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a0+1a1+1a2+1+1an{displaystyle a_{0}+{cfrac {1}{a_{1}+{cfrac {1}{a_{2}+{cfrac {1}{ddots +{cfrac {1}{a_{n}}}}}}}}}}

Thus it is natural to look at the continued fraction expansion: log 2 3 = 1:2:::= 1;1;1;2;2;3;1;5;:::: Theapproximation log 2 3 ˇ1;1;1;2 = 8 5 means that 28 = 256 is not too far from 35 = 243, that is, 5 fifths producealmost3 octaves.Thenextapproximation log 2 3 ˇ1;1;1;2;2 = 19 12 tellsusthat219 = 524288 iscloseto312 = 531441,thatis 3 2! CONTINUED FRACTIONS, PELL’S EQUATION, AND OTHER APPLICATIONS JEREMY BOOHER Continued fractions usually get short-changed at PROMYS, but they are interesting in their own right and useful in other areas of number theory. For example, they given a way to write a prime congruent to 1 modulo 4 as a sum of two squares.

A finite continued fraction, where n{displaystyle n} is a non-negative integer, a0{displaystyle a_{0}} is an integer, and ai{displaystyle a_{i}} is a positive integer, for i=1,,n{displaystyle i=1,ldots ,n}.

CONTINUED FRACTIONS. In Chapter 3 we were looking for rational approximations to various real numbers as a means of proving irrationality or transcendence. There is in fact a standard procedure for obtaining all of the best' rational approximations. Two ”different” continued fraction representations for 1/2: 1 2 = 0;2 = 0;1,1. However, we require that an 1, where an is the last element of a finite continued fraction. Then the answer is ”yes”. Prove that under the assumption an 1 the continued fraction representation given in Proposition 1 is unique.

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{displaystyle a_{i}} 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 numberp{displaystyle p}/q{displaystyle 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){displaystyle (p,q)}. The numerical value of an infinite continued fraction is 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{displaystyle alpha } is the value of a unique infinite continued fraction, whose coefficients can be found using the non-terminating version of the Euclidean algorithm applied to the incommensurable values α{displaystyle alpha } and 1. This way of expressing real numbers (rational and irrational) is called their continued fraction representation.

It is generally assumed that the numerator of all of the fractions is 1. 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.

  • 7Infinite continued fractions and convergents
  • 9Best rational approximations
  • 13Other continued fraction expansions
  • 14Applications

Motivation and notation[edit]

Consider, for example, the rational number415/93, which is around 4.4624. As a first approximation, start with 4, which is the integer part; 415/93 = 4 + 43/93. 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 obtain a second approximation of 4 + 1/2 = 4.5; 93/43 = 2 + 7/43.The remaining fractional part, 7/43, is the reciprocal of 43/7, and 43/7 is around 6.1429. Use 6 as an approximation for this to obtain 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, 1/7, 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.

The expression 4 + 1/2 + 1/6 + 1/7 is called the continued fraction representation of 415/93. This can be represented by the abbreviated notation 415/93 = [4; 2, 6, 7]. (It is customary to replace only the first comma by a semicolon.) Some older textbooks use all commas in the (n + 1)-tuple, for example, [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,..] (sequence A010124 in the OEIS). The pattern repeats indefinitely with a period of 6.
  • e = [2;1,2,1,1,4,1,1,6,1,1,8,..] (sequence A003417 in the OEIS). The pattern repeats indefinitely with a period of 3 except that 2 is added to one of the terms in each cycle.
  • π = [3;7,15,1,292,1,1,1,2,1,3,1,..] (sequence A001203 in the OEIS). No pattern has ever been found in this representation.
  • ϕ = [1;1,1,1,1,1,1,1,1,1,1,1,..] (sequence A000012 in the OEIS). The golden ratio, the irrational number that is the 'most difficult' to approximate rationally. See: A property of the golden ratio φ.

Continued fractions are, in some ways, more 'mathematically natural' representations of a 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..
  • Every rational number has an essentially unique continued fraction representation. Each rational can be represented in exactly two ways, since [a0;a1,.. an−1,an] = [a0;a1,.. an−1,(an−1),1]. Usually the first, shorter one is chosen as the canonical representation.
  • The continued fraction representation of an irrational number is unique.
  • The real numbers whose continued fraction eventually repeats are precisely the quadratic irrationals.[5] For example, the repeating continued fraction [1;1,1,1,..] is the golden ratio, and the repeating continued fraction [1;2,2,2,..] is the square root of 2. In contrast, the decimal representations of quadratic irrationals are apparently random. The square roots of all (positive) integers, that are not perfect squares, are quadratic irrationals, hence are unique periodic continued fractions.
  • The successive approximations generated in finding the continued fraction representation of a number, that is, by truncating the continued fraction representation, are in a certain sense (described below) the 'best possible'.

Basic formula[edit]

A continued fraction is an expression of the form

a0+b1a1+b2a2+b3a3+{displaystyle a_{0}+{cfrac {b_{1}}{a_{1}+{cfrac {b_{2}}{a_{2}+{cfrac {b_{3}}{a_{3}+{_{ddots }}}}}}}}}

where ai and bi can be any complex numbers. Usually they are required to be integers.If bi = 1 for all i the expression is called a simple continued fraction.If the expression contains a finite number of terms, it is called a finite continued fraction.If the expression contains an infinite number of terms, it is called an infinite continued fraction.[6]

Thus, all of the following illustrate valid finite simple continued fractions:

Examples of finite simple continued fractions
FormulaNumericRemarks
a0{displaystyle a_{0}}2{displaystyle 2}All integers are a degenerate case
a0+1a1{displaystyle a_{0}+{cfrac {1}{a_{1}}}}2+13{displaystyle 2+{cfrac {1}{3}}}Simplest possible fractional form
a0+1a1+1a2{displaystyle a_{0}+{cfrac {1}{a_{1}+{cfrac {1}{a_{2}}}}}}3+12+118{displaystyle -3+{cfrac {1}{2+{cfrac {1}{18}}}}}First integer may be negative
a0+1a1+1a2+1a3{displaystyle a_{0}+{cfrac {1}{a_{1}+{cfrac {1}{a_{2}+{cfrac {1}{a_{3}}}}}}}}115+11+1102{displaystyle {cfrac {1}{15+{cfrac {1}{1+{cfrac {1}{102}}}}}}}First integer may be zero

Calculating continued fraction representations[edit]

Consider a real number r.Let i=r{displaystyle i=lfloor rrfloor } be the integer part of r and letf=ri{displaystyle f=r-i} be the fractional part of r.Then the continued fraction representation of r is[i;a1,a2,]{displaystyle [i;a_{1},a_{2},ldots ]}, where [a1;a2,]{displaystyle [a_{1};a_{2},ldots ]} is the continued fraction representation of 1/f{displaystyle 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. This process can be efficiently implemented using the Euclidean algorithm when the number is rational. The table below shows an implementation of this procedure for the number 3.245, resulting in the continued fraction expansion [3; 4,12,4].

Find the continued fraction for 3.245=649200{displaystyle 3.245={frac {649}{200}}}
StepReal
Number
Integer
part
Fractional
part
SimplifiedReciprocal
of f
1r=649200{displaystyle r={frac {649}{200}}}i=3{displaystyle i=3}f=6492003{displaystyle f={frac {649}{200}}-3}=49200{displaystyle ={frac {49}{200}}}1f=20049{displaystyle {frac {1}{f}}={frac {200}{49}}}
2r=20049{displaystyle r={frac {200}{49}}}i=4{displaystyle i=4}f=200494{displaystyle f={frac {200}{49}}-4}=449{displaystyle ={frac {4}{49}}}1f=494{displaystyle {frac {1}{f}}={frac {49}{4}}}
3r=494{displaystyle r={frac {49}{4}}}i=12{displaystyle i=12}f=49412{displaystyle f={frac {49}{4}}-12}=14{displaystyle ={frac {1}{4}}}1f=41{displaystyle {frac {1}{f}}={frac {4}{1}}}
4r=4{displaystyle r=4}i=4{displaystyle i=4}f=44{displaystyle f=4-4}=0{displaystyle =0}STOP
Continued fraction form for 3.245=649200=[3;4,12,4]{displaystyle 3.245={frac {649}{200}}=[3;4,12,4]}

= 3 + 1/4 + 1/12 + 1/4

Notations[edit]

The integers a0{displaystyle a_{0}}, a1{displaystyle a_{1}} etc., are called the coefficients or terms of the continued fraction.[2] One can abbreviate the continued fraction

x=a0+1a1+1a2+1a3{displaystyle x=a_{0}+{cfrac {1}{a_{1}+{cfrac {1}{a_{2}+{cfrac {1}{a_{3}}}}}}}}

in the notation of Carl Friedrich Gauss

x=a0+K3i=11ai{displaystyle x=a_{0}+{underset {i=1}{overset {3}{mathrm {K} }}}~{frac {1}{a_{i}}}}

or as

x=[a0;a1,a2,a3]{displaystyle x=[a_{0};a_{1},a_{2},a_{3}]},

or in the notation of Pringsheim as

x=a0+1a1+1a2+1a3,{displaystyle x=a_{0}+{frac {1mid }{mid a_{1}}}+{frac {1mid }{mid a_{2}}}+{frac {1mid }{mid a_{3}}},}

or in another related notation as

Continued Fractions Book Pdf

x=a0+1a1+1a2+1a3.{displaystyle x=a_{0}+{1 over a_{1}+{}}{1 over a_{2}+{}}{1 over a_{3}{}}.}

Sometimes angle brackets are used, like this:

x=a0;a1,a2,a3.{displaystyle x=leftlangle a_{0};a_{1},a_{2},a_{3}rightrangle .}

The semicolon in the square and angle bracket notations is sometimes replaced by a comma.[3][4]

One may also define infinite simple continued fractions as limits:

[a0;a1,a2,a3,]=limn[a0;a1,a2,,an].{displaystyle [a_{0};a_{1},a_{2},a_{3},ldots ]=lim _{nto infty }[a_{0};a_{1},a_{2},ldots ,a_{n}].}

This limit exists for any choice of a0{displaystyle a_{0}} and positive integers a1,a2,{displaystyle a_{1},a_{2},ldots }[7][8]

Finite continued fractions[edit]

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, with the conditions that the first coefficient is an integer and other coefficients being positive integers. 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:

[a0; a1, a2, .., an − 1, an, 1] = [a0; a1, a2, .., an − 1, an + 1].
[a0; 1] = [a0 + 1].

Of reciprocals[edit]

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[a0;a1,a2,,an]{displaystyle [a_{0};a_{1},a_{2},ldots ,a_{n}]} and [0;a0,a1,,an]{displaystyle [0;a_{0},a_{1},ldots ,a_{n}]} are reciprocals. For instance if a{displaystyle a} is an integer and x<1{displaystyle x<1} then

x=0+1a+1b{displaystyle x=0+{frac {1}{a+{frac {1}{b}}}}} and 1x=a+1b{displaystyle {frac {1}{x}}=a+{frac {1}{b}}}.

If x>1{displaystyle x>1} then

x=a+1b{displaystyle x=a+{frac {1}{b}}} and 1x=0+1a+1b{displaystyle {frac {1}{x}}=0+{frac {1}{a+{frac {1}{b}}}}}.

The last number that generates the remainder of the continued fraction is the same for both x{displaystyle x} and its reciprocal.

For example,

2.25=94=[2;4]{displaystyle 2.25={frac {9}{4}}=[2;4]} and 12.25=49=[0;2,4]{displaystyle {frac {1}{2.25}}={frac {4}{9}}=[0;2,4]}.

Infinite continued fractions and convergents[edit]

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 useful because its initial segments provide rational approximations to the number. These rational numbers are called the convergents of the continued fraction.[9][10] The larger a term is in the continued fraction, the closer the corresponding convergent is to the irrational number being approximated. Numbers like π have occasional large terms in their continued fraction, which makes them easy to approximate with rational numbers. Other numbers like e have only small terms early in their continued fraction, which makes them more difficult to approximate rationally. The golden ratio ϕ has terms equal to 1 everywhere—the smallest values possible—which makes ϕ the most difficult number to approximate rationally. In this sense, therefore, it is the 'most irrational' of all irrational numbers. Even-numbered convergents are smaller than the original number, while odd-numbered ones are larger.

For a continued fraction [a0; a1, a2, ..], the first four convergents (numbered 0 through 3) are

a0/1, a1a0 + 1/a1, a2(a1a0 + 1) + a0/a2a1 + 1, a3(a2(a1a0 + 1) + a0) + (a1a0 + 1)/a3(a2a1 + 1) + a1

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:

hn = anhn − 1 + hn − 2,
kn = ankn − 1 + kn − 2.

The successive convergents are given by the formula

hn/kn = anhn − 1 + hn − 2/ankn − 1 + kn − 2

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 01 and 10. For example, here are the convergents for [0;1,5,2,2].

n−2−101234
an01522
hn010151127
kn101161332

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, .. , 2k−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:

n−2−101234567
an11212121
hn01125719267197
kn10113411154156
x0 = 1 = 1/1
x1 = 1/2(1 + 3/1) = 2/1 = 2
x2 = 1/2(2 + 3/2) = 7/4
x3 = 1/2(7/4 + 3/7/4) = 97/56

Properties[edit]

Baire space is a topological space on infinite sequences of natural numbers. The infinite continued fraction provides a homeomorphism from Baire space to the space of irrational real numbers (with the subspace topology inherited from the usual topology on the reals). The infinite continued fraction also provides a map between the quadratic irrationals and the dyadic rationals, and from other irrationals to the set of infinite strings of binary numbers (i.e. the Cantor set); this map is called the Minkowski question mark function. The mapping has interesting self-similar fractal properties; these are given by the modular group, which is the subgroup of Möbius transformations having integer values in the transform. Roughly speaking, continued fraction convergents can be taken to be Möbius transformations acting on the (hyperbolic) upper half-plane; this is what leads to the fractal self-symmetry.

Some useful theorems[edit]

If a0{displaystyle a_{0}}, a1{displaystyle a_{1}}, a2{displaystyle a_{2}}, {displaystyle ldots } is an infinite sequence of positive integers, define the sequences hn{displaystyle h_{n}} and kn{displaystyle k_{n}} recursively:

hn=anhn1+hn2{displaystyle h_{n}=a_{n}h_{n-1}+h_{n-2}}h1=1{displaystyle h_{-1}=1}h2=0{displaystyle h_{-2}=0}
kn=ankn1+kn2{displaystyle k_{n}=a_{n}k_{n-1}+k_{n-2}}k1=0{displaystyle k_{-1}=0}k2=1{displaystyle k_{-2}=1}

Theorem 1. For any positive real number z{displaystyle z}

[a0;a1,,an1,z]=zhn1+hn2zkn1+kn2.{displaystyle left[a_{0};a_{1},dots ,a_{n-1},zright]={frac {zh_{n-1}+h_{n-2}}{zk_{n-1}+k_{n-2}}}.}

Theorem 2. The convergents of [a0{displaystyle a_{0}}; a1{displaystyle a_{1}}, a2{displaystyle a_{2}}, {displaystyle ldots }] are given by

[a0;a1,,an]=hnkn.{displaystyle left[a_{0};a_{1},dots ,a_{n}right]={frac {h_{n}}{k_{n}}}.}

Theorem 3. If the n{displaystyle n}th convergent to a continued fraction is hn{displaystyle h_{n}}/kn{displaystyle k_{n}}, then

knhn1kn1hn=(1)n.{displaystyle k_{n}h_{n-1}-k_{n-1}h_{n}=(-1)^{n}.}

Corollary 1: Each convergent is in its lowest terms (for if hn{displaystyle h_{n}} and kn{displaystyle k_{n}} had a nontrivial common divisor it would divide knhn1kn1hn{displaystyle k_{n}h_{n-1}-k_{n-1}h_{n}}, which is impossible).

Corollary 2: The difference between successive convergents is a fraction whose numerator is unity:

hnknhn1kn1=hnkn1knhn1knkn1=(1)n+1knkn1.{displaystyle {frac {h_{n}}{k_{n}}}-{frac {h_{n-1}}{k_{n-1}}}={frac {h_{n}k_{n-1}-k_{n}h_{n-1}}{k_{n}k_{n-1}}}={frac {(-1)^{n+1}}{k_{n}k_{n-1}}}.}

Corollary 3: The continued fraction is equivalent to a series of alternating terms:

a0+n=0(1)nknkn+1.{displaystyle a_{0}+sum _{n=0}^{infty }{frac {(-1)^{n}}{k_{n}k_{n+1}}}.}

Corollary 4: The matrix

[hnhn1knkn1]{displaystyle {begin{bmatrix}h_{n}&h_{n-1}k_{n}&k_{n-1}end{bmatrix}}}

has determinant plus or minus one, and thus belongs to the group of2×2{displaystyle 2times 2}unimodular matricesGL(2,Z){displaystyle mathrm {GL} (2,mathbb {Z} )}.

Theorem 4. Each (s{displaystyle s}th) convergent is nearer to a subsequent (n{displaystyle n}th) convergent than any preceding (r{displaystyle r}th) convergent is. In symbols, if the n{displaystyle n}th convergent is taken to be [a0;a1,,an]=xn{displaystyle [a_{0};a_{1},ldots ,a_{n}]=x_{n}}, then

xrxn>xsxn{displaystyle left x_{r}-x_{n}right >left x_{s}-x_{n}right }

for all r<s<n{displaystyle r<s<n}.

Corollary 1: The even convergents (before the n{displaystyle n}th) continually increase, but are always less than xn{displaystyle x_{n}}.

Corollary 2: The odd convergents (before the n{displaystyle n}th) continually decrease, but are always greater than xn{displaystyle x_{n}}.

Theorem 5.

1kn(kn+1+kn)<xhnkn<1knkn+1.{displaystyle {frac {1}{k_{n}(k_{n+1}+k_{n})}}<left x-{frac {h_{n}}{k_{n}}}right <{frac {1}{k_{n}k_{n+1}}}.}

Corollary 1: A convergent is nearer to the limit of the continued fraction than any fraction whose denominator is less than that of the convergent.

Corollary 2: A convergent obtained by terminating the continued fraction just before a large term is a close approximation to the limit of the continued fraction.

Semiconvergents[edit]

If

hn1kn1,hnkn{displaystyle {frac {h_{n-1}}{k_{n-1}}},{frac {h_{n}}{k_{n}}}}

are consecutive convergents, then any fractions of the form

hn1+mhnkn1+mkn,{displaystyle {frac {h_{n-1}+mh_{n}}{k_{n-1}+mk_{n}}},}

where m{displaystyle m} is an integer such that 0man+1{displaystyle 0leq mleq a_{n+1}}, are called semiconvergents, secondary convergents, or intermediate fractions. The (m+1){displaystyle (m+1)}-st semiconvergent equals the mediant of the m{displaystyle m}-th one and the convergent hnkn{displaystyle {tfrac {h_{n}}{k_{n}}}}. It follows that semiconvergents represent a monotonic sequence of fractions between the convergents hn1kn1{displaystyle {tfrac {h_{n-1}}{k_{n-1}}}} (corresponding to m=0{displaystyle m=0}) and hn+1kn+1{displaystyle {tfrac {h_{n+1}}{k_{n+1}}}} (corresponding to m=an+1{displaystyle m=a_{n+1}}). Sometimes the term is taken to mean that being a semiconvergent excludes the possibility of being a convergent (i.e., 0<m<an+1{displaystyle 0<m<a_{n+1}}), rather than that a convergent is a kind of semiconvergent.

The semiconvergents to the continued fraction expansion of a real number x{displaystyle x} include all the rational approximations that are better than any approximation with a smaller denominator. Another useful property is that consecutive semiconvergents ab{displaystyle {tfrac {a}{b}}} and cd{displaystyle {tfrac {c}{d}}} are such thatadbc=±1{displaystyle ad-bc=pm 1}.

Best rational approximations[edit]

One can choose to define a best rational approximation to a real number x as a rational number n/d, d > 0, that is closer to x than any approximation with a smaller or equal denominator. The simple continued fraction for x can be used to generate all of the best rational approximations for x by applying these three rules:

  1. Truncate the continued fraction, and reduce its last term by a chosen amount (possibly zero).
  2. The reduced term cannot have less than half its original value.
  3. If the final term is even, half its value is admissible only if the corresponding semiconvergent is better than the previous convergent. (See below.)

For example, 0.84375 has continued fraction [0;1,5,2,2]. Here are all of its best rational approximations.

Continued fraction [0;1]  [0;1,3]  [0;1,4]  [0;1,5]  [0;1,5,2]  [0;1,5,2,1]  [0;1,5,2,2] 
Rational approximation13/44/55/611/1316/1927/32
Decimal equivalent10.750.8~0.83333~0.84615~0.842110.84375
Error+18.519%−11.111%−5.1852%−1.2346%+0.28490%−0.19493%0%

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 requires that when ak is even, the halved term ak/2 is admissible if and only if x − [a0 ; a1, .., ak − 1] > x − [a0 ; a1, .., ak − 1, ak/2] [11] This is equivalent[11] to:[12]

[ak; ak − 1, .., a1] > [ak; ak + 1, ..].

The convergents to x are 'best approximations' in a much stronger sense than the one defined above. Namely, n/d is a convergent for x if and only if dxn has the smallest value among the analogous expressions for all rational approximations m/c with cd; that is, we have dxn < cxm so long as c < d. (Note also that dkxnk → 0 as k → ∞.)

Best rational within an interval[edit]

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

x = [a0; a1, a2, .., ak − 1, ak, ak + 1, ..]
y = [a0; a1, a2, .., ak − 1, bk, bk + 1, ..]

where x and y have identical continued fraction expansions up through ak−1, a rational that falls within the interval (x, y) is given by the finite continued fraction,

z(x,y) = [a0; a1, a2, .., ak − 1, min(ak, bk) + 1]

This rational will be best in the sense that no other rational in (x, y) will have a smaller numerator or a smaller denominator.[citation needed]

If x is rational, it will have two continued fraction representations that are finite, x1 and x2, and similarly a rational y will have two representations, y1 and y2. The coefficients beyond the last in any of these representations should be interpreted as +∞; and the best rational will be one of z(x1, y1), z(x1, y2), z(x2, y1), or z(x2, y2).

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

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]

and the best rational between these two is

[3; 7, 16] = 355/113 = 3.1415929..

Thus, 355/113 is the best rational number corresponding to the rounded decimal number 3.1416, in the sense that no other rational number that would be rounded to 3.1416 will have a smaller numerator or a smaller denominator.

Interval for a convergent[edit]

A rational number, which can be expressed as finite continued fraction in two ways,

z = [a0; a1, .., ak − 1, ak, 1] = [a0; a1, .., ak − 1, ak + 1]

will be one of the convergents for the continued fraction expansion of a number, if and only if the number is strictly between

x = [a0; a1, .., ak − 1, ak, 2] and
y = [a0; a1, .., ak − 1, ak + 2]

The numbers x and y are formed by incrementing the last coefficient in the two representations for z. It is the case 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

[3; 7, 15, 2]=688/219 ≈ 3.1415525
[3; 7, 17]=377/120 ≈ 3.1416667

Comparison[edit]

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(akbk) < 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 ≤ in, then x < y if n is even and y < x if n is odd.

Continued fraction expansions of π[edit]

To calculate the convergents of π we may set a0 = ⌊π⌋ = 3, define u1 = 1/π − 3 ≈ 7.0625 and a1 = ⌊u1⌋ = 7, u2 = 1/u1 − 7 ≈ 15.9966 and a2 = ⌊u2⌋ = 15, u3 = 1/u2 − 15 ≈ 1.0034. Continuing like this, one can determine the infinite continued fraction of π as

[3;7,15,1,292,1,1,..] (sequence A001203 in the OEIS).

The fourth 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.

Volume 2: Psychological and Biological Models, Cambridge, MA: MIT Press. Neural network software reviews.

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:

3/1, 22/7, 333/106, 355/113, ..

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 22/7π is less than 1/7 × 106 = 1/742 (in fact, 22/7π is just more than 1/791 = 1/7 × 113).

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:

3/1 + 1/1 × 71/7 × 106 + 1/106 × 113 − ..

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.

Generalized continued fraction[edit]

A generalized continued fraction is an expression of the form

x=b0+a1b1+a2b2+a3b3+a4b4+{displaystyle 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:

π=[3;7,15,1,292,1,1,1,2,1,3,1,]{displaystyle pi =[3;7,15,1,292,1,1,1,2,1,3,1,ldots ]}

or

π=3+17+115+11+1292+11+11+11+12+11+13+11+{displaystyle 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:

π=41+122+322+522+722+922+=41+123+225+327+429+=3+126+326+526+726+926+{displaystyle 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 }}}}}}}}}}}
π=2+21+11/2+11/3+11/4+=2+21+121+231+341+{displaystyle displaystyle pi =2+{cfrac {2}{1+{cfrac {1}{1/2+{cfrac {1}{1/3+{cfrac {1}{1/4+ddots }}}}}}}}=2+{cfrac {2}{1+{cfrac {1cdot 2}{1+{cfrac {2cdot 3}{1+{cfrac {3cdot 4}{1+ddots }}}}}}}}}
π=2+43+134+354+574+{displaystyle displaystyle pi =2+{cfrac {4}{3+{cfrac {1cdot 3}{4+{cfrac {3cdot 5}{4+{cfrac {5cdot 7}{4+ddots }}}}}}}}}

The first two of these are special cases of the arctangent function with π = 4 arctan (1).

π=3+136+13+236+13+23+33+436+13+23+33+43+53+636+13+23+33+43+53+63+73+836+{displaystyle pi =3+{cfrac {1^{3}}{6+{cfrac {1^{3}+2^{3}}{6+{cfrac {1^{3}+2^{3}+3^{3}+4^{3}}{6+{cfrac {1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}}{6+{cfrac {1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}+7^{3}+8^{3}}{6+ddots }}}}}}}}}}}

The continued fraction of π{displaystyle pi } above consisting of cubes uses the Nilakantha series and an exploit from Leonhard Euler.[13]

Other continued fraction expansions[edit]

Periodic continued fractions[edit]

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.

A property of the golden ratio φ[edit]

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[14] states that any irrational number k can be approximated by infinitely many rational m/n with

kmn<1n25.{displaystyle 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 1n25{displaystyle {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 adbc = ±1, shares this property with the golden ratio φ; and that all other real numbers can be more closely approximated.

Regular patterns in continued fractions[edit]

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=e1=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,1,1,],{displaystyle 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:

e1/n=[1;n1,1,1,3n1,1,1,5n1,1,1,7n1,1,1,].{displaystyle e^{1/n}=[1;n-1,1,1,3n-1,1,1,5n-1,1,1,7n-1,1,1,dots ],!.}

Another, more complex pattern appears in this continued fraction expansion for positive odd n:

e2/n=[1;n12,6n,5n12,1,1,7n12,18n,11n12,1,1,13n12,30n,17n12,1,1,],{displaystyle e^{2/n}=left[1;{frac {n-1}{2}},6n,{frac {5n-1}{2}},1,1,{frac {7n-1}{2}},18n,{frac {11n-1}{2}},1,1,{frac {13n-1}{2}},30n,{frac {17n-1}{2}},1,1,dots right],!,}

with a special case for n = 1:

e2=[7;2,1,1,3,18,5,1,1,6,30,8,1,1,9,42,11,1,1,12,54,14,1,1,3k,12k+6,3k+2,1,1].{displaystyle e^{2}=[7;2,1,1,3,18,5,1,1,6,30,8,1,1,9,42,11,1,1,12,54,14,1,1dots ,3k,12k+6,3k+2,1,1dots ],!.}

Other continued fractions of this sort are

tanh(1/n)=[0;n,3n,5n,7n,9n,11n,13n,15n,17n,19n,]{displaystyle tanh(1/n)=[0;n,3n,5n,7n,9n,11n,13n,15n,17n,19n,dots ]}

where n is a positive integer; also, for integer n:

tan(1/n)=[0;n1,1,3n2,1,5n2,1,7n2,1,9n2,1,],{displaystyle tan(1/n)=[0;n-1,1,3n-2,1,5n-2,1,7n-2,1,9n-2,1,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,].{displaystyle 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

Simple Continued Fractions

S(p/q)=Ip/q(2/q)I1+p/q(2/q),{displaystyle 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,],{displaystyle 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,].{displaystyle 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.

Typical continued fractions[edit]

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 tends to 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, approximately 3.27582, 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.

Applications[edit]

Square roots[edit]

Generalized continued fractions are used in a method for computing square roots.

The identity

x=1+x11+x{displaystyle {sqrt {x}}=1+{frac {x-1}{1+{sqrt {x}}}}}

(1)

leads via recursion to the generalized continued fraction for any square root:[15]

x=1+x12+x12+x12+{displaystyle {sqrt {x}}=1+{cfrac {x-1}{2+{cfrac {x-1}{2+{cfrac {x-1}{2+{ddots }}}}}}}}

(2)

Pell's equation[edit]

Continued fractions play an essential role in the solution of Pell's equation. For example, for positive integers p and q, and non-square n, it is true that if p2nq2 = ±1, then p/q is a convergent of the regular continued fraction for n. The converse holds if the period of the regular continued fraction for n is 1, and in general the period describes which convergents give solutions to Pell's equation.[16]

Dynamical systems[edit]

Continued fractions also play a role in the study of dynamical systems, 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 − ⌊1/x called the Gauss map, which lops off digits of a continued fraction expansion: h([0; a1, a2, a3, ..]) = [0; a2, a3, ..]. 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.

Eigenvalues and eigenvectors[edit]

The Lanczos algorithm uses a continued fraction expansion to iteratively approximate the eigenvalues and eigenvectors of a large sparse matrix.[17]

Networking applications[edit]

Continued fractions have also been used in modelling optimization problems for wireless network virtualization to find a route between a source and a destination.[18]

Examples of rational and irrational numbers[edit]

Numberr012345678910
123ar123
ra123
12.3ar1233
ra1237/3123/10
1.23ar14217
ra15/411/916/13123/100
0.123ar087125
ra01/87/578/6523/187123/1 000
ϕ =
5 + 1/2
ar11111111111
ra123/25/38/513/821/1334/2155/3489/55144/89
ϕ =
5 + 1/2
ar−22111111111
ra−23/25/38/513/821/1334/2155/3489/55144/89233/144
2ar12222222222
ra13/27/517/1241/2999/70239/169577/4081 393/9853 363/2 3788 119/5 741
12ar01222222222
ra012/35/712/1729/4170/99169/239408/577985/1 3932 378/3 363
3ar11212121212
ra125/37/419/1126/1571/4197/56265/153362/209989/571
13ar01121212121
ra011/23/54/711/1915/2641/7156/97153/265209/362
32ar01626262626
ra016/713/1584/97181/2091 170/1 3512 521/2 91116 296/18 81735 113/40 545226 974/262 087
32ar13151141181
ra14/35/429/2334/2763/50286/227349/277635/5045 429/4 3096 064/4 813
ear21211411611
ra238/311/419/787/32106/39193/711 264/4651 457/5362 721/1 001
πar37151292111213
ra322/7333/106355/113103 993/33 102104 348/33 215208 341/66 317312 689/99 532833 719/265 3811 146 408/364 9134 272 943/1 360 120
Numberr012345678910

ra: rational approximant obtained by expanding continued fraction up to ar

History[edit]

  • 300 BCE Euclid's Elements contains an algorithm for the greatest common divisor which generates a continued fraction as a by-product
  • 499 The Aryabhatiya contains the solution of indeterminate equations using continued fractions
  • 1572 Rafael Bombelli, L'Algebra Opera – method for the extraction of square roots which is related to continued fractions
  • 1613 Pietro Cataldi, Trattato del modo brevissimo di trovar la radice quadra delli numeri – first notation for continued fractions
Cataldi represented a continued fraction as a0{displaystyle a_{0}} & n1d1{displaystyle {frac {n_{1}}{d_{1}cdot }}} & n2d2{displaystyle {frac {n_{2}}{d_{2}cdot }}} & n3d3{displaystyle {frac {n_{3}}{d_{3}cdot }}} with the dots indicating where the following fractions went.
  • 1695 John Wallis, Opera Mathematica – introduction of the term 'continued fraction'
  • 1737 Leonhard Euler, De fractionibus continuis dissertatio – Provided the first then-comprehensive account of the properties of continued fractions, and included the first proof that the number e is irrational.[19]
  • 1748 Euler, Introductio in analysin infinitorum. Vol. I, Chapter 18 – proved the equivalence of a certain form of continued fraction and a generalized infinite series, proved that every rational number can be written as a finite continued fraction, and proved that the continued fraction of an irrational number is infinite.[20]
  • 1761 Johann Lambert – gave the first proof of the irrationality of π using a continued fraction for tan(x).
  • 1768 Joseph-Louis Lagrange – provided the general solution to Pell's equation using continued fractions similar to Bombelli's
  • 1770 Lagrange – proved that quadratic irrationals expand to periodic continued fractions.
  • 1813 Carl Friedrich Gauss, Werke, Vol. 3, pp. 134–138 – derived a very general complex-valued continued fraction via a clever identity involving the hypergeometric function
  • 1828 Évariste Galois proved the periodicity of continued fractions for quadratic irrationals.[21]
  • 1892 Henri Padé defined Padé approximant
  • 1972 Bill Gosper – First exact algorithms for continued fraction arithmetic.

See also[edit]

Notes[edit]

  1. ^'Continued fraction - mathematics'.
  2. ^ abPettofrezzo & Byrkit (1970, p. 150)
  3. ^ abLong (1972, p. 173)
  4. ^ abPettofrezzo & Byrkit (1970, p. 152)
  5. ^Weisstein, Eric W.'Periodic Continued Fraction'. MathWorld.
  6. ^Collins, Darren C. 'Continued Fractions'(PDF). MIT Undergraduate Journal of Mathematics. Archived from the original(PDF) on 2001-11-20.
  7. ^Long (1972, p. 183)
  8. ^Pettofrezzo & Byrkit (1970, p. 158)
  9. ^Long (1972, p. 177)
  10. ^Pettofrezzo & Byrkit (1970, pp. 162–163)
  11. ^ abM. Thill (2008), 'A more precise rounding algorithm for rational numbers', Computing, 82: 189–198, doi:10.1007/s00607-008-0006-7
  12. ^Shoemake, Ken (1995), 'I.4: Rational Approximation', in Paeth, Alan W. (ed.), Graphic Gems V, San Diego, California: Academic Press, pp. 25–31, ISBN0-12-543455-3
  13. ^Foster, Tony (June 22, 2015). 'Theorem of the Day: Theorem no. 203'(PDF). Robin Whitty. Retrieved June 25, 2015.
  14. ^Theorem 193: Hardy, G.H.; Wright, E.M. (1979). An Introduction to the Theory of Numbers (Fifth ed.). Oxford.
  15. ^Ben Thurston, 'Estimating square roots, generalized continued fraction expression for every square root', The Ben Paul Thurston Blog
  16. ^Niven, Ivan; Zuckerman, Herbert S.; Montgomery, Hugh L. (1991). An introduction to the theory of numbers (Fifth ed.). New York: Wiley. ISBN0-471-62546-9.
  17. ^Martin, Richard M. (2004), Electronic Structure: Basic Theory and Practical Methods, Cambridge University Press, p. 557, ISBN9781139643658.
  18. ^Afifi, Haitham; et al. (April 2018). 'MARVELO: Wireless Virtual Network Embedding for Overlay Graphs with Loops'. 2018 IEEE Wireless Communications and Networking Conference (WCNC).
  19. ^Sandifer, Ed (February 2006). 'How Euler Did It: Who proved e is irrational?'(PDF). MAA Online.
  20. ^'E101 – Introductio in analysin infinitorum, volume 1'. Retrieved 2008-03-16.
  21. ^Wolfram, Stephen (2002). A New Kind of Science. Wolfram Media, Inc. p. 915. ISBN1-57955-008-8.

References[edit]

  • Siebeck, H. (1846). 'Ueber periodische Kettenbrüche'. J. Reine Angew. Math. 33. pp. 68–70.
  • Heilermann, J. B. H. (1846). 'Ueber die Verwandlung von Reihen in Kettenbrüche'. J. Reine Angew. Math. 33. pp. 174–188.
  • Magnus, Arne (1962). 'Continued fractions associated with the Padé Table'. Math. Z. 78. pp. 361–374.
  • Chen, Chen-Fan; Shieh, Leang-San (1969). 'Continued fraction inversion by Routh's Algorithm'. IEEE Trans. Circuit Theory. 16 (2). pp. 197–202. doi:10.1109/TCT.1969.1082925.
  • Gragg, William B. (1974). 'Matrix interpretations and applications of the continued fraction algorithm'. Rocky Mount. J. Math. 4 (2). p. 213. doi:10.1216/RJM-1974-4-2-213.
  • Jones, William B.; Thron, W. J. (1980). Continued Fractions: Analytic Theory and Applications. Encyclopedia of Mathematics and its Applications. 11. Reading. Massachusetts: Addison-Wesley Publishing Company. ISBN0-201-13510-8.
  • Khinchin, A. Ya. (1964) [Originally published in Russian, 1935]. Continued Fractions. University of Chicago Press. ISBN0-486-69630-8.
  • Long, Calvin T. (1972), Elementary Introduction to Number Theory (2nd ed.), Lexington: D. C. Heath and Company, LCCN77-171950
  • Perron, Oskar (1950). Die Lehre von den Kettenbrüchen. New York, NY: Chelsea Publishing Company.
  • Pettofrezzo, Anthony J.; Byrkit, Donald R. (1970), Elements of Number Theory, Englewood Cliffs: Prentice Hall, LCCN77-81766
  • Rockett, Andrew M.; Szüsz, Peter (1992). Continued Fractions. World Scientific Press. ISBN981-02-1047-7.
  • H. S. Wall, Analytic Theory of Continued Fractions, D. Van Nostrand Company, Inc., 1948 ISBN0-8284-0207-8
  • Cuyt, A.; Brevik Petersen, V.; Verdonk, B.; Waadeland, H.; Jones, W. B. (2008). Handbook of Continued fractions for Special functions. Springer Verlag. ISBN978-1-4020-6948-2.
  • Rieger, G. J. (1982). 'A new approach to the real numbers (motivated by continued fractions)'. Abh. Braunschweig.Wiss. Ges. 33. pp. 205–217.

External links[edit]

Continued Fractions Pdf Download

  • Hazewinkel, Michiel, ed. (2001) [1994], 'Continued fraction', Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN978-1-55608-010-4
  • Linas Vepstas Continued Fractions and Gaps (2004) reviews chaotic structures in continued fractions.
  • Continued Fractions on the Stern-Brocot Tree at cut-the-knot
  • Continued fraction calculator, WIMS.
  • Continued Fraction Arithmetic Gosper's first continued fractions paper, unpublished. Cached on the Internet Archive's Wayback Machine
  • Weisstein, Eric W.'Continued Fraction'. MathWorld.
  • Continued Fractions by Stephen Wolfram and Continued Fraction Approximations of the Tangent Function by Michael Trott, Wolfram Demonstrations Project.
  • OEISsequence A133593 ('Exact' continued fraction for Pi)
Look up continued fraction in Wiktionary, the free dictionary.
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