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The Big Internet Math-Off 2019, Group 1 – Grant Sanderson vs Alaric Stephen

This is the fifth match in our group stage: from Group 1, it’s Grant Sanderson up against Alaric Stephen. The pitches are below, and at the end of this post there’s a poll where you can vote for your favourite bit of maths.

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!

Grant Sanderson – Keep/kill/divide

Grant Sanderson makes videos on the YouTube channel 3blue1brown centered on visualizing math. Once upon a time, he also used to make content for Khan Academy. He’s @3blue1brown on Twitter.

In a lot of the math I put out online, the focus tends to be on visual intuitions.  Here I’d like to appeal to a different sense, your arithmetic sense, where the landscape of possible perceptions is no less replete beauty or artistic delights.
Let’s start with an infinite fraction related to $\pi$.

\[ \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \ldots = \frac{\pi^2}{6} \]

We could use many other series for what’s about to follow, but if the goal is to make a compelling pitch, we might as well choose one of the biggest celebrity series, right?  We’re going to take a hammer and chisel to this series, knocking out some terms and carving down others in a seemingly random manner.  Here’s how it works:

  • If you’re the $n$th term, and $n$ is prime, you get to stay as you are.  For example, the terms $\frac{1}{2^2}$ and $\frac{1}{3^2}$ each stay as they are.
  • If you’re the $n$th term, and $n$ is a power of a prime, $p^k$ for some prime $p$ and positive integer $k$…well you’re not as cool as a prime, so we don’t like you as much, but your loose resemblance to a prime is enough to keep you in the club.  On one condition!  You’ll get scaled down by $\frac{1}{k}$, where $k$ is the power of the prime. For example, $\frac{1}{8^2}$ is the $8$th term, and $8 = 2^3$, so that term will turn into $\frac{1}{3} \cdot \frac{1}{8^2}$. Similarly $\frac{1}{729^2}$ is the $(3^6)$th term, so it gets scaled down by $\frac{1}{6}$.
  • If you’re the $n$th term, and $n$ is neither a prime nor a power of a prime, you get kicked out of the series entirely.  So $\frac{1}{1^2}$, $\frac{1}{6^2}$, $\frac{1}{1728^2}$, etc. are all kicked out.

If you prefer, you could be even cleaner and write only the second and third rule, since the second rule encompasses the first.
After the alteration, here’s what we’re left with:

\[ \color{red}{\underbrace{ \frac{1}{1^2}}_{\text{kill}}} + \color{green}{\underbrace{ \frac{1}{2^2}}_{\text{keep}}} + \color{green}{\underbrace{ \frac{1}{3^2}}_{\text{keep}}} + \color{blue}{\underbrace{ \frac{1}{4^2}}_{\times(1/2)}} + \color{green}{\underbrace{ \frac{1}{5^2}}_{\text{keep}}} + \color{red}{\underbrace{ \frac{1}{6^2}}_{\text{kill}}} + \color{green}{\underbrace{ \frac{1}{7^2}}_{\text{keep}}} + \color{blue}{\underbrace{ \frac{1}{8^2}}_{\times(1/3)}} + \color{blue}{\underbrace{ \frac{1}{9^2}}_{\times(1/2)}} + \color{red}{\underbrace{ \frac{1}{10^2}}_{\text{kill}}} + \color{green}{\underbrace{ \frac{1}{11^2}}_{\text{keep}}} + \color{red}{\underbrace{ \frac{1}{12^2}}_{\text{kill}}} + \ldots \]

\[ \color{green}{\frac{1}{2^2}} + \color{green}{\frac{1}{3^3}} + \color{blue}{\frac{1}{2}} \cdot \color{blue}{\frac{1}{4^2}} + \color{green}{\frac{1}{5^2}} + \color{green}{\frac{1}{7^2}} + \color{blue}{\frac{1}{3}} \cdot \color{blue}{\frac{1}{8^2}} + \color{blue}{\frac{1}{2}} \cdot \color{blue}{\frac{1}{9^2}} + \color{green}{\frac{1}{11^2}} + \color{green}{\frac{1}{13^2}} + \color{blue}{\frac{1}{4}} \cdot \color{blue}{\frac{1}{16^2}} + \ldots \]

So what does this equal?  Well, it’s certainly not $\frac{\pi^2}{6} \approx 1.644934 \ldots$ anymore. In fact, you know it must be meaningfully smaller since we kicked out and scaled down many terms. The value it approaches is $0.4977003 \ldots$, which is actually still closely related to $\frac{\pi^2}{6}$.  In fact, if we raise $e$ to the power of this new-fangled series, we get back $\frac{\pi^2}{6}$!

\[ e^{\left( \color{green}{\frac{1}{2^2}} + \color{green}{\frac{1}{3^3}} + \color{blue}{\frac{1}{2}} \cdot \color{blue}{\frac{1}{4^2}} + \color{green}{\frac{1}{5^2}} + \color{green}{\frac{1}{7^2}} + \color{blue}{\frac{1}{3}} \cdot \color{blue}{\frac{1}{8^2}} + \color{blue}{\frac{1}{2}} \cdot \color{blue}{\frac{1}{9^2}} + \color{green}{\frac{1}{11^2}} + \color{green}{\frac{1}{13^2}} + \color{blue}{\frac{1}{4}} \cdot \color{blue}{\frac{1}{16^2}} + \ldots \right)} = \frac{\pi^2}{6} \]

How about that?  A formula with $e$, all the primes, and $\pi$, all wrapped together.  It really feels like we should tell all those people tattooing $e^{\pi i} = -1$ to themselves that they’re missing out.

What’s shocking is that you can play this same game with many other series, and get the same result. This chiseling process is, at least for some series, the world’s weirdest way to take logarithms. For example, here’s what it looks like with another celebrity in math, the Leibnitz formula

\[ \color{red}{\underbrace{\frac{1}{1}}_{\text{kill}}} + \color{green}{\underbrace{\frac{-1}{3}}_{\text{keep}}} + \color{green}{\underbrace{\frac{1}{5}}_{\text{keep}}} + \color{green}{\underbrace{\frac{-1}{7}}_{\text{keep}}} + \color{blue}{\underbrace{\frac{1}{9}}_{\times (1/2)}} + \color{green}{\underbrace{\frac{-1}{11}}_{\text{keep}}} + \color{green}{\underbrace{\frac{1}{13}}_{\text{keep}}} + \color{red}{\underbrace{\frac{-1}{15}}_{\text{kill}}} + \color{green}{\underbrace{\frac{1}{17}}_{\text{keep}}} + \color{green}{\underbrace{\frac{-1}{19}}_{\text{keep}}} + \color{red}{\underbrace{\frac{1}{21}}_{\text{kill}}} + \color{green}{\underbrace{\frac{-1}{23}}_{\text{keep}}} + \color{blue}{\underbrace{\frac{1}{25}}_{\times (1/2)}} + \color{blue}{\underbrace{\frac{-1}{27}}_{\times (1/3)}} + \ldots = \frac{\pi}{4} \]

\[ e^{\left(\color{green}{\frac{-1}{3}} + \color{green}{\frac{1}{5}} + \color{green}{\frac{-1}{7}} + \color{blue}{\frac{1}{2}} \cdot \color{blue}{\frac{1}{9}} + \color{green}{\frac{-1}{11}} + \color{green}{\frac{1}{13}} + \color{green}{\frac{1}{17}} + \color{green}{\frac{-1}{19}} + \color{green}{\frac{-1}{23}} + \color{blue}{\frac{1}{2}} \cdot \color{blue}{\frac{1}{25}} + \color{blue}{\frac{1}{3}} \cdot \color{blue}{\frac{-1}{27}} + \cdots \right)} = \frac{\pi}{4} \]

This will work whenever the terms are multiplicative.  That is, if your sum looks like $\sum_{n=1}^{\infty} a_n$, where these terms satisfy $(a_n)(a_m) = a_{(mn)}$.

At this point, you may have two questions: Why, and why.  Why on earth does this work? And why would anyone care?  Needless to say, mathematicians don’t typically make discoveries by applying totally random procedures to the objects they work with and seeing what pops out. If you’re curious, what’s going on with all these series follows from taking an Euler product, summing many geometric series, and applying the Taylor expansion of $\ln(x)$, steps which I’d highly encourage you to retrace for yourself. As to why we’d care, if you’ve ever heard of the Riemann zeta function (the subject of the famous Riemann hypothesis) or its cousins the L-functions, the relationship outlined above is deeply intertwined with why these functions are related to the distribution of primes, and prime counting functions.


Alaric Stephen – Bidding Game

Alaric Stephen is one of the two hosts of the recreational maths podcast Odds and Evenings. He’s currently teaching Maths and Engineering at Hereford Sixth Form College.

In an experiment you are given £25 and the opportunity to bet on a coin which you know comes up heads with a probability of 0.6. You can flip it however many times you like within half an hour and if you manage to make it to £250 your winnings are capped there. What strategies are there for this?

One of my students wrote me a programme so you can have a go yourself (thanks Ed!). Once you’ve had a play around you can hear Alex and me discuss what happened when this experiment was run for real and debate the best strategy in a mini episode of our podcast below. 

Check out the paper mentioned in the show, and our podcast website.


So, which bit of maths do you want to win? Vote now!

Match 5: Group 1 - Grant Sanderson vs Alaric Stephen

  • Grant Sanderson with unexpected logarithms (69%, 336 Votes)
  • Alaric Stephen with the bidding game (31%, 150 Votes)

Total Voters: 486

This poll is closed.

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The poll closes at 9am BST on the 6th. Whoever wins the most votes will win the match, and once the group stages are over, the number of wins will determine who goes through to the semi-final.

Come back tomorrow for our fifth match of the group stages, featuring Jorge Nuno Silva and Sameer Shah. Or check out the announcement post for your follow-along wall chart!

10 Responses to “The Big Internet Math-Off 2019, Group 1 – Grant Sanderson vs Alaric Stephen”

  1. Avatar Reasu

    The series at the beginning of Grant’s pitch is wrong. It should be 1/6^2, not 6/1^2 etc.

    Reply
  2. Avatar Jeremy

    Is it possible that Grant’s pitch has something wrong in his first equation? The 5 and 6 should be in the denominator, right?

    Reply
  3. Avatar Reasu

    @mscroggs, does finding and pointing out this mistake earn me some bonus stickers? :•)

    Reply
  4. Avatar Ed Ceney

    I and other friends currently not able to see the vote button but can see the number of votes cast and who the votes were for. Is there a way to fix this?

    Reply

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