Welcome to Rise to the Equation! I hope you enjoy my video for the Summer of Math Exposition contest
hosted by Three Blue One Brown. Ah, yes, the famous Basel Problem,
first proved by Leonhard Euler in 1734. Its solution never ceases to amaze:
the sum of the reciprocals of the squares of all the natural numbers from one to infinity
magically conspire to equal pi squared over 6. There’s just one problem, though. We’ve seen
this before, haven’t we? Three Blue One Brown provided a spectacular visual proof using light.
Our friend Steve from Black Pen Red Pen proved it using an integral in the complex domain.
Mathologer gave us a sketch of Euler’s classic 1734 proof, and we even have
me – Rise to the Equation – using a Fourier Series. Plus a gazillion others on YouTube.
So why do we need yet another proof in this video? Because Cauchy’s proof is surprisingly accessible.
It doesn’t require any advanced math degrees, just some high school precalculus. Plus,
not only it is beautiful, but for some strange reason, Cauchy’s proof is extremely rare. So,
we’re going to brave some uncharted territory here and provide a complete proof from head to toe.
My goal is to break the proof down into five bite-sized steps, or ACTS, just as if you were
watching the action unfold in a Shakespearean play. So, let’s get started with Act One,
Setting up a trig inequality using the squeeze theorem.
Cauchy’s solution begins, fittingly, with a circle. We’ll put it at center O and draw an
interior triangle with two vertices A and B on the circle, and the other vertex at O. The circle will
have radius r and central angle theta. We’ll also construct right triangle OAC. Our goal is
to establish an inequality relating the areas of triangle OAB, sector OAB, and triangle OAC.
Let’s focus on triangle OAB first. If we assign a height h, we can calculate the area. It’s just the
radius times height over two , and since h = r sin theta, the area is r squared sine theta over two.
Next is the area of the sector OAB. Its area is just a textbook formula: r squared theta over two.
For triangle OAC, if we assign a height big H we can get its area using these two relationships,
from which we get r squared tan theta over 2. Now let’s focus on those boxed areas in red.
Dividing through by r squared over two we get this nifty little trig inequality,
which is valid for theta between 0 and pi over 2. Finally, if we square the reciprocals we arrive
at our desired inequality. This expression is going to be our launchpad for Act 2.
For Act 2 we’re going to take our trig inequality from Act 1 and put some limits on it.
Let’s rewrite our trig inequality using a familiar trig identity.
Next let’s replace theta with n pi over 2 N plus 1, where little n is an integer between 1 and
some fixed integer big N. Let’s multiply all three expressions by pi squared over 2 N plus 1 squared.
That’s now given us the exact expression we’re interested in:
1 over n squared. Now we’re going to sum all three expressions from 1 to big N.
Now this term is a constant added up N times, so we can replace it with N pi squared
over 2N plus 1 squared. And this is the same constant, which I’ll pull in front of the sum.
Since the series is infinite, we’ll take limits as N goes to infinity.
We can do a little simplifying. This limit just goes to zero
since there’s an N on top but N squared on the bottom. So now our expression looks like this.
The top and bottom expressions are identical. If we can figure out what the cotangent square
sum is equal to, we can determine the limits and show that our series sums to pi squared over six.
For our third Act we’re going to start from a clean slate applying De Moivre’s theorem.
According to De Moivre’s Theorem, if we start with a complex number, then when we raise it
to the nth power, it simplifies to this. But, of course, the reverse is also true.
Our goal is to get this into a cotangent only expression so we can relate back to sum of
cotangents squared. Now let’s divide both sides by sine theta to the n. Rewriting the right side
as a fraction to the nth power we get this. And dividing through by sin theta we’re left with
cotangent theta plus i to the n. Now Cauchy applies the binomial theorem. I’ll just show
the first two and last two terms. The expressions in parentheses are the binomial coefficients,
n choose 0, n choose 1, and so on. Notice that the imaginary number i is being raised
to increasing integer exponents, so some terms will be imaginary and others will be real.
Next we rearrange by grouping together the real terms followed by the imaginary terms.
And now comes a common trick when dealing with complex numbers:
Cauchy is going to do the equating of the imaginary parts, which I’ve highlighted in blue.
The final result is this rather unusual identity which will serve as the linchpin to our solution.
Let’s convert that identity into a polynomial.
If we can get the LHS equal to zero then well have an expression strictly in terms
of cotangent. With a clever substitution, it actually does equal zero! Remember earlier
we made the substitution n pi over 2 N plus 1. We’re also going to replace n with 2 times big N
plus 1. Notice what happens if I substitute those two expressions in for sin n theta, the 2N + 1’s
cancel out, leaving us with sine n pi. And since n is an integer, sine n pi is always zero.
Sweet. So let’s go ahead and set that Left Hand Side equal to zero. But we need to carry through
our substitutions for theta and n on the right hand side of the expression as well. Which leaves
us with this. Now one more substitution. Let’s let t equal cotangent squared theta.
Plugging it in leaves us with the following polynomial. Oh wow. Now we have this beautiful
polynomial as a function of t, from which we can analyze the roots much more readily.
For some reason Act 5 is my favorite. It’s like a supporting actor comes out
of nowhere to receive an Oscar, and his name happens to be Vieta.
We know that for a quadratic the sum of the roots is negative b over a,
right? Well, Vieta showed that this little rule can be extended to any polynomial, not
just quadratics. Cauchy realized this of course and applied it to the polynomial we just derived.
So in this case the sum of the N roots of P of t
would be the negative of the second coefficient divided by the leading coefficient.
That expression can easily be simplified using the formula for n choose k.
I’ll leave it to you as a fun algebra exercise to show that it equals 2N times 2N-1 over 6.
Now we’re really close – recall that we made the substitution t equals cotangent squared theta
and theta equals n pi over 2N plus 1. Let’s insert them into the sum and see what we get.
Yes! We get a beautiful and simple formula for the sum of cotangent squared!
Which is exactly what we were looking for to complete the proof. So let’s swap those
boxed sums in red for their more streamlined counterparts and start attacking those limits!
We’ll pull pi squared over 6 out in front of the limit since it’s a constant,
and we’ll expand the expressions containing N. Those limits go to 1
since the leading term N squared has identical coefficients in the numerator and denominator.
So that leaves us with the sum squeezed between the same values of
pi squared over six from either side, and that of course means we proved the Basel Problem.
Let’s add Cauchy to our list of Solutions!