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.

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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.

maths students

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.

Enjoy!

elliptical1

Elliptical Curves

Elliptical curves are a very important new area of mathematics which have been greatly explored over the past few decades.  They have shown tremendous potential as a tool for solving complicated number problems and also for use in cryptography.

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Andrew Wiles, who solved one of the most famous maths problems of the last 400 years, Fermat’s Last Theorem, using elliptical curves.  In the last few decades there has also been a lot of research into using elliptical curves instead of RSA encryption to keep data transfer safe online.  So, what are elliptical curves?  On a simple level they can be regarded as curves of the form:

y² = x³ +ax + b

So for example the following is an elliptical curve:

elliptical1

If we’re being a bit more accurate, we also need 4a³ + 27b² ≠ 0.  This stops the graph having “singular points” which cause problems with the calculations.  We also have a “point at infinity” which can be thought of as an extra point added on to the usual x,y plane representing infinity.  This also helps with calculations – though we don’t need to go into this in any more detail here!

What makes elliptical curves so useful is that we can create a group structure using them.  Groups are very important mathematical structures – again because of their usefulness in being applied to problem solving.  A group needs to have certain properties.  For example, we need to be able to combine 2 members of the group to create a 3rd member which is also in the group.  This is how it is done with elliptical curves:

elliptical2

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.

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

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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).

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

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We can also have elliptical curves as a collection of discrete points.  We start with the curve y² = x³ +x+1, above, and just look at the positive integer solutions mod 7.  For example, when x = 1,

y² = 1³ +1 + 1

y² = 3

So this has no integer solution.

Next, when x  = 2 we have:

y² = 2³ +2 +1 = 11.

However when we are working mod 7 we look at the remainder when 11 is divided by 7 (which is 4).  So:

y² = 4 (mod 7)

y = 2

When x = 3 we have:

y² = 3³ +3 +1 = 31

y² = 3 (mod 7)

which has no integer solutions.

In fact, all the following coordinate points satisfy the equation (mod 7):

(2,2), (0,1), (0,6), (2,5).

Even though all this might seem very abstract, these methods of calculating points on elliptical curves form the basis of elliptical cryptography.  The basic idea is that it takes computers a very long time to make these sorts of calculations – and so they can be used very effectively to encrypt data.

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.

 

 

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Hyperbolic Geometry

The usual geometry taught in school is that of Euclidean geometry – in which angles in a triangle add up 180 degrees.  This is based on the idea that the underlying space on which the triangle is drawn is flat.  However, if the underlying space in curved then this will no longer be correct.  On surfaces of constant curvature triangles will have angles greater than 180 degrees, and on surfaces of constant negative curvature triangles will have angles less than 180 degrees.  Hyperbolic geometry is based on a geometry in which the underlying space is negatively curved.

Pseudosphere

pseudosphere

The pseudosphere pictured above is an example of a surface of constant negative curvature.  It was given the name pseudosphere as a usual sphere is the opposite – a surface of constant positive curvature.  So the question arises, what would the geometry on a surface such as a pseudosphere look like?  If we imagine 2 dimensional beings who live on the surfaces of a pseudosphere, what geometry would they believe their reality to be based on?

Poincare Disc

The Poincare disc is one way we can begin to understand what hyperbolic geometry would be like for its inhabitants.  The Poincare disc is a map which translates points in the hyperbolic plane to points in the Euclidean disc.  For example, we translate points from the spherical Earth onto a flat atlas – and this helps us understand our spherical reality.  In the same way, we can arrive at a model of the hyperbolic plane that helps us to understand it, without actually having to picture a pseudosphere everytime we want to do some maths.

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The Poincare disc maps the entire hyperbolic plane onto a finite disc.  The shortest line between 2 points (which we would call a straight line in Euclidean geometry) is represented in the disc model by both circular arcs and diameters of the circle.  These circular arcs are orthogonal (at 90 degrees) to the boundary circle edge.  So our model shows that if we wanted to travel between points E and F on a pseudosphere, that the shortest route would be the curved path of the arc shown.  Indeed, being able to calculate the shortest distance between 2 points is an integral part of understanding the underlying geometry.

Distance on the Poincare Disc

The easiest case to look at is the distance from point A on the centre, to a point x on the diameter.  Remember that diameters are also straight lines in the Poincare model.  What we want to calculate is what this distance from A to x actually represents in the hyperbolic plane.  The formula to calculate this is:

Hyperbolic distance = 2tanh-1x

Where tanh-1x is the inverse of the hyperbolic tangent function.  This can be calculated with a calculator, or we can use the alternative definition for tanh-1x

2tanh-1x =  ln(1+x) – ln(1-x)

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So, if our point F is 0.5 units away from A, then this relates to a hyperbolic distance of 2tanh-10.5 = 1.0986.

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As we get further to the boundary of the unit circle this represents ever larger distances in the hyperbolic plane.  For example if our point F is 0.9 units away from A, then this relates to a hyperbolic distance of 2tanh-10.9 = 2.9444

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And if our point is 0.999999 units away from A we get a hyperbolic distance of  2tanh-10.999999 = 14.5087

In fact, as we get closer and closer to the edge of the disc (say x = 1) then the hyperbolic distance gets closer and closer to infinity.  In this way the Poincare Disc, even though it has a radius of 1, is able to represent the entire (infinite) hyperbolic plane.  We can see this behavior of 2tanh-1x by considering the graph of tanh-1x:

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y = tanh-1x has 2 vertical asymptotes at x =1 and x = -1, so as we approach these values, the graph tends to infinity.

So, remarkably the Poincare disc is able to represent an infinite amount of information within a finite space.  This disc model was hugely helpful for mathematicians as they started to investigate hyperbolic geometry.

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Mathematical art

Escher was a Dutch artist who worked in the 20th century.  He used mathematical concepts within his pieces.  The above piece is one of his most famous, which is art based on the Poincare disc.  It consists of tessellations of the same shape – but because distance is not the same as in Euclidean geometry, those fish which are closer to the boundary edge appear smaller.  It is a fantastic illustration of the strange nature of hyperbolic geometry and a pictorial representation of infinity.

If you enjoyed this post you might also like:

Non Euclidean Geometry V – The Shape of the Universe – a look at how non Euclidean geometry is part of the physics of understanding the universe.

Circular Inversion – Reflecting in a Circle – study another transformation which is used in hyperbolic geometry.

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Plotting Stewie Griffin from Family Guy

Computer aided design gets ever more important in jobs – and with graphing software we can  create art using maths functions.  For example the above graph was created by a user, Kara Blanchard on Desmos.  You can see the original graph here, by clicking on each part of the function you can see which functions describe which parts of Stewie.  There are a total of 83 functions involved in this picture.  For example, the partial ellipse:

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when x is bounded between 3.24 and 0.9, and y is bounded as less than 1.5 generates Stewie’s left cheek.  This is what he looks like without it:

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By clicking on the various functions you can discover which ones are required to complete the full drawing.  Other artwork designed by users includes:

A minion from Despicable Me 

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A sunflower:

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And Hello Kitty:

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See if you can create some designs of your own!  This could make an interesting maths investigation for anyone thinking about a career in computer design or art – as it is a field which will grow in importance in the coming years.

You might also like to look at a similar post on using Wolfram Alpha to plot the Batman and Superman logos.  

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Modeling Volcanoes – When will they erupt?

A recent post by the excellent Maths Careers website looked at how we can model volcanic eruptions mathematically. This is an important branch of mathematics – which looks to assign risk to events and these methods are very important to statisticians and insurers. Given that large-scale volcanic eruptions have the potential to end modern civilisation, it’s also useful to know how likely the next large eruption is.

The Guardian has recently run a piece on the dangers that large volcanoes pose to humans.  Iceland’s Eyjafjallajökull volcano which erupted in 2010 caused over 100,000 flights to be grounded and cost the global economy over $1 billion – and yet this was only a very minor eruption historically speaking.  For example, the Tombora eruption in Indonesia (1815) was so big that the explosion could be heard over 2000km away, and the 200 million tones of sulpher that were emitted spread across the globe, lowering global temperatures by 2 degrees Celsius.  This led to widespread famine as crops failed – and tens of thousands of deaths.

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Super volcanoes

Even this destruction is insignificant when compared to the potential damage caused by a super volcano.  These volcanoes, like that underneath Yellowstone Park in America, have the potential to wipe-out millions in the initial explosion and and to send enough sulpher and ash into the air to cause a “volcanic winter” of significantly lower global temperatures.  The graphic above shows that the ash from a Yellowstone eruption could cover the ground of about half the USA. The resultant widespread disruption to global food supplies and travel would be devastating.

So, how can we predict the probability of a volcanic eruption?  The easiest model to use, if we already have an estimated probability of eruption is the Poisson distribution:

 P(X{=}k)= \frac{\lambda^k e^{-\lambda}}{k!},

 This formula calculates the probability that X equals a given value of k.  λ is the mean of the distribution.  If X represents the number of volcanic eruptions we have Pr(X ≥1) = 1 – Pr(x = 0).  This gives us a formula for working out the probability of an eruption as 1 -e.  For example, the Yellowstone super volcano erupts around every 600,000 years.  Therefore if λ is the number of eruptions every year, we have λ = 1/600,000  ≈ 0.00000167 and 1 -e also ≈ 0.00000167. This gets more interesting if we then look at the probability over a range of years. We can do this by modifying the formula for probability as 1 -e-tλ where t is the number of years for our range.

So the probability of a Yellowstone eruption in the next 1000 years is 1 -e-0.00167 ≈ 0.00166, and the probability in the next 10,000 years is 1 -e-0.0167 ≈ 0.0164. So we have approximately a 2% chance of this eruption in the next 10,000 years.

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A far smaller volcano, like Katla in Iceland has erupted 16 times in the past 1100 years – giving a average eruption every ≈ 70 years. This gives λ = 1/70 ≈ 0.014. So we can expect this to erupt in the next 10 years with probability 1 -e-0.14 ≈ 0.0139. And in the next 30 years with probability 1 -e-0.42 ≈ 0.34.

The models for volcanic eruptions can get a lot more complicated – especially as we often don’t know the accurate data to give us an estimate for the λ.  λ can be estimated using a technique called Maximum Likelihood Estimation – which you can read about here.

If you enjoyed this post you might also like:

Black Swans and Civilisation Collapse. How effective is maths at guiding government policies?

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Can you find a sequence of consecutive integers that add up to 1000?

This puzzle is based on the excellent book A First Step to Mathematical Olympiad Problems – which is full of problems that could be extended to become exploration ideas.

Step 1 – arithmetic formula

Our first step is to write out what we want:

a + (a+1) + (a+2) + … (a +n) = 1000

next we notice that the LHS is an arithmetic series with first term a, last term a+n and n+1 terms.  Therefore we can use the sum of an arithmetic sequence formula:

Sn = 0.5n(u1 + un)

Sn = 0.5(n+1)(a + a+n) = 1000

Sn = (n+1)(2a+n) = 2000

Step 2 – logic

However, we currently have 2 unknowns, n and a, and only 1 equation – so we can’t solve this straight away.  However we do know that both a and n are integers – and n can be taken as positive.

The next step is to see that one of the brackets (n+1)(2a+n) must be odd and the other even (if  n is odd then 2a + n is odd.  If n is even then n+1 is odd).   Therefore we can look at the odd factors of 2000:

Step 3 – prime factorisation

Using prime factorisation: 2000 = 24 x 5³

Therefore any odd factors must solely come from the prime factor combinations of 5 – i.e 5, 25 and 125.

Step 4 – trial and error

So we now know that either (n+1) or (2a+n) must be 5, 25, 125.  And therefore the other bracket must be 400, 80 or 16 (as 5 x 400 = 2000 etc).  Next we can equate the (n+1) bracket to one of these 6 values, find the value of n and hence find a.  For example:

If one bracket is 5 then the other bracket is 400.

So if (n+1) = 5 and (2a+n) = 400 then n = 4 and a = 198.

This means that the sequence: 198+199+200+201+202 = 1000.

If (n+1) = 400 and (2a+n) = 5 then n = 399 and a = -197.

This means the sequence: -197 + -196+ -195 … + 201 + 202 = 1000.

We follow this same method for brackets 25, 80 and 125,16.  This gives the following other sequences:

28+29+30+…+51+52 = 1000

-54+-53+-52+…+69+70 = 1000

-27+-26+-25+…+51+52 = 1000

55+56+57+…+69+70 = 1000

So with a mixture of mathematical formulae, prime factorisation, logic and trial and error we have our solutions.  A good example of how mathematics is often solved in reality!

Mandelbrot and Julia Sets – Pictures of Infinity

The above video is of a Mandelbrot zoom.  This is a infinitely large picture – which contains fractal patterns no matter how far you enlarge it.  To put this video in perspective, it would be like starting with a piece of A4 paper and enlarging it to the size of the universe – and even at this magnification you would still see new patterns emerging.

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To understand how to make the Mandelbrot set, we first need to understand Julia sets.  Julia sets are formed by the iterative process:

Zn+1 = Zn2 + c

Here Z is a complex number (of the form a + bi) and c is a constant that we choose.  So, for example if we choose Z1 = 1+i and c = 1 then:

Z2 = Z12 + 1
Z2 =(1+i)2 + 1
Z2 = 2i + 1

We then repeat this process:
Z3 = Z22 + 1
Z3 = (2i+1)2 + 1
Z3 = 4i-3

and so on – what we are looking for is whether this iterated Z value will diverge to infinity (i.e get larger and larger) or if it will remain bounded. If diverges to infinity we colour the initial point 1+i as red on a complex axis. If it remains bounded we will colour it in black. In this case our initial point 1 + i will diverge to infinity and so it will be coloured in red.

Next we do this for every single point in the complex plane – each time seeing what happens when we iterate it many times. Each time we colour it in as red if it diverges and black if it remains bounded. Once we have done that we will have a picture which represents what happens to every point in the complex plane. This then is our Julia set.

For example the Julia set for c = 1 looks like this:

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This is because every single complex number when iterated by  Zn+1 = Zn2 + 1 will diverge to infinity (get infinitely big).

Not very interesting so far, but different values of c provide some amazing patterns.

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This above pattern is generated by c = 0.376 – 0.1566i.

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and this pattern is for c = 0.376 – 0.1566i.

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and this one is c = -0.78 + 0.1i.

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This last one for c = 0.4 + 0.1i looks different to the others – this one has patterns but they are not connected together as in the other examples.

Mandelbrot Set

This brings us on to how to calculate the Mandelbrot set.  We calculate every possible Julia set for all complex numbers c, and then for every Julia set which is connected then we colour the c value in black, and every value of c which the Julia set is disconnected we colour the c value in red.  We then have a new plot in the complex plane of c values.  This gives us the Mandelbrot set shown below:

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Don’t worry if this seem a bit complicated – it is!  You can play around making your own Julia sets by choosing a c value at this online generator.  You might also like towatch the Numberphile video on the same topic:

 If you enjoyed this post you might also like Dan Pearcy’s post on this topic which explains how Geogebra can be used to generate these sets.  Also PlusMaths have a number of posts on this amazing subject

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Reaction times – How fast are you?

Go to the Human Benchmark site and test your reaction times.  You have five attempts to press the mouse as soon as you see the screen turn green.  You can then see how your reaction times compare with people around the world.  According to the site over there have been over 15 million clicks – with a median reaction time of 251 milliseconds and a mean reaction time of 262 milliseconds.

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We can see how this data looks plotted on a chart.  As we can see this is quite a good approximation of a bell curve – but with a longer tail to the right (some people have much longer reaction times than we would expect from a pure normal distribution).  In a true normal distribution we would have the mean and the median the same.  Nevertheless this is close enough to model our data using a normal distribution.

From the data we can take the mean time as 255 milliseconds, and a standard deviation of around 35 (just by looking at the points where around 68% are within 1s.d)

So, with X ∼ N(255, 35²) we can then see how we compare with people around the world.  Reaction times significantly faster than average would suggest an ability to do well in sports such as baseball or cricket where batters need to react to the ball in a fraction of a second.

I just tried this, and got an average of 272.  I can work out what percentage of the population I’m faster than by doing the normal distribution calculation – which gives 31% of people slower than this.  Trying it again gives an average of 261 – this time 43% of people would be slower than this.

Have a go yourselves and see how you get on!

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Are you Psychic?

There have been people claiming to have paranormal powers for thousands of years.  However, scientifically we can say that as yet we still have no convincing proof that any paranormal abilities exist.  We can show this using some mathematical tests – such as the binomial or normal distribution.

ESP Test 

You can test your ESP powers on this site (our probabilities will be a little different than their ones).  You have the chance to try and predict what card the computer has chosen.  After repeating this trial 25 times you can find out if you possess psychic powers.  As we are working with discrete data and have a fixed probability of guessing (0.2) then we can use a binomial distribution.  Say I got 6 correct, do I have psychic powers?

We have the Binomial model B(25, 0.2), 25 trials and 0.2 probability of success.  So we want to find the probability that I could achieve 6 or more by luck.

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The probability of getting exactly 6 right is 0.16.  Working out the probability of getting 6 or more correct would take a bit longer by hand (though could be simplified by doing 1 – P(x ≤ 5).  Doing this, or using a calculator we find the probability is 0.38.  Therefore we would expect someone to get 6 or more correct just by guessing 38% of the time.

So, using this model, when would we have evidence for potential ESP ability?  Well, a minimum bar for our percentages would probably be 5%.  So how many do you need to get correct before there is less than a 5% of that happening by chance?

Using our calculator we can do trial and error to see that the probability of getting 9 or more correct by guessing is only 4.7%.  So, someone getting 9 correct might be showing some signs of ESP.  If we asked for a higher % threshold (such as 1%) we would want to see someone get 11 correct.

Now, in the video above, one of the Numberphile mathematicians manages to toss 10 heads in a row.  Again, we can ask ourselves if this is evidence of some extraordinary ability.  We can calculate this probability as 0.510 = 0.001. This means that such an event would only happen 0.1% of the time. But, we’re only seeing a very small part of the total video. Here’s the full version:

Suddenly the feat looks less mathematically impressive (though still an impressive endurance feat!)

You can also test your psychic abilities with this video here.

Medical Data Mining

It’s worth watching the video above, where Derren Brown manages to flip 10 heads in a row.  With Derren being a professional magician, you might expect some magic or sleight of hand – but no, it’s all filmed with a continuous camera, and no tricks.  So, how does he achieve something which should only occur with probability (0.5)10 ≈ 0.001, or 1 time in every thousand?  Understanding this trick is essential to understanding the dangers of accepting data presented to you without being aware of how it was generated.

At 7 minutes in Derren reveals the trick – it’s very easy, but also a very persuasive way to convince people something unusual is happening.  The trick is that Derren has spent the best part of an entire day tossing coins – and only showed the sequence in which he achieved 10 heads in a row.  Suddenly with this new information the result looks much less remarkable.

Scientific tests are normally performed to a 5% confidence interval – that is, if there is a less than 5% chance of something happening by chance then we regard the data as evidence to reject the null hypothesis and to accept the alternate hypothesis.   In the case of the coin toss, we would if we didn’t know better, reject the null hypothesis that this is a fair coin and conjecture that Derren is somehow affecting the results.

Selectively presenting results from trials is called data mining – and it’s a very powerful way to manipulate data.  Unfortunately it is also a widespread technique in the pharmaceutical industry when they release data on new drugs.  Trials which show a positive effect are published, those which show no effect (or negative effects) are not.  This is a massive problem – and one which has huge implications for people’s health.  After all, we are prescribed drugs based on scientific trials which attest to their efficiency.  If this data is being mined to skew results in the drug company’s favour then we may end up taking drugs that don’t work – or even make us worse.

Dr Ben Goldacre has written extensively on this topic – and an extract from his article “The Drugs Don’t Work” is well worth a read:

The Drugs Don’t Work

Reboxetine is a drug I have prescribed. Other drugs had done nothing for my patient, so we wanted to try something new. I’d read the trial data before I wrote the prescription, and found only well-designed, fair tests, with overwhelmingly positive results. Reboxetine was better than a placebo, and as good as any other antidepressant in head-to-head comparisons. It’s approved for use by the Medicines and Healthcare products Regulatory Agency (the MHRA), which governs all drugs in the UK. Millions of doses are prescribed every year, around the world. Reboxetine was clearly a safe and effective treatment. The patient and I discussed the evidence briefly, and agreed it was the right treatment to try next. I signed a prescription.

But we had both been misled. In October 2010, a group of researchers was finally able to bring together all the data that had ever been collected on reboxetine, both from trials that were published and from those that had never appeared in academic papers. When all this trial data was put together, it produced a shocking picture. Seven trials had been conducted comparing reboxetine against a placebo. Only one, conducted in 254 patients, had a neat, positive result, and that one was published in an academic journal, for doctors and researchers to read. But six more trials were conducted, in almost 10 times as many patients. All of them showed that reboxetine was no better than a dummy sugar pill. None of these trials was published. I had no idea they existed.

It got worse. The trials comparing reboxetine against other drugs showed exactly the same picture: three small studies, 507 patients in total, showed that reboxetine was just as good as any other drug. They were all published. But 1,657 patients’ worth of data was left unpublished, and this unpublished data showed that patients on reboxetine did worse than those on other drugs. If all this wasn’t bad enough, there was also the side-effects data. The drug looked fine in the trials that appeared in the academic literature; but when we saw the unpublished studies, it turned out that patients were more likely to have side-effects, more likely to drop out of taking the drug and more likely to withdraw from the trial because of side-effects, if they were taking reboxetine rather than one of its competitors.

The whole article is a fantastic (and worrying) account of regulatory failure.  At the heart of this problem lies a social and political misunderstanding of statistics which is being manipulated by drug companies for profit.  A proper regulatory framework would ensure that all trials were registered in advance and their data recorded.  Instead what happens is trials are commissioned by drugs companies, published if they are favourable and quietly buried if they are not.  This data mining would be mathematically rejected in an IB exploration coursework, yet these statistics still governs what pills doctors do and don’t prescribe.

When presented data therefore, your first question should be, “Where did this come from?” shortly followed by, “What about the data you’re not showing me?”  Lies, damn lies and statistics indeed!

If you enjoyed this post, you might also like:

How contagious is Ebola? – how we can use differential equations to model the spread of the disease.

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Tetrahedral Numbers – Stacking Cannonballs

This is one of those deceptively simple topics which actually contains a lot of mathematics – and it involves how spheres can be stacked, and how they can be stacked most efficiently.  Starting off with the basics we can explore the sequence:

1, 4, 10, 20, 35, 56….

These are the total number of cannons in a stack as the stack gets higher.  From the diagram we can see that this sequence is in fact a sum of the triangular numbers:

S1 = 1

S2 1+3

S3 1+3+6

S4 1+3+6+10

So we can sum the first n triangular numbers to get the general term of the tetrahedral numbers. Now, the general term of the triangular numbers is 0.5n2 + 0.5n therefore we can think of tetrahedral numbers as the summation:

\bf \sum_{k=1}^{n}0.5k+0.5k^2 = \sum_{k=1}^{n}0.5k+\sum_{k=1}^{n}0.5k^2

But we have known results for the 2 summations on the right hand side:

\bf \sum_{k=1}^{n}0.5k =\frac{n(n+1)}{4}

and

\bf \huge \sum_{k=1}^{n}0.5k^2 = \frac{n(n+1)(2n+1)}{12}

and when we add these two together (with a bit of algebraic manipulation!) we get:

\bf S_n= \frac{n(n+1)(n+2)}{6}

This is the general formula for the total number of cannonballs in a stack n rows high. We can notice that this is also the same as the binomial coefficient:

\bf S_n={n+2\choose3}

cannonball2

Therefore we also can find the tetrahedral numbers in Pascals’ triangle (4th diagonal column above).

The classic maths puzzle (called the cannonball problem), which asks which tetrahedral number is also a square number was proved in 1878. It turns out there are only 3 possible answers. The first square number (1) is also a tetrahedral number, as is the second square number (4), as is the 140th square number (19,600).

We can also look at something called the generating function of the sequence. This is a polynomial whose coefficients give the sequence terms. In this case the generating function is:

\bf \frac{x}{(x-1)^4} = x + 4x^2 + 10x^3 + 20x^4 ...

canonball

Having looked at some of the basic ideas behind the maths of stacking spheres we can look at a much more complicated mathematical problem. This is called Kepler’s Conjecture – and was posed 400 years ago. Kepler was a 17th century mathematician who in 1611 conjectured that there was no way to pack spheres to make better use of the given space than the stack above. The spheres pictured above fill about 74% of the given space. This was thought to be intuitively true – but unproven. It was chosen by Hilbert in the 18th century as one of his famous 23 unsolved problems. Despite much mathematical efforts it was only finally proved in 1998.

If you like this post you might also like:

The Poincare Conjecture – the search for a solution to one of mathematics greatest problems.

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