Movatterモバイル変換


[0]ホーム

URL:


Jump to content
WikipediaThe Free Encyclopedia
Search

Continued fraction

From Wikipedia, the free encyclopedia
Mathematical expression
An infinite continued fraction is defined by the sequences{ai},{bi}{\displaystyle \{a_{i}\},\{b_{i}\}}, fori=0,1,2,{\displaystyle i=0,1,2,\ldots }, witha0=0{\displaystyle a_{0}=0}.

Acontinued fraction is amathematical expression written as afraction whosedenominator contains a sum involving another fraction, which may itself be a simple or a continued fraction.[1] If thisiteration (repetitive process) terminates with a simple fraction, the result is afinite continued fraction; if it continues indefinitely, the result is aninfinite continued fraction. The special case in which all numerators are equal to one is referred to as asimple (or regular) continued fraction. Anyrational number can be expressed as a finite simple continued fraction, and anyirrational number can be expressed as an infinite simple continued fraction.

Different areas ofmathematics use different terminology andnotation for continued fractions. Innumber theory, the unqualified termcontinued fraction usually refers to simple continued fractions, whereas the general case is referred to asgeneralized continued fractions. Incomplex analysis andnumerical analysis, the general case is usually referred to by the unqualified termcontinued fraction.

The numerators and denominators of continued fractions can besequences{ai},{bi}{\displaystyle \{a_{i}\},\{b_{i}\}} ofconstants orfunctions.

Formulation

[edit]

A 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 thean (n > 0) are thepartial numerators, thebn are thepartial denominators, and the leading termb0 is called theinteger part of the continued fraction.

The successiveconvergents of the continued fraction are formed by applying thefundamental recurrence formulas:

x0=A0B0=b0,x1=A1B1=b1b0+a1b1,x2=A2B2=b2(b1b0+a1)+a2b0b2b1+a2, {\displaystyle {\begin{aligned}x_{0}&={\frac {A_{0}}{B_{0}}}=b_{0},\\x_{1}&={\frac {A_{1}}{B_{1}}}={\frac {b_{1}b_{0}+a_{1}}{b_{1}}},\\x_{2}&={\frac {A_{2}}{B_{2}}}={\frac {b_{2}(b_{1}b_{0}+a_{1})+a_{2}b_{0}}{b_{2}b_{1}+a_{2}}},\ \dots \end{aligned}}}

whereAn is the numerator andBn is the denominator, calledcontinuants,[2][3] of thenth convergent. They are given by thethree-term recurrence relation[4]

An=bnAn1+anAn2,Bn=bnBn1+anBn2for n1{\displaystyle {\begin{aligned}A_{n}&=b_{n}A_{n-1}+a_{n}A_{n-2},\\B_{n}&=b_{n}B_{n-1}+a_{n}B_{n-2}\qquad {\text{for }}n\geq 1\end{aligned}}}

with initial values

A1=1,A0=b0,B1=0,B0=1.{\displaystyle {\begin{aligned}A_{-1}&=1,&A_{0}&=b_{0},\\B_{-1}&=0,&B_{0}&=1.\end{aligned}}}

If the sequence of convergents{xn} approaches alimit, the continued fraction is convergent and has a definite value. If the sequence of convergents never approaches a limit, the continued fraction is divergent. It may diverge by oscillation (for example, the odd and even convergents may approach two different limits), or it may produce an infinite number of zero denominatorsBn.

History

[edit]

The story of continued fractions begins with theEuclidean algorithm,[5] a procedure for finding thegreatest common divisor of two natural numbersm andn. That algorithm introduced the idea of dividing to extract a new remainder – and then dividing by the new remainder repeatedly.

Nearly two thousand years passed beforeBombelli (1579) devised atechnique for approximating the roots of quadratic equations with continued fractions in the mid-sixteenth century. Now the pace of development quickened. Just 24 years later, in 1613,Pietro Cataldi introduced the first formal notation for the generalized continued fraction.[6] Cataldi represented a continued fraction as

a0&n1d1&n2d2&n3d3{\displaystyle {a_{0}\cdot }\,\&\,{\frac {n_{1}}{d_{1}\cdot }}\,\&\,{\frac {n_{2}}{d_{2}\cdot }}\,\&\,{\frac {n_{3}}{d_{3}}}}

with the dots indicating where the next fraction goes, and each& representing a modern plus sign.

Late in the seventeenth centuryJohn Wallis introduced the term "continued fraction" into mathematical literature.[7] New techniques for mathematical analysis (Newton's andLeibniz'scalculus) had recently come onto the scene, and a generation of Wallis' contemporaries put the new phrase to use.

In 1748Euler published a theorem showing that a particular kind of continued fraction is equivalent to a certain very generalinfinite series.[8]Euler's continued fraction formula is still the basis of many modern proofs ofconvergence of continued fractions.

In 1761,Johann Heinrich Lambert gave the firstproof thatπ is irrational, by using the following continued fraction fortanx:[9]

tan(x)=x1+x23+x25+x27+{\displaystyle \tan(x)={\cfrac {x}{1+{\cfrac {-x^{2}}{3+{\cfrac {-x^{2}}{5+{\cfrac {-x^{2}}{7+{}\ddots }}}}}}}}}

Continued fractions can also be applied to problems innumber theory, and are especially useful in the study ofDiophantine equations. In the late eighteenth centuryLagrange used continued fractions to construct the general solution ofPell's equation, thus answering a question that had fascinated mathematicians for more than a thousand years.[10] Lagrange's discovery implies that the canonical continued fraction expansion of thesquare root of every non-square integer is periodic and that, if the period is of lengthp > 1, it contains apalindromic string of lengthp − 1.

In 1813Gauss derived from complex-valuedhypergeometric functions what are now calledGauss's continued fractions.[11] They can be used to express many elementary functions and some more advanced functions (such as theBessel functions), as continued fractions that are rapidly convergent almost everywhere in the complex plane.

Notation

[edit]

The long continued fraction expression displayed in the introduction is easy for an unfamiliar reader to interpret. However, it takes up a lot of space and can be difficult to typeset. So mathematicians have devised several alternative notations. One convenient way to express a generalized continued fraction sets each nested fraction on the same line, indicating the nesting by dangling plus signs in the denominators:

x=b0+a1b1+a2b2+a3b3+{\displaystyle x=b_{0}+{\frac {a_{1}}{b_{1}+}}\,{\frac {a_{2}}{b_{2}+}}\,{\frac {a_{3}}{b_{3}+\cdots }}}

Sometimes the plus signs are typeset to vertically align with the denominators but not under the fraction bars:

x=b0+a1b1+a2b2+a3b3+{\displaystyle x=b_{0}+{\frac {a_{1}}{b_{1}}}{{} \atop +}{\frac {a_{2}}{b_{2}}}{{} \atop +}{\frac {a_{3}}{b_{3}}}{{} \atop \!{}+\cdots }}

Pringsheim wrote a generalized continued fraction this way:

x=b0+|a1b1|+|a2b2|+|a3b3|+{\displaystyle x=b_{0}+{{} \atop {{\big |}\!}}\!{\frac {a_{1}}{\,b_{1}\,}}\!{{\!{\big |}} \atop {}}+{{} \atop {{\big |}\!}}\!{\frac {a_{2}}{\,b_{2}\,}}\!{{\!{\big |}} \atop {}}+{{} \atop {{\big |}\!}}\!{\frac {a_{3}}{\,b_{3}\,}}\!{{\!{\big |}} \atop {}}+\cdots }

Carl Friedrich Gauss evoked the more familiarinfinite productΠ when he devised this notation:

x=b0+Ki=1aibi.{\displaystyle x=b_{0}+{\underset {i=1}{\overset {\infty }{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}.\,}

Here the "K" stands forKettenbruch, the German word for "continued fraction". This is probably the most compact and convenient way to express continued fractions; however, it is not widely used by English typesetters.

Some elementary considerations

[edit]

Here are some elementary results that are of fundamental importance in the further development of the analytic theory of continued fractions.

Partial numerators and denominators

[edit]

If one of the partial numeratorsan+1 is zero, the infinite continued fraction

b0+Ki=1aibi{\displaystyle b_{0}+{\underset {i=1}{\overset {\infty }{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}\,}

is really just a finite continued fraction withn fractional terms, and therefore arational function ofa1 toan andb0 tobn+1. Such an object is of little interest from the point of view adopted in mathematical analysis, so it is usually assumed that allai ≠ 0. There is no need to place this restriction on the partial denominatorsbi.

The determinant formula

[edit]

When thenth convergent of a continued fraction

xn=b0+Kni=1aibi{\displaystyle x_{n}=b_{0}+{\underset {i=1}{\overset {n}{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}\,}

is expressed as a simple fractionxn =An/Bn we can use thedeterminant formula

An1BnAnBn1=(1)na1a2an=i=1n(ai){\displaystyle A_{n-1}B_{n}-A_{n}B_{n-1}=\left(-1\right)^{n}a_{1}a_{2}\cdots a_{n}=\prod _{i=1}^{n}(-a_{i})}1

to relate the numerators and denominators of successive convergentsxn andxn − 1 to one another. The proof for this can be easily seen byinduction.

Proof

Base case

The casen = 1 results from a very simple computation.

Inductive step

Assume that (1) holds forn − 1. Then we need to see the same relation holding true forn. Substituting the value ofAn andBn in (1) we obtain:
=bnAn1Bn1+anAn1Bn2bnAn1Bn1anAn2Bn1=an(An1Bn2An2Bn1){\displaystyle {\begin{aligned}&=b_{n}A_{n-1}B_{n-1}+a_{n}A_{n-1}B_{n-2}-b_{n}A_{n-1}B_{n-1}-a_{n}A_{n-2}B_{n-1}\\&=a_{n}(A_{n-1}B_{n-2}-A_{n-2}B_{n-1})\end{aligned}}}
which is true because of our induction hypothesis.
An1BnAnBn1=(1)na1a2an=i=1n(ai){\displaystyle A_{n-1}B_{n}-A_{n}B_{n-1}=\left(-1\right)^{n}a_{1}a_{2}\cdots a_{n}=\prod _{i=1}^{n}(-a_{i})\,}
Specifically, if neitherBn norBn − 1 is zero (n > 0) we can express the difference between the(n − 1)th andnth convergents like this:
xn1xn=An1Bn1AnBn=(1)na1a2anBnBn1=i=1n(ai)BnBn1.{\displaystyle x_{n-1}-x_{n}={\frac {A_{n-1}}{B_{n-1}}}-{\frac {A_{n}}{B_{n}}}=\left(-1\right)^{n}{\frac {a_{1}a_{2}\cdots a_{n}}{B_{n}B_{n-1}}}={\frac {\prod _{i=1}^{n}(-a_{i})}{B_{n}B_{n-1}}}.\,}

The equivalence transformation

[edit]

If{ci} = {c1,c2,c3, ...} is any infinite sequence of non-zero complex numbers we can prove, by induction, that

b0+a1b1+a2b2+a3b3+a4b4+=b0+c1a1c1b1+c1c2a2c2b2+c2c3a3c3b3+c3c4a4c4b4+{\displaystyle b_{0}+{\cfrac {a_{1}}{b_{1}+{\cfrac {a_{2}}{b_{2}+{\cfrac {a_{3}}{b_{3}+{\cfrac {a_{4}}{b_{4}+\ddots \,}}}}}}}}=b_{0}+{\cfrac {c_{1}a_{1}}{c_{1}b_{1}+{\cfrac {c_{1}c_{2}a_{2}}{c_{2}b_{2}+{\cfrac {c_{2}c_{3}a_{3}}{c_{3}b_{3}+{\cfrac {c_{3}c_{4}a_{4}}{c_{4}b_{4}+\ddots \,}}}}}}}}}

where equality is understood as equivalence, which is to say that the successive convergents of the continued fraction on the left are exactly the same as the convergents of the fraction on the right.

The equivalence transformation is perfectly general, but two particular cases deserve special mention. First, if none of theai are zero, a sequence{ci} can be chosen to make each partial numerator a 1:

b0+Ki=1aibi=b0+Ki=11cibi{\displaystyle b_{0}+{\underset {i=1}{\overset {\infty }{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}=b_{0}+{\underset {i=1}{\overset {\infty }{\operatorname {K} }}}{\frac {1}{c_{i}b_{i}}}\,}

wherec1 =1/a1,c2 =a1/a2,c3 =a2/a1a3, and in generalcn+1 =1/an+1cn.

Second, if none of the partial denominatorsbi are zero we can use a similar procedure to choose another sequence{di} to make each partial denominator a 1:

b0+Ki=1aibi=b0+Ki=1diai1{\displaystyle b_{0}+{\underset {i=1}{\overset {\infty }{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}=b_{0}+{\underset {i=1}{\overset {\infty }{\operatorname {K} }}}{\frac {d_{i}a_{i}}{1}}\,}

whered1 =1/b1 and otherwisedn+1 =1/bnbn+1.

These two special cases of the equivalence transformation are enormously useful when the generalconvergence problem is analyzed.

Notions of convergence

[edit]

As mentioned in the introduction, the continued fraction

x=b0+Ki=1aibi{\displaystyle x=b_{0}+{\underset {i=1}{\overset {\infty }{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}\,}

converges if the sequence of convergents{xn} tends to a finite limit. This notion of convergence is very natural, but it is sometimes too restrictive. It is therefore useful to introduce the notion of general convergence of a continued fraction. Roughly speaking, this consists in replacing theKi=naibi{\displaystyle \operatorname {K} _{i=n}^{\infty }{\tfrac {a_{i}}{b_{i}}}} part of the fraction bywn, instead of by 0, to compute the convergents. The convergents thus obtained are calledmodified convergents. We say that the continued fractionconverges generally if there exists a sequence{wn}{\displaystyle \{w_{n}^{*}\}} such that the sequence of modified convergents converges for all{wn}{\displaystyle \{w_{n}\}} sufficiently distinct from{wn}{\displaystyle \{w_{n}^{*}\}}. The sequence{wn}{\displaystyle \{w_{n}^{*}\}} is then called anexceptional sequence for the continued fraction. See Chapter 2 ofLorentzen & Waadeland (1992) for a rigorous definition.

There also exists a notion ofabsolute convergence for continued fractions, which is based on the notion of absolute convergence of a series: a continued fraction is said to beabsolutely convergent when the series

f=n(fnfn1),{\displaystyle f=\sum _{n}\left(f_{n}-f_{n-1}\right),}

wherefn=Ki=1naibi{\displaystyle f_{n}=\operatorname {K} _{i=1}^{n}{\tfrac {a_{i}}{b_{i}}}} are the convergents of the continued fraction,converges absolutely.[12] TheŚleszyński–Pringsheim theorem provides a sufficient condition for absolute convergence.

Finally, a continued fraction of one or more complex variables isuniformly convergent in anopen neighborhoodΩ when its convergentsconverge uniformly onΩ; that is, when for everyε > 0 there existsM such that for alln >M, for allzΩ{\displaystyle z\in \Omega },

|f(z)fn(z)|<ε.{\displaystyle |f(z)-f_{n}(z)|<\varepsilon .}

Even and odd convergents

[edit]

It is sometimes necessary to separate a continued fraction into its even and odd parts. For example, if the continued fraction diverges by oscillation between two distinct limit pointsp andq, then the sequence{x0,x2,x4, ...} must converge to one of these, and{x1,x3,x5, ...} must converge to the other. In such a situation it may be convenient to express the original continued fraction as two different continued fractions, one of them converging top, and the other converging toq.

The formulas for the even and odd parts of a continued fraction can be written most compactly if the fraction has already been transformed so that all its partial denominators are unity. Specifically, if

x=Ki=1ai1{\displaystyle x={\underset {i=1}{\overset {\infty }{\operatorname {K} }}}{\frac {a_{i}}{1}}\,}

is a continued fraction, then the even partxeven and the odd partxodd are given by

xeven=a11+a2a2a31+a3+a4a4a51+a5+a6a6a71+a7+a8{\displaystyle x_{\text{even}}={\cfrac {a_{1}}{1+a_{2}-{\cfrac {a_{2}a_{3}}{1+a_{3}+a_{4}-{\cfrac {a_{4}a_{5}}{1+a_{5}+a_{6}-{\cfrac {a_{6}a_{7}}{1+a_{7}+a_{8}-\ddots }}}}}}}}\,}

and

xodd=a1a1a21+a2+a3a3a41+a4+a5a5a61+a6+a7a7a81+a8+a9{\displaystyle x_{\text{odd}}=a_{1}-{\cfrac {a_{1}a_{2}}{1+a_{2}+a_{3}-{\cfrac {a_{3}a_{4}}{1+a_{4}+a_{5}-{\cfrac {a_{5}a_{6}}{1+a_{6}+a_{7}-{\cfrac {a_{7}a_{8}}{1+a_{8}+a_{9}-\ddots }}}}}}}}\,}

respectively. More precisely, if the successive convergents of the continued fractionx are{x1,x2,x3, ...}, then the successive convergents ofxeven as written above are{x2,x4,x6, ...}, and the successive convergents ofxodd are{x1,x3,x5, ...}.[13]

Conditions for irrationality

[edit]

Ifa1,a2,... andb1,b2,... are positive integers withakbk for all sufficiently largek, then

x=b0+Ki=1aibi{\displaystyle x=b_{0}+{\underset {i=1}{\overset {\infty }{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}\,}

converges to an irrational limit.[14]

Fundamental recurrence formulas

[edit]

The partial numerators and denominators of the fraction's successive convergents are related by thefundamental recurrence formulas:

A1=1B1=0A0=b0B0=1An+1=bn+1An+an+1An1Bn+1=bn+1Bn+an+1Bn1{\displaystyle {\begin{aligned}A_{-1}&=1&B_{-1}&=0\\A_{0}&=b_{0}&B_{0}&=1\\A_{n+1}&=b_{n+1}A_{n}+a_{n+1}A_{n-1}&B_{n+1}&=b_{n+1}B_{n}+a_{n+1}B_{n-1}\,\end{aligned}}}

The continued fraction's successive convergents are then given by

xn=AnBn.{\displaystyle x_{n}={\frac {A_{n}}{B_{n}}}.\,}

These recurrence relations are due toJohn Wallis (1616–1703) andLeonhard Euler (1707–1783).[15]These recurrence relations are simply a different notation for the relations obtained by Pietro Antonio Cataldi (1548-1626).

As an example, consider the simple continued fraction in canonical form that represents thegolden ratioφ:

φ=1+11+11+11+11+{\displaystyle \varphi =1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+\ddots \,}}}}}}}}}

Applying the fundamental recurrence formulas we find that the successive numeratorsAn are{1, 2, 3, 5, 8, 13, ...} and the successive denominatorsBn are{1, 1, 2, 3, 5, 8, ...}, theFibonacci numbers. Since all the partial numerators in this example are equal to one, the determinant formula assures us that the absolute value of the difference between successive convergents approaches zero quite rapidly.

Linear fractional transformations

[edit]

Alinear fractional transformation (LFT) is acomplex function of the form

w=f(z)=az+bcz+d,{\displaystyle w=f(z)={\frac {az+b}{cz+d}},\,}

wherez is a complex variable, anda,b,c,d are arbitrary complex constants such thatcz +d ≠ 0. An additional restriction thatadbc is customarily imposed, to rule out the cases in whichw =f(z) is a constant. The linear fractional transformation, also known as aMöbius transformation, has many fascinating properties. Four of these are of primary importance in developing the analytic theory of continued fractions.

  • Ifc ≠ 0 the LFT has one or twofixed points. This can be seen by considering the equation
f(z)=zaz+b=cz2+dzcz2+(da)zb=0,{\displaystyle f(z)=z\Rightarrow az+b=cz^{2}+dz\Rightarrow cz^{2}+(d-a)z-b=0,}
which is clearly aquadratic equation inz. The roots of this equation are the fixed points off(z). If thediscriminant(da)2 + 4bc is zero the LFT fixes a single point; otherwise it has two fixed points.
z=g(w)=+dwbcw+a{\displaystyle z=g(w)={\frac {{\phantom {+}}dw-b}{-cw+a}}\,}
such thatf(g(z)) =g(f(z)) =z for every pointz in the extended complex plane, and bothf andg preserve angles and shapes at vanishingly small scales. From the form ofz =g(w) we see thatg is also an LFT.
  • Thecomposition of two different LFTs for whichadbc is itself an LFT for whichadbc. In other words, the set of all LFTs for whichadbc is closed under composition of functions. The collection of all such LFTs, together with the "group operation" composition of functions, is known as theautomorphism group of the extended complex plane.
  • Ifa = 0 the LFT reduces to
w=f(z)=bcz+d,{\displaystyle w=f(z)={\frac {b}{cz+d}},\,}
which is a very simplemeromorphic function ofz with onesimple pole (atd/c) and aresidue equal tob/c. (See alsoLaurent series.)

The continued fraction as a composition of LFTs

[edit]

Consider a sequence of simple linear fractional transformations

τ0(z)=b0+z,τ1(z)=a1b1+z,τ2(z)=a2b2+z,τ3(z)=a3b3+z,{\displaystyle {\begin{aligned}\tau _{0}(z)&=b_{0}+z,\\[4px]\tau _{1}(z)&={\frac {a_{1}}{b_{1}+z}},\\[4px]\tau _{2}(z)&={\frac {a_{2}}{b_{2}+z}},\\[4px]\tau _{3}(z)&={\frac {a_{3}}{b_{3}+z}},\\&\;\vdots \end{aligned}}}

Here we useτ to represent each simple LFT, and we adopt the conventional circle notation for composition of functions. We also introduce a new symbolΤn to represent the composition ofn + 1 transformationsτi; that is,

T1(z)=τ0τ1(z)=τ0(τ1(z)),T2(z)=τ0τ1τ2(z)=τ0(τ1(τ2(z))),{\displaystyle {\begin{aligned}{\boldsymbol {\mathrm {T} }}_{\boldsymbol {1}}(z)&=\tau _{0}\circ \tau _{1}(z)=\tau _{0}{\big (}\tau _{1}(z){\big )},\\{\boldsymbol {\mathrm {T} }}_{\boldsymbol {2}}(z)&=\tau _{0}\circ \tau _{1}\circ \tau _{2}(z)=\tau _{0}{\Big (}\tau _{1}{\big (}\tau _{2}(z){\big )}{\Big )},\,\end{aligned}}}

and so forth. By direct substitution from the first set of expressions into the second we see that

T1(z)=τ0τ1(z)=b0+a1b1+zT2(z)=τ0τ1τ2(z)=b0+a1b1+a2b2+z{\displaystyle {\begin{aligned}{\boldsymbol {\mathrm {T} }}_{\boldsymbol {1}}(z)&=\tau _{0}\circ \tau _{1}(z)&=&\quad b_{0}+{\cfrac {a_{1}}{b_{1}+z}}\\[4px]{\boldsymbol {\mathrm {T} }}_{\boldsymbol {2}}(z)&=\tau _{0}\circ \tau _{1}\circ \tau _{2}(z)&=&\quad b_{0}+{\cfrac {a_{1}}{b_{1}+{\cfrac {a_{2}}{b_{2}+z}}}}\,\end{aligned}}}

and, in general,

Tn(z)=τ0τ1τ2τn(z)=b0+Kni=1aibi{\displaystyle {\boldsymbol {\mathrm {T} }}_{\boldsymbol {n}}(z)=\tau _{0}\circ \tau _{1}\circ \tau _{2}\circ \cdots \circ \tau _{n}(z)=b_{0}+{\underset {i=1}{\overset {n}{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}\,}

where the last partial denominator in the finite continued fractionK is understood to bebn +z. And, sincebn + 0 =bn, the image of the pointz = 0 under the iterated LFTΤn is indeed the value of the finite continued fraction withn partial numerators:

Tn(0)=Tn+1()=b0+Kni=1aibi.{\displaystyle {\boldsymbol {\mathrm {T} }}_{\boldsymbol {n}}(0)={\boldsymbol {\mathrm {T} }}_{\boldsymbol {n+1}}(\infty )=b_{0}+{\underset {i=1}{\overset {n}{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}.\,}

A geometric interpretation

[edit]

Defining a finite continued fraction as the image of a point under the iterated linear fractional transformationΤn(z) leads to an intuitively appealing geometric interpretation of infinite continued fractions.

The relationship

xn=b0+Kni=1aibi=AnBn=Tn(0)=Tn+1(){\displaystyle x_{n}=b_{0}+{\underset {i=1}{\overset {n}{\operatorname {K} }}}{\frac {a_{i}}{b_{i}}}={\frac {A_{n}}{B_{n}}}={\boldsymbol {\mathrm {T} }}_{\boldsymbol {n}}(0)={\boldsymbol {\mathrm {T} }}_{\boldsymbol {n+1}}(\infty )\,}

can be understood by rewritingΤn(z) andΤn+1(z) in terms of thefundamental recurrence formulas:

Tn(z)=(bn+z)An1+anAn2(bn+z)Bn1+anBn2Tn(z)=zAn1+AnzBn1+Bn;Tn+1(z)=(bn+1+z)An+an+1An1(bn+1+z)Bn+an+1Bn1Tn+1(z)=zAn+An+1zBn+Bn+1.{\displaystyle {\begin{aligned}{\boldsymbol {\mathrm {T} }}_{\boldsymbol {n}}(z)&={\frac {(b_{n}+z)A_{n-1}+a_{n}A_{n-2}}{(b_{n}+z)B_{n-1}+a_{n}B_{n-2}}}&{\boldsymbol {\mathrm {T} }}_{\boldsymbol {n}}(z)&={\frac {zA_{n-1}+A_{n}}{zB_{n-1}+B_{n}}};\\[6px]{\boldsymbol {\mathrm {T} }}_{\boldsymbol {n+1}}(z)&={\frac {(b_{n+1}+z)A_{n}+a_{n+1}A_{n-1}}{(b_{n+1}+z)B_{n}+a_{n+1}B_{n-1}}}&{\boldsymbol {\mathrm {T} }}_{\boldsymbol {n+1}}(z)&={\frac {zA_{n}+A_{n+1}}{zB_{n}+B_{n+1}}}.\,\end{aligned}}}

In the first of these equations the ratio tends towardAn/Bn asz tends toward zero. In the second, the ratio tends towardAn/Bn asz tends to infinity. This leads us to our first geometric interpretation. If the continued fraction converges, the successive convergentsAn/Bn are eventuallyarbitrarily close together. Since the linear fractional transformationΤn(z) is acontinuous mapping, there must be a neighborhood ofz = 0 that is mapped into an arbitrarily small neighborhood ofΤn(0) =An/Bn. Similarly, there must be a neighborhood of the point at infinity which is mapped into an arbitrarily small neighborhood ofΤn(∞) =An−1/Bn−1. So if the continued fraction converges the transformationΤn(z) maps both very smallz and very largez into an arbitrarily small neighborhood ofx, the value of the continued fraction, asn gets larger and larger.

For intermediate values ofz, since the successive convergents are getting closer together we must have

An1Bn1AnBnAn1AnBn1Bn=k{\displaystyle {\frac {A_{n-1}}{B_{n-1}}}\approx {\frac {A_{n}}{B_{n}}}\quad \Rightarrow \quad {\frac {A_{n-1}}{A_{n}}}\approx {\frac {B_{n-1}}{B_{n}}}=k\,}

wherek is a constant, introduced for convenience. But then, by substituting in the expression forΤn(z) we obtain

Tn(z)=zAn1+AnzBn1+Bn=AnBn(zAn1An+1zBn1Bn+1)AnBn(zk+1zk+1)=AnBn{\displaystyle {\boldsymbol {\mathrm {T} }}_{\boldsymbol {n}}(z)={\frac {zA_{n-1}+A_{n}}{zB_{n-1}+B_{n}}}={\frac {A_{n}}{B_{n}}}\left({\frac {z{\frac {A_{n-1}}{A_{n}}}+1}{z{\frac {B_{n-1}}{B_{n}}}+1}}\right)\approx {\frac {A_{n}}{B_{n}}}\left({\frac {zk+1}{zk+1}}\right)={\frac {A_{n}}{B_{n}}}\,}

so that even the intermediate values ofz (except whenz ≈ −k−1) are mapped into an arbitrarily small neighborhood ofx, the value of the continued fraction, asn gets larger and larger. Intuitively, it is almost as if the convergent continued fraction maps the entire extended complex plane into a single point.[16]

Notice that the sequence{Τn} lies within theautomorphism group of the extended complex plane, since eachΤn is a linear fractional transformation for whichabcd. And every member of that automorphism group maps the extended complex plane into itself: not one of theΤn can possibly map the plane into a single point. Yet in the limit the sequence{Τn} defines an infinite continued fraction which (if it converges) represents a single point in the complex plane.

When an infinite continued fraction converges, the corresponding sequence{Τn} of LFTs "focuses" the plane in the direction ofx, the value of the continued fraction. At each stage of the process a larger and larger region of the plane is mapped into a neighborhood ofx, and the smaller and smaller region of the plane that's left over is stretched out ever more thinly to cover everything outside that neighborhood.[17]

For divergent continued fractions, we can distinguish three cases:

  1. The two sequences{Τ2n−1} and{Τ2n} might themselves define two convergent continued fractions that have two different values,xodd andxeven. In this case the continued fraction defined by the sequence{Τn} diverges by oscillation between two distinct limit points. And in fact this idea can be generalized: sequences{Τn} can be constructed that oscillate among three, or four, or indeed any number of limit points. Interesting instances of this case arise when the sequence{Τn} constitutes asubgroup of finite order within the group of automorphisms over the extended complex plane.
  2. The sequence{Τn} may produce an infinite number of zero denominatorsBi while also producing a subsequence of finite convergents. These finite convergents may not repeat themselves or fall into a recognizable oscillating pattern. Or they may converge to a finite limit, or even oscillate among multiple finite limits. No matter how the finite convergents behave, the continued fraction defined by the sequence{Τn} diverges by oscillation with the point at infinity in this case.[18]
  3. The sequence{Τn} may produce no more than a finite number of zero denominatorsBi. while the subsequence of finite convergents dances wildly around the plane in a pattern that never repeats itself and never approaches any finite limit either.
Visual interpretation of a simple continued fraction

Interesting examples of cases 1 and 3 can be constructed by studying the simple continued fraction

x=1+z1+z1+z1+z1+{\displaystyle x=1+{\cfrac {z}{1+{\cfrac {z}{1+{\cfrac {z}{1+{\cfrac {z}{1+\ddots }}}}}}}}\,}

wherez is any real number such thatz < −1/4.[19]

Euler's continued fraction formula

[edit]
Main article:Euler's continued fraction formula

Euler proved the following identity:[8]

a0+a0a1+a0a1a2++a0a1a2an=a01a11+a1a21+a2an1+an.{\displaystyle a_{0}+a_{0}a_{1}+a_{0}a_{1}a_{2}+\cdots +a_{0}a_{1}a_{2}\cdots a_{n}={\frac {a_{0}}{1-{\frac {a_{1}}{1+a_{1}-{\frac {a_{2}}{1+a_{2}-\cdots {\frac {a_{n}}{1+a_{n}}}}}}}}}.\,}

From this many other results can be derived, such as

1u1+1u2+1u3++1un=1u1u12u1+u2u22u2+u3un12un1+un,{\displaystyle {\frac {1}{u_{1}}}+{\frac {1}{u_{2}}}+{\frac {1}{u_{3}}}+\cdots +{\frac {1}{u_{n}}}={\frac {1}{u_{1}-{\frac {u_{1}^{2}}{u_{1}+u_{2}-{\frac {u_{2}^{2}}{u_{2}+u_{3}-\cdots {\frac {u_{n-1}^{2}}{u_{n-1}+u_{n}}}}}}}}},\,}

and

1a0+xa0a1+x2a0a1a2++xna0a1a2an=1a0a0xa1+xa1xa2+xan1xan+x.{\displaystyle {\frac {1}{a_{0}}}+{\frac {x}{a_{0}a_{1}}}+{\frac {x^{2}}{a_{0}a_{1}a_{2}}}+\cdots +{\frac {x^{n}}{a_{0}a_{1}a_{2}\ldots a_{n}}}={\frac {1}{a_{0}-{\frac {a_{0}x}{a_{1}+x-{\frac {a_{1}x}{a_{2}+x-\cdots {\frac {a_{n-1}x}{a_{n}+x}}}}}}}}.\,}

Euler's formula connecting continued fractions and series is the motivation for thefundamental inequalities[link or clarification needed], and also the basis of elementary approaches to theconvergence problem.

Examples

[edit]

Transcendental functions and numbers

[edit]

Here are two continued fractions that can be built viaEuler's identity.

ex=x00!+x11!+x22!+x33!+x44!+=1+x11x2+x2x3+x3x4+x{\displaystyle e^{x}={\frac {x^{0}}{0!}}+{\frac {x^{1}}{1!}}+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+{\frac {x^{4}}{4!}}+\cdots =1+{\cfrac {x}{1-{\cfrac {1x}{2+x-{\cfrac {2x}{3+x-{\cfrac {3x}{4+x-\ddots }}}}}}}}}
log(1+x)=x11x22+x33x44+=x10x+12x21x+22x32x+32x43x+{\displaystyle \log(1+x)={\frac {x^{1}}{1}}-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+\cdots ={\cfrac {x}{1-0x+{\cfrac {1^{2}x}{2-1x+{\cfrac {2^{2}x}{3-2x+{\cfrac {3^{2}x}{4-3x+\ddots }}}}}}}}}

Here are additional generalized continued fractions:

arctanxy=xy1y2+(1xy)23y21x2+(3xy)25y23x2+(5xy)27y25x2+=x1y+(1x)23y+(2x)25y+(3x)27y+{\displaystyle \arctan {\cfrac {x}{y}}={\cfrac {xy}{1y^{2}+{\cfrac {(1xy)^{2}}{3y^{2}-1x^{2}+{\cfrac {(3xy)^{2}}{5y^{2}-3x^{2}+{\cfrac {(5xy)^{2}}{7y^{2}-5x^{2}+\ddots }}}}}}}}={\cfrac {x}{1y+{\cfrac {(1x)^{2}}{3y+{\cfrac {(2x)^{2}}{5y+{\cfrac {(3x)^{2}}{7y+\ddots }}}}}}}}}
exy=1+2x2yx+x26y+x210y+x214y+x218y+e2=7+25+17+19+111+{\displaystyle e^{\frac {x}{y}}=1+{\cfrac {2x}{2y-x+{\cfrac {x^{2}}{6y+{\cfrac {x^{2}}{10y+{\cfrac {x^{2}}{14y+{\cfrac {x^{2}}{18y+\ddots }}}}}}}}}}\quad \Rightarrow \quad e^{2}=7+{\cfrac {2}{5+{\cfrac {1}{7+{\cfrac {1}{9+{\cfrac {1}{11+\ddots }}}}}}}}}
log(1+xy)=xy+1x2+1x3y+2x2+2x5y+3x2+=2x2y+x(1x)23(2y+x)(2x)25(2y+x)(3x)27(2y+x){\displaystyle \log \left(1+{\frac {x}{y}}\right)={\cfrac {x}{y+{\cfrac {1x}{2+{\cfrac {1x}{3y+{\cfrac {2x}{2+{\cfrac {2x}{5y+{\cfrac {3x}{2+\ddots }}}}}}}}}}}}={\cfrac {2x}{2y+x-{\cfrac {(1x)^{2}}{3(2y+x)-{\cfrac {(2x)^{2}}{5(2y+x)-{\cfrac {(3x)^{2}}{7(2y+x)-\ddots }}}}}}}}}

This last is based on an algorithm derived by Aleksei Nikolaevich Khovansky in the 1970s.[20]

Example: thenatural logarithm of 2 (=[0; 1, 2, 3, 1, 5,2/3, 7,1/2, 9,2/5,..., 2k − 1,2/k,...] ≈ 0.693147...):[21]

log2=log(1+1)=11+12+13+22+25+32+=2312922153221{\displaystyle \log 2=\log(1+1)={\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{3+{\cfrac {2}{2+{\cfrac {2}{5+{\cfrac {3}{2+\ddots }}}}}}}}}}}}={\cfrac {2}{3-{\cfrac {1^{2}}{9-{\cfrac {2^{2}}{15-{\cfrac {3^{2}}{21-\ddots }}}}}}}}}

π

[edit]

Here are three ofπ's best-known generalized continued fractions, the first and third of which are derived from their respectivearctangent formulas above by settingx =y = 1 and multiplying by 4. TheLeibniz formula forπ:

π=41+122+322+522+=n=04(1)n2n+1=4143+4547+{\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+\ddots }}}}}}}}=\sum _{n=0}^{\infty }{\frac {4(-1)^{n}}{2n+1}}={\frac {4}{1}}-{\frac {4}{3}}+{\frac {4}{5}}-{\frac {4}{7}}+-\cdots }

converges too slowly, requiring roughly3 × 10n terms to achieven correct decimal places. The series derived byNilakantha Somayaji:

π=3+126+326+526+=3n=1(1)nn(n+1)(2n+1)=3+11231235+1347+{\displaystyle \pi =3+{\cfrac {1^{2}}{6+{\cfrac {3^{2}}{6+{\cfrac {5^{2}}{6+\ddots }}}}}}=3-\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(n+1)(2n+1)}}=3+{\frac {1}{1\cdot 2\cdot 3}}-{\frac {1}{2\cdot 3\cdot 5}}+{\frac {1}{3\cdot 4\cdot 7}}-+\cdots }

is a much more obvious[why?] expression but still converges quite slowly, requiring nearly 50 terms for five decimals and nearly 120 for six. Both convergesublinearly toπ. On the other hand:

π=41+123+225+327+=41+16134+16314544551+16601138341+{\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{3+{\cfrac {2^{2}}{5+{\cfrac {3^{2}}{7+\ddots }}}}}}}}=4-1+{\frac {1}{6}}-{\frac {1}{34}}+{\frac {16}{3145}}-{\frac {4}{4551}}+{\frac {1}{6601}}-{\frac {1}{38341}}+-\cdots }

convergeslinearly toπ, adding at least three digits of precision per four terms, a pace slightly faster than thearcsine formula forπ:

π=6sin1(12)=n=03(2nn)16n(2n+1)=31601+61613+181625+601637+{\displaystyle \pi =6\sin ^{-1}\left({\frac {1}{2}}\right)=\sum _{n=0}^{\infty }{\frac {3\cdot {\binom {2n}{n}}}{16^{n}(2n+1)}}={\frac {3}{16^{0}\cdot 1}}+{\frac {6}{16^{1}\cdot 3}}+{\frac {18}{16^{2}\cdot 5}}+{\frac {60}{16^{3}\cdot 7}}+\cdots \!}

which adds at least three decimal digits per five terms.[22]

  • Note: this continued fraction'srate of convergenceμ tends to3 −8 ≈ 0.1715729, hence1/μ tends to3 +8 ≈ 5.828427, whosecommon logarithm is0.7655... ≈13/17 >3/4. The same1/μ = 3 +8 (thesilver ratio squared) also is observed in theunfolded general continued fractions of both thenatural logarithm of 2 and thenth root of 2 (which works for any integern > 1) if calculated using2 = 1 + 1. For thefolded general continued fractions of both expressions, the rate convergenceμ = (3 −8)2 = 17 −288 ≈ 0.02943725, hence1/μ = (3 +8)2 = 17 +288 ≈ 33.97056, whose common logarithm is1.531... ≈26/17 >3/2, thus adding at least three digits per two terms. This is because the folded GCF folds each pair of fractions from the unfolded GCF into one fraction, thus doubling the convergence pace. The Manny Sardina reference further explains "folded" continued fractions.
  • Note: Using the continued fraction forarctanx/y cited above with the best-knownMachin-like formula provides an even more rapidly, although still linearly, converging expression:
π=16tan1154tan11239=165+1215+2225+3235+4v+12717+221195+321673+.{\displaystyle \pi =16\tan ^{-1}{\cfrac {1}{5}}\,-\,4\tan ^{-1}{\cfrac {1}{239}}={\cfrac {16}{5+{\cfrac {1^{2}}{15+{\cfrac {2^{2}}{25+{\cfrac {3^{2}}{35+\ddots }}}}}}}}\,-\,{\cfrac {4}{v+{\cfrac {1^{2}}{717+{\cfrac {2^{2}}{1195+{\cfrac {3^{2}}{1673+\ddots }}}}}}}}.}

Roots of positive numbers

[edit]

Thenth root of any positive numberzm can be expressed by restatingz =xn +y, resulting in

zmn=(xn+y)mn=xm+mynxnm+(nm)y2xm+(n+m)y3nxnm+(2nm)y2xm+(2n+m)y5nxnm+(3nm)y2xm+{\displaystyle {\sqrt[{n}]{z^{m}}}={\sqrt[{n}]{\left(x^{n}+y\right)^{m}}}=x^{m}+{\cfrac {my}{nx^{n-m}+{\cfrac {(n-m)y}{2x^{m}+{\cfrac {(n+m)y}{3nx^{n-m}+{\cfrac {(2n-m)y}{2x^{m}+{\cfrac {(2n+m)y}{5nx^{n-m}+{\cfrac {(3n-m)y}{2x^{m}+\ddots }}}}}}}}}}}}}

which can be simplified, by folding each pair of fractions into one fraction, to

zmn=xm+2xmmyn(2xn+y)my(12n2m2)y23n(2xn+y)(22n2m2)y25n(2xn+y)(32n2m2)y27n(2xn+y)(42n2m2)y29n(2xn+y).{\displaystyle {\sqrt[{n}]{z^{m}}}=x^{m}+{\cfrac {2x^{m}\cdot my}{n(2x^{n}+y)-my-{\cfrac {(1^{2}n^{2}-m^{2})y^{2}}{3n(2x^{n}+y)-{\cfrac {(2^{2}n^{2}-m^{2})y^{2}}{5n(2x^{n}+y)-{\cfrac {(3^{2}n^{2}-m^{2})y^{2}}{7n(2x^{n}+y)-{\cfrac {(4^{2}n^{2}-m^{2})y^{2}}{9n(2x^{n}+y)-\ddots }}}}}}}}}}.}

Thesquare root ofz is a special case withm = 1 andn = 2:

z=x2+y=x+y2x+y2x+3y6x+3y2x+=x+2xy2(2x2+y)y13y26(2x2+y)35y210(2x2+y){\displaystyle {\sqrt {z}}={\sqrt {x^{2}+y}}=x+{\cfrac {y}{2x+{\cfrac {y}{2x+{\cfrac {3y}{6x+{\cfrac {3y}{2x+\ddots }}}}}}}}=x+{\cfrac {2x\cdot y}{2(2x^{2}+y)-y-{\cfrac {1\cdot 3y^{2}}{6(2x^{2}+y)-{\cfrac {3\cdot 5y^{2}}{10(2x^{2}+y)-\ddots }}}}}}}

which can be simplified by noting that5/10 =3/6 =1/2:

z=x2+y=x+y2x+y2x+y2x+y2x+=x+2xy2(2x2+y)yy22(2x2+y)y22(2x2+y).{\displaystyle {\sqrt {z}}={\sqrt {x^{2}+y}}=x+{\cfrac {y}{2x+{\cfrac {y}{2x+{\cfrac {y}{2x+{\cfrac {y}{2x+\ddots }}}}}}}}=x+{\cfrac {2x\cdot y}{2(2x^{2}+y)-y-{\cfrac {y^{2}}{2(2x^{2}+y)-{\cfrac {y^{2}}{2(2x^{2}+y)-\ddots }}}}}}.}

The square root can also be expressed by aperiodic continued fraction, but the above form converges more quickly with the properx andy.

Example 1

[edit]

Thecube root of two (21/3 or32 ≈ 1.259921...) can be calculated in two ways:

Firstly, "standard notation" ofx = 1,y = 1, and2zy = 3:

23=1+13+22+49+52+715+82+1021+112+=1+21912427574581063111381.{\displaystyle {\sqrt[{3}]{2}}=1+{\cfrac {1}{3+{\cfrac {2}{2+{\cfrac {4}{9+{\cfrac {5}{2+{\cfrac {7}{15+{\cfrac {8}{2+{\cfrac {10}{21+{\cfrac {11}{2+\ddots }}}}}}}}}}}}}}}}=1+{\cfrac {2\cdot 1}{9-1-{\cfrac {2\cdot 4}{27-{\cfrac {5\cdot 7}{45-{\cfrac {8\cdot 10}{63-{\cfrac {11\cdot 13}{81-\ddots }}}}}}}}}}.}

Secondly, a rapid convergence withx = 5,y = 3 and2zy = 253:

23=54+0.550+25+4150+55+7250+85+10350+115+=54+2.512531247595712658101771.{\displaystyle {\sqrt[{3}]{2}}={\cfrac {5}{4}}+{\cfrac {0.5}{50+{\cfrac {2}{5+{\cfrac {4}{150+{\cfrac {5}{5+{\cfrac {7}{250+{\cfrac {8}{5+{\cfrac {10}{350+{\cfrac {11}{5+\ddots }}}}}}}}}}}}}}}}={\cfrac {5}{4}}+{\cfrac {2.5\cdot 1}{253-1-{\cfrac {2\cdot 4}{759-{\cfrac {5\cdot 7}{1265-{\cfrac {8\cdot 10}{1771-\ddots }}}}}}}}.}

Example 2

[edit]

Pogson's ratio (1001/5 or5100 ≈ 2.511886...), withx = 5,y = 75 and2zy = 6325:

1005=52+3250+125+18750+275+331250+425+=52+5312653121837952733632542488855.{\displaystyle {\sqrt[{5}]{100}}={\cfrac {5}{2}}+{\cfrac {3}{250+{\cfrac {12}{5+{\cfrac {18}{750+{\cfrac {27}{5+{\cfrac {33}{1250+{\cfrac {42}{5+\ddots }}}}}}}}}}}}={\cfrac {5}{2}}+{\cfrac {5\cdot 3}{1265-3-{\cfrac {12\cdot 18}{3795-{\cfrac {27\cdot 33}{6325-{\cfrac {42\cdot 48}{8855-\ddots }}}}}}}}.}

Example 3

[edit]

Thetwelfth root of two (21/12 or122 ≈ 1.059463...), using "standard notation":

212=1+112+112+1336+232+2560+352+3784+472+=1+213611113108232518035372524749324.{\displaystyle {\sqrt[{12}]{2}}=1+{\cfrac {1}{12+{\cfrac {11}{2+{\cfrac {13}{36+{\cfrac {23}{2+{\cfrac {25}{60+{\cfrac {35}{2+{\cfrac {37}{84+{\cfrac {47}{2+\ddots }}}}}}}}}}}}}}}}=1+{\cfrac {2\cdot 1}{36-1-{\cfrac {11\cdot 13}{108-{\cfrac {23\cdot 25}{180-{\cfrac {35\cdot 37}{252-{\cfrac {47\cdot 49}{324-\ddots }}}}}}}}}}.}

Example 4

[edit]

Equal temperament'sperfect fifth (27/12 or1227 ≈ 1.498307...), withm = 7:

With "standard notation":

2712=1+712+52+1936+172+3160+292+4384+412+=1+27367519108173118029432524155324.{\displaystyle {\sqrt[{12}]{2^{7}}}=1+{\cfrac {7}{12+{\cfrac {5}{2+{\cfrac {19}{36+{\cfrac {17}{2+{\cfrac {31}{60+{\cfrac {29}{2+{\cfrac {43}{84+{\cfrac {41}{2+\ddots }}}}}}}}}}}}}}}}=1+{\cfrac {2\cdot 7}{36-7-{\cfrac {5\cdot 19}{108-{\cfrac {17\cdot 31}{180-{\cfrac {29\cdot 43}{252-{\cfrac {41\cdot 55}{324-\ddots }}}}}}}}}}.}

A rapid convergence withx = 3,y = −7153, and2zy = 219 + 312:

2712=12312715312=320.5715343121171536137153123122371536257153203123571536377153283124771536{\displaystyle {\sqrt[{12}]{2^{7}}}={\cfrac {1}{2}}{\sqrt[{12}]{3^{12}-7153}}={\cfrac {3}{2}}-{\cfrac {0.5\cdot 7153}{4\cdot 3^{12}-{\cfrac {11\cdot 7153}{6-{\cfrac {13\cdot 7153}{12\cdot 3^{12}-{\cfrac {23\cdot 7153}{6-{\cfrac {25\cdot 7153}{20\cdot 3^{12}-{\cfrac {35\cdot 7153}{6-{\cfrac {37\cdot 7153}{28\cdot 3^{12}-{\cfrac {47\cdot 7153}{6-\ddots }}}}}}}}}}}}}}}}}
2712=323715312(219+312)+715311137153236(219+312)23257153260(219+312)35377153284(219+312).{\displaystyle {\sqrt[{12}]{2^{7}}}={\cfrac {3}{2}}-{\cfrac {3\cdot 7153}{12(2^{19}+3^{12})+7153-{\cfrac {11\cdot 13\cdot 7153^{2}}{36(2^{19}+3^{12})-{\cfrac {23\cdot 25\cdot 7153^{2}}{60(2^{19}+3^{12})-{\cfrac {35\cdot 37\cdot 7153^{2}}{84(2^{19}+3^{12})-\ddots }}}}}}}}.}

More details on this technique can be found inGeneral Method for Extracting Roots using (Folded) Continued Fractions.

Higher dimensions

[edit]

Another meaning forgeneralized continued fraction is a generalization to higher dimensions. For example, there is a close relationship between the simple continued fraction in canonical form for the irrational real numberα, and the wayinteger lattice points in two dimensions lie to either side of the liney =αx. Generalizing this idea, one might ask about something related to lattice points in three or more dimensions. One reason to study this area is to quantify themathematical coincidence idea; for example, formonomials in several real numbers, take thelogarithmic form and consider how small it can be. Another reason is to find a possible solution toHermite's problem.

There have been numerous attempts to construct a generalized theory. Notable efforts in this direction were made byFelix Klein (theKlein polyhedron),Georges Poitou andGeorge Szekeres.

See also

[edit]

Notes

[edit]
  1. ^Wall 1948, p. 13.
  2. ^Cusick & Flahive 1989.
  3. ^Chrystal 1999.
  4. ^Jones & Thron 1980, p. 20.
  5. ^Euclid (2008) - The Euclidean algorithm generates a continued fraction as a by-product.
  6. ^Cataldi 1613.
  7. ^Wallis 1699.
  8. ^abEuler 1748, Chapter 18.
  9. ^Havil 2012, pp. 104–105.
  10. ^Brahmagupta (598–670) was the first mathematician to make a systematic study of Pell's equation.
  11. ^Gauss 1813.
  12. ^Lorentzen & Waadeland 1992.
  13. ^Oskar Perron derives even more general extension and contraction formulas for continued fractions. SeePerron (1977a),Perron (1977b).
  14. ^Angell 2021.
  15. ^Porubský 2008.
  16. ^This intuitive interpretation is not rigorous because an infinite continued fraction is not a mapping: it is thelimit of a sequence of mappings. This construction of an infinite continued fraction is roughly analogous to the construction of an irrational number as the limit of aCauchy sequence of rational numbers.
  17. ^Because of analogies like this one, the theory ofconformal mapping is sometimes described as "rubber sheet geometry".
  18. ^One approach to theconvergence problem is to constructpositive definite continued fractions, for which the denominatorsBi are never zero.
  19. ^This periodic fraction of period one is discussed more fully in the articleconvergence problem.
  20. ^An alternative way to calculate log(x)
  21. ^Borwein, Crandall & Fee 2004, p. 278, 280.
  22. ^Beckmann 1971.

References

[edit]
  • Angell, David (2021).Irrationality and Transcendence in Number Theory. Chapman and Hall/CRC.ISBN 9780367628376.
  • Chrystal, George (1999).Algebra, an Elementary Text-book for the Higher Classes of Secondary Schools and for Colleges: Pt. 1. American Mathematical Society. p. 500.ISBN 0-8218-1649-7.
  • Lorentzen, Lisa; Waadeland, Haakon (1992).Continued Fractions with Applications. Reading, MA: North Holland.ISBN 978-0-444-89265-2. (Covers primarily analytic theory and some arithmetic theory.)
  • Perron, Oskar (1977a) [1954].Die Lehre von den Kettenbrüchen. Vol. Band I: Elementare Kettenbrüche (3 ed.). Vieweg + Teubner Verlag.ISBN 9783519020219.
  • Perron, Oskar (1977b) [1954].Die Lehre von den Kettenbrüchen. Vol. Band II: Analytisch-funktionentheoretische Kettenbrüche (3 ed.). Vieweg + Teubner Verlag.ISBN 9783519020226.
  • Porubský, Štefan (2008)."Basic definitions for continued fractions".Interactive Information Portal for Algorithmic Mathematics. Prague, Czech Republic: Institute of Computer Science of the Czech Academy of Sciences. Retrieved2 May 2022.
  • Szekeres, George (1970). "Multidimensional continued fractions".Ann. Univ. Sci. Budapest. Eötvös Sect. Math.13:113–140.

External links

[edit]
Look upcontinued fraction in Wiktionary, the free dictionary.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Continued_fraction&oldid=1321239837"
Category:
Hidden categories:

[8]ページ先頭

©2009-2025 Movatter.jp