The Continued Fraction Algorithm

Let $x\in\mathbb{R}$ and write

$$x = a_0 + t_0$$

with $a_0\in\mathbb{Z}$ and $0\leq t_0 < 1$. If $t_0\neq 0$, write

$$\frac{1}{t_0} = a_1 + t_1$$

with $a_1\in\mathbb{N}$ and $0\leq t_1 < 1$. Thus $t_0 = \frac{1}{a_1 + t_1}=[0,a_1+t_1]$, which is a (nonintegral) continued fraction expansion of $t_0$. Continue in this manner so long as $t_n\neq 0$ writing

$$\frac{1}{t_n} = a_{n+1} + t_{n+1}$$

with $a_{n+1}\in\mathbb{N}$ and $0\leq t_{n+1}<1$. This process, which associates to a real number $x$ the sequence of integers $a_0, a_1, a_2, \ldots$, is called the continued fraction algorithm.

Example 1.1   Let $x=\frac{8}{3}$. Then $x=2+\frac{2}{3}$, so $a_0=2$ and $t_0=\frac{2}{3}$. Then $\frac{1}{t_0}=\frac{3}{2} = 1+\frac{1}{2}$, so $a_1=1$ and $t_1=\frac{1}{2}$. Then $\frac{1}{t_1} = 2$, so $a_2=2$, $t_2=0$, and the sequence terminates. Notice that

$$\frac{8}{3} = [2,1,2],$$

so the continued fraction algorithm produces the continued fraction of $\frac{8}{3}$.

Proposition 1.2   For every $n$ such that $a_n$ is defined, we have

$$x = [a_0, a_1, \ldots, a_{n}+t_n],$$

and if $t_{n}\neq 0$ then $x = [a_0, a_1, \ldots, a_{n}, \frac{1}{t_n}].$

Proof. Use induction. The statements are both true when $n=0$. If the second statement is true for $n-1$, then

$$x = [a_0,a_1, \ldots, a_{n-1},\frac{1}{t_{n-1}}] =[a_0,a_1, \ldots, a_{n-1},a_n + t_n] =[a_0,a_1, \ldots, a_{n-1},a_n, \frac{1}{t_n}].$$

Similarly, the first statement is true for $n$ if it is true for $n-1$.

$\qedsymbol$

Example 1.3   Let $x = \frac{1+\sqrt{5}}{2}.$ Then

$$x = 1 + \frac{-1 + \sqrt{5}}{2},$$

so $a_0 = 1$ and $t_0 = \frac{-1+\sqrt{5}}{2}$. We have

$$\frac{1}{t_0} = \frac{2}{-1+\sqrt{5}} = \frac{-2-2\sqrt{5}}{-4} = \frac{1+\sqrt{5}}{2}$$

so again $a_1=1$ and $t_1 = \frac{-1+\sqrt{5}}{2}$. Likewise, $a_n = 1$ for all $n$. Does the following crazy-looking equality makes sense??

$$\frac{1+\sqrt{5}}{2} = 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \cdots}}}}}$$

Example 1.4   Next suppose $x = e$. Then

$$a_0, a_1, a_2, \ldots = 2,1,2,1,1,4,1,1,6,1,1,8,1,1,\ldots$$

? contfrac(exp(1))
%1 = [2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 
       12, 1, 1, 14, 1, 1, 16, 1, 1, 18, 1, 1, 20, 2]
? \\ to get more terms, increase the real precision:
? \p60
? contfrac(exp(1),[])
%12 = [2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 
   12, 1, 1, 14, 1, 1, 16, 1, 1, 18, 1, 1, 20, 1, 1, 22, 1, 
   1, 24, 1, 1, 26, 1, 1, 28, 1, 1, 30, 1, 1, 32, 1, 1, 34, 
   1, 1, 36, 1, 1, 38, 1, 1, 40, 1, 1, 42, 2]

The following program uses a proposition we proved yesterday to compute the partial convergents of a continued fraction:

{convergents(v)=
    local(pp,qq,p,q,tp,tq,answer);
    pp=1; qq=0; p=v[1]; q=1;    \\ pp is p_{n-1} and p is p_n.
    answer = vector(length(v)); \\ put answer in this vector
    answer[1] = p/q;
    for(n=2,length(v),
       tp=p; tq=q; p=v[n]*p+pp; q=v[n]*q+qq; pp=tp; qq=tq;
       answer[n] = p/q;
    );
    return(answer);
}

Let's try this with $\pi$:

? contfrac(Pi)
%26 = [3, 7, 15, 1, 292, 1, 1, ...]
? convergents([3,7,15])
%27 = [3, 22/7, 333/106]
? convergents([3,7,15,1,292])
%28 = [3, 22/7, 333/106, 355/113, 103993/33102]
? %[5]*1.0
%29 = 3.1415926530119026040...
? % - Pi
%30 = -0.000000000577890634...