This series of videos is about periodic continued fractions. Before watching the videos, remember the following facts and definitions.

Definition - A nearly simple continued fraction is

\[[a_0,a_1,a_2,\ldots,a_n,z] = a_0 + \frac{1}{a_1 + \frac{1}{a_2+\frac{1}{\ddots + \frac{1}{z}}}}\]

and if \(\frac{p_n}{q_n} = [a_0,\ldots,a_n]\) and \(\frac{p_{n-1}}{q_{n-1}} = [a_0,\ldots,a_{n-1}]\) are the \(n\)-th and \(n-1\)-st convergents, then we have

\[[a_0,a_1,\ldots,a_n,z] = \frac{p_{n-1} + p_nz}{q_{n-1} + q_n z}\]

In this first video, we will prove the following lemma.

Lemma - If \([a_0,a_1,\ldots,a_n,z] = \frac{A+Bz}{C+Dz}\) then \(\frac{1}{[a_n,a_{n-1},\ldots,a_0,\frac{1}{w}]} = \frac{A+Cw}{B+Dw}\).

Now using the lemma above, we are able to establish that if \(a_0,a_1,\ldots,a_n\) are positive integers such that

\[\alpha = [\overline{a_0,a_1,\ldots,a_n}]\]

and

\[\beta = [\overline{a_n,a_{n-1},\ldots,a_0}]\]

then \(\alpha\) is the root of a quadratic polynomial and the conjugate root \(\alpha'\) satisfies \(\alpha' = \frac{-1}{\beta}\), so that \(-1<\alpha'<0\).

The following video illustrates the process.

In fact, there is a converse to the above result.

Theorem (Converse to above). Suppose that \(\alpha>1\) is the root of a quadratic polynomial \(Ax^2 + Bx + C = 0\) whose conjugate root \(\alpha'\) is such that \(-1<\alpha'<0\). Then the continued fraction of \(\alpha\) is purely periodic.

One of the ingredients in the proof of the converse is to prove that for fixed \(D>0\) there are only finitely many \(P,Q\) such that \(\frac{P+\sqrt{D}}{Q}>1\) and \(-1 <\frac{P-\sqrt{D}}{Q}<0\).

This is discussed briefly in the following video.

Finally, we discuss the continued fraction expansion of \(\sqrt{N}\) in the below video.