An explanation of how Euler solved the Basel problem

NOTE: I did not make this explanation. I based it on the explanation from the book “Journey through Genius” by William Dunham, and changed a few parts – I liked how much clearer this book’s explanation was than the other, more formal proofs I have seen.

What is the Basel problem?

The Basel problem had stumped mathematicians for years before Euler came along. It asked a question about this infinite sum:

$\frac{1}{1}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...$

essentially, add the reciprocals of the squares. It was demonstrated by the Bernoullis that the sum, if you took it to all of the infinite terms, was less than 2. The Basel problem was simply to find the exact value of the infinite sum. Not approximate – it had to be exact. You can add together a few terms by hand and see it’s around 1.645, but that’s only an approximate answer.

How did Euler solve it?

Step 1: Establishing some ideas

Idea 1

Euler started with the Taylor expansion of the sine function (a technique that was already well established). This gave him a more useful formula for sin:

$sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!}-...$

so now the sin function was in terms of the sum of an infinite number of polynomials that followed a simple pattern.

Idea 2

The other idea was one of factorisation. Suppose you have some polynomial P(x), of nth degree. it has n roots at x=a,x=b, x=c, and so on, where P(a)=P(b)=P(c)=…=0. Suppose also that P(0)=1. then you can factorise P(x) like this:
$P(x) = \bigg(1-\frac{x}{a}\bigg)\bigg(1-\frac{x}{b}\bigg)\bigg(1-\frac{x}{c}\bigg)...$

If you substitute any root in, you see it will make one term 0, which means the whole product becomes 0. Also, if you substitute in 0, all the x terms vanish, and you just get 1 multiplied by itself a lot.

The proof

Euler started with the function $f(x) = \frac{sin(x)}{x} = 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+...$

So now we have an infinite polynomial, where f(0)=1. This means we can factorise it, in the way suggested by Euler’s second idea. We know this polynomial is equal to $\frac{sin(x)}{x}$ , so to find the roots, we solve $\frac{sin(x)}{x}=0, x \neq 0$ (note that we are ignoring the case of x=0, since we already established f(0)=1). Obviously, the only time $\frac{sin(x)}{x}=0$ is when $sin(x)=0$ . as such, the solutions to f(x) are at $x=\pm \pi,x= \pm 2\pi,$ and so on. Now we can factor f(x) according to that second idea:

$f(x) = \bigg[\bigg(1-\frac{x}{\pi}\bigg)\bigg(1-\frac{x}{-\pi}\bigg)\bigg]\bigg[\bigg(1-\frac{x}{2\pi}\bigg)\bigg(1-\frac{x}{-2\pi}\bigg)\bigg]\bigg[\bigg(1-\frac{x}{3\pi}\bigg)\bigg(1-\frac{x}{-3\pi}\bigg)\bigg]...$

Those terms are grouped together so you can expand them with some basic algebra, the difference of two squares:

$\bigg[\bigg(1-\frac{x}{\pi}\bigg)\bigg(1-\frac{x}{-\pi}\bigg)\bigg] = \bigg(1-\frac{x^2}{\pi^2}\bigg)$

So now we can rewrite that infinite product with a slightly neater one.

$f(x) = \bigg(1-\frac{x^2}{\pi^2}\bigg)\bigg(1-\frac{x^2}{2^2\pi^2}\bigg)\bigg(1-\frac{x^2}{3^2\pi^2}\bigg)\bigg(1-\frac{x^2}{4^2\pi^2}\bigg)...$

ok, so now we see the reciprocals of the squares starting to climb back into the equation. Let’s step back for a second and see what we’ve figured out here:

$\frac{sin(x)}{x} = 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+... = \bigg(1-\frac{x^2}{\pi^2}\bigg)\bigg(1-\frac{x^2}{2^2\pi^2}\bigg)\bigg(1-\frac{x^2}{3^2\pi^2}\bigg)\bigg(1-\frac{x^2}{4^2\pi^2}\bigg)...$

So we’ve equated an infinite sum with an infinite product that contains the reciprocals of the squares. How do we untangle them? We need to expand the product, at least in part.This bit is confusing, so i’ll jump back to a simple expansion first.

Quick diversion

$(1+x^2)(1+2x^2)(1+3x^2)$

This expansion will obviously have a whole number term, an $x^2$ term, an $x^4$ term, and an $x^6$ term. How do we find out what, say, the $x^2$ term looks like? We could just expand the whole thing out by hand, and find that it’s $x^2+2x^2+3x^2$ . This doesn’t help much if you have infinite terms though.

Instead, let’s try this – to get one term, you have to multiply together one number from every bracket, right? This, for example, gets you all the x² terms:

(1+x²)(1+2x²)(1+3x²) -> 3x²

(1+x²)(1+2x²)(1+3x²) -> 2x²

(1+x²)(1+2x²)(1+3x²) -> x²

x²+2x²+3x²=(1+2+3)x²

So to get the coefficient of the x² term, you just add all the coefficients of the x² terms in the brackets.

Ok, back to it

Now that we’ve established how to expand for the x² term, we can do it on our more complex infinite product:

$f(x) =\bigg(1-\frac{x^2}{\pi^2}\bigg)\bigg(1-\frac{x^2}{2^2\pi^2}\bigg)\bigg(1-\frac{x^2}{3^2\pi^2}\bigg)\bigg(1-\frac{x^2}{4^2\pi^2}\bigg)... = 1 - \bigg( \frac{1}{\pi^2}+ \frac{1}{2^2\pi^2}+ \frac{1}{3^2\pi^2}+...\bigg)x^2 + \bigg(...\bigg)x^4 + ...$

We’re almost there. Now we can equate the original infinite sum with our new one:

$1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+... = 1 - \bigg( \frac{1}{\pi^2}+ \frac{1}{2^2\pi^2}+ \frac{1}{3^2\pi^2}+...\bigg)x^2 + \bigg(...\bigg)x^4 + ...$

We can equate the coefficients in front of the $x^2$ terms:

$\frac{1}{3!} = \bigg( \frac{1}{\pi^2}+ \frac{1}{2^2\pi^2}+ \frac{1}{3^2\pi^2}+...\bigg)$

$\frac{1}{1}+ \frac{1}{2^2}+ \frac{1}{3^2}+...=\frac{\pi^2}{6}$

QED