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:
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.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!
Best Answer
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 $1000$.