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IntMath Newsletter: Bubble sort, symbols, study skills, stress

By Murray Bourne, 21 Jun 2019

21 Jun 2019

In this Newsletter:

1. New on IntMath: Bubble sort rainbow
2. Resources: Symbols, study skills
3. Math in the news: Number sense
4. Math movies: Stress and learning
5. Math puzzle: Pencils
6. Final thought: No environment, no economy

1. New on IntMath: Bubble sort rainbow

Bubble sort rainbow

Bubble sorting is a simple sort technique that all coders learn about. Here are two visualizations of the bubble sort process, one involving numbers, and the other colors, in the Math Art Code series. See:

Bubble sort rainbow

2. Resources

(a) What math symbol is that?

One of the beauties of mathematics is a downside for new learners. Mathematicians love to be precise and concise, so they make use of a vast range of symbols, which can make things very confusing for the novice.

math symbols

In the image at left, there are 12 math symbols. Can you name each one, and give an example of its use?

Here are some places you can find out the meaning of all these symbols (and more), with examples of their use:

Basic Math Symbols (RapidTables)

List of mathematical symbols (Wikipedia)

Exhaustive List of Mathematical Symbols and Their Meaning (ScienceStruck)

Common Symbols in Mathematics: Maths Glossary (SkillsYouNeed)

You may also be interested in my discussion here:

Towards more meaningful math notation

(b) Study skills tips


Many students are ineffective and inefficient when it comes to study time. It gets worse as the number of distractions — mostly from devices — gets worse.

Here's a list of study tips I came across recently. We can't expect too many students to read and act on such a list (that's not realistic), but they could form a good starting point for a class activity.

25 Scientifically Proven Tips For More Effective Studying

Exercise: The title says the tips are "scientifically proven". Do you agree there's enough evidence in the article to conclude this?

3. Math in the news:

'Number sense' in humans and animals

4 dots

Number sense graph

Researchers from the University of Tübingen were studying how an artificial neural network classified objects from images, and they noticed a curious and unexpected outcome — the network began to develop a 'number sense'.

For example, while examining the image of 4 dots (top), the neural network had peak activity at the number 4 (bottom)

According to the original paper, Number detectors spontaneously emerge in a deep neural network designed for visual object recognition:

Although our model was merely trained to classify natural images in a task that was unrelated to numerosity, its spontaneously emergent numerosity-tuned units allowed reliable categorization of the number of items in dot displays. These findings suggest that the spontaneous emergence of the number sense is based on mechanisms inherent to the visual system.

For an accessible summary of the research, see: 'Number sense' arises from the recognition of visible objects

4. Math Movies

It's summer holidays for many of you now and your stress levels are probably lower now than the last time you took a math test.

It's a good time to think about how to address your stress, before the next semester starts. Here are 2 videos covering the issue, each from a different point of view.

(a) The surprising link between stress and memory

study and stress

Continuing on the study skills theme, it's important to be able to recognize why and when we get stressed over math and other tests — and what to do about it. This short TED-Ed video explores the issue.

See: The surprising link between stress and memory

(b) Why we choke under pressure and how to avoid it

Why we choke under pressure and how to avoid it

Cognitive scientist Sian Leah Beilock reveals what happens in your brain and body when you choke in stressful situations, sharing psychological tools that can help you perform at your best when it matters most.

See: Why we choke under pressure and how to avoid it

5. Math puzzles

The puzzle in the last IntMath Newsletter asked about averages in a bowling game.

Correct answers with sufficient reasons were submitted by Rebecca, Michael, Tom, Gerard, Nicola (who rightly questioned one of the assumptions of the question), Eamon, and Deborah.

New math puzzle: pencils


We have two cylindrical pencils where the ratio of the height to the diameter is the same for the first pencil as for the second. If the sum of their heights is 1, the sum of their surface areas is and the sum of their volumes is 2π, find all possibilities for the dimensions of each pencil.

You can leave your response here.

Request: If you present your answer with a full explanation of how you obtained it, then others can follow it and learn something from it.

6. Final thought - economies and ecosystems

There is understandable concern that jobs (mining, manufacturing, energy) will disappear if nations actually get serious about mitigating climate change.

In We Can't Have A Thriving Economy Without Thriving Ecosystems, David Festa makes the point that destroying ecosystems destroys jobs, and more importantly, destroys food sources.

It seems to me we all need to make our lifestyles more sustainable. That means some jobs will disappear, but it's possible that many more — ones better for the planet and our enjoyment of our planet — will be created.

Are we willing to lose a bit, or do we all lose the lot eventually?

Until next time, enjoy whatever you learn.

See the 5 Comments below.

5 Comments on “IntMath Newsletter: Bubble sort, symbols, study skills, stress”

  1. Eva says:

    Note for 2.b

    Dear Murray,

    I have found this website very useful when advising studnets about effective and proven stundy skills techniques:
    For a start they only talk about 6 stategies which are managable.
    I hope you find it useful too.


  2. Murray says:

    @Eva: Thanks for the resource!

  3. Georgios Varelas says:




    {2}\cdot\pi\cdot\frac{{{{\left({d}_{1}\right)}}^{2}}}{4}+{2}\cdot\pi\cdot\frac{{{{\left({d}_{2}\right)}}^{2}}}{4}+\pi\cdot{\left({d}_{1}\right)}\cdot{\left({h}_{1}\right)}+\pi\cdot{\left({d}_{1}\right)}\cdot{\left({h}_{1}\right)} ={8}\cdot\pi\Rightarrow{\left({m}^{2}+{1}\right)}\cdot{\left({d}_{1}\right)}\cdot{\left({\left({d}_{1}\right)}+{2}\cdot{\left({h}_{1}\right)}\right)}={16}{\left({B}\right)}



    {m}={2}-\sqrt{3}; {\left({h}_{1}\right)}=\frac{{{\left({1}+\sqrt{3}\right)}\cdot\sqrt{3}}}{6}, {\left({h}_{2}\right)}=\frac{{{\left(-{1}+\sqrt{3}\right)}\cdot\sqrt{3}}}{6}; {\left({d}_{1}\right)}=\frac{{{\left({1}+\sqrt{3}\right)}\cdot{2}\cdot\sqrt{3}}}{3};

    {\left({d}_{2}\right)}=\frac{{{\left(-{1}+\sqrt{3}\right)}\cdot{2}\cdot\sqrt{3}}}{3}; or:

    {m}={2}+\sqrt{3}; {\left({h}_{1}\right)}=\frac{{{\left(-{1}+\sqrt{3}\right)}\cdot\sqrt{3}}}{6}, {\left({h}_{2}\right)}={\left({1}+\sqrt{3}\right)}\cdot\frac{\sqrt{3}}{6};{\left({d}_{1}\right)}={\left(-{1}+\sqrt{3}\right)}\cdot{2}\cdot\frac{\sqrt{3}}{3}; {\left({d}_{2}\right)}={\left({1}+\sqrt{3}\right)}\cdot{2}\cdot\frac{\sqrt{3}}{3};

  4. Nicola says:

    The setup of the problem gives us
    \frac{{h}_{1}}{{d}_{1}}=\frac{{h}_{2}}{{d}_{2}}={c} where h is height and d is diameter of the cylinder while c is a constant parameter.
    We can also write it in terms of radius
    \frac{{h}_{1}}{2}{\left({r}_{1}\right)}=\frac{{h}_{2}}{2}{\left({r}_{2}\right)}={c} or \frac{{h}_{1}}{{r}_{1}}=\frac{{h}_{2}}{{r}_{2}}={2}{c}

    We know that the surface area of a cylinder is the area of top and bottom parts plus the area of the "folded" part: if we unfold it, it is like the area of a rectangle.
    So the formulas are {A}{1}={2}\cdot\pi\cdot{{\left({r}_{1}\right)}}^{2}+\cdot\pi\cdot{\left({r}_{1}\right)}\cdot{\left({h}_{1}\right)} and similarly for A2.
    By doing some algebra (grouping and substituting {\left({h}_{1}\right)}={2}\cdot{c}\cdot{\left({r}_{1}\right)}, we can rewrite the formula for Area {A}_{1}={2}\cdot\pi\cdot{{\left({r}_{1}\right)}}^{2}\cdot{\left({1}+{2}\cdot{c}\right)} and similarly for {A}_{2}.

    We then substitute in the given equation {A}{1}+{A}{2}={8}\pi and find that {{\left({r}_{1}\right)}}^{2}+{{\left({r}_{2}\right)}}^{2}=\frac{4}{{1}+{2}{c}}.
    Graphically this is the equation of a circle centred at (0,0) with perimeter depending on the value of c (which has to be c>-1/2).
    The bigger the c, the smaller the circle. Note that with {c}={0}, the radius is 2, and with {c}={1} it is \frac{2}{\sqrt{3}}.

    If we consider the other given equation {\left({h}_{1}\right)}+{\left({h}_{2}\right)}={1} and substitute the h calculated at the beginning we obtain {\left({r}_{1}\right)}+{\left({r}_{2}\right)}=\frac{1}{2}{c}

    Finally, we consider the volume formula: it is just the area of the top circle multiplied by the height of the cylinder. {V}_{1}=\pi\cdot{{\left({r}_{1}\right)}}^{2}\cdot{\left({h}_{1}\right)}. We then substitute {h}_{1} and get {V}{1}=\pi\cdot{2}\cdot{c}\cdot{{\left({r}_{1}\right)}}^{3} and similarly for {V}_{2}.
    We then substitute these into the third given equation {V}_{1}+{V}_{2}={2}\pi, and obtain {{\left({r}_{1}\right)}}^{3}+{{\left({r}_{2}\right)}}^{3}=\frac{1}{c}.

    We can now put together the last equation with the second one and obtain \frac{{{{\left({r}_{1}\right)}}^{3}+{{\left({r}_{2}\right)}}^{3}}}{{{\left({r}_{1}\right)}+{\left({r}_{2}\right)}}}={2}.
    Graphically this like an ellipse centred at {\left({0},{0}\right)} with major axis {y}={x} (in our case {\left({r}_{1}\right)}={\left({r}_{2}\right)}).

    Note that this ellipse no longer depends on c.

    All the possible values of {\left({r}_{1}\right)} and {\left({r}_{2}\right)} are the ones for which c is 0<=c<=1.
    When c=0 the ellipse is inscribed in the circle, touching it at the two extreme points on the major axis; whilst when c=1 the circle is inscribed in the ellipse, touching it on the two extreme points along the minor axis {y}=-{x}{\left({\left({r}_{2}\right)}=-{\left({r}_{1}\right)}\right.} in our case).

    Since we want to find all the possible pairs of {\left({r}_{1}\right)}, {\left({r}_{2}\right.}) in R, we need to find for what values of the parameter c the two equations (graphs) have a solution with the same coordinates {\left({r}_{1},{r}_{2}\right)}.
    This happens for 0<=c<=1. In fact, when c is smaller than zero the circle becomes bigger and no longer touches the ellipse (disappearing after c<0.5 because the right part of the equation becomes negative while the left is a sum of squares). When c is bigger the circle becomes smaller than the ellipse, without touching it at any point.

  5. Nicola says:

    I wanted to expand a little bit on my initial solution.

    I determined that the equation \frac{{{{\left({r}_{1}\right)}}^{3}+{{\left({r}_{2}\right)}}^{3}}}{{r}_{1}+{r}_{2}} is an ellipse by plotting its graph (with desmos by the way).
    It can be shown more algebraically by factoring the numerator as follows {\left({{\left({r}_{1}\right)}}^{3}+{{\left({r}_{2}\right)}}^{3}\right)}={\left({\left({r}_{1}\right)}+{\left({r}_{2}\right)}\right)}{\left({{\left({r}_{1}\right)}}^{2}-{\left({r}_{1}\right)}{\left({r}_{2}\right)}+{{\left({r}_{2}\right)}}^{2}\right)}; the part (r_1)+(r_2) simplifies with the denominator and we are left with {\left({r}{1}^{2}-{r}{1}{\left({r}_{2}\right)}+{{\left({r}_{2}\right)}}^{2}\right)}={2} which is the equation of an ellipse inclined by 45 degrees.

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