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| In the metrical theory of [[continued fraction|regular continued fractions]], the ''k''th '''complete quotient''' ζ<sub> ''k''</sub> is obtained by ignoring the first ''k'' partial denominators ''a''<sub>''i''</sub>. For example, if a regular continued fraction is given by
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| :<math>
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| x = [a_0; a_1, a_2, a_3, \dots] = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{\ddots}}}},
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| </math>
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| then the successive complete quotients ζ<sub> ''k''</sub> are given by
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| :<math>
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| \begin{align}
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| \zeta_0 & = [a_0; a_1, a_2, a_3, \dots]\\
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| \zeta_1 & = [a_1; a_2, a_3, a_4, \dots]\\
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| \zeta_2 & = [a_2; a_3, a_4, a_5, \dots]\\
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| \zeta_k & = [a_k; a_{k+1}, a_{k+2}, a_{k+3}, \dots]. \,
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| \end{align}
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| </math>
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| ==A recursive relationship==
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| From the definition given above we can immediately deduce that
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| :<math>
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| \zeta_k = a_k + \frac{1}{\zeta_{k+1}} = [a_k; \zeta_{k+1}], \,
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| </math>
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| or, equivalently,
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| :<math>
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| \zeta_{k+1} = \frac{1}{\zeta_k - a_k}.\,
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| </math>
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| ==Complete quotients and the convergents of ''x''==
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| Denoting the successive [[convergent (continued fraction)|convergent]]s of the regular continued fraction ''x'' = [''a''<sub>0</sub>; ''a''<sub>1</sub>, ''a''<sub>2</sub>, …] by ''A''<sub>0</sub>, ''A''<sub>1</sub>/''B''<sub>1</sub>, ''A''<sub>2</sub>/''B''<sub>2</sub>, … (as explained more fully in the article [[fundamental recurrence formulas]]), it can be shown that
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| :<math>
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| x = \frac{A_k \zeta_{k+1} + A_{k-1}}{B_k \zeta_{k+1} + B_{k-1}}\,
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| </math>
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| for all ''k'' ≥ 0.
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| This result can be better understood by recalling that the successive convergents of an infinite regular continued fraction approach the value ''x'' in a sort of zig-zag pattern:
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| :<math>
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| A_0 < \frac{A_2}{B_2} < \frac{A_4}{B_4} < \cdots < \frac{A_{2n}}{B_{2n}} < x
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| < \frac{A_{2n+1}}{B_{2n+1}} < \cdots < \frac{A_5}{B_5} < \frac{A_3}{B_3} < \frac{A_1}{B_1}.\,
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| </math>
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| so that when ''k'' is even we have ''A''<sub>''k''</sub>/''B''<sub>''k''</sub> < ''x'' < ''A''<sub>''k''+1</sub>/''B''<sub>''k''+1</sub>, and when ''k'' is odd we have ''A''<sub>''k''+1</sub>/''B''<sub>''k''+1</sub> < ''x'' < ''A''<sub>''k''</sub>/''B''<sub>''k''</sub>. In either case, the ''k'' + 1st complete quotient ζ<sub> ''k''+1</sub> is the unique real number that expresses ''x'' in the form of a [[continued fraction#Semiconvergents|semiconvergent]].
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| ==Complete quotients and equivalent real numbers==
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| ===An equivalence relation defined by LFTs===
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| Consider the set of [[linear fractional transformation]]s (LFTs) defined by
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| :<math>
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| f(x) = \frac{a + bx}{c + dx}\,
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| </math> | |
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| where ''a'', ''b'', ''c'', and ''d'' are [[integer]]s, and ''ad'' − ''bc'' = ±1. Since this set of LFTs contains an identity element (0 + ''x'')/1, and since it is closed under [[composition of functions]], and every member of the set has an inverse in the set, these LFTs form a [[group (mathematics)|group]] (the group operation being composition of functions), [[Modular group|GL(2,'''Z''')]].
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| We can define an [[equivalence relation]] on the set of [[real number]]s by means of this group of linear fractional transformations. We will say that two real numbers ''x'' and ''y'' are equivalent (written ''x'' ~ ''y'') if
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| :<math> | |
| y = f(x) = \frac{a + bx}{c + dx}\,
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| </math>
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| for some integers ''a'', ''b'', ''c'', and ''d'' such that ''ad'' − ''bc'' = ±1.
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| Clearly this relation is symmetric, reflexive, and transitive, so it is an equivalence relation and it can be used to separate the real numbers into [[equivalence class]]es. All the [[rational number]]s are equivalent, because each rational number is equivalent to zero. What can be said about the [[irrational number]]s? Do they also fall into a single equivalence class?
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| ===A theorem about "equivalent" irrational numbers===
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| Amazingly enough, two irrational numbers ''x'' and ''y'' are equivalent under this scheme if and only if the infinitely long "tails" in their expansions as regular continued fractions are exactly the same. More precisely, the following theorem can be proved.
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| Let ''x'' and ''y'' be two irrational (real) numbers, and let the ''k''th complete quotient in the regular continued fraction expansions of ''x'' and ''y'' be denoted by ζ<sub> ''k''</sub> and ψ<sub> ''k''</sub>, respectively, Then ''x'' ~ ''y'' (under the equivalence defined in the preceding section) if and only if there are positive integers ''m'' and ''n'' such that ζ<sub> ''m''</sub> = ψ<sub> ''n''</sub>.
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| ===An example===
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| The [[golden ratio]] φ is the irrational number with the very simplest possible expansion as a regular continued fraction: φ = [1; 1, 1, 1, …]. The theorem tells us first that if ''x'' is any real number whose expansion as a regular continued fraction contains the infinite string
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| [1, 1, 1, 1, …], then there are integers ''a'', ''b'', ''c'', and ''d'' (with ''ad'' − ''bc'' = ±1) such that
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| :<math>
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| x = \frac{a + b\phi}{c + d\phi}.\,
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| </math>
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| Conversely, if ''a'', ''b'', ''c'', and ''d'' are integers (with ''ad'' − ''bc'' = ±1), then the regular continued fraction expansion of every real number ''y'' that can be expressed in the form
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| :<math>
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| y = \frac{a + b\phi}{c + d\phi}\,
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| </math>
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| eventually reaches a "tail" that looks just like the regular continued fraction for φ.
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| ==References==
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| *{{cite book|last1 = Rockett|first1 = Andrew M.|last2 = Szüsz | first2 = Peter|title = Continued Fractions|publisher = World Scientific|year = 1992|pages = 4–8|isbn = 981-02-1052-3}}
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| [[Category:Continued fractions]]
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Wilber Berryhill is what his spouse enjoys to call him and he totally enjoys this name. North Carolina is where we've been living for years and will by no means move. It's not a typical thing but what I like doing is to climb but I don't have the time recently. Credit authorising is how she makes a living.
Here is my webpage: clairvoyance (simply click the following web site)