Pi square is nearly 10

Tau day 2026: Pi square is nearly 10

In the US and countries with a similar date format, today is the τ day (τ = 2π). I still think that τr and τr²/2 are better formulas than 2πr and πr², since they match the mv and mv²/2 ones (and many other reasons). But that ship has sailed, so τ is relegated to just being the double of π.

Well, addition is trivial, but did you know that π² ≈ 10 and π² ≈ g (where g is the acceleration due to gravity at sea level on Earth)? How did we get to these coincidences? For today, let's just check the first fact.

We have π² ≈ 9.8696 which is close to 10 (for certain definitions of 10). Let's start with that famous formula where π² 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 ζ 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 4/(4n²) ≤ 4/(4n² − 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 makes the last sum telescope, so we have:

[ \zeta(2) \le 1 + \frac{2}{3} = \frac{5}{3} ]

This is why we get π² ≤ 6 × 5/3 = 10.

The difference between π² and 10

Next, looking at the difference between π² 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 O(n⁻⁴), which means that they tend to 0 pretty fast. The first few values are:

• 1/60
• 1/315
• 1/1008

Since the error between π² and 10 is 6δ, summing these terms and multiplying by 6 we get 0.125. So, we could approximate that π² is almost 10, up to an eighth of a unit.

Is this useful?

I recently had to determine very fast if the perimeter of a circle of radius 1/10 is above 1 or not. Knowing that 10 is approximately π² told me that this is approximately 2/π 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 π ≤ 4 so 2π ≤ 10, that is the perimeter is already less than 1. Probably a better example would be if we have to compute log π very fast (where the logarithm is in base 10). Since π² is approximately 10, the log in question is slightly less than 0.5.

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