IB Maths and GCSE Maths Resources from British International School Phuket. Theory of Knowledge (ToK). Maths explorations and investigations. Real life maths. Maths careers. Maths videos. Maths puzzles and Maths lesson resources.

British International School Phuket

Welcome to the British International School Phuket’s maths website. I am currently working at BISP and so I am running my site as the school’s maths resources website for both our students and students around the world.

We are a British international school located on the tropical island of Phuket in Southern Thailand. We offer a number of scholarships each year, catering for a number of national and international standard sports stars as well as for academic excellence. You can find out more about our school here.

BISP has a very proud tradition in mathematical excellence. Our students have achieved the top in Thailand awards for the Cambridge Maths IGCSEs 3 years in a row. Pictured above are our world-class maths students for this year, Minjin Kang and Natchongrat (Oy) Terdkiatkhachorn.

Please explore the site – there is a huge amount of content! Some of the most popular includes:

A large “Flipping the classroom” videos section for IB students. This covers the entire IB HL, SL and Studies syllabus.

A new School Code Challenge activity which allows students to practice their code breaking skills – each code hides the password needed to access the next level.

Over 200 ideas to help with students’ Maths Explorations – many with links to additional information to research.

This is a classic puzzle which is discussed in some more detail by the excellent Wired article. The puzzle is best represented by the picture below. We have a hunter who whilst in the jungle stumbles across a monkey on a tree branch. However he knows that the monkey, being clever, will drop from the branch as soon as he hears the shot being fired. The question is therefore, at what angle should the hunter aim so that he still hits the monkey?

(picture from the Wired article – originally from a UCLA physics textbook)

The surprising conclusion is that counter to what you would expect, you should actually still aim at the monkey on the branch – and in this way your bullet’s trajectory will still hit the monkey as it falls. You can see a video of this experiment at the top of the page.

You can use tracking software (such as the free software tracker ) to show this working graphically. Tracker provides a video demo with the falling monkey experiment:

As you can see from the still frame, we have the gun in the bottom left corner, lined up with the origin, the red trace representing the bullet and the blue trace representing the falling monkey.

We can then generate a graph to represent this data. The red line is the height of the bullet with respect to time. The faint blue line (with yellow dots) is the height of the monkey with respect to time. We can see clearly that the red line can be modeled as a quadratic. The blue line should in theory also be a quadratic (see below):

but in our model, the blue line is so flat as to be better modeled as a linear approximation – which is shown in pink. Now we can use regression technology to find the equation of both of these lines, to show not only that they do intersect, but also the time of that intersection.

We have the linear approximation as y = -18.5t + 14.5
and the quadratic approximation as y = -56t^{2}+39t +0.1

So the 2 graphs will indeed intersect when -18.5t + 14.6 = -56t^{2}+39t +0.1

which will be around 0.45 seconds after the gun is fired.

(A more humane version, also from Wired – where we can throw the monkey a banana)

Newtonian Mathematics

The next question is can we prove this using some algebra? Of course! The key point is that the force of gravity will affect both the bullet and the falling monkey equally (it will not be affected by the different weights of the two – see the previous post here about throwing cannonballs from the Leaning Tower of Pisa). So even thought the bullet deviates from the straight line path lined up in the gun sights, the distance the bullet deviates will be exactly the same distance that the monkey falls. So they still collide. Mathematically we have:

The vertical height of the bullet given by:

y = V_{0}t – 0.5gt^{2}

Where V_{0} is the initial vertical speed, t is the time, g is the gravitational force (9.8)

The vertical height of the monkey is given by:

y = h – 0.5gt^{2}

where h is the initial vertical height of the monkey.

Therefore these will intersect when:

V_{0}t – 0.5gt^{2} = h – 0.5gt^{2}
V_{0}t = h
V_{0}/h = t

And for any given non-zero value of V_{0} we will have a t value – which represents the time of collision.

Well done – you have successfully shot the monkey!

Manchester University are running their 5th Alan Turing Cryptography Competition this January. It’s aimed at secondary and post 16 students. If you are in the UK and in year 11 or below you can register for the official prizes, for everyone else you can still register and see if you make it onto the leaderboard.

Read some of the introduction to the competition below:

Do you like breaking codes and solving ciphers?

Can you, and your friends, unravel the mystery of the Artificial Adventure?

Would you like the chance to use your mathematical skills to win some great prizes?

The competition starts on Monday 25th January, and you can register your team (or join an existing team) here. A team consists of at most 4 members. It is also possible to register as a `non-competing’ team, for instance if you’re a teacher who would like to follow the competition or if some members of your team are too old to take part. Registration opens on Monday 30th November.

The competition follows the story of two young cipher sleuths, Mike and Ellie, as they get caught up in a crptographic adventure `The Tale of the Artificial Adventure’. Every week or two weeks a new chapter of the story is released, each with a fiendish code to crack. There are six chapters in total (plus an epilogue to conclude the story). Points can be earned by cracking each code and submitting your answer. The leaderboard keeps track of how well each team is doing.

The competition starts on January 25th. Click on the competition website to register – and good luck!

Crack the Beale Papers and find a $65 Million buried treasure?

The story of a priceless buried treasure of gold, silver and jewels (worth around $65 million in today’s money) began in January 1822. A stranger by the name of Thomas Beale walked into the Washington Hotel Virginia with a locked iron box, which he gave to the hotel owner, Robert Morriss. Morriss was to look after the box for Beale as he went off on his travels.

In May 1822 Morriss received a letter from Beale which stated that the box contained papers of huge value – but that they were encoded for protection. Beale went on to ask that Morriss continue to look after the box until his return. He added that if he did not return in the next 10 years then he had instructed a close friend to send the cipher key on June 1832. After that time Morriss would be able to decipher the code and learn of the box’s secrets.

Well, Beale never returned, nor did Morriss receive the promised cipher key. Eventually he decided to open the box. Inside were three sheets of paper written in code, and an explanatory note. The note detailed that Beale had, with a group of friends discovered a seam of gold and other precious metals in Santa Fe. They had mined this over a number of years – burying the treasure in a secret location for safe keeping. The note then explained that the coded messages would give the precise location of the treasure as well as detailing which men were due a share.

Morriss devoted many years to trying to decipher the code in vain – before deciding at the age of 84 in 1862 that he should share his secret with a close friend. That friend would later publish the Beale Papers in 1885. The pamphlet that was published stirred huge interest in America – inspiring treasure hunters and amateur cryptographers to try and crack the code. The second of the 3 coded messages was cracked by the author of the pamphlet using what is known as a book code. The United States Declaration of Independence was used as the book to encode the message above.

The first number 115 refers to the 115th word in the Declaration of Independence, which is the word “instituted”. Therefore the first letter of the decoded message is “I”. The second number is 73, which refers to the 73rd word in the declaration – which is “hold”, so the second letter of the decoded message is “h”. Following this method, the following message was revealed:

I have deposited in the county of Bedford, about four miles from Buford’s, in an excavation or vault, six feet below the surface of the ground, the following articles, belonging jointly to the parties whose names are given in number three, herewith:

The first deposit consisted of ten hundred and fourteen pounds of gold, and thirty-eight hundred and twelve pounds of silver, deposited Nov. eighteen nineteen. The second was made Dec. eighteen twenty-one, and consisted of nineteen hundred and seven pounds of gold, and twelve hundred and eighty-eight of silver; also jewels, obtained in St. Louis in exchange for silver to save transportation, and valued at thirteen thousand dollars.

The above is securely packed in iron pots, with iron covers. The vault is roughly lined with stone, and the vessels rest on solid stone, and are covered with others. Paper number one describes the exact locality of the vault, so that no difficulty will be had in finding it. Source

After the pamphlet was published there was great interest in cracking the 2 remaining papers, an interest which has persisted into modern times. One of the uncracked papers is shown below:

In 1983 2 amateur treasure hunters were jailed for trying to dig up graves in Bedford, sure that they were about to find the missing gold. In 1989 a professional treasure hunter called Mel Fisher secretly bought a large plot of land after believing that the treasure was buried underneath. However nothing was found. Up until now all efforts to crack the code above have ended in failure. Perhaps the pamphlet was a giant hoax? Or perhaps the treasure is still waiting to be found.

The town of Bedford still receives visitors from around the world, keen to try and crack this centuries old puzzle. You can hire metal detectors and go looking for it yourself. The map above from 1891 shows the 4 mile radius from Buford’s tavern which is thought to contain the treasure. Maybe one day Beale’s papers will finally be cracked.

For more information on this topic read Simon Singh’s excellent The Code Book – which has more details on this case and many other code breaking puzzles throughout history.

If you want to try your own codebreaking skills, head over to our Schoolcodebreaking site – to test your wits against students from schools around the world!

This post is also inspired by the excellent book by Robert Banks – Towing Icebergs. This book would make a great investment if you want some novel ideas for a maths investigation.

The challenge is to design a parachute with a big enough area to make sure that someone can land safely on the ground. How can we go about doing this? Let’s start (as in the last post) with some Newtonian maths.

Newton’s Laws:

For an object falling through the air we have:

p_{s}gV – p_{a}gV – F_{D} = p_{s}Va

p_{s} = The density of the falling object
p_{a} = The density of the air it’s falling in
F_{D} = The drag force
g = The gravitational force
V = The volume of the falling object
a = The acceleration of the falling object

Time to simplify things

Things look a little complicated at the moment – luckily we can make our lives easier through a little simplification. p_{a} will be many magnitudes smaller than than p_{s} – as the density of air is much smaller than the density of objects like cannonballs. Therefore we ignore this part of the equation, giving an approximate equation:

p_{s}gV – F_{D} ≈ p_{s}Va

We now rewrite things to make it easier to substitute values in later.

p_{s}V = m, where m = mass of an object (as density x volume = mass)
This gives:

mg – F_{D} ≈ ma

and as mg = W (mass x gravitation force = weight) we can rewrite this as:

W – F_{D} ≈ (w/g)a

Now, the key information to know when looking at a parachute design is the terminal velocity that will be reached when the parachute is open – that means the maximum velocity that a parachutist will potentially be hitting the ground traveling.

Now, when a person is traveling at terminal velocity their acceleration is 0, so we can set a = 0 in the equation above to give:

W – F_{D} = 0

Now we need an equation for F_{D} (the drag force).
F_{D} = 0.5p_{a}C_{D}AU^{2}

where
p_{a} = density of the air
C_{D} = the drag coefficient
A = area of parachute
U = velocity

So

when the parachutist is traveling at their terminal velocity with the parachute open we have:

W – F_{D} = 0
W = 0.5p_{a}C_{D}AU^{2}

OK, nearly there. Next thing to consider is what is the maximum velocity we want someone to be traveling when they hit the ground. This is advised to be around 5 m/s – similar to jumping from a 2 metre ladder. Much more than this and you would risk breaking a bone (or worse!)

So we are finally ready to solve our equation. We want to find what value of A (the area of the parachute) will make us land safely.

We have:

p_{a} = 0.6kg/m^{3} (approximate density of air at 3000m)
C_{D} = 1.40 (a calculated drag coefficient for an open parachute)
U = velocity = 5m/s (this is the maximum velocity we want to want to avoid injury)
W = 100kg (we will have this as the combined weight of the parachutist and the parachute)

So,

W = 0.5p_{a}C_{D}AU^{2}
100 = 0.5(0.6)(1.40)A(5)^{2}
A = 9.5m^{2}

So if we had a circular parachute with radius 1.7m it should slow us down sufficiently for us to land safely.

Galileo: Throwing cannonballs off The Leaning Tower of Pisa

This post is inspired by the excellent book by Robert Banks – Towing Icebergs. This book would make a great investment if you want some novel ideas for a maths investigation.

Galileo Galilei was an Italian mathematician and astronomer who (reputedly) conducted experiments from the top of the Tower of Pisa. He dropped various objects from in order to measure how long it took for them to reach the bottom, coming to the remarkable conclusion that the objects’ weight did not affect the speed at which it fell. But does that really mean that a feather and a cannonball would fall at the same speed? Well, yes – as long as they were dropped in a vacuum. Let’s have a look at how we can prove that.

Newton’s Laws:

For an object falling through the air we have:

p_{s}gV – p_{a}gV – F_{D} = p_{s}Va

p_{s} = The density of the falling object
p_{a} = The density of the air it’s falling in
F_{D} = The drag force
g = The gravitational force
V = The volume of the falling object
a = The acceleration of the falling object

To understand where this equation comes from we note that Newton second law (Force = mass x acceleration) is

F = ma

The LHS of our equation (p_{s}gV – p_{a}gV – F_{D}) represents the forces acting on the object and the RHS (p_{s}Va) represents mass x acceleration.

Time to simplify things

Things look a little complicated at the moment – luckily we can make our lives easier through a little simplification. p_{a} will be many magnitudes smaller than than p_{s} – as the density of air is much smaller than the density of objects like cannonballs. Therefore we ignore this part of the equation, giving an approximate equation:

p_{s}gV – F_{D} ≈ p_{s}Va

Next, we can note that in a vacuum F_{D} (the drag force) will be 0 – as there is no air resistance. Therefore this can also be ignored to get:

p_{s}gV ≈ p_{s}Va

g ≈ a

But we have a = dU/dt where U = velocity, therefore,

g ≈ a
g ≈ dU/dt
g dt ≈ dU

and integrating both sides will give:

gt ≈ U

Therefore the velocity (U) of the falling object in a vacuum is only dependent on time and the gravitational force. In other words it is independent of the object’s mass. Amazing!

This might be difficult to believe – as it is quite unintuitive. So if you’re not convinced you can watch the video below in which Brian Cox tests this out in the world’s largest vacuum chamber.

If you liked this post you might also like:

War maths – how modeling projectiles plays an essential part in waging wars.

British International School, Phuket (BISP) and Rangsit University, one of Thailand’s largest higher education institutions, are proud to announce the launch of the 2015-16 Thailand Maths Challenge.

This is a competition aimed at students aged between 16 and 18, currently studying at school. Over 500 of Thailand’s top schools have been invited to participate: international schools, bilingual schools, Thai private schools, and state schools in order to find the best young mathematicians in the country.

BISP Headmaster Neil Richards said, “I am immensely proud that the school has initiated the Thailand Maths Challenge, most particularly because of the partnership between the international and Thai communities, and the close collaboration with Rangsit University, who are to be congratulated for their foresight in offering this wonderful opportunity.

“Mathematical ability is just as relevant now as it was before the onset of the computer age, and I hope that the Thailand Maths Challenge will inspire young people to pursue mathematics at the highest possible level.”

The initial round will be conducted at schools where teachers will choose the students who will represent them. The best students from this round will then be invited to a finalists day at Rangsit University, during which the overall winner of the Thailand Maths Challenge will be announced.

Taking part in this competition will offer a very impressive addition to a student’s CV, and the overall prize for the competition will be the incredible offer of a full scholarship to Rangsit University. We look forward to your participation.

You can see further information about the competition here:

For those students studying the IB Maths Higher Level Calculus option, I’ve just finished putting together video playlists to cover the whole option syllabus. These include both videos teaching the course content and also worked past paper solutions. Hopefully this should make what is a demanding unit a little bit more manageable.

Calculus Option Part 1:

Limits, limits at infinity, determining if limits exist, differentiating from first principles, L’Hopital’s Rule, Squeeze theorem, Rolle’s theorem, Mean Value theorem.

Calculus Option Part 2:

differential equations, solving differential equations through separating variables, substitution and integrating factor, sketching slope fields.

Calculus Option Part 3:

Improper integrals, Comparison test, Riemann sums, divergence test, comparison test, limit comparison test, alternating series test, absolute convergence, power series, Ratio test.

Calculus Option Part 4:

Power series, Taylor and Maclaurin series,

Other playlists

You can find all the playlists for the:
HL core maths content here.
SL maths here
Studies maths here.

The coastline paradox arises from the difficulty of measuring shapes with complicated edges such as those of countries like the Britain. As we try and be ever more accurate in our measurement of the British coastline, we get an ever larger answer! We can see this demonstrated below:

This first map represents an approximation of the British coastline with each line representing 200km. With this scale we arrive at an estimation of around 2400km. Yet if we take each line with length 50km we get the following:

This map now has a length of around 3400km. Indeed by choosing ever smaller measuring lengths we can make it much larger still. Coastlines have similar attributes to fractals – which are shapes which exhibit self similarity on ever smaller scales.

We can attempt to classify the dimension of fractals by using decimals. Just as 1 dimension represents a straight line and 2 dimensions represents a surface, we can have a pattern with dimension (say) 1.32. These dimensions make sense in terms of classifying fractal. A fractal with dimension close to 1 will be close to a straight line, one with a dimension close to 2 will be very “crinkly” indeed.

We can use the graph above, which was used by one of the founding fathers of fractal mathematics – Mandelbrot – to help expand his early ideas on the subject. The x axis is a log base 10 scale of the length chosen to measure the coastline in. The y axis is a log base 10 scale of the subsequent coastline length. So for example if we take our first estimate of the British coastline, i.e measurements of 200km, which achieved an estimate of 2400km – then we would plot the coordinate ( log(200), log(2400) ) For our second estimate this achieves the point (log(50), log(3400) ).

We can see that countries with steeper slopes (i.e those whose coastline greatly increases with ever smaller measuring scales) will have a more jagged coastline and so can be regarded as having a higher dimension. Mandelbrot assigned the coastline dimension as related to the gradient of the slope.

Finding the gradient of a log-log graph

However to find the gradient of the lines above is slightly complicated by the fact that we have a log-log plot. There is a formula we should use:

In the formula above, m is the gradient and F_{1} and F_{2} are the corresponding y values to x_{1} and x_{2}. So using our coordinate values ( log(200), log(2400) ) and (log(50), log(3400) ) we would get a slope of:

log(2400/3400)/log(200/50) = -0.251

We then take the absolute value of this and add 1 – which gives a coastline dimension of 1.251 for Britain’s West coast.

We can also read off the approximate values from the graph. If we take the points (1.5, 3.3) and (2.7, 3) then we have a slope of:

log(3/3.3)/log(2.7/1.5) = -0.162 which gives a coastline dimension of 1.162.

Actually, with a more accurate reading of this scale Mandelbrot arrived at a coastline dimension of 1.25 for Britain – agreeing with our previous working out.

The coastline dimensions of other countries

The coastline of the German land frontier was assigned a dimension of 1.15 – i.e it is not as jagged as that of Britain. Meanwhile below we can see the South African coast:

This has a very smooth coastline – and as such the log-log graph looks to have an almost flat gradient. As such it has a dimension of 1.02.

You might think that winning at rock, paper, scissors was purely a matter of chance – after all mathematically each outcome has the same probability. We can express the likelihood of winning in terms of a game theory grid:

It is clear that in theory you would expect to win, draw and lose with probability 1/3. However you can actually exploit human psychology to give yourself a significant edge at this game. Below is a report of a Chinese study on the psychology of game players:

Zhijian and co carried out their experiments with 360 students recruited from Zhejiang University and divided into 60 groups of six players. In each group, the players played 300 rounds of Rock-Paper-Scissors against each other with their actions carefully recorded.

As an incentive, the winners were paid in local currency in proportion to the number of their victories. To test how this incentive influenced the strategy, Zhijian and co varied the payout for different groups. If a loss is worth nothing and a tie worth 1, the winning payout varied from 1.1 to 100.

The results reveal a surprising pattern of behavior. On average, the players in all the groups chose each action about a third of the time, which is exactly as expected if their choices were random.

But a closer inspection of their behavior reveals something else. Zhijian and co say that players who win tend to stick with the same action while those who lose switch to the next action in a clockwise direction (where R → P → S is clockwise).

So, for example if person A chooses Rock and person B chooses Paper, then person B wins. Human nature therefore seems to mean that person B is more likely to stick to a winning strategy and choose Paper again, whilst person A is more likely to copy that previous winning behaviour and also choose Paper. A draw.

So you can exploit this by always moving anticlockwise i.e R → S → P. To look at our example again, person A chooses Rock and person B chooses Paper, then person B wins. This time person A follows his previous pattern and still chooses Paper, but person B exploits this new knowledge to choose Rock. Player B wins.

You can play against a Wolfram Alpha AI player here. This program will track your win percentage, and will also adapt its behavior to exploit any non-random behavior that you exhibit. Even though you may not be conscious of your biases, they probably will still be there – and the designers of this simulator are confident that the program will be beating you after about 50 games. Have a go!

There are some additional tips for winning at rock paper scissors – if you are in a single game competition then choose paper. This is because men are most likely to choose rock, and scissors are the least popular choice. Also you should try some reverse psychology and announce what you will throw. Most opponents will not believe you and modify their throw as a result.

Rock, Paper, Scissors, Lizard, Spock

You can of course make the game as complicated as you wish – the version above was popularised (though not invented by) The Big Bang Theory. The grid below shows the possible outcomes for this game:

And of course, why stop there? Below is a 15 throw version of the game

If you’ve honed your strategy then maybe you could compete in the a professional rock, paper, scissors tournament – here you can watch the final of a $50,000 Las Vegas competition.

If you liked this post you might also like:

Game Theory and Tic Tac Toe – Tic Tac Toe has already been solved using Game Theory – this topic also brings in an introduction to Group Theory.

This post builds on some of the ideas in the previous post on elliptical curves. This blog originally appeared in a Plus Maths article I wrote here. The excellent Numberphile video above expands on some of the ideas below.

On a (slightly simplified) level elliptical curves they can be regarded as curves of the form:

y² = x³ +ax + b

So for example the curve below is an elliptical curve. This curve also has an added point at infinity though we don’t need to worry about that here. Elliptical curve cryptography is based on the difficulty in solving arithmetic problems on these curves. If you remember from the last post, we have a special way of defining the addition of 2 points.

Let’s say take 2 points A and B on y² = x³ -4x + 1. In the example we have A = (2,1) and B = (-2,1). We now want to find an answer for A + B which also is on the elliptical curve. If we add them as we might vectors we get (0,2) – but unfortunately this is not on the curve. So, we define the addition A + B through the following geometric steps.

We join up the points A and B. This line intersects the curve in one more place, C.

We then reflect the point C in the x axis. We then define this new point C’ = A + B. In this case this means that (2,1) + (-2,1) = (1/4, -1/8).

Trying another example, y² = x³ -5x + 4 (below), with A = (1,0) and B = (0,2) we have C = (3,-4) and C’ = (3,4). Therefore (1,0) + (0,2) = (3,4).

We also need to have a definition when A and B define the same point on the curve. This will give us the definition of 2A. In this case we take the tangent to the curve at that point, and then as before find the intersection of this line and the curve, before reflecting the point. This probably is easier to understand with another graph:

Here we used the graph y² = x³ -5x + 4 again. This time point A = B = (-1.2, 2.88) and we have drawn the tangent to the curve at this point, which gives point D, which is then reflected in the x axis to give D’. D’ = (2.41, -2.43). Therefore we can say 2A = D’, or 2(-1.2, 2.88) = (2.41, -2.43).

Now addition of points is defined we can see how elliptical curve cryptography works. The basic idea is that given 2 points on the curve, say A and B, it takes a huge amount of computing power to work out the value a such that aA = B. For example, say I use the curve y² = x³ -25x to encrypt, and the 2 points on the curve are A = (-4,6) and B = (1681/144 , -62279/1728). Someone who wanted to break my encryption would need to find the value a such that a(-4,6) = (1681/144 , -62279/1728). The actual answer is a =2 which we can show graphically. As we want to show that 2(-4,6) = (1681/144 , -62279/1728) , we can use the previous method of finding the tangent at the point (-4,6):

We can then check with Geogebra which shows that B’ is indeed (1681/144 , -62279/1728). When a is chosen so that it is very large, this calculation becomes very difficult to attack using brute force methods – which would require checking 2(4,-6), 3(4,-6), 4(4,-6)… until the solution (1681/144 , -62279/1728) was found.

NSA and hacking data

Elliptical curve cryptography has some advantages over RSA cryptography – which is based on the difficulty of factorising large primes – as less digits are required to create a problem of equal difficulty. Therefore data can be encoded more efficiently (and thus more rapidly) than using RSA encryption. Currently the digital currency Bitcoin uses elliptical curve cryptography, and it is likely that its use will become more widespread as more and more data is digitalised. However, it’s worth noting that as yet no-one has proved that it has to be difficult to crack elliptical curves – there may be a novel approach which is able to solve the problem in a much shorter time. Indeed many mathematicians and computer scientists are working in this field.

Government digital spy agencies like the NSA and GCHQ are also very interested in such encryption techniques. If there was a method of solving this problem quickly then overnight large amounts of encrypted data would be accessible – and for example Bitcoin currency exchange would no longer be secure. It also recently transpired that the NSA has built “backdoor” entries into some elliptical curve cryptography algorithms which have allowed them to access data that the people sending it thought was secure. Mathematics is at the heart of this new digital arms race.

If you enjoyed this post you might also like:

RSA Encryption – the encryption system which secures the internet.

Circular inversion – learn about some other geometrical transformations used in university level mathematics.