Algebra

Continued Fractions and Convergents

Convergent

To find convergents to 3.14159=314159100000. Proceed as in the rule for H.C.F.

7|100000|314159| 3
         | 99113|300000| 
         |------|------| 
        1|   887| 14159|15
         |   854|  887| 
         |------|------| 
        1|    33|  5289| 
         |    29|  4435| 
         |------|------| 
        4|     4|   854|25
         |     4|   66| 
         |------|------| 
         | |   194| 
         | |   165| 
         |------|------| 
         | |    29|7
         | |    28| 
         |------|------| 
         | |     1| 

The continued fraction is

3+1
              7+1
                15+⋯

Or, as it is more conveniently written, 3+17+ 115+ ⋯ The convergents are formed as follows:-

3715125174
            3223333559208956376149314159
            1'7'106'113'2931'3044'24239'100000'

160 Rule: Write the quotients in a row, and the first two convergents at sight (in the example 3 and 3+17). Multiply the numerator of any convergent by the next quotient, and add the previous numerator. The result is the numerator of the next convergent. Proceed in the same way to determine the denominator. The last convergent should be the original fraction in its lowest terms.161

Formula

Formula for forming the convergents. If 𝑝_𝑛−2𝑞_𝑛−2, 𝑝_𝑛−1𝑞_𝑛−1, 𝑝_𝑛𝑞_𝑛 are any consecutive convergents, and 𝑎_𝑛−2, 𝑎_𝑛−1, 𝑎_𝑛, the corresponding quotients; then 𝑝_𝑛=𝑎_𝑛𝑝_𝑛−1+𝑝_𝑛−2, 𝑞_𝑛=𝑎_𝑛𝑞_𝑛−1+𝑞_𝑛−2 The 𝑛^th convergent is therefore 𝑝𝑛𝑞𝑛=𝑎𝑛𝑝𝑛−1+𝑝𝑛−2𝑎𝑛𝑞𝑛−1+𝑞𝑛−2≡𝐹𝑛162 The true value of the continued fraction will be expressed by 𝐹=𝑎'𝑛𝑝𝑛−1+𝑝𝑛−2𝑎'𝑛𝑞𝑛−1+𝑞𝑛−2 in which 𝑎^'_𝑛 is the complete quotient or value of the continued fraction commencing with 𝑎_𝑛. Alternately, by (162)163 𝑝𝑛𝑞𝑛−1−𝑝𝑛−1𝑞𝑛=±1 The convergents are alternately greater and less than the original fraction, and are always in their lowest terms.164 The difference between 𝐹_𝑛 and the true value of the continued fraction is <1𝑞_𝑛𝑞_𝑛+1 and >1𝑞_𝑛(𝑞_𝑛+𝑞_𝑛+1) and this difference therefore diminishes as 𝑛 increases. Proof: with (163) By taking the difference, 𝑝_𝑛𝑞_𝑛=𝑎'𝑝_𝑛+1+𝑝_𝑛𝑎'𝑞_𝑛+1+𝑞_𝑛 Also 𝐹 is nearer the true value than any other fraction with a less denominator.165 𝐹_𝑛𝐹_𝑛+1 is greater or less than 𝐹² according as 𝐹_𝑛 is greater or less than 𝐹_𝑛+1.166

General Theory of Continued Fractions

First Class of Continued Fraction

𝐹=𝑏1𝑎1+ 𝑏2𝑎2+ 𝑏3𝑎3+ ⋯

Second Class of Continued Fraction

𝐹=𝑏1𝑎1− 𝑏2𝑎2− 𝑏3𝑎3− ⋯ where 𝑎₁, 𝑏₁, ⋯ are taken as positive quantities. 𝑏₁𝑎₁, 𝑏₂𝑎₂, ⋯ are termed components of the continued fraction. If the components be infinite in number, the continued fraction is said to be infinite.
Let the successive convergents be denoted by 𝑝₁𝑞₁=𝑏₁𝑎₁; 𝑝₂𝑞₂=𝑏₁𝑎₁+ 𝑏₂𝑎₂; 𝑝₃𝑞₃=𝑏₁𝑎₁+ 𝑏₂𝑎₂+ 𝑏₃𝑎₃; and so on.167

The Law of Formation of the Convergents

The law of formation of the convergents is For 𝐹, {𝑝𝑛=𝑎𝑛𝑝𝑛−1+𝑏𝑛𝑝𝑛−2𝑞𝑛=𝑎𝑛𝑞𝑛−1+𝑏𝑛𝑞𝑛−2 For 𝑉, {𝑝𝑛=𝑎𝑛𝑝𝑛−1−𝑏𝑛𝑝𝑛−2𝑞𝑛=𝑎𝑛𝑞𝑛−1−𝑏𝑛𝑞𝑛−2 168 The relation between the successive differences of the convergents is, by (168), 𝑝𝑛+1𝑞𝑛+1−𝑝𝑛𝑞𝑛=∓𝑏𝑛+1𝑞𝑛−1𝑞𝑛+1𝑝𝑛𝑞𝑛−𝑝𝑛−1𝑞𝑛−1 Take the − sign for 𝐹, and the + for 𝑉. 169 By (168) 𝑝𝑛𝑞𝑛−1−𝑝𝑛−1𝑞𝑛=(−1)𝑛−1𝑏1𝑏2𝑏3⋯𝑏𝑛 170 The odd convergents for 𝐹, 𝑝₁𝑞₁, 𝑝₃𝑞₃, ⋯, continually decrease, and the even convergents, 𝑝₂𝑞₂, 𝑝₄𝑞₄, ⋯, continually increase. by (167)
Every odd convergent is greater, and every even convergent is less, than all following convergents. by (169).171

Definition

If the difference between consecutive convergents diminishes without limit, the infinite continued fraction is said to be definite. If the same difference tends to a fixed value greater than zero, the infinite continued fraction is indefinite; the odd convergents tending to one value, and the even convergents to another.172 𝐹 is definite if the ratio of every quotient to the next component is greater than a fixed quantity. Proof: Apply (169) successively.173 𝐹 is incommensurable when the components are all proper fractions and infinite in number. Proof: Indirectly, and by (168).174 If 𝑎 be never less than 𝑏+1, the convergents of 𝑉 are all positive proper fractions, increasing in magnitude, 𝑝_𝑛 and 𝑞_𝑛 also increasing with 𝑛. by (167) and (168).175 If, in this case, 𝑉 be infinite, it is also definite, being =1, if 𝑎 always =𝑏+1 while 𝑏 is less than 1, (175); and being less than 1, if 𝑎 is ever greater than 𝑏+1. by (180).176 𝑉 is incommensurable when it is less than 1, and the components are all proper fractions and infinite in number.177 If in the continued fraction 𝑉 (167), we have 𝑎_𝑛=𝑏_𝑛+1 always; then, by (168), 𝑝_𝑛=𝑏₁+𝑏₁𝑏₂+𝑏₁𝑏₂𝑏₃+⋯ to 𝑛 terms, and 𝑞_𝑛=𝑝_𝑛+1.180 If, in the continued fraction 𝐹, 𝑎_𝑛 and 𝑏_𝑛 are constant and equal, say, to 𝑎 and 𝑏 respectively; then 𝑝_𝑛 and 𝑞_𝑛 are respectively equal to the coefficients of 𝑥^𝑛−1 in the expansions of 𝑏1−𝑎𝑥−𝑏𝑥² and 𝑎+𝑏𝑥1−𝑎𝑥−𝑏𝑥². Proof. 𝑝_𝑛 and 𝑞_𝑛 are the 𝑛^th terms of two recurring series. See (168) and (251). 181

To convert a Series into a Continued Faaction

To convert a Series into a Continued Fraction, The series: 1𝑛+𝑥𝑛₁+𝑥²𝑛₂+⋯𝑥^𝑛𝑛_𝑛 is equal to a continued fraction 𝑉 (167), with 𝑛+1 components; the first, second, and 𝑛+1^th components being, 1𝑛, 𝑛²𝑥𝑛₁+𝑛𝑥, ⋯, 𝑛²_𝑛−1𝑥𝑛_𝑛+𝑛_𝑛−1𝑥. Proved by induction. 182 The series, 1𝑟+𝑥𝑟𝑟₁+𝑥²𝑟𝑟₁𝑟₂+⋯𝑥^𝑛𝑟𝑟₁𝑟₂⋯𝑟_𝑛 is equal to a continued fraction 𝑉 (167), with 𝑛+1 components, the first, second, and 𝑛+1^th components being, 1𝑟, 𝑟𝑥𝑟₁+𝑥, ⋯, 𝑟_𝑛−1𝑥𝑟_𝑛+𝑥. Proved by Induction.183 The sign of 𝑥 may be changed in either of the statements in (182) or (183).184 Also, if any of these series are convergent and infinite, the continued fraction become infinite.185

To Find the Value of a Continued Fraction with recurring quotients

Let the continued fraction be 𝑥=𝑏₁𝑎₁+  ⋯+ 𝑏_𝑛𝑎_𝑛+𝑦 where 𝑦=𝑏_𝑛+1𝑎_𝑛+1+  ⋯+ 𝑏_𝑛+𝑚𝑎_𝑛+𝑚+𝑦 so that there are 𝑚 recurring quotients. Form the 𝑛^th convergent for 𝑥, and the 𝑚^th for 𝑦. Then, by substituting the complete quotients 𝑎_𝑛+𝑦 for 𝑎_𝑛, and 𝑎_𝑛+𝑚+𝑦 for 𝑎_𝑛+𝑚 in (168), two equations are obtained of the forms. 𝑥=𝐴𝑦+𝐵𝐶𝑦+𝐷 and 𝑦=𝐸𝑦+𝐹𝐺𝑦+𝐻 from which, by eliminating 𝑦, a quadratic equation for determining 𝑥 is obtained.186 If 𝑏₁𝑎₁+  ⋯+ 𝑏_𝑛𝑎_𝑛+ be a continued fraction, and 𝑝₁𝑞₁, ⋯, 𝑝_𝑛𝑞_𝑛 the corresponding first 𝑛 convergents; then 𝑞_𝑛−1𝑞_𝑛, developed by (168), produces the continued fraction 1𝑎_𝑛 +𝑏_𝑛𝑎_𝑛−1+ 𝑏_𝑛−1𝑎_𝑛−2+ ⋯+𝑏₃𝑎₂ +𝑏₂𝑎₁ the quotients being the same but in reversed order.187

Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive