Just a quick message saying that I’m going off on holiday for a week so there will be no posts!

M x

Just a quick message saying that I’m going off on holiday for a week so there will be no posts!

M x

So I have a confession to make… I have completely neglected my blog for the past month. Although it really upset me that I couldn’t upload regular (or in fact any) content, I have just been so busy with work that it was impossible. In these last few months I have been in complete exam mode, and although it’s been absolutely exhausting, it has also been extremely rewarding.

But now, I am finally finished and can concentrate on uploading more regularly. Thank you all so much for your patience, new blog posts coming soon!

M x

The Polymath Project is a collaboration among mathematicians to solve important problem in mathematics by providing a platform for mathematicians to communicate with each other on how to find the best route to the solution.

It began in January 2009 when Tim Gowers posted a problem on his blog and asked readers to reply with partial ideas or answers. This experiment resulted in a new answer to a difficult problem, proving the benefits of collaboration.

Previous Polymath projects that have successfully led to proofs incude the density version of the Hales-Jewett theorem and the Erdös discrepancy problem, as well as famously reducing the bound on the smallest gaps between primes.

Recently the 12th Polymath Project has started; Timothy Chow of MIT has proposed a new Polymath Project – resolve Rota’s basis conjecture.

**What is the Rota’s Basis Conjecture?**

The Rota’s Basis Conjecture states:

“If *B1*,* B2*,….,* Bn* are *n* bases of an n-dimensional vector space *V* (not necessarily distinct or disjoint), then there exists an *n x n* grid of vectors (*vij*) such that:

- the
*n*vectors in row*i*are the members of the*i*th basis*Bi*(in some order), and - in each column of the matrix, the
*n*vectors in that column form a basis of*V*.”

Although easy to state, this conjecture has revealed itself hard to prove (*like Fermat’s Last Theorem*)!

M x

Today I wanted to discuss the geometry of curves and surfaces.

First let us consider a curve **r**(s) which is parameterised by s, the arc length.

Now, **t**(s) = is a unit tangent vector and so **t**^{2} = 1, thus **t**.**t** = 1. If we differentiate this, we get that **t**.**t**‘ = 0, which specifies a direction normal to the curve, provided **t**‘ is not equal to zero. This is because if the dot product of two vectors is zero, then those two vectors are perpendicular to each other.

Let us define **t’** = K**n **where the unit vector **n**(s) is called the *principal normal* and K(s) is called the *curvature*. Note that we can always make K positive by choosing an appropriate direction for **n**.

Another interesting quantity is the *radius of curvature*, a, which is given by

a = 1/curvature

Now that we have **n** and **t** we can define a new vector **b **= **t **x** n**, which is orthonormal to both **t** and **n**. This is called the *binormal*. Using this, we can then examine the *torsion* of the curve, which is given by

T(s) = –**b’**.**n**

As the plane is rotated about **n** we can find a range

where and are the *principal curvatures*. Then

is called the *Gaussian curvature.*

Gauss’ *Theorema Egregium* (which literally translates to ‘Remarkable Theorem’!) says that K is **intrinsic** to the surface. This means that it can be expressed in terms of lengths, angles, etc. which are measured entirely on the surface!

For example, consider a geodesic triangle on a surface S.

Let θ1, θ2, θ3 be the interior angles. Then the *Gauss-Bonnet theorem* tells us that

which generalises the angle sum of a triangle to curved space.

Let us check this when S is a sphere of radius a, for which the geodesics are great circles. We can see that == 1/a, and so K = 1/a^{2}, a constant. As shown below, we have a family of geodesic triangles D with θ1 = α, θ2 = θ3 = π/2.

Since K is constant over S,

Then θ1 + θ2 + θ3 = π + α, agreeing with the prediction of the theorem.

M x

The Banach-Tarski Paradox is a theorem in geometry which states that:

“It is possible to decompose a ball into five pieces which can be reassembled by rigid motions to form two balls of the same size as the original.”

It was first stated in 1924, and is called a paradox as it contradicts basic geometric intuition.

An alternate version of this theorem tells us that:

“It is possible to take a solid ball the size of a pea, and by cutting it into afinitenumber of pieces, reassemble it to forma solid ball the size of the sun.”

Below is an awesome video explaining how this paradox works:

M x

In today’s post I wanted to quickly highlight a cool relationship between Mandelbrot and Julia sets.

Consider the function, which depends of complex parameter *z*:

**f( z) = x^{2} + z **

Fixing this *z*, f(*z*) defines a map from the complex plane to itself. We can start from any value of *x* and apply this function over and over, which would give us a sequence of numbers. This sequence can either go off to infinity, or not. The boundary of the set of values of *x* where it doesn’t is the Julia set for this particular *z*, which we fixed initially.

Conversely, starting with *x* = 0, we can draw the set of numbers for which the resulting sequence does not go off to infinity. This is called the Mandelbrot set. (*Note the subtle difference between the two*).

Okay, so the cool relationship is that, near the number *z*, the Mandelbrot looks like the Julia set for the number *z*, or as Wikipedia describes:

“There is a close correspondence between the geometry of the Mandelbrot set at a given point and the structure of the corresponding Julia set.”

To illustrate this, consider the following Julia set:

Zooming into the Mandelbrot set at the same value of *z* gives us this image:

They are extremely similar! So, essentially, the Mandelbrot sets looks like a lot of Julia sets! (Click here to explore this in more detail).

This amazing result is used in lots of results on the Mandelbrot set, for example, it was exploited by Shishikura to prove that “*for a dense set of parameters in the boundary of the Mandelbrot set, the Julia set has Hausdorff dimension two, and then transfers this information to the parameter plane*“.

M x

Starting a new term in university comes with 4 brand new topics to learn. One of these is Probability. I’d like to share a little bit of (simple) maths that I learnt in my lecture last week on calculating probabilities when there are equally likely outcomes.

I will not define what a probability measure or space is rigorously, as it is quite irrelevant for the simple examples I’m considering today. But, I will introduce the following notation:

Let Ω be a set and *F* be a set of all subsets of Ω. The elements of Ω are called **outcomes** and the elements of* F* are called **events**. As we are using this as a probability model, Ω will be an abstraction of a real set of outcomes, and *F* will model observable events.

Then let us write, **P(A)** as the probability of the event A occurring. Note that:

P(A) = |A|/|Ω|.

Now, let us consider a few examples of how we would use this. In the following examples we use symmetry and randomness to argue that there are equally likely outcomes.

The throw of a die has six possible outcomes, and so

Ω = {1, 2, 3, 4, 5, 6} and P(A) = |A|/6 for A ⊆ Ω.

Therefore, for example P({2, 4, 6}) = **1/2**.

A bag contains *r* balls. Say we draw *k* balls from this bag without looking. If we consider that the balls are labelled by {1, . . . , n}, we have selected a subset of {1, . . . , n} of size *k*.

Therefore, Ω is the set of subsets of {1, . . . , n} of size *k*, so

with the probability of a individual outcome being 1/|Ω|.

In an idealised well-shuffled pack of cards, every possible order is equally likely, so it is modelled by the set Ω of permutations of {1, . . . , 52}.

Because of this, we can see that |Ω| = 52!

**What is the probability of the event A that the first two cards are aces?**

There are 4 choices for the first ace, 3 choices for the second, and the rest can come in any order. So,

|A| = 4 × 3 × 50!

and

P(A) = |A|/|Ω| = 12/(52 × 51) = **1/221**.

Suppose that *r* digits are chosen from a table of random numbers. **What is the probability that k is the greatest digit drawn?**

Firstly, model the set of possible outcomes by

Ω = {0, 1, . . . , 9}^{n}

Consider the event Ak = [no digit exceeds *k*], and the event Bk = [the largest digit is *k*] So:

|Ω| = 10^{n} ;

|Ak| = (k + 1)^{n} ;

|Bk| = |Ak \ Ak−1| = (k + 1)^{n} − k^{n}.

Therefore, P(Bk) = **[(k + 1) ^{n} − k^{n}]/10^{n}**.

Hope you enjoyed today’s post. M x