Tau day 2026: Pi square is nearly 10 · mihai.page
In the US and countries with a similar date format, today is the \(\tau\) day<br>(\(\tau = 2 \pi\)). I still think that \(\tau r\) and \(\frac{\tau r^2}{2}\) are<br>better formulas than \(2\pi r\) and \(\pi r^2\), since they match the \(mv\) and<br>\(\frac{mv^2}{2}\) ones (and many other reasons). But that ship has sailed, so<br>\(\tau\) is relegating to just being the double of \(\pi\).
Well, addition is trivial, but did you know that \(\pi^2 \approx 10\) and \(\pi^2<br>\approx g\) (where \(g\) is the acceleration due to gravity at sea level on<br>Earth)? How did we get to these coincidences?
For today, let’ just check the first fact. We have \(\pi^2 \approx 9.8696\)<br>which is close to 10 (for certain definitions of 10).
Let’s start with that famous formula where \(\pi^2\) shows up,<br>the Basel problem: what is the value of the sum of the reciprocal of<br>the squares of natural numbers? We know the answer from Euler:
\[<br>\sum_{n=1}^{\infty}{\frac{1}{n^2}} = \frac{\pi^2}{6}<br>\]
That is
\[<br>\pi^2 = 6\zeta(2)<br>\]
where \(\zeta\) is the Riemann zeta function. In our case we can do this<br>manipulation
\[<br>\zeta(2) = \sum_{n=1}^{\infty}{\frac{1}{n^2}} = 1 + \sum_{n=2}^{\infty}{\frac{4}{4n^2}}<br>\]
But \(\frac{4}{4n^2} \le \frac{4}{4n^2 - 1}\) and the denominator here is a<br>difference of squares. That is
\[<br>\frac{4}{4n^2-1} = \frac{4}{(2n - 1)(2n + 1)} = \frac{2}{2n - 1} - \frac{2}{2n + 1}<br>\]
This make the last sum telescope, so we have
\[<br>\zeta(2) \le 1 + \frac{2}{3} = \frac{5}{3}<br>\]
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:
\[<br>\delta = \frac{5}{3} - \zeta(2) = \sum_{n=2}^{\infty}{\left(\frac{4}{4n^2 - 1} - \frac{1}{n^2}\right)} = \sum_{n=2}^{\infty}{\frac{1}{n^2(4n^2 - 1)}}<br>\]
The terms in the sum are of the order \(\mathcal{O}\left(n^{-4}\right)\), which<br>means that they tend to 0 pretty fast. The first few values are<br>\(\frac{1}{60}\), \(\frac{1}{315}\) and \(\frac{1}{1008}\). Since the error between<br>\(\pi^2\) and \(10\) is \(6\delta\), summing these terms and multiplying by 6 we get<br>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<br>circle of radius \(\frac{1}{10}\) is above 1 or not. Knowing that \(10\) is<br>approximately \(\pi^2\) told me that this is approximately \(\frac{2}{\pi}\)<br>without actually doing any math.
In the future, we will look at the other approximation, which might be just a<br>coincidence? Let’s see.
PS: I just realized that the example above is trivial. We know \(\pi \le 4\) so<br>\(2 \pi \le 10\), that is the perimeter is aready less than 1. Probably a better<br>example would be if we have to compute \(\log \pi\) very fast (where the<br>logarithm is in base 10). Since \(\pi^2\) is approximately 10, the log in<br>question is slightly less than 0.5.
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