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. My name is Andrew Chambers and I am currently working at BISP.  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.

There are a huge amount of resources to explore – especially for students doing their IAs and for students looking for revision videos.  You may also like to try our school code breaking site – where you can compete with over 10,000 students from around the world who have made it onto our school leaderboard.

 

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IB Revision with Revision Village

There’s a really great website been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams.  I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL, SL and Studies students.

You choose your subject (HL/SL/Studies) and get the following screen:

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The Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions like:

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What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial.  Really useful!

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions.  This also has some harder exams for those students aiming for 6s and 7s.

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The Past IB Exams section takes you to full worked solutions to each full paper.

and lastly you can also get a prediction exam for the upcoming year.

For HL students you can click here

For SL students you can click here

For Studies students you can click here

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

telephone2

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:

telephone

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:

 

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 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:

telephone3

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,

telephone5

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

The fifth derivative of the function is:

telephone6

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 number

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.

perfectnumber 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

friendlynumber

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.

solitarynumber

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!

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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:

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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!

penalties2

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.

Shoot left when ahead

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.

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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:

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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.

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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.

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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.

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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.

This carries on the previous investigation into Farey sequences, and is again based on the current Nrich task Ford Circles.  Below are the Farey sequences for F2, F3 and F4. You can read about Farey sequences in the previous post.

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This time I’m going to explore the link between Farey sequences and circles.  First we need the general equation for a circle:

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This has centre (p,q) and radius r.  Therefore

Circle 1:

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has centre:

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and radius:

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Circle 2:

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has centre:

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and radius:

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Now we can plot these circles in Geogebra – and look for the values of a,b,c,d which lead to the circles touching at a point.

When a = 1, b = 2, c = 2, d = 3:

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Do we notice anything about the numbers a/b and c/d ?  a/b = 1/2 and c/d = 2/3 ?  These are consecutive terms in  the Fsequence.  So do other consecutive terms in the Farey sequence also generate circles touching at a point?

a = 1, b = 1, c = 2, d = 3

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Again we can see that the fractions 1/1 and 2/3 are consecutive terms in the Fsequence. So by drawing some more circle we can graphically represent all the fractions in the Fsequence:

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So these four circles represent the four non-zero fractions of in the Fsequence!

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and this is the visual representation of the non-zero fractions of in the Fsequence.  Amazing!

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Modelling more Chaos

This post was inspired by Rachel Thomas’ Nrich article on the same topic.  I’ll carry on the investigation suggested in the article.  We’re going to explore chaotic behavior – where small changes to initial conditions lead to widely different outcomes.  Chaotic behavior is what makes modelling (say) weather patterns so complex.

f(x) = sin(x)

This time let’s do the same with f(x) = sin(x).

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Starting value of x = 0.2

 

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Starting value of x = 0.2001

 

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Both graphs superimposed 

 

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This time the graphs do not show any chaotic behavior over the first 40 iterations – a small difference in initial condition has made a negligible difference to the output.  Even after 200 iterations we get the 2 values x = 0.104488151 and x = 0.104502319.

f(x) = tan(x)

Now this time with f(x) = tan(x).

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Starting value of x = 0.2

 

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Starting value of x = 0.2001

 

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Both graphs superimposed 

 

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This time both graphs remained largely the same up until around the 38th data point – with large divergence after that.  Let’s see what would happen over the next 50 iterations:

 

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Therefore we can see that tan(x) is much more susceptible to small initial state changes than sin(x).  This makes sense by considering the graphs of tan(x) and sin(x).  Sin(x) remains bounded between -1 and 1, whereas tan(x) is unbounded with asymptotic behaviour as we approach pi/2.

This is a mini investigation based on the current Nrich task Farey Sequences.

As Nrich explains:

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I’m going to look at Farey sequences (though I won’t worry about rearranging them in order of size).  Here are some of the first Farey sequences.  The missing fractions are all ones which simplify to a fraction already on the list (e.g. 2/4 is missing because this is the same as 1/2)

You should be able to notice that the next Farey sequence always contains the previous Farey sequence, so the problem becomes working out which of the new fractions added will not cancel down to something already on the list.

Highest Common Factors

Fractions will not cancel down (simplify) if the numerator and denominator have a highest common factor (HCF) of 1.  For example 2/4 simplifies because the highest common factor of 2 and 4 is 2.  Therefore both top and bottom can be divided by 2.  4/5 does not simplify because the HCF of 4 and 5 is 1.

We call 2 numbers which have a HCF of 1 relatively prime.

for example for the number 4: 1 and 3 are both relatively prime (HCF of 1 and 4  =1, HCF of 3 and 4 = 1).

Relatively prime numbers

2: 1

3: 1,2

4: 1,3

5: 1,2,3,4

6: 1,5

7: 1,2,3,4,5,6

8: 1,3,5,7

9: 1,2,4,5,7,8

You might notice that these give the required numerators for any given denominator – i.e when the denominator is 9, we want a numerator of 1,2,4,5,7,8.

Euler totient function

Euler’s totient function is a really useful function in number theory – which counts the number of relatively prime numbers a given number has.  For example from our list we can see that 9 has 6 relatively prime numbers.

Euler’s totient function is defined above – it’s not as complicated as it looks!  The strange symbol on the right hand side is the product formula – i.e we multiply terms together.  It’s easiest to understand with some examples.  To find Euler’s totient function we first work out the prime factors of a number.  Say we have the number 8.  The prime factors of 8 are 23. Therefore the only unique prime factor is 2.

Therefore the Euler totient function tells me to simply do 8 (1 – 1/2) = 4.  This is how many relatively prime numbers 8 has.

Let’s look at another example – this time for the number 10.  10 has the prime factorisation 5 x 2.  Therefore it has 2 unique primes, 2 and 5.  Therefore the Euler totient function tells me to do 10(1-1/2)(1-1/5) = 4.

One more example, this time with the number 30.  This has prime factorisation 2 x 3 x 5.  This has unique prime factors 2,3,5 so I will do 30(1 -1/2)(1-1/3)(1-1/5) =8.

An equation for the number of fractions in the Farey sequence

Therefore I can now work out how many fractions will appear in a given Farey sequence.  I notice that for (say) F5 I will add Euler’s totient for n = 2, n = 3, n = 4 and n = 5. I then add 2 to account for 0/1 and 1/1. Therefore I have:

For example to find F6

There are lots of things to investigate about Farey functions – could you prove why all Farey sequences have an odd number of terms? You can also look at how well the Farey sequence is approximated by the following equation:

For example when n = 10 this gives:

and when n = 1000 this gives:

These results compare reasonably well as an estimation to the real answers of 33 and 304,193 respectively.

 

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Modelling Chaos

This post was inspired by Rachel Thomas’ Nrich article on the same topic.  I’ll carry on the investigation suggested in the article.  We’re going to explore chaotic behavior – where small changes to initial conditions lead to widely different outcomes.  Chaotic behavior is what makes modelling (say) weather patterns so complex.

Let’s start as in the article with the function:

f(x) = 4x(1-x)

We can then start an iterative process where we choose an initial value, calculate f(x) and then use this answer to calculate a new f(x) etc. For example when I choose x = 0.2, f(0.2) = 0.64. I then use this value to find a new value f(0.64) = 0.9216. I used a spreadsheet to plot 40 iterations for the starting values of x = 0.2 and x = 0.2001. This generated the following spreadsheet (cut to show the first 10 terms):

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I then imported this table into Desmos to map how the change in the starting value from 0.2 to 0.2001 affected the resultant graph.

Starting value of x = 0.2

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Starting value of x = 0.2001

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Both graphs superimposed 

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We can see that for the first 10 terms the graphs are virtually the same – but then we get a wild divergence, before the graphs seem to synchronize more closely again.  One thing we notice is that the data is bounded between 0 and 1.  Can we prove why this is?

If we start with a value of x such that:

0<x<1.

then when we plot f(x) = 4x – 4x2 we can see that the graph has a maximum at x = 1/2:
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Therefore any starting value of x between 0 and 1 will also return a new value bounded between 0 and 1.  Starting values of x > 1 and x < -1 will tend to negative infinity because x2 grows much more rapidly than x.

f(x) = ax(1-x)

Let’s now explore what happens as we change the value of a whilst keeping our initial starting values of x = 0.2 and x = 0.2001

a = 0.8

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both graphs are superimposed but are identical at the scale we are using.  We can see that both values are attracted to 0 (we can say that 0 is an attractor for our system).

a = 1.2

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Again both graphs are superimposed but are identical at the scale we are using.  We can see that both values are attracted to 1/6 (we can say that 1/6 is an attractor for our system).

In general, for f(x) = ax(1-x) with -1≤x≤1, the attractors are given by x = 0 and x = 1 – 1/a, but it depends on the starting conditions as to whether we will end up being attracted to this point.

f(x) = 0.8x(1-x)

So, let’s look at f(x) = 0.8x(1-x) for different starting values 1≤x≤1.  Our attractors are given by x = 0 and x = 1 – 1/0.8 = -0.25.

When our initial value is x = 0 we remain at the point x = 0.

When our initial value is x = -0.25 we remain at the point x = -0.25.

When our initial value is x < -0.25 we tend to negative infinity.

When our initial value is  -0.25 < x ≤ 1 we tend towards x = 0.

Starting value of x = -0.249999:

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Therefore we can say that x = 0 is a stable attractor, initial values close to x = 0 will still tend to 0.

However x = -0.25 is a fixed point rather than a stable attractoras

x = -0.250001 will tend to infinity very rapidly,

x = -0.25 stays at x = -0.25.

x = -0.249999 will tend towards 0.

Therefore there is a stable equilibria at x = 0 and an unstable equilibria at x = -0.25.

 

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Modelling tides: What is the effect of a full moon?

Let’s have a look at the effect of the moon on the tides in Phuket.  The Phuket tide table above shows the height of the tide (meters) on given days in March, with the hours along the top.  So if we choose March 1st (full moon) we get the following graph:

Phuket tide at full moon:

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If I use the standard sine regression on Desmos I get the following:

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This doesn’t look a very useful graph – but the R squared value is very close to one – so what’s gone wrong?  Well, Desmos has done what we asked it to do – found a sine curve that goes through the points, it’s just that it’s chosen a b value of close to 120 – meaning that the curve has a very small period.  So to prevent Desmos doing this, we need to fix the period first.   If we are in radians the we use the formula period = 2pi / b.  Therefore looking at the original graph we can see that this period is around 12.  Therefore we have:

period = 2pi/b

12 = 2pi/b

b = 2pi/12 or pi/6.

Plotting this new graph gives something that looks a lot nicer:

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Phuket tide at new moon:

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Screen Shot 2018-03-25 at 5.26.40 PM

Analysis:

Both graphs show a very close fit to the original data – though both under-value the tide at 2300.  We can see that the full moon has indeed had an effect on the amplitude of the sine curves – with the amplitude of 1.21m for the full moon and only 1.03m for the new moon.

Further study:

We could then see if this relationship holds throughout the year – is there a general formula to explain the moons effect on the amplitude?  We could also see how we have to modify the sine wave to capture the tidal height over an entire week or month.  Can we capture it with a single equation (perhaps a damped sine wave?) or is it only possible as a piecewise function?  We could also use some calculus to find the maximum and minimum points.

There is a very nice pdf which goes into more detail on the maths behind modeling tides here.  There we go – a nice simple investigation which can be expanded in a number of directions.

 

Circular Motion: Modelling a ferris wheel

This is a nice simple example of how the Tracker software can be used to demonstrate the circular motion of a Ferris wheel.  This is sometimes asked in IB maths exams – so it’s nice to get a visual representation of what is happening.

First I took a video from youtube of a Ferris wheel, loaded it into Tracker, and then used the program to track the position of a single carriage as it moved around the circle.  I then used Tracker’s graphing capabilities to plot the height of the carriage (y) against time (t).  This produces the following graph:

As we can see this is a pretty good fit for a sine curve. So let’s use the regression tool to find what curve fits this:

The pink curve with the equation:

y = -116.1sin(0.6718t+2.19)

fits reasonably well.  If we had the original dimensions of the wheel we could scale this so the y scale represented the metres off the ground of the carriage.

There we go!  Short and simple, but a nice starting point for an investigation on circular motion.

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