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Given that the continued fraction expansion of $\pi$ is $[3, 7, 15, 1, 292, ...]$. Prove that $\pi ≈ [3, 7, 15, 1]= 355/113$ is the best approximation of $\pi$ with a 3-digit denominator.

From the following theorem:

  1. Any convergent $x/y$ of an irrational number $\alpha$ satisfies $\lvert\frac{x}{y} - \alpha\rvert < \frac{1}{y^2}$. $x$, $y$ are positive integers and gcd(x, y) = 1.

  2. If $\alpha$ is an irrational number and x/y satisfies $\lvert \frac{x}{y} - \alpha\rvert < \frac{1}{2y^2}$, then $\frac{x}{y}$ is a convergent of $\alpha$.

Have $\lvert\frac{355}{113} - \pi\rvert < \frac{1}{113^2}$.

I tried to consider $n = \frac{a}{b}$ where $\lvert \frac{a}{b} - \pi \rvert < \lvert \frac{355}{113} - \pi \rvert$ and hope to raise a contradiction.

I considered the cases when $\frac{a}{b} - \pi$ is positive and negative but wasn't able to conclude anything from the results.

Could someone please point me in the right direction? Thank you!

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  • $\begingroup$ How do you define best approximation? I guess through $|q\alpha-p|$ or $|q^2\alpha-pq|$, but it is not stated anywhere, so the initial question is not really meaningful at the moment. $\endgroup$ – Jack D'Aurizio Dec 10 '18 at 22:39
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    $\begingroup$ It's not stated in the original problem but I'm assuming it means an approximation with the smallest error? $\endgroup$ – Lin Dec 10 '18 at 22:52
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    $\begingroup$ Hint: Contrapositive to 2 is: If $\frac{x}{y}$ is not a convergent of $\alpha$ then $|\frac{x}{y} - \alpha| \ge \frac{1}{2y^2}$. $\endgroup$ – Daniel Schepler Dec 10 '18 at 23:27
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The approximation $\pi\approx\frac{355}{113}$ is not only good, it is exceptionally good, since $$\left|\pi - \frac{355}{113}\right|<\frac{1}{\color{red}{291}\cdot 113^2}<\frac{1}{2\cdot 1000^2}.$$ In particular if some approximation $\frac{p}{q}$ with $q\leq 1000$ achieves an absolute error which is less than the absolute error achieved by $\frac{355}{113}$, such approximation is a convergent of the continued fraction of $\pi$.
$\frac{355}{113}$ is a convergent, the next convergent is $\frac{103993}{33102}$, with a denominator much larger than $1000$.
It follows that the Chinese approximation $\pi\approx\frac{355}{113}$ is the best rational approximation (in the absolute error sense) of $\pi$ with a denominator less than $1000$.

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  • $\begingroup$ Thanks! So I tried to compare $\lvert \frac{355}{113} - \pi \rvert$ and the difference between the 3rd and 4th convergents and got $\lvert \frac{355}{113} - \pi \rvert<\frac{1}{113*33102}$, which eventually leads to $\lvert \frac{355}{113} - \pi \rvert<\frac{1}{2*1000^2}$. But not sure if that's how you got 291? $\endgroup$ – Lin Dec 11 '18 at 14:17
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    $\begingroup$ @Lin: $$\pi=[3;7,15,1,\color{red}{292},1,\ldots]$$ $\endgroup$ – Jack D'Aurizio Dec 11 '18 at 18:07

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