Continued fraction


In mathematics, continued or complex fractions are expressed as follows:

x = a 0 + 1 a 1 + 1 a 2 + 1 a 3 + 1 ⋱   {\displaystyle x=a_{0}+{\cfrac {1}{a_{1}+{\cfrac {1}{a_{2}+{\cfrac {1}{a_{3}+{\cfrac {1}{\ddots \,}}}}}}}}} x=a0+a1+a2+a3+1111

Here a 0 a_{0}a0is some integer, and all other numbers an a_{n}anBoth are positive integers, and longer expressions can be defined accordingly. If the partial numerator and partial denominator are allowed to assume arbitrary values, including functions in some contexts, the resulting expression is a Generalized continued fraction . When it is necessary to distinguish the above standard forms from generalized continued fractions, it may be called simple or normal continued fractions, or is said to be in canonical form.

1. Common applications

Continued fractions are often used to approximate irrational numbers, for example:

2 = 1 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 + ⋱ {\displaystyle {\sqrt {2}}=1+{\frac {1}{2+{\cfrac {1}{2+{\cfrac {1}{2+{\cfrac {1}{2+{\cfrac {1}{2+{\cfrac {1}{2+\ddots }}}}}}}}}}}}} 2 =1+2+2+2+2+2+2+111111

This gives 2 \sqrt{2}2 The asymptotic fraction of :

1 1 , 3 2 , 7 5 , 17 12 、 … {\displaystyle {\frac {1}{1}},{\frac {3}{2}},{\frac {7}{5}},{\frac {17}{12}}}、… 11,23,57,1217
5 + 1 2 = 1 + 1 1 + 1 1 + 1 1 + 1 1 + 1 1 + 1 1 + ⋱ {\displaystyle {\frac { {\sqrt {5}}+1}{2}}=1+{\frac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+\ddots }}}}}}}}}}}}} 25 +1=1+1+1+1+1+1+1+111111

This gives the asymptotic fraction of the golden section:

1 1 , 2 1 , 3 2 , 5 3 , 8 5 , 13 8 、 … {\displaystyle {\frac {1}{1}},{\frac {2}{1}},{\frac {3}{2}},{\frac {5}{3}},{\frac {8}{5}},{\frac {13}{8}}}、… 11,12,23,35,58,813

Note that the denominators of the above series are 1 , 1 , 2 , 3 , ⋯ 1,1,2,3,\cdots1,1,2,3, The same numbers can be arranged in order to get the Fibonacci sequence.

π = 3 + 1 7 + 1 15 + 1 1 + 1 292 + 1 1 + 1 1 + ⋱ {\displaystyle \pi =3+{\frac {1}{7+{\cfrac {1}{15+{\cfrac {1}{1+{\cfrac {1}{292+{\cfrac {1}{1+{\cfrac {1}{1+\ddots }}}}}}}}}}}}} Pi=3+7+15+1+292+1+1+111111

From this we get the asymptotic fraction of pi 3 1 , 22 7 \frac {3}{1},\frac {22}{7}13,722(about rate), 333 106 , 355 113 {\frac {333}{106}},{\frac {355}{113}}106333,113355(density rate), 103993 33102 103993 33102 {\frac {103993}{33102}}\frac{103993}{33102}3310210399333102103993 ⋯ \cdots

It can be shown mathematically that an asymptotic fraction obtained from a (narrowly defined) continued fraction has a value that is the closest approximation to the exact value in a fraction whose numerator or denominator is smaller than the next asymptotic fraction.

2. Research motivation

The motivation for studying continued fractions stems from the desire to have a "mathematically pure" representation of real numbers.

Most people are familiar with the decimal representation of real numbers:

r = ∑ i = 0 ∞ a i 1 0 − i {\displaystyle r=\sum _{i=0}^{\infty }a_{i}10^{-i}} r=i=0ai10i

Here a 0 a_{0}a0Can be any integer, other ai a_{i}aiAll are { 0 , 1 , 2 , … , 9 } \{0,1,2,\ldots ,9\}{ 0,1,2,,9 } an element. In this representation, for example the numberπ \piπ is represented as a sequence of integers{ 3 , 1 , 4 , 1 , 5 , 9 , 2 , … } \{3,1,4,1,5,9,2,\ldots \}{ 3,1,4,1,5,9,2,}

This kind of decimal means something is wrong. For example, the constant 10 is used in this case because we used 10 10Decimal system. We can also use octal or binary systems. Another problem is that many rational numbers lack finite representation in this system. For example, the number1 3 {\frac {1}{3}}31is represented as an infinite sequence { 0 , 3 , 3 , 3 , 3 , … } \{0,3,3,3,3,\ldots \}{ 0,3,3,3,3,}

Continued fraction notation avoids these two problems of real number representation. Let us consider how to describe a number such as 415 93 {\frac {415}{93}}93415, about 4.4624 4.46244.4624 . approximately4 44 , which is actually better than4 44 is a little more, about4 + 1 2 4+{\frac {1}{2}}4+21. but the 2 2 in the denominator2 is inaccurate; a more accurate denominator is2 than 22 a little more, about2 + 1 6 2+{\frac {1}{6}}2+61, so 415 93 {\frac {415}{93}}93415Approximately 4 + 1 2 + 1 6 4+{\frac {1}{2+{\frac {1}{6}}}}4+2+611. but the 6 in the denominator 66 is inaccurate; a more accurate denominator is 6 than6A little more than 6 , actually6 + 1 7 6+{\frac {1}{7}}6+71. So 415 93 {\frac {415}{93}}93415Actually 4 + 1 2 + 1 6 + 1 7 4+{\frac {1}{2+{\frac {1}{6+{\frac {1}{7}}}}}}4+2+6+7111. This is accurate.

Remove the expression 4 + 1 2 + 1 6 + 1 7 4+{\frac {1}{2+{\frac {1}{6+{\frac {1}{7}}}}}}4+2+6+7111The redundant part in can get short notation [ 4 ; 2 , 6 , 7 ] [4;2,6,7][4;2,6,7]

The continued fraction representation of real numbers can be defined in this way. It has some desirable properties:

The continued fraction representation of a rational number is finite.

  • Continued fraction representations of "simple" rational numbers are short.
  • The continued fraction representation of any rational number is unique if it has no trailing 1s. ( [ a 0 ; a 1 , … , an , 1 ] = [ a 0 ; a 1 , … , an + 1 ] [a_{0};a_{1},\ldots ,a_{n},1]= [a_{0};a_{1},\ldots ,a_{n}+1][a0;a1,,an,1]=[a0;a1,,an+1]
  • The continued fraction representation of irrational numbers is unique.
  • The term of a continued fraction loops if and only if it is a continued fraction representation of a quadratic irrational number (i.e., a real solution to a quadratic equation with integer coefficients).
  • number xxThe truncated continued fraction of x means that xxThe "best possible" rational approximation of x in a specific sense (see Corollary 1 of Theorem 5 below).

This last property is very important, and cannot be the case with traditional decimal point representation. A truncated decimal representation of a number yields a rational approximation to the number, but usually not a very good approximation. For example, truncation 1 7 = 0.142 857 … {\frac {1}{7}}=0.142\ 857\ldots71=0.142 857... produces approximation ratios at various positions such as142 1000 {\frac {142}{1000}}1000142 14 100 {\frac {14}{100}} 10014and 1 10 {\frac {1}{10}}101. But the obvious best rational approximation is " 1 7 {\frac {1}{7}}71" itself. π \piThe truncated decimal representation of π yields approximation ratios such as31415 10000 {\frac {31415}{10000}}1000031415and 314 100 {\frac {314}{100}}100314π \piThe continued fraction representation of π starts at[ 3 ; 7 , 15 , 1 , 292 , … ] [3;7,15,1,292,\ldots ][3;7,15,1,292,... ] . Truncating this representation yields an excellent rational approximation to3 33 22 7 {\frac {22}{7}} 722 333 106 {\frac {333}{106}} 106333 355 113 {\frac {355}{113}} 113355 103   993 33   102 {\frac {103\ 993}{33\ 102}} 33 102103 993 ⋯ \cdots 314 100 {\frac {314}{100}} 100314and 333 106 {\frac {333}{106}}106333The denominators of are quite close, but the approximate value 314 100 {\frac {314}{100}}100314The error is much higher than 333 106 {\frac {333}{106}}10633319 1919 times. as pairπ \piApproximation of π , [ 3 ; 7 , 15 , 1 ] [3;7,15,1][3;7,15,1 ] than3.1416 3.14163.1416 Exact100 100100 times.

3. Algorithm for representation of continued fractions

Consider the real number rrr . ii_i isrrthe integer part of r , whilefff is its fractional part. thenrrThe continued fraction representation of r is [ i ; … ] [i;\ldots ][i;] , where “… \ldots " is1 f {\frac {1}{f}}f1The continued fraction representation of . It is customary to replace the first comma with a semicolon.

To calculate the real number rrThe continued fraction representation of r , first writerrThe integer part of r (floored), then fromrrr subtracts this integer part. If the difference is0 00 to stop; otherwise find the inverse of this difference and repeat. The process will terminate if and only ifrrr is a rational number.

find 3.245 3.245The continued fraction of 3.245
3 3 3 3.245 − 3 3.245 - 3 3.2453 = 0.245 1 0.245 {\frac {1}{0.245}} 0.2451 = 4.082... =4.082... =4.082...
4 4 4 4.082... − 4 4.082...-4 4.082...4 = 0.082... =0.082... =0.082... 1 0.082... {\frac {1}{0.082...}} 0.082...1 = 12.250 =12.250 =12.250
12 12 12 12.250 − 12 12.250-12 12.25012 = 0.250 =0.250 =0.250 1 0.250 {\frac {1}{0.250}} 0.2501 = 4.000 =4.000 =4.000
4 4 4 4.000 − 4 4.000 - 4 4.0004 = 0.000 = 0.000 =0.000 stop
The continued fraction of 3.245 is [ 3 ; 4 , 12 , 4 ] [3;4,12,4][3;4,12,4]
3.245 = 3 + 1 4 + 1 12 + 1 4 3.245=3+{\cfrac {1}{4+{\cfrac {1}{12+{\cfrac {1}{4}}}}}} 3.245=3+4+12+4111

Number 3.245 3.2453.245 can also be expressed as a continued fraction expansion[ 3 ; 4 , 12 , 3 , 1 ] [3;4,12,3,1][3;4,12,3,1 ] ; see finite continued fractions below.

This algorithm works well for real numbers, but can lead to numerical disaster if implemented with floating point numbers. Instead, any floating-point number is an exact rational number (on modern computers the denominator is usually 2 2power of 2 , usually 10 10on electronic calculatorspowers of 10 ), so variants of Euclid's algorithm can be used to give exact results.

4. Representation of continued fractions

Continued fractions can be abbreviated as:

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

Or, written in Pringsheim's notation:

x = a 0 + 1 ∣ ∣ a 1 + 1 ∣ ∣ a 2 + 1 ∣ ∣ a 3 {\displaystyle x=a_{0}+{\frac {1\mid }{\mid a_{1}}}+{\frac {1\mid }{\mid a_{2}}}+{\frac {1\mid }{\mid a_{3}}}} x=a0+a11+a21+a31

There is also a related notation:

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

Angle brackets are sometimes used, as in:

x = ⟨ a 0 ; a 1 , a 2 , a 3 ⟩    {\displaystyle x=\left\langle a_{0};a_{1},a_{2},a_{3}\right\rangle \;} x=a0;a1,a2,a3

Semicolons are optional when using angle brackets.

It is also possible to define infinite simple continued fractions as limits:

[ a 0 ; a 1 , a 2 , a 3 ,   … ] = lim ⁡ n → ∞ [ a 0 ; a 1 , a 2 ,   … , a n ] {\displaystyle [a_{0};a_{1},a_{2},a_{3},\,\ldots ]=\lim _{n\to \infty }[a_{0};a_{1},a_{2},\,\ldots ,a_{n}]} [a0;a1,a2,a3,]=nlim[a0;a1,a2,,an]

For positive integers a 1 , a 2 , a 3 ⋯ a_{1}, a_{2}, a_{3}\cdotsa1,a2,a3Any choice of ... has this limit.

Or you can use Gaussian notation

x = a 0 + K 3 i = 1   1 a i    {\displaystyle x=a_{0}+{\underset {i=1}{\overset {3}{\mathrm {K} }}}~{\frac {1}{a_{i}}}\;} x=a0+i=1K3 ai1

5. Finite continued fractions

All finite continued fractions represent a rational number, and all rational numbers can be represented as finite continued fractions in two different ways. Both representations are identical except for the final term. expressed in longer continued fractions whose final term is 1 11 ; the shorter means the last1 is removed 11 , and add 1to the new terminal term1 . The final term in the short representation is thus greater than1 11 if short means at least two terms. Its symbol means:

[ a 0 ; a 1 , a 2 , a 3 ,   … , a n , 1 ] = [ a 0 ; a 1 , a 2 , a 3 ,   … , a n + 1 ] [a_{0};a_{1},a_{2},a_{3},\,\ldots ,a_{n},1]=[a_{0};a_{1},a_{2},a_{3},\,\ldots ,a_{n}+1] [a0;a1,a2,a3,,an,1]=[a0;a1,a2,a3,,an+1]

For example:

2.25 = 9 4 = [ 2 ; 3 , 1 ] = [ 2 ; 4 ]    − 4.2 = − 21 5 = [ − 5 ; 1 , 3 , 1 ] = [ − 5 ; 1 , 4 ]    2.25={\frac {9}{4}}=[2;3,1]=[2;4]\;\\ -4.2=-{\frac {21}{5}}=[-5;1,3,1]=[-5;1,4]\; 2.25=49=[2;3,1]=[2;4]4.2=521=[5;1,3,1]=[5;1,4]

6. Reciprocal of continued fractions

The continued fraction representation of a rational number is the same as its reciprocal except that it is shifted left or right by one bit depending on whether the number is less than or greater than 1, respectively. In other words, [ a 0 ; a 1 , a 2 , a 3 , … , an ] [a_{0};a_{1},a_{2},a_{3},\ldots ,a_{n}][a0;a1,a2,a3,,an] [ 0 ; a 0 , a 1 , a 2 , … , a n ] [0;a_{0},a_{1},a_{2},\ldots ,a_{n}] [0;a0,a1,a2,,an] are reciprocals of each other. This is because if $a$ is an integer, then ifx < 1 x<1x<1 ,则 x = 0 + 1 a + 1 b x=0+{\tfrac {1}{a+{\frac {1}{b}}}} x=0+a+b11 1 x = a + 1 b {\tfrac {1}{x}}=a+{\tfrac {1}{b}} x1=a+b1, and if x > 1 x>1x>1 ,x = a + 1 bx=a+{\tfrac {1}{b}}x=a+b1 1 x = 0 + 1 a + 1 b {\tfrac {1}{x}}=0+{\tfrac {1}{a+{\frac {1}{b}}}} x1=0+a+b11Generate pair xx with last numberx and its reciprocal are remainders of the same continued fraction.

For example:

2.25 = 9 4 = [ 2 ; 4 ]    1 2.25 = 4 9 = [ 0 ; 2 , 4 ]    2.25 = \frac{9}{4} = [2; 4] \;\\ \frac{1}{2.25} = \frac{4}{9} = [0; 2, 4] \; 2.25=49=[2;4]2.251=94=[0;2,4]

7. Infinitely continued fractions

All infinitely continued fractions are irrational numbers, and all irrational numbers can be represented in an exact way as infinitely continued fractions.

The infinite continued fraction representation of irrational numbers is very useful because its initial segment provides an excellent rational approximation to this number. These rational numbers can be called the convergence of this continued fraction (convergent, also translated as "asymptotic"). All even numbers converge less than the original number, and odd numbers converge greater than it.

For continued fractions [ a 0 ; a 1 , a 2 , … ] [a_{0};a_{1},a_{2},\ldots ][a0;a1,a2,] , the first four convergent (number0 00 to3 33 is

a 0 1 , a 1 a 0 + 1 a 1 , a 2 ( a 1 a 0 + 1 ) + a 0 a 2 a 1 + 1 , a 3 [ a 2 ( a 1 a 0 + 1 ) + a 0 ] + ( a 1 a 0 + 1 ) a 3 ( a 2 a 1 + 1 ) + a 1 {\displaystyle {\frac {a_{0}}{1}},\qquad {\frac {a_{1}a_{0}+1}{a_{1}}},\qquad {\frac {a_{2}(a_{1}a_{0}+1)+a_{0}}{a_{2}a_{1}+1}},\qquad {\frac {a_{3}[a_{2}(a_{1}a_{0}+1)+a_{0}]+(a_{1}a_{0}+1)}{a_{3}(a_{2}a_{1}+1)+a_{1}}}} 1a0,a1a1a0+1,a2a1+1a2(a1a0+1)+a0,a3(a2a1+1)+a1a3[a2(a1a0+1)+a0]+(a1a0+1)

In plain language, number 3 3The 3 convergent molecules are obtained by the 3rd33 quotients (a 2 a_2a2) times the 2nd 22 convergent numerators, plus the 1st11 convergent molecule. The formation of the denominator is similar.

If a continuous convergence is found, with molecules h 1 , h 2 , … h_{1},h_{2},\ldotsh1,h2, and denominatorsk 1 , k 2 , … k_{1},k_{2},\ldotsk1,k2, , then the relevant recurrence relation is:

h n = a n h n − 1 + h n − 2 , k n = a n k n − 1 + k n − 2 {\displaystyle h_{n}=a_{n}h_{n-1}+h_{n-2},\qquad k_{n}=a_{n}k_{n-1}+k_{n-2}} hn=anhn1+hn2,kn=ankn1+kn2

Successive convergence is given by

h n k n = a n h n − 1 + h n − 2 a n k n − 1 + k n − 2 {\displaystyle {\frac {h_{n}}{k_{n}}}={\frac {a_{n}h_{n-1}+h_{n-2}}{a_{n}k_{n-1}+k_{n-2}}}} knhn=ankn1+kn2anhn1+hn2

8. Some useful theorems

如果 a 0 , a 1 , a 2 , … a_{0},a_{1},a_{2},\ldots a0,a1,a2,... is an infinite sequence of positive integers, recursively defined sequencehn h_{n}hnand kn k_{n}kn

h n = a n h n − 1 + h n − 2   h − 1 = 1   h − 2 = 0 k n = a n k n − 1 + k n − 2   k − 1 = 0   k − 2 = 1 h_{n}=a_nh_{n-1}+h_{n-2}\,\quad h_{-1}=1\,\quad h_{-2}=0\\ k_{n}=a_nk_{n-1}+k_{n-2}\,\quad k_{-1}=0\,\quad k_{-2}=1 hn=anhn1+hn2h1=1h2=0kn=ankn1+kn2k1=0k2=1

Theorem 1

For any positive number x ∈ R x\in \mathbb {R}xR

[ a 0 ; a 1 ,   … , a n − 1 , x ] = x h n − 1 + h n − 2 x k n − 1 + k n − 2 {\displaystyle \left[a_{0};a_{1},\,\dots ,a_{n-1},x\right]={\frac {xh_{n-1}+h_{n-2}}{xk_{n-1}+k_{n-2}}}} [a0;a1,,an1,x]=xkn1+kn2xhn1+hn2

Theorem 2

[ a 0 , a 1 , a 2 , … ] [a_{0},a_{1},a_{2},\ldots ] [a0,a1,a2,] converge to

[ a 0 ; a 1 ,   … , a n ] = h n k n {\displaystyle \left[a_{0};a_{1},\,\dots ,a_{n}\right]={\frac {h_{n}}{k_{n}}}} [a0;a1,,an]=knhn

give.

Theorem 3

If the nnth of the continued fractionn convergence ishn / kn h_{n}/k_{n}hn/kn,but

k n h n − 1 − k n − 1 h n = ( − 1 ) n   k_nh_{n-1}-k_{n-1}h_n=(-1)^n \, knhn1kn1hn=(1)n

  • Corollary 1: Every convergence is in its lowest terms (if hn h_{n}hnand kn k_{n}knhas unusual common divisor, then it divides knhn − 1 − kn − 1 hn k_{n}h_{n-1}-k_{n-1}h_{n}knhn1kn1hn, which is of course impossible).

  • Corollary 2: The difference between successive convergences is a unit fraction:

∣ h n k n − h n − 1 k n − 1 ∣ = ∣ h n k n − 1 − k n h n − 1 k n k n − 1 ∣ = 1 k n k n − 1 \left|\frac{h_n}{k_n}-\frac{h_{n-1}}{k_{n-1}} \right|= \left|\frac{h_nk_{n-1}-k_nh_{n-1}}{k_nk_{n-1}}\right|= \frac{1}{k_nk_{n-1}} knhnkn1hn1 = knkn1hnkn1knhn1 =knkn11

  • Corollary 3: Continued fractions are equivalent to series of alternating terms:

a 0 + ∑ n = 0 ∞ ( − 1 ) n k n + 1 k n a_0 + \sum_{n=0}^\infty \frac{(-1)^{n}}{k_{n+1}k_{n}} a0+n=0kn+1kn(1)n

  • Corollary 4: Matrix

[ h n h n − 1 k n k n − 1 ] \begin{bmatrix} h_n & h_{n-1} \\ k_n & k_{n-1} \end{bmatrix} [hnknhn1kn1]

The determinant value of is positive 1 11 or minus1 11 , thus belonging to2 × 2 2\times22×2 unimodular matrixS ∗ L ( 2 , Z ) S^{*}L(2,\mathbb {Z} )SL(2,Z ) group.

Theorem 4

Each (th sss ) are greater than any preceding (rrr ) converges closer to the subsequent (nthn ) converge. In notational terms, if thennthn convergence is[ a 0 ; a 1 , a 2 , … an ] = xn [a_{0};a_{1},a_{2},\ldots a_{n}]=x_{n}[a0;a1,a2,an]=xn,but

∣ x r − x n ∣ > ∣ x s − x n ∣ \left|x_{r}-x_{n}\right|>\left|x_{s}-x_{n}\right| xrxn>xsxn

For all r < s < n r<s<nr<s<n

  • Corollary 1: Odd convergence (at nnn before) keeps increasing and is always less thanxn x_{n}xn

  • Corollary 2: Convergence in even numbers (at nnn before) keeps decreasing and is always greater thanxn x_{n}xn

Theorem 5

1 k n ( k n + 1 + k n ) < ∣ x − h n k n ∣ < 1 k n k n + 1 \frac{1}{k_n(k_{n+1}+k_n)}< \left|x-\frac{h_n}{k_n}\right|< \frac{1}{k_nk_{n+1}} kn(kn+1+kn)1< xknhn <knkn+11

  • Corollary 1: Any convergent fraction is closer to this continued fraction than any other fraction whose denominator is smaller than the convergent denominator.

  • Corollary 2: Any convergence immediately preceding a large quotient is a close approximation to this continued fraction.

9. Semi-convergent

If hn − 1 kn − 1 {\frac {h_{n-1}}{k_{n-1}}}kn1hn1and hnkn {\frac {h_{n}}{k_{n}}}knhnis continuous convergence, then any fraction of the form

hn − 1 + ahnkn − 1 + akn \frac{h_{n-1} + ah_n}{k_{n-1}+ak_n}kn1+to knhn1+ahn

aa herea is a non-negative integer, and the numerator and denominator are innnn andn + 1 n+1n+1 term (inclusive), called "semi-convergent", sub-convergent or intermediate fractions. The term often means that ruling out the possibility is convergence, not that convergence is a kind of semi-convergence.

For the real number xxThe semiconvergence of the continued fraction expansion of x includes all rational approximations that are better than any approximation with a smaller denominator. Another useful property is that the continuous semiconvergenceab {\frac {a}{b}}baand cd {\frac {c}{d}}dcWith ad − bc = ± 1 ad-bc=\pm 1adbc=±1

10. Best Rational Number Approximation

The theory of continued fractions plays a fundamental role in the field of Diophantine approximation, and can solve the problem of optimal approximation of real numbers. For details, please refer to the corresponding main page. In fact, the initial motivation for developing the theory of continued fractions was precisely to solve the problem of the best approximation of real numbers.

11. Continued fraction history

  • 300 BC - Euclid, "Elements" - The algorithm for the greatest common divisor produces a continued fraction as a by-product.
  • Aryabhatiya 499 contains the solution of uncertain equations using continued fractions.
  • 1572 - Rafael Bombelli, L'Algebra Opera - Method for extracting square roots related to continued fractions.
  • 1613 - Pietro Cataldi, "Trattato del modo brevissimo di trovar la radice quadra delli numeri" - The first notation for continued fractions
    Cataldi expresses continued fractions as a 0 . & n 1 d 1 . & n 2 d 2 . & n 3 d 3 a_{0}.\, \& \frac{n_{1}}{d_{1}}. \&{n_{2} \over d_{2}}. \&{n_{3} \over d_{3}}a0.&d1n1&d2n2&d3n3with dots indicating where subsequent continued fractions go.
  • 1695 - John Wallis, Opera Mathematica - Introduces the term "continued fractions".
  • 1737 Leonhard Euler, De fractionibus continuis dissertatio - provides the first comprehensive account of the properties of continued fractions and is the first to demonstrate the number eee is an irrational number.
  • 1748 Leonhard Euler, "Introductio in analysis in infinitorum". Vol. I, Chapter 18 - Proves the equivalence of continued fractions of a particular form and generalized infinite series. Proved that every rational number can be written as a finite continued fraction, and proved that the continued fraction of irrational numbers is infinite.
  • 1768 Joseph-Louis Lagrange - Provides a general solution to Pell's equations using continued fractions similar to Bombelli.
  • 1761 Johann Lambert - using tan ⁡ ( x ) \tan(x)The continued fraction of tan ( x ) proved for the first time thatπ \piThe irrationality of π .
  • 1770 Lagrange - proved that quadratic irrational numbers extend to periodic continued fractions.
  • 1813 - Carl Friedrich Gauss, Werke, Book III, pp. 134-138 - Deduces very general complex-valued continued fractions by a clever identity involving hypergeometric series.
  • In 1892 Henri Padé defined the Padé approximation
  • 1972 Bill Gosper - First exact algorithm for continued fraction arithmetic.

  • references

(Former Soviet Union) A. Ya. Khinchin (A. Ya. Khinchin), translated by Liu Shijun and Liu Shaoyue. Continued Fractions. Shanghai: Shanghai Science and Technology Press. 1965.

Wikipedia: Continued fractions

wiki: Continued fraction

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