# Why does $63725\pi$ give four approximations to $\pi$?

The fraction $\frac{355}{113}$ was first used for approximating $\pi$ by the Chinese mathematician and astronomer Zu Chongzhi in the 5th century (see Milü).

An isolated case?

The almost-integer $113\pi$ takes the approximate value $354.99997$ and the double of $113$ is one unit over a square,

$2\times113=226=1+225=1+15^2=1+(2\times7+1)^2,$

which raises the question whether other numbers of the form $\frac{1+(2n+1)^2}{2}\pi$ may be close to an integer as well.

Four almost-integers in one

Among the first few hundreds of cases, $n=178$ gives the curious

$$\frac{1+(2\times178+1)^2}{2}\pi=63725\pi\approx 200197.991850009574121.$$

The immediate result is $\pi\approx\frac{200198}{63725}$. Moreover, two zeros and the almost $98\approx100$ in the integer part, besides the three zeros in the decimal part, allow for easily building a set of four consecutive approximations to $\pi$ with increasing accuracy.

$$\pi\approx \frac{2\times10^5}{5^2(50^2+7^2)} \approx 3.138$$

$$\pi\approx \frac{2\times(10^5+10^2)}{5^2(50^2+7^2)} \approx 3.14162$$

$$\pi\approx \frac{2\times(10^5+10^2-1)}{5^2(50^2+7^2)} \approx 3.14159278$$

$$\pi\approx \frac{2\times\left(10^5+10^2-1-\frac{163}{4\times10^4}\right)}{5^2(50^2+7^2)} \approx 3.14159265358964$$

These fractions give one, three, six and twelve correct decimal digits of $\pi$ respectively.

Besides its inner structure, the appearance of $163$, the largest Heegner number, seems to increase the likelihood of some underlying reason that explains this number.

In particular,

Q Is there a series related to $63725\pi$ that explains these four approximations?

• What kind of series do you want ? – Peter Apr 3 '16 at 22:14
• The closest convergent is 208341/66317, by the way. ("Closest" in the sense of "closest denominator".) – Patrick Stevens Apr 3 '16 at 22:16
• @Peter I don't know, maybe that $(10^5+10^2-10\times10^{-1}-\frac{163}{4}10^{-4})$ seems to suggest a $10^{3k}$ in the denominator. – Jaume Oliver Lafont Apr 3 '16 at 22:24

The obtained representations partly illustrate the properties of the $\pi$ number, but to a much greater extent - the possibilities of information compression. In this case, an infinite decimal fraction is considered. If the base of the number system and the multiplier are changed, other decompositions may be found.

Therefore, statements about the uniqueness of this case are hardly correct.

• I find your answer useful, so I wonder why that downvote without commenting... Do you have a similar example of several approximations in a row using another number system? – Jaume Oliver Lafont Apr 25 '17 at 16:56
• @JaumeOliverLafont The choice of the base of the number system seems too random for me. So I've compared continued fraction for $\pi$ and for $63725\pi$. Fraction for $\pi$ contains the fragment 2,2,2,2 which can be related with the square roots representations. – Yuri Negometyanov Apr 25 '17 at 17:30

From Kronecker's Approximation Theorem, $\{k+m\pi\mid k,m \in \mathbb{Z}\}$ is dense in $\mathbb{R}$. As a result, any natural number can be approximated by that sequence with any precision, thus it's easy to show that $$\left|\pi - \frac{q}{m}\right|<\frac{1}{m10^t}$$ where $q \in \mathbb{Z}$, $m,t \in \mathbb{N}$. Then $$\left|\frac{1+(2n+1)^2}{2}\pi - \frac{1+(2n+1)^2}{2}\frac{q}{m}\right|<\frac{1+(2n+1)^2}{2}\frac{1}{m10^t} \Leftrightarrow$$ $$\left|\frac{1+(2n+1)^2}{2}\pi - (2n^2+2n+1)\frac{q}{m}\right|<\frac{2n^2+2n+1}{m10^t}$$ The next question is if $m \mid 2n^2+2n+1$ or: $$2n^2+2n+1=m\cdot k$$ which is true when $$n=\frac{\sqrt{2mk-1}-1}{2}$$ is integer, like for the following $(m,k)$ pairs $(1,1)$, $(5,1)$, $(13,1)$, $(25,1)$, $(41,1)$, $(5,5)$, $(17,5)$ and $\color{red}{(63725,1)}$. In this configuration $$\left|\frac{1+(2n+1)^2}{2}\pi - kq\right|<\frac{k}{10^t}$$ which for large $t$ gives a good approximation. This can be exploited to answer whether other numbers of the form $\frac{1+(2n+1)^2}{2}\pi$ may be close to an integer as well.
But then, this property isn't specific to $\pi$ only, for example, from:
$$\left|63725\cdot n - 63725\cdot \sqrt{n^2-1}\right|=\frac{63725}{n+\sqrt{n^2-1}}<\frac{63725}{2\sqrt{n^2-1}}$$ for large $n$, it can be made arbitrarily small. For example: $$63725\sqrt{999999999^2 - 1}\approx 63724999936274.99996813...$$
To answer why $63725\pi$ gives four approximations, it's probably because of $$0<\frac{200198}{63725}-\pi<\frac{1.278931}{10^{7}}<\frac{1}{10^{6}}$$ which is $$\frac{200198}{63725}- \frac{2\cdot(10^5+10^2)}{63725} + \frac{2\cdot(10^5+10^2)}{63725} -\pi < \frac{1}{10^{6}} \Leftrightarrow \\ \frac{2\cdot(10^5+10^2-1)}{63725}- \frac{2\cdot(10^5+10^2)}{63725} + \frac{2\cdot(10^5+10^2)}{63725} -\pi < \frac{1}{10^{6}} \Leftrightarrow \\ \frac{2\cdot(10^5+10^2)}{63725} -\pi < \frac{1}{10^{6}}+\frac{2}{63725}=0.00003238485680...$$ which means both $\frac{2\cdot(10^5+10^2)}{63725}$ and $\frac{200198}{63725}$ are close to each other enough to "support" their approximations of $\pi$. The other values can be similarly checked. Alternatively, apply all of the above to $\{k+m\alpha \mid k,m \in \mathbb{Z}\}$, where $\alpha=63725\pi$-irrational.