# Pi and the Basel Problem

25-Mar-2022

So, following up on my previous post What is Pi, Non-geometrically?. I just want to make a separate post about some extra things.

At the end of the post I mentioned this relationship:

$\pi^2=6\sum_{n=1}^\infty \frac{1}{n^2}$

If you are familiar with it, this is the solution of the famous Basel Problem. I feel like I want to talk about it as a separate posts, due to how special it was for me.

# The story about Basel Problem… for me

I’m not going to explain again about the problem, since you can just read it in the wiki. I just want express “my feelings” regarding the problem.

For a quite long time, I never knew that this problem is very special and has it’s own famous name. Initially, I encountered the problem in high school when I am learning a totally different subject. It is about gravity.

As you may know, the Newtonian Universal Gravitation Force is expressed as:

$F = \frac{G m_1 m_2}{r^2}$

With $G$ the Universal Gravitation constant, $m_i$ is the interacting mass, and $r$ is the distance between each point of mass. What made me so fascinated about this is the physical meaning of the force expression and intuitive reasoning with additive properties of force.

Before we dive further, let me remind you that this is me in high school. So I obviously had no sophisticated understanding for mathematical rigor. It is just based on a kid’s mind, that if the reality around you are considered as truth, then a particular mathematical equation must be true as well if it describes the same phenomenon.

Let’s take a look at gravity for instance (in the classical sense).

Any object with mass will have gravitational potential. It will interact with another mass and exert force, no matter how far they would be. However, the force is linearly additive. In which, if we have 3 objects, being A as the observer, the following is true according to Newton’s Law:

$\sum F_A=F_{AB}+F_{AC} = F_R$

If A and B is close, usually the force is greater. If A and C is far, it will have lesser force. However, they add up together. Interestingly, since force is additive, we can “imagine” that both B and C doesn’t exist, and instead a single object exists called R. We can find the correct distance between A and R, in such a way that the resulting gravitational force is equivalent to $F_{AB}+F_{AC}$.

This was a very mindblowing concept for me. We can easily decompose or combine many objects into one single objects. Suppose that we have 100 objects, we can eliminate all of them and replace it with just one object with the correct distance, and it will have the same influence to A the same way as the original condition.

The relation between this and the Basel problem is that, if we consider a single kind of object, but there are infinitely many of them, we can place them each having distance 1 meter apart. Then all of the force added up together is exactly the same way as if just one single object placed at certain distance R.

If we describe it in expression:

$\frac{1}{R^2} = \sum_{n=1}^\infty \frac{1}{n^2}$

There must exists a value R where this would solve the equation. Also, R would be really close to the origin.

What is R?

That is how I solved Basel problem the first time. But of course at that moment, the steps is not rigour enough. I will not going to talk about it today. What I’m going to talk about is that at time I don’t know about internet, wikipedia, or wolfram alpha yet. So I never knew about the name of the problem.

Then it really excites me when I finally learned about the name from this 3Blue1Brown video about the Basel problem From that, I found the wikipedia link. I’m really impressed at rediscovering the original problem and reading various approach to solve the problem.

There are many ways and angles to solve the same problem. I think it is one of the beauty of math.

# The Euler’s approach of solving the Basel Problem

In the previous post, our goal is to find a non-geometrical approach to solve a particular problem, then hoping a useful or fundamental constant appears. This turns out to be Pi.

Meanwhile, Basel problem start from a simple problem description, then it was revealed that the solution contains a mention of $\pi$ in it. Which is kind of mindblowing at first. How come some geometrical integer sums gives raise to a circular constant?

As you might have guessed, the previous post is like a reverse-engineered direction of Basel Problem. Which is not intentional from my side. I was just trying to find out how to define Pi so it raises naturally (like the $e$ constant from exponential function). What I have instead, is the expression of the Basel problem. So, it is the other way around if we compare it with Basel problem.

Basically the difference between Euler’s method and my previous post is that Euler started with the Taylor series of $\sin(x)$. Thus he needs to find an expression of $\sin(x)$ as sums and infinite products to compare the coefficient.

Alternatively, in my previous post, we didn’t assume any trigonometry function. We just start with the Taylor series of exponential in complex plane. From there we can pick which series to proceed, the real part or imaginary part.

If we pick the real part, which in formal sense is the $\cos(x)$ function. We will have:

$\pi^2=8\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}$

Which is Basel problem, but for odd terms.

If we pick the imaginary part, which in formal sense is the $\sin(x)$ function. We will recover the original Basel function:

$\pi^2=6\sum_{n=1}^\infty \frac{1}{n^2}$

Another thing to note between Euler’s method and my previous posts, is how we define the zeroes/roots. My roots is just a simple integer multiple. It doesn’t assume anything about $\pi$. Euler define the roots using $\pi$ due to the nature of the trigonometry function for radian units.

Despite using a different set of roots, it turns out that the result is still the same. The difference is the usage. I am using Basel expression to approximate the value of $\pi$. Meanwhile, Euler uses $\pi$ to approximate the result of Basel expression.

Hahaha… wonderful isn’t it?

# The odd and even terms of the Basel problem

You might wonder if the real and imaginary part approach produce a consistent result. We can show that they are in fact consistent.

From the real part, we got the odd terms. Surely the full Basel expressions is the sums of the odd and even terms if they are both convergent.

Now how do we got the even part? The even part is actually the same indexing but scaled twice. Let the result of Basel problem as $B$. As you can see below:

$B=\sum_{n=1}^\infty \frac{1}{n^2} \\ \frac{B}{4}=\sum_{n=1}^\infty \frac{1}{2^2n^2} \\ \frac{B}{4}=\sum_{n=1}^\infty \frac{1}{(2n)^2}$

So, if we add both odd and even terms:

$B=B_{odd} + B_{even} \\ B=B_{odd}+ \frac{B}{4} \\ B_{odd}=\frac{3}{4} B$

From these we can easily subsitute the value to see if it match:

$B_{odd}=\frac{3}{4} B \\ \frac{\pi^2}{8}=\frac{3}{4}\frac{\pi^2}{6}$

They are indeed the same.

# Final Remark

The reason why I can intuitively know that this is going to be consistent is because of the position of the roots, if it is an integer multiples.

Remember that the first time I noticed about Basel problem is how I identify it as a gravitational bodies. The original Basel problem is expressed by placing an object in each integer multiple of meters till infinity. If we scale the unit twice. The roots will just scale twice. For example, object A in 1 meter distance becomes 2 meters. 2 becomes 4, etc. It is intuitive enough to imagine that the distance of the replacement object R, is going to scale the same way, twice. That means the series and the result for even numbers is just a scaled version of the same thing.

In other words, we can instead imagine if we replace all the objects into two representatives $R_{odd}$ and $R_{even}$, as opposed to just one $R$ object. The result has to be consistent with existing gravitational laws. Then the force caused by this $R_{odd}$ and $R_{even}$, has to be the same with $R$.

It is really interesting how these infinites added up together nicely.