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Sphere packing problem: Pyramid design

Sphere packing problems are a maths problems which have been considered over many centuries – they concern the optimal way of packing spheres so that the wasted space is minimised.  You can achieve an average packing density of around 74% when you stack many spheres together, but today I want to explore the packing density of 4 spheres (pictured above) enclosed in a pyramid.

Considering 2 dimensions

 

First I’m going to consider the 2D cross section of the base 3 spheres.  Each sphere will have a radius of 1.  I will choose A so that it is at the origin.  Using some basic Pythagoras this will give the following coordinates:

Finding the centre

 

Next I will stack my single sphere on top of these 3, with the centre of this sphere directly in the middle.  Therefore I need to find the coordinate of D.  I can use the fact that ABC is an equilateral triangle and so:

3D coordinates

Next I can convert my 2D coordinates into 3D coordinates.  I define the centre of the 3 base circles to have 0 height, therefore I can add z coordinates of 0.  E will be the coordinate point with the same x and y coordinates as D, but with a height, a, which I don’t yet know:

In order to find I do a quick sketch, seen below:

Here I can see that I can find the length AD using trig, and then the height DE (which is my a value) using Pythagoras:

Drawing spheres

The general equation for spheres with centre coordinate (a,b,c) and radius 1 is:

Therefore the equation of my spheres are:

Plotting these on Geogebra gives:

 

 

Drawing a pyramid

Next I want to try to draw a pyramid such that it encloses the spheres.  This is quite difficult to do algebraically – so I’ll use some technology and a bit of trial and error.

First I look at creating a base for my pyramid.  I’ll try and construct an equilateral triangle which is a tangent to the spheres:

This gives me an equilateral triangle with lengths 5.54. I can then find the coordinate points of F,G,H and plot them in 3D.  I’ll choose point E so that it remains in the middle of the shape, and also has a height of 5.54 from the base. This gives the following:

As we can see, this pyramid does not enclose the spheres fully.  So, let’s try again, this time making the base a little bit larger than the 3 spheres:

 

 

This gives me an equilateral triangle with lengths 6.6.  Taking the height of the pyramid to also be 6.6 gives the following shape:

 

This time we can see that it fully encloses the spheres.  So, let’s find the density of this packing.  We have:

Therefore this gives:

and we also have:

Therefore the density of our packaging is:

Given our diagram this looks about right – we are only filling less than half of the available volume with our spheres.

Comparison with real data

 

[Source: Minimizing the object dimensions in circle and sphere packing problems]

We can see that this task has been attempted before using computational power – the table above shows the average density for a variety of 2D and 3D shapes.  The pyramid here was found to have a density of 46% – so our result of 44% looks pretty close to what we should be able to achieve.  We could tweak our measurements to see if we could improve this density.

So, a nice mixture of geometry, graphical software, and trial and error gives us a nice result.  You could explore the densities for other 2D and 3D shapes and see how close you get to the results in the table.

Screen Shot 2020-04-08 at 9.56.08 AM

Projectile Motion III: Varying gravity

We can also do some interesting things with projectile motion if we vary the gravitational pull when we look at projectile motion.  The following graphs are all plotted in parametric form.

Screen Shot 2020-04-06 at 1.17.08 PM

Here t is the parameter, v is the initial velocity which we will keep constant, theta is the angle of launch which we will vary, and g is the gravitational constant which on Earth we will take as 9.81 m/s2.

Earth 

Say we take a projectile and launch it with a velocity of 10 m/s.  When we vary the angle of launch we get the folowing graphs:

Screen Shot 2020-04-08 at 9.56.08 AM

On the y axis we have the vertical height, and on the x axis the horizontal distance.  Therefore we can see that the maximum height that we achieve is around 5m and the maximum horizontal distance is around 10m.

Other planets and universal objects

We have the following values for the gravitational pull of various objects:

Enceladus (Moon of Saturn): 0.111 m/s2, The Moon: 1.62 m/s2,  Jupiter: 24.8 m/s2, The Sun: 274 m/s2, White dwarf black hole surface gravity: 7×1012m/s2.

So for each one let’s see what would happen if we launched a projectile with a velocity of 10 m/s.  Note that the mass of the projectile is not relevant (though it would require more force to achieve the required velocity).

Enceladus:

Screen Shot 2020-04-08 at 10.24.18 AM

The Moon:

Screen Shot 2020-04-08 at 10.25.23 AM

Jupiter:

Screen Shot 2020-04-08 at 10.26.10 AM

The Sun:

Screen Shot 2020-04-08 at 10.26.56 AM

Black hole surface gravity:

This causes some issues graphically!  I’ll use the equations derived in the last post to find the coordinates of the maximum point for a given launch angle theta:

Screen Shot 2020-04-08 at 10.35.01 AM

Here we have v = 10 and g = 7×1012m/s2.  For example if we take our launch angle (theta) as 45 degrees this will give the coordinates of the maximum point at:

(7.14×10-12, 3.57×10-12).

Summary:

We can see how dramatically life would be on each surface!  Whilst on Earth you may be able to throw to a height of around 5m with a launch velocity of 10 m/s., Enceladus  would see you achieve an incredible 450m.  If you were on the surface of the Sun then probably the least of your worries would be how hight to throw an object, nevertheless you’d be struggling to throw it 20cm high.  And as for the gravity at the surface of a black hole you wouldn’t get anywhere close to throwing it a nanometer high (1 billionth of a meter).

 

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Projectile Motion Investigation II

Another example for investigating projectile motion has been provided by fellow IB teacher Ferenc Beleznay.  Here we fix the velocity and then vary the angle, then to plot the maximum points of the parabolas.  He has created a Geogebra app to show this (shown above).  The locus of these maximum points then form an ellipse.

Screen Shot 2020-04-06 at 8.34.58 PM

We can see that the maximum points of the projectiles all go through the dotted elliptical line.  So let’s see if we can derive this equation.

Let’s start with the equations for projectile motion, usually given in parametric form:

Screen Shot 2020-04-06 at 1.17.08 PM

Here v is the initial velocity which we will keep constant, theta is the angle of launch which we will vary, and g is the gravitational constant which we will take as 9.81.

We can plot these curves parametrically, and for each given value of theta (the angle of launch) we will create a projectile motion graph.  If we plot lots of these graphs for different thetas together we get something like this:

Screen Shot 2020-04-06 at 9.36.30 PM

We now want to see if the maximum points are in any sort of pattern.  In order to find the maximum point we want to find when the gradient of dy/dx is 0.  It’s going to be easier to keep things in parametric form, and use partial differentiation.  We have:

Screen Shot 2020-04-06 at 9.12.58 PM

Therefore we find the partial differentiation of both x and y with respect to t.  (This simply means we can pretend theta is a constant).

Screen Shot 2020-04-06 at 9.13.43 PM

We can then say that:

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We then find when this has a gradient of 0:

Screen Shot 2020-04-06 at 9.16.25 PM

We can then substitute this value of t back into the original parametric equations for x:

Screen Shot 2020-04-06 at 9.17.50 PM

and also for y:

Screen Shot 2020-04-06 at 9.18.30 PM

We now have the parametric equations in terms of theta for the locus of points of the maximum points.  For example, with g= 9.81 and v =1 we have the following parametric equations:

Screen Shot 2020-04-06 at 9.21.37 PM

This generates an ellipse (dotted line), which shows the maximum points generated by the parametric equations below (as we vary theta):

Screen Shot 2020-04-06 at 9.26.39 PM

And here is the graph:

Screen Shot 2020-04-06 at 8.34.58 PM

We can vary the velocity to create a new ellipse.  For example the ellipse generated when v =4 creates the following graph:

Screen Shot 2020-04-06 at 9.34.04 PM

So, there we go, we have shown that different ellipses will be created by different velocities.  If you feel like a challenge, see if you can algebraically manipulate the parametric equations for the ellipse into the Cartesian form!

Classical Geometry Puzzle: Finding the Radius

This is another look at a puzzle from Mind Your Decisions.  The problem is to find the radius of the following circle:

We are told that line AD and BC are perpendicular and the lengths of some parts of chords, but not much more!  First I’ll look at my attempt to solve this.  It’s not quite as “nice” as the solution in the video as it requires the use of a calculator, but it still does the job.

Method 1, extra construction lines:

These are the extra construction lines required to solve this problem.  Here is the step by step thought process:

  1. Find the hypotenuse of triangle AGC.
  2. Use the circle theorem angles in the same segment are equal to show that angle CBD = angle CAG.
  3. Therefore triangle AGC and GBD are similar, so length BG = 4.  We can now use Pythagoras to find length BD.
  4. We can find length CD by Pythagoras.
  5. Now we have 3 sides of a triangle, CDB.  This allows use to find angle BDC using the cosine rule.
  6. Now we the circle theorem angles in the same segment are equal to show that angle BDC = angle BEC.
  7. Now we use the circle theorem angles in a semi circle are 90 degrees to show ECB = 90.
  8. Now we have a right angled triangle BCE where we know both an angle and a side, so can use trigonometry to find the length of BE.
  9. Therefore the radius is approximately 4.03.

Method 2, creating a coordinate system

This is a really beautiful solution – which does not require a calculator (and which is discussed in the video above).  We start by creating a coordinate system based around point G at (0,0).  Because we have perpendicular lines we can therefore create coordinates for A, B and C.  We also mark the centre of the circle as (p,q).

First we start with the equation of a circle centre (p.q):

Next we create 3 equations by substituting in our coordinates:


Next we can do equation (3) – equation (1) to give:

Next we can substitute this value for p into equations (1) and (3) and equate to get:

Lastly we can substitute both values for p and q into equation (1) to find r:

We get the same answer as before – though this definitely feels like a “cleaner” solution.  There are other ways to solve this – but some of these require the use of equations you may not already know (such as the law of sines in a circumcircle, or the equation for perpendicular chords and radius).  Perhaps explore any other methods for solving this – what are the relative merits of each approach?

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?

Ramanujan’s Taxi Cabs and the Sum of 2 Cubes

The Indian mathematician Ramanujan (picture cite: Wikipedia) is renowned as one of great self-taught mathematical prodigies.  His correspondence with the renowned mathematician G. H Hardy led him to being invited to study in England, though whilst there he fell sick.  Visiting him in hospital, Hardy remarked that the taxi that had brought him to the hospital had a very “rather dull number” – number 1729.  Ramanujan remarked in reply, ” No Hardy, it’s a very interesting number!  It’s the smallest number expressible as the sum of 2 cubes in 2 different ways!”

Ramanujan was profoundly interested in number theory – the study of integers and patterns inherent within them.  The general problem referenced above is finding integer solutions to the below equation for given values of A:

In the case that A = 1729, we have 2 possible ways of finding distinct integer solutions:

The smallest number which can be formed through 3 distinct (positive) integer solutions to the equation is A = 87, 539, 319.

Although this began as a number theory problem it has close links with both graphs and group theory – and it is from these fields that mathematicians have gained a deeper understanding as to the nature of its solutions.  The modern field of elliptical curve cryptography is closely related to the ideas below and provides a very secure method of encrypting data.

We start by sketching the graph of:

For some given integer value of A. We will notice that the graph has a line of symmetry around y = x and also an asymptote at y = -x.  If we plot:

We can see that both our integer solutions to this problem (1,12) and (9,10) lie on the curve:

Group theory

Groups can be considered as sets which follow a set number of rules with regards to operations like multiplication, addition etc.  Establishing that a set is a group then allows certain properties to be inferred.  If we can establish the following rules hold then we can create an Abelian group.  If we start with a set A and and operation Θ.

1) Identity. For an element e in A, we have a Θ e = a for all a in A.

(for example 0 is the identity element for the addition operation for the set of integers numbers. a+0 = a for all a in the real numbers).

2) Closure.  For all elements a,b in A, a Θ b = c, where c is also in A.

(For example with the addition operation, the addition of 2 integers numbers is still an integer)

3) Associativity. For all elements a,b,c in A, (a Θ b) Θ c = a Θ (b Θ c)

(For example with the addition operation, (1+2) + 3 = 1 + (2+3) )

4) Inverse.  For each a in A there exists a b in A such that a Θ b = b Θ a = e.  Where e is the identity.

(For example with the addition operation, 4+-4 = -4+4 = 0.  0 is the identity element for addition)

5) Commutativity. For all elements a,b in A, a Θ b = b Θ a

(For example with the addition operation 1+2 = 2+1).

As we have seen, the set of integers under the operation addition forms an abelian group.

Establishing a group

So, let’s see if we can establish a Abelian group based around the rational coordinates on our graph.  We can demonstrate with the graph:

We then take 2 coordinate points with rational coordinates (i.e coordinates that can be written as a fraction of integers).  In this case A (1,12) and B (9,10).

We then draw the line through A and B.  This will intersect the graph in a 3rd point, C (except in a special case to be looked at in a minute).

We then reflect this new point C in the line y = x, giving us C’.

In this case C’ is the point  (46/3, -37/3)

We therefore define addition (our operation Θ) in this group as:

A + B = C’.

(1,12) + (9,10) = (46/3, -37/3).

We now need to deal with the special case when a line joining 2 points on the curve does not intersect the curve again.  This will happen whenever the gradient of this line is -1, which will make it parallel to the graph’s asymptote y = -x.

In this case we affix a special point at infinity to the Cartesian (x,y) plane.  We define this point as the point through which all lines with gradient -1 intersect.  Therefore in our expanded geometry, the line through AB will intersect the curve at this point at infinity. Let’s call our special point  Φ.  Now we have a new geometry,  the (x,y) plane affixed with Φ.

We can now create an Abelian group.  For any 2 rational points P(x,y), Q(x,y) we will have:

1) Identity.  P + Φ = Φ + P = P

2) Closure.  P + Q = R. (Where R(x,y) is also a rational point on the curve)

3) Associativity. (P+Q) + R = P+(Q+R)

4) Inverse.  P + (-P) = Φ

5) Commutativity.  P+Q = Q+P

Understanding the identity

Let’s see if we can understand some of these.  For the identity, if we have a point A on the line and the point at infinity then this will contain the line with gradient -1.  Therefore the line between the point at infinity and A will intersect the curve again at B.  Our new point, B’ will be created by reflecting this point in the line y = x.  This gets us back to point A.  Therefore P + Φ = P as required.

Understanding the inverse

With the inverse of our point P(x,y) given as -P = (-x,-y) we can see that this is the reflection in the line y = x.  We can see that we we join up the 2 points reflected in the line y = x we will have a line with slope -1, which will intersect with the curve at our point at infinity.  Therefore P + (-P) = Φ.

Through our graphical understanding the commutativity rule also follows immediately, It doesn’t matter which of the 2 points come first when we draw a line that connects them, therefore P+Q = Q+P.

Understanding associativity and closure

Neither associativity nor closure are obvious from our graph.  We could check individual points to show that (P+Q) + R = P+(Q+R), but it would be harder to explain why this always held.  Equally whilst it’s clear that P+Q will always create a point on the curve it’s not obvious that this will be a rational point.

In fact we do have both associativity and closure for our group as we have the following algebraic definition for our addition operation:

For a given point:

On a given curve of the form:

The addition of 2 points is given by:

In the case of our curve:

If we take P = (1,12).  P + P will be given by:

We can check this result graphically.  If P and Q are the same point, then the line that passes through both P and Q has to be the tangent to the curve at that point.  Therefore we would have:

Here the tangent at A does indeed meet the curve again – at point C, which does reflect in y = x to give us the coordinates above.

We could also find this intersection point algebraically.  If we differentiate the original curve to find the gradient when x = 1 we can find the equation of the tangent when x=1 and then substitute this back into the equation of the curve to find the intersection point.  This would give us:

We would then reverse the x and y coordinates to reflect in the line y = x.  This also gives us the same coordinates.

More generally if we have the 2 rational coordinates on the curve:

We have the algebraic formula for addition as:

 

If P = (1,12) and Q = (9,10), P + Q would give (after much tedious substitution!):

This agrees with the coordinates we found earlier using the much easier geometrical approach. As we can see from this formula, both coordinate points will always be rational – as they will be composed of combinations of our original rational coordinates.  For any given curve there will be a generator set of coordinates through which we can generate all other rational coordinates on the curve through our addition operation.

So, we seem to have come a long way from our original goal – finding integer solutions to an algebraic equation. Instead  we seem to have got sidetracked into studying graphs and establishing groups.  However by reinterpreting this problem as one in group theory then this then opens up many new mathematical techniques to help us understand the solutions to this problem.

A fuller introduction to this topic is the very readable, “Taxicabs and the Sum of Two Cubes” by Joseph Silverman (from which the 2 general equations were taken) .

The Shoelace Algorithm to find areas of polygons

This is a nice algorithm, formally known as Gauss’s Area formula, which allows you to work out the area of any polygon as long as you know the Cartesian coordinates of the vertices.  The case can be shown to work for all triangles, and then can be extended to all polygons by first splitting them into triangles and following the same approach.

Let’s see if we can work out the algorithm ourselves using the construction at the top of the page.  We want the area of the triangle (4), and we can see that this will be equivalent to the area of the rectangle minus the area of the 3 triangles (1) (2) (3).

Let’s start by adding some other coordinate points for the rectangle:

Therefore the area of the rectangle will be:

(1) + (2) +(3) +(4): (x3-x2)(y1-y3)

And the area of triangles will be:

(1): 0.5(x3-x2)(y2-y3)

(2): 0.5(x1-x2)(y1-y2)

(3): 0.5(x3-x1)(y1-y3)

Therefore the area of triangle (4) will be:

Area = (x3-x2)(y1-y3) – 0.5(x3-x2)(y2-y3) – 0.5(x1-x2)(y1-y2) – 0.5(x3-x1)(y1-y3)

Therefore we have our algorithm!  Let’s see if it works with the following coordinates added:

x1 = 2  x = 1  x = 3
y1 = 3 y = 2  y = 1

Area = (x3-x2)(y1-y3) – 0.5(x3-x2)(y2-y3) – 0.5(x1-x2)(y1-y2) – 0.5(x3-x1)(y1-y3)

Area = (3-1)(3-1) – 0.5(3-1)(2-1) – 0.5(2-1)(3-2) – 0.5(3-2)(3-1)

Area = 4 – 1 – 0.5 – 1 = 1.5 units squared

We could check this using Pythagoras to find all 3 sides of the triangle, followed by the Cosine rule to find an angle, followed by the Sine area of triangle formula, but let’s take an easier route and ask Wolfram Alpha (simply type “area of a triangle with coordinates (1,2) (2,3) (3,1)).  This does indeed confirm an area of 1.5 units squared.  Our algorithm works.  We can of course simplify the area formula by expanding brackets and simplifying.  If we were to do this we would get the commonly used version of the area formula for triangles.

The general case for finding areas of polygons

The general formula for the area of an n-sided polygon is given above.

For a triangle this gives:

For a quadrilateral this gives:

For a pentagon this gives:

You might notice a nice shoelace like pattern (hence the name) where x coordinates criss cross with the next y coordinate along.  To finish off let’s see if it works for an irregular pentagon.

If we arbitrarily assign our (x1, y1) as (1,1) and then (x2, y2) as (3,2), and continue in a clockwise direction we will get the following:

area = absolute of 0.5( 1×2 + 3×4 + 3×1 + 4×0 + 2×1 – 3×1 – 3×2 – 4×4 – 2×1 – 1×0)

area = 4.

Let’s check again with Wolfram Alpha – and yes it does indeed have an area of 4.

It could be a nice exploration task to take this further and to explore how many different methods there are to find the area of polygons – and compare their ease of use, level of mathematics required and aesthetic appeal.

Soap Bubbles and Catenoids

Soap bubbles form such that they create a shape with the minimum surface area for the given constraints.  For a fixed volume the minimum surface area is a sphere, which is why soap bubbles will form spheres where possible.  We can also investigate what happens when a soap film is formed between 2 parallel circular lines like in the picture below: [Credit Wikimedia Commons, Blinking spirit]


In this case the shape formed is a catenoid – which provides the minimum surface area (for a fixed volume) for a 3D shape connecting the two circles.  The catenoid can be defined in terms of parametric equations:

Where cosh() is the hyperbolic cosine function which can be defined as:

For our parametric equation, t and u are parameters which we vary, and c is a constant that we can change to create different catenoids.  We can use Geogebra to plot different catenoids.  Below is the code which will plot parametric curves when c =2 and t varies between -20pi and 20 pi.

 

We then need to create a slider for u, and turn on the trace button – and for every given value of u (between 0 and 2 pi) it will plot a curve.  When we trace through all the values of u it will create a 3D shape – our catenoid.

Individual curve (catenary)


Catenoid when c = 0.1

Catenoid when c = 0.5

Catenoid when c = 1

Catenoid when c = 2

Wormholes

For those of you who know your science fiction, the catenoids above may look similar to a wormhole.  That’s because the catenoid is a solution to the hypothesized mathematics of wormholes.  These can be thought of as a “bridge” either through curved space-time to another part of the universe (potentially therefore allowing for faster than light travel) or a bridge connecting 2 distinct universes.

Screen Shot 2019-09-29 at 7.03.34 PM

Above is the Morris-Thorne bridge wormhole [Credit The Image of a Wormhole].

Further exploration:

This is a topic with lots of interesting areas to explore – the individual curves (catenary) look similar to, but are distinct from parabola.  These curves appear in bridge building and in many other objects with free hanging cables.  Proving that catenoids form shapes with minimum surface areas requires some quite complicated undergraduate maths (variational calculus), but it would be interesting to explore some other features of catenoids or indeed to explore why the sphere is a minimum surface area for a given volume.

If you want to explore further you can generate your own Catenoids with the Geogebra animation I’ve made here.

Plotting the Mandelbrot Set

The video above gives a fantastic account of how we can use technology to generate the Mandelbrot Set – one of the most impressive mathematical structures you can imagine. The Mandelbrot Set can be thought of as an infinitely large picture – which contains fractal patterns no matter how far you enlarge it.  Below you can see a Mandelbrot zoom – which is equivalent to starting with a piece of A4 paper and enlarging it to the size of the universe!  Even at this magnification you would still see new patterns emerging.

The way the Mandelbrot set is formed in the first video is by using the following iterative process:

Zn+1 = Zn2 + c

Here Z is a complex number (of the form a + bi) and c is a constant that we choose.  We choose our initial Z value as 0.  Z1 = 0. We then choose a value of c (which is also a complex number) and see what happens when we follow the iterative process.

Let’s choose c = 2i +1. Z1 = 0

Zn+1 = Zn2 + 2i +1
Z2 = (0)2 + 2i +1
Z2 = 2i + 1

We then repeat this process:
Z3 = Z22 + 2i +1
Z3 = (2i+1)2 + 2i +1
Z3 = (2i)(2i) + 2i + 2i + 1 + 2i +1
Z3 = 6i-2 (as i.i = -1)

As we continue this process Zn spirals to infinity.

What we are looking for is whether this iterated Z value will diverge to infinity (i.e get larger and larger) or if it will remain bounded. If diverges to infinity we colour the initial point 2i+1 as blue on a complex axis. If it remains bounded we will colour it in black. In this case our initial point 2i+1 will diverge to infinity and so it will be coloured in blue.

So, let’s use Geogebra to see this is action.  The Geogrebra online program for this is here.

We choose a value for c.  Let’s say c = 0.23 + 0.42i.  Z1 = 0
Zn+1 = Zn2 + 0.23 + 0.42i.
Z2 = (0)2 + 0.23 + 0.42i.
Z2 = 0.23 + 0.42i.

Z3 = Z22 + 0.23 + 0.42i.
Z3 = (0.23 + 0.42i.)2 + 0.23 + 0.42i.
Z3 = 0.1065 + 0.6132i
Z4 = (0.1065 + 0.6132i)2 + 0.23 + 0.42i.
Z4 = -0.13467199 + 0.5506116i

We carry on with this iterative process and plot the points that we get each time.  We can see the (0.23, 0.42), (0.1065, 0.42) and (-0.13467199, 0.5506116) correspond to the first coordinates on the spiral after (0,0).  We can see that as this process continues we see a convergence to a point close to (0.05, 0.45).

Screen Shot 2019-04-25 at 3.09.17 PM

If we choose another starting value for c: c = 0.17 + 0.56i we get the following diagram:

Screen Shot 2019-04-25 at 3.09.33 PM

Again we have a stable spiral which spirals around a geometric shape and does not diverge to infinity.

If we choose another starting value for c: c = -0.25 + 0.64i we get the following diagram:

Screen Shot 2019-04-25 at 3.10.04 PM

If we choose another starting value for c: c = 0.11 + 0.59i we get the following diagram:

Screen Shot 2019-04-25 at 3.09.43 PM

However, If we choose another starting value for c: c = 0.3 + 0.68i we get the following diagram:

Screen Shot 2019-04-25 at 3.10.12 PM

This time we can see that the orbit of points does not converge, but instead it diverges to infinity.

We can then colour in each point – simply categorising whether the value of c leads to an orbit which diverges or remains bounded.  Black means it remains bounded, blue that it has escaped to infinity.  So, below we can see that when we do the iterative process with c = 0.39+ 0.63i our orbit will escape to infinity (as it is coloured blue)

Screen Shot 2019-04-25 at 3.13.45 PM

If we do this exercise in much finer detail we arrive at the following picture:

Screen Shot 2019-04-25 at 3.14.02 PM

This is the Mandelbrot Set – and will keep producing fractal patterns as you zoom in to infinity.

IB Revision

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

Screen Shot 2018-03-19 at 4.42.05 PM.pngThe 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. 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!

Screen Shot 2019-07-27 at 10.02.40 AM

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 and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

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.
 

 

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.

Screen Shot 2018-05-26 at 7.42.08 PM

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

Screen Shot 2018-05-26 at 7.51.54 PM

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

Circle 1:

Screen Shot 2018-05-26 at 7.51.57 PM

has centre:

Screen Shot 2018-05-26 at 7.50.54 PM

and radius:

Screen Shot 2018-05-26 at 7.50.58 PM

Circle 2:

Screen Shot 2018-05-26 at 7.53.22 PM

has centre:

Screen Shot 2018-05-26 at 7.53.28 PM

and radius:

Screen Shot 2018-05-26 at 7.53.31 PM

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:

Screen Shot 2018-05-26 at 4.29.09 PM

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

Screen Shot 2018-05-26 at 8.02.27 PM

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:

Screen Shot 2018-05-26 at 8.10.26 PM

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

Screen Shot 2018-05-26 at 8.15.35 PM

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

IB Revision

Screen Shot 2018-03-19 at 4.35.19 PM

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

Screen Shot 2018-03-19 at 4.42.05 PM.pngThe 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. 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!

Screen Shot 2019-07-27 at 10.02.40 AM

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 and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

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.

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IB Maths Exploration Guide

IB Maths Exploration Guide

A comprehensive 63 page pdf guide to help you get excellent marks on your maths investigation. Includes:

  1. Investigation essentials,
  2. Marking criteria guidance,
  3. 70 hand picked interesting topics
  4. Useful websites for use in the exploration,
  5. A student checklist for top marks
  6. Avoiding common student mistakes
  7. A selection of detailed exploration ideas
  8. Advice on using Geogebra, Desmos and Tracker.

Available to download here.

IB HL Paper 3 Practice Questions (120 page pdf)

IB HL Paper 3 Practice Questions 

Seventeen full investigation questions – each one designed to last around 1 hour, and totaling around 40 pages and 600 marks worth of content.  There is also a fully typed up mark scheme.  Together this is around 120 pages of content.

Available to download here.

IB Exploration Modelling and Statistics Guide


IB Exploration Modelling and Statistics Guide

A 60 page pdf guide full of advice to help with modelling and statistics explorations – focusing in on non-calculator methods in order to show good understanding. Includes:

  1. Pearson’s Product: Height and arm span
  2. How to calculate standard deviation by hand
  3. Binomial investigation: ESP powers
  4. Paired t tests and 2 sample t tests: Reaction times
  5. Chi Squared: Efficiency of vaccines
  6. Spearman’s rank: Taste preference of cola
  7. Linear regression and log linearization.
  8. Quadratic regression and cubic regression.
  9. Exponential and trigonometric regression.

Available to download here.

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