In the US and countries with a similar date format, today is the \(\tau\) day (\(\tau = 2 \pi\)). I still think that \(\tau r\) and \(\frac{\tau r^2}{2}\) are better formulas than \(2\pi r\) and \(\pi r^2\), since they match the \(mv\) and \(\frac{mv^2}{2}\) ones (and many other reasons). But that ship has sailed, so \(\tau\) is relegating to just being the double of \(\pi\).
Well, addition is trivial, but did you know that \(\pi^2 \approx 10\) and \(\pi^2 \approx g\) (where \(g\) is the acceleration due to gravity at sea level on Earth)? How did we get to these coincidences?
For today, let’ just check the first fact. We have \(\pi^2 \approx 9.8696\) which is close to 10 (for certain definitions of 10).
Let’s start with that famous formula where \(\pi^2\) shows up, the Basel problem: what is the value of the sum of the reciprocal of the squares of natural numbers? We know the answer from Euler:
\[ \sum_{n=1}^{\infty}{\frac{1}{n^2}} = \frac{\pi^2}{6} \]
That is
\[ \pi^2 = 6\zeta(2) \]
where \(\zeta\) is the Riemann zeta function. In our case we can do this manipulation
\[ \zeta(2) = \sum_{n=1}^{\infty}{\frac{1}{n^2}} = 1 + \sum_{n=2}^{\infty}{\frac{4}{4n^2}} \]
But \(\frac{4}{4n^2} \le \frac{4}{4n^2 - 1}\) and the denominator here is a difference of squares. That is
\[ \frac{4}{4n^2-1} = \frac{4}{(2n - 1)(2n + 1)} = \frac{2}{2n - 1} - \frac{2}{2n + 1} \]
This make the last sum telescope, so we have
\[ \zeta(2) \le 1 + \frac{2}{3} = \frac{5}{3} \]
This is why we get \(\pi^2 \le 6\times\frac{5}{3} = 10\).
Next, looking at the difference between \(\pi^2\) and \(10\) we have:
\[ \delta = \frac{5}{3} - \zeta(2) = \sum_{n=2}^{\infty}{\left(\frac{4}{4n^2 - 1} - \frac{1}{n^2}\right)} = \sum_{n=2}^{\infty}{\frac{1}{n^2(4n^2 - 1)}} \]
The terms in the sum are of the order \(\mathcal{O}\left(n^{-4}\right)\), which means that they tend to 0 pretty fast. The first few values are \(\frac{1}{60}\), \(\frac{1}{315}\) and \(\frac{1}{1008}\). Since the error between \(\pi^2\) and \(10\) is \(6\delta\), summing these terms and multiplying by 6 we get 0.125.
So, we could approximate that \(\pi^2\) is almost \(10\), up to an eight of a unit.
Is this useful? I recently had to determine very fast if the perimeter of a circle of radius \(\frac{1}{10}\) is above 1 or not. Knowing that \(10\) is approximately \(\pi^2\) told me that this is approximately \(\frac{2}{\pi}\) without actually doing any math.
In the future, we will look at the other approximation, which might be just a coincidence? Let’s see.
PS: I just realized that the example above is trivial. We know \(\pi \le 4\) so \(2 \pi \le 10\), that is the perimeter is aready less than 1. Probably a better example would be if we have to compute \(\log \pi\) very fast (where the logarithm is in base 10). Since \(\pi^2\) is approximately 10, the log in question is slightly less than 0.5.