Hello! My name is Colin and I am a mathematician. Welcome to issue 0 of Double Maths First Thing, in which I highlight some of the mathematical things that have caught my eye this week.

First up, a nod to physicists Arnab Priya Saha and Aninda Sinha for doing something with no real application: they “accidentally discovered a new formula for pi”. There’s a bit about it in Scientific American, a Numberphile video, and a paper in Physical Review Letters (open access). I’ve not worked through it in detail, but it’s got a Pochhammer symbol in it, so it must be good.

I promise this isn’t *always* going to be about pi, but I also stumbled on a proof that pi is irrational — again, I’ve not worked through the details, but it looks like it would be accessible to a good A-level class with a bit of hand-holding.

Via reddit, a surprisingly tricky problem with a lovely twist in the tail: show that \( 3^k + 5^k = n^3 \) has no solutions for \( k > 1 \). (There’s a hint and a spoiler over on mathstodon.)

I’ve recently been on holiday in Northern Ireland. We visited W5 in Belfast, which is a pretty cool science museum — lots of hands-on stuff, including a build-your-own Scalextric-style car, bottle rockets and a green-screen bit where you can present the news about the alien invasion. On the minus side… there are lots of missed opportunities for highlighting the maths that underpins it all. Still, it’s a fun half-day if you’re all Titanic-ed out.

In the proper news, the Guardian had a long read about Field’s Medallist Alexander Grothendieck; although it too is a bit maths-light, it’s understandable given quite how heavy Grothendieck’s maths is. Katie Steckles also pointed me at the devastating news that UK railcard discounts are dropping from 34% to 33.4%, which strikes me as the sort of thing that probably costs more to implement than it could possibly save the train operators.

If you’re in the market for more maths, I can heartily recommend both the Finite Group, whose next livestream is on Friday September 13th, and Big MathsJam, which is a gathering of amazing geeks the first weekend of November. Early-bird tickets are (just about) still available; I have mine already. There’s also a day of recreational maths lectures in memory of David Singmaster on Saturday September 21st in London or online, and a New Scientist event about how maths explains the world the following Saturday, also in London.

That’s all for this week! If there’s something I should know about, you can find me on Mathstodon as @icecolbeveridge, or at my personal website.

Until next time,

C

]]>The Royal Society has announced its award winners for 2024, which include mathematicians Ingrid Daubechies (Bakerian Medal/lecture for the physical sciences), Hannah Fry (David Attenborough Award/lecture for public engagement) and Philip Maini (Sylvester medal for mathematical research).

And the joint IMA/LMS Christopher Zeeman Medal for 2024 has been awarded to Brady Haran for his work in communicating mathematics via the Numberphile channel on YouTube.

The Protect Pure Maths campaign group has relaunched as the Campaign for Mathematical Sciences, encompassing a broader remit to promote and support mathematical activity in the UK.

In research news, an elliptic curve with rank at least 29 has been found by researchers Noam Elkies and Zev Klagsbrun. The previous record was rank ≥28, found by Elkies in 2006. *(via Robin Houston)*

\[y^{2} + xy = x^{3} – 27006183241630922218434652145297453784768054621836357954737385x + 55258058551342376475736699591118191821521067032535079608372404779149413277716173425636721497\]

Pierre Cartier, mathematician and Bourbaki member, has died aged 92.

]]>We spoke to Jon Chase, aka Oort Kuiper the Science Rapper, about his TikTok channel and how he’s been using it to share mathematical raps.

**Channel** **title**: Science Rapper (Sci Comm)**Link**: tiktok.com/@science.rapper.sc**Topics covered:** KS4 maths raps, plus more general STEM**Average video length**: 45-75 seconds**Recommended videos**: Factorising to solve a quadratic, 8 Circle theorems

It mostly started as a way to share some of the stuff I have done in STEM, to get an idea of what worked on TikTok. After trying out some other avenues on TikTok, I soon discovered that it was maths raps that gained the most interest, and so I decided to focus on that.

I’m a science communicator that makes raps to communicate STEM ideas and information. The focus on maths is predominantly targeted towards education-based TikTok audiences, as this platform seems to be where those audiences are engaging with the content most. Other than maths on TikTok, I do plenty of other science communication ranging from stage shows, workshops and presenting on screen, to writing articles – as well as a few books.

The videos are just short raps about a particular topic in maths – either overviewing it, or providing rapped worked examples. It’s quick and easy and catchy (depending on your preferences, of course). It’s a very unique way to teach/share maths.

It was exciting to see my factorising video going viral! It’s now got over 400,000 likes and over 3,000 comments – and it’s always fun watching the comments unfold: essentially, the audience educates each other through their comments, as well as arguing about methods.

I’m hoping to cover more content from the GCSE curriculum, to provide greater access to engaging and quick maths revision!

]]>If you do some of these, you might spot what’s funny about them. For example.

\[ \Large \begin{bmatrix}

\color{navy}{4} & \color{navy}{8}\\

\color{navy}{2} & \color{navy}{3}

\end{bmatrix} \begin{bmatrix}

\color{cyan}{8} & \color{cyan}{8}\\

\color{cyan}{2} & \color{cyan}{7}

\end{bmatrix} = \begin{bmatrix}

\color{navy}{4}\color{cyan}{8} & \color{navy}{8}\color{cyan}{8}\\

\color{navy}{2}\color{cyan}{2} & \color{navy}{3}\color{cyan}{7}

\end{bmatrix} \]

That is, the answer to each question can be made by treating the element in the first matrix as the first digit and the corresponding element in the second matrix as the second digit in the answer element. This is not how matrix multiplication works, and ought to be funny if I hadn’t totally over-explained the joke!

I saw one of these in a meme that Katie posted in the Finite Group chat and it got me thinking about how these work.

If we set up the matrices like this

\[ \begin{bmatrix}

a & b\\

c & d

\end{bmatrix} \begin{bmatrix}

e & f\\

g & h

\end{bmatrix} = \begin{bmatrix}

10a+e & 10b+f\\

10c+g & 10d+h

\end{bmatrix} \]

Then we establish four equations with eight unknowns.

\[ \begin{align*}

ae + bg &= 10a+e\\

af+bh &= 10b+f\\

ce+dg &= 10c+g\\

cf+dh &= 10d+h

\end{align*}\]

Since there are more unknowns than equations, these don’t have a single solution. What I wanted was to find integer solutions with all values single-digits. I wrote some Python code to find these. I removed some that look overly symmetrical – either the rows of the matrix are identical, or the same matrix is repeated. This left 73 items.

From these 73 items, I wrote a second Python script that picks 20 of them at random and builds these into a LaTeX worksheet. For the Mastodon post I reformatted this into the shape and size that I thought would display better on social media, and added in one of the squared matrices for an extra hint something weird is up, hoping people might notice this isn’t just a boring post about matrix multiplication practice!

You can view these scripts and associated files on GitHub.

]]>In computation news, the fifth Busy Beaver number has been found. This Quanta article gives a good writeup. *(via TheHigherGeometer)*

In “are we nearly there yet” news, 202 trillion digits of pi have been calculated, breaking the previous record. The computation used the Chudnovsky algorithm (pictured below) and took around 100 days to crunch. *(via Robin Houston)*

A new James Maynard paper potentially rules out some exceptions to Riemann Hypothesis, making a big step forward in understanding the structure of the prime numbers.

And finally, the Antikythera mechanism has been theorised to be connected to the lunar rather than the solar calendar, based on (possibly sketchy) gravitational wave research. Dubious!

Sarah Hart’s excellent maths/literature book Once Upon a Prime has been awarded the 2024 Euler Book Prize, which is awarded annually to authors of “exceptional mathematics books that significantly impact public perception and understanding of mathematics”.

In case there weren’t enough books about maths and literature (we’re looking at you, Rob Eastaway) the incredible Ben Orlin of Math With Bad Drawings has announced his next book will also touch on the overlap of words and numbers, and will be out on 3rd September.

And earlier this month, Tim Harford gave the inaugural Vicky Neale Public Lecture at Oxford University. The lecture recognises the invaluable contribution to mathematical education of the late Vicky Neale, and Tim spoke about “how data built the modern world – and how we can use it to build a better one”. The lecture can be re-watched on the Oxford Mathematics YouTube Channel.

]]>The Carnival rounds up maths blog posts from all over the internet, including some from our own Aperiodical. See our Carnival of Mathematics page for more information.

]]>If you don’t know about the OEIS, then congratulations! You’re one of today’s lucky 10,000. Except for possibly Wikipedia, the OEIS is probably the most important and useful mathematical community resource on the internet. The main use case is, you’re doing some maths and you find a sequence. You wonder whether it’s something new or interesting. So you type it in to the OEIS search bar, and if it exists, you’re told what it is, whether there are formulas for it, different suggestions of where it crops up, lists of terms… it’s like a Who’s Who of number sequences.

There’s seemingly no limit to what it contains. For example, it contains the decimal form of the n-th color mentioned in the song “I Can Sing a Rainbow”, which I would call borderline “not maths at all”. At the other end of the spectrum (ahahaha), it also contains the all 1s sequence, which is borderline “not a sequence at all”. The idea seems to be “if it’s a reasonable sequence someone has ever had cause to think about, it belongs here”.

Now, I’m very fond of saying I’m a mathematician. I have two degrees, several academic papers, a number of published books, an equation named after me and no proper job. But at the same time, my impostor syndrome insists that a *proper* mathematician would have at least *something* in the OEIS.

I am now doubting whether I really qualify as an impostor.

This all started — as most of my maths seems to these days — from a question posed by my 10-year-old, Bill.

“Does Pascal’s triangle go

upthe way, too?”

If you’re not familiar with Pascal’s triangle, then congratulations! Etc. You might want to visit Matt Enlow’s piece on it in the Math-off final.

Meanwhile, here are the first few rows:

\[ \begin{array}{ccc ccc ccc ccc ccc}

&&&&&&& 1\\

&&&&&& 1 && 1 \\

&&&&& 1 && 2 && 1 \\

&&&& 1 && 3 && 3 && 1 \\

&&& 1 && 4 && 6 && 4 && 1 \\

&& 1 && 5 && 10 && 10 && 5 && 1 \\

&1 && 6 && 15 && 20 && 15 && 6 && 1 \\

1 && 7 && 21 && 35 && 35 && 21 && 7 && 1 \\

\end{array} \]

Translating that into how boring grown-up mathematicians speak, he was asking whether you can add rows *above* the 0th row — what would the -1st row look like? The -5th?

We had a good chat about this — how we could change the rules slightly to make it work, but remain consistent, what patterns he could find, and eventually he wandered off with a pad of paper to see what he could fill in for himself.

I *was* going to ask him about what would happen with fractions, before I nerdsniped myself asking “what would happen with complex powers? What’s the \( i\)th row of Pascal’s Triangle?”

There’s a trick for generating the \( n \)th row of Pascal’s triangle:

- Let your current entry be 1 and your fraction be \( \frac{n}{1} \).
- Write down your current entry.
- Multiply it by your fraction to get a new current entry.
- Subtract one from the top of your fraction and add one to its bottom.
- Go back to step 2 until either your entry becomes 0 or you get bored (or both).

So, for example, the fourth row would be:

- Write down 1, and multiply by \( \frac{4}{1} \).
- Write down 4, and multiply by \( \frac{3}{2} \).
- Write down 6, and multiply by \( \frac{2}{3} \).
- Then 4, 1, and 0 — so we stop.

We get the numbers 1, 4, 6, 4, 1, as we would have hoped.

This also works for non-natural numbers. We could find the -3rd row to answer Bill’s question:

- Write down 1, and multiply by \( \frac{-3}{1} \).
- Write down -3, and multiply by \( \frac{-4}{2} \).
- Write down 6, and multiply by \( \frac{-5}{3} \).
- Then -10, 15, -21, and we get bored so we stop.

We get the numbers 1, -3, 6, -10, 15, -21 and can hypothesise or prove that these are the triangular numbers with alternating signs.

Well, it works just the same way, only with slightly trickier calculations.

- Write down 1, and multiply by \( \frac{i}{1} \)
- Write down ( i ), and multiply by \( \frac{i-1}{2} \)
- Write down \( \frac{-1 – i}{2} \), and multiply by \( \frac{i-2}{3} \)
- Honestly, I’m bored already. We have machines for this sort of thing.

It’s also a lot easier if we just ignore the bottom of the fraction — it’s trivial to say “we need to divide the \( k \)th term by \( k!\),” and it doesn’t affect the working out at all.

And if we do that, we get a Gaussian integer sequence:

\[ 1, i, (-1-i), (3+i), -10, (40-10i), (-190+90i), \dots \]

Now, the OEIS, like any good hip-hop artist, prefers to keep it real. You are as likely to find Gaussian integers in the OEIS as in the work of Grandmaster Flash (who is surprisingly mediocre at chess). However, I reasoned, the real and imaginary parts might be there separately.

And boom, there they were:

The OEIS is useful not just as a research tool, but as an educational tool: it’s forever throwing up unexpected links and ideas and threads to pull at. Here, for example, it raises the question “what on earth is an e.g.f.?”

Luckily, Wikipedia is also a useful research and educational tool: the e.g.f. is the exponential generating function. Let’s start with a sidetrack into (common or garden) generating functions, which are the sort of dark magic it’s worth messing about with.

A generating function is a way of turning a sequence of numbers into a function, the better to understand its properties. For exampe, if you take the Fibonacci sequence, with terms 0, 1, 1, 2, 3, 5, …, the associated generating function is

\[ g(x) = 0 + 1x + 1x^2 + 2x^3 + 3x^4 + 5x^5 + \dots \]

Generating functions make certain manipulations very simple: for example, you can “move the sequence along” if you multiply by a power of \( x\). As a result, you can use the definition of the Fibonacci sequence to find that \( x^2 g(x) = x g(x) + g(x) \) — each term is the sum of the previous two — and rearrange to find that \( g(x) = \frac{1}{x^2 – x – 1} \). As a result, you can say “Let’s put \( x = 100 \) into that” and find that \( \frac{1}{9,899} = 0.00 01 01 02 03 05 08 13 \dots \) . (After a while, the carries get in the way and break the sequence.) There are all sorts of cool applications in number theory and probability.

In particular, the binomial expansion of \( (1 + x)^n \) is the generating function for the \( n \)th row of Pascal’s triangle — that is, the coefficient of \( x^k \) in the generating function of the \( n \)th row is \( nCr(n, k) \).

So what about the exponential generating function?

Rather than taking the coefficients from \( a_0 + a_1 x + a_2 x^2 + \dots \), the e.g.f. takes its coefficients from \( b_0 + b_1 x + b_2 \frac{x^2}{2!} + b_3 \frac{x^3}{3!} + \dots \). It’s the same idea, but the manipulations are slightly different (you shift by differentiating or integrating, for example.)

When we’re looking at the \( k \)th element of a row in Pascal’s triangle, it’s the same as multiplying the coefficient by \( k! \) — or rather, you get \( nPr(n, k) \) .

So: what I worked out above, the Gaussian integer sequence, is the e.g.f. of \( (1 + x)^i \).

It wasn’t immediately obvious to me how that was linked to the OEIS sequence, until I took a few minutes to work it through.

If we combined the sequences to make the Gaussian integers, we’d get \( \cos( \ln(1+x)) + i \sin(\ln(1+x))\). And that’s a familiar form: \( \cos(z) + i \sin(z) \equiv e^{iz} \).

So our sequences combine to make \( e^{i \ln(1+x)}) \), which works out to be \( (1+x)^i \). Lovely!

And lastly, the permutations observation from before means that each term can be written as \( nPr(i, k) \).

There’s a slight wrinkle, though: it’s not exactly reasonable to pick \( k \) items from a set containing \( i \) of them. Even the formula of \( nPr(n, k) = \frac{n!}{(n-k)!}) \) isn’t well-defined: you can’t *really* take the factorial of something that isn’t an integer.

However, there’s a handy generalisation of the factorial function to all complex numbers, the gamma function, defined such that \( \Gamma(n+1) \equiv n! \) on the natural numbers.

As a consequence, we can extend the permutation function in a similar way: on the natural numbers, \( nPr(n, k ) \equiv \frac{\Gamma(n+1)}{\Gamma(n+1-k)}) \), and (having checked with the Mathematical Authorities), we’re allowed to use that definition for all complex numbers (apart from the unnatural integers, where \( \Gamma(x) \) is undefined. But that’s for another day.)

Where were we? Oh, yes. Each term of our Gaussian integer sequence is \( nPr(i, k) \), which we can write as \( \frac{\Gamma(i+1)}{\Gamma(i+1-k)}) \); the two sequences correspond to the real and imaginary parts of this.

And *that* formula, or anything like it, was missing from the OEIS.

Now it isn’t, and I can finally cut up my impostor card and say with confidence: my name is Colin, and I am a mathematician.

]]>Over the past month, we’ve heard from 16 interesting mathematicians and whittled them down to just 2. Today, we’re pitting Matt Enlow against Angela Tabiri to determine **The World’s Most Interesting Mathematician ***(2024, of the people who I asked to take part and were available)*.

Take a look at both pitches, vote for the bit of maths that made you do the loudest “Aha!”, and if *you* know any more cool facts about either of the topics presented here, please write a comment below!

I would like to dedicate today’s excursion to my 11th- and 12th-grade math teacher, John Barrow, who passed away on June 20th of this year. It’s safe to say that I wouldn’t be here, doing this, if it weren’t for him. He encouraged me to double up in math my junior year, and then in my senior year, he taught me AP Calculus BC one-on-one. I don’t think I appreciated at the time what a privilege that was—particularly to be doing it with such a kind-hearted, caring, and inspirational teacher. Rest in peace, Mr. Barrow.

Today we will be looking at a mathematical “object” that has been a favorite of mathematicians for millenia: Pascal’s Triangle.

A lot of people already know that the sums of the entries in the rows of Pascal’s Triangle are the powers of two:

But what you may *not* know is that if, instead, you *multiply* the entries in each row together, you get…

… uh… numbers that get really big, really fast!

Okay, so maybe that’s not so much of a surprise. Particularly since the entries in the triangle increase in size rather quickly themselves.

But suppose we wanted to try to get a better handle on just *how quickly* those products are increasing. One way we could do that is by calculating the *ratios of successive products*. In other words: How many times greater is each row’s product than the previous row’s product? If, for example, those ratios were to approach some constant, we could say that the products were increasing approximately exponentially.

First, let’s get a little more data. (By convention, we call the “row” with a single 1 in it the \(0^{\text{th}}\) row, and the next row, with two 1’s in it, the \(1^{\text{st}}\) row, etc.)

Whoa. Those are indeed some big numbers. In the interest of conserving space, I’ll write them using scientific notation.

But before I do that… I’m noticing something. Look at the curve formed by the lead digits of the products. I can tell just by looking at that curve that these numbers are increasing *more quickly* than merely exponentially. If the numbers were increasing exponentially, then that curve would not be curved at all; it would look more like a straight line. For example, check out the list of the first 50 powers of 162 below.

Technically, *that* is what exponential growth looks like: a nice, straight line!

Okay. So. Anyway. Scientific notation.

Okay. Now let’s calculate the ratios of successive products, as mentioned earlier:

As expected, these ratios themselves are increasing rather quickly. Ah, but… Notice the steep slope of the lead digits. That looks *much* straighter than the curve in the earlier list of products! So maybe… the sequence of *ratios* is growing exponentially? Let’s look at the ratios… *of the ratios!* (We’ll call these the “second ratios.”)

Aha! (… Again, this is where you yell “Aha!” really loudly.) These second ratios are much better behaved. They are increasing, but it looks like the *rate* at which they are increasing… is decreasing. So a natural thing to wonder at this point is, **are these second ratios approaching a specific value, or will they increase without bound?**

(… I think I just felt a few shivers of ecstasy from some of you out there who see where this might be going…)

Let’s introduce some notation to help us try to get an answer to this question. Define \(p(n)\) as the product of the entries in the \(n\)th row of Pascal’s Triangle. In other words, \[p(n):=\prod_{k=0}^{n}\binom{n}{k}=\prod_{k=0}^{n}\frac{n!}{(n-k)!k!}.\]

Now. How can we create an expression for those second ratios using \(p(n)\)? Well, the sequence of first ratios begins \[\frac{p(1)}{p(0)}, \frac{p(2)}{p(1)}, \frac{p(3)}{p(2)}, \frac{p(4)}{p(3)},\ldots,\frac{p(n+1)}{p(n)},\ldots\] which means that the sequence of second ratios begins \[\frac{p(0)p(2)}{p(1)^2}, \frac{p(1)p(3)}{p(2)^2}, \frac{p(2)p(4)}{p(3)^2},\ldots,\frac{p(n-1)p(n+1)}{p(n)^2},\ldots\]

So it looks like what we’re hoping to determine is the value (if it exists) of \[\lim_{n\to\infty}\frac{p(n-1)\,p(n+1)}{p(n)^2}.\]

The prospect of using our definition of \(p(n)\) to evaluate this limit seems daunting; the notation-wrangling alone is enough to make us want to leave it to someone else.

Fortunately, this \(p\) function of ours has a nifty property (proven in the Postscript): For all values of \(n\), \[\frac{p(n+1)}{p(n)}=\frac{(n+1)^n}{n!}.\]

You may recognize the expression on the left as representing our first ratios. So if you plug values of 0, 1, 2, 3, etc. in for \(n\) in the right-hand expression, you will see that you get the values of those first ratios calculated earlier: \[\frac{1^0}{0!}=1,\qquad \frac{2^1}{1!}=2,\qquad \frac{3^2}{2!}=4.5,\qquad \frac{4^3}{3!} \approx 10.7,\qquad \text{etc.}\]

Now let’s take another look at that expression for which we’re trying to find a limit, and make use of that nifty property: \[\frac{p(n-1)p(n+1)}{p(n)^2} = \frac{\ \frac{p(n+1)}{p(n)}\ }{\ \frac{p(n)}{p(n-1)}\ } = \frac{\frac{(n+1)^n}{n!}}{\frac{n^{n-1}}{(n-1)!}}\] And this expression actually simplifies quite nicely: \[\frac{\frac{(n+1)^n}{n!}}{\frac{n^{n-1}}{(n-1)!}} = \frac{(n-1)!\,(n+1)^n}{n!\;n^{n-1}} \\ = \frac{n+1}{n}\left(\frac{n+1}{n}\right)^{n-1} = \left(1+\frac{1}{n}\right)^n.\] And, as most calculus students know, \[\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n \approx 2.71828,\] better known as Euler’s number, or simply \(e\).

I was scandalized when I first encountered this. What on earth is \(e\) doing in Pascal’s Triangle?!? Even though we just proved it, it’s still hard to make sense of.

Showing *that* something is true does not necessarily illuminate *why* it’s true. I think I’ve done the former here, but I think it would take some more contemplation on my part before I’m able to do the latter.

I would love to hear *your* attempts at such illumination!

Oh, and by the way: Those of you who read the title of my pitch and thought, “Wait a minute—there’s no ‘e’ in ‘Pascal’!”… Well, now you know the truth!

\[\begin{aligned} \frac{p(n+1)}{p(n)} &= \frac{\prod_{k=0}^{n+1}\frac{(n+1)!}{(n+1-k)!k!}}{\prod_{k=0}^{n}\frac{n!}{(n-k)!k!}} & & \text{by definition}\\[0.5em] &= \frac{\prod_{k=0}^{n}\frac{(n+1)!}{(n+1-k)!k!}}{\prod_{k=0}^{n}\frac{n!}{(n-k)!k!}} & & \text{remove a “1” from the top product} \\[0.5em] &= \prod_{k=0}^{n}\frac{\ \frac{(n+1)!}{(n+1-k)!k!}\ }{\ \frac{n!}{(n-k)!k!}\ } & & \text{combine products} \\[0.5em] &= \prod_{k=0}^{n}\frac{(n+1)!(n-k)!k!}{(n+1-k)!k!n!} & & \text{invert and multiply} \\[0.5em] &= \prod_{k=0}^{n}\frac{n+1}{n+1-k} & & \text{simplify} \\[0.5em] &= \frac{\prod_{k=0}^{n}(n+1)}{\prod_{k=0}^{n}(n+1-k)} & & \text{separate products} \\[0.5em] &= \frac{(n+1)^{n+1}}{(n+1)!} & & \text{simplify} \\[0.5em] &= \frac{(n+1)^n}{n!} & & \text{cancel an }n+1 \end{aligned}\]

**Matt Enlow** teaches mathematics at the Dana Hall School in Wellesley, MA. You can follow him on X, BlueSky and Mathstodon.

We start off today’s pitch with a hands-on activity. For this, you need three circular objects, a thread, a rule, a paper and a pen. Using the thread, find the length of the circular object by wrapping the thread around the object and measuring it on the rule. Record this in one column on the paper. Next, locate the centre of the circular object and measure the diameter (from one end of the circle through the centre to the opposite end). Record this diameter in another column on the paper. In a third column, divide the length of the circular object by the diameter. Repeat this process for the two remaining circular objects. What do you observe? Irrespective of the size of the circular object, the ratio of the length (circumference) to the diameter is a constant, called Pi (π).

In a quest to find the area of a circle, we unwrap the circle to get a right angled triangle with base \(2\pi r\) and height \(r\). Then the area of this right angled triangle is \(\frac{1}{2} \times 2 \pi r \times r \) which is equal to \(\pi r ^2\). Hence, the area of the circle is \(\pi r^2\).

Let us continue to the world of prime numbers to learn some interesting facts. The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, … They seem to be random with no pattern. However, if we take any prime number greater than or equal to 5 and square it, the result is divisible by 24 with a remainder of 1. For instance \(5^2 = 24 + 1\), \(7^2 = 2 \times 24 + 1\), \(11^2 = 5 \times 24 + 1\), …

Recall that the first few digits of the number \(\pi\) are

\[ \pi = 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59230 78164 06286 20899 8628 03482 53421 17067 \ldots \]

Recall the prime chunks from Katie’s pitch. Then, the first prime chunk after 3 is 14159. Squaring this results in \( 14159^2 = 8,353,220 \times 24 + 1\). As an assignment, you can verify this for the other prime chunks in \(\pi\).

For our third interesting fact, we consider the sequence 1, 1, 2, 3, 5, 8, … called the Fibonacci sequence. If we draw golden rectangles as pitched by Matt, we can find the ratio of the sides of the rectangles. This ratio gives the golden ratio \(\Phi\). The golden ratio is common in nature, in particular on the body. The proportion of the length of the body from the head to toe, to the length from the head to the navel exhibits the golden ratio. This is evident in Leonardo Da Vinci’s The Vitruvian Man. Also, the ratio of the length from the top of the nose of an adult to the centre of the lip to the length from the centre of the lip to the chin exhibits the golden ratio. Cosmetic surgeons use the golden ratio in reconstructing the face.

Why should we care about all these interesting facts? They make maths fun and rewarding as diverse careers can be pursued after a strong foundation in mathematics. Example: Data scientist, software engineering, sound engineer, accounting, … Thus, your love for mathematics can be translated into a rewarding career.

If you found this pitch interesting and would like me to be crowned the World’s Most Interesting Mathematician, vote for me 😇 and subscribe to the Femafricmaths YouTube channel.

Medaase! Asante Sana! A dupe! Na gode! Merci! Mweebale! Murakoze Urakoze! Thank you!

**Angela Tabiri** is a mathematician and youth mentoring in STEM expert from Ghana. She is the founder of Femafricmaths, a non profit organisation that promotes female African mathematicians to highlight the diversity in careers after a degree in mathematics. You can follow Femafricmaths on YouTube, Instagram, Facebook and X.

So, which bit of maths has tickled your fancy the most? Vote now!

Note: There is a poll embedded within this post, please visit the site to participate in this post's poll.Voting is now closed: **Angela Tabiri is the World’s Most Interesting Mathematician** *(2024, of the people who I asked to take part and were available)*!

Congratulations Angela, and thanks to Matt and everyone else for playing this year!

]]>The Carnival rounds up maths blog posts from all over the internet, including some from our own Aperiodical. See our Carnival of Mathematics page for more information.

]]>Take a look at both pitches, vote for the bit of maths that made you do the loudest “Aha!”, and if *you* know any more cool facts about either of the topics presented here, please write a comment below!

Mathematics is difficult. Mathematics is abstract. What will I use maths for after high school? In this contest, we demonstrate six applications of mathematics in everyday life.

Starting your morning in Ghana usually involves a breakfast of “koko” and “koose”. Koko is a porridge made from millet while koose is made from beans. The koko seller stores the porridge in an aluminium cylindrical object. Using the mathematics of computing the volume, the koko seller can compute the quantity of porridge and how much the portions need to be sold in order to make profit. The volume of the cylinder can be explained as each layer of the cylinder consisting of circular surfaces so the sum of all these circular surfaces gives the volume of \(\pi r^2 h\).

When joining a road from a side street, you estimate the speed of the driver in the lane you are joining to see if the vehicle will be a safe distance from you by the time you join the road. If the estimation shows that the speed is high and you are not safe to join the road, you need to wait for the road to clear up in order to safely join. This involves speed, calculated as the ratio of distance to time.

In buildings which are tiled, mathematics is used to compute the area of the floor to estimate the quantity of tiles needed to tile the floor. If there are special designs to be incorporated, maths is used to compute the precise location where the designs need to be in order for the artwork of the tiles to come out beautifully.

When window frames are not fixed properly, the smallest space will allow water to seep in when it rains. If the right angle is accurately fixed at the corners of the window frame, this will reduce the chances of rain seeping through the window.

Malaria is a challenge especially in developing countries. Anytime you test positive for malaria, immediate treatment is recommended. The treatment prescribed is the stage of malaria. Mathematics is used in determining the dosages to ensure that they are taken in the right quantities and at the right time.

The wall clock shows a twelve hour time when we have twenty four hours. With modular arithmetic, we can convert times from 13:00 to 24:00 to times between 0.00 t0 12:00. We do this by computing the remainder when divided by 12. For instance: 13:00 becomes 1:00pm since 13 leaves a remainder of 1 when divided by 12.

In conclusion, mathematics is the foundation for most professions due to its universality. A deep understanding of mathematical concepts and availability of interesting learning materials will make it accessible to all.

#mathsiseverywhere #mathsisfun

**Angela Tabiri** is a mathematician and youth mentoring in STEM expert from Ghana. She is the founder of Femafricmaths, a non profit organisation that promotes female African mathematicians to highlight the diversity in careers after a degree in mathematics. You can follow Femafricmaths on YouTube, Instagram, Facebook and X.

**Ayliean** (noun): Mathsy, arty, crochet crafty, origami, activist, zine author known for making badges and trouble. You can follow them on YouTube, as @Ayliean on all social media, or look at their homepage.

So, which bit of maths has tickled your fancy the most? Vote now!

Note: There is a poll embedded within this post, please visit the site to participate in this post's poll.The poll closes at 08:00 BST tomorrow. Whoever wins the most votes will get the chance to tell us about more fun maths in the *final!*

We’re giving the competitors a break before the final round. Come back on the 23rd for the grand final, or check out the announcement post for your follow-along wall chart!

]]>