Math 2.0

Number theory Definition

Convergents of a continued fraction

Builds on Simple continued fraction

Definition

Let [a0;a1,a2,… ][a_0; a_1, a_2, \dots] be a simple continued fraction. For each nn such that ana_n exists, the nn-th convergent is the rational number pnqn=[a0;a1,…,an]\dfrac{p_n}{q_n} = [a_0; a_1, \dots, a_n]. Here pnp_n and qnq_n are given by p−2=0p_{-2} = 0, p−1=1p_{-1} = 1, q−2=1q_{-2} = 1, q−1=0q_{-1} = 0 and pn=anpn−1+pn−2,qn=anqn−1+qn−2(n≥0).p_n = a_n p_{n-1} + p_{n-2}, \qquad q_n = a_n q_{n-1} + q_{n-2} \quad (n \ge 0).

Remarks

The recurrences give pnqn−1−pn−1qn=(−1)n−1p_n q_{n-1} - p_{n-1} q_n = (-1)^{n-1}. So gcd⁡(pn,qn)=1\gcd(p_n, q_n) = 1, and the fraction pn/qnp_n / q_n is in lowest terms. For 2=[1;2,2,… ]\sqrt{2} = [1; 2, 2, \dots], the convergents are 1,32,75,1712,4129,…1, \frac{3}{2}, \frac{7}{5}, \frac{17}{12}, \frac{41}{29}, \dots. For a general continued fraction, the same recurrences hold with pn=bnpn−1+anpn−2p_n = b_n p_{n-1} + a_n p_{n-2} and qn=bnqn−1+anqn−2q_n = b_n q_{n-1} + a_n q_{n-2}.

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  1. Continued fraction definition
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Further reading

  1. Web https://en.wikipedia.org/wiki/Simple_continued_fraction, Infinite continued fractions and convergents
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DefinesConvergents of a continued fraction
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