Regular Continued Fractions and Diophantine Approximation — Examples

Direct calculation gives. [given, algebra] $[1;1,1,2]=1+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}=1+\frac{1}{1+\frac{2}{3}}=1+\frac{3}{5}=\frac{8}{5}\mathrm{.}$

Likewise. [F1, step 1.1, algebra] $[1;1,1,1,1]=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1}}}}=1+\frac{3}{5}=\frac{8}{5},$ so the same rational has two finite expansions. The last digit of $[1;1,1,2]$ is $2$, so [F1] identifies it as the normalized one and shows there are no further finite expansions.

The Euclidean quotient digits are $3,2,1,3$, so [F1] gives. [F1, given, algebra] $\frac{37}{11}=[3;2,1,3]\mathrm{.}$ Its convergents are $3,3+\frac{1}{2}=\frac{7}{2},3+\frac{1}{2+\frac{1}{1}}=\frac{10}{3},3+\frac{1}{2+\frac{1}{1+\frac{1}{3}}}=\frac{37}{11}\mathrm{.}$

The penultimate convergent is $10\mathrm{/}3$, and. [F2, step 1.1, algebra] $37\cdot 3-11\cdot 10=111-110=1.$ So $1$ is an explicit integer linear combination of $37$ and $11$, which is exactly the Bezout identity for $\gcd (37,11)=1$ in the sense of [F2].

Since $1<\sqrt{2}<2$, [F3] gives $a_0=1$. Then [F3, F4, given, algebra] ${\alpha }_1=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1,$ so $2<{\alpha }_1<3$ and $a_1=2$. Moreover $\frac{1}{{\alpha }_1-2}=\frac{1}{\sqrt{2}-1}={\alpha }_1,$ so every later complete quotient is again ${\alpha }_1$. Thus the algorithm never terminates and produces the digits $1,2,2,\dots$; by [F4] its continued fraction converges to the original number. Hence $\sqrt{2}=[1;\bar{2}]\mathrm{.}$

Applying [F5] to the digits from step 1.1 gives [F5, step 1.1, algebra] $(p_0,q_0)=(1,1),(p_1,q_1)=(3,2),(p_2,q_2)=(7,5),(p_3,q_3)=(17,12),$ so the convergents begin $1,\frac{3}{2},\frac{7}{5},\frac{17}{12}\mathrm{.}$ The same recurrence gives $q_4=29$. For the displayed pairs one checks $3\cdot 1-1\cdot 2=1,7\cdot 2-3\cdot 5=-1,17\cdot 5-7\cdot 12=1,$ exactly as [F1] predicts.

The error formula [F2] now gives [F2, step 2.1, algebra] $\mid \sqrt{2}-\frac{3}{2}\mid =\frac{3-2\sqrt{2}}{2}<\frac{1}{10},\mid \sqrt{2}-\frac{7}{5}\mid =\frac{5\sqrt{2}-7}{5}<\frac{1}{60},$ and similarly $\mid \sqrt{2}-\frac{17}{12}\mid =\frac{17-12\sqrt{2}}{12}<\frac{1}{348}\mathrm{.}$ So the concrete convergents alternate around $\sqrt{2}$ with the expected quality of approximation.

Since $3<\sqrt{14}<4$, the first digit is $a_0=3$. Then [given, algebra] ${\alpha }_1=\frac{1}{\sqrt{14}-3}=\frac{\sqrt{14}+3}{5},{\alpha }_2=\frac{1}{{\alpha }_1-1}=\frac{\sqrt{14}+2}{2},$ ${\alpha }_3=\frac{1}{{\alpha }_2-2}=\frac{\sqrt{14}+2}{5},{\alpha }_4=\frac{1}{{\alpha }_3-1}=\sqrt{14}+3,$ and finally ${\alpha }_5=\frac{1}{{\alpha }_4-6}=\frac{1}{\sqrt{14}-3}={\alpha }_1\mathrm{.}$ So the digits from ${\alpha }_1$ onward are $1,2,1,6$ and then repeat.

Step 1.1 shows that ${\alpha }_5={\alpha }_1$. By the determinism in [F1], the same four digits $1,2,1,6$ therefore repeat from ${\alpha }_1$ onward. Hence $\sqrt{14}=[3;\overline{1,2,1,6}]\mathrm{.}$

Let $x_m=[1;\underset{m\text{ further digits}}{\underbrace{1,\dots ,1}}]$. By [F2], $x_m\to x$ and $x\ge 1$, while [F3] gives $x_{m+1}=1+1\mathrm{/}{x}_m$ and $x_m\ge 1$. Hence $\mid \frac{1}{x_m}-\frac{1}{x}\mid =\frac{\mathrm{\mid }{x}_m-x\mathrm{\mid }}{x_{m}x}\le \mathrm{\mid }{x}_m-x\mathrm{\mid }⟶0,$ so taking limits in the recursion gives $x=1+\frac{1}{x}\mathrm{.}$ Multiplication by $x$ now gives $x^2-x-1=0,$ whose positive solution is $x=\frac{1+\sqrt{5}}{2}\mathrm{.}$

The continued fraction is purely periodic, so [F1] says its value is a quadratic irrational. Step 1.1 exhibits the quadratic equation explicitly, and its positive root is the golden ratio $\phi$.

Direct decimal comparison gives. [given, algebra] $\mid \pi -\frac{22}{7}\mid \approx 1.264\times {10}^{-3},\mid \pi -\frac{333}{106}\mid \approx 8.322\times {10}^{-5},$ $\mid \pi -\frac{355}{113}\mid \approx 2.668\times {10}^{-7}<\frac{1}{2\cdot {113}^2}\approx 3.916\times {10}^{-5}\mathrm{.}$ Moreover $355=3\cdot 113+16$ and $113=7\cdot 16+1$, so $\gcd (355,113)=1$. Thus $355\mathrm{/}113$ is reduced and satisfies Legendre's criterion.

By [L1] and [F1], the fraction $355\mathrm{/}113$ is a convergent of $\pi$. Then [F2] says no. [L1, F1, F2, step 1.1] rational with denominator at most $113$ approximates $\pi$ more closely. Since $22\mathrm{/}7$ and $333\mathrm{/}106$ have denominators $7$ and $106$, neither can beat $355\mathrm{/}113$.

The direct errors from step 1.1 also show that $333\mathrm{/}106$ improves on. [step 1.1, algebra] $22\mathrm{/}7$, but that both are far worse than $355\mathrm{/}113$.

The first few convergents of $[1;\bar{2}]=\sqrt{2}$ are. [given, algebra] $1,\frac{3}{2},\frac{7}{5},\frac{17}{12},$ so $10\mathrm{/}7$ is not a convergent of $\sqrt{2}$.

Nevertheless. [given, algebra] $\frac{10}{7}-\sqrt{2}=\frac{100-98}{7(10+7\sqrt{2})}=\frac{2}{7(10+7\sqrt{2})}<\frac{3}{4\cdot 7^2}\mathrm{.}$ Thus $10\mathrm{/}7$ satisfies the displayed $3\mathrm{/}(4s^2)$ bound.

Step 1.1 and step 1.2 together refute the statement. In the light of [F1],. [F1, F2, step 1.1, step 1.2] this shows that the sharp constant $1\mathrm{/}2$ in Legendre's criterion cannot simply be replaced by the larger constant $3\mathrm{/}4$.

Since $-2<-\sqrt{2}<-1$, the first digit is $a_0=-2$. Then. [F1, given, algebra] ${\alpha }_1=\frac{1}{-\sqrt{2}+2}=\frac{2+\sqrt{2}}{2},{\alpha }_2=\frac{1}{{\alpha }_1-1}=\sqrt{2},$ so $a_1=1$ and $a_2=1$.

One more step gives. [F1, F2, step 1.1, algebra] ${\alpha }_3=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1,$ so every later digit is $2$. Hence the digit string is $[-2;1,1,\bar{2}],$ and [F2] identifies its value with $-\sqrt{2}$. In particular the negative sign is absorbed entirely into the first digit, while every later digit stays positive as required by [F1].