Solving the Basel problem, using an elementary introduction to Fourier series!

Hello and welcome!

We are going to be tackling a *beautiful *problem, the ‘Basel Problem’. It was posed in 1650 by Pietro Mengoli, and it took *84 years *to solve it, when the brilliant Euler did so.

Euler’s proof was brilliant, but also not strictly speaking complete! He used manipulations which couldn’t be justified with the mathematics developed at that time. It would take *100 years *before the gaps in Euler’s proof were completely filled. …

Here we crack a tricky yet simple problem at the intersection of computer science and mathematics. It also gives a lovely insight into how to prove something is impossible.

Let’s take a look at the chessboard below.

Here we investigate and understand the brilliant formula which unlocks the power of complex numbers.

Here’s a fun 2-minute guide to the infamous Axiom of Choice, one of the stranger mathematical curiosities.

The short answer, not at all rigorous answer, is that the axiom of choice allows mathematicians to extract elements from an infinite number of infinitely large sets at once. This turns out to be very important, because mathematicians are *very fond indeed *of making infinite-sized things, and even more fond of being able to formalise their manipulations of infinite-sized things.

But let’s dig a little deeper.

The axiom of choice allows us to pick elements from ‘indexed sets’. When dealing with ‘finite things’, this seems kinda obvious. For instance, if A={1,2,3}, B={3,4,5}, and C={5,6}, then it is easy to pick an element from each. Just pick, say, 1 from A, 3 from B, and 6 from C. In fact, if you only every deal with finite sets (e.g. dealing with finite numbers, finite graphs, a finite number of people, etc…), then you will never need the Axiom of Choice. That includes lots of interesting maths, from computer science to graph theory. …

A sequence of real numbers is simply a *list *of real numbers, where for any integer n, you are allowed to receive the nth number in the list. Fortunately, mathematicians deal in theoretical objects, so we never run out of memory, and never get stack overflows from recursive definitions.

A famous example of a sequence of real numbers is:

0.9, 0.99, 0.999, 0.9999, 0.99999, …

where the Nth number in the list has N nines following the zero.

We say this sequence ‘converges’ to 1. Why? The short and cheeky answer is to read the answer I wrote here. But I’ll explain again. The first term is only 0.1 away from 1, the second term is only 0.01 away from 1, and the Nth term is 0.00…01 away from one, where there are N zeros preceding the 1, i.e. …

Let’s take a look at an unusual proof of the infinity of prime numbers.

By the *Fundamental Theorem of Arithmetic*, we can write any number as the product of primes. For example, 45 = 5*3², and 100 = 2²5²

A variation of this is that any number can be written as the product of a *square-free *number *s *and a square, *r²*, and this can be done uniquely. A *square-free *number is one where no perfect square divides it. I will call this the number’s *sr² *form.

For instance, 12 is *not *square-free, because 4 divides it. To get 12 into its *sr² *form we write 12 = 3*(2)² = *sr², *where s=3, and…

Over 2,300 years ago, Euclid proved the Fundamental Theorem of Arithmetic. Now it is our turn.

The Fundamental Theorem of Arithmetic states that we can decompose any number uniquely into the product of prime numbers. For example, 350 = 2*7*5², and there is *no other way *to write 350 as the product of prime numbers.

Bezout’s lemma tells us that, if two numbers *a *and *b *share no prime factors other than 1 and -1, then we can find *q *and *s *such that *qa+sb=1*.

Ok, let’s take some time to digest that with a worked example. Consider 7, and 9. The only numbers with divide 7 are {7,1,-1,-7}. The only numbers which divide 9 are {9,3,1,-1,-3,-9}. So, 7 and 9 share no prime factors. …

A visual proof of the Gauss Bonnet Theorem for triangles on spheres!

Spherical geometry is a beautiful, and very visual, area of mathematics, with weird properties (such as that the angles of triangles don’t sum to 180 !!!).

The Gauss-Bonnet Theorem for triangles on spheres is a special, but rather beautiful, case of the more general Gauss-Bonnet Theorem.

The theorem can be understood visually

First, let’s visualise what a triangle on a sphere is. Below are two visualisations of the same triangle.

This is the code heavy addendum to the more mathsy article I wrote on the Erdos-Szekeres Theorem. However, this is also self-contained, and is focused on finding monotone sequences rather than proving things about their size.

Our task is to write a program to find the longest increasing subsequence in a sequence of numbers. And then to visualise the results!

For instance, the longest increasing subsequence in [1,2,3,0,4] is [1,2,3,4].

The mathsy result is quite astonishing. …

A theorem by two mathematical greats, *which anybody, no maths experience required, *can understand! An amazing and beautiful result accompanied by some pretty hand-coded visualisations!

Regardless of your experience with math, this theorem can be understood, and the elegance of the solution appreciated. Throughout I will both use diagrams to explain concepts where possible. Where I use algebra notation like ‘x’ and ‘y’, I will also give concrete examples using real numbers so you can understand what is going on even if you aren’t comfortable with algebra.

Before I state the problem, I want to build some understanding. That’s because the terms can be confusing to people who haven’t done much math, but by taking a little more time we can understand the problem without previous knowledge of any complicated language. If you are familiar with the terms used (in **bold**), then you can skip to the next section. …

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