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How does a Curve have an Equation?

November 1, 2017 in A1 Leaving Cert Maths, General, MATH6015, MATH6040, MATH6055 | 4 comments

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

A 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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Solving ‘Easy’ Equations: Part II

October 12, 2017 in A1 Leaving Cert Maths, General, Grinds, Leaving Cert Maths | Leave a comment

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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Why Do We Multiply Matrices Like We Do?

September 29, 2017 in General, MATH6040, MATH8009 | 3 comments

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 vectors. Sets 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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The Area of a Pentagon

September 22, 2017 in General | Leave a comment

A nice little question:

Given a regular pentagon with side length $s$ , what 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: A First Look

September 20, 2017 in A1 Leaving Cert Maths, General, Grinds, MATH6000, MATH6014, MATH6015, MATH6037, MATH6040, MATH6055, MATH7021 | 1 comment

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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Why Can’t I Divide by Zero?

September 13, 2017 in A1 Leaving Cert Maths, General, Grinds, MATH6000, MATH6015, MATH6037, MATH6038, MATH6040, MATH6055, MATH7016, MATH7019, MATH7021 | 3 comments

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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PhD — Finished!

May 9, 2017 in General, Quantum Groups, Random Walks on Finite Quantum Groups, Research, Research Updates | Leave a comment

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.

Tangents to Circles

December 15, 2016 in General, Grinds | Leave a comment

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:

Find the tangents to a circle $\mathcal{C}$ of a given slope.

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):

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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Differential Calculus: An Introduction

December 6, 2016 in A1 Leaving Cert Maths, General, Grinds, Leaving Cert Maths, MATH6015, MATH6037 | 2 comments

Arguably, the three central concepts in the theory of differential calculus are that of a function, that of a tangent and that of a limit. Here we introduce functions and tangents.

Functions

When looking at differential calculus, two good ways to think about functions are via algebraic geometry and interdependent variables. Neither give the proper, abstract, definition of a function, but both give a nice way of thinking about them.

Algebraic Geometry Approach

Let us set up the plane, $\Pi$ . We choose a distinguished point called the origin and a distinguished direction which we call ‘positive $x$ ‘. Draw a line through the origin in the direction of positive $x$ . This is the $x$ -axis. Choose a unit distance for the $x$ -direction.

Now, perpendicular to the $x$ -axis, draw a line through the origin. This is the $y$ -axis. By convention positive $y$ is anti-clockwise of positive $x$ . Choose a unit distance for the $y$ -direction.

This is the plane, $\Pi$ :

Now points on the plane can be associated with a pair of numbers $(a,b)$ . For example, the point a distance one along the positive $x$ and five along the negative $y$ can be denoted by the coordinates (1,-5):

Similarly, I can take a pair of numbers, say (-1,3), and this corresponds to a point on the plane.

This gives a duality:

points on the plane $\Leftrightarrow$ pairs of numbers

Now consider the completely algebraic objects

$x=3\,,\,\,y=2\,,\,\,y=x\,,\,\,y=x^2\,,\,\,y=x+2\,,\,\,x^2+y^2=1$ .

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Solving ‘Easy’ Equations: Part I

October 28, 2016 in A1 Leaving Cert Maths, General, Grinds, Leaving Cert Maths, MATH6000 | 2 comments

This post follows on from this post where the logic for the below is discussed. I am not going to define here what easy means!

Here is the strategy/guiding principle:

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.

Read the rest of this entry »

	J.P. McCarthy on MATH6040: Spring 2019, Week…
	Student on MATH6040: Spring 2019, Week…
	J.P. McCarthy on MATH6040: Spring 2019, Week…
	Student on MATH6040: Spring 2019, Week…
	J.P. McCarthy on MATH6040: Spring 2019, Week…

Category Archive

How does a Curve have an Equation?

Set Theory

Solving ‘Easy’ Equations: Part II

Why Do We Multiply Matrices Like We Do?

The Area of a Pentagon

Quadratics: A First Look

Geometry

Why Can’t I Divide by Zero?

Any Set of Numbers Collapses to a Single Number

PhD — Finished!