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Convergents constant

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If you use all the convergents of the simple continued fraction of a positive real constant x(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 (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

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Cf. Iterated continued fractions from convergents.

In order to get from iterate xn to iterate xn+1

  1. Express xn with the convergents ci(n1) of xn1 as a continued fraction [c0(n);c1(n),c2(n),]=1q0(n)[p0(n);q0(n)q1(n)/p1(n),q1(n)q2(n)/p2(n),...];
  2. Compute the convergents c0(n+1)p0(n+1)q0(n+1)=1q0(n)(p0(n)),c1(n+1)p1(n+1)q1(n+1)=1q0(n)(p0(n)+q0(n)q1(n)p1(n)),c2(n+1)p2(n+1)q2(n+1)=1q0(n)(p0(n)+q0(n)q1(n)p1(n)+q1(n)q2(n)p2(n)),
    (using the efficient recursive method shown on generalized continued fractions convergents;)
  3. The next iterate is xn+1=[c0(n+1);c1(n+1),c2(n+1),].

For example, starting with x0x, where x is a positive real constant, first obtain the simple continued fraction

x0=[a0(0);a1(0),a2(0),]=a0(0)+1a1(0)+1a2(0)+1a3(0)+1a4(0)+1a5(0)+1a6(0)+1a7(0)+1,

giving convergents

{p0(1)q0(1),p1(1)q1(1),p2(1)q2(1),p3(1)q3(1),p4(1)q4(1),p5(1)q5(1),p6(1)q6(1),p7(1)q7(1),...}

then

x1=p0(1)q0(1)+1p1(1)q1(1)+1p2(1)q2(1)+1p3(1)q3(1)+1p4(1)q4(1)+1p5(1)q5(1)+1p6(1)q6(1)+1p7(1)q7(1)+1
=1q0(1){p0(1)+q0(1)q1(1)p1(1)+q1(1)q2(1)p2(1)+q2(1)q3(1)p3(1)+q3(1)q4(1)p4(1)+q4(1)q5(1)p5(1)+q5(1)q6(1)p6(1)+q6(1)q7(1)p7(1)+q7(1)q8(1)}


=1q0(1){a0(1)+b1(1)a1(1)+b2(1)a2(1)+b3(1)a3(1)+b4(1)a4(1)+b5(1)a5(1)+b6(1)a6(1)+b7(1)a7(1)+b8(1)},

giving convergents

{p0(2)q0(2),p1(2)q1(2),p2(2)q2(2),p3(2)q3(2),p4(2)q4(2),p5(2)q5(2),p6(2)q6(2),p7(2)q7(2),...}



xn=p0(n)q0(n)+1p1(n)q1(n)+1p2(n)q2(n)+1p3(n)q3(n)+1p4(n)q4(n)+1p5(n)q5(n)+1p6(n)q6(n)+1p7(n)q7(n)+1
=1q0(n){p0(n)+q0(n)q1(n)p1(n)+q1(n)q2(n)p2(n)+q2(n)q3(n)p3(n)+q3(n)q4(n)p4(n)+q4(n)q5(n)p5(n)+q5(n)q6(n)p6(n)+q6(n)q7(n)p7(n)+q7(n)q8(n)}


=1q0(n){a0(n)+b1(n)a1(n)+b2(n)a2(n)+b3(n)a3(n)+b4(n)a4(n)+b5(n)a5(n)+b6(n)a6(n)+b7(n)a7(n)+b8(n)},

giving convergents

{p0(n+1)q0(n+1),p1(n+1)q1(n+1),p2(n+1)q2(n+1),p3(n+1)q3(n+1),p4(n+1)q4(n+1),p5(n+1)q5(n+1),p6(n+1)q6(n+1),p7(n+1)q7(n+1),...}

Convergence of iterated continued fractions from convergents

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We define the limit of iterated continued fractions from convergents for a constant x as

xlimnxn

Convergents constants

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Table of convergents constants

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Cf. Table of convergents constants.

See also

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