Introduction
Hello everyone. Today I’ll be walking through the solution to the the Basel Problem proposed Pietro Mengoli in 1650 and solved by Leonard Euler in 1734. The problem asks for the precise summation of the inverse of squares, as listed below.
\[1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{25} + ...\]
Leonard Euler (1707 - 1783) was a Swiss mathematician, physicist, astronomer, geogra-
pher, logician, and engineer who founded the studies of graph theory and topology and
made discoveries in analytic number theory, complex analysis, and infinitesimal calculus.
The summation was found to be \(\frac{\pi}{6}\), but how did Euler find this?
Proof
To start, let us take the McLaurin Expansion of sin(x).
A Maclaurin Series is a Taylor series expansion of a function about 0. It can be represented as:
\[\sum_{n=1}^{\infty} \frac{f^{(n)}(0)}{n!}(x^{n})\]
When expanding the sum, it looks like this:
\(f(x) = f(0) + f^\prime(0)x + \frac{f^{\prime \prime}(0)}{2!}x^{2} + \frac{f^{\prime \prime \prime}(0)}{3!}x^{3} + ... + \frac{f^{n}(0)}{n!}x^{n} + ...\)
The MacLaurin Expansion of sin(x) is
\[x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...\]
Now, we divide both sides by x
\[\frac{sin(x)}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + ...\]
Let’s set this aside for a moment. Let’s now rewrite sin(x) as a product of its roots. The roots of sin(x) are 0, \(\pi\), \(2\pi\), or basically \(n\pi\).
\[sin(x) = a(x)(x+\pi)(x-\pi)(x+2\pi)(x-2\pi)(x+3\pi)(x-3\pi)...\]
Now isolating \(a\)
\[a = \frac{sin(x)}{x} \cdot \frac{1}{[x^{2}-\pi^{2}][x^{2}-(2\pi)^{2}][x^{2}-(3\pi)^{2}]...}\]
The \(\lim_{x\to\\0} \frac{sin(x)}{x} = 1\), so we can substitute that in
\[a = \frac{1}{[x^{2}-\pi^{2}][x^{2}-(2\pi)^{2}][x^{2}-(3\pi)^{2}]...}\]
Since it’s \(x\) approaching 0, we can plug in 0
\[a = \frac{1}{[-\pi^{2}][-(2\pi)^{2}][-(3\pi)^{2}]...}\]
We originally started off with
\[sin(x) = a(x)(x+\pi)(x-\pi)(x+2\pi)(x-2\pi)(x+3\pi)(x-3\pi)...\]
Since we have the value of \(a\) above, we can plug in that value
\[sin(x) = x \cdot \frac{[x^2-\pi^2]}{-\pi^2} \cdot \frac{[x^2-(2\pi)^2]}{[(-2\pi)^2]} \cdot \frac{[x^2-(3\pi)^2]}{[(-3\pi)^2]}\]
It is quite hard to multiply all this out an infinite number of times, so we can simply take the coefficients of the first few terms of both series. One is a power series, and the other is a product series, but they’re both for sin(x), so we can equate those coefficients
\[-\frac{1}{6} = -\frac{1}{\pi^2} - \frac{1}{(2\pi)^2} - \frac{1}{(3\pi)^2} - ...\]
We can take out the negative and factor out the \(\pi^{2}\)
\[\frac{1}{6} = \frac{1}{\pi^2} \left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + ...\right)\]
Finally, we can multiply by \(\pi ^ {2}\) on both sides and we are left with our original inverse of squares, now defined!
\[\frac{\pi^2}{6} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + ...\]
Conclusion
That concludes the proof and I hope you guys found this interesting. I thought it was pretty neat how Euler manipulated the equations and did those substitutions to get the summation.