Find two rational numbers $\frac ab$ satisfying $$\mid \pi - \frac ab\mid < \frac{1}{\sqrt 5 b^2}.$$
I dont know how to find such rationals. IS there a method? or trial?
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2That's Hurwitz's theorem. For a solution, look up continued fractions. – lhf Oct 29 '16 at 16:14
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2What about $\frac{a}{b}=\frac{355}{113}$, that is a well-known convergent of the continued fraction of $\pi$? – Jack D'Aurizio Oct 29 '16 at 16:17
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1Also the even better know approximation $\tfrac{a}{b}=\tfrac{22}{7}$ for $\pi$ works. – Servaes Oct 29 '16 at 16:25
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@Servaes: that is the convergent before $\frac{355}{113}$ :D – Jack D'Aurizio Oct 29 '16 at 17:57
2 Answers
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See theorem 20 on page 31 in Khinchin's little book, out of three consecutive convergents to any fixed irrational number, at least one satisfies $|\mbox{error}| < 1 / q^2 \sqrt 5.$
$$ \pi : 3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, 1, 84, $$
$$ \begin{array}{cccccccccccc} & & 3 & & 7 & & 15 & & 1 & & 292 & \\ \frac{0}{1} & \frac{1}{0} & & \frac{3}{1} & & \frac{22}{7} & & \frac{333}{106} & & \frac{355}{113} & & \frac{103993}{33102} \end{array} $$
3/1: $1/ \sqrt 5 \approx 0.447, |\mbox{error}| \approx 0.14159$ is smaller
22/7: $1/(49 \sqrt 5) \approx 0.0091268, |\mbox{error}| \approx 0.00126$ is smaller
333/106: $1/(11236 \sqrt 5) \approx 0.0000398, |\mbox{error}| \approx $0.0000832$ is BIGGER
355/113: $1/(12769 \sqrt 5) \approx 0.0000350, |\mbox{error}| \approx $0.000000266$ is smaller
103993/33102: $1/(1095742404 \sqrt 5) \approx 4.08 \cdot 10^{-10},|\mbox{error}| \approx 5.7789 \cdot 10^{-10}$ is BIGGER
The better known result is simply that, if the next digit $a_{k+1} \geq 3,$ then the inequality holds because $$ \left| \alpha - \frac{p_k}{q_k} \right| \leq \frac{1}{q_k^2 \; a_{k+1}} $$ is first written down in the proof of Theorem 22, bottom of page 35. So, except for the pretend convergent $1/0$ that just starts the setup, we see how the convergents just before the digits $7,15,292$ all satisfy the desired inequality, and the one just before 292 is the best ( ratio of error to $1/q^2$).
Will Jagy
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The usual way is to get the continued fraction for $\pi$. See How to find continued fraction of pi . Then work out as many convergents as necessary. There is a recursive method for generating the numerators and denominators of the convergents, also explained in the answers in the link above.
