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Algebra
 Continued Fractions and Convergents
 Convergent
  Formula
 General Theory of Continued Fractions
  First Class of Continued Fraction
  Second Class of Continued Fraction
  The Law of Formation of the Convergents
  Definition
 To convert a Series into a Continued Faaction
  To Find the Value of a Continued Fraction with recurring quotients
 Sources and References

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 𝑝𝑛−2𝑞𝑛−2, 𝑝𝑛−1𝑞𝑛−1, 𝑝𝑛𝑞𝑛 are any consecutive convergents, and 𝑎𝑛−2, 𝑎𝑛−1, 𝑎𝑛, the corresponding quotients; then 𝑝𝑛=𝑎𝑛𝑝𝑛−1+𝑝𝑛−2, 𝑞𝑛=𝑎𝑛𝑞𝑛−1+𝑞𝑛−2 The 𝑛th 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 𝑎𝑛. 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 𝐹𝑛 and the true value of the continued fraction is <1𝑞𝑛𝑞𝑛+1 and >1𝑞𝑛(𝑞𝑛+𝑞𝑛+1) and this difference therefore diminishes as 𝑛 increases. Proof: with (163) By taking the difference, 𝑝𝑛𝑞𝑛=𝑎'𝑝𝑛+1+𝑝𝑛𝑎'𝑞𝑛+1+𝑞𝑛 Also 𝐹 is nearer the true value than any other fraction with a less denominator.165 𝐹𝑛𝐹𝑛+1 is greater or less than 𝐹2 according as 𝐹𝑛 is greater or less than 𝐹𝑛+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+1𝑝𝑛𝑞𝑛=∓𝑏𝑛+1𝑞𝑛−1𝑞𝑛+1𝑝𝑛𝑞𝑛𝑝𝑛−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, 𝑝𝑛 and 𝑞𝑛 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 𝑎𝑛=𝑏𝑛+1 always; then, by (168), 𝑝𝑛=𝑏1+𝑏1𝑏2+𝑏1𝑏2𝑏3+⋯ to 𝑛 terms, and 𝑞𝑛=𝑝𝑛+1.180 If, in the continued fraction 𝐹, 𝑎𝑛 and 𝑏𝑛 are constant and equal, say, to 𝑎 and 𝑏 respectively; then 𝑝𝑛 and 𝑞𝑛 are respectively equal to the coefficients of 𝑥𝑛−1 in the expansions of 𝑏1−𝑎𝑥−𝑏𝑥2 and 𝑎+𝑏𝑥1−𝑎𝑥−𝑏𝑥2. Proof. 𝑝𝑛 and 𝑞𝑛 are the 𝑛th 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+⋯𝑥𝑛𝑛𝑛 is equal to a continued fraction 𝑉 (167), with 𝑛+1 components; the first, second, and 𝑛+1th components being, 1𝑛, 𝑛2𝑥𝑛1+𝑛𝑥, ⋯, 𝑛2𝑛−1𝑥𝑛𝑛+𝑛𝑛−1𝑥. Proved by induction. 182 The series, 1𝑟+𝑥𝑟𝑟1+𝑥2𝑟𝑟1𝑟2+⋯𝑥𝑛𝑟𝑟1𝑟2⋯𝑟𝑛 is equal to a continued fraction 𝑉 (167), with 𝑛+1 components, the first, second, and 𝑛+1th components being, 1𝑟, 𝑟𝑥𝑟1+𝑥, ⋯, 𝑟𝑛−1𝑥𝑟𝑛+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+  ⋯+ 𝑏𝑛𝑎𝑛+𝑦 where 𝑦=𝑏𝑛+1𝑎𝑛+1+  ⋯+ 𝑏𝑛+𝑚𝑎𝑛+𝑚+𝑦 so that there are 𝑚 recurring quotients. Form the 𝑛th convergent for 𝑥, and the 𝑚th for 𝑦. Then, by substituting the complete quotients 𝑎𝑛+𝑦 for 𝑎𝑛, and 𝑎𝑛+𝑚+𝑦 for 𝑎𝑛+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+  ⋯+ 𝑏𝑛𝑎𝑛+ be a continued fraction, and 𝑝1𝑞1, ⋯, 𝑝𝑛𝑞𝑛 the corresponding first 𝑛 convergents; then 𝑞𝑛−1𝑞𝑛, developed by (168), produces the continued fraction 1𝑎𝑛 +𝑏𝑛𝑎𝑛−1+ 𝑏𝑛−1𝑎𝑛−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

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References

  1. B. Joseph, 1978, University Mathematics: A Textbook for Students of Science &amp; Engineering
  2. Wheatstone, C., 1854, On the Formation of Powers from Arithmetical Progressions
  3. Stroud, K.A., 2001, Engineering Mathematics
  4. Coolidge, J.L., 1949, The Story of The Binomial Theorem
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