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# Convergents constant

(Redirected from Convergents constants)

This page is based on empirical evidence.

This (probably original) research topic requires further investigation.

If you use all the convergents of the simple continued fraction of a positive real constant $\scriptstyle x \,\in\, (n, n+1) \,$ as the terms of a generalized continued fraction, then likewise use the new convergents in another generalized continued fraction, and so on... ad infinitum, for all numbers in the unit interval $\scriptstyle (n, n+1) \,$, you get the convergents constant for all numbers of that interval. (Cf. Talk:Table_of_convergents_constants#Open_Problem)

## Iterated continued fractions from convergents

In order to get from iterate $\scriptstyle x_n \,$ to iterate $\scriptstyle x_{n+1} \,$

1. Express $\scriptstyle x_n \,$ with the convergents $\scriptstyle c_i(n-1) \,$ of $\scriptstyle x_{n-1} \,$ as a continued fraction $\scriptstyle [c_0(n);\, c_1(n),\, c_2(n),\, \ldots] \,=\, \frac{1}{q_0(n)}\,[p_0(n);\, q_0(n)\,q_1(n)\,/\,p_1(n),\, q_1(n)\,q_2(n)\,/\,p_2(n),\, ...]; \,$
2. Compute the convergents $\scriptstyle c_0(n+1) \,\equiv\, \frac{p_0(n+1)}{q_0(n+1)} \,=\, \frac{1}{q_0(n)} \, \left(p_0(n)\right),\, c_1(n+1) \,\equiv\, \frac{p_1(n+1)}{q_1(n+1)} \,=\, \frac{1}{q_0(n)} \left( \, p_0(n) + \frac{q_0(n)\,q_1(n)}{p_1(n)} \right),\, c_2(n+1) \,\equiv\, \frac{p_2(n+1)}{q_2(n+1)} \,=\, \frac{1}{q_0(n)} \left( \, p_0(n) + \frac{q_0(n)\,q_1(n)}{p_1(n) + \frac{q_1(n)\,q_2(n)}{p_2(n)}} \right),\, \ldots \,$
(using the efficient recursive method shown on generalized continued fractions convergents;)
3. The next iterate is $\scriptstyle x_{n+1} \,=\, [c_0(n+1);\, c_1(n+1),\, c_2(n+1),\, \ldots]. \,$

For example, starting with $\scriptstyle x_0 \,\equiv\, x \,$, where $\scriptstyle x \,$ is a positive real constant, first obtain the simple continued fraction

$x_0 = [a_0(0);a_1(0),a_2(0),\dots]= a_0(0) + \cfrac{1}{a_1(0) + \cfrac{1}{a_2(0) + \cfrac{1}{a_3(0) + \cfrac{1}{a_4(0) + \cfrac{1}{a_5(0) + \cfrac{1}{a_6(0) + \cfrac{1}{a_7(0) + \cfrac{1}{\ddots}}}}}}}}, \,$

giving convergents

$\Bigg\{ \frac{p_0(1)}{q_0(1)}, \frac{p_1(1)}{q_1(1)}, \frac{p_2(1)}{q_2(1)}, \frac{p_3(1)}{q_3(1)}, \frac{p_4(1)}{q_4(1)}, \frac{p_5(1)}{q_5(1)}, \frac{p_6(1)}{q_6(1)}, \frac{p_7(1)}{q_7(1)}, ... \Bigg\} \,$

then

$x_1 = \tfrac{p_0(1)}{q_0(1)} + \cfrac{1}{\tfrac{p_1(1)}{q_1(1)} + \cfrac{1}{\tfrac{p_2(1)}{q_2(1)} + \cfrac{1}{\tfrac{p_3(1)}{q_3(1)} + \cfrac{1}{\tfrac{p_4(1)}{q_4(1)} + \cfrac{1}{\tfrac{p_5(1)}{q_5(1)} + \cfrac{1}{\tfrac{p_6(1)}{q_6(1)} + \cfrac{1}{\tfrac{p_7(1)}{q_7(1)} + \cfrac{1}{\ddots}}}}}}}} \,$
$= \frac{1}{q_0(1)} \left\{ p_0(1) + \cfrac{q_0(1) q_1(1)}{p_1(1) + \cfrac{q_1(1) q_2(1)}{p_2(1) + \cfrac{q_2(1) q_3(1)}{p_3(1) + \cfrac{q_3(1) q_4(1)}{p_4(1) + \cfrac{q_4(1) q_5(1)}{p_5(1) + \cfrac{q_5(1) q_6(1)}{p_6(1) + \cfrac{q_6(1) q_7(1)}{p_7(1) + \cfrac{q_7(1) q_8(1)}{\ddots}}}}}}}} \right\} \,$

$= \frac{1}{q_0(1)} \left\{ a_0(1) + \cfrac{b_1(1)}{a_1(1) + \cfrac{b_2(1)}{a_2(1) + \cfrac{b_3(1)}{a_3(1) + \cfrac{b_4(1)}{a_4(1) + \cfrac{b_5(1)}{a_5(1) + \cfrac{b_6(1)}{a_6(1) + \cfrac{b_7(1)}{a_7(1) + \cfrac{b_8(1)}{\ddots}}}}}}}} \right\}, \,$

giving convergents

$\Bigg\{ \frac{p_0(2)}{q_0(2)}, \frac{p_1(2)}{q_1(2)}, \frac{p_2(2)}{q_2(2)}, \frac{p_3(2)}{q_3(2)}, \frac{p_4(2)}{q_4(2)}, \frac{p_5(2)}{q_5(2)}, \frac{p_6(2)}{q_6(2)}, \frac{p_7(2)}{q_7(2)}, ... \Bigg\} \,$

$\cdots \,$

$x_n = \tfrac{p_0(n)}{q_0(n)} + \cfrac{1}{\tfrac{p_1(n)}{q_1(n)} + \cfrac{1}{\tfrac{p_2(n)}{q_2(n)} + \cfrac{1}{\tfrac{p_3(n)}{q_3(n)} + \cfrac{1}{\tfrac{p_4(n)}{q_4(n)} + \cfrac{1}{\tfrac{p_5(n)}{q_5(n)} + \cfrac{1}{\tfrac{p_6(n)}{q_6(n)} + \cfrac{1}{\tfrac{p_7(n)}{q_7(n)} + \cfrac{1}{\ddots}}}}}}}} \,$
$= \frac{1}{q_0(n)} \left\{ p_0(n) + \cfrac{q_0(n) q_1(n)}{p_1(n) + \cfrac{q_1(n) q_2(n)}{p_2(n) + \cfrac{q_2(n) q_3(n)}{p_3(n) + \cfrac{q_3(n) q_4(n)}{p_4(n) + \cfrac{q_4(n) q_5(n)}{p_5(n) + \cfrac{q_5(n) q_6(n)}{p_6(n) + \cfrac{q_6(n) q_7(n)}{p_7(n) + \cfrac{q_7(n) q_8(n)}{\ddots}}}}}}}} \right\} \,$

$= \frac{1}{q_0(n)} \left\{ a_0(n) + \cfrac{b_1(n)}{a_1(n) + \cfrac{b_2(n)}{a_2(n) + \cfrac{b_3(n)}{a_3(n) + \cfrac{b_4(n)}{a_4(n) + \cfrac{b_5(n)}{a_5(n) + \cfrac{b_6(n)}{a_6(n) + \cfrac{b_7(n)}{a_7(n) + \cfrac{b_8(n)}{\ddots}}}}}}}} \right\}, \,$

giving convergents

$\Bigg\{ \frac{p_0(n+1)}{q_0(n+1)}, \frac{p_1(n+1)}{q_1(n+1)}, \frac{p_2(n+1)}{q_2(n+1)}, \frac{p_3(n+1)}{q_3(n+1)}, \frac{p_4(n+1)}{q_4(n+1)}, \frac{p_5(n+1)}{q_5(n+1)}, \frac{p_6(n+1)}{q_6(n+1)}, \frac{p_7(n+1)}{q_7(n+1)}, ... \Bigg\} \,$

### Convergence of iterated continued fractions from convergents

We define the limit of iterated continued fractions from convergents for a constant $\scriptstyle x \,$ as

$x_{\infty} \equiv \lim_{n \to \infty} x_n \,$