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This follows on from this post.

Recall the Doubling Mapping D:[0,1)\rightarrow [0,1) given by:

\displaystyle D(x)=\begin{cases} 2x & \text{ if }x<1/2 \\ 2x-1 & \text{ if }x\geq 1/2 \end{cases}

At the end of the last post we showed that this dynamical system displays sensitivity to initial conditions. Now we show that it displays topological mixing (a chaotic orbit) and density of periodic points.

First we must talk about periodic points.

Periodic Points

Consider, for example, the initial state \displaystyle x_0=\frac{1}{9}. The orbit of x_0 is given by:

\displaystyle \text{orb}(x_0)=\left\{\frac{1}{9},\frac29,\frac49,\frac89,\frac79,\frac59,\frac19,\frac29,\dots\right\}

Here we see \frac19 repeats itself and so gets ‘stuck’ in a repeating pattern:

graph1

The orbit of x_0=1/9.

The orbit of any fraction, e.g. \displaystyle x_0=\frac{4}{243}, must be periodic, because \displaystyle D\left(\frac{i}{243}\right) is either equal to \displaystyle \frac{2i}{243} of \displaystyle \frac{2i-243}{243} and so the orbit consists only of states of the form:

\displaystyle \frac{i}{243},

and there are only 243 of these and so after 244 iterations, some state must be repeated and so we get locked into a periodic cycle.

If we accept the following:

Proposition

A fraction \frac{p}{q} has a recurring binary expansion:

\displaystyle \frac{p}{q}=0.b_1\dots b_m\overline{a_1a_2\dots a_n}_2,

then this is another way to see that fractions are (eventually) periodic. Take for example,

\displaystyle x_0=0.101,101,101,101,\dots_2=0.\overline{101}_2=\frac{5}{7}.

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Dynamical Systems

A dynamical system is a set of states S together with an iterator function f:S\rightarrow S which is used to determine the next state of a system in terms of the previous state. For example, if x_0\in S is the initial state, the subsequent states are given by:

x_1=f(x_0),

x_2=f(x_1)=f(f(x_0))=(f\circ f)(x_0)=:f^2(x_0)

x_3=f(x_2)=f(f^2(x_0))=f^3(x_0),

and in general, the next state is got by applying the iterator function:

x_{i}=f(x_{i-1})=f^i(x_0).

The sequence of states

\{x_0,x_1,x_2,\dots\}

is known as the orbit of x_0 and the x_i are known as the iterates.

Such dynamical systems are completely deterministic: if you know the state at any time you know it at all subsequent times. Also, if a state is repeated, for example:

\text{orb}(x_0)=\{x_0,x_1,x_2,x_3,x_4=x_2,x_5\dots,\}

then the orbit is destined to repeated forever because

x_5=f(x_4)=f(x_2)=x_3,

x_6=f(x_5)=f(x_3)=x_4=x_2, etc:

\Rightarrow \text{orb}(x_0)=\{x_0,x_1,x_2,x_3,x_2,x_3,x_2,\dots\}

Example: Savings

Suppose you save in a bank, where monthly you receive 0.1\%=0.001 interest and you throw in 50 per month, starting on the day you open the account.

This can be modeled as a dynamical system.

Let S=\mathbb{R} be the set of euro amounts. The initial amount of savings is x_0=50. After one month you get interest on this: 0.001\times50, you still have your original 50 and you are depositing a further €50, so the state of your savings, after one month, is given by:

x_1=50+0.001\times 50+50=(1+0.001)50+50.

Now, in the second month, there is interest on all this:

interest in second month 0.001\times((1+0.001)50+50)=0.001x_1,

we also have the x_1=(1+0.001)50+50 from the previous month and we are throwing in an extra €50 so now the state of your savings, after two months, is:

x_2=x_1+0.001x_1+50=(1+0.001)x_1+50,

and it shouldn’t be too difficult to see that how you get from x_i\longrightarrow x_{i+1} is by applying the function:

f(x)=(1+0.001)x+50.

Exercise

Use geometric series to find a formula for x_n.

Weather

If quantum effects are neglected, then weather is a deterministic system. This means that if we know the exact state of the weather at a certain instant (we can even think of the state of the universe – variations in the sun affecting the weather, etc), then we can calculate the state of the weather at all subsequent times.

This means that if we know everything about the state of the weather today at 12 noon, then we know what the weather will be at 12 noon tomorrow…

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In school, we learn how a line has an equation… and a circle has an equation… what does this mean?

The short answer is

points (x_0,y_0) on curve \longleftrightarrow solutions (x_0,y_0) of equation

however this note explains all of this from first principles, with a particular emphasis on the set-theoretic fundamentals.

Set Theory

set is a collection of objects. The objects of a set are referred to as the elements or members and if we can list the elements we include them in curly-brackets. For example, call by S the set of whole numbers (strictly) between two and nine. This set is denoted by

S=\{3,4,5,6,7,8\}.

We indicate that an object x is an element of a set X by writing x\in X, said, x in X or x is an element of X. We use the symbol \not\in to indicate non-membership. For example, 2\not\in S.

Elements are not duplicated and the order doesn’t matter. For example:

\{x,x,y\}=\{x,y\}=\{y,x\}.

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This post follows on from this post where the following principle was presented:

Fundamental Principle of Solving ‘Easy’ Equations

Identify what is difficult or troublesome about the equation and get rid of it. As long as you do the same thing to both numbers (the “Lhs” and the “Rhs”), the equation will be replaced by a simpler equation with the same solution.

There are a number of subtleties here: basically sometimes you get extra ‘solutions’ (that are not solutions at all), and sometimes you can lose solutions.

Let us write the squaring function, e.g. 6\mapsto 36, x\mapsto x^2 by f(x)=x^2 and the square-rooting function by x\mapsto \sqrt{x}. It appears that (x^2,\sqrt{x}) are an inverse pair but not quite exactly. While

\displaystyle 81\overset{\sqrt{x}}{\mapsto} 9\overset{x^2}{\mapsto}81 and

\displaystyle 7\overset{x^2}{\mapsto}49\overset{\sqrt{x}}{\mapsto}7,

check out O.K. note that

-4\overset{x^2}{\mapsto}+16\overset{\sqrt{x}}{\mapsto}=+4\neq -4,

does not bring us back to where we started.

This problem can be fixed by restricting the allowable inputs to x^2 to positive numbers only but for the moment it is better to just treat this as a subtlety, namely while (\sqrt{x})^2=x, \sqrt{x^2}=\pm x… in fact I recommend that we remember that with an x^2 there will generally be two solutions.

The other thing we look out for as much as possible is that we cannot divide by zero.

There are other issues around such as the fact that \sqrt{x}>0, so that the equation \sqrt{x}=-2 has no solutions (no, x=4 is not a solution! Check.). This equation has no solutions.

Often, in context, these subtleties are not problematic. For example, equations with no solutions rarely arise and quantities might be positive so that if we have \pm\sqrt{a}, only +\sqrt{a} need be considered (for example, a might be a length).

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In this short note we will explain why we multiply matrices in this “rows-by-columns” fashion. This note will only look at 2\times 2 matrices but it should be clear, particularly by looking at this note, how this generalises to matrices of arbitrary size.

First of all we need some objects. Consider the plane \Pi. By fixing an origin, orientation (x– and y-directions), and scale, each point P\in\Pi can be associated with an ordered pair (a,b), where a is the distance along the x axis and b is the distance along the y axis. For the purposes of linear algebra we denote this point P=(a,b) by

\displaystyle P=\left(\begin{array}{c}a\\ b\end{array}\right).

graph7

We have two basic operations with points in the plane. We can add them together and we can scalar multiply them according to, if Q=(c,d) and \lambda\in\mathbb{R}:

P+Q=\left(\begin{array}{c}a\\ b\end{array}\right)+\left(\begin{array}{c}c\\ d\end{array}\right)

\displaystyle=\left(\begin{array}{c}a+c\\ b+d\end{array}\right), and

\lambda\cdot P=\lambda\cdot \left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{c}\lambda\cdot a\\ \lambda\cdot b\end{array}\right).

Objects in mathematics that can be added together and scalar-multiplied are said to be vectorsSets of vectors are known as vector spaces and a feature of vector spaces is that all vectors can be written in a unique way as a sum of basic vectors. 

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A nice little question:

Given a regular pentagon with side length swhat is the relationship between the area and the side-length?

First of all a pentagon:

pentagon

We use triangulation to cut it into a number of triangles:

pentagon1

With 180^\circ in each of the three triangles, there 3\times 180^\circ=540^\circ in those angles around the edges, and, as there are five of them, they are each 108^\circ.

Next triangulate from the centre. With a plain oul pentagon we might not be sure that such a centre exists but if you start with a circle and inscribe five equidistant points along the circle, the centre of the circle serves as this centre:

pentagon2

As everything is symmetric, each of these triangles are the same and the ‘rays’ are also the same as they are all radii. The angle at the centre is equal to 360^\circ/5=72^\circ, and furthermore, by symmetry, the rays bisect the larger angles 108^\circ/2=54^\circ and so each of these triangles are 72^\circ,54^\circ,54^\circ.

pentagon3

Using radians, because they are nicer, \displaystyle 54^\circ=\frac{3\pi}{10}. Note that, where h is the perpendicular height:

\displaystyle \tan\left(\frac{3\pi}{10}\right)=\frac{h}{s/2}\Rightarrow h=\frac{s}{2}\tan(3\pi/10).

A problem for another day is finding the exact value of \tan\left(3\pi/10\right). It is

\displaystyle \frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}

\displaystyle \Rightarrow h=\frac{\sqrt{5}+1}{2\sqrt{10-2\sqrt{5}}}\cdot s.

Therefore the area of one such triangle is:

\displaystyle A(\Delta)=\frac{1}{2}s\cdot \frac{\sqrt{5}+1}{2\sqrt{10-2\sqrt{5}}}\cdot s=\frac{\sqrt{5}+1}{4\sqrt{10-2\sqrt{5}}}\cdot s^2,

Therefore the area of the pentagon is five times this:

\displaystyle A=5\cdot\frac{\sqrt{5}+1}{4\sqrt{10-2\sqrt{5}}}\cdot s^2

\displaystyle=\underbrace{\frac{5\sqrt{5}+5}{4\sqrt{10-2\sqrt{5}}}}_{=:\alpha}\cdot s^2,

with \alpha\approx 1.721. It might be possible to simply \alpha further.

 

Quadratics are ubiquitous in mathematics. For the purposes of this piece a quadratic is a real-valued function q:\mathbb{R}\rightarrow \mathbb{R} of the form

q(x)=ax^2+bx+c,

where a,\,b,\,c\in \mathbb{R} such that a\neq 0. There is a little bit more to be said — particularly about the differences between a quadratic and a quadratic function but for those this piece is addressed to (third level: non-maths; all second level), the distinction is unimportant.

Geometry

The basic object we study is the square function, s:\mathbb{R}\rightarrow \mathbb{R}, x\mapsto x^2:

graph1

All quadratics look similar to x^2. If a>0 then the quadratic has this \bigcup geometry. Otherwise it looks like y=-x^2 and has \bigcap geometry

The geometry dictates that quadratics can have either zero, one or two real roots. A root of a function is an input x such that f(x)=0. As the graph of a function is of the form y=f(x), roots are such that y=f(x)=0\Rightarrow y=0, that is where the graph cuts the x-axis. With the geometry of quadratics they can cut the x-axis no times, once (like s(x)=x^2), or twice.

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There are a number of ways of explaining why you cannot divide by zero. Here are my two favourites.

Any Set of Numbers Collapses to a Single Number

How old are you? Zero years old.

How tall are you? Zero metres old.

How many teeth do you have? Zero.

How many Superbowls has Tom Brady won? Zero

Yep, if you allow division by zero you only end up with one number to measure everything with.

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After a long time I have finally completed my PhD studies when I handed in my hardbound thesis (a copy of which you can see here).

It was a very long road but thankfully now the pressure is lifted and I can enjoy my study of quantum groups and random walks thereon for many years to come.

As I said in the previous post, there is a duality:

Points on a Curve (Geometry) \Leftrightarrow Solutions of an Equation (Algebra)

This means we can answer geometric questions using algebra and answer algebraic questions using geometry.

Problem

Consider the following two questions:

  1. Find the tangents to a circle \mathcal{C} of a given slope.
  2. Find the tangents to a circle \mathcal{C} through a given point.

Both can be answered using the duality principle.

Example

Find the tangents to the circle

\mathcal{C}\equiv x^2+y^2-4x+6y-12=0

that are

(a) parallel to the line L\equiv 4x+3y+20=0

(b) through the point (-10,-5) [caution: the numbers here are disgusting]

Solution (a) i:

 First of all a sketch (and the remark that a tangent is a line):

circle.jpg

Here we see the circle \mathcal{C} and the line L on the bottom left. The two tangents we are looking for are as shown. They have the same slope as L and have only one intersection with \mathcal{C}. These two pieces of information will allow us to find the equations of the tangents.

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