Here’s the final match of The Big Internet Math-Off.

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!

Matt Enlow – Hidden in Pascale’s Triangle

Dedication

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.

Excursion

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…)

Limits

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!

Postscript: Proof that \(\frac{p(n+1)}{p(n)}=\frac{(n+1)^n}{n!}\)

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

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

Angela Tabiri – #MathsMotivation

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!

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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!