Proof Without Words #1

I decided to start a new series called ‘Proof Without Words’: a collection of pictures which prove a mathematical fact with an image. Remember, these are in no way rigorous and are just meant to give an idea of why the fact is true!

1/2+1/4+1/8+1/16+1/32+ … = 1

Sum of the first n positive odd numbers = n^2

arctan(1) + arctan(2) + arctan(3) = π

Screen Shot 2017-02-21 at 9.21.36 PM.png

Screen Shot 2017-02-21 at 9.22.18 PM.png

Viviani’s Theorem

The sum of the distances from any interior point to the sides of an equilateral triangle equals the length of the triangle’s altitude.

Screen Shot 2017-02-21 at 9.24.11 PM.png

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MATHS BITE: The Cantor Set

The Cantor Set is constructed in the following way:

Start with the interval [0,1]. Next, remove the open middle third interval, which gives you two line segments [0,1/3] and [2/3,1]. Again, remove the middle third for each remaining interval, which leaves you now with 4 intervals. Repeat this final step ad infinitum.

Cantor_set_binary_tree.svg.png

The points in [0,1] that do not eventually get removed in the procedure form the Cantor set.

How many points are there in the Cantor Set?

Consider the diagram below:

Screen Shot 2017-02-21 at 8.05.41 PM.png

An interval from each step has been coloured in red, and each red interval (apart from the top one) lies underneath another red interval. This nested sequence shrinks down to a point, which is contained in every one of the red intervals, and hence is a member of the Cantor set. In fact, each point in the Cantor set corresponds to a unique infinite sequence of nested intervals.

To label a point in the Cantor set according to the path of red intervals that is taken to reach it, label each point by an infinite sequence consisting of 0s and 1s.

A 0 in the nth position symbolises that the point lies in the left hand interval after the nth stage in the Cantor process.

A in the nth position symbolises that the point lies in the right hand interval after the nth stage in the Cantor process.

For example, the point 0 in [0,1] is represented by the sequence 0000…., the point 1 is represented by the sequence 1111…. and the point 1/3 is represented by the sequence 01111….

So, as there are infinite sequences consisting of 0s and 1s, there are an infinite number of elements in the Cantor set. If we place a point before any one of these infinite sequences, for example 0100010… becomes .0100010…, then we convert an infinite sequence of 0s and 1s to the binary expansion of a real number between 0 and 1. This means that the number of points in the Cantor set is the same as the number of points in the interval [0,1]. We conclude that the infinite process of removing middle thirds from the interval [0,1] has no effect on the number of points in [0,1]!

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Intrinsic Geometry

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

Curves, Curvature and Normals

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

Now, t(s) = Screen Shot 2017-02-14 at 8.05.47 AM.png is a unit tangent vector and so t2 = 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’ = Kwhere 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 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

Image result for binormal tangent

Intrinsic Geometry

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

Screen Shot 2017-02-14 at 8.16.01 AM.pngwhereScreen Shot 2017-02-14 at 8.16.08 AM.png and Screen Shot 2017-02-14 at 8.16.14 AM.png are the principal curvatures. Then

Screen Shot 2017-02-14 at 8.17.34 AM.png

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.

Screen Shot 2017-02-14 at 8.20.37 AM.png

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

Screen Shot 2017-02-14 at 8.22.10 AM.png

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 Screen Shot 2017-02-14 at 8.16.08 AM.png=Screen Shot 2017-02-14 at 8.16.14 AM.png= 1/a, and so K = 1/a2, a constant. As shown below, we have a family of geodesic triangles D with θ1 = α, θ2 = θ3 = π/2.

screen-shot-2017-02-14-at-8-26-15-am

Since K is constant over S,

Screen Shot 2017-02-14 at 8.27.09 AM.png

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

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VIDEO: Map of Mathematics

Today I wanted to share with you a video which I came across the other day on the Map of Mathematics.

Although many people view maths as synonymous with pain, boredom or frustration, one must appreciate its diversity and vast implications on other subjects; you may not have the background to see beauty in a particular equation, but virtually anyone can appreciate the amazing advancements humans have made from basic counting to creating full-on artificial intelligence.

While an artistic temperament is often considered the exact opposite of the kind of personality that loves complicated equations, pure mathematicians are really just a bunch of lunatics endlessly working with abstraction and beauty.

– Rhett Jones

In the video below, Dominic Walliman takes viewers through the major fields of math starting at the beginning and shows us how they inform and relate to each other. Of course many details have been left out, as to properly connect the various disciplines of math we would need a 3D web! Also, in reality, “the study of math’s foundations has yet to discover a complete and consistent set of axioms.

Hope you enjoyed the video! M x

Mazes

What’s the difference between a maze and a labyrinth?

It is generally accepted that a labyrinth contains only one path, which often spirals around folding back in on itself in ever-decreasing loops. On the other hand, a maze contains branching paths, which lead to nowhere, and therefore present the person with choices and the potential for them to get lost.

Mathematicians have created maze-generating algorithms, which tend to fall into two main types:

  • Ones that start with a single bounded space and then sub-divide it with walls to produce smaller sub-spaces;
  • Ones that start with a bunch of disconnected rooms and then demolish walls to create paths between them.

Escaping Mazes

Most methods work for ‘simple mazes’, i.e. mazes with no short-cuts via bridges or passage loops (circular paths that lead back to where they started).

So, if we assume the maze is simple, the most common method to escape the maze is wall-following. Here, you place a hand on the wall of the maze, keep walking maintaining contact with the wall and eventually you will get out.

Why does this work? If you imagine picking the wall of the maze and stretching its perimeter to remove any corners. You will eventually create a circle-like shape, which must form part of the maze’s outer border.

Trémaux’s Algorithm

The problem is that most mazes aren’t simple, for example the Escot Gardens’ beech hedge maze in Devon.

Beech Hedge Maze in Devon | lotstodo.co.uk

So, another method of maze escape is known as Trémaux’s algorithm, which works is all cases.

Here’s how it works. Basically, with this algorithm, you leave a trail as you navigate your way through the maze. Then follow these rules:

  • If you arrive at a junction you have not previously encountered, randomly select a way to go;
  • If that leads you to a junction where one path is new but the other is not, select the path you have not been on;
  • If you are choosing between a once or twice-used path, choose the path that has only be used once, then leave a second trail behind you;
  • Never select a path that already has two trails.

This method works for all mazes and is guaranteed to get you out!

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Stirling’s Formula

Today I wanted to discuss something I learnt last week in my Probability course: Stirling’s Formula. Stirling’s Formula is an approximation for factorials, and leads to quite accurate results even for small values of n.

The formula can be written in two ways:

{\displaystyle \ln n!=n\ln n-n+O(\ln n)}

or

{\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n},}

where the ~ sign means that the two quantities are asymptotic (i.e. their ratios tend to 1 as n tends to infinity).

File:Mplwp factorial gamma stirling.svg

Comparison of Factorial with Stirling’s Approximation | Source: Wikipedia

Proof of Stirling’s Formula

The following identity arises using integration by parts:

Screen Shot 2017-02-06 at 8.11.00 AM.png

Taking f(x) = log x, we obtain

Screen Shot 2017-02-06 at 8.11.48 AM.png

Next, sum over n, and by recalling that log x + log y = log xy we get the following expression:

Screen Shot 2017-02-06 at 8.12.23 AM.png

where

Screen Shot 2017-02-06 at 8.13.23 AM.png

Next, define

Screen Shot 2017-02-06 at 8.13.38 AM.png

which allows us to rearrange the above expression to:

Screen Shot 2017-02-06 at 8.13.41 AM.png

So as n tends to infinity we get

Screen Shot 2017-02-06 at 8.13.46 AM.png

(*)

How do we show that Screen Shot 2017-02-06 at 8.15.19 AM.png ?

Firstly, note that from (*) it follows that

Screen Shot 2017-02-06 at 8.16.31 AM.png

So, we need to show that

Screen Shot 2017-02-06 at 8.16.57 AM.png

Let’s set

Screen Shot 2017-02-06 at 8.17.25 AM.png

Note that I0=π/2 and I1 = 1. Then for n≥2, we can integrate by parts to see that

Screen Shot 2017-02-06 at 8.19.17 AM.png

And so, we obtain the following two expressions:

Screen Shot 2017-02-06 at 8.19.36 AM.png

In is decreasing in n and In/In-2 → 1, so it follows that I2n/I2n+1 → 1. Therefore,

Screen Shot 2017-02-06 at 8.21.18 AM.png

as required.

Although the end result is satisfying, I find that some steps in this proof are like ‘pulling-a-rabbit-out-of-a-hat’! What do you think? Mx

 

 

VIDEO: Banach–Tarski Paradox

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 a finite number of pieces, reassemble it to form a solid ball the size of the sun.”

Below is an awesome video explaining how this paradox works:

 

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