Proof without words: Periodic continued fractions

Let x := x₀ be a real number. Set a₀ = [x], the greatest integer in x and $\frac{1}{x_{0} - a_{0}}$ $\frac{1}{{{x}_{0}}-{{a}_{0}}}$its complete quotients.

Set a_i = [x_i], the greatest integer in x_i and $x_{i} + 1 = \frac{1}{x_{i} - a} for all i \geq 1.$ ${{x}_{i}}+1=\frac{1}{{{x}_{i}}-a}\text{for}\,\text{all}\,\,i\ge 1.$Then,

x = a_{0} + \frac{1}{a_{1} + \frac{1}{a_{2} + \frac{1}{\dots}}}

$$x={{a}_{0}}+\frac{1}{{{a}_{1}}+\frac{1}{{{a}_{2}}+\frac{1}{\ldots }}}$$

The algorithm stops after finitely many steps if and only if x is rational. The above expansion is called The simple continued fraction of x. It is customarily written x = [a₀,a₁, . . . ,a_n,. ].

We call convergents of x the reduced fractions difined by:

\frac{p_{0}}{q_{0}} = a_{0},

$$\frac{{{p}_{0}}}{{{q}_{0}}}={{a}_{0}},$$

\begin{array}{l} \frac{p_{1}}{q_{1}} = a_{0} + \frac{1}{a_{1}}, \\ \dots, \\ \frac{p_{n}}{q_{n}} = a_{0} + \frac{1}{a_{1} +} \frac{1}{a_{2} +} \frac{1}{a_{3} +} \dots \frac{1}{a_{n}}, \dots . \end{array}

$$\begin{align}& \frac{{{p}_{1}}}{{{q}_{1}}}={{a}_{0}}+\frac{1}{{{a}_{1}}}, \\ & \ldots \,\,\,, \\ & \frac{{{p}_{n}}}{{{q}_{n}}}={{a}_{0}}+\frac{1}{{{a}_{1}}+}\frac{1}{{{a}_{2}}+}\frac{1}{{{a}_{3}}+}\cdots \frac{1}{{{a}_{n}}},\cdots . \\ \end{align}$$

If there exists k ≥ 0 and m > 0 such that whenever r > k, we have a_r = a_r+_m, the continued fraction is said periodic, with period (b₁, . . . ,b_m) = (a_k+₁, . . . ,a_k+_m) and pre-period (a₀,a₁, . . . ,a_k), which can be written for simplicity $x = [a_{0}, a_{1}, \dots a_{k}, \bar{b_{1}, \dots b_{m}}] .$ $x=\left[ {{a}_{0}},{{a}_{1}},\cdots {{a}_{k}},\overline{{{b}_{1}}\,,\cdots {{b}_{m}}} \right].$These so-called periodic continued fractions are precisely those that represent quadratic irrationalities.

We find a closed form expression for $x = [a_{0}, a_{1}, \dots a_{k}, \bar{b_{1}, \dots b_{m}}]$ $x=\left[ {{a}_{0}},{{a}_{1}},\cdots {{a}_{k}},\overline{{{b}_{1}}\,,\cdots {{b}_{m}}} \right]$, which generalized a previous resut of Roger B. Nelsen.

Main result

Lemma 1

Let x > 0 such that $x = a + \frac{b}{c x}, t h e n x = \frac{1}{2} (a^{2} + \sqrt{a^{2} + 4 \frac{b}{c}})$ $x=a+\frac{b}{cx},then\,x\,=\frac{1}{2}\left( {{a}^{2}}+\sqrt{{{a}^{2}}+4\frac{b}{c}} \right)$

Proof. Consider the Following figure:

We have ${(2 x - a)}^{2} = a^{2} + 4 \frac{b}{c}, then x = \frac{1}{2} (a^{2} + \sqrt{a^{2} + 4 \frac{b}{c}}) .$ ${{\left( 2x-a \right)}^{2}}={{a}^{2}}+4\frac{b}{c},\,\text{then }x=\frac{1}{2}\left( {{a}^{2}}+\sqrt{{{a}^{2}}+4\frac{b}{c}} \right)\,\,.$

Lemma 2

$I f x = [\bar{a_{0}, a_{1}, \dots, a_{n}}], t h e n x = \frac{p_{n} - q_{n} - 1}{q_{n}} + \frac{p_{n} - 1}{q_{n} x} .$ $If\,x=\left[ \overline{{{a}_{0}},{{a}_{1}},\cdots ,{{a}_{n}}} \right],\,then\,x=\frac{{{p}_{n}}-{{q}_{n}}-1}{{{q}_{n}}}+\frac{p_n-1}{{{q}_{n}}x}.$

Proof. We have $x = [\bar{a_{0}, a_{1}, \dots, a_{n}}] = [a_{0}, a_{1}, \dots a_{n}, x] = \frac{p_{n} x - p_{n - 1}}{q_{n} x - q_{n - 1}} . Then, q_{n} x^{2} = (p_{n} - q_{n - 1}) + p_{n - 1} . which$ $\,x=\left[ \overline{{{a}_{0}},{{a}_{1}},\cdots ,{{a}_{n}}} \right]=\left[ {{a}_{0}},{{a}_{1}},\cdots {{a}_{n}},x \right]=\frac{{{p}_{n}}x-{{p}_{n-1}}}{{{q}_{n}}x-{{q}_{n-1}}}.\text{Then}\,\text{,}\,{{q}_{n}}{{x}^{2}}=\left( {{p}_{n}}-{{q}_{n-1}} \right)+{{p}_{n-1}}.\,\text{which}$

gives $x = \frac{p_{n} - q_{n - 1}}{q_{n}} + \frac{p_{n - 1}}{q_{n} x} .$ $x=\frac{{{p}_{n}}-{{q}_{n-1}}}{{{q}_{n}}}+\frac{{{p}_{n-1}}}{{{q}_{n}}x}.$Completing the proof.

Theorem 3

The periodic continued fraction $[\bar{a_{0}, a_{1}, \dots, a_{n}}]$ $\left[ \overline{{{a}_{0}},{{a}_{1}},\cdots ,{{a}_{n}}} \right]$equals

$\frac{1}{2} [{(\frac{p_{n} - q_{n - 1}}{q_{n}})}^{2} + \sqrt{{(\frac{p_{n} - q_{n - 1}}{q_{n}})}^{2} + 4 \frac{p_{n - 1}}{q_{n}}}] .$ $$\frac{1}{2}\left[ {{\left( \frac{{{p}_{n}}-{{q}_{n-1}}}{{{q}_{n}}} \right)}^{2}}+\sqrt{{{\left( \frac{{{p}_{n}}-{{q}_{n-1}}}{{{q}_{n}}} \right)}^{2}}+4\frac{{{p}_{n-1}}}{{{q}_{n}}}} \right].$$

Corollary 4

(Theorem [1] ). . The periodic continued fraction $[\bar{a, b}]$ $\left[ \overline{a,b} \right]$equals

$\frac{1}{2} (a^{2} + \sqrt{a^{2} + 4 \frac{a}{b}}) .$ $$\frac{1}{2}\left( {{a}^{2}}+\sqrt{{{a}^{2}}+4\frac{a}{b}} \right).$$

Corollary 5

The periodic continued fraction $[\bar{a, b, c}]$ $\left[ \overline{a,b,c} \right]$equals

$\frac{1}{2} [{(a + \frac{c - b}{b c + 1})}^{2} + \sqrt{{(a + \frac{c - b}{b c + 1})}^{2} + 4 \frac{a b + 1}{b c + 1}}] .$ $$\frac12\left[\left(a+\frac{c-b}{bc+1}\right)^2+\sqrt{\left(a+\frac{c-b}{bc+1}\right)^2+4\frac{ab+1}{bc+1}}\right].$$

Example 6

As examples, notice that $[\bar{1}] = [\bar{1, 1, 1}] = \frac{1}{2} (1 + \sqrt{5}), [\bar{a}] = [\bar{a, a}] = [\bar{a, a, a}] = \frac{1}{2} (a^{2} + \sqrt{a^{2} + 4}),$ $\left[ {\bar{1}} \right]=\left[ \overline{1,1,1} \right]=\frac{1}{2}\left( 1+\sqrt{5} \right),\left[ {\bar{a}} \right]=\left[ \overline{a,a} \right]=\left[ \overline{a,a,a} \right]=\frac{1}{2}\left( {{a}^{2}}+\sqrt{{{a}^{2}}+4} \right),$ $[\bar{3, 1, 2}] = \frac{1}{2} (\frac{100}{9} + \sqrt{\frac{148}{9}}) .$ $\left[ \overline{3,1,2} \right]=\frac{1}{2}\left( \frac{100}{9}+\sqrt{\frac{148}{9}} \right).$

Corollary 7

Let $x = [a_{0}, a_{1}, \dots, a_{k}, \bar{b_{1}, \dots b_{m}}],$ $x=\left[ {{a}_{0}},{{a}_{1}},\ldots ,{{a}_{k}},\overline{{{b}_{1}},\cdots {{b}_{m}}} \right],$be a periodic continued fraction, with period (b₁, . . . ,b_m) and pre-period (a₀,a₁, . . . ,a_k).

Note $\frac{p_{i}}{q_{i}} = [a_{0}, a_{1}, \dots, a_{i}], f o r a l l 0 \leq i \leq k a n d \frac{{p^{'}}_{j}}{{q^{'}}_{j}} = [b_{1}, \dots b_{j}] f o r a l l 0 \leq j \leq m .$ $\frac{{{p}_{i}}}{{{q}_{i}}}=\left[ {{a}_{0}},{{a}_{1}},\cdots ,{{a}_{i}} \right],for\,all\,0\le i\le k\,and\,\frac{{{{{p}'}}_{j}}}{{{{{q}'}}_{j}}}=\left[ {{b}_{1}},\cdots {{b}_{j}} \right]for\,all\,0\,\le j\le m.$Then,

$x = \frac{p_{k} (\frac{1}{2} [{(\frac{{p^{'}}_{m} - {q^{'}}_{m - 1}}{{q^{'}}_{m}})}^{2} + \sqrt{{(\frac{{p^{'}}_{m} - {q^{'}}_{m - 1}}{q_{n}})}^{2} + 4 \frac{{p^{'}}_{m - 1}}{{q^{'}}_{m}}}]) + p_{k - 1}}{q_{k} (\frac{1}{2} [{(\frac{{p^{'}}_{m} - {q^{'}}_{m - 1}}{{q^{'}}_{m}})}^{2} + \sqrt{{(\frac{{p^{'}}_{m} - {q^{'}}_{m - 1}}{q_{n}})}^{2} + 4 \frac{{p^{'}}_{m - 1}}{{q^{'}}_{m}}}]) + q_{k - 1}} .$ $$x=\frac{{{p}_{k}}\left( \frac{1}{2}\left[ {{\left( \frac{{{{{p}'}}_{m}}-{{{{q}'}}_{m-1}}}{{{{{q}'}}_{m}}} \right)}^{2}}+\sqrt{{{\left( \frac{{{{{p}'}}_{m}}-{{{{q}'}}_{m-1}}}{{{{{q}}}_{n}}} \right)}^{2}}+4\frac{{{{{p}'}}_{m-1}}}{{{{{q}'}}_{m}}}} \right] \right)+{{p}_{k-1}}}{{{q}_{k}}\left( \frac{1}{2}\left[ {{\left( \frac{{{{{p}'}}_{m}}-{{{{q}'}}_{m-1}}}{{{{{q}'}}_{m}}} \right)}^{2}}+\sqrt{{{\left( \frac{{{{{p}'}}_{m}}-{{{{q}'}}_{m-1}}}{{{{{q}}}_{n}}} \right)}^{2}}+4\frac{{{{{p}'}}_{m-1}}}{{{{{q}'}}_{m}}}} \right] \right)+{{q}_{k-1}}}.$$

Example 8

As examples, notice that

$\begin{array}{l} [1, 2, 3, 4, 5, 2, \bar{1, 1, 1, 4}] = \frac{225 \sqrt{7} + 43}{157 \sqrt{7} + 30}, \\ [1, 2, 2, n, \bar{1, 2 n}] = \frac{7 \sqrt{n^{2} + 2 n} + 3}{5 \sqrt{n^{2} + 2 n + 2}}, \\ [1, 2, 2, 1, 4, n, \bar{n, 2 n}] = \frac{57 \sqrt{n^{2} + 2} + 10}{33 \sqrt{n^{2} + 2 + 7}} . \end{array}$ $$\begin{align}& \left[ 1,2,3,4,5,2,\overline{1,1,1,4} \right]=\frac{225\sqrt{7}+43}{157\sqrt{7}+30}, \\ & \left[ 1,2,2,n,\overline{1,2n} \right]=\frac{7\sqrt{{{n}^{2}}+2n}+3}{5\sqrt{{{n}^{2}}+2n+2}}, \\ & \left[ 1,2,2,1,4,n,\overline{n,2n} \right]=\frac{57\sqrt{{{n}^{2}}+2}+10}{33\sqrt{{{n}^{2}}+2+7}}. \\ \end{align}$$

Conclusions

We find a closed form expression $for x = [a_{0}, a_{1}, \dots, a_{k}, \bar{b_{1}, \dots, b_{m}}],$ $\text{for}x=\left[ {{a}_{0}},{{a}_{1}},\cdots ,{{a}_{k}},\overline{{{b}_{1}},\cdots ,{{b}_{m}}} \right],$which generalized a previous resut of Roger B. Nelsen.

eISSN:: 2444-8656
Lingua:: Inglese

Frequenza di pubblicazione:: Volume Open
Argomenti della rivista:: Life Sciences, other, Mathematics, Applied Mathematics, General Mathematics, Physics

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Proof without words: Periodic continued fractions

Pubblicato online: 26 apr 2019

Pagine: 57 - 60

Ricevuto: 06 feb 2019

Accettato: 22 mar 2019

DOI: https://doi.org/10.2478/AMNS.2019.1.00006

Parole chiaveContinued fractions, periodic, proof without words

© 2019 Amara Chandoul, published by Sciendo

This work is licensed under the Creative Commons Attribution 4.0 International License.

Parole chiave
Continued fractions, periodic, proof without words