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**Life on the Beach with Markov Chains**

Markov chains are exceptionally useful tools for calculating probabilities – and are used in fields such as economics, biology, gambling, computing (such as Google’s search algorithm), marketing and many more. They can be used when we have the probability of a future event dependent on a current event. The image above is of the Russian mathematician Andrey Markov [public domain picture from here] who was the first mathematician to work in this field (in the late 1800s).

**Beach life**

The picture above is an example of a situation which can be modeled using a Markov chain. We have two states: Beach (state 1) and Home (state 2). In our happy life we spend all the hours of the day in either one of these two states. If we are on the beach then the probability we remain on the beach in one hour is 0.6 and the probability we go home in one hour is 0.4. If we are at home the probability we remain at home in one hour is 0.8 and the probability we go to the beach is 0.2. Hopefully you can see how this information is represented above.

**Using matrices**

First we need to represent our information in a matrix form. A general 2×2 matrix is written as:

Where the subscript tells you the row and column (e.g. a_12 tells you it is in the first row and 2nd column).

For our Markov chain we define the following matrix:

Here m_11 is the situation of starting in state 1 and moving to state 1. m_12 is the situation of starting in state 1 and moving to state 2. Therefore p(m_12) is the probability of starting in state 1 and moving to state 2. So for our beach existence we have:

The 0.6 shows that if we are already on the beach we have a 0.6 chance of still being on the beach in one hour.

**Where will we be in the future?**

The benefit of Markov chains is that they allow us to utilise computer power to now calculate where someone will be in the future – simply by taking the power of the matrix. To find the probabilities after 2 hours I can square my matrix:

Using the rules of matrix multiplication this then gives:

Which we can the calculate as:

I’ve used arrow notation here to represent the start and end state so p(m_1 arrow 1) means starting at 1 and ending at 1 after 2 hours. We can see that for someone who started in the beach, the chance of them being on the beach in 2 hours is 0.44. Equally the probability of someone who started in the house being on the beach in 2 hours is 0.28.

I can then carry on with matrix multiplication to work out where someone will likely be for any given number of hours in the future.

p(m_1 arrow 1) means starting at 1 and ending at 1 after n hours. So for example if I want to see where someone will be in 24 hours I simply do:

We can see that now it doesn’t actually matter (to 3sf anyway) where someone started – if they started on the beach there is a 0.333 chance they are on the beach in 24 hours, if they started in the house there is also a 0.333 chance they are on the beach in 24 hours. So I can conclude that as the number of hours increase towards infinity that the person in this scenario would spend 1/3 of their time on the beach and 2/3 of their time at home – not a bad life!

**A more demanding beach life**

We can see that things get much more complicated, even by adding an extra state. Now we have 3 possible states, Beach (state 1), Home (state 2) and SCUBA (state 3). This time we need a 3×3 matrix:

and as before we define our probability matrix as:

So from our diagram we have the following:

For example the 0.8 in row 2 column 2 shows that there is a 0.8 chance of starting in state 2 (Home) and ending in state 2 (Home) in one hour.

Then using our same notation we have:

Which shows that after 2 hours there is (for example) a 0.19 chance that someone who started in state 2 (Home) is now found in state 1 (Beach).

After 24 hours we have the following matrix:

So we notice the same situation as last time – as the number of hours increase it gets less important where we started from – we can see that to 3sf it doesn’t now matter where we started – the probability after 24 hours of being found on the beach is 0.407, the probability of being found at home is 0.333 and the probability of being found diving is 0.260.

Hopefully this is a quick example to demonstrate the power of Markov chains in working with probabilities. There is a lot more to explore (maybe in another post!)

**Spotting fake data with Benford’s Law**

In the current digital age it’s never been easier to fake data – and so it’s never been more important to have tools to detect data that has been faked. Benford’s Law is an extremely useful way of testing data – because when people fake data they tend to do so in a predictable way. Benford’s Law looks at the probability that a number in certain data set (many measurements, street address, stock prices etc.) begins with a given number (its leading digit). Whilst we might expect the leading digits (d) would be equally likely occur, in reality they follow the following equation:

So for example we can see that a leading digit of 1 is much more likely than a leading digit of a 9:

**Testing some data**

I wanted to test some data to see if it did indeed follow Benford’s Law. So, I downloaded an Excel file with 531 data points from the CDC website. This gave the moving 7-day average Covid cases per 100,000 people for every day from 12th March 2020 to 3rd October 2021. I then used the nice Excel techniques shown above in the video to manipulate the data into a useful form. Once this had been done I could then use Desmos to plot this data (dot plot and left aligned frequency histogram). You can see this data below:

The red curve is the continuous (rather than discrete) curve created by working out the expected frequencies for each digit. On Desmos I generated this by the following equation:

We can see that our data largely follows our expected curve – so we would not have any evidence to suggest faked data! We could conduct a Chi-Squared test to measure the goodness of fit of our data (this is also explained in the video).

**Conclusion**

This is a simple but effective method to test for faked data – if data fails this test it doesn’t necessarily mean it was faked (eg. data on heights of men in cm will clearly have nearly all 1s as leading digits!) but most non-random real life data measurements do follow this rule. Try to find your own data (try to do this with a large data set) and try for yourself.

**Weaving a Spider Web**

I often see some beautiful spider webs near my house, similar to the one pictured above (picture from here). They clearly have some sort of mathematical structure, so I decided to have a quick go at creating my own.

Looking at the picture above there are 2 main parts, an inner spiral, then a structure of hanging threads from lines which radiate from the centre.

Firstly I will use the general parametric equation of a hypocycloid:

and take the special case when a = 10 and b = 9:

This gives the following graph:

I can then vary the value of *n* in the following equations:

Which generates the following:

Next, I can generate the spiral in the centre by using an Archimedean spiral, plotting the curve in polar form as:

Which now gives:

Lastly, I want to have straight lines radiating from the centre going through the vertices of the graphs. I can notice that at these vertices the gradient will be undefined (as we can’t define the gradient at a sharp point). Therefore I can differentiate and look for when the gradient will be undefined.

I can see that this will be undefined when the denominator is zero. Therefore:

I can notice that all the vertices are are on the same lines, therefore I can simply choose n =9 to make my life easier, and then solve for t. I use the fact that sine is an odd function to help here.

Here p is an integer. I’ll then rearrange the first of these two equations for t to show how I can then find my equations of the lines.

If I now substitute this value of t back into my parametric equations I get:

So, this will tell me the coordinates of the vertices of the “sharp points” of the graph. Therefore the equation of the straight lines through these points and also through the origin are given by the first equation below. I can then choose my values of p (with p an integer) to get specific solutions. For example when I choose p = 1 above I get the equation of a line which will pass through one of these vertices:

Let’s check that this works:

Yes! So, we can use this method to find the other lines radiating from the centre. This gives us our final spider web:

So, there we go, a quick go at making a spider web – quite a simplistic pattern, but still utilising parametric equations, polar coordinates and also calculus and trigonometric equations.

**Elliptical Curve Cryptography**

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

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

y² = x³ +ax + b

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

**Addition of two **points **A and B**

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

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

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

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

**Addition of 2 points when A = B**

We have to also be able to cope with the situation when the point A and B are the same. Here we create the line through A which is the tangent to the curve at that point:

We then use the same transformation as before to say that A+B = C’. For example with the curve y² = x³ -12x, if we start with the point A(-2,4) then this transformation tells us that A + A = (4,-4).

**Elliptical curves over finite fields**

For the purposes of cryptography we often work with elliptical curves over finite fields. This means we (say) only consider integer coordinate solutions and work in modulo arithmetic (mod prime).

Say we start with the curve y² = x³ +x+1, and just look at the positive integer solutions mod 7. (Plotted using the site here).

When x = 1,

y² = 1³ +1 + 1

y² = 3

So this has no integer solution.

Next, when x = 2 we have:

y² = 2³ +2 +1 = 11.

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

y² = 4 (mod 7)

y = 2 or y = -2 = 5 (mod 7)

When x = 3 we have:

y² = 3³ +3 +1 = 31

y² = 3 (mod 7)

which has no integer solutions.

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

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

**Addition under modulo arithmetic**

Let’s look at the coordinate points we calculated before for the elliptical curve y² = x³ +x+1 (integers solutions and mod 7) – they form a group under addition. (Table generated here)

In order to calculate addition of points when dealing with elliptical curves with integer points mod prime we use the same idea as expressed above for general graphs.

The table tells us that (0,1) + (0,1) = (2,5). If we were doing this from the graph we would draw the tangent to the curve at (0,1), find where it intersects the graph again, then reflect this point in the x axis. We can do all this algebraically.

First we find the gradient of the tangent when x = 0:

Next we have to do division modulo 7 (you can use a calculator here, and you can also read more about division modulo p here).

Next we find the equation of the tangent through (0,1):

Next we find where this tangent intersects the curve again (I used Wolfram Alpha to solve this mod 7)

We then substitute the value x = 2 into the original curve to find the y coordinates:

(2,2) is the point where the tangent would touch the curve and (2,5) is the equivalent of the reflection transformation. Therefore our answer is (2,5). i.e (0,1) + (0,1) = (2,5) as required.

When adding points which are not the same we use the same idea – but have to find the gradient of the line joining the 2 points rather than the gradient of the tangent. We can also note that when we try and add points such as (2,5) and (2,2) the line joining these does not intersect the graph again and hence we affix the point an infinity as (2,5) + (2,2).

**Using elliptical codes for cryptography**

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

Say for example two people wish to create an encryption key.

They decide on an elliptical curve and modulo. Let’s say they decide on y² = x³ +x+1 for integers, mod 7.

This creates the addition group

Next they choose a point of the curve. Let’s say they choose P(1,1).

Person 1 chooses a secret number n and then sends nP (openly). So say Person 1 chooses n = 2. 2(1,1) = (1,1) + (1,1) = (0,2). Person 1 sends (0,2).

Person 2 chooses a secret number m and then sends mP (openly). So say Person 2 chooses m = 3. 3(1,1) = (1,1) + (1,1) + (1,1) = (0,2) + (1,1) = (0,5). Person 2 sends (0,5).

Both Person 1 and Person 2 can easily calculate mnP (the secret key).

Person 1 receives (0,5) and so does 2(0,5) = (0,5) + (0,5) = (1,1). This is the secret key.

Person 2 receives (0,2) and so does 3(0,2) = (0,2) + (0,2) +(0,2) = (1,1). This is the same secret key.

But for a person who can see mP and nP there is no quick method for working out mnP – with a brute force approach extremely time consuming. Therefore this method can be successfully used to encrypt data.

**Prime Spirals – Patterns in Primes**

One of the fundamental goals of pure mathematicians is gaining a deeper understanding of the distribution of prime numbers – hence why the Riemann Hypothesis is one of the great unsolved problems in number theory and has a $1 million prize for anyone who can solve it. Prime numbers are the the building blocks of our number system and are essential to our current encryption methods such as RSA encryption. Hence finding patterns in the primes is one of the great mathematical pursuits.

**Polar coordinates**

The beautiful prime spiral was generated above on Desmos using polar coordinates. We can see a clear spiral pattern – so let’s see how to create this. Polar coordinates (r, θ) need a length (r) from the origin and an angle of anti-clockwise rotation from the origin (θ). So for example in polar coordinates (2,2) means a length of 2 from the origin and a rotation of 2 radians. By considering trigonometry and the unit circle we can say that the polar coordinates (r, θ) are equivalent to the Cartesian coordinate (r.cosθ, r.sinθ).

**Plotting prime pairs**

So we plot the first few prime pairs:

Polar: (2,2). Cartesian: (2cos2, 2sin2).

Polar: (3,3). Cartesian: (3cos3, 3sin3).

Polar: (5,5). Cartesian: (5cos5, 5sin5).

In Desmos (making sure we are in radians) we input:

We can then change the Desmos graph view to polar (first click on the spanner on the right of the screen). This gives the first 3 points of our spirals. Note I have labeled the points as polar coordinates.

I then downloaded the first 1000 prime numbers from here. I then copied this list of comma separated values and pasted it into an empty part of square brackets M = [ ] in Desmos to create a list.

I can then plot every point in the list as a prime pair by doing the following:

We can then generate our prime spiral for the first 1000 prime pairs:

Just to see how powerful Desmos really is, I then downloaded all the prime numbers less than or equal to 100,000 from here. This time we see the following graph:

We can see that we lose the clear definition of the spiral – though there are still circular spirals with higher densities of primes than others. Also we can see that there are higher densities of the primes on some of the radial lines out from the origin – and other radial lines where no primes appear.

**Prime Number Theorem**

We can also use our Desmos result to investigate another (more fundamental) result about the distribution of prime numbers. The prime number theorem states:

Here pi(N) is the number of prime numbers less than or equal to N. The little squiggle means that as N gets large pi(N) becomes better and better approximated by the function on the RHS.

For our purple “spiral” above we downloaded all the primes less than or equal to 100,000 – and Desmos tells us that there were 9,592 of them. So let’s see how close the prime number theorem gets us:

We can see that we are off by an error of around 9.46% – not too bad, though still a bit out. As we make N larger we will find that we get a better and better approximation.

Let’s look at what would happen if we took N as 1,000,000,000. From Wikipedia we can see that there are 50,847,534 primes less than or equal to 1,000,000,000. Therefore:

This time we are off by an error of only 5.10%. Have a look at the table of values in Wikipedia to find how large N has to be to be within 1% accuracy.

So this is a nice introduction to looking for patterns in the primes – and a good chance to explore some of the nice graphical capabilities of Desmos. See if you can find any more patterns of your own!

**Hailstone Numbers**

Hailstone numbers are created by the following rules:

**if n is even:** divide by 2

**if n is odd:** times by 3 and add 1

We can then generate a sequence from any starting number. For example, starting with 10:

10, 5, 16, 8, 4, 2, 1, 4, 2, 1…

we can see that this sequence loops into an infinitely repeating 4,2,1 sequence. Trying another number, say 58:

58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1…

and we see the same loop of 4,2,1.

Hailstone numbers are called as such because they fall, reach one (the ground) before bouncing up again. The proper mathematical name for this investigation is the Collatz conjecture. This was made in 1937 by a German mathematian, Lothar Collatz.

One way to investigate this conjecture is to look at the length of time it takes a number to reach the number 1. Some numbers take longer than others. If we could find a number that didn’t reach 1 even in an infinite length of time then the Collatz conjecture would be false.

The following graphic from wikipedia shows how different numbers (x axis) take a different number of iterations (y axis) to reach 1. We can see that some numbers take much longer than others to reach one. Some numbers take over 250 iterations – but every number checked so far does eventually reach 1.

For example, the number 73 has the following pattern:

73, 220, 110, 55, 166, 83, 250, 125, 376, 188, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1…

**No proof yet**

Investigating what it is about certain numbers that leads to long chains is one possible approach to solving the conjecture. This conjecture has been checked by computers up to a staggering 5.8 x 10^{18} numbers. That would suggest that the conjecture could be true – but doesn’t prove it is. Despite looking deceptively simple, Paul Erdos – one of the great 20th century mathematicians stated in the 1980s that “mathematics is not yet ready for such problems” – and it has remained unsolved over the past few decades. Maybe you could be the one to crack this problem!

**Exploring this problem with Python.**

We can plot this with Python – such that we also generate a nice graphical representation of these numbers. The graph above shows what happens to the number 500 when we follow this rule – we “bounce” up to close to 10,000 before falling back into the closed loop after around 100 iterations.

**Numbers with large iterations:**

871 takes 178 steps to reach 1:

77,031 takes 350 steps to reach 1:

9,780,657,630 takes 1132 steps to reach 1:

If you want to explore this code yourself, the following code has been written to run on repl.it. You can see the code yourself here, and I have also copied it below:

Have a play – and see what nice graphs you can draw!

**Galileo’s Inclined Planes**

*This post is based on the maths and ideas of Hahn’s Calculus in Context – which is probably the best mathematics book I’ve read in 20 years of studying and teaching mathematics. Highly recommended for both students and teachers!*

Hahn talks us though the mathematics, experiments and thought process of Galileo as he formulates his momentous theory that in free fall (ignoring air resistance) an object falling for *t* seconds will fall a distance of ct² where c is a constant. This is counter-intuitive as we would expect the mass of an object to be an important factor in how far an object falls (i.e that a heavier object would fall faster). Galileo also helped to overturned Aristotle’s ideas on motion. Aristotle had argued that any object in motion would eventually stop, Galileo instead argued that with no friction a perfectly spherical ball once started rolling would roll forever. Galileo’s genius was to combine thought experiments and real data to arrive at results that defy “common sense” – to truly understand the universe humans had to first escape from our limited anthropocentric perspective, and mathematics provided an opportunity to do this.

**Inclined Planes**

Galileo conducted experiments on inclined planes where he placed balls at different heights and then measured their projectile motion when they left the ramp, briefly ran past the edge of a flat surface and then fell to the ground. We can see the set up of one ramp above. The ball starts at O, and we mark as h this height. At an arbitrary point P we can see that there are 2 forces acting on the ball, F which is responsible for the ball rolling down the slope, and f, which is a friction force in the opposite direction. At point P we can mark the downwards force mg acting on the ball. We can then use some basic rules of parallel lines to note that the angles in triangle PCD are equal to triangle AOB.

Galileo’s t**imes squared law of fall**

We have the following equation for the total force acting on the ball at point P:

We also have the following relationship from physics, where m is the mass and a(t) the acceleration:

This therefore gives:

Next we can use trigonometry on triangle PCD to get an equation for F:

Next we can use another equation from physics which gives us the frictional force on a perfectly spherical, homogenous body rolling down a plane is:

So this gives:

We can then integrate to get velocity (our constant of integration is 0 because the velocity is 0 when t = 0)

and integrate again to get the distance travelled of the ball (again our constant of integration is 0):

When Galileo was conducting his experiments he did not know *g, *instead he noted that the relationship was of the form;

where c is a constant related to a specific incline. This is a famous result called the *times squared law of fall. * It shows that the distance travelled is independent of the mass and is instead related to the time of motion squared.

**Velocity also independent of the angle of incline**

Above we have shown that the distance travelled is independent of the mass – but in the equation it is still dependent on the angle of the incline. We can go further and then show that the velocity of the ball is also independent of the angle of incline, and is only dependent on the height at which the ball starts from.

If we denote as t_b as the time when the ball reaches point A in our triangle we have:

This is equal to the distance from AO, so we can use trigonometry to define:

This can then be rearranged to give:

this is the time taken to travel from O to A. We can the substitute this into the velocity equation we derived earlier to give the velocity at point A. This is:

This shows that the velocity of the ball at point A is only dependent on the height and not the angle of incline or mass. The logical extension of this is that if the angle of incline has no effect on the velocity, that this result would still hold as the angle of incline approaches and then reaches 90 degrees – i.e when the ball is in free fall.

Galileo used a mixture of practical experiments on inclined planes, mathematical calculations and thought experiments to arrive at his truly radical conclusion – the sign of a real genius!

Essential resources for IB students:

Revision Village has 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 and SL students for both Analysis and Applications.

There is a comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions. 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 a large number of ready made quizzes, exams and predicted papers. These all have worked solutions and allow you to focus on specific topics or start general revision. This also has some excellent challenging questions for those students aiming for 6s and 7s.

Each course also has a dedicated video tutorial section which provides 5-15 minute tutorial videos on every single syllabus part – handily sorted into topic categories.

2) Exploration Guides and Paper 3 Resources

I’ve put together four comprehensive pdf guides to help students prepare for their exploration coursework and Paper 3 investigations. The exploration guides talk through the marking criteria, common student mistakes, excellent ideas for explorations, technology advice, modeling methods and a variety of statistical techniques with detailed explanations. I’ve also made 17 full investigation questions which are also excellent starting points for explorations. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.

**Finding focus with Archimedes**

*This post is based on the maths and ideas of Hahn’s Calculus in Context – which is probably the best mathematics book I’ve read in 20 years of studying and teaching mathematics. Highly recommended for both students and teachers!*

Hard as it is to imagine now, for most of the history of mathematics there was no coordinate geometry system and therefore graphs were not drawn using algebraic equations but instead were constructed. The ancient Greeks such as Archimedes made detailed studies of conic sections (parabolas, ellipses and hyperbola) using ideas of relationships in constructions. The nice approach to this method is that it makes clear the link between conic sections and their properties in reflecting light – a property which can then be utilized when making lenses. A parabolic telescope for example uses the property that all light collected through the scope will pass through a single focus point.

Let’s see how we can construct a parabola without any algebra – simply using the constructions of the Greeks. We start with a line and a focus point F not on the line. This now defines a **unique parabola**.

This unique parabola is defined as all the points A such that the distance from A to F is equal as the perpendicular distance from A to the line.

We can see above that point A must be on our parabola because the distance AB is the same as the distance AF.

We can also see that point C must be on our parabola because the length CD is the same as CF. Following this same method we could eventually construct every point on our parabola. This would finally create the following parabola:

**Focus point of a parabolic **mirror

We can now see how this parabola construction gives us an intrinsic understanding of reflective properties. If we have a light source entering parallel to the perpendicular though the focus then we can use the fact that this light will pass through the focus to find the path the light traces before it is reflected out.

Newton made use of this property when designing his parabolic telescope. It’s interesting to note how a different method leads to a completely different appreciation of the properties of a curve.

**Finding the area under a quadratic curve without calculus**

Amazingly a method for finding the area under a quadratic curve was also discovered by the Greek scientist and mathematician Archimedes around 2200 years ago – and nearly 2000 years before calculus. Archimedes’ method was as follows.

Choose 2 points on the curve, join them to make 2 sides of a triangle. Choose the 3rd point of the triangle as the point on the quadratic with the same gradient as the chord. This is best illustrated as below. Here I generated a parabola with focus at (0,1) and line with the x axis.

Here I chose points B and C, joined these with a line and then looked for the point on the triangle with the same gradient. This then gives a triangle with area 4. Archimedes then discovered that the area of the parabolic segment (i.e the total area enclosed by the line BC and the parabola) is 4/3 the area of the triangle. This gives 4/3 of 4 which is 5 1/3. Once we have this we can find the area under the curve (i.e the integral) using simple areas of geometric shapes.

**Using calculus**

We can check that Archimedes’ method does indeed work. We want to find the area enclosed by the 2 following equations:

This is given by:

It works! Now we can try a slightly more difficult example. This time I won’t choose 2 points parallel to the x-axis.

This time I find the gradient of the line joining B and C and then find the point on the parabola with the same gradient. This forms my 3rd point of the triangle. The area of this triangle is approximately 1.68. Therefore Archimedes’ method tells us the area enclosed between the line and the curve will be approximately 4/3 (1.68) = 2.24. Let’s check this with calculus:

Again we can see that this method works – our only error was in calculating an approximate area for the triangle rather than a more precise answer.

So, nearly 2000 years before the invention of calculus the ancient Greeks were already able to find areas bounded by line and parabolic curves – and indeed Archimedes was already exploring the ideas of the limit of sums of areas upon which calculus in based.

Essential resources for IB students:

Revision Village has 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 and SL students for both Analysis and Applications.

There is a comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions. 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 a large number of ready made quizzes, exams and predicted papers. These all have worked solutions and allow you to focus on specific topics or start general revision. This also has some excellent challenging questions for those students aiming for 6s and 7s.

Each course also has a dedicated video tutorial section which provides 5-15 minute tutorial videos on every single syllabus part – handily sorted into topic categories.

2) Exploration Guides and Paper 3 Resources

I’ve put together four comprehensive pdf guides to help students prepare for their exploration coursework and Paper 3 investigations. The exploration guides talk through the marking criteria, common student mistakes, excellent ideas for explorations, technology advice, modeling methods and a variety of statistical techniques with detailed explanations. I’ve also made 17 full investigation questions which are also excellent starting points for explorations. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.

**Find the average distance between 2 points on a square**

This is another excellent mathematical puzzle from the MindYourDecisions youtube channel. I like to try these without looking at the answer – and then to see how far I get. This one is pretty difficult (and the actual solution exceptionally difficult!) The problem is to take a square and randomly choose 2 points somewhere inside. If you calculate the distance between the 2 points, then do this trial approaching an infinite number of times what will the average distance be? Here is what I did.

**Simplify the situation: 1×1 square**

This is one of the most important strategies in tackling difficult maths problems. You simplify in order to gain an understanding of the underlying problem and possibly either develop strategies or notice patterns. So, I started with a unit square and only considered the vertices. We can then list all the possible lengths:

We can then find the average length by simply doing:

**2×2 square**

We can then follow the same method for a 2×2 square. This gives:

Which gives an average of:

**Back to a 1×1 square**

Now, we can imagine that we have a 1 x 1 square with dots at every 0.5. This is simply a scaled version of the 2×2 square, so we can divide our answer by 2 to give:

**3×3 square**

Following the same method we have:

This gives an average of:

and if we imagine a 1×1 square with dots at every 1/3. This is simply a scaled version of the 3×3 square, so we can divide our answer by 3 to give:

We can then investigate what happens as we consider more and more dots inside our 1×1 square. When we have considered an infinite number then we will have our average distance – so we are looking the limit to infinity. This suggests using a graph. First I calculated a few more terms in the sequence:

Then I plotted this on Desmos. The points looked like they fit either an exponential or a reciprocal function – both which have asymptotes, so I tried both. The reciprocal function fit with an R squared value of 1. This is a perfect fit so I will use that.

This was plotted using the regression line:

And we can find the equation of the horizontal asymptote by seeing what happens when x approaches infinity. This will give a/c. Using the values provided by Desmos’ regression I got 0.515004887. Because I have been using approximate answers throughout I’ll take this as 0.52 (2sf). **Therefore I predict that the average distance between 2 points in a 1×1 square will be approximately 0.52**. And more generally, the average distance in an n x n square will be 0.52(n). This is somewhat surprising as a result – it’s not obvious why it would be a little over half the distance from 0 to 1.

**Brute forcing using Python**

We can also write a quick code to approximate this answer using Python (This is a Monte Carlo method). I generate 4 random numbers to represent the 2 x-coordinates and 2-y coordinates of 2 random points. I then work out the distance between them and repeat this 10 million times, then calculate the average distance. This gives:

**Checking with the actual answer**

Now for the moment of truth – and we watch the video to find out how accurate this is. The correct answer is indeed 0.52 (2sf) – which is great – our method worked! The exact answer is given by:

Our graphical answer is not quite accurate enough to 3 sf – probably because we relied on rounded values to plot our regression line. Our Python method with 10 million trials was accurate to 4 sf. Just to keep my computer on its toes I also calculated this with 100 million trials. This gave 0.5214126210834646 (now accurate to 5 sf).

We can also find the percentage error when using our graphical method. This is only:

Overall this is a decent result! If you are feeling *extremely* brave you might want to look at the video to see how to do this using calculus.

**Extension: The average distance between 2 points in a unit circle**

I modified the Python code slightly to now calculate the average distance between 2 points in a unit circle. This code is:

which returns an answer of 0.9054134561871364. I then looked up what the exact answer is. For the unit circle it is 128/(45 pi). This is approximately 0.9054147874. We can see that our computer method was accurate to 5 sf here. Again, the actual mathematical proof is extremely difficult.

**Reflection**

This is a nice example of important skills and techniques useful in mathematics – simplification of a problem, noticing patterns, graphical methods, computational power and perseverance!

Essential resources for IB students:

Revision Village has 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 and SL students for both Analysis and Applications.

There is a comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions. 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 a large number of ready made quizzes, exams and predicted papers. These all have worked solutions and allow you to focus on specific topics or start general revision. This also has some excellent challenging questions for those students aiming for 6s and 7s.

Each course also has a dedicated video tutorial section which provides 5-15 minute tutorial videos on every single syllabus part – handily sorted into topic categories.

2) Exploration Guides and Paper 3 Resources

I’ve put together four comprehensive pdf guides to help students prepare for their exploration coursework and Paper 3 investigations. The exploration guides talk through the marking criteria, common student mistakes, excellent ideas for explorations, technology advice, modeling methods and a variety of statistical techniques with detailed explanations. I’ve also made 17 full investigation questions which are also excellent starting points for explorations. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.