## Sale Prices: Fifty Seven Point Five Percent Off on Streets of Canterbury!

Whilst wandering through the streets of Canterbury this sign caught my eye. With the mathematics department’s favourite coffee shop in the back-ground, the sign promises a further drop of prices for outdoor enthusiasts.

‘Half price’ and ‘15%’ are both eye catching figures which are immediately understandable to the shopper, but the sign promises something else.

Will anyone think this is a ‘65% off’ sale and that bargains can be picked up for 35% of their recommended retail price(RRP)?

In fact a 15% reduction applied to a 50% sale is compound percentage problem best dealt with by percentage multipliers:

$0.5 \times 0.85 =0.425$

that is, half price with a further 15% applied amounts to a reduction of 57.5% with shoppers paying 42.5% of the RRP rather than 35%. Are the marketeers messing with our heads?

Perhaps nicer for the maths teachers in the coffee shop to see calculate in fractioons without electronic assistance:

$\dfrac{50}{100} \times \dfrac{85}{100}= \dfrac{1}{2} \times \dfrac{17}{20}=\dfrac{17}{40}$

Much nicer perhaps, but a ‘$\frac{23}{40}$ off’ sale doesn’t have the same ring about it I suppose.

## Edexcel Practice Paper G – Statistics and Mechanics

This paper felt a little rough around the edges as a finished product but good revision practice nevertheless. Only Word versions available with the parsing of symbols and format presenting small problems.

Notes:
– Statistics section seemed very chatty, had to write a lot for the marks on offer.
-Q4, not sure whether ‘p-value’ is defined anywhere for pupils who have done this course, context makes it clear what it is I think but could be confusing.
-Q3, as an ex-weather forecaster I didn’t like the premise of this question. It doesn’t seem at reasonable that same pressure ranges would relate to same weather at different locations. Use of the term hurricane could trouble the pedantic – Hurricanes in US, Typhoons in China.
-A lot of marks came very quickly in the Mechanics section, easy?
-I didn’t like the premise of Q7, two see-saws ‘joined’ seems to me to make one long rod. In any case the question assumes that they can move independently and that no moment is transferred through the ‘join’. They have to be side by side but able to move freely.

## Edexcel Mathematics Practice Paper A

Practice Paper A – Questions

Practice Paper A – Mark Scheme

Practice Paper A – JPED Write Out

My write out included to show students how I would have answered the questions using the style of mathematics taught in my lessons.

This paper was one of a number of revision and preparation resources published by the board.  I could only find Word originals which did not handle the typesetting of the mathematical symbols; pdf versions of these Word documents are added here.

I wonder if this paper would have passed the quality tests of the real exams.

Notes on the paper in no particular order:

• Q6 – initially confusing.  The figure is bounded by ‘arcs’, would have preferred ‘circular arcs’, there are, after all, many types of arcs and this one looks like an ellipse.  At first sight it looks like an integration questions, but once you have figured that the arcs are circular together with the presence of $\pi$ it has to be all about radians and sector area.
• Q11 – there must be an error in the mark scheme here, I am happier with my answers than Edexcel’s.

## Counting Pennies, Generating Functions and STEP

STEP I 1997, Question 1
Show that you can make up 10 pence in eleven ways using 10p, 5p, 2p and 1p coins.

In how many ways can you make up 20 pence using 20p,10p 5p, 2p and 1p coins?

There is an interesting way of approaching this question using generating functions.

A generating function is of the form:

$a_{0}+a_{1}t+a_{2}t^{2}+a_{3}t^{3}+\dots$

where the coefficients of $t$ count various problems and issues of convergence are unimportant.

Show that you can make up 10 pence in eleven ways using 10p, 5p, 2p and 1p coins.

This will be the co-efficient of $t^{10}$ in,

$(1+t+t^{2}+t^{3}+\dots+t^{10})(1+t^{2}+t^{4}+\dots+t^{10})(1+t^{5}+t^{10})(1+t^{10})$

by way of explanation:

$\overset{\textnormal{the 1p s}}{\overbrace{(1+t+t^{2}+t^{3}+\dots+t^{10})}}\overset{\textnormal{the 2p s}}{\overbrace{(1+t^{2}+t^{4}+\dots+t^{10})}}\overset{\textnormal{the 5p s}}{\overbrace{(1+t^{5}+t^{10})}}\overset{\textnormal{the 10p s}}{\overbrace{(1+t^{10})}}$

The candidate still needs to expand the function to get the answer, but having changed the problem to one of algebra, they might enjoy the benefits of practice and familiarity in the manual computation.

For those of at liberty to explore the question at leisure, a Computer Algebra System (CAS) is useful. In fact, the website Wolfram Alpha will expand the brackets for us, computing the answers to all similar counting problems in the process. The Wolfram Alpha query box understands the latex code if this is just pasted in, but only if it is presented in an explicit form, that is without the dots.  In order to simplify the question for the CAS notice that each bracket is a geometric progression, which can be summed using standard A level theory reducing the function to:

$\dfrac{(1-t^{11})}{(1-t)}\dfrac{(1-t^{12})}{(1-t^{2})}\dfrac{(1-t^{15})}{(1-t^{5})}\dfrac{(1+t^{10})}{1}$

Use of Wolfram Alpha then gives the following:

which contains our answer to the first part: 11.

Equipped with some methods and machinery we can now proceed to the second part of the question.  In fact, we can adapt the method and functions to count any similar loose change problem.

In how many ways can you make up 20 pence using 20p,10p 5p, 2p and 1p coins?

This will be the coefficient of $t^{20}$ in,

$(1+t+t^{2}+t^{3}+\dots+t^{20})(1+t^{2}+t^{4}+\dots+t^{20})(1+t^{5}+t^{10}+\dots+t^{20})(1+t^{10}+t^{20})(1+t^{20})$

$\overset{\textnormal{the 1 p s}}{\overbrace{(1+t+t^{2}+t^{3}+\dots+t^{20})}}\overset{\textnormal{the 2 p s}}{\overbrace{(1+t^{2}+t^{4}+\dots+t^{20})}}\overset{\textnormal{the 5 p s}}{\overbrace{(1+t^{5}+t^{10}+\dots+t^{20})}}\overset{\textnormal{the 10p s}}{\overbrace{(1+t^{10}+t^{20})}}\overset{\textnormal{the 20p s}}{\overbrace{(1+t^{20})}}$

which is equal to

$\dfrac{(1-t^{21})}{(1-t)}\dfrac{(1-t^{22})}{(1-t^{2})}\dfrac{(1-t^{25})}{(1-t^{5})}\dfrac{(1+t^{10}+t^{20})(1+t^{20})}{1}$

and which Wolfram Alpha gives and expansion of:

and our answer: 41.

## Rabbits, Matrices and the Golden Section – nth term of the Fibonacci Sequence by diagonalising a matrix

### Fibonacci Sequence

Leonardo Pisano, or Leonardo Fibonacci, studied rabbit populations in 1202 in the following way.

Rabbit couples (male and female) inhabit an island.  Each rabbit couple becomes fertile 2 months after being born and then begets a male-female pair every month thereafter.  If the population of the island starts with one couple, how many couples, $f_{n}$, are there after $n$ months?

To work this out one needs to add the number of rabbit couples alive after $n-1$ months, $f_{n-1}$ (since there are no deaths), with the new-born couples.  The number of new-born couples is equal to the number of  fertile rabbit couples, which is just the number of rabbit couples alive two months previously, $f_{n-2}$.  Hence,

$f_{n}=f_{n-1}+f_{n-2}$,

resulting in the numbers,

0, 1, 1, 2, 3, 5, 8, … .

Some quote the first term as 1, but let’s say that $f_{0}=0$ and start from there instead.

### Matrices

After teaching matrices as a Further Mathematics topic for many years I had always concentrated on geometric interpretations to illustrate the topic.  The topic of diagonalisation, was restricted to symmetric matrices, which produce mutually perpendicular eigen-vectors.  The denationalization process could be visualised as a rotation to a new set of axes, a readable transformation (stretch etc.) followed by a rotation back the original basis.

Recently, whilst reading a a text book on Number Theory and Cryptography (Baldoni,Ciliberto,Piacentini Cattaneo) I came across the following example, which should be within the reach of Further Mathematics students.

### Fibonacci Sequence by Matrices

The Fibonacci Sequence can be expressed in matrices:

$A=\begin{pmatrix} 0 & 1\\ 1 & 1 \end{pmatrix}$

then,

$A \begin{pmatrix} f_{n-2} \\f_{n-1} \end{pmatrix}= \begin{pmatrix} f_{n-1} \\f_{n-2} +f_{n-1}\end{pmatrix}=\begin{pmatrix} f_{n-1} \\f_{n} \end{pmatrix}$.

This is a recursive definition.  A good questions is: Is there a formula for of $f_{n}$, which does not involve calculating intermediate values?

Well, each stage of this calculation involves a matrix multiplication by $A$, thus

$A \begin{pmatrix} f_{n-2} \\f_{n-1} \end{pmatrix}=A^{n} \begin{pmatrix} f_{0} \\f_{1} \end{pmatrix}$

and all that is needed is to calculate $A^{n}$.

A little bit of matrix multiplication yields the following matrix power series for $A$:

$A^{1}=\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$

$A^{2}=\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$

$A^{3}=\begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}$

$A^{4}=\begin{pmatrix} 2 & 3 \\ 3 & 5 \end{pmatrix}$

and so on, where the Fibonacci numbers appear as entries in successive matrices.

Interesting, but finding a formula for the $n^{th}$ Fibonacci number looks to be no closer.

Had the matrix $A$ been a diagonal matrix, things would have been different because if

$B=\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$

then,

$B^{n}=\begin{pmatrix} a^{n} & 0 \\ 0 & b^{n}\end{pmatrix}$.

However, it is possible to diagonalise the matrix $A$.

An $n \times n$ matrix can be diagonalised if and only if it has $n$ distinct eigen-values.

Eigen values are given by the characteristic equation,

$\begin{vmatrix} 0-t & 1 \\ 1 & 1-t \end{vmatrix}=0$

that is,

$-t(1-t)-1=0$ or $t^{2}-t-1=0$.

The solution to this quadratic is the Golden Ratio $\Phi$,

$t=\Phi=\dfrac{ 1+ \sqrt{5}}{2}$ and $\dfrac{ 1- \sqrt{5}}{2}$

As it has distinct roots, $A$ can be diagonalised using it’s eigen-vectors,

$\begin{pmatrix} \frac{-1+\sqrt{5}}{2} \\ 1\end{pmatrix}$ and $\begin{pmatrix} \frac{-1-\sqrt{5}}{2} \\ 1\end{pmatrix}$,

to get matrix $C$,

$C=\begin{pmatrix} \frac{-1-\sqrt{5}}{2} & \frac{-1+\sqrt{5}}{2} \\ 1 & 1 \end{pmatrix}$,

in which case,

$C^{-1}AC=D=\begin{pmatrix} \frac{1-\sqrt{5}}{2} & 0 \\ 0 & \frac{1+\sqrt{5}}{2} \end{pmatrix}$

or

$A=CDC^{-1}$.

It is now an easy matter to find successive powers of $A$:

$A^{n}=(CDC^{-1})^{n}=CD^{n}C^{-1}$.

Hence,

$\begin{pmatrix} f_{n-1} \\f_{n} \end{pmatrix}=A \begin{pmatrix} f_{n-2} \\f_{n-1} \end{pmatrix}=A^{n} \begin{pmatrix} f_{0} \\f_{1} \end{pmatrix}= CD^{n}C^{-1}\begin{pmatrix} f_{0} \\f_{1} \end{pmatrix}$

where,

$\begin{pmatrix} f_{n-1} \\f_{n} \end{pmatrix}=A^{n} \begin{pmatrix} f_{0} \\f_{1} \end{pmatrix}=\begin{pmatrix} \frac{-1-\sqrt{5}}{2} & \frac{-1+\sqrt{5}}{2} \\ 1 & 1 \end{pmatrix} \begin{pmatrix} \left(\frac{1-\sqrt{5}}{2}\right)^{n} & 0 \\ 0 & \left(\frac{1+\sqrt{5}}{2}\right)^{n} \end{pmatrix}\begin{pmatrix} -\frac{1}{\sqrt{5}} & \frac{5-\sqrt{5}}{10}\\\frac{1}{\sqrt{5}} & \frac{5+\sqrt{5}}{10}\end{pmatrix}\begin{pmatrix} f_{0} \\f_{1} \end{pmatrix}$

thus,

$\begin{pmatrix} f_{n-1} \\f_{n} \end{pmatrix}=\begin{pmatrix} \frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right) ^{n-1}- \left( \frac{1-\sqrt{5}}{2} \right) ^{n-1}\right] \\ \frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right) ^{n}- \left( \frac{1-\sqrt{5}}{2} \right) ^{n}\right]\end{pmatrix}$,

and hence the formula for the $n^{th}$ Fibonacci number is,

$f_{n}=\frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right) ^{n}- \left( \frac{1-\sqrt{5}}{2} \right) ^{n}\right]$.

### Matrices and Wolfram Alpha

Tricky calculations are need to verify the above by hand.  Help is at hand from the Wolfram Alpha website.  Other computational engines and environments exist but this is free and readily available.

Encoding matrix $A$ as [[0,1],[1,1]] etc. a long string of characters can be prepared separately and then pasted into the command line as follows.

## Area of a circle – proof by exhuastion

Once $\pi$ has been defined as the ratio circumference to diameter, the area of a circle must be $\pi r^{2}$.

A proof relies on an infinite, limiting process which paves the way to some calculus-like ideas.

A circle is cut up into 6 sectors which are then rearranged into a near rectangle,

What doesn’t look too close to a rectangle at 6 sectors looks better at 26:

As the number of sectors becomes very big the shape becomes indistinguishable from a rectangle and the argument is complete.

I think that Archimedes used such arguments, or, proof by exhaustion, in some of his solid geometry work.

## Integration by Substitution

Current UK exam textbooks pass over proofs and mathematical discussions in a hurry to show the ‘how to’ of exam questions.

Integration by substitution is a little more than just backwards chain-rule and deserves a fuller treatment.

Try this,

Let,

$y=\displaystyle \int \textnormal{f}(x) \ \textnormal{d}x$

then,

$\dfrac{\textnormal{d}y}{\textnormal{d}x}=\textnormal{f}(x)$

Suppose that there exists a function g, of another variable $u$, such that $x=\textnormal{g}(u)$ and let, $\textnormal{f}(x)=\textnormal{f}(\textnormal{g}(u))=\textnormal{F}(u)$. So that,

$\dfrac{\textnormal{d}y}{\textnormal{d}x}=\textnormal{F}(u)$

Now, by the chain rule,

$\dfrac{\textnormal{d}y}{\textnormal{d}u}=\dfrac{\textnormal{d}y}{\textnormal{d}x}\times \dfrac{\textnormal{d}x}{\textnormal{d}u}=\textnormal{F}(u)\dfrac{\textnormal{d}x}{\textnormal{d}u}$

Hence,

$y=\displaystyle \int \dfrac{\textnormal{d}y}{\textnormal{d}u} \ \textnormal{d}u=\displaystyle \int \textnormal{F}(u)\dfrac{\textnormal{d}x}{\textnormal{d}u} \ \textnormal{d}u$

i.e.

$y=\displaystyle \int \textnormal{f}(x) \ \textnormal{d}x=\displaystyle \int \textnormal{F}(u)\dfrac{\textnormal{d}x}{\textnormal{d}u} \ \textnormal{d}u$

## Differentiation From First Principles

The gradient of a smooth curve, $\textnormal{f}(x)$, at a point $x$ is the gradient of the tangent to the curve at the point $x$. Point $P$ is on the curve and $Q$ is a neighbouring point whose $x$ value is displaced a small quantity, $\delta x$.

The idea behind differentiation is that as $\delta x$ becomes very small, the gradient of $PQ$ tends towards the gradient of the curve. In the limit as $\delta x$ becomes infinitesimally close to zero, the gradient $PQ$ becomes the gradient of the curve.

We write:

$\textnormal{gradient f}(x)=\dfrac{\textnormal{d}y}{\textnormal{d}x}=\lim_{\delta x \rightarrow 0}\left(\dfrac{\delta y}{\delta x}\right)=\lim_{\delta x \rightarrow 0}\left(\dfrac{\textnormal{f}(x+\delta x)-\textnormal{f}(x)}{\delta x}\right)$

there is a fair bit of analytic work missing (higher education) to make these ideas sound.

We also write:

$\dfrac{\textnormal{d}y}{\textnormal{d}x}=\textnormal{f}'(x).$

STANDARD RESULTS

Standard results can be proved for different functions.

If $\textnormal{f}(x)=x^{n}$ then

If $\textnormal{f}(x)=\sin x$, then we need to consider the small angle approximation that is if $\delta x$ radians is very small (infinitesimal), then $\delta x\approx\sin \delta x$ and $\cos \delta x \approx 1$, and compound trigonometry from which follows,

The differentiation process described above is linear and extends to more complicated functions. That is to say that if, $y=a\textnormal{f}(x)+b\textnormal{g}(x)$ where $a,b \in \mathbb{R}$,
$\dfrac{\textnormal{d}y}{\textnormal{d}x}=a\textnormal{f}'(x)+b\textnormal{g}'(x)$

## The Fundamental Theorem of Calculus

Integration is introduced as the reversal of differentiation i.e. in solving a differential equation, $\dfrac{\textnormal{d}y}{\textnormal{d}x}=\textnormal{g}(x)$. The link between integration and area is often passed over and is the subject of the Fundamental Theorem of Calculus. [The following discussion can be adapted for a decreasing function or, piece-wise, a function which successively increases or decreases.]

Consider and area function, $A(x)$, defined by the area under $\textnormal{f}(x)$ between $a$ and and a general point, $x$. If a small increment, $\delta x$, is applied to $x$ giving a small element, $\delta A$ of area. Now,

$\textnormal{f}(x)\delta x \leqslant \delta A \leqslant \textnormal{f}(x+\delta x)\delta x$

dividing though by $\delta x$, gives,

$\textnormal{f}(x) \leqslant \dfrac{\delta A}{\delta x} \leqslant \textnormal{f}(x+\delta x),$

a limit sandwich where, as $\delta x \rightarrow 0$,

$\dfrac{\textnormal{d}A}{\textnormal{d}x}=\textnormal{f}(x)$

The curve function, $\textnormal{f}(x)$ is the derivative of the area function; hence the area function is the anti-derivative of the curve function and,

$\displaystyle\int \textnormal{f}(x) \textnormal{d}x=A(x).$