IB Maths and GCSE Maths Resources. 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.

Enjoy!

Stellar Numbers Investigation

This is an old IB internal assessment question and so can not be used for the new IB exploration – however it does give a good example of the sort of pattern investigation that is possible.

The task starts off with the fairly straightforward problem of trying to find the nth term formula for the triangular numbers:

Method 1

There are a number of ways to do this, probably the easiest is to notice that the second differences are always constant (+1 each time).  Therefore we have a quadratic sequence in the form an2 + bn + c

We can now substitute the known values when n = 1, 2, 3 into this to find 3 equations:

a(1) + b(1) + c = 1

a(2)2 + b(2) + c = 3

a(3)2 + b(3) + c = 6

this gives us:

a + b + c = 1

4a + 2b + c = 3

9a + 3b + c = 6

We can then eliminate using simultaneous equations to find a, b, c.  In fact our job is made easier by knowing that if the second difference is a constant, then the a in our formula will be half that value.  Therefore as our second difference was 1, the value of a will be 1/2.  We then find that b = 1/2 and c = 0.  So the formula for the triangular numbers is:

0.5n2 + 0.5n

Method 2

We can also derive this formula by breaking down triangular numbers into the following series:

1

1+2

1+2+3

1+2+3+4

Therefore we have the sum of an arithmetic series, with first term 1, common difference 1 and last term n, and so we can use the sum of an arithmetic series formula:

Sn = 0.5n(a1 + an)

Sn = 0.5n(1 + n) = 0.5n2 + 0.5n

Once this is done, we are asked to find the nth term for the 6-stellar numbers (with 6 vertices) below:

which give the pattern 1, 13, 37, 73

Method 1

Once again we can use the method for quadratic sequences.  The second difference is 12, giving us an2 + bn + c with a = 12/2 = 6. Substituting values gives us:

1 = 6(1)2 + b(1) + c
13 = 6(2)2 + b(2) + c

This simplifies to:

1 = 6 + b + c
13 = 24 + 2b + c

Therefore we can eliminate to find that b = -6 and c = 1.

which gives 6n2 – 6n + 1

Method 2

A more interesting method makes use of the triangular numbers.  We can first note a recurrence relationship in the stellar numbers – each subsequent pattern contains all the previous patterns inside.  In fact we can state the relationship as:

S1

S2 = S1 + outside star edge

S3 = S2 + outside star edge

S4 = S3 + outside star edge

The outside star edge of S2 can be thought of as 6 copies of the 2nd triangular number

The outside star edge of S3 can be thought of as 6 copies of the 3rd triangular number, but where we subtract 6×1 (the first triangular number) because we double count one of the internal points six times. We also subtract 6 as we double count each vertex.

The outside star edge of S4 can be thought of as 6 copies of the 4th triangular number, but where we subtract 6 x 3 (the second triangular number) because we double count three of the internal points six times. We also subtract 6 as we double count each vertex.

The outside star edge of S5 can be thought of as 6 copies of the 5th triangular number, but where we subtract 6 x 6 (the third triangular number) because we double count six of the internal points six times. We also subtract 6 as we double count each vertex.

Therefore we can form a formula for the outside star:

6(0.5n2 + 0.5n) – 6(0.5(n-2)2 + 0.5(n-2)) – 6

which simplifies to:

12(n -1)

We can now put this into our recurrence relationship:

S1 = 1

S2 = 1 + 12(n -1)

S3 = 1 + 12((n-1) -1) + 12(n -1)

S4 = 1 + 12((n-2) -1) + 12((n-1) -1) + 12(n -1)

Note that when we substituted the nth term formula for S2 into S3 we had to shift the n value to become n-1 as we were now on the 3rd term rather than 2nd term.

So:

S1 = 1

S2 = 1 + 12(n -1)

S3 = 1 + 12(n-1) + 12(n-2)

S4 = 1 + 12(n-1) + 12(n-2) + 12(n-3)

So:

S1 = 1 + 0

S2 = 1 + 12

S3 = 1 + 12+ 24

S4 = 1 + 12 + 24 + 36

So using the formula for the sum of an arithmetic Sn = 0.5n(a1 + an) we have

Sn = 1 + 0.5(n-1)(12 + 12(n-1))

Sn = 6n2 – 6n + 1

Quite a bit more convoluted – but also more interesting, and also more clearly demonstrating how the sequence is generated.

Generalising for p-stellar numbers

We can then generalise to find stellar number formulae for different numbers of vertices.  For example the 5-stellar numbers pictured above have the formula 5n2 – 5n + 1.  In fact the p-stellar numbers will have the formula pn2 – pn + 1.

We can prove this by using the same recurrence relationship before:

S1

S2 = S1 + outside star edge

S3 = S2 + outside star edge

S4 = S3 + outside star edge

and by noting that the outside star edge is found in the same way as before for a p-stellar shape – only this time we subtract p for the number of vertices counted twice.  This gives:

p(0.5n2 + 0.5n) – p(0.5(n-2)2 + 0.5(n-2)) – p

which simplifies to

2p(n-1)

and so substituting this into our recurrence formula:

S1 = 1

S2 = 1 + 2p(n-2)

S3 = 1 + 2p(n-2) + 2p(n-1)

S4 = 1 + 2p(n-3) + 2p(n-2) + 2p(n-1)

We have the same pattern as before – an arithmetic series in terms of 2p, and using Sn = 0.5n(a1 + an) we have:

Sn= 1 + 0.5(n-1)(2p + 2p(n-1) )

Sn = pn2 – pn + 1

Therefore, although our second method was slower, it allowed us to spot the pattern in the progression – and this then led very quickly to a general formula for the p-stellar numbers.

If you like this you might also like:

The Goldbach Conjecture – The Goldbach Conjecture states that every even integer greater than 2 can be expressed as the sum of 2 primes.  No one has ever managed to prove this.

Making Music With Sine Waves

Sine and cosine waves are incredibly important for understanding all sorts of waves in physics.  Musical notes can be thought of in terms of sine curves where we have the basic formula:

y = sin(bt)

where t is measured in seconds.  b is then connected to the period of the function by the formula period = 2π/b.

When modeling sound waves we normally work in Hertz – where Hertz just means full cycles (periods) per second.  This is also called the frequency.  Sine waves with different Hertz values will each have a distinct sound – so we can cycle through scales in music through sine waves of different periods.

For example the sine wave for 20Hz is:

20Hz means 20 periods per second (i.e 1 period per 1/20 second) so we can find the equivalent sine wave by using

period = 2π/b.

1/20 = 2π/b.

b = 40π

So, 20Hz is modeled by y = sin(40πt)

You can plot this graph using Wolfram Alpha, and then play the sound file to hear what 20Hz sounds like.  20Hz is regarded as the lower range of hearing spectrum for adults – and is a very low bass sound.

The middle C on a piano is modeled with a wave of 261.626Hz.  This gives the wave

which has the equation, y = sin(1643.84πt).  Again you can listen to this sound file on Wolfram Alpha.

At the top end of the sound spectrum for adults is around 16,000 – 20,000Hz.  Babies have a ability to hear higher pitched sounds, and we gradually lose this higher range with age.  This is the   sine wave for 20,000Hz:

which has the equation, y = sin(40,000πt). See if you can hear this file - warning it’s a bit painful!

As well as sound waves, the whole of the electromagnetic spectrum (radio waves, microwaves, infrared, visible light, ultraviolet, x rays and gamma rays) can also be thought of in terms of waves of different frequencies.  So, modelling waves using trig graphs is an essential part of understanding the physical world.

If you enjoyed this post you might also like:
Fourier Transforms – the most important tool in mathematics? - how we can use advanced mathematics to understand waves – with applications for everything from WIFI, JPEG compression, DNA analysis and MRI scans.

Surviving the Zombie Apocalypse

This is part 2 in the maths behind zombies series. See part 1 here

We have previously looked at how the paper from mathematicians from Ottawa University discuss the mathematics behind surviving the zombie apocalypse – and how the mathematics used has many other modelling applications – for understanding the spread of disease and the diffusion of gases. In the previous post we saw how the zombie diffusion rate could be predicted by the formula:

In this equation Z(x,t) stands for the density of zombies at point x and time t. Z0 stands for the initial zombie density – where all zombies are starting at the same point (x between 0 and 1). L stands for the edge of the domain.  This is a 1 dimensional model – where zombies only travel in a straight line.  For modelling purposes, this would be  somewhat equivalent to being trapped in a 50 metre by 1 metre square fenced area – with (0,0) as the bottom left corner of the fence. L would be 50 in this case, and all zombies would initially be in the 1 metre square which went through the origin.

We saw that as the time, t gets large this equation can be approximated by:

Which means that after a long length of time our 50 metre square fenced area will have an equal density of zombies throughout. If we started with 100 zombies in our initial 1 metre square area (say emerging from a tomb), then Z0 = 100 and with L = 50 we would have an average density of 100/2 = 2 zombies per metre squared.

When will the zombies arrive?

So, say you have taken the previous post’s advice and run as far away as possible.  So, you’re at the edge of the 50 metre long fence.  The next question to ask therefore, how long before the zombies reach you? To answer this we need to solve the initial equation Z(x,t) to find t when x = 50 and Z(50,t) = 1.  We solve to find Z(50,t) = 1 because this represents the time t when there is a density of 1 zombie at distance 50 metres from the origin.  In other words when a zombie is standing where you are now!  Solving this would be pretty tough, so we do what mathematicians like to do, and take an approximation.  This approximate solution for t is given by:

where L is the distance we’re standing away (50 metres in this case) and D is the diffusion rate.  D can be altered to affect the speed of the zombies.  In the study they set D as 100 – which is claimed to be consistent with a slow, shuffling zombie walk.  Therefore the time the zombies will take to arrive is approximately t = 0.32(50)2/100 = 8 minutes. If we are a slightly further distance away (say we are trapped along a 100 metre fence) then the zombies will arrive in approximately t = 0.32(100)2/100 = 32 minutes.

Fight or flight?

Fighting (say by lobbing missiles at the oncoming hordes) would slow the diffusion rate D, but would probably be less effective than running – as the time is rapidly increased by the L2 factor.  Let’s look at a scenario to compare:

You are 20 metres from the zombies.  You can decide to spend 1 minute running an extra 30 metres away (you’re not in good shape) to the edge of the fence (no rocks here) or can spend your time lobbing rocks with your home-made catapult to slow the advance.  Which scenario makes more sense?

Scenario 1

You get to the edge of the fence in 1 minute.  The zombies will get to the edge of the fence in t = 0.32(50)2/100 = 8 minutes.  You therefore have an additional 7 minutes to sit down, relax, and enjoy your last few moments before the zombies arrive.

Scenario 2

You successfully manage to slow the diffusion rate to D = 50 as the zombies are slowed by your sharp-shooting.  The zombies will arrive in 0.32(20)2/50 = 2.6 minutes.  If only you’d paid more attention in maths class.

If you liked this post you might also like:

How contagious is Ebola? – using differential equations to model infections.

Modelling for Zombies

Some mathematicians at the University of Ottawa have just released a paper looking at the mathematics behind a zombie apocalypse.  What are the best strategies for avoiding being eaten?  How quickly would zombies spread through the population?  This may seem a little silly as zombies aren’t real – but actually the mathematics behind how diseases spread through a population is very useful – and, well, zombies are as good a way as any to introduce this.

The graphic above from the paper shows how zombie movement can be modelled.  Given that zombies randomly move around, and any bumping would lead to a tendency towards finding space, they are modelled in the same way that we model the diffusion of gas.  If you start with a small concentrated number of particles they will spread out to fill the given space like shown above.

Diffusion can be modelled by the diffusion equation above.  We have:

t: time (in specified units)

x: position of the x axis.

w: the density of zombies at time t and point x.  We could also write w(t,x) in function notation.

a: a is a constant.

The “curly d” in the equation means the partial differential.  This works the same as normal differentiation but when we differentiate we are only interested in differentiating the denominator letter – and act as though all other letters are constants.  This is easier to show with an example.

z = 3xy2

The partial differential of z with respect to x is 3y2
The partial differential of z with respect to y is 6xy

So, going back to our diffusion equation, we need to find a function w(x,t) which satisfies this equation – and then we can use this function to model the spread of zombies through an area.  There are lots of different solutions to this equation (see a list here).  One of the easiest is:

w(x,t) = A(x2 + 2at) + B

where we have introduced 2 new constants, A and B.

We can check that this works by finding the left handside and right handside of the diffusion equation:

Therefore as the LHS and RHS are equal, the diffusion equation is satisfied. Therefore we have the following zombie density model:

w(x,t) = A(x2 + 2at) + B

this will tell us at point x and time t what the zombie density is.  We would need particular values to then find A, B and a.  For example, we can restrict x between 0 and 1 and t between 1 and 5, then set A = -1, B = 21, a = 2 to give:

w(x,t) = (-x2 + -4t) + 21

This begins to fit the behavior we want – at any fixed point x the density will decrease with time, and as we move further away from the initial point (x = 0) we have lower density.  This is only very rough however.

A more complicated solution to the diffusion equation is given above.  In this equation Z(x,t) stands for the density of zombies at point x and time t. Z0 stands for the initial zombie density – where all zombies are starting at the same point (x between 0 and 1). L stands for the edge of the domain.  This is a 1 dimensional model – where zombies only travel in a straight line.  For modelling purposes, this would be  somewhat equivalent to being trapped in a 50 metre by 1 metre square fenced area – with (0,0) as the bottom left corner of the fence. L would be 50 in this case, and all zombies would initially be in the 1 metre square which went through the origin.

Luckily as t gets large this equation can be approximated by:Which means that after a long length of time our 50 metre square fenced area will have an equal density of zombies throughout. If we started with 100 zombies in our initial 1 metre square area (say emerging from a tomb), then with Z0 = 100 and with L = 50 we would have an average density of 100/2 = 2 zombies per metre squared.  In other words zombies would be evenly spaced out across all available space.

So, what advice can you take from this when faced with a zombie apocalypse? Well if zombies move according to diffusion principles then initially you have a good advantage to outrun them – after-all they will be moving randomly and you will be running linearly as far away as possible. That will give you some time to prepare your defences for when the zombies finally reach you. As long as you get far enough away, when they do reach your corner their density will be low and therefore much easier to fight.

Good luck!

If you liked this post you might also like:

Surviving the Zombie Apocalypse – more zombie maths.  How long before the zombies arrive?

How contagious is Ebola? – using differential equations to model infections.

Maths Puzzles

These should all be accessible for top sets in KS4 and post 16.  See if you can manage to get all 3 correct.

Puzzle Number 1

Why is xx undefined when x = 0 ?

Puzzle Number 2

I multiply 3 consecutive integers together. My answer is 8 times the smallest of the 3 integers I multiplied. What 3 numbers could I have chosen?

Puzzle Number 3

You play a game as follows:

1 point for a prime number
2 points for an even number
-3 points for a square number

(note if you choose (say) the number 2 you get +1 for being a prime and +2 for being an even number giving a total of 3 points)

You have the numbers 1-9 to choose from. You need to choose 6 numbers such that their score adds to zero. How many different ways can you find to win this game?

When you have solved all 3 puzzles, click here to find out the solutions.

If you like this post, you might also like:

A Maths Snooker Puzzle. A great little puzzle which tests logic skills.

Visualising Algebra Through Geometry.  How to use geometry to simplify puzzles

The Chinese Postman Problem

There is a fantastic pdf resource from Suffolk Maths which goes into a lot of detail on this topic – and I will base my post on their resource.   Visit their site for a more in-depth treatment.

The Chinese Postman Problem was first posed by a Chinese mathematician in 1962.  It involved trying to calculate how a postman could best choose his route so as to mimise his time.  This is the problem that Kuan Mei-Ko tried to solve:

How could a postman visit every letter on the graph in the shortest possible time?

Solving this requires using a branch of mathematics called graph theory, created by Leonard Euler.  This mathematics looks to reduce problems to network graphs like that shown above.  Before we can solve this we need to understand some terminology:

Above we have 3 graphs.  A graph which can be drawn without taking the pen off the paper or retracing any steps is called traversable (and has a Euler trail).  Graph 1 is not traversable.  Graph 2 is traversable as long as you start at either A or D, and Graph 3 is traversable from any point that you start.  It turns out that what dictates whether a graph is traversable or not is the order of their vertices.

Looking at each letter we count the number of lines at the vertex.  This is the order.  For graph 1 we have 3 lines from A so A has an order of 3.  All the vertices on graph 1 have an order of 3.  For graph 2 we have the orders (from A, B, C, D, E in turn) 3, 4, 4, 3, 2.  For graph 3 we have the orders 4,4,4,4,2,2.

This allows us to arrive at a rule for working out if a graph is traversable.

If all orders are even then a graph is traversable.  If there are 2 odd vertices then we can find a traversable graph by starting at one of the odd vertices and finishing at the other.  We need therefore to pair up any odd vertices on the graph.

Next we need to understand how to pair the odd vertices.  For example if I have 2 odd vertices, they can only be paired in one way.  If I have 4 vertices (say A,B,C,D) they can be paired in 3 different ways (either AB and CD or AC and BD or AD and BC) .  The general term rule to calculate how many ways n odd vertices can be paired up is (n-1) x (n-3) x (n-5) … x 1.

So now we are ready to actually solve the Chinese Postman Problem.  Here is the algorithm:

So, we can now apply this to the Chinese Postman Problem below:

Step 1:  We can see that the only odd vertices are A and H.

Step 2:  We can only pair these one way (AH)

Step 3 and 4: The shortest way to get from A to H is ABFH which is length 160.  This is shown below:

Step 5 and 6: The total distance along all the lines in the initial diagram is 840m.  We add our figure of 160m to this.  Therefore the optimum (minimum) distance it is possible to complete the route is 1000m.

Step 7:  We now need to find a route of distance 1000m which includes the loop ABFH (or in reverse) which starts and finishes at one of the odd vertices.  One solution provided by Suffolk Maths is ADCGHCABDFBEFHFBA.  There are others!

The Bridges of Konigsburg

Graph theory was invented as a method to solve the Bridges of Konigsburg problem by Leonard Euler.  This was a puzzle from the 17oos – Konigsburg was a Russian city with 7 bridges, and the question was, could anyone walk across all 7 without walking over any bridge twice.  By simplifying the problem into one of connected lines, Euler was able to prove that this was in fact impossible.

If you like this post you might also like:

Knight’s Tour – This puzzles dates over 1000 years and concerns the ways in which a knight can cover all squares on a chess board.

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.

Analytic Continuation and the Riemann Zeta Function

Analytic Continuation is a very important mathematical technique which allows us to extend the domain of functions.  It is essential in higher level mathematics and physics and leads to some remarkable results. For example, by using analytic continuation we can prove that the sum of the natural numbers (1 + 2 + 3 + ….) is -1/12. Results don’t get more surprising than that!

Analytic continuation concerns functions of the form:

f(z) where z is a complex number and f(z) is (complex) differentiable.

Remember complex numbers are of the form a + bi and can be thought of as coordinate points in an x,y axis.  For the purposes of this post we will only look at real values of z (real numbers are still a subset of complex numbers).

The idea of analytic continuation is to take an original function with a restricted domain, then to find another function which is the same within that restricted domain, but also is valid outside that domain.  This sounds very complicated – but let’s look at a couple of examples:

$f(z) = \frac{(z+1)(z+2)}{(z+1)}$

This is a function which is defined for all values except for z = -1. When z = -1 we have zero on the denominator so the function doesn’t exist. However we can write a new function:

$g(z) = (z+2)$

Now, g(z) = f(z) for all z when z is not -1, but g(z) also exists when z = -1. Therefore we can regard g(z) as the analytic continuation of f(z), and we have extended the domain of f(z) from all values except -1, to all values of z.

A more interesting example is the following:

$f(z) = \sum_{n=0}^\infty z^{n}$

This is the infinite series:

$1 + z + z^{2}+ ...$

This function is analytic (complex differentiable) only when   -1 < z< 1.  (Don’t worry about how this is calculated – though it is related to the domain of convergence). Therefore this is our restricted domain.

But we can notice that the sum of a geometric sequence formula allows us to calculate f(z) in a different way:

$\sum_{n=0}^\infty z^{n} = \frac{1}{(1-z)}$

Here we have used the formula for summing a geometric, with the first term 1 and common ratio z.

Therefore we could write:

$g(z) = \frac{1}{(1-z)}$

f(z) = g(z) when -1 < z< 1 , but g(z) is complex differentiable for all values except for z = 1 (when the denominator is 0).  Therefore g(z) is the analytic continuation of f(z) from  -1 < z< 1 to all values of z except z = 1.

One example of analytic continuation that I’ve written about before is the Riemann Sphere. This extends by analytic continuation the complex plane into the complex plane plus infinity.

Another example is used in showing that the sum of natural numbers is -1/12. There are a few different methods to show this – some discussed previously here. I’m going to try and talk through another proof of this result. It’s a bit difficult, but try and understand the general method!

The proof revolves around the Riemann Zeta function, (Riemann is pictured above). This is defined as:

$\zeta(z)= \sum_{n=1}^{\infty}n^{-z}$

This can also be written as:

$\zeta(z)=\frac{1}{1^{z}} +\frac{1}{2^{z}} +\frac{1}{3^{z}}..$

So, if we want to find the sum of 1 + 2 + 3 … then we need to substitute z = -1 into the above summation. However this formula for the zeta function is only valid for the domain z > 1, so we first need to extend the function through analytic continuation.

Through analytic continuation (where we extend the domain from z > 1 to all complex numbers apart from -1) we can rewrite the zeta function as:

$\zeta(1-z)=2^{1-z}\pi^{-z}cos(0.5\pi z)\Gamma(z)\zeta(z)$

and substituting z = 2 into this formula, so that we end up with zeta(-1) we get:

$\zeta(-1)=2^{-1}\pi^{-2}cos(\pi)\Gamma(2)\zeta(2)$

Now,

$\zeta(2) = \frac{\pi^{2}}{6}$

$\Gamma(2) = 1$

$cos(\pi) = -1$

Therefore

$\zeta(-1)=-\frac{1}{12}$

We have proved that 1 + 2 + 3 … = -1/12 !

If you enjoyed this post you might also like:

The Riemann Hypothesis Explained. What is the Riemann Hypothesis – and how solving it can win you \$1 million.

Unbelievable: 1+2+3+4…. = -1/12 ? A result that at first glance looks ridiculous – and yet can be shown to be correct.  How?

Murder in the Maths Department

A murder has been committed in the maths department! A body has been discovered surrounded by mathematical objects and only the hardworking maths teachers were in school, doing long division sums for fun at the weekend. One of them must be the murderer.  (The wall of fame of successful detectives will be posted here)

1) The murderer
2) The room
3) The murder weapon

The murder suspects are:

1) Al Jabra – who was wearing a white, T-shirt with 2 stripes and ripped jeans on the day of the murder.
2) Polly Gon – who was wearing a knee-length green skirt, white blouse and gold watch.
3) Lisa Perbound- who was wearing a blue Adidas T-shirt with 3 stripes on the sleeves, Bermuda shorts and a baseball cap.
4) May Trix- who was wearing a black and white pin-stripe suit with shiny black shoes.
5) Ella Ment- who was wearing a blue knitted jumper with a picture of pi on the front, and brown cords.

The possible rooms are:

The possible rooms are:

1) The Canteen
2) The Tuck-shop
3) Room 20
4) Room 18
5) Room 17
6) Room 7

The possible murder weapons are:

1) A wooden metre ruler
2) A large metal stapler
3) A dusty trundle wheel
4) A sharp compass
5) A large maths textbook
6) An oversized calculator

Clue Number 1

When you have solved this clue – click here, and enter the last word only of the decoded message (no capital letters).

How Infectious is Ebola?

Ebola is the latest virus to warrant global fears over a pandemic which infects large numbers of people.  Throughout history we have seen pandemic diseases such as the Black Death in Middle Ages Europe and the Spanish Flu at the beginning of the 20th century. More recently we have seen HIV responsible for millions of deaths.  In the last few years there have been scares over bird flu and SARS – yet neither fully developed into a major global health problem.  So, how contagious is Ebola, and how can we use mathematics to predict its spread?

The basic model is based on the SIR model.  The SIR model looks at how much of the population is susceptible to infection, how many of these go on to become infectious, and how many of these go on to recover (and in what timeframe).  However given the nature of modelling diseases with very high mortality rates like Ebola, for our Ebola model the SIR stands for Susceptible, Infectious and Dead.

Another important parameter is R0, this is defined as how many people an infectious person will pass on their infection to in a totally susceptible population.  Some of the R0 values for different diseases are shown above.  Studies into Ebola estimate the R0 value at somewhere between 1.7 and 8.6.  Therefore whilst Ebola is contagious, it is nowhere near as contagious as a fully airbourne disease like measles.

The Guardian datablog have an excellent graphic to show the contagiousness relative to deadliness of different diseases. You can notice that we have nothing in the top right hand corner (very deadly and very contagious). This is just as well as that could be enough to seriously dent the human population. Most diseases we worry about fall into 2 categories – contagious and not very deadly or not very contagious and deadly. Ebola is in the latter category.

The equations above represent a SIR (susceptible, infectious, dead) model which can be used to model the spread of Ebola.

dS/dt represents the rate of change of those who are susceptible to the illness with respect to time.  dI/dt represents the rate of change of those who are infected with respect to time.  dR/dt represents the rate of change of those who have died with respect to time.

For example, if dI/dt is high then the number of people becoming infected is rapidly increasing.  When dI/dt is zero then there is no change in the numbers of people becoming infected (number of infections remain steady).  When dI/dt is negative then the numbers of people becoming infected is decreasing.

The constants β and μ are chosen depending on the type of disease being modelled.  β represents the contact rate – which is how likely someone will get the disease when in contact with someone who is ill.  ν is the recovery rate which is how quickly people recover (and become immune.

N is the total population

μ is the per capita death rate (Calculated by μ = 1/(duration of illness) )

N – let’s take as the population of Sierra Leone (6 million)

In the case of Ebola we have the following estimated values:

μ between 1/4 and 1/10 (because it takes an infected person between 4 and 10 days to die).  Let’s take it as 1/7 ≈ 0.14

β as approximately 0.6

N – let’s take as the population of Sierra Leone (6 million)

Therefore our 3 equations for rates of change become:

dS/dt = -0.6 I S/6,000,000

dI/dt = 0.6 I S/6,000,000 – 0.14 I

dR/dt = 0.14 I

Unfortunately these equations are very difficult to solve – but luckily we can use a computer program to plot what happens.   We need to assign starting values for S, I and R – the numbers of people susceptible, infectious and dead. We have 6 million people in Sierra Leone, and currently around 8000 reported cases of Ebola.  If we assume all 6 million are susceptible, then putting this all into the program gives the following outcome:

This graph is pretty incredible –  though it clearly shows some of our assumptions were wrong!  Given a starting point of 6 million people all susceptible to Ebola, and 8000 infectious individuals, then within 20 days the population would have crashed from 6 million to less than 1 million and within 60 days you would have nearly everyone dead.

Clearly therefore this graph is very sensitive to our initial assumed values.  Say for example Ebola was less contagious than we previous assumed – and so we had β = 0.15 with the other values the same.  Then we get the following:

This graph is very drastically different to the last one – you have infections remaining low – though this would still be enough to see a big population drop over the 3 years of the simulation.

Modelling disease outbreaks with real accuracy is therefore an incredibly important job for mathematicians.  Understanding how diseases spread and how fast they can spread through populations is essential to developing effective medical strategies to minimise deaths.  If you want to save lives maybe you should become a mathematician rather than a doctor!

If you enjoyed this post you might also like:

Differential Equations in Real Life – some other uses of differential equations in modelling predator-prey relationships between animal populations.

Modelling Infectious Diseases –  How we can use computer modelling to look at other infectious diseases like measles.

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