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If we consider continued fraction expansion as a
machine for producing digits, it is natural to
compare it with decimal expansion.

Recall
\[ N(x) = \left\lfloor \frac{1}{x} \right\rfloor,
\qquad Tx = \frac{1}{x} - \left\lfloor \frac{1}{x}
\right\rfloor \]
for continued fractions. In comparison, we have
\[ D x = \left\lfloor 10x \right\rfloor, \qquad Sx
= 10x - \left\lfloor 10x \right\rfloor \]
($0 < x \le 1$) for decimal expansions.

\begin{remark*}
  If we put the uniform density on $(0, 1]$, and
  $X$ chosen at random from $(0, 1]$ then $DX$ has
  the same uniform distribution.

  Indeed, $X, SX, S^2X$ all have the same
  distribution and so $DX, DSX, DS^2X, \ldots$ are
  IID random variables.
\end{remark*}

\vspace{-1em}
\begin{flashcard}[gauss-cfe-density]
Gauss observed that if take the density
\[ f(x) = \cloze{\frac{1}{\log 2} \frac{1}{1 + x}} \]
then $X$ and $TX$ have the same distribution.

\begin{proof}
  \cloze{Let $X$ have density function $f(x) =
  \frac{1}{\log 2} \frac{1}{1 + x}$. Then
  \begin{align*}
    \PP(Tx \le t)
    &= \PP \left( \frac{1}{x} - \left\lfloor
    \frac{1}{x} \right\rfloor \le t \right) \\
    &= \sum_{n = 1}^\infty \PP \left( 0 \le
    \frac{1}{x} - n \le t \right) \\
    &= \sum_{n = 1}^\infty \int_{(t +
    n)^{-1}}^{n^{-1}} f(x) \dd x \\
    &= \sum_{n = 1}^\infty \int_{\frac{1}{n +
    t}}^{\frac{1}{n}} \frac{1}{\log 2} \frac{1}{x
    + 1} \dd x \\
    &= \frac{1}{\log 2} \sum_{n = 1}^\infty [\log
    (1 + x)]_{\frac{1}{t + n}}^{\frac{1}{n}} \\
    &= \frac{1}{\log 2} \sum_{n = 1}^\infty \log
    \left( 1 + \frac{1}{n} \right) - \log \left( 1
    + \frac{1}{t + n}\right) \\
    &= \frac{1}{\log 2} \sum_{n = 1}^\infty
    (\log(n + 1) - \log(n)) - (\log(n + t + 1) -
    \log(n + t)) \\
    &= \frac{1}{\log 2} \lim_{N \to \infty}
    \sum_{n = 1}^N (\log(n + 1) - \log(n) -
    \log(n + t + 1) + \log(n + 1)) \\
    &= \frac{1}{\log 2} \lim_{N \to \infty}
    (\log(N + 1) - \log(N + t + 1) + \log(1 + t))
    \\
    &= \frac{1}{\log 2} \lim_{N \to \infty}
    \frac{\log(N + 1)}{(N + t + 1)} + \log(1 + t)
    \\
    &= \frac{1}{\log 2} \log(1 + t)
  \end{align*}
  so $TX$ has density
  \[ \frac{1}{\log 2} \frac{1}{t + 1} . \qedhere
  \]}
\end{proof} 
\end{flashcard}

Thus if we use the density function
\[ f(t) = \frac{1}{\log 2} \frac{1}{1 + t} \]
then
\begin{align*}
  \PP(T^n X = j)
  &= \PP(X = j) \\
  &= \frac{1}{\log 2} \int_{\frac{1}{j +
  1}}^{\frac{1}{j}} \frac{1}{1 + x} \dd x \\
  &= \frac{1}{\log 2} [\log(1 + x)]_{\frac{1}{j +
  1}}^{\frac{1}{j}} \\
  &= \frac{1}{\log 2} \left[ -\log \left( \frac{j
  + 2}{j + 1} \right) + \log \left( \frac{j +
  1}{j} \right) \right] \\
  &= \frac{1}{\log 2} \log \left( \frac{(j +
  1)^2}{j(j + 2)} \right) \\
  &= \frac{1}{\log 2} \log \left( 1 + \frac{1}{j(j
  + 2)} \right) \\
  &\approx \frac{1}{\log 2} \frac{1}{j(j + 1)} \\
  &\approx \frac{1}{\log 2} \frac{1}{j^2}
\end{align*}
where the approximations are assuming $j$ is
large (using the fact that $\log(1 + x) \approx x$
for $x$ small.

\subsubsection*{* Non-examinable material}

Using the above observation and some more work,
one can prove that if $a_j$ is the $j$-th term of
the continued fraction expansion, then
\[ \log(a_1 a_2 \cdots a_n)^{1/n} \]
converges as $n \to \infty$ almost everywhere.

Uses ergodic theory -- see \courseref[Probability
and Measure]{PM}.

** This is the end of the non-examinable comments.

\subsubsection*{What about convergence}

We show that if $a_0 \in \ZZ$, $a_j \in \ZZ$, $a_j
\ge 1$, then writing 
\[ \frac{p_n}{q_n} = a_0 + \frac{1}{a_1 +
\frac{1}{a_2 + \frac{1}{\ddots + \frac{1}{a_n}}}} \]
we have that $\frac{p_n}{q_n}$ converges as $n \to
\infty$.

It is then easy to show that for the continued
fraction expansion algorithm applied to $x$ will
yield a sequence which converges to $x$.

Note we shall always take $p_n, q_n$ coprime.
$\frac{p_n}{q_n}$ is called the $n$-th convergent.

Our discussion starts from the fact that we
usually produce continuous fractions downwards but
we evaluate them upwards.
\[ \frac{r_k}{s_k} = a_k + \frac{1}{\frac{r_{k +
1}}{s_{k + 1}}} = a_k + \frac{s_{k + 1}}{r_{k +
1}} = \frac{a_k r_{k + 1} + s_{k + 1}}{r_{k + 1}} \]
and since $r_{k + 1}, s_{k + 1}$ are coprime, we
have
\begin{align*}
  r_k
  &= a_k r_{k + 1} + s_{k + 1} \\
  s_k
  &= r_{k + 1}
\end{align*}
We write our result in matrix form:
\[ \begin{pmatrix}
  r_k \\
  s_k
\end{pmatrix}
= \begin{pmatrix}
  a_k & 1 \\
  1 & 0
\end{pmatrix}
\begin{pmatrix}
  r_{k + 1} \\
  s_{k + 1}
\end{pmatrix} \]
with
\[ \begin{pmatrix}
  r_n \\
  s_n
\end{pmatrix} =
\begin{pmatrix}
  a_n \\
  1
\end{pmatrix} .\]
We find that
\[ \begin{pmatrix}
  p_n \\
  q_n
\end{pmatrix}
= \begin{pmatrix}
  r_0 \\
  s_0
\end{pmatrix} =
\begin{pmatrix}
  a_0 & 1 \\
  1 & 0
\end{pmatrix}
\begin{pmatrix}
  a_1 & 1 \\
  1 & 0
\end{pmatrix} \cdots
\begin{pmatrix}
  a_{n - 1} & 1 \\
  1 & 0
\end{pmatrix}
\begin{pmatrix}
  a_n \\
  1
\end{pmatrix} .\]
\begin{align*}
  \begin{pmatrix}
    p_{n - 1} \\
    q_{n - 1}
  \end{pmatrix}
  &= \begin{pmatrix}
    a_0 & 1 \\
    1 & 0
  \end{pmatrix}
  \begin{pmatrix}
    a_1 & 1 \\
    1 & 0
  \end{pmatrix} \cdots
  \begin{pmatrix}
    a_{n - 2} & 1 \\
    1 & 0
  \end{pmatrix}
  \begin{pmatrix}
    a_{n - 1} \\
    1
  \end{pmatrix} \\
  &= \begin{pmatrix}
    a_0 & 1 \\
    1 & 0
  \end{pmatrix}
  \begin{pmatrix}
    a_1 & 1 \\
    1 & 0
  \end{pmatrix} \cdots
  \begin{pmatrix}
    a_{n - 2} & 1 \\
    1 & 0
  \end{pmatrix}
  \begin{pmatrix}
    a_{n - 1} & 1 \\
    1 & 0
  \end{pmatrix}
  \begin{pmatrix}
    1 \\
    0
  \end{pmatrix} \\
  \begin{pmatrix}
    p_n & p_{n - 1} \\
    q_n & q_{n - 1}
  \end{pmatrix}
  &= \begin{pmatrix}
    a_0 & 1 \\
    1 & 0
  \end{pmatrix}
  \begin{pmatrix}
    a_1 & 1 \\
    1 & 0
  \end{pmatrix}
  \cdots
  \begin{pmatrix}
    a_n & 1 \\
    1 & 0
  \end{pmatrix}
\end{align*}
Thus
\[ \begin{pmatrix}
  p_n & p_{n - 1} \\
  q_n & q_{n - 1}
\end{pmatrix} =
\begin{pmatrix}
  p_{n - 1} & p_{n - 2} \\
  q_{n - 1} q_{n - 2}
\end{pmatrix}
\begin{pmatrix}
  a_n & 1 \\
  1 & 0
\end{pmatrix} \]
so
\begin{align*}
  p_n
  &= a_n p_{n - 1} + p_{n - 2} \\
  q_N
  &= a_n q_{n - 1} + q_{n - 2}
\end{align*}

\begin{remark*}
  $q_0 = 1$, $q_1 \ge 1$ and $q_n \ge q_{n - 1} +
  q_{n - 2}$ (since $a_n \ge 1$, $n \ge 1$). So
  $q_n \to \infty$.
\end{remark*}
