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!

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $1000$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @Lin: $$\pi=[3;7,15,1,\color{red}{292},1,\ldots]$$ $\endgroup$ – Jack D'Aurizio Dec 11 '18 at 18:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.