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Every formula that involves $\pi$ has an underlying trigonometric interpretation but it is not usually obvious.

I wonder if there are any formula like the Gaussian integral $f_{n}(x)=\sqrt{\frac{1}{2\pi n^2}} \exp\left(\frac{-x^2}{2n^2}\right)$ which contains $\pi$ but does not have any obvious trigonometric interpretation.

George V. Williams
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Fraïssé
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2 Answers2

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The Basel problem was one of the earlier examples of this. $$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.$$

angryavian
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  • Not quite obvious, but many of the solutions of the Basel problem do involve trigonometric functions, e.g. Euler's original solution using an infinite product representation of $\sin(x)/x$. Does that count as "trigonometric interpretation"? – Robert Israel Jan 01 '15 at 21:40
  • Also $\sum n^{-4}=\pi^4/90$, $\sum n^{-6}=\pi^6/945$, .... – Gerry Myerson Jan 02 '15 at 00:51
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The number of squarefree numbers (numbers not divisible by any square number other than 1) less than $n$ is asymptotically equal to $6n/\pi^2$.

The number of pairs $a,b$ with $0\le a\le n$, $0\le b\le n$, $\gcd(a,b)=1$ is asymptotically $6n^2/\pi^2$.

I wrote an expository paper on the number of three-term geometric progressions with all terms less than $n$, and the answer came out to involve $\pi$. I'll come back with more details.

EDIT: Here's the paper. The number of three-term geometric progressions $a,b,c$ with $1\le a\le b\le c\le n$ is asymptotically $(3/\pi^2)n\log n$.

Gerry Myerson
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