Cartoon from here

The Gini Coefficient – Measuring Inequality

The Gini coefficient is a value ranging from 0 to 1 which measures inequality. 0 represents perfect equality – i.e everyone in a population has exactly the same wealth.  1 represents complete inequality – i.e 1 person has all the wealth and everyone else has nothing.  As you would expect, countries will always have a value somewhere between these 2 extremes.  The way its calculated is best seen through the following graph (from here):

The Gini coefficient is calculated as the area of A divided by the area of A+B.  As the area of A decreases then the curve which plots the distribution of wealth (we can call this the Lorenz curve) approaches the line y = x.  This is the line which represents perfect equality.

Inequality in Thailand

The following graph will illustrate how we can plot a curve and calculate the Gini coefficient.  First we need some data.  I have taken the following information on income distribution from the 2002 World Bank data on Thailand where I am currently teaching:

Thailand:

The bottom 20% of the population have 6.3% of the wealth
The next 20% of the population have 9.9% of the wealth
The next 20%  have 14% of the wealth
The next 20% have 20.8% of the wealth
The top 20% have 49% of the wealth

I can then write this in a cumulative frequency table (converting % to decimals):

Here the x axis represents the cumulative percentage of the population (measured from lowest to highest), and the y axis represents the cumulative wealth.  This shows, for example that the the bottom 80% of the population own 51% of the wealth.  This can then be plotted as a graph below (using Desmos):

From the graph we can see that Thailand has quite a lot of inequality – after all the top 20% have just under 50% of the wealth.  The blue line represents how a perfectly equal society would look.

To find the Gini Coefficient we first need to find the area between the 2 curves.  The area underneath the blue line represents the area A +B.  This is just the area of a triangle with length and perpendicular height 1, therefore this area is 0.5.

The area under the green curve can be found using the trapezium rule, 0.5(a+b)h.  Doing this for the first trapezium we get 0.5(0+0.063)(0.2) = 0.0063.  The second trapezium is 0.5(0.063+0.162)(0.2) and so on.  Adding these areas all together we get a total trapezium area of 0.3074.  Therefore we get the area between the two curves as 0.5 – 0.3074  ≈ 0.1926

The Gini coefficient is then given by 0.1926/0.5  = 0.3852.

The actual World Bank calculation for Thailand’s Gini coefficient in 2002 was 0.42 – so we have slightly underestimated the inequality in Thailand.  We would get a more accurate estimate by taking more data points, or by fitting a curve through our plotted points and then integrating.  Nevertheless this is a good demonstration of how the method works.

In this graph (from here) we can see a similar plot of wealth distribution – here we have quintiles on the x axis (1st quintile is the bottom 20% etc).  This time we can compare Hungary – which shows a high level of equality (the bottom 80% of the population own 62.5% of the wealth) and Namibia – which shows a high level of inequality (the bottom 80% of the population own just 21.3% of the wealth).

How unequal is the world?

We can apply the same method to measure world inequality.  One way to do this is to calculate the per capita income of all the countries in the world and then to work out the share of the total global per capita income the (say) bottom 20% of the countries have.  This information is represented in the graph above (from here).  It shows that there was rising inequality (i.e the richer countries were outperforming the poorer countries) in the 2 decades prior to the end of the century, but that there has been a small decline in inequality since then.

If you want to do some more research on the Gini coefficient you can use the following resources:

The intmaths site article on this topic – which goes into more detail and examples of how to calculate the Gini coefficient

The ConferenceBoard site which contains a detailed look at world inequality

The World Bank data on the Gini coefficients of different countries.

Is Intergalactic space travel possible?

The Andromeda Galaxy is around 2.5 million light years away – a distance so large that even with the speed of light at traveling as 300,000,000m/s it has taken 2.5 million years for that light to arrive.  The question is, would it ever be possible for a journey to the Andromeda Galaxy to be completed in a human lifetime?  With the speed of light a universal speed limit, it would be reasonable to argue that no journey greater than around 100 light years would be possible in the lifespan of a human – but this remarkably is not the case.  We’re going to show that a journey to Andromeda would indeed be possible within a human lifespan.   All that’s needed (!) is a rocket which is able to achieve constant acceleration and we can arrive with plenty of time to spare.

Time dilation

To understand how this is possible, we need to understand that as the speed of the journey increases, then time dilation becomes an important factor.  The faster the rocket is traveling the greater the discrepancy between the internal body clock of the astronaut on the rocket and an observer on Earth.  Let’s see how that works in practice by using the above equation.

Here we have

t(T): The time elapsed from the perspective of an observer on Earth

T: The time elapsed from the perspective of an astronaut on the rocket

c: The speed of light approx 300,000,000 m/s

a: The constant acceleration we assume for our rocket.  For this example we will take a = 9.81 m/s2 which is the same as the gravity experienced on Earth. This would be the most natural for a human environment.  The acceleration is measured relative to an inert observer.

Sinh(x): This is the hyperbolic sine function which can be defined as:

We should note that all our units are in meters, seconds and m/s2 therefore when the astronaut experiences 1 year passing on this rocket, we first need to convert this to seconds:  1 year = 60 x 60 x 24 x 365 = 31,536,000 seconds.  Therefore T = 31,536,000 and:

which would give us the time experienced on Earth in seconds, therefore by dividing by (60 x 60 x 24 x 365) we can arrive at the time experienced on Earth in years:

Using either Desmos or Wolfram Alpha this gives an answer  of 1.187.  This means that 1 year experienced on the rocket is experienced as 1.19 years on Earth.  Now we have our formula we can easily calculate other values.  Two years is:

which gives an answer of 3.755 years.  So 2 years on the rocket is experienced as 3.76 years on Earth.  As we carry on with the calculations, and as we see the full effects of time dilation we get some startling results:

After 10 years on the space craft, civilization on Earth has advanced (or not) 15,000 years.  After 20 years on the rocket, 445,000,000 years have passed on Earth and after 30 years 13,500,000,000,000 years, which around 1000 times greater than the age of the Universe post Big Bang.  So, as we can see, time is no longer a great concern.

Distance travelled

Next let’s look at how far we can reach from Earth.  This is given by the following equation:

Here we have

x(T): The distance travelled from Earth

T, c and a as before.

Cosh(x): This is the hyperbolic cosine function which can be defined as:

Again we note that we are measuring in meters and seconds.  Therefore to find the distance travelled in one year we convert 1 year to seconds as before:

Next we note that this will give an answer in meters, so we can convert to light years by dividing by 9.461×1015

Again using Wolfram Alpha or Desmos we find that after one year the spacecraft will be 0.563 light years from Earth.  After two years we have:

which gives us 2.91 light years from Earth.  Calculating the next values gives us the following table:

We can see that as our spacecraft approaches the speed of light, we will travel the expected number of light years as measured by an observer on Earth.

So we can see that we would easily reach the Andromeda Galaxy within 20 years on a spacecraft and could have spanned the size of the observable universe within 30 years.   Now, all we need is to build a spaceship capable of constant acceleration, and resolve how a human body could cope with such forces and we’re there!

How likely is this?

Well, the technology needed to build a spacecraft capable of constant acceleration to get close to light speed is not yet available – but there are lots of interesting ideas about how these could be designed in theory.  One of the most popular ideas is to make a “solar sail” – which would collect light from the Sun (or any future nearby stars) to propel it along on its journey.  Another alternative would be a laser sail – which rather than relying on the Sun, would receive pin-point laser beams from the Earth.

Equally we are a long way from being able to send humans – much more likely is that the future of spaceflight will be carried out by machines.  Scientists have suggested that if the spacecraft payload was around 1 gram (say either a miniaturized robot or digital data depending on the mission’s aim), a solar sail or laser sail could be feasibly built which would be sufficient to achieve 25% the speed of light.

NASA have begun launching continuous acceleration spacecraft powered by the Sun.  In 2018 they launched the  Near-Earth Asteroid Scout.  This will unfurl a solar sail and be propelled to a speed of 28,600 m/s.  Whilst this is a long way from near-light speeds, it is a proof of concept and does show one potential way that interstellar travel could be achieved.

You can read more about the current scientific advances on solar sails here, and some more on the mathematics of space travel here.

This is a nice example of using some maths to solve a puzzle from the mindyourdecisions youtube channel (screencaptures from the video).

How to Avoid The Troll: A Puzzle

In these situations it’s best to look at the extreme case first so you get some idea of the problem.  If you are feeling particularly pessimistic you could assume that the troll is always going to be there.  Therefore you would head to the top of the barrier each time.  This situation is represented below:

The Pessimistic Solution:

Another basic strategy would be the optimistic strategy.  Basically head in a straight line hoping that the troll is not there.  If it’s not, then the journey is only 2km.  If it is then you have to make a lengthy detour.  This situation is shown below:

The Optimistic Solution:

The expected value was worked out here by doing 0.5 x (2) + 0.5 x (2 + root 2) = 2.71.

The question is now, is there a better strategy than either of these?  An obvious possibility is heading for the point halfway along where the barrier might be.  This would make a triangle of base 1 and height 1/2.  This has a hypotenuse of root (5/4).  In the best case scenario we would then have a total distance of 2 x root (5/4).  In the worst case scenario we would have a total distance of root(5/4) + 1/2 + root 2.  We find the expected value by multiply both by 0.5 and adding.  This gives 2.63 (2 dp).  But can we do any better?  Yes – by using some algebra and then optimising to find a minimum.

The Optimisation Solution:

To minimise this function, we need to differentiate and find when the gradient is equal to zero, or draw a graph and look for the minimum.  Now, hopefully you can remember how to differentiate polynomials, so here I’ve used Wolfram Alpha to solve it for us.  Wolfram Alpha is incredibly powerful -and also very easy to use.  Here is what I entered:

and here is the output:

So, when we head for a point exactly 1/(2 root 2) up the potential barrier, we minimise the distance travelled to around 2.62 miles.

So, there we go, we have saved 0.21 miles from our most pessimistic model, and 0.01 miles from our best guess model of heading for the midpoint.  Not a huge difference – but nevertheless we’ll save ourselves a few seconds!

This is a good example of how an exploration could progress – once you get to the end you could then look at changing the question slightly, perhaps the troll is only 1/3 of the distance across?  Maybe the troll appears only 1/3 of the time?  Could you even generalise the results for when the troll is y distance away or appears z percent of the time?

Zeno’s Paradox – Achilles and the Tortoise

This is a very famous paradox from the Greek philosopher Zeno – who argued that a runner (Achilles) who constantly halved the distance between himself and a tortoise  would never actually catch the tortoise.  The video above explains the concept.

There are two slightly different versions to this paradox.  The first version has the tortoise as stationary, and Achilles as constantly halving the distance, but never reaching the tortoise (technically this is called the dichotomy paradox).  The second version is where Achilles always manages to run to the point where the tortoise was previously, but by the time he reaches that point the tortoise has moved a little bit further away.

The first version we can think of as follows:

Say the tortoise is 2 metres away from Achilles.  Initially Achilles halves this distance by travelling 1 metre.  He halves this distance again by travelling a further 1/2 metre.  Halving again he is now 1/4 metres away.  This process is infinite, and so Zeno argued that in a finite length of time you would never actually reach the tortoise.  Mathematically we can express this idea as an infinite summation of the distances travelled each time:

1 + 1/2 + 1/4 + 1/8 …

Now, this is actually a geometric series – which has first term a = 1 and common ratio r = 1/2.  Therefore we can use the infinite summation formula for a geometric series (which was derived about 2000 years after Zeno!):

sum = a/(1-r)

sum = 1/(1-0.5)

sum = 2

This shows that the summation does in fact converge – and so Achilles would actually reach the tortoise that remained 2 metres away.  There is still however something of a sleight of hand being employed here however – given an infinite length of time we have shown that Achilles would reach the tortoise, but what about reaching the tortoise in a finite length of time?  Well, as the distances get ever smaller, the time required to traverse them also gets ever closer to zero, so we can say that as the distance converges to 2 metres, the time taken will also converge to a finite number.

There is an alternative method to showing that this is a convergent series:

S = 1+ 1/2 + 1/4 + 1/8 + 1/16 + …

0.5S = 1/2+ 1/4 + 1/8 + 1/16 + …

S – 0.5S = 1

0.5S = 1

S = 2

Here we notice that in doing S – 0.5S all the terms will cancel out except the first one.

Achilles and the Tortoise

The second version also makes use of geometric series.  If we say that the tortoise has been given a 10 m head start, and that whilst the tortoise runs at 1 m/s, Achilles runs at 10 m/s, we can try to calculate when Achilles would catch the tortoise.  So in the first instance, Achilles runs to where the tortoise was (10 metres away).  But because the tortoise runs at 1/10th the speed of Achilles, he is now a further 1m away.  So, in the second instance, Achilles now runs to where the tortoise now is (a further 1 metre).  But the tortoise has now moved 0.1 metres further away.  And so on to infinity.

This is represented by a geometric series:

10 + 1 + 0.1 + 0.01 …

Which has first time a = 10 and common ratio r = 0.1.  So using the same formula as before:

sum = a/(1-r)

sum = 10/(1-0.1)

sum = 11.11m

So, again we can show that because this geometric series converges to a finite value (11.11), then after a finite time Achilles will indeed catch the tortoise (11.11m away from where Achilles started from).

We often think of mathematics and philosophy as completely distinct subjects – one based on empirical measurement, the other on thought processes – but back in the day of the Greeks there was no such distinction.  The resolution of Zeno’s paradox by use of calculus and limits to infinity some 2000 years after it was first posed is a nice reminder of the power of mathematics in solving problems across a wide range of disciplines.

The Chess Board Problem

The chess board problem is nothing to do with Zeno (it was first recorded about 1000 years ago) but is nevertheless another interesting example of the power of geometric series.  It is explained in the video above.  If I put 1 grain of rice on the first square of a chess board, 2 grains of rice on the second square, 4 grains on the third square, how much rice in total will be on the chess board by the time I finish the 64th square?

The mathematical series will be:

1+ 2 + 4 + 8 + 16 +……

So a = 1 and r = 2

Sum = a(1-r64)/(1-r)

Sum = (1-264)/(1-2)

Sum = 264 -1

Sum = 18, 446,744, 073, 709, 551, 615

This is such a large number that, if stretched from end to end the rice would reach all the way to the star Alpha Centura and back 2 times.

Fourier Transform

The Fourier Transform and the associated Fourier series is one of the most important mathematical tools in physics. Physicist Lord Kelvin remarked in 1867:

“Fourier’s theorem is not only one of the most beautiful results of modern analysis, but it may be said to furnish an indispensable instrument in the treatment of nearly every recondite question in modern physics.”

The Fourier Transform deals with time based waves – and these are one of the fundamental building blocks of the natural world. Sound, light, gravity, radio signals, Earthquakes and digital compression are just some of the phenomena that can be understood through waves. It’s not an exaggeration therefore to see the study of waves as one of the most important applications of mathematics in our modern life.

Here are some real life applications in a wide range of fields:

JPEG picture and MP3 sound compression – to allow data to reduced in size.

Analysing DNA sequences – to allow identification of specific regions of genetic code

Apps like Shazam which can recognise a song from a sample of music

Processing mobile phone network data and WIFI data

Signal processing – in everything from acoustic guitar amps or electrical currents through capacitors

Radio telescopes – used to construct images of the night sky

Building’s natural frequencies – architects can design buildings to better withstand earthquakes.

Medical imaging such as MRI scans

There are many more applications – this Guardian article is a good introduction to some others.

So, what is the Fourier Transform? It takes a graph like the graph f(t) = cos(at) below:

and transforms it into:

From the above cosine graph we can see that it is periodic time based function. Time is plotted on the x axis, and this graph will tell us the value of f(t) at any given time. The graph below with 2 spikes represents this same information in a different way. It shows the frequency (plotted on the x axis) of the cosine graph. Now the frequency of a function measures how many times it repeats per second. So for a graph f(t) = cos(at) it can be calculated as the inverse of the period. The period of cos(at) is 2pi/a so it has a frequency of a/2pi.

Therefore the frequency graph for cos(ax) will have spikes at a/2pi and -a/2pi.

But how does this new representation help us? Well most real life waves are much more complicated than simple sine or cosine waves – like this trumpet sound wave below:

But the remarkable thing is that every continuous wave can be modelled as the sum of sine and cosine waves. So we can break-down the very complicated wave above into (say) cos(x) + sin(2x) + 2cos(4x) . This new representation would be much easier to work with mathematically.

The way to find out what these constituent sine and cosine waves are that make up a complicated wave is to use the Fourier Transform. By transforming a function into one which shows the frequency peaks we can work out what the sine and cosine parts are for that function.

For example, this transformed graph above would show which frequency sine and cosine functions to use to model our original function. Each peak represents a sine or cosine function of a specific frequency. Add them all together and we have our function.

The maths behind this does get a little complicated. I’ll try and talk through the method using the function f(t) = cos(at).

$\\1.\ f(t) = cosat\\$

So, the function we want to break down into its constituent cosine and sine waves is cos(at). Now, obviously this function can be represented just with cos(at) – but this is a good demonstration of how to use the maths for the Fourier Transform. We already know that this function has a frequency of a/2pi – so let’s see if we can find this frequency using the Transform.

$\\2.\ F(\xi) = \int_{-\infty}^{\infty} f(t)(e^{-2\pi i\xi t})dt\\$

This is the formula for the Fourier Transform. We “simply” replace the f(t) with the function we want to transform – then integrate.

$\\3.\ f(t)= 0.5({e}^{iat}+ {e}^{-iat})\\$

To make this easier we use the exponential formula for cosine. When we have f(t) = cos(at) we can rewrite this as the function above in terms of exponential terms.

$\\4.\ F(\xi) = 0.5\int_{-\infty}^{\infty} (e^{iat}+e^{-iat})(e^{-2\pi i\xi t})dt\\$

We substitute this version of f(t) into the formula.

$\\5.\ F(\xi) = \frac{1}{2} \int_{-\infty}^{\infty} e^{it(a-2\pi \xi) }dt + \frac{1}{2} \int_{-\infty}^{\infty}e^{it(-a-2\pi \xi)}dt\\$

Next we multiply out the exponential terms in the bracket (remember the laws of indices), and then split the integral into 2 parts. The reason we have grouped the powers in this way is because of the following step.

$\\6.\ \delta (a-2\pi \xi) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{it(a-2\pi \xi)}\\$

This is the delta function – which as you can see is very closely related to the integrals we have. Multiplying both sides by pi will get the integral in the correct form. The delta function is a function which is zero for all values apart from when the domain is zero.

$\\7.\ F(\xi) =\pi [ \delta (a-2\pi \xi ) + \delta (a+2\pi \xi ) ]\\$

So, the integral can be simplified as this above.

$\\8.\ a-2\pi \xi = 0 \ or \ a+2\pi \xi = 0\\$

So, our function F will be zero for all values except when the delta function is zero. This gives use the above equations.

$\\9.\ \xi = \pm\frac{a}{2\pi }\\$

Therefore solving these equations we get an answer for the frequency of the graph.

$\\10.\ frequency\ of\ cosat = \frac{a}{2\pi }$

This frequency agrees with the frequency we already expected to find for cos(at).

A slightly more complicated example would be to follow the same process but this time with the function f(t) = cos(at) + cos(bt). If the Fourier transform works correctly it should recognise that this function is composed of one cosine function with frequency a/2pi and another cosine function of b/2pi. If we follow through exactly the same method as above (we can in effect split the function into cos(at) and cos(bt) and do both separately), we should get:

$\\7.\ F(\xi) =\pi [ \delta (a-2\pi \xi ) + \delta (a+2\pi \xi ) + \delta (b-2\pi \xi ) + \delta (b+2\pi \xi ) ]\\$

This therefore is zero for all values except for when we have frequencies of a/2pi and b/2pi. So the Fourier Transform has correctly identified the constituent parts of our function.

If you want to read more about Fourier Transforms, then the Better Explained article is an excellent start.

Non Euclidean Geometry – An Introduction

It wouldn’t be an exaggeration to describe the development of non-Euclidean geometry in the 19th Century as one of the most profound mathematical achievements of the last 2000 years.  Ever since Euclid (c. 330-275BC) included in his geometrical proofs an assumption (postulate) about parallel lines, mathematicians had been trying to prove that this assumption was true.  In the 1800s however, mathematicians including Gauss started to wonder what would happen if this assumption was false – and along the way they discovered a whole new branch of mathematics.  A mathematics where there is an absolute measure of distance, where straight lines can be curved and where angles in triangles don’t add up to 180 degrees.  They discovered non-Euclidean geometry.

Euclid’s parallel postulate (5th postulate)

Euclid was a Greek mathematician – and one of the most influential men ever to live.  Through his collection of books, Elements, he created the foundations of geometry as a mathematical subject.  Anyone who studies geometry at secondary school will still be using results that directly stem from Euclid’s Elements – that angles in triangles add up to 180 degrees, that alternate angles are equal, the circle theorems, how to construct line and angle bisectors.  Indeed you might find it slightly depressing that you were doing nothing more than re-learn mathematics well understood over 2000 years ago!

All of Euclid’s results were based on rigorous deductive mathematical proof – if A was true, and A implied B, then B was also true.  However Euclid did need to make use of a small number of definitions (such as the definition of a line, point, parallel, right angle) before he could begin his first book  He also needed a small number of postulates (assumptions given without proof) – such as:  “(It is possible) to draw a line between 2 points” and “All right angles are equal”

Now the first 4 of these postulates are relatively uncontroversial in being assumed as true.  The 5th however drew the attention of mathematicians for centuries – as they struggled in vain to prove it.  It is:

If a line crossing two other lines makes the interior angles on the same side less than two right angles, then these two lines will meet on that side when extended far enough.

This might look a little complicated, but is made a little easier with the help of the sketch above.  We have the line L crossing lines L1 and L2, and we have the angles A and B such that A + B is less than 180 degrees.  Therefore we have the lines L1 and L2 intersecting.  Lines which are not parallel will therefore intersect.

Euclid’s postulate can be restated in simpler (though not quite logically equivalent language) as:

At most one line can be drawn through any point not on a given line parallel to the given line in a plane.

In other words, if you have a given line (l) and a point (P), then there is only 1 line you can draw which is parallel to the given line and through the point (m).

Both of these versions do seem pretty self-evident, but equally there seems no reason why they should simply be assumed to be true.  Surely they can actually be proved?  Well, mathematicians spent the best part of 2000 years trying without success to do so.

Why is the 5th postulate so important?

Because Euclid’s proofs in Elements were deductive in nature, that means that if the 5th postulate was false, then all the subsequent “proofs” based on this assumption would have to be thrown out.  Most mathematicians working on the problem did in fact believe it was true – but were keen to actually prove it.

As an example, the 5th postulate can be used to prove that the angles in a triangle add up to 180 degrees.

The sketch above shows that if A + B are less than 180 degrees the lines will intersect.  Therefore because of symmetry (if one pair is more than 180 degrees, then other side will have a pair less than 180 degrees), a pair of parallel lines will have A + B = 180.  This gives us:

This is the familiar diagram you learn at school – with alternate and corresponding angles.   If we accept the diagram above as true, we can proceed with proving that the angles in a triangle add up to 180 degrees.

Once, we know that the two red angles are equal and the two green angles are equal, then we can use the fact that angles on a straight line add to 180 degrees to conclude that the angles in a triangle add to 180 degrees.  But it needs the parallel postulate to be true!

In fact there are geometries in which the parallel postulate is not true  – and so we can indeed have triangles whose angles don’t add to 180 degrees.  More on this in the next post.

If you enjoyed this you might also like:

Non-Euclidean Geometry II – Attempts to Prove Euclid – The second part in the non-Euclidean Geometry series.

The Riemann Sphere – The Riemann Sphere is a way of mapping the entire complex plane onto the surface of a 3 dimensional sphere.

Circular Inversion – Reflecting in a Circle The hidden geometry of circular inversion allows us to begin to understand non-Euclidean geometry.

The Telephone Numbers – Graph Theory

The telephone numbers are the following sequence:

1, 1, 2, 4, 10, 26, 76, 232, 764, 2620, 9496…

(where we start from n=0).

This pattern describes the total number of ways which a telephone exchange with n telephones can place a connection between pairs of people.

To illustrate this idea, the graph below is for n=4.  This is when we have 10 telephones:

Each red line represents a connection.  So the first diagram is for when we have no connections (this is counted in our sequence).  The next five diagrams all show a single connection between a pair of phones.  The last three diagrams show how we could have 2 pairs of telephones connected at the same time.  Therefore the 4th telephone number is 10.   These numbers get very large, very quickly.

Finding a recursive formula

The formula is given by the recursive relationship:

T(n) = T(n-1) + (n-1)T(n-2)

This means that to find (say) the 5th telephone number we do the following:

T(5) = T(5-1) + (5-1)T(5-2)

T(5) = T(4) + (4)T(3)

T(5) = 10 + (4)4

T(5) = 26

This is a quick way to work out the next term, as long as we have already calculated the previous terms.

Finding an nth term formula

The telephone numbers can be calculated using the nth term formula:

This is going to be pretty hard to derive!  I suppose the first step would start by working out the total number of connections possible between n phones – and this will be the the same as the graphs below:

These clearly follow the same pattern as the triangular numbers which is 0.5(n² +n) when we start with n = 1.  We can also think of this as n choose 2 – because this gives us all the ways of linking 2 telephones from n possibilities.  Therefore n choose 2 also generates the triangular numbers.

But then you would have to work out all the permutations which were allowed – not easy!

Anyway, as an example of how to use the formula to calculate the telephone numbers, say we wanted to find the 5th number:

We have n = 5.  The summation will be from k = 0 and k = 2 (as 5/2 is not an integer).

Therefore T(5) = 5!/(20(5-0)!0!) + 5!/(21(5-2)!1!) + 5!/(22(5-4)!2!)

T(5) = 1 + 10 + 15 = 26.

Finding telephone numbers through calculus

Interestingly we can also find the telephone numbers by using the function:

y = e0.5x2+x

and the nth telephone number (starting from n = 1)  is given by the nth derivative when x = 0.

For example,

So when x = 0, the third derivative is 4.  Therefore the 3rd telephone number is 4.

The fifth derivative of the function is:

So, when x =0 the fifth derivative is 26.  Therefore the 5th telephone number is 26.

If you liked this post you might also like:

Fermat’s Theorem on the Sum of two Squares – A lesser known theorem from Fermat – but an excellent introduction to the idea of proof.

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

Happy Numbers

Happy numbers are defined by the rule that you start with any positive integer, square each of the digits then add them together.  Now do the same with the new number.  Happy numbers will eventually spiral down to a number of 1.  Numbers that don’t eventually reach 1 are called unhappy numbers.

As an example, say we start with the number 23.  Next we do 2²+3² = 13.  Now, 1²+3² = 10.  Now 1²+o² = 1.  23 is therefore a happy number.

There are many things to investigate.  What are the happy numbers less than 100?  Is there a rule which dictates which numbers are happy?  Are there consecutive happy numbers?  How about prime happy numbers?  Can you find the infinite cycle of sadness?

Nrich has a discussion on some of the maths behind happy numbers.  You can use an online tool to test if numbers are happy or sad.

Perfect Numbers

Perfect numbers are numbers whose proper factors (factors excluding the number itself) add to the number.  This is easier to see with an example.

6 is a perfect number because its proper factors are 1,2,3 and 1+2+3 = 6

8 is not a perfect number because its proper factors are 1,2,4 and 1+2+4 = 7

Perfect numbers have been known about for about 2000 years – however they are exceptionally rare.  The first 4 perfect numbers are 6, 28, 496, 8128.  These were all known to the Greeks.  The next perfect number wasn’t discovered until around 1500 years later – and not surprisingly as it’s 33,550,336.

The next perfect numbers are:

8,589,869,056 (discovered by Italian mathematician Cataldi in 1588)

137,438,691,328 (also discovered by Cataldi)

2,305,843,008,139,952,128 (discovered by Euler in 1772).

and they keep getting bigger.  The next number to be discovered has 37 digits are was discovered over 100 years later.  Today, even with vast computational power, only a total of 48 perfect numbers are known.  The largest has 34,850,340 digits.

There are a number of outstanding questions about perfect numbers.  Are there an infinite number of perfect numbers?  Is there any odd perfect number?

Euclid in around 300BC proved that that 2p−1(2p−1) is an even perfect number whenever 2p−1 is prime.  Euler (a rival with Euclid for one of the greatest mathematicians of all time), working on the same problem about 2000 years later went further and proved that this formula will provide every even perfect number.

This links perfect numbers with the search for Mersenne Primes – which are primes in the form 2p−1.  These are themselves very rare, but every new Mersenne Prime will also yield a new perfect number.

The first Mersenne Primes are

(22−1) = 3

(23−1) = 7

(25−1) = 31

(27−1) = 127

Therefore the first even perfect numbers are:

21(22−1) = 6

22(23−1) = 28

24(25−1) = 496

26(27−1) = 8128

Friendly Numbers

Friendly numbers are numbers which share a relationship with other numbers.  They require the use of σ(a) which is called the divisor function and means the addition of all the factors of a.  For example σ(7) = 1 + 7 = 8 and σ(10) = 1 +2 +5 + 10 = 18.

Friendly numbers therefore satisfy:

σ(a)/a = σ(b)/b

As an example (from Wikipedia)

σ(6) / 6 = (1+2+3+6) / 6 = 2,

σ(28) / 28 = (1+2+4+7+14+28) / 28 = 2

σ(496)/496 = (1+2+4+8+16+31+62+124+248+496)/496 = 2

Therefore 28 and 6 are friendly numbers because they share a common relationship.  In fact all perfect numbers share the same common relationship of 2.  This is because of the definition of perfect numbers above!

Numbers who share the same common relationship are said to be in the same club.  For example, 30,140, 2480, 6200 and 40640 are all in the same club – because they all share the same common relationship 12/5.

(eg. σ(30) /30  = (1+2+3+5+6+10+15+30) / 30 = 12/5 )

Are some clubs of numbers infinitely big?  Which clubs share common integer relationships?  There are still a number of unsolved problems for friendly numbers.

Solitary Numbers

Solitary numbers are numbers which don’t share a common relationship with any other numbers.  All primes, and prime powers are solitary.

Additionally all number that satisfy the following relationship:

HCF of σ(a) and  a = 1.

are solitary.  All this equation means is that the highest common factor (HCF) of σ(a) and a is 1.  For example lets choose the number 9.

σ(9)= 1+3+9 = 13.  The HCF of 9 and 13 = 1.  So 9 is solitary.

However there are some numbers which are not prime, prime powers or satisfy HCF (σ(a) and  a) = 1, but which are still solitary.   These numbers are much harder to find!  For example it is believed that the following numbers are solitary:

10, 14, 15, 20, 22, 26, 33, 34, 38, 44, 46, 51, 54, 58, 62, 68, 69, 70, 72, 74, 76, 82, 86, 87, 88, 90, 91, 92, 94, 95, 99

But no-one has been able to prove it so far.  Maybe you can!

Give your university applications a headstart on other students with Coursera.

Applying for university as an international student is incredibly competitive – for the top universities you’ll be competing with the best students from around the world, and so giving yourself a competitive advantage to make your university application stand out is really important.   One way to do this is by completing a course run by some of the world’s top Universities.

Universities offering courses:

Examples of the top universities offering courses include: The University of Tokyo, Caltec, University of Manchester, Imperial College London, Duke, Stanford, Yale, University of Sydney, National University of Singapore, amongst many others, alongside major companies such as IBM, Google, Intel and Goldman Sachs.

Courses on offer

You can sign up for free and access modules run by these universities and companies, with the possibility of obtaining a certificate at the end of the course which can then go towards your university application.

Some of the courses on offer include:

Biological science courses such as Genetics and Evolution from Duke University, Understaning the brain from the University of Chicago, Astrobiology and the search for Extraterrestrial life from Edinburgh University and Medical Neuroscience from Duke University,

Business courses such as Business foundations from University of Pennsylvania, Digital Marketing from the University of Illinois, Viral Marketing and How to Create Contagious Content with the University of Pennsylvania, the Math Behind Moneyball with the University of Houston and studying the Global Financial Crisis with Yale University.

Physical science courses such as Welcome to Game Theory with Tokyo University, Science Literacy with Erasmus University Rotterdam and The Journey of the Universe with Yale University.

Arts and humanities courses such as Creative writing from Wesleyan University, Graphics design from Californian Institute of Arts, Music Production from Berklee College of Music, Introduction to Philosophy from the University of Edinburgh.

Overall there are over 3000 courses from 170 universities and partners – so almost certainly there’ll be something worth investigating.  Have a look and give your University application a boost over everyone else!

Statistics to win penalty shoot-outs

With the World Cup upon us again we can perhaps look forward to yet another heroic defeat on penalties by England. England are in fact the worst country of any of the major footballing nations at taking penalties, having won only 1 out of 7 shoot-outs at the Euros and World Cup. In fact of the 35 penalties taken in shoot-outs England have missed 12 – which is a miss rate of over 30%. Germany by comparison have won 5 out of 7 – and have a miss rate of only 15%.

With the stakes in penalty shoot-outs so high there have been a number of studies to look at optimum strategies for players.

One study published in Psychological Science looked at all the penalties taken in penalty shoot-outs in the World Cup since 1982. What they found was pretty incredible – goalkeepers have a subconscious bias for diving to the right when their team is behind.

As is clear from the graphic, this is not a small bias towards the right, but a very strong one. When their team is behind the goalkeeper apparently favours his (likely) strong side 71% of the time. The strikers’ shot meanwhile continues to be placed either left or right with roughly the same likelihood as in the other situations. So, this built in bias makes the goalkeeper much less likely to help his team recover from a losing position in a shoot-out.

Shoot high

Analysis by Prozone looking at the data from the World Cups and European Championships between 1998 and 2010 compiled the following graphics:

The first graphic above shows the part of the goal that scoring penalties were aimed at. With most strikers aiming bottom left and bottom right it’s no surprise to see that these were the most successful areas.

The second graphic which shows where penalties were saved shows a more complete picture – goalkeepers made nearly all their saves low down. A striker who has the skill and control to lift the ball high makes it very unlikely that the goalkeeper will save his shot.

The last graphic also shows the risk involved in shooting high. This data shows where all the missed penalties (which were off-target) were being aimed. Unsurprisingly strikers who were aiming down the middle of the goal managed to hit the target! Interestingly strikers aiming for the right corner (as the goalkeeper stands) were far more likely to drag their shot off target than those aiming for the left side. Perhaps this is to do with them being predominantly right footed and the angle of their shooting arc?

Win the toss and go first

The Prozone data also showed the importance of winning the coin toss – 75% of the teams who went first went on to win. Equally, missing the first penalty is disastrous to a team’s chances – they went on to lose 81% of the time. The statistics also show a huge psychological role as well. Players who needed to score to keep their teams in the competition only scored a miserable 14% of the time. It would be interesting to see how these statistics are replicated over a larger data set.

Don’t dive

A different study which looked at 286 penalties from both domestic leagues and international competitions found that goalkeepers are actually best advised to stay in the centre of the goal rather than diving to one side. This had quite a significant affect on their ability to save the penalties – increasing the likelihood from around 13% to 33%. So, why don’t more goalkeepers stay still? Well, again this might come down to psychology – a diving save looks more dramatic and showcases the goalkeeper’s skill more than standing stationary in the centre.

So, why do England always lose on penalties?

There are some interesting psychological studies which suggest that England suffer more than other teams because English players are inhibited by their high public status (in other words, there is more pressure on them to perform – and hence that pressure is harder to deal with).  One such study noted that the best penalty takers are the ones who compose themselves prior to the penalty.  England’s players start to run to the ball only 0.2 seconds after the referee has blown – making them much less composed than other teams.

However, I think you can put too much analysis on psychology – the answer is probably simpler – that other teams beat England because they have technically better players.  English footballing culture revolves much less around technical skill than elsewhere in Europe and South America – and when it comes to the penalty shoot-outs this has a dramatic effect.

As we can see from the statistics, players who are technically gifted enough to lift their shots into the top corners give the goalkeepers virtually no chance of saving them.  England’s less technically gifted players have to rely on hitting it hard and low to the corner – which gives the goalkeeper a much higher percentage chance of saving them.

Test yourself

You can test your penalty taking skills with this online game from the Open University – choose which players are best suited to the pressure, decide what advice they need and aim your shot in the best position.

If you liked this post you might also like:

Championship Wages Predict League Position? A look at how statistics can predict where teams finish in the league.

Premier League Wages Predict League Positions? A similar analysis of Premier League teams.

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