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Finding the average distance in a polygon
Over the previous couple of posts I’ve looked at the average distance in squares, rectangles and equilateral triangles. The logical extension to this is to consider a regular polygon with sides 1. Above is pictured a regular pentagon with sides 1 enclosed in a 2 by 2 square. The points N and O represent 2 randomly chosen points which we find the distance between. On average what is the distance between these randomly chosen points N and O?
Starting with a hexagon
It’s a little easier to start with a hexagon as we get some nicer coordinate points. So, our first challenge is to find the coordinates of a regular hexagon with sides 1. Luckily we can use the complex roots of unity to do this. We start by finding the 6th roots of unity and then converting these to coordinates in an Argand diagram:
This then allows us to plot the following:
We can then work out the inequalities which define the inside of the hexagon when we generate points within the 2×2 square centred at (0,0). This gives:
We can then run the following code to find the average distance:
This gives the following result:
We can check this result as the exact value is:
which is 0.8262589495. So we can see we are accurate here to 3 sf.
Pentagon
For the pentagon we can find the coordinates by finding the 5th roots of unity:
We then need to scale all coordinate points by a factor, because in a pentagon the distance from the centre to the points is not 1 (as is the case in roots of unity). We can find the distance from the centre to the edge of a pentagon by the following trigonometry:
So, when we scale all coordinate points by this factor we get:
And we can then do the same method as before and run the following Python code:
This gives:
n-sided polygon
We can now consider an n-sided polygon with sides 1. Let’s start with the values we’ve found for an equilateral triangle (0.364), a square (0.522), a pentagon (0.697) and a hexagon (0.826.
When we plot these they appear to follow a linear relationship:
average distance = 0.14n
We can check that this is correct by considering the fact that an n sided polygon will approximate a circle when n gets large. So an n sided polygon with sides length 1 can be approximated by a circle with circumference n. This allows us to work out the radius.
We can then substitute this into the equation for the average distance of 2 points in a circle.
So we would expect the average distance between 2 points in a regular polygon of sides 1 to approach the equation (as n gets large):
average distance = 0.144101239n
And we’ve finished! Everything cross-checks and works nicely. We’ve been able to use a mixture of complex numbers, geometry, coding and trigonometry to achieve this result.
Finding the average distance in an equilateral triangle
In the previous post I looked at the average distance between 2 points in a rectangle. In this post I will investigate the average distance between 2 randomly chosen points in an equilateral triangle.
Drawing a sketch.
The first step is to start with an equilateral triangle with sides 1. This is shown above. I sketched this using Geogebra – and used some basic Pythagoras to work out the coordinates of point C.
I can then draw a square of sides 1 around this triangle as shown above. I’m then going to run a Python program to randomly select points and then work out the distance between them – but I need to make sure that the 2 points chosen are both inside this triangle. For this I need to work out the equation of the line AC and CB.
Using basic coordinate geometry we can see that the line AC has equation y = √3x. We want the inequality y < √3x so that we are on the correct side of this line.
The line BC has equation y = -√3x + √3. Therefore the triangle must also satisfy the inequality y < -√3x + √3.
I can then run the following code on Python, with finds the average distance between points (a,c) and (b,d) both within the unit square but also subject to the 2 inequality constraints above.
When this is run it performs 999,999 trials and then finds the average distance. This returns the following value:
So we can see that the average distance is just over a third of a unit.
Finding the average distance of an equilateral triangle of length n.
We can then draw the sketch above to find the equation of lines AC and CB for an equilateral triangle with lengths n. This leads to the following inequalities:
y < √3x
y < -√3x + √3n
So we can then modify the code as follows:
This then returns the average distances for equilateral triangles of sizes 1 to 10.
And when we plot this on Desmos we can see that there is a linear relationship:
The regression line has gradient 0.36 (2sf) so we can hypothesise that for an equilateral triangle of size n, the average distance between 2 points is approximately 0.36n.
Checking the maths
I then checked the actual equation for the average distance between 2 points in an equilateral triangle of sides n:
This gives us:
So we can see that we were accurate to 2 significant figures. So this is a nice mixture of geometry, graphing and computational power to provide a result which would be otherwise extremely difficult to calculate.
Plotting Pi and Searching for Mona Lisa
This is a very nice video from Numberphile – where they use a string of numbers (pi) to write a quick Python Turtle code to create some nice graphical representations of pi. I thought I’d quickly go through the steps required for people to do this by themselves.
Firstly you can run the Turtle code on trinket.io. If you type the above code this will take the decimal digits of pi one at a time and for each one move forward 10 steps and then turn by 36 degrees times by that digit. So for example the 1 will lead to a right turn of 36 degrees and the 4 will lead to a right turn of 36 x 4 = 144 degrees.
Next it would be nice to have more digits of pi to paste in rather than type. So we can go to the onlinenumbertools website and generate as many digits of pi as we want. Select them to be comma separated and also to not include the first digit 3. You can then copy and paste this string in place of the 1,4,1 in the code above.
1000 digits of pi
If we run this program after pasting the first 1000 digits of pi we get (after waiting a while!) the above picture. There are a number of questions that they then raise in the video – if this program was ran infinitely long would the whole screen eventually be black? Would this create every possible image that can be created by 36 degree turns? Would it be possible to send this picture (say to an alien civilization) and for the recipient to be able to reverse engineer the digits of pi?
2000 digits of pi
If you increase the digits of pi to around 2000 you get the above picture. The graph spends a large time in the central region before finally “escaping” to the left. It then left my screen at the top.
3000 digits of pi
We can see that the turtle “returned” from off the top of the screen and then joined back up with the central region. This starts to look like a coastline – maybe the south of the UK!
Different bases: Base 3
We can consider the digits of pi in base three – which means that they are all equivalent to 0,1,2. This means that we can use these to specify either 0 degree, 120 degree or 240 degree turns. We can change the code as shown above to achieve this. Note the i%3 gives i mod 3. For example if the digit is 8, then 8 mod 3 is 2 (the remainder when 8 is divided by 3) and so this would turn 120 x 2 = 240 degrees.
This then creates a pattern which looks similar to the Sierpinski triangle fractal design:
Base 4
Using a similar method, we can create the following using a base 4 design:
This creates what looks like a map layout.
Base 5:
In base 5 the turtle quickly departed from my screen! With turns of 72 we don’t see the tessellating shapes that we do with base 3 and 4.
Base 6:
With a 60 degree turn we can now see a collection of equilateral triangles and hexagons.
You can explore with different numbers and different bases to see what patterns you can create!
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
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 1018 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!
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Chaos and strange Attractors: Henon’s map
Henon’s map was created in the 1970s to explore chaotic systems. The general form is created by the iterative formula:
The classic case is when a = 1.4 and b = 0.3 i.e:
To see how points are generated, let’s choose a point near the origin. If we take (0,0) the next x coordinate is given by:
We would then continue this process over several thousands iterations. If we do this then we get the very strange graph at the top of the page – the points are attracted to a flow like structure, which they then circulate round. The graph above was generated when we took our starting coordinate as (0.1,0.1), let’s take a different starting point. This time let’s have (1.1, 1.1):
We can see that exactly the same structure appears. All coordinates close to the origin will get attracted to this strange attractor – except for a couple of fixed points near the origin which remain where they are. Let’s see why. First we can rewrite the iterative formula just in terms of x:
Next we use the fact that when we have a fixed point the x coordinate (and y coordinate) will not change. Therefore we can define the following:
This allows us to then make the following equation:
Which we can then solve using the quadratic formula:
Which also gives y:
So therefore at these 2 fixed points the coordinates do not get drawn to the strange attractor.
Above we can see the not especially interesting graph of the repeated iterations when starting at this point!
But we can also see the chaotic behavior of this system by choosing a point very close to this fixed point. Let’s choose (0.631354477, 0.631354477) which is correct to 9 decimal places as an approximation for the fixed point.
We can see our familiar graph is back. This is an excellent example of chaotic behavior – a very small change in the initial conditions has created a completely different system.
This idea was suggested by the excellent Doing Maths With Python – which is well worth a read if you are interested in computer programing to solve mathematical problems.
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
The Barnsley Fern: Mathematical Art
This pattern of a fern pictured above was generated by a simple iterative program designed by mathematician Michael Barnsely. I downloaded the Python code from the excellent Tutorialspoint and then modified it slightly to run on repl.it. What we are seeing is the result of 40,000 individual points – each plotted according to a simple algorithm. The algorithm is as follows:
Transformation 1: (0.85 probability of occurrence)
xi+1 = 0.85xi +0.04yi
yi+1= -0.04xi+0.85yi+1.6
Transformation 2: (0.07 probability of occurrence)
xi+1 = 0.2xi -0.26yi
yi+1= 0.23xi+0.22yi+1.6
Transformation 3: (0.07 probability of occurrence)
xi+1 = -0.15xi -0.28yi
yi+1= 0.26xi+0.24yi+0.44
Transformation 4: (0.01 probability of occurrence)
xi+1 = 0
yi+1= 0.16yi
So, I start with (0,0) and then use a random number generator to decide which transformation to use. I can run a generator from 1-100 and assign 1-85 for transformation 1, 86-92 to transformation 2, 93-99 for transformation 3 and 100 for transformation 4. Say I generate the number 36 – therefore I will apply transformation 1.
xi+1 = 0.85(0)+0.04(0)
yi+1= -0.04(0)+0.85(0)+1.6
and my new coordinate is (0,1.6). I mark this on my graph.
I then repeat this process – say this time I generate the number 90. This tells me to do transformation 2. So:
xi+1 = 0.2(0) -0.26(1.6)
yi+1= 0.23(0)+0.22(1.6)+1.6
and my new coordinate is (-0.416, 1.952). I mark this on my graph and carry on again. The graph above was generated with 40,000 iterations – let’s see how it develops over time:
1000 iterations:
10,000 iterations:
100,000 iterations:
500,000 iterations:
If we want to understand what is happening here we can think of each transformation as responsible for a different part of our fern. Transformation 1 is most likely and therefore this fills in the smaller leaflets. Transformations 2 and 3 fill in the bottom left and right leaflet (respectively) and transformation 4 fills in the stem.
It’s quite amazing to think that a simple computer program can create what looks like art – or indeed that is can replicate what we see in nature so well. This fern is an example of a self-similar pattern – i.e one which will look the same at different scales. You could zoom into a detailed picture and see the same patterns repeating. You might want to explore the idea of fractals in delving into this topic in more detail.
Changing the iterations
We can explore what happens when we change the iterations very slightly.
Christmas tree
Crazy spiral
Modern art
You can modify the code to run this here. Have a go!
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Sierpinski Triangle: A picture of infinity
This pattern of a Sierpinski triangle pictured above was generated by a simple iterative program. I made it by modifying the code previously used to plot the Barnsley Fern. You can run the code I used on repl.it. What we are seeing is the result of 30,000 iterations of a simple algorithm. The algorithm is as follows:
Transformation 1:
xi+1 = 0.5xi
yi+1= 0.5yi
Transformation 2:
xi+1 = 0.5xi + 0.5
yi+1= 0.5yi+0.5
Transformation 3:
xi+1 = 0.5xi +1
yi+1= 0.5yi
So, I start with (0,0) and then use a random number generator to decide which transformation to use. I can run a generator from 1-3 and assign 1 for transformation 1, 2 for transformation 2, and 3 for transformation 3. Say I generate the number 2 – therefore I will apply transformation 2.
xi+1 = 0.5(0) + 0.5
yi+1= 0.5(0)+0.5
and my new coordinate is (0.5,0.5). I mark this on my graph.
I then repeat this process – say this time I generate the number 3. This tells me to do transformation 3. So:
xi+1 = 0.5(0.5) +1
yi+1= 0.5(0.5)
and my new coordinate is (1.25, 0.25). I mark this on my graph and carry on again. The graph above was generated with 30,000 iterations.
Altering the algorithm
We can alter the algorithm so that we replace all the 0.5 coefficients of x and y with another number, a.
a = 0.3 has disconnected triangles:
When a = 0.7 we still have a triangle:
By a = 0.9 the triangle is starting to degenerate
By a = 0.99 we start to see the emergence of a line “tail”
By a = 0.999 we see the line dominate.
And when a = 1 we then get a straight line:
When a is greater than 1 the coordinates quickly become extremely large and so the scale required to plot points means the disconnected points are not visible.
If I alternatively alter transformations 2 and 3 so that I add b for transformation 2 and 2b for transformation 3 (rather than 0.5 and 1 respectively) then we can see we simply change the scale of the triangle.
When b = 10 we can see the triangle width is now 40 (we changed b from 0.5 to 10 and so made the triangle 20 times bigger in length):
Fractal mathematics
This triangle is an example of a self-similar pattern – i.e one which will look the same at different scales. You could zoom into a detailed picture and see the same patterns repeating. Amazingly fractal patterns don’t fit into our usual understanding of 1 dimensional, 2 dimensional, 3 dimensional space. Fractals can instead be thought of as having fractional dimensions.
The Hausdorff dimension is a measure of the “roughness” or “crinkley-ness” of a fractal. It’s given by the formula:
D = log(N)/log(S)
For the Sierpinski triangle, doubling the size (i.e S = 2), creates 3 copies of itself (i.e N =3)
This gives:
D = log(3)/log(2)
Which gives a fractal dimension of about 1.59. This means it has a higher dimension than a line, but a lower dimension than a 2 dimensional shape.
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Square Triangular Numbers
Square triangular numbers are numbers which are both square numbers and also triangular numbers – i.e they can be arranged in a square or a triangle. The picture above (source: wikipedia) shows that 36 is both a square number and also a triangular number. The question is how many other square triangular numbers we can find?
The equation we are trying to solve is:
a2 = 0.5(b2+b)
for some a, b as positive integers. The LHS is the formula to generate square numbers and the RHS is the formula to generate the triangular numbers.
We can start with some simple Python code (which you can run here):
for c in range(1,10001):
for d in range(1,10001):
if c**2 == (d**2+d)/2:
print(c**2, c,d)
This checks the first 10000 square numbers and the first 10000 triangular numbers and returns the following:
1 1 1
36 6 8
1225 35 49
41616 204 288
1413721 1189 1681
48024900 6930 9800
i.e 1225 is the next square triangular number after 36, and can be formed as 352 or as 0.5(492+49). We can see that there are very few square triangular numbers to be found in the first 50 million numbers. The largest we found was 48,024,900 which is made by 69302 or as 0.5(98002+9800).
We can notice that the ratio between each consecutive pair of square triangular numbers looks like it converges as it gives:
36/1 = 36
1225/36 = 34.027778
41616/1225 = 33.972245
1413721/41616 = 33.970612
48024900/1413721 = 33.970564
So, let’s use this to predict that the next square triangular number will be around
48024900 x 33.9706 = 1,631,434,668.
If we square root this answer we get approximately 40391
If we solve 0.5(b2+b) = 1,631,434,668 using Wolfram we get approximately 57120.
Therefore let’s amend our code to look in this region:
for c in range(40380,40400):
for d in range(57100,57130):
if c**2 == (d**2+d)/2:
print(c**2, c,d)
This very quickly finds the next solution as:
1631432881 40391 57121
This is indeed 403912 – so our approximation was very accurate. We can see that this also gives a ratio of 1631432881/48024900 = 33.97056279 which we can then use to predict that the next term will be 33.970563 x 1631432881 = 55,420,693,460. Square rooting this gives a prediction that we will use the 235,416 square number. 235,4162 gives 55,420,693,056 (using Wolfram Alpha) and this is indeed the next square triangular number.
So, using a mixture of computer code and some pattern exploration we have found a method for finding the next square triangular numbers. Clearly we will quickly get some very large numbers – but as long as we have the computational power, this method should continue to work.
Using number theory
The ever industrious Euler actually found a formula for square triangular numbers in 1778 – a very long time before computers and calculators, so let’s have a look at his method:
We start with the initial problem, and our initial goal is to rearrange it into the following form:
Next we make a substitution:
Here, when we get to the equation 1 = x2 – 2y2 we have arrived at a Pell Equation (hence the rearrangement to get to this point). This particular Pell Equation has the solution quoted above where we can define Pk as
Therefore we have
Therefore for any given k we can find the kth square triangular number. The a value will give us the square number required and the b value will give us the triangular number required. For example with k = 3:
This tells us the 3rd square triangular number is the 35th square number or the 49th triangular number. Both these give us an answer of 1225 – which checking back from our table is the correct answer.
So, we have arrived at 2 possible methods for finding the square triangular numbers – one using modern computational power, and one using the skills of 18th century number theory.
Essential resources for IB students:
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
When do 2 squares equal 2 cubes?
Following on from the hollow square investigation this time I will investigate what numbers can be written as both the sum of 2 squares, 2 cubes and 2 powers of 4. i.e a2+b2 = c3+d3 = e4+f4.
Geometrically we can think of this as trying to find an array of balls such that we can arrange them into 2 squares, or we can rearrange them and stack them to form 2 cubes, or indeed we can arrange them into 2 4-dimensional cubes. I’ll add the constraints that all of a,b,c,d,e,f should be greater than 1 and that the pair of squares or cubes (etc) must be distinct. Therefore we can’t for example have 2 squares the same size.
Infinite solutions
Let’s look at why we can easily find infinite solutions if the squares or cubes (etc) can be the same size.
We want to find solutions to:
a2+b2 = c3+d3 = e4+f4.
so we look at the powers 2,3,4 which have LCM of 12. Therefore if we choose powers with the same base we can find a solution. For example we chose to work with base 2. Therefore we choose
a = 26, b = 26, which gives 212+212
c = 24, d = 24, which gives 212+212
e = 23, f = 23, which gives 212+212
Clearly these will be the same. So we can choose any base we wish, and make the powers into the same multiples of 12 to find infinite solutions.
Writing some code
Here is some code that will find some other solutions:
list1=[]
for a in range(2, 200):
for b in range(2,200):
list1.append(a**2+b**2)
list2=[]
for j in list1:
for c in range(2,200):
for d in range(2,200):
if c**3+d**3 == j:
list2.append(c**3+d**3)
print(list2)
for k in list2:
for e in range(2,200):
for f in range(2,200):
if k == e**4+f**4:
print(k,e,f)
This returns the following solutions: 8192, 18737, 76832. Of these we reject the first as this is the solution 212+212 which we found earlier and which uses repeated values for the squares, cubes and powers of 4. The 3rd solution we also reject as this is formed by 14 4 + 14 4. Therefore the only solution up to 79202 (we checked every value up to and including 1992 + 1992) is:
18737 = 642+1212 = 173+243 = 114+84.
Therefore if we had 18,737 balls we could arrange them into 2 squares, a 64×64 square and a 121×121 square. Alternatively we could rearrange them into 2 cubes, one 17x17x17 and one 24x24x24. Or we could enter a higher dimensional space and create 2 tesseracts one with sides 11x11x11x11 and the other with 14x14x14x14.
With only 1 solution for around the first 80,000 numbers it looks like these numbers are quite rare – could you find another one? And could you find one that also satisfies g5+h5?
Essential resources for IB students:
1) 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.
Hollow Cubes investigation
Hollow cubes like the picture above [reference] are an extension of the hollow squares investigation done previously. This time we can imagine a 3 dimensional stack of soldiers, and so try to work out which numbers of soldiers can be arranged into hollow cubes.
Therefore what we need to find is what numbers can be formed from a3-b3
Python code
We can write some Python3 code to find this out (this can be run here):
for k in range(1,200):
for a in range(0, 100):
for b in range(0,100):
if a**3-b**3 == k :
print(k,a,b)
This gives the following: (the first number is the number of soldiers and the 2 subsequent numbers are the 2 cubes).
1 1 0
7 2 1
8 2 0
19 3 2
26 3 1
27 3 0
37 4 3
56 4 2
61 5 4
63 4 1
64 4 0
91 6 5
98 5 3
117 5 2
124 5 1
125 5 0
127 7 6
152 6 4
169 8 7
189 6 3
We could perhaps investigate any patterns in these numbers, or explore how we can predict when a hollow cube has more than one solution. I’ll investigate which numbers can be written as both a hollow square and also a hollow cube.
Hollow squares and hollow cubes
list1=[]
for a in range(2, 50):
for b in range(2,50):
if a**2-b**2 !=0:
if a**2-b**2 > 0:
list1.append(a**2-b**2)
list2=[]
for j in list1:
for c in range(2,50):
for d in range(2,50):
if c**3-d**3 == j:
list2.append(c**3-d**3)
print(list2)
This returns the following numbers which can all be written as both hollow squares and hollow cubes.
[56, 91, 19, 117, 189, 56, 208, 189, 217, 37, 279, 152, 117, 448, 513, 504, 448, 504, 387, 665, 504, 208, 875, 819, 936, 817, 61, 999, 988, 448, 728, 513, 189, 1216, 936, 784, 335, 469, 1323, 819, 1512, 1352, 1197, 992, 296, 152, 1519, 1512, 1197, 657, 1664, 1323, 1647, 1736, 1701, 1664, 936, 504, 2107, 1387, 1216, 1027, 91, 2015, 279, 2232]
Hollow squares, cubes and hypercubes
Taking this further, can we find any number which can be written as a hollow square, hollow cube and hollow hypercube (4 dimensional cube)? This would require our soldiers to be able to be stretch out into a 4th dimensional space – but let’s see if it’s theoretically possible.
Here’s the extra code to type:
list1=[]
for a in range(2, 200):
for b in range(2,200):
if a**2-b**2 !=0:
if a**2-b**2 > 0:
list1.append(a**2-b**2)
list2=[]
for j in list1:
for c in range(2,200):
for d in range(2,200):
if c**3-d**3 == j:
list2.append(c**3-d**3)
print(list2)
for k in list2:
for e in range(2,200):
for f in range(2,200):
if k == e**4-f**4:
print(k)
Very pleasingly this does indeed find some solutions:
9919: Which can be formed as either 1002-92 or 223-93 or 104-34.
14625: Which can be formed as either 1212-42 or 253-103 or 114-24.
Given that these took some time to find, I think it’ll require a lot of computer power (or a better designed code) to find any number which is a hollow square, hollow cube, hollow hypercube and hollow 5-dimensional cube, but I would expect that there is a number out there that satisfies all criteria. Maybe you can find it?
IB Maths Resources from Intermathematics 
