The eloquence of... Maths

Proving the Chain Rule

2013-02-21T11:17:00.000Z

The chain rule is a method of differentiating a function within a function, for example $\sin x^2$. If you have studied maths to a sufficient (but not necessary) level it is likely that you will know that $\begin{align*}\frac{d}{dx}\left(fg(x)\right) = g'(x)f'g(x) \end{align*}$, but you may not know why this is the case.

We know from basic differentiation that $\begin{align*}\frac{d}{dx}(g(x)) = \lim_{h\to 0}\left(\frac{g(x+h) - g(x)}{h}\right) \end{align*}$

Let $\begin{align*}v = \frac{g(x+h) - g(x)}{h} - g'(x)\ (1)\end{align*}$ clearly $v\to 0\ as\ h\to 0$

This idea can be extended to a function of a function, as long as the function is differentiable for some function of $x,\ y$ then as $\begin{align*}\ k\to 0\ ,\ \frac{f(y+k)-f(y)}{k} \to f'(y)\end{align*}$

Let $\begin{align*}w = \frac{f(y+k)-f(y)}{k} - f'(y)\ (2)\end{align*}$ clearly $w\to 0\ as\ k\to 0$

Rearranging $(1)$ and $(2)$ we get:

$\begin{align*}g(x+h) = g(x) + \{g'(x)+v\}h\ (3) \end{align*}$

$\begin{align*}f(y+k) = f(y) + \{f'(y)+w\}k\ (4) \end{align*}$

$\begin{align*}(3) \Rightarrow fg(x+h) = f(g(x) + \{g'(x)+v\}h) \end{align*}$

If we let $k = \{g'(x)+v\}h$ and $y=g(x)$, clearly $k\to0$ as$\ h\to0$

This reduces $(4)$ to$\begin{align*}\ \ f(g(x) + \{g'(x)+v\}h) = fg(x) + \{f'g(x)+w\}\{g'(x)+v\}h \end{align*}$

The left hand side of this statement is equivalent to $fg(x+h)$, we are now in a position to simplify $\begin{align*}\frac{fg(x+h)-fg(x)}{h}\end{align*}$

$\begin{align*}\frac{fg(x+h)-fg(x)}{h} \equiv \frac{fg(x)+\{f'g(x)+w\}\{g'(x)+v\}h - fg(x)}{h} \equiv \{f'g(x)+w\}\{g'(x)+v\}\end{align*}$

We now have a reasonably familiar expression that needs a bit of tweaking to get to the final result.

$\begin{align*}LHS\to\frac{d}{dx}\left(fg(x)\right)\ as\ h\to0\ \therefore\ RHS\to\frac{d}{dx}\left(fg(x)\right)\ as\ h\to0\end{align*}$

As $h\to0\ k,v\to0, w\to0$ as $k\to0$

$\begin{align*}\Rightarrow \lim_{h\to0}\{f'g(x)+w\}\{g'(x)+v\} = g'(x)f'g(x) \end{align*}$

$\begin{align*}\frac{d}{dx}(fg(x)) = g'(x)f'g(x) \end{align*}$

And that is the chain rule! If at any point you do not understand what I have done please leave a comment.

Triangle Properties: STEP I 2004 Question 6 Solution

2013-02-18T15:52:00.002Z

This is an interesting STEP question (as many of them are!) but it gets you to, unwillingly, prove some interesting properties of triangles. I also hope that this post can act as a train of thought needed throughout a STEP question.

At first glance there are probably a few viable ways of attacking the problem, some neater than others. The most obvious one is to go at it algebraically and try to achieve the desired result.

$\begin{align*}Midpoint\ of\ BC,\ M_{BC}\left(\frac{p_2 + p_3}{2}, \frac{q_2 + q_3}{2}\right)\end{align*}$

$\begin{align*}Midpoint\ of\ AC,\ M_{AC}\left(\frac{p_1 + p_3}{2}, \frac{q_1 + q_3}{2}\right)\end{align*}$

$\begin{align*}Gradient\ of\ line\ connecting\ M_{BC}\ and\ A,\ m_1 = \frac{q_2 + q_3 - 2q_1}{p_2 + p_3 - 2p_1} \end{align*}$

$\begin{align*}Gradient\ of\ line\ connecting\ M_{AC}\ and\ B,\ m_1 = \frac{q_1 + q_3 - 2q_2}{p_1 + p_3 - 2p_2} \end{align*}$

As you can see, this turns very messy, very quickly. I could persevere, find the equation of both the lines equate them and solve for x but that is going to be a humongous algebraic slog. There are then two options, persevere or go at it via a different method. Given that this is the first part of the question looking for a more concise method is undoubtedly the way to go.

Given that we are thinking about geometrical properties of lines and their midpoints it makes sense to use vectors to solve this problem. Most importantly it will likely cut down on the algebra a lot due to their concise notation.

$\begin{align*}\ \ Let\ \overrightarrow{OA} = \mathbf{a} = \begin{bmatrix} p_1 \\ q_1 \end{bmatrix},\ \overrightarrow{OB} = \mathbf{b} = \begin{bmatrix} p_2 \\ q_2 \end{bmatrix},\ \overrightarrow{OC} = \mathbf{c} = \begin{bmatrix} p_3 \\ q_3 \end{bmatrix} \end{align*}$

$\begin{align*}Let\ the\ midpoints\ of\ AB,\ AC,\ BC\ be\ denoted\ D,\ E,\ F \end{align*}$

$\begin{align*}\ \ \therefore \overrightarrow{OD} = \mathbf{\frac{1}{2}(a+b)},\ \overrightarrow{OE} = \mathbf{\frac{1}{2}(a+c)},\ \overrightarrow{OF} = \mathbf{\frac{1}{2}(b+c)}\end{align*}$

That has set up all of the required details to begin actually answering the question. The vector equations for the line joining A and the midpoint of BC and the line joining B and the midpoint AC, are the lines $AF$ and $BE$:

$\begin{align*}\mathbf{r_{AF}} = \mathbf{a} + \lambda (\mathbf{b + c - 2a}) \end{align*}$

$\begin{align*}\mathbf{r_{BE}} = \mathbf{b} + \mu (\mathbf{a+c-2b}) \end{align*}$

They intersect when $\mathbf{r_{AF}} = \mathbf{r_{BE}}$.

$\Rightarrow \mathbf{a} - \mathbf{b} = (\mu - \lambda)\mathbf{c} + (\mu + 2\lambda)\mathbf{a} - (\lambda + 2\mu)\mathbf{b}$

$\mathbf{c}$ is not parallel to $\mathbf{a}$ or $\mathbf{b} \Rightarrow \mu - \lambda = 0$

$\Rightarrow \mathbf{a} - \mathbf{b} = 3\lambda\mathbf{a} - 3\lambda\mathbf{b}$

This must then mean that $\begin{align*}\lambda = \frac{1}{3}\end{align*}$. Denote the point of intersection $Q$, it lies on $\begin{align*}\mathbf{r_{AF}},\ \lambda = \frac{1}{3}\end{align*}$.

$\begin{align*}\ \ \ \overrightarrow{OQ} = \mathbf{a} + \frac{1}{3}(\mathbf{b} + \mathbf{c} - 2\mathbf{a}) = \frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c}) = \frac{1}{3}\begin{bmatrix} p_1+p_2+p_3 \\ q_1+q_2+q_3 \end{bmatrix}\end{align*}$

$\begin{align*}\therefore Q\left(\frac{p_1+p_2+p_3}{3},\frac{q_1+q_2+q_3}{3}\right)\end{align*}$

We have found the point of intersection we now need to show that this lies on the line connecting C to the midpoint of AB, this is the vector equation $CD$.

$\begin{align*}\mathbf{r_{CD}} = \mathbf{c} + \omega (\mathbf{a + b - 2c})\end{align*}$

$\begin{align*}\ \ \overrightarrow{OQ} = \mathbf{r_{CD}} \Rightarrow \frac{1}{3}\mathbf{a} + \frac{1}{3}\mathbf{b} - \frac{2}{3}\mathbf{c} = \omega(\mathbf{a} + \mathbf{b} - 2\mathbf{c})\end{align*}$

This is satisfied by $\omega = \frac{1}{3}$ $\therefore\ Q\ lies\ on\ CD$.

Let us think about what we have actually shown here. We have shown that for any triangle $ABC$, if we connect the vertices to the midpoint of the opposite side they all meet at a common point (they are concurrent), this point is called the centroid. This has useful implications to finding the centre of mass of an object.

To answer the next part of the question we will also use vectors, this part seems almost tailor made to do so given that we are looking for when two lines are perpendicular.

$\begin{align*}\ \ \overrightarrow{OH} = \mathbf{h} = \begin{bmatrix} p_1 + p_2 + p_3 \\ q_1 + q_2 + q_3 \end{bmatrix} = \mathbf{a}+\mathbf{b}+\mathbf{c}\end{align*}$

$\begin{align*}\mathbf{r_{AH}} = \mathbf{a} + \lambda (\mathbf{h} - \mathbf{a}) = \mathbf{a} + \lambda (\mathbf{b} + \mathbf{c}) \end{align*}$

$\begin{align*}\mathbf{r_{BC}} = \mathbf{b} + \mu (\mathbf{c} - \mathbf{b}) \end{align*}$

If $AH \perp BC,\ (\mathbf{b} + \mathbf{c}) \cdot (\mathbf{c} - \mathbf{b}) = 0 \Rightarrow \mathbf{b} \cdot \mathbf{b} = \mathbf{c} \cdot \mathbf{c} \Rightarrow |\mathbf{b}| = |\mathbf{c}| \therefore p_2^2+q_2^2 = p_3^2+q_3^2 (1)$ as required.

A virtually identical method is taken for $BH \perp AC$.

$\begin{align*}\mathbf{r_{BH}} = \mathbf{b} + \lambda (\mathbf{a} + \mathbf{c}) \end{align*}$

$\begin{align*}\mathbf{r_{AC}} = \mathbf{a} + \mu (\mathbf{c} - \mathbf{a}) \end{align*}$

If $BH \perp AC,\ (\mathbf{a} + \mathbf{c}) \cdot (\mathbf{c} - \mathbf{a}) = 0 \Rightarrow \mathbf{a} \cdot \mathbf{a} = \mathbf{c} \cdot \mathbf{c} \Rightarrow |\mathbf{a}| = |\mathbf{c}| \therefore p_1^2+q_1^2 = p_3^2+q_3^2 (2)$.

The final part can be easily deduced if we subtract the two statements we have already shown to be true, eliminating $p_3,\ q_3$.

$\begin{align*}\ \ (1) - (2) \Rightarrow p_1^2 + q_1^2 = p_2^2 + q_2^2 \Rightarrow(\mathbf{a} + \mathbf{b}) \cdot (\mathbf{b} - \mathbf{a}) = 0 \therefore AC \perp BH,\ BC \perp AH \Rightarrow AB \perp CH \end{align*}$

And that is the entirety of the question completed!

The line through a vertex which is perpendicular to the opposite side is called an altitude. We have shown that the three altitudes of a triangle meet at a common point (they are concurrent) if, this point is called the orthocentre.

This question is very nice for forcing you to prove interesting properties of triangles unwillingly, I highly recommend that you do this question yourself. If you do not understand what I have done at any point comment and I will explain.

Logarithmic Differentiation: Proving the Product Rule

2013-02-17T00:15:00.002Z

If you have standard maths to a pre-undergraduate level then it is pretty likely that you have met the product rule for differentiating two functions of $x$:$\ u,\ v$

$\begin{align*}\frac{d}{dx}(uv) = u'v+v'u\end{align*}$

But, I hear you scream, why?! Well there is a very neat proof for the product rule, but it can be extended to the quotient rule, or the product of three functions, or eighteen.

Consider $y = uv$ for some functions of $x$: $u,\ v$, this then means that:

$\ln y = \ln uv$
$\ln y = \ln u + \ln v$

We can then differentiate both sides implicitly to get:

$\begin{align*}\frac{1}{y} \frac{dy}{dx} = \frac{u'}{u} + \frac{v'}{v}\end{align*}$

$\begin{align*}\frac{1}{y} \frac{dy}{dx} = \frac{u'v + v'u}{uv}\end{align*}$

$y = uv$ so we can then multiply through by $uv$ to get:

$\begin{align*}\frac{dy}{dx} = u'v + v'u\end{align*}$

Which is the product rule! This is a very concise little proof of the product rule, but it does assume that you can differentiate implicitly and also that $\begin{align*}\frac{d}{dx}(\ln f(x)) = \frac{f'(x)}{f(x)}\end{align*}$.

Logarithmic differentiation also has useful applications to more complicated derivatives, by reducing more complex functions to ones that are much simpler to differentiate.

$\begin{align*}\ e.g\ Given\ y = xe^x\cos (1+x^2)\ find \frac{dy}{dx}\end{align*}$

Taking logs of both sides and rearranging gives:

$\begin{align*}\ \ln y = \ln x + x\ln e + \ln cos (1+x^2)\end{align*}$

We can differentiate both sides implicitly as we did earlier and we get:

$\begin{align*}\ \frac{1}{y}\frac{dy}{dx} = \frac{1}{x} + 1 - \frac{2x\sin(1+x^2)}{\cos(1+x^2)}\end{align*}$

$\begin{align*}\ \frac{1}{y}\frac{dy}{dx} = \frac{\cos(1+x^2) + x\cos(1+x^2) - 2x^2\sin(1+x^2)}{x\cos(1+x^2)} \end{align*}$

Multiplying through by $y$ we get:

$\begin{align*}\ \frac{dy}{dx} = e^x\left((x+1)\cos(1+x^2) - 2x^2\sin(1+x^2)\right) \end{align*}$

It only took a few lines to evaluate a pretty complex derivative rather than having to break it down and use the product rule on three products.

See if you can prove that $\begin{align*}\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v-v'u}{v^2}\end{align*}$also evaluate$\begin{align*}\frac{d}{dx}\left(\frac{3x^2sinx}{\sqrt{x^2+1}}\right)\end{align*}$

Fibonacci Numbers and Continued Fractions

2012-12-19T17:14:00.001Z

I feel I needed an updated post on the Fibonacci sequence with some of the more core ideas behind it, you can still access the old post.

The definition of the Fibonacci sequence is:

$\begin{align}F_{n+2}=F_{n+1}=F_n\ where\ F_0=1,\ F_1=1\end{align}$

You can see quickly that the first few values of the sequence are 1, 1, 2, 3, 5, 8, 13, ..., etc. this is a weakly increasing sequence, that is ${F}_{n+1}\ge F_n \forall\ n \ge 1$.

This clearly means that the sequence does not tend towards a limit, but does the ratio between terms tend towards a limit?

Consider that the $\lim_{n\to\infty}(\frac{F_{n+1}}{F_n})=x$. If we divide $(1)$ through by $F_{n+1}$ we get:

$\begin{align*}\frac{F_{n+2}}{F_{n+1}}=1 + \frac{F_n}{F_{n+1}}\end{align*}\tag{2}$

Clearly from our definition of the $\lim_{n\to\infty}(\frac{F_{n+1}}{F_n})=x$ we get that $(2)\equiv x=1+\frac{1}{x}$. This is easy enough to solve, we simply multiply through by $x$ and rearrange to get a quadratic we can solve.

$\begin{align*}x^2-x-1=0
\\\Rightarrow x = \frac{1\pm\sqrt{5}}{2}\end{align*}$

Only one of these two possible x values are logical, as the sequence is weakly increasing and all of the terms are greater than or equal to 0 clearly the ratio between terms is positive $\therefore\ x = \frac{1+\sqrt{5}}{2}$. This is the golden ratio, $\phi$.

Our definition that $x=1+\frac{1}{x}$ has interesting implications when we represent it as a recurrence relation, $x_{n+1}=1+\frac{1}{x_n}$. If we set $x_0 = 1$ then $x_1 = 1+\frac{1}{1},\ x_2 = 1 + \frac{1}{1+\frac{1}{1}}$, etc. This continues on such that $\lim_{n\to\infty}(x_{n+1})\equiv\phi = 1 + \frac{1}{1+\frac{1}{1+\frac{1}{\ddots}}}$, this is an example of an infinite continued fraction.

A continued fraction is essentially fractions within fractions (frac-cep-tions?), $a_1 + \frac{1}{a_2 + \frac{1}{\ddots \, + \frac{1}{a_m}}}$.

An infinite continued fraction is one that does not terminate after a finite sequence of iterations, $a_1 + \frac{1}{a_2 + \frac{1}{\ddots}}$. An irrational number can be continually, more accurately, approximated using infinite continued fractions.

If we want to represent any integers as an infinite continued fractions we can begin by thinking of the general continued fraction:

$\begin{align*}a=\frac{k}{1+\frac{k}{1+\frac{k}{\ddots}}} \Rightarrow a=\frac{k}{1+a} \Rightarrow a^2+a-k=0\end{align*}\tag{3}$

For a to be rational in this quadratic $1+4k$ must be a square number, clearly this square will be odd (odd + even = odd) from this you can quickly calculate the first few values of k to be $k = 2,\ 6,\ 12,\ 20,\ ...$, it can be quickly shown that the nth term $k_n=n^2+n$. Using the quadratic formula on $(3)$ we find that $a=\frac{-1+\sqrt{4(n^2+n)+1}}{2}\equiv\frac{-1\pm\sqrt{(2n+1)^2}}{2}\equiv\frac{2n}{2}=n$.

Therefore for all positive integers, a, we can represent them as an infinite continued fraction if and only if $k=a^2+a$. Or to put it mathematically:

$a=\frac{k}{1+\frac{k}{1+\frac{k}{\ddots}}},\ a\in\mathbb{N} \Leftrightarrow k=a^2+a$

This is one example of how useful (and fun!) continued fractions can be, they are also much more mathematically relevant than decimals as they provide a lot more useful information about the nature of the number itself.

See if you can express $\sqrt{2}$ as an infinite continued fraction. Leave any answers in the comment section.

Modular Arithmetic: Rule of 3

2012-10-26T21:09:00.001+01:00

Did you know that if a number is divisible by 3 then the sum of its digits are also divisible by 3? Have you ever wondered why? The answer to this problem lies within the field of modular arithmetic.

To begin we need to very precisely decide what our question really is. We start by defining some integer, n, with digits a_ra_r-1...a₁a₀. This then means that: n = a₀ + 10a₁ + ... + 10^r-1a_r-1 + 10^ra_r.

In order to proceed with our proof we must first prove a simpler fact, that aⁿ - bⁿ is always divisible by a - b. We will use induction to prove this lemma.

What this lemma means is that if a ≡ b mod m that aⁿ ≡ bⁿ mod m, for any integer n, too. With this fact ready we can now proceed with our proof.

It is clear that 10 ≡ 1 mod 3 (10 - 1 can be perfectly divided by 3), utilising our lemma it is clear that 10^k ≡ 1 mod 3. If a ≡ b mod m and c ≡ d mod m then ac ≡ bd mod m (the multiplicative law) and that (a + c) ≡ (b + d) mod m (the additive law), read the proofs here.

0 is obviously divisible by 3, hence a_k - a_k (the k^kt n minus the same digit) is divisible by 3 too - this means that a_k ≡ a_k mod 3. Utilising the rules of modular arithmetic we have that 10^ka_k ≡ a_k mod 3. It then follows from the additive law that n ≡ (a_r + a_r-1 + ... + a₁ + a₀) mod 3.

Therefore if 3|n then (a_r + a_r-1 + ... + a₁ + a₀) ≡ 0 mod 3, or in English, the sum of the digits of a number that is divisible by 3 will also be divisible by 3. The rule of 3 is proven.

There are also similar laws for division by 9 and 11 but I will leave these proofs up to the reader of this post, you use the exact same techniques and facts as I have used here. Leave any comments on any problems (or successes!) you have with this.

Sets: An Introduction to Sets

2012-10-06T23:42:00.000+01:00

A set is a very simple idea on the face of it, it is merely a collection or group; that is all. Sets are not objects in the real world; they are created within our minds not by our hands. An example of a set is all even numbers or all of the children in Europe or even the set of green bumblebees. But that is obviously not the sort of set we are interested in, we are only interested in ones that have interesting mathematical properties.

Sets by themselves may seem pretty uninteresting but when used in other areas of mathematics they show how powerful they truly are. They are used as a foundation from which the majority of mathematics can be derived, that is why they're so fundamentally important to mathematicians! Sets have to be defined in such a way that creates some list of numbers.

By convention, sets are denoted by capital letters, elements (read more in the next paragraph) are labelled by lower case letters. Beyond this the notation for sets is basic: you list each element, separated by a comma, and surround this list by curly braces, {}. For example, some set, Q, contains elements {p, q, r}.

There are a range of ways in which a set may be defined, either the items are defined using a semantic description for example: all of the odd integers or all prime numbers divisible by 4. Or you can define a set by listing the elements individually, for example: {2, 4, 8, 16}, with this version an ellipsis may be used to show that the set continues indefinitely in the same manner {3, 9, 27, 81, ...} like so.

A set constitutes of elements, an element is one distinct item that makes up a set. For example 1 is an element of the set of integers, the notation for an element, x, belonging to a set R is: x ∈ R. x ∉ R means x is not an element of R.

A subset is a set that entirely lies within another set, for example all prime numbers are integers. A ⊆ B means every element of A is also an element of B and this is read as "A is a subset of B". An interesting point from this definition is that for any set, S, every element of S is clearly also in S thus S is a subset of S (S ⊆ S). This is a bit strange so we introduce something more thorough than subsets and that is proper subsets: A is a proper subset of B if every element in A is also in B and there is at least one element in B that is not in A; this has the slightly different notation of A ⊂ B.

The next two you may be more familiar with if you have done maths past GCSE level you may have encountered them, they are union and intersection. A union B is essentially A and B, the notation for this is A ∪ B, it is everything that is in A and B. A intersection B is where A and B overlap and the notation for this is A ∩ B.

The union of all of the natural numbers {1, 2, 3, ...} and all of the integers {-2, -1, 0, 1, 2, ...} will result in just the set of integers, the reason for this is that it does not matter if a set contains duplicate identical elements it is the same as long as the same elements are present, not how many. {1, 2, 3} is the same set as {1, 1, 2, 3, 2, 3} as the same elements are present in both. It then follows that A ∪ B = A + B - A ∩ B, which is a vitally important fact for combining sets.

You may also be interested in the size of infinite sets, which investigates that there are some infinities larger than others!

Why Does -1 × -1 = 1?

2012-10-03T19:47:00.000+01:00

It may seem obvious that -1 × -1 = 1, but is it really as intuitive as it seems? In fact, really, it isn't nearly as obvious as it may appear and the proof can be nowhere near as rigorous as you would normally require one to be but it has to suffice because of how 'basic' the idea is.

The proof relies on the distributive law of arithmetic, which is that a(b + c) = ab + ac, this is so obvious to most of you that it may seem redundant even to state it but it is from this simple fact that the proof lies.

We will show that -1 × -1 = 1 by contradiction. We can obviously state that -1 × -1 = 1 or -1, then we assume that -1 × -1 = -1 is true. Consider -1(1 - 1) then by the distributive law we have that this equals to: -1 × 1 + -1 × -1 which from our assumption is -1 - 1 = -2. But this would then imply that -1 × 0 = -2 which is contradictory therefore -1 × -1 = 1 (which would make the equation we considered hold true, try it yourself).

This is a rather difficult proof to fully trust because we have made a pretty huge assumption, why must -1 × -1 = 1 or -1? Well, essentially, it doesn't have to but it just seems logical that it would be. As it is such a fundamental part of mathematics it is essentially defined by us in order for fundamental laws to continue to work. So if you find that the 'proof' I have given you is not enough then that's fine, it is true simply because it functions correctly..

Creating a Prime Checker

2012-08-28T19:47:00.003+01:00

For this post I will be utilising Fermat's Little Theorem to check for primes from a certain number to another number, of course you can make it check just one number if you so wish. If you do not want to program it yourself you can feel free to just download my .exe prime checker, download my source code or you may want to download my list of the first 7,000,000 prime numbers created with this prime checker.

I will be using Python to create this for a few reasons but the most important of these is that Python is very useful in the sense that it does not cap the size of a number you wish to store - which is vitally important for this program as the numbers will get very, very large. Before I start you may want to read some of my posts on basic programming in Python:

Introduction to Python

Python: Mathematical Terminology

Python: Interaction and Variables

Fermat's little theorem is that a^p-1/p always has a remainder of 1 for a prime p that is co-prime to a. And utilising this fact is how we check for a number being prime or not. One problem with Fermat's little theorem is that it can also occasionally work for when p is not actually prime (this is called a Poulet number), these are rare but to minimise the probability that the number isn't actually prime is to check the number with a different value of a multiple times.

Before we begin we need to be able to check what the highest common factor of two numbers actually is (in this case it will be a and p). To do this we will use the Euclidean algorithm, if you read that post it will tell you how to create that function and also what each part means. For the purposes of this program I will simply give you the code needed.

def gcd(a,b):
   while b != 0:
       a, b = b, a%b
   return a

Now we get down to the good bit and properly begin programming the prime checker!

def primecheck(num):
count = 0
a = 2
prime = True

We begin by creating a new user-defined function called 'primecheck', it will require one variable, which we will call 'num'. Python uses indentation to distinguish between different code segments and functions, so everything contained within the function will be indented. Three local variables will be required for the function, one to be used as a counter for when repeating Fermat's prime check to reduce the chance that the number is a Poulet number, one to be used as the 'a' in Fermat's little theorem and one to return whether the number is prime or not.

if (num - 1)/6 == int((num-1)/6) and (num - 5)/6 == int(num - 5)/6):
{Fermat's prime check}
else:
prime = False
return prime

Now that we have the local variables for the function we can begin writing the code needed to check if a number is prime or not. The algorithm itself takes a fairly long time, so we want to begin by trying to get rid of numbers that are obviously not prime without needing to do very much to the number. All prime numbers (other than 2 and 3) can be written in the form 6n + 1 or 6n + 5 (why?), so we can check that our number satisfies this before we proceed with the more processor heavy check.

If the number cannot be expressed in the form 6n + 1 with n as an integer then the check is stopped there and the number is returned as not being prime, if it does meet that criteria then the Fermat primality check is performed. After the prime check is performed the outcome is returned (whether prime is True or False).

while (count < 10):
count = count + 1
while gcd(a, num) != 1: a = a + 1 if pow(a, num - 1, num) != 1: count = 10 prime = False a = a + 1

The check is performed 10 times with a different value of a to ensure that the number is not a Poulet prime, that is why there is a loop while count is less than 10. The loop starts by making a note that the check has been performed by increasing count by 1, a check is then performed to ensure that a and num are coprime, if they are not a is changed and the check is reperformed. Once a and num are coprime we utilise the pow function, pow(a,b,c) returns the remainder of a^b/c; so in this code we are utilising this function to utilise Fermat's little theorem to check that num is prime by finding the remainder of a^{num - 1}/num and if this is not 1 then the test has failed and num is immediately returned as false. If all the checks come back with no problems then num is returned as prime.

This is it for the coding of the actual prime checker, the rest of the programming is to give the program structure, variables and how to save the outcome of each check.

num = input("First prime to check ")
lastnum = input("Last prime to check ")
while num <= lastnum:
  prime = primecheck(num)
  if prime == True:
      FILE = open("primes.txt", "a")
      FILE.write(str(num))
      FILE.write(", ")
      FILE.close()
      print num," is prime."
  else:
      print num, " is not prime."
  num = num + 2

To begin two variables are created for the first prime number to check and the last prime number to check in order to give the program an end point. While the number to be checked is less than the last number a check will be performed. A boolean variable prime is assigned to the output of the primecheck when performed on num. If prime is true then a text document called primes is opened to append and the number is added to it, read more on editing files from within Python. A message is displayed to the user to state whether the number is or is not prime. 2 is added to the previous number to ensure that the next number is odd, this relies on the fact that the first number entered was also odd, if it was not then you would be checking whether or not only even numbers are prime which obviously they will not be (except 2).

And that is it for the whole program! Hopefully yours is now working correctly, if you have any problems please leave a comment and I will help you as soon as I can. If you haven't already you may wish to download my .exe prime checker, download my source code or you may want to download my list of the first 7,000,000 prime numbers created with this prime checker.

Highest Common Factor: Proving Euclid's Algorithm

2012-08-27T19:03:00.002+01:00

The Euclidean algorithm will find the highest common factor of two numbers (the largest number that will divide both numbers). It is a rather simple algorithm to understand and implement having been discovered by the great mathematician Euclid of Alexandria 300 BC making it one of the oldest algorithms still applicable and in use in modern times.

Overview of Euclidean Algorithm

1.) Begin by inputting two numbers m and n
2.) If m < n then swap m and n (the larger number should be set to m)
3.) Divide m by n then get the remainder from this, r. If r = 0 then return n as the highest common factor and stop the algorithm.
4.) Let m = n and n = r. Repeat step 3.

The obvious questions here are "why does this work?" and "how do you know that the answer is always correct?". The best way to answer these questions is through an example.

Example: Find the highest common factor of 216 and 38.
From the algorithm we must set m = 216 and n = 38, m/n gives a remainder of 26. We now set m = 38 and n = 26, m/n gives a remainder of 12. Set m = 26 and n = 12, m/n gives a remainder of 2. Set m = 12 and n = 2, m/n gives a remainder of 0 so the highest common factor of 216 and 38 is 2.

It is clear from this algorithm that the important element from one step to the next relies on the fact that hcf(m,n) = hcf(n,r) is true. We will write this as a lemma.

This is how the algorithm works but it is not a proof as to that it will always work on any two integers, m and n. To prove this we will utilise how the algorithm in a more formal sense and then utilise those facts to prove the algorithm is consistent and thus correct.

And that is how and why the Euclidean algorithm works! It is relatively simple to program this algorithm and because of this and that it is very efficient it has many uses in mathematical programs, most importantly (to me!) is for prime checkers.

Understanding the Higgs Boson

2012-07-15T00:01:00.001+01:00

Unless you have been hiding under a rock somewhere you will almost certainly have seen the news that the Higgs boson has been 'discovered', I say this tentatively as it has not officially been confirmed just a 'Higgs-like' particle has been observed. But what even is the Higgs boson?

By the 1970s physicists had devised what is known as the "Standard Model" of particle physics - it is a theory of the elementary particles, how they interact with each other and what these interactions mean. The elementary particles can be broken down into two core categories: fermions and bosons. Fermions are matter particles, they are what make protons, atoms, stars and us, whereas bosons are what helps the fermions to "communicate", this communication is called a force.

The standard model predicted the existence of a lot of particles far before they were discovered: the bottom quark, top quark, tau neutrino and the Higgs boson. All of these particles were required for the standard model to work in it's current form. In 1964 Peter Higgs (as well as Robert Brout and Francois Englert) devised a theory (that was later proven) for how elementary particles are able to have mass, this was called the Higgs mechanism.

The Higgs mechanism requires a field pervading through the whole universe and depending on how particles interact with this field dictates what mass they have, this is called the Higgs field. In quantum theory when a field interacts with another object it acts as a particle. In the electromagnetic field this particle is a photon, in the Higgs field the particle is the Higgs boson. The discovery of this particle means that the equations that are currently in place are correct and able to describe every particle we know about and how these particles interact together via the forces incorporated in the standard model (all except gravity), so essentially we now certainly understand the quantum world to a great deal of precision!

The reason the Higgs boson has eluded us for so long is that it is very, very unstable and decays incredibly quickly - in fact it only exists for approximately one zeptosecond, a thousandth of a billionth of a billionth of a second. Because it exists for such a short amount of time you have to observe the after effects of a collision and map how much energy the produced particles have and compare this to what you would expect if a Higgs boson had decayed, the problem is that it can decay into a lot of things. So you have to perform a lot of high energy collisions between particles, observe the data and see if it fits the predictions about how the Higgs boson should behave.

Properties of Higgs Boson:

Mass: 126 GeV/c2 which is equivalent to 2.25 × 10-25kg or 134 protons.
It has no electric charge or spin
Read about how the Higgs interacts with other particles

But is this it? Do we understand everything now that we have discovered the long elusive 'God particle' (a terrible, terrible term coined by the media)? The answer is a wonderfully fanatic, no! Thankfully. In fact, the discovery really means that we can now raise more questions and further advance our understanding of everything. At the smallest scale gravity is still a total mystery to us, it is not incorporated into the current standard model at all - one of the most primitive every day forces is still a complete mystery to us! Quantum mechanics and general relativity need to be united before we can even begin to ponder that we understand everything, we are some way off from a theory of everything but the best hat in the ring currently is string theory.

Just as one final note, the Higgs boson has not been officially confirmed as to existing but that a new, previously unknown boson with a mass between 125-127 GeV/c2 that has behaviour "consistent with" a Higgs particle. Further analysis is required to fully confirm it's existence but a very cautious tick is currently next to it!

Fermat's Little Theorem

2012-07-09T00:50:00.000+01:00

Before reading through this post it is useful if you have some knowledge of modular arithmetic.

Fermat's little theorem (it's 'little' because of Fermat's comparatively more difficult to prove last theorem, not because it is less useful, in fact it is more useful) is a vital result from the field of number theory as it provides a method of checking (within reason) whether a number is prime or not. Also it is a very interesting result that utilises modular arithmetic.

Fermat's little theorem is that given a prime number, p, and any integer, a, a^p - a will divide by p to give an integer value. Or, more mathematically, p|a^p - a. So from the definitions in modular arithmetic this means that a^p ≡ a (mod p); it can also be wrote as a^p-1 ≡ 1 (mod p) - both are equally valid but the second is used more often mainly because it has a 1 on the right hand side and this is 'neater' and mathematicians love things to be neat!

There is a condition for the integer a and that is that it is coprime to p. Coprime means that the highest common factor between two numbers is 1; so for example 3 and 46 are coprime as the only factor they have in common is 1. The reason for why a must be coprime to p will come later.

When Pierre de Fermat first stated this theorem in a letter to his friend (in 1640) he did not include a proof of why this is the case as it was "too long". To attempt to 'one up' one of the greatest mathematicians in the past 400 years I will include a proof of this fact.

There are three points of contention within this proof that I feel I should justify/explain further. (A) is that these numbers are chosen intelligently in order to prove Fermat's little theorem, this was not the original proof of the theorem as it is difficult to be able to choose these numbers initially but in hindsight it is clear that these work very well.

(B) requires further explanation. The reason that when a, 2a, 3a,..., (p - 1)a are divided by p will give remainders 1, 2, 3,..., (p - 1) in some order is because of Euclid's lemma, if a prime number divides the product of two numbers then the prime number must divide at least one of the factors. Clearly none of the numbers in list A can be divided by p (a is coprime to p), so some integer in the range of list B, n, is coprime to p too - this means that na can not be divided by p so this means that a, 2a, 3a,..., (p - 1)a when divided by p must have remainders in the region of 1, 2, 3,..., (p - 1). Read more on this.

For (C) why can we cancel out (p - 1)! in modular arithmetic? It isn't an equals sign after all. If we consider ax ≡ ay (mod n), this means that ax - ay can be divided by n, factorising this we get a(x - y) can be divided by n implying that either a can be divided by n or x - y can be. In our case a is (p - 1)!, x is a^{p - 1} and y is 1; but (p - 1)! is clearly coprime to p so that means that a^{p - 1} - 1 can be divided by p so we can in fact cancel the (p - 1)!s.

And that is Fermat's little theorem fully proved! You will have almost certainly encountered Fermat's little theorem whether you knew it or not, almost all online security and encryption utilises prime numbers to stop data being intercepted and Fermat's little theorem is used to check the primality of numbers to be used.

If you do not understand anything in this post, feel I have made a mistake or need extra explanation on anything covered here leave a comment and I will get back to you as soon as I can.

Challenge Question 1: Solutions

2012-06-10T14:01:00.000+01:00

This is the worked solution to Challenge Question 1.

To begin thinking about the problem it is a must to visualise or draw out what is actually going on and begin labelling what we know.

As all that matters in this question is the ratio from side to side we can say that the largest square has a side length of 1, the next is 1/2, the next 1/4, etc. This means that the area of the first square is 1, then next is 1/4, the next 1/16, etc.

You can see that there is a ratio of 1/4 between the area of a square and the square before it. So to find the total area of all the squares you can use the formula for the sum of an infinite geometric series a/(1 - r) (explained here), multiply it by 4 (for each branch) and then add 1 (for the central square). So we get [4 × 0.25/(1 - 0.25)] + 1, which gives a total area for the squares as 7/3.

Now for the area of the shaded squares. The first shaded square has a side length of 1/4, so it has an area of 1/16, the next shaded square has a side length of 1/8 and an area of 1/64. You can see that the ratio from the area of one square to the next is 1/16. The total area of the shaded squares uses the sum of an infinite geometric series again and multiplies it by 4 (for each branch). So we get 4 × 0.125/(1 - 1/16), which equals 8/15.

So the area of all the overlapping squares is 8/15 and the total area of all the squares is 7/3, so the proportion of the shaded squares is 8/15 ÷ 7/3 = 0.22857142857... So the final answer is that the shaded squares occupy 22.86% of the total area of the squares.

SOLVED: Challenge Question 2

2012-06-08T15:02:00.000+01:00

If you are the first to solve this problem I will give you complete set of the Richard Feynman Lectures on Physics (read up upon the Feynman Lectures on Physics).

Question:

A man, who is 1.65m tall, wishes to find the height of a tree with a shadow 36.52m long. He walks 24.12m from the base of the tree along the shadow of the tree until his head is in a position where the tip of his shadow exactly overlaps the end of the tree top's shadow. How tall is the tree? Round to the nearest hundredth.

Leave answers either in the comments box or contact me with the answer through our contact page or email me at rouge.ray@gmail.com.

AQA C2 2012 Unofficial Worked Answers

2012-06-07T14:01:00.000+01:00

A lot of people found this paper very difficult as some of the questions required more thought than usual. I didn't actually sit this paper, so I couldn't wait to get the paper to see why people found it so difficult. I managed to get hold of the paper so I thought I would share my answers.

You may wish to first download the June C2 2012 paper.

a.) d = 9

b.) U₁₀₀ = 23 + 99(9)

U₁₀₀ = 914

c.) S_n = 0.5n[2a + (n-1)d]

S₂₈₀ = 140[2(23) + 279(9)]

S₂₈₀ = 357,980

a.) Area = 0.5 × 26 × 31.5 × 5/13

Area = 157.5cm²

b.) cos(sin^-1(5/13))

12/13

c.) AC² = 26² + 31.5² - (2 × 26 × 31.5 × 12/13)

AC² = 156.25

AC = 12.5cm

a.) x³ - 2x^3/2 + 1

b.) 0.25x⁴ - 0.8x^5/2 + x + c

c.) (0.25(4)⁴ - 0.8(4)^5/2 + 4) - (0.25(1)⁴ - 0.8(1)^5/2 + (1))

42.4 - 0.45

41.95

a.) U₁ = 12

U₂ = 3

b.) r = 0.25

c.) S∞ = a/(1-r)

S∞ = 12/(1-0.25)

S∞ = 16

d.) U₄ = 48(0.25)⁴ = 0.1875

S∞ = 0.1875/(1-0.25)

S∞ = 0.25

a.) rθ = 18(2π/3)
rθ = 12π

b.) i.) α = 2(π - π/3 - π/2)

α = π/3

ii.) PT = QT = 18tan(π/3) = 18√3

Area of two triangles = 2 × 0.5 × 18 × 18√3 = 324√3

Area of sector = 0.5 × 18² × 2π/3 = 108π

Shaded area = 324√3 - 108π = 221.8924551m²

Shaded area = 222m² to 3 significant figures

a.) i.) f'(2) = 3(2)² - 4/2² - 11

12 - 1 - 11 = 0

ii.) f''(x) = 6x + 8x^-3

f''(2) = 6(2) + 8(2)^-3

f''(2) = 13

iii.) f''(x) > 0, so it is a minimum point

b.) y = x³ + 4x^-1 -11x + c

1 = 2³ + 4(2)^-1 -11(2) + c

1 = 8 + 2 - 22 + c

c = 13, so the equation is:

y = x³ + 4x^-1 -11x + 13

a.) tanθ + 1 = 0 or sin²θ - 3cos²θ = 0

tanθ = -1

sin²θ = 3cos²θ, divide through by cos²θ to get:

tan²θ = 3

tanθ = ±√3

tanθ = -√3, -1, √3

b.) θ = -45°, 135°, -60°, 120°, 60°

θ = 60°, 120°, 135°

a.) It must go through (0,1)

b.) i.) 7^x = 7^2x - 12

Let a = 7^x

a = a² - 12

a² - a - 12 = 0

(a - 4)(a + 3) = 0

7^x = 4 or 7^x = -3; this is illogical so they meet at y = 4

ii.) 7^x = 4

log7^x = log4

xlog7 = log4

x = log4/log7

x = 0.712 to three significant figures

a.) h = 0.25

x = 0, y = 0; x = 0.25, y = 0.0263; x = 0.5, y = 0.0969; x = 0.75, y = 0.1938; x = 1, y = 0.3010

0.25 × 0.5[(0 + 0.3010) + 2(0.0263 + 0.0969 + 0.1938)]

= 0.117 to 3 significant figures

b.) Translated by:

c.) i.) log₁₀10 + log₁₀x²

log₁₀10 + 2log₁₀x

1 + 2log₁₀x

ii.) 1 + 2log₁₀x = log₁₀(10x²) = log₁₀(√10x)² = 2log₁₀(√10x)

Stretch parallel to the x-axis scale factor of 1/√10

iii.) 1 + 2log₁₀x = log₁₀(x² + 1)

log₁₀(10x²) = log₁₀(x² + 1)

10x² = x² + 1

x² = 1/9

x = 1/3 as x > 0

m_OP = log₁₀(10(1/3)²) ÷ 1/3

m_OP = 3log₁₀(10/9)

m_OP = log₁₀(1000/729)

Deriving the Quadratic Formula

2012-06-04T19:55:00.000+01:00

If you have studied Maths up to GCSE level then it is likely that you will have encountered a method of solving any quadratic equation (an equation of the form ax² + bx + c = 0) using the quadratic formula.

This formula is used to find where the quadratic equation crosses the x-axis (this is at y = 0). Although you have encountered this formula at GCSE it is unlikely that you will have encountered the proof as to why the formula works, which is a shame as it ties in some other GCSE Maths nicely and is in no way complex.

To solve ax² + bx + c = 0 for x you begin by completing the square, rearrange and find what x equals, simple!

And there it is, the quadratic formula! As you can see, it really isn't very difficult to derive the formula and it is a great shame that teachers do not take to the time to show students why the formula works rather than just letting them blindly accept it.

Do You Exist?

2012-06-02T15:01:00.000+01:00

This may seem somewhat ludicrous to even consider that you may not exist. I mean, you do things, you interact with people, you feel things so obviously you exist, right? Well, not necessarily...

The problem with proving whether or not you exist is that it is very hard to even pin down what existence actually is. It is defined in the dictionary as "objective reality or being", but that poses yet more questions; if we do not know what existence is how can we begin to comprehend reality?

Within mathematics in order to prove something you must only use something that is an already proved axiom. Multiplication works because it is essentially multiple additions (2 × 3 is the same as 2 + 2 + 2). And in fact addition can be proved to be true using something simpler, something more fundamental than itself; logic. You can prove that one plus one equals two (or if you prefer a well explained video), and from this simple fact all other vastly complex mathematics can be used and proved in the knowledge that it is correct.

But existence can not be tackled in the same manner, it is the most fundamental property. Existence precedes even the most basic mathematical principles. We must exist for anything to hold true, including maths. It has to be taken as a given for our principles of logic, maths, science, everything to be true. But this does not prove that we exist. Unfortunately everything else functioning so well because of one assumption does not prove the assumption, it could just be that everything else is radically wrong.

And the main problem of trying to understand our existence - you cannot attack it with maths. You cannot break existence down into a mathematical problem so ones perception of what does, or does not constitute existence is merely subjective. When opinions are the dominant factors of an argument you can never reach a correct answer because essentially, there is none.

I can feel the philosophers amongst you grinding at your teeth, beginning to pound at your keyboard reciting Descartes quote as if it were the ultimate truth. For those of you who do not know of Descartes, he also pondered whether or not he existed but then saw that if he was able to actually ponder his existence then there must on some level be something that exists to do the pondering; hence his quote "I think, therefore I am". It is a well constructed argument, it concisely answers the question using a very logical approach. But that in itself is it's downfall, it uses logic. Logic is something that can only function as a tool if in fact there is existence on some level, so actually he uses the fact that we exist to prove that we exist.

Also, why does the property of thinking create existence? This plays on the difficulty on defining what existence actually is. It almost defines existence as thinking, then citing the fact that we do think to be a proof of existence, which is fundamentally wrong.

So, do you exist? Probably. There is no real means of actually answering the question when we cannot truly define existence. But if you do not exist and your whole life and world is not truly there, you are none the wiser and will never truly know, so why does it matter? Ignorance is bliss as they say. If you wish to believe that you exist, then you exist.

SOLVED: Challenge Question 1

2012-05-17T08:45:00.000+01:00

If you are the first to solve this problem I will give you complete set of the Richard Feynman Lectures on Physics (read up upon the Feynman Lectures on Physics).

Question:
Take a square of arbitrary side length and construct four new squares with a side length half of the original positioned in the corners a quarter of the way down and across the two sides creating the corner. This would look like:

Then in each of the new squares in the furthest corner from the first square create a new square with half the side length a quarter of the way down and across the two sides creating the corner. Repeat this step for each newly constructed square to it infinity. To help visualise this I have created another graphic:

The question is, what proportion of the total area of the squares are overlapping with another square?

Leave answers either in the comments box or contact me with the answer through our contact page or email me at rouge.ray@gmail.com.

Introductory Mechanics: Maths of Snooker

2012-04-07T00:12:00.000+01:00

Whether you like to think of it or not, Maths is a part of every day life and snooker exhibits a lot of mechanical properties that can explain everything about snooker.

You may have never thought exactly why a ball actually stops moving, just that it loses speed and stops. But this means that there is a deceleration (in fact this is incorrect terminology, it is better to say negative acceleration). Newton's laws of motion is where the mechanics begins to come into play, his first law is that a body remains a constant velocity or at rest unless an external force is applied.

As there is a negative acceleration when a snooker ball is hit it means that there must be another external force on the ball besides the force that was applied when it was hit, in fact there are three forces acting upon it. The weight of the ball (note that weight and mass are different!), the reaction force of the object, and the friction between the ball and the cloth. You may be thinking what about when the ball is hit, surely that is a force? And it is! But, that is a non-constant force, it is applied just once and as it is not constant it does not affect the acceleration of the ball.

To explain what these forces are I first need to explain a common misconception. Weight is not mass. Weight changes depending on the gravitational strength of a planet (this is why astronauts jump higher on the moon), Weight = Mass × Gravitational Field Strength. The reaction force is what stops an object falling through a surface, in order to remain at rest or terminal velocity all forces must be in equilibrium (equal), so if the ball is not moving off the table or through it, the reaction force is equal to its weight. Friction is proportional to the reaction force when it is about to move, it is this that allows us to work out friction easily. Friction (F) = Coefficient of Friction (μ) × Reaction Force (R); the coefficient of friction is a property between that is unique between two materials.

And we do know what the coefficient of friction is between a snooker ball and the cloth, it is 0.5 and the mass of each ball is roughly 0.15kg. The gravitational field strength of Earth is about 9.81 ms-2, this means that we can work out what the weight and the friction is between the ball and the cloth. W = 0.15kg × 9.81 ms^-2, which gives W = 1.47N. Now that we have the weight we can work out what the friction between the ball and the cloth with the friction coefficient, F = 0.5 × 1.47N, which gives F = 0.736N. We now know the three constant forces acting on the snooker ball, only one actually moves the ball in anyway, and that is the friction.

So a person supplies an initial force to the ball, which gives the ball an instantaneous velocity in the direction it was hit. But this force is not constant so friction kicks in immediately and slows the ball down, but at what rate? Using Newton's second law of motion (Force = Mass × Acceleration), a = 0.736N ÷ 0.15kg so the acceleration is 4.9ms^-2.

So we could get an estimate of when the ball would finish up if we knew where the ball started and what it's initial velocity was. But, as you will know if you have played, the cue ball slows considerably when it hits cushions or other balls and there is a reason for this; energy. The kinetic energy of the ball can be calculated initially from the formula ½mv², where m is the mass of the ball and v is the initial velocity. When the ball hits a cushion or another ball it slows down as the energy of the ball is converted into heat and sound energy, the mass of the ball will not change (much) in a collision, which means the velocity has to decrease to compensate for the decrease in energy. There is no simple formula for working out the loss in energy from contact, which is why snooker players tend to avoid going into balls when not necessary because of the unpredictability of it.

That is all well and good, but what about actually playing a shot? Well it depends what sort of shot you wish to play, but I will talk about playing a pot. If you know the distance between the cue ball and the object ball, the object ball and the pocket and the angle between the two distances, you can work out a variety of things; I will illustrate this with an image.

The dotted lines are lengths and angles that can be worked out, the solid
lines are ones that would have to be measured.

But what does all of this actually mean? The angle θ created by the two balls and the pocket is the angle that has to be adhered to in order for the object ball to be potted. The angle α can be worked out first by working out N using the cosine rule and then using the sine rule with N and θ; ϕ can be worked out using SOHCAHTOA. Then by adding them: α + ϕ is the angle the cue ball has to be played at in order to hit the red ball at the correct angle to direct it towards to pocket. We can work out everything from here now, but first I need to introduce to SUVAT equations.

Where u is the initial velocity of
an object, v is the final velocity,
a is the constant acceleration, t is the
time taken to do the motion and s is
the displacement travelled.

So if the cue ball is hit with some velocity, what will the velocity of the ball be when it comes into contact with the object ball? Well we know the acceleration of the ball, it is decelerating due to friction, -4.9ms^-2 (both vertically and horizontally) and the distance from the cue ball to the object ball (L × cos(α + ϕ) horizontally and L × sin(α + ϕ) vertically).

So we have u, a and s and want to find v, so we need to use the equation v² = u² + 2as. Inputting this information into the equation gives that (horizontally) v_h = √[u² - 9.8Lcos(α + ϕ)] and vertically v_v = √[u² - 9.8Lsin(α + ϕ)]. You may notice that this means if the initial velocity of the cue ball is struck at less than √[9.8Lcos(α + ϕ)] horizontally and √[9.8Lsin(α + ϕ)] vertically it will not reach the object ball.

How fast the object ball will travel depends on how fast the cue ball is struck, obviously. But what velocity does the object ball need to be struck at initially in order for the ball to go in? To work this out is pretty simple, we need to know the distance the ball has to travel, the angle created from the line between then object ball and the pocket and the horizontal line from the ball (we'll call it λ) is equal to θ + α + ϕ - 180. We know that the final velocity has to be greater than 0 (v > 0) and the acceleration is -4.9ms^-2 (both vertically and horizontally). So we know v, a and s and want to find u, so we need to use v² = u² + 2as again. Inputting the values we have we get horizontally u_h > √[v² + 9.8Mcos(λ)] and vertically u_v > √[v² + 9.8Msin(λ)]. This tells you how fast the object ball has to go, but not how fast the cue ball has to go initially.

To work this out we need to use the conservation of momentum (Momentum = Mass × Velocity), the conservation of momentum states that momentum before a collision = momentum after the collision. Both balls have a mass of 0.15kg. The initial momentum is 0.15 × u, after the collision the balls move at right angles to each other with the same speed, this means that:

Where λ = θ + α + ϕ - 180, M is the distance between the
object ball and the pocket and v is the velocity you want the
object ball to finish with.

If you want the minimum velocity the cue ball has to be struck at for the object ball to go into the pocket you simply let v = 0 which gives you:

As you can see, this is a pretty tough calculation and not something you want to have to do before every shot! But the human brain is a remarkable thing and it is able to not do calculations of this magnitude by using past experiences to predict about where to hit the cue ball to hit the object ball in the correct way with the correct pace.

Some of the maths here would not work for an angle on the different side, or behind the ball, etc. You would need to remodel what I have done slightly, but it would just mean creating relationships between triangles slightly differently. I did skim over some mathematical steps in this post because it would have required a lot more text and potentially disengaged readers, if you have any problems with any of the maths here please leave a comment and I will explain.

Solving Trigonometric Equations

2012-04-04T23:26:00.000+01:00

If you have done maths up to any level past the age of 14 it is likely that you have encountered trigonometry in some shape or form. You may remember SOHCAHTOA, meaning sin(x) = opposite/adjacent in a right angled triangle; and this is the simplest form of trigonometry you will have encountered.

It is likely that you have also encountered some basic trigonometric equations. A basic equation would be cos(x) = 1, and you may know that this means that x = 0; but this is not the only value of x that solves the equation. If you have ever gone far enough as to draw the graphs then you may have noticed that there are multiple solutions to every trigonometric equation. This is because trigonometric functions are periodic, so they repeat values of y for different values of x.

To illustrate this consider cos(x) = a, let cos(α) = a, this means that α is a solution to the equation. But it isn't the only one, I will show this on the graph of y = cos(x); note that the graph is in radians and I will be explaining in terms of radians, if you do not understand please leave a comment.

Each red dot represents a solution to an equation
involving cos(x). Notice how they all lay on a
horizontal line.

So we know that the first solution is for x = α, but what are the next ones? You have to look at when the graph repeats itself. The cosine graph has a period of 2π and it is symmetrical at x = 0, 2π, ..., 2nπ; this means that cos(x) = cos(-x). This implies that x = ±α at least, and as the function repeats every 2π you can add that on to x again and again to get the same result. This means that if cos(x) = cos(α) then x = 2πn ± α. This gives us a general solution to cosine function!

Now to do with sin. Suppose that we have a solution to some equation involving sin where sin(x) = sin(α). Again you have to look at the sine graph for when it repeats itself and where it has lines of symmetry. In fact the sine graph is the same as the cosine graph, just it has been translated to the right by π/2, this means it still repeats every 2π but the line of symmetries are at x = π/2, 3π/2, ...(2n-1)π/2. The first of these mean that when it is an even multiple of π you can find another solution by adding α that is: x = 2πn + α. The second of these mean that whenever it is an odd multiple of π subtracting α gives another solution, that is: x = (2n+1)π + α. To find a range of solutions for sin(x) = sin(α) then x = 2πn + α or 2πn + π + α. This gives us a general solution to sine function!

And the easiest one of all is finding solutions to tan(x) = tan(α). The tan graph repeats every π and has no lines of symmetry. This simply means that if x = α, it also equals π + α and 2π + α, etc. To find a range of solutions for tan(x) = tan(α) then x = πn + α. This gives a general solution to the tangent function!

That is fine for when you have simple trigonometric equations, but what if you have one that contains more than one trigonometric function? Well there are trigonometric identities that help you to rearrange equations into a form that is easier to solve. I will simply list them in this post, but I will prove them in a later post. For UK readers, next to the identity I will state in which module you need to utilise them in.

- sinӨ/cosӨ = tanӨ; this is required for Core 2 and beyond.
- cos2Ө + sin2Ө = 1; this is required for Core 2 and beyond.
- cosӨ/sinӨ = cotӨ; this is required for Core 3 and beyond.
- tan2Ө + 1 = sec2Ө ; this is required for Core 3 and beyond.
- cot2Ө + 1 = cosec2Ө; this is required for Core 3 and beyond.
- sin(A + B) = sinAcosB + cosAsinB; this is required for Core 4.
- cos(A + B) = cosAcosB - sinAsinB; this is required for Core 4.
- sin(A - B) = sinAcosB - cosAsinB; this is required for Core 4.
- cos(A - B) = cosAcosB + sinAsinB; this is required for Core 4.
- tan(A ± B) = (tanA ± tanB) / (1 ∓ tanAtanB); this is required for Core 4.
- sin2Ө = 2sinӨcosӨ; this is required for Core 4.
- cos2Ө = 1 - 2sin2Ө; this is required for Core 4.
- tan2Ө = 2tanӨ / (1 - 2tan2Ө); this is required for Core 4.
- sin3Ө = 3sinӨ - 4sin3Ө; this is required for Core 4.
- cos3Ө = 4cos3Ө - 3cosӨ; this is required for Core 4.
- tan3Ө = (3tanӨ - tan3Ө) / (1 - 3tan2Ө); this is required for Core 4.

Logarithms

2012-03-16T23:59:00.001Z

The idea of logarithms is that they are the reverse operation of exponents, that is the purpose of them. That is that if a^b = x, then, log_ax = b. John Napier was the first person to introduce logarithms in the 1600s and they rapidly came into use, they were especially useful for one fact about logarithms.

That is that log_a(xy) = log_a(x) + log_a(y) . This made large multiplications a lot simpler and simply turned it into a problem of addition. And in fact we can prove this fact relatively simply, and I will do after proving one other property. That is that log_a(x^c) = c log_a(x).

Let log_a(x) = b, if we multiply both sides by c and write log_a(x) = b in exponent form we get:
c log_a(x) = bc and a^b = x. If we raise both sides of a^b = x to the power of c we get:
a^bc = x^cturning this into a log gives:
log_a(x^c) = bc, and we know that bc = c log_a(x), so this means that:
log_a(x^c) = c log_a(x), proving what we needed to know.

Now to prove that log_a(xy) = log_a(x) + log_a(y):
Let m = log_a(x) and n = log_a(y) and then write these in exponent form

x = a^mand y = aⁿthen we multiply these together to give:

x × y = a^m× aⁿ = a^m+n, we can now take log_a of both sides and evaluate

log_a(xy) = log_a(a)^m+n, then we use what we proved previously to get:

log_a(xy) = (m + n) log_aa, note that log_aa = 1 for all a.
log_a(xy) = m + n, we know from our first line that m = log_a x and n = log_a y so:
log_a(xy) = log_a(x) + log_a(y), proving what we want!

We can utilise the last two properties to also prove another property (we can also use the last method to prove it too, but this is far easier).

Consider log_a(x/y) and notice that this can be wrote as:

log_a(x × y^-1), which from using our last fact means that:
log_a(x × y^-1) = log_a(x) + log_a(y^-1), using the first property (log_a(x^c) = c log_a(x)) this means that:
log_a(x/y) = log_a(x) - log_a(y), providing another identity.

And that really is it for the basics of logarithms, you can do a lot with them so make sure you can understand these simply properties, they will make other problems a lot, lot simpler.
W4JUG97V4GUR

Introduction: Differential Equations

2012-03-11T19:58:00.001Z

Differential Equations have an absolute massive range of uses and are one of the most useful applications from maths to the real world, they have massive influences in: quantum mechanics, electromagnetic equations, biology, economics and an absolute ton more in all of physics; read more about notable differential equations. First I need to explain what a differential equation exactly is. A differential equation is any equation that involves a derivative of any degree, examples of differential equations are:

Differential equations can have different orders, when only the first derivative is present it is a first order differential equation. When there is a second derivative is involved it is said to be a differential equation of the second order. Differential equations of higher orders follow the same pattern for what order it is.

Differential equations can be either linear or non-linear. A differential equation is said to be linear if the highest order derivative of the dependent variable y can be expressed as a linear function of y and the
lower order derivatives. Hence a second order differential equation is said to be linear it must be possible to express it in the form:

To solve a differential equation you wish to find y in terms of x. Some of these are very easy to solve and can be solved by algebra and calculus methods (analytical methods), and you may notice that the first of these examples is very easy to solve and can be solved with calculus. All we need to do is integrate and we will get the answer for all values of y as y = x³+ c. But often it is not that easy to solve and you can not easily express y in terms of x, when this is the case numerical methods are very, very useful.

A numerical method is one that will find approximate solutions to the equation at a point you wish to find. There are a lot of numerical methods to solving differential equations, some more effective that others. I will introduce just the one of these in this post.

Euler's Method for Finding Solutions to Differential Equations
This method was created by the mathematician who seems to have influenced all areas of maths in some way or another, Leonard Euler. If we have some unknown function, but we have what its derivative equals and where one point is on the unknown function then we can find approximate y-values for a point further along the curve.

This method is most effective for small values of h. It can of course be repeated to give a more accurate value for y. If we know a point on some function is (3,5) and we want to find the point 3.4, rather than using a step size of 0.4, we could use one 0.1 and just repeated the algorithm 4 times. This will be more accurate because it will map the graph and it's gradient at each step of 0.1, rather than assuming the graph will have the same gradient the whole time.

This is just an introduction to what differential equations are and one relatively simple method of solving them. If you liked this please like us on Facebook.

What is Anti-Matter?

2012-03-04T20:10:00.001Z

To start with, antimatter really isn't that different from ordinary matter, it is composed of particles (anti-particles, but still particles nonetheless) just like normal matter is. In fact the only real difference between matter and antimatter is that (generally) they have the opposite (but exact same magnitude) charge to their matter counterparts.

Antimatter is, however, a relatively new concept for science. In fact the first real, serious suggestion that antimatter could even exist was in 1928, by the English theoretical physicist Paul Dirac. He started out with pretty modest intentions, he wanted to discover how an electron spins, something that was not known at the time. And he managed it, the original equation he came up with was a slightly messy looking one, but still rather neat for a problem that had alluded physicists for so long.

But Dirac wasn't satisfied there, he felt that the equation could be simplified further and he felt that notation was the key, he worked and eventually, he managed to create an equation to explain the spin of an electron with just a few terms.

He then decided that this was ready for full publication, but when he did he encountered a few problems. The solutions to the equation implied that if it was how the electron behaves it must have a particle that is identical to it, exact for it having opposite charge. It wasn't generally accepted that this could be true, but Dirac believed that the equation was constructed so elegantly and beautifully that it must be true, and in fact this particle was discovered to exist in 1932 by Carl Anderson, this was the positron.

The main useful property on anti-matter is that when it collides with matter, it annihilates. This means that all of the mass of the two particles is converted into energy, and it is a lot of energy. Using Einstein's famous equation, E = mc², we can work out how much energy is produced when annihilating an electron and positron. They each have a mass of 9.11 × 10^-31kilograms and c is the speed of light, 299,792,458 metres per second. So the energy produced in the annihilation of an electron and positron is, E = 2 × 9.11 × 10^-31 kg × 299,792,458² m/s = 1.64 × 10^-13joules.

Now this may seem incredibly low, but this is for one pair of tiny particles. If we had a collection of electrons and positrons the mass of a grain of sand (about 2.3× 10^-5 kilograms) this energy level jumps to 4.13 × 10¹²joules. This is about 50 times more than the average energy a car uses in a year. From matter no greater than the size of a grain of sand.

The energy released is in the form of gamma radiation which although it can be harmful it has very useful properties too. It is only through annihilation that PET scans (positron emission tomography) are possible. Electrons is fired at where a scan is needed to be taken from, and with the aid of a radioactive isotope it collides with a positron and produces a gamma ray. The gamma radiation is picked up via detectors and the data that is received can be reconstructed to show a detailed 'slice' of the area.

Because of the massive amounts of energy that are produced from a relatively small mass it would seem reasonable to try and harness this energy to power, well, everything. But here lies the problem. It is very difficult to produce large amounts of antimatter artificially. CERN when fully operational can produce 10 million anti-matter particles a minute, but even at this rate it would take over 100 billion years to produce just 1 gram of anti-hydrogen.

As you might imagine it is also very difficult to store anti-matter, because when it interacts with matter it annihilates into energy. In fact anti-matter has only been stored for 16 minutes at most in the whole of history, this is yet another stumbling block as to why anti-matter is very hard to utilise.

Because of both of these factors it is estimated (by NASA in 1999) that one gram of anti-hydrogen would cost $62,500,000,000,000 (that is 62.5 trillion dollars!). Because of this massive sum of money for such small amounts of anti-matter, research into it is a big thing in modern physics.

Proof of Differentiation

2012-03-03T00:00:00.000Z

This post is essentially to prove to you why differentiation gives the gradient of a curve, this is known as differentiation from first principles. This does not require you to understand how differentiation works, this is a proof of the well known principle of differentiation and why it is true for all values. If you do want to read more on the basics of differentiation you may want to download my Core 1 revision guide.

To begin explaining why differentiation works we have to consider what the gradient of something actually is, and that is the rate of change of a graph. The rate of change is the: change in y/change in x (which is what dy/dx means). So if we have some function, f(x) and we want to find the gradient of it at any given point it may be sensible to think that to find the gradient of a curve at any point (x, f(x)) we do:

But, this means that the denominator is 0, so the gradient of the curve according to this is undefined, which makes no sense at all. So we need to think of another way to try and find the gradient of a curve.

Let us consider the graph y = 2x² + 3x - 1, if we want to find the general gradient for any point x, let us consider the line created by the two x co-ordinates x and x + h, the y co-ordinates of these points are 2x² + 3x - 1 and 2x² + 4xh + 3x + 2h² + 3h - 1 respectively. So the gradient of this line is:

So this says that the gradient of y = 2x² + 3x - 1 is 4x + 3, which agrees with differentiation as we know it. But this is just an example that differentiation works, not a proof that it is true for all values. But, applying the same principles we used in our example we can prove it for all values.

Let f(x) = axⁿ, consider the line from connecting two points x and x + h, the y co-ordinates for these points is axⁿand a(x+h)ⁿ, the expansion for this may not seem immediately apparent but if you have encountered Pascal's triangle and the binomial expansion you will understand that a(x+h)ⁿcan be expanded to an extent. To fully understand what I will be next doing you may want to read up on the binomial expansion. If you really do not understand this next step, leave a comment and I will explain to the best of my ability.

And that is it! Hopefully you managed to understand what I have done, but if any points have you confused, please leave a comment. If you liked this, why not like us on Facebook? Go on, you know you want to.

Introduction to Modular Arithmetic and Congruence

2012-02-26T22:25:00.000Z

Modular arithmetic is an arithmetic system for the integers where the numbers wrap around, like on a clock for example. The numbers start at one, they go round to twelve and then start again at one, this would be an example of modulo 12. However generally in modular arithmetic we start at 0 and go to 11, before starting at 0 again. This means that 7 o'clock would be 6 mod 12 (mod is often used to shorten modulo). In fact because the number line "wraps around" it means that 19 o'clock = 7 o'clock (as you will know if you have used a 24 hour clock), this means that 18 mod 12 = 6 mod 12; because of this a mod m is wrote where a < m.

However it does not have to be just modulo 12, it can be modulo anything (as long at is a positive integer), so let us jump straight into the definition. So that you are aware of the notation I will use a|b means b divides a (this means that b/a is an integer). Let m be a positive integer and let a, b both be integers if m|a-b then a is congruent to b modulo m, wrote mathematically this is: a ≡ b mod m. Make sure you do not think "≡" means equivalent in this case, because it does not!

You can add in modular arithmetic (provided they have the same modulo). Let m be a positive integer and let a, b,c and d be integers if a ≡ b mod m and c ≡ d mod m then a + c ≡ (b+d) mod m. So basically you add the parts preceding the modulo together and then find modulo m of that.
You may also notice that a/m has a remainder of b, this is how we can quickly work out that 19 o'clock is the same as 7 o'clock, or the same as 103 o'clock. You simple divide the number by the modulo and work out the remainder.

An example of why we would use this using our clock as an example is if it is 4 o'clock now, what will the time be in 157 hours? So what we have is 3 mod 12 and 157 mod 12, adding these together we get 160 mod 12, 160/12 = 13 remainder 4. This means that 160 mod 12 = 4 mod 12, so it will be 5 o'clock 157 hours from 4 o'clock.

This should give you an idea of how to do basic addition with congruences, if you do not understand fully read what I have wrote again and then if you still do not understand, post a comment. Now we have the basic principles in place we can begin to go further. Now we will prove that a mod m + b mod m = (a+b) mod m and also how multiplication works in modular arithmetic.

Multiplication between two congruences is just as easy as addition with congruences, I will first provide a definition to you, before proving it. Let m be a positive integer and let a, b,c and d be integers if a ≡ b mod m and c ≡ d mod m then ac ≡ (bd) mod m; this is the same sort of principle as addition. Now to prove them.

First to prove addition:

Now to prove multiplication:

These are the very basics of congruences, but the are integral to modular arithmetic (I mean, think how important addition and multiplication is normally), they are very, very powerful and I will begin to explore them more and more in my upcoming posts. I will prove just one more result as it is pretty easy and uses multiplication.

The proposition is that aⁿ ≡ bⁿ mod m is also true provided a ≡ b mod m and n is a positive integer. We will prove this by induction. Let P(n) be be the statement of the proposition, then P(1) is obviously true as we already know that a ≡ b mod m. So let us assume that P(n) is true, then we have that aⁿ ≡ bⁿ mod m and that a ≡ b mod m, multiplying the two of them together (using our proved method) we get a × aⁿ ≡ b × bⁿ mod m, this then simplifies to: aⁿ⁺¹ ≡ bⁿ⁺¹ mod m, which is P(n+1). Hence P(n) is true for all values of n!

I hope that you like this post and that is helps you. If you did like it then please also like us on Facebook.

Basic Introduction to Topology

2012-02-24T12:11:00.000Z

Topology is the study of properties that are preserved within objects through twisting, stretching and deformations of that object. However you are not allowed to tear the object, this is key.

In more formal terms, topology is the study of shapes without reference to distance. Or in yet more formal terms it is the study of continuous functions. The study of continuous functions requires the knowledge of range and domain of a function, it studies functions with domains and ranges that make the function continuous. But if you knew that, it is unlikely you would be reading this blog post for absolute beginners as to what topology is!

Also a slight detour from the topic, a continuous function is one where a small change in the input results in a small change in the output too, if it does not follow this then it is said to be discontinuous. Also, just to note, if the inverse function of a continuous function is also continuous it is said to be bicontinuous.

A good way to picture the rules of topology is to picture plasticine, if you have a shape that can be twisted, stretched or deformed in anyway without being torn and put together into a new form then your original shape is topologically equivalent to the new shape you would make.

The reason that topology is useful is that it examines the properties of a shape or region and investigates the properties that are independent of geometric properties the shape or region has. This is the great advantage that topology utilises is that it does not rely on angles or distances, the only thing you need to know is the type of "continuous information" of the object.

Two objects with the same topological properties are said to be homeomorphic. The most well known example of this is that a torus and coffee cup are homeomorphic, they are (topologically speaking) the same shape. As a lot of objects are homeomorphic, it means that if we have a theorem that applies to one object in topology then it also applies to every other object that is homeomorphic to it.

An animation depicting how you
create a coffee cup from a torus.

The point of topology is not to be confusing or just strange, like all Maths it serves some purpose. Be that to merely solve more problems in Maths with a greater ease or to answer problems in the "real world", there is always a purpose to a Mathematical field. And of course, topology is no different. Topology was created to investigate the relationships between objects, and that is exactly what topology does! In fact the origins of topology stemmed from the famed Leonhard Euler when he tried to solve the problem Seven Bridges of Konigsberg, however it truly took off when George Cantor and Henri Poincaré. Poincaré (the famous Mathematician) also famously said "mathematics is not the study of the objects themselves, but the relations between them".

Previously unanswerable questions are now possible, because of the aid of topology. Some that you may wonder why anyone ever wanted to know and others that are pretty useful and interesting. This list should show you what I mean by that.

If a coffee is stirred gently then is there always a drop of coffee that finishes in the same place as it started.
It is not possible for the velocity of the wind to be non-zero at every point on Earth. This means that at some point at Earth, this means that at any time at all there is no wind whatsoever.
There are always two points on the equator that are 180° apart in longitude that have the same temperature.
A tennis ball can never be combed in such a way that a cowlick does not appear.
It is always possible to cut two irregular pancakes on a plate in half using just one slice.

And that is it for a basic introduction to topology, I will try to get more posts done on topology in the future, so please stay tuned for that! Be sure to comment any questions, tell your friends about us and like us on Facebook!