators, and the leading term b0 is called the integer part of the continued fraction. To illustrate the use of generalized continued fractions, consider the following example. The sequence of partial denominators of the simple continued fraction of π does not show any obvious pattern: \[ \pi=[3;7,15,1,292,1,1,1,2,1,3,1,\ldots]\,\! \] or \[ \pi=3+\cfrac{1}{7+\cfrac{1}{15+\cfrac{1}{1+\cfrac{1}{292+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{3+\cfrac{1}{1+\ddots}}}}}}}}}}}\] However, several generalized continued fractions for π have a perfectly regular structure, such as: \[ \pi=\cfrac{4}{1+\cfrac{1^2}{2+\cfrac{3^2}{2+\cfrac{5^2}{2+\cfrac{7^2}{2+\cfrac{9^2}{2+\ddots}}}}}} =\cfrac{4}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\cfrac{4^2}{9+\ddots}}}}} =3+\cfrac{1^2}{6+\cfrac{3^2}{6+\cfrac{5^2}{6+\cfrac{7^2}{6+\cfrac{9^2}{6+\ddots}}}}} \] The first two of these are special cases of the arctangent function with \pi = 4 arctan 1. The numbers with periodic continued fraction expansion are precisely the irrational solutions of quadratic equations with rational coefficients (rational solutions have finite continued fraction expansions as previously stated). The simplest examples are the golden ratio φ = [1;1,1,1,1,1,…] and √2 = [1;2,2,2,2,…]; while √14 = [3;1,2,1,6,1,2,1,6…] and √42 = [6;2,12,2,12,2,12…]. All irrational square roots of integers have a special form for the period; a symmetrical string, like the empty string (for √2) or 1,2,1 (for √14), followed by the double of the leading integer. Because the continued fraction expansion for φ doesn't use any integers greater than 1, φ is one of the most "difficult" real numbers to approximate with rational numbers. Hurwitz's theorem[8] states that any real number k can be approximated by infinitely many rational m/n with \[ \left k  {m \over n}\right < {1 \over n^2 \sqrt 5}.\] While virtually all real numbers k will eventually have infinitely many convergents m/n whose distance from k is significantly smaller than this limit, the convergents for φ (i.e., the numbers 5/3, 8/5, 13/8, 21/13, etc.) consistently "toe the boundary", keeping a distance of almost exactly {\scriptstyle{1 \over n^2 \sqrt 5}} away from φ, thus never producing an approximation nearly as impressive as, for example, 355/113 for π. It can also be shown that every real number of the form (a + bφ)/(c + dφ) – where a, b, c, and d are integers such that ad − bc = ±1 – shares this property with the golden ratio φ; and that all other real numbers can be more closely approximated. While there is no discernable pattern in the simple continued fraction expansion of π, there is one for e, the base of the natural logarithm: \[ e = e^1 = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, 1, 1, \dots] \,\!,\] which is a special case of this general expression for positive integer n: \[ e^{1/n} = [1; n1, 1, 1, 3n1, 1, 1, 5n1, 1, 1, 7n1, 1, 1, \dots] \,\!.\] Another, more complex pattern appears in this continued fraction expansion for positive odd n: \[ e^{2/n} = \left[1; \frac{n1}{2}, 6n, \frac{5n1}{2}, 1, 1, \frac{7n1}{2}, 18n, \frac{11n1}{2}, 1, 1, \frac{13n1}{2}, 30n, \frac{17n1}{2}, 1, 1, \dots \right] \,\!,.\] with a special case for n = 1: \[ e^2 = [7; 2, 1, 1, 3, 18, 5, 1, 1, 6, 30, 8, 1, 1, 9, 42, 11, 1, 1, 12, 54, 14, 1, 1 \dots, 3k, 12k+6, 3k+2, 1, 1 \dots] \,\!.\] Other continued fractions of this sort are \[ \tanh(1/n) = [0; n, 3n, 5n, 7n, 9n, 11n, 13n, 15n, 17n, 19n, \dots] \,\!\] where n is a positive integer; also, for integral n: \[ \tan(1/n) = [0; n1, 1, 3n2, 1, 5n2, 1, 7n2, 1, 9n2, 1, \dots]\,\!,\] with a special case for n = 1: \[ \tan(1) = [1; 1, 1, 3, 1, 5, 1, 7, 1, 9, 1, 11, 1, 13, 1, 15, 1, 17, 1, 19, 1, \dots]\,\!.\] If In(x) is the modified, or hyperbolic, Bessel function of the first kind, we may define a function on the rationals p/q by \[ S(p/q) = \frac{I_{p/q}(2/q)}{I_{1+p/q}(2/q)},\] which is defined for all rational numbers, with p and q in lowest terms. Then for all nonnegative rationals, we have \[ S(p/q) = [p+q; p+2q, p+3q, p+4q, \dots]\,\!\] with similar formulas for negative rationals; in particular we have \[ S(0) = S(0/1) = [1; 2, 3, 4, 5, 6, 7, \dots]\,\!.\] Many of the formulas can be proved using Gauss's continued fraction. Most irrational numbers do not have any periodic or regular behavior in their continued fraction expansion. Nevertheless Khinchin proved that for almost all real numbers x, the ai (for i = 1, 2, 3, ...) have an astonishing property: their geometric mean is a constant (known as Khinchin's constant, K ≈ 2.6854520010...) independent of the value of x. Paul Lévy showed that the nth root of the denominator of the nth convergent of the continued fraction expansion of almost all real numbers approaches an asymptotic limit, which is known as Lévy's constant. Lochs' theorem states that nth convergent of the continued fraction expansion of almost all real numbers determines the number to an average accuracy of just over n decimal places. Continued fractions play an essential role in the solution of Pell's equation. For example, for positive integers p and q, p2 − 2q2 = ±1 only if p/q is a convergent of √2. Continued fractions also play a role in the study of chaos, where they tie together the Farey fractions which are seen in the Mandelbrot set with Minkowski's question mark function and the modular group Gamma. The backwards shift operator for continued fractions is the map \[ h(x)=1/x  \lfloor 1/x \rfloor, \] called the Gauss map, which lops off digits of a continued fraction expansion: \[ h([0;a_1,a_2,a_3,\dots]) = [0;a_2,a_3,\dots] \]. The transfer operator of this map is called the Gauss–Kuzmin–Wirsing operator. The distribution of the digits in continued fractions is given by the zero'th eigenvector of this operator, and is called the Gauss–Kuzmin distribution. The Lanczos algorithm uses a continued fraction expansion to iteratively approximate the eigenvalues and eigenvectors of a large sparse matrix. 300 BC Euclid's Elements contains an algorithm for the greatest common divisor which generates a continued fraction as a byproduct Cataldi represented a continued fraction as a_0.\, & n_1 \over d_1. & n_2 \over d_2. & {n_3 \over d_3} with the dots indicating where the following fractions went. 1695 John Wallis, Opera Mathematica – introduction of the term "continued fraction" See also Stern–Brocot tree Notes ^ http://www.britannica.com/EBchecked/topic/135043/continuedfraction References Jones, William B.; Thron, W. J. (1980). Continued Fractions: Analytic Theory and Applications. Encyclopedia of Mathematics and its Applications.. 11. AddisonWesley Publishing Company. ISBN 0201135108. 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