Yesterday morning, my son and I did something similar with our cat, Tabby. This is in response to Matt Parker’s latest initiative, Psychic Pets. Matt is hoping to get thousands of pet owners to make predictions, in order that the odds are good a pet can be found which predicted all prior results for both teams in the final. The good news is it’s fairly straightforward to take part.

Matt’s made a video which explains.

And, for readers who are teachers, there are teaching resources related to the project from Think Maths.

I figure there are teams people Matt reaches are more likely to pick and some that are likely to be less well-picked, and so chose Peru to help Matt get good coverage and because their first match was later the same day. I wrote out sheets with team names and ‘draw’, and offered Tabby the choice of each combination of sheets with a cat biscuit on it. (I’m not completely on top of the details of the World Cup, but the Psychic Pets website has all the info.) Tabby is the sort of cat who will do almost anything for a cat biscuit, even take part in one of Matt’s silly projects. She made her basic three predictions, which I videoed and uploaded to the site.

Made a contribution for @standupmaths’ latest stupid project, @PsychicPets.

You can do it too! https://t.co/yGgW5rI6YB pic.twitter.com/mmHRrmFH9Z— Peter Rowlett (@peterrowlett) June 16, 2018

Well, it didn’t take long to learn that her first prediction, Peru for the win, wasn’t to be. Pat Parslow pointed out on Twitter that this doesn’t prove my cat isn’t psychic — she could be toying with me. But, either way, it doesn’t help advance Matt’s cause.

But, while it’s too late for Tabby, do you have a pet and could get involved? Get started at Psychic-Pets.com.

]]>Enrico Fermi apparently had a knack for making rough estimates with very little data. Fermi problems are problems which ask for estimations for which very little data is available. Some standard Fermi problems:

- How many piano tuners are there in New York City?
- How many hairs are there on a bear?
- How many miles does a person walk in a lifetime?
- How many people in the world are talking on their mobile phones right now?

Hopefully you get the idea. These are problems for which little data is available, but for which intelligent guesses can be made. I have used problems of this type with students as an exercise in estimation and making assumptions. Inspired by a tweet from Alison Kiddle, I have set these up as a comparison of which is bigger from two unknowable things. Are there more cats in Sheffield or train carriages passing through Sheffield station every day? That sort of thing.

The point of these is not to look up information or make wild guesses, but instead to come up with a back-of-the-envelope, ‘wrong, but useful‘, orders of magnitude estimate. Some ‘rules’, if you want to play with these the way I would:

- don’t look up information;
- don’t make precise calculations using calculator or computer;
- be imprecise — there are 400 days in a year, people are 2m tall, etc.;
- round numbers where possible and calculate in your head.

One approach is to estimate by bounding – come up with numbers that are definitely too small and too large, and then use an estimate that is an average of these. But which average?

Say I think some quantity is bigger than 2 but smaller than 400. The arithmetic mean would be $\mathrm{AM}(2,400)=\frac{2+400}{2}=201$. The geometric mean would be $ \mathrm{GM}(2,400)=\sqrt{2\times 400} = 28.28\!\ldots$.

Which is a better estimate? The arithmetic mean is half the upper bound, but 100 times the lower bound. On this basis, for an ‘order of magnitude’-type estimate, you might agree that the geometric mean is a better average to use here. Following my Maths Jam talk, Rob Low said that the geometric mean makes more sense for an order of magnitude estimate, since it corresponds to the arithmetic mean of logs. To see this, consider \[

\begin{align*}

\log(\mathrm{GM}(A, B)) &= \log(\sqrt{AB}) \\

&= \log((AB)^{\frac{1}{2}}) \\

&= \frac{1}{2}\log(AB) \\

&= \frac{1}{2}(\log(A) + \log(B)) = \mathrm{AM}(\log(A), \log(B)) \text{.}

\end{align*}

\]

So, geometric mean it is. However, taking a square root is not usually easy in your head, and we want to avoid making precise calculations by calculator or computer. Enter the approximate geometric mean.

For the approximate geometric mean, take $2=2 \times 10^0$ and $400=4 \times 10^2$, then the AGM of $2$ and $400$ is: \[ \begin{align*}

\frac{2+4}{2} \times 10^{\frac{0+2}{2}} &= 3 \times 10^1\\

&= 30 \approx 28.28\!\ldots = \sqrt{2\times 400} = \mathrm{GM}(2,400) \text{.}

\end{align*} \]

Why does this work? Let $A=a \times 10^x$ and $B=b \times 10^y$. Then \[

\begin{align*}

\mathrm{GM}(A,B)=\sqrt{AB}&=\sqrt{ab \times 10^{x+y}}\\

&=\sqrt{ab} \times 10^{\frac{x+y}{2}} \text{,}

\end{align*}

\]

and \[\mathrm{AGM}(A,B) = \frac{a+b}{2} \times 10^{\frac{x+y}{2}}\text{.}\]

Setting aside the $10^{\frac{x+y}{2}}$ term, which appears in both averages, is it obvious that, for single digit numbers $>0$, \[\mathrm{GM}(a,b)=\sqrt{ab} \approx \frac{a+b}{2}=\mathrm{AM}(a,b) \text{?} \]

There is a standard result that says \[ \begin{align*}

0 \le (x-y)^2 &= x^2 – 2xy + y^2\\

&= x^2 + 2xy + y^2 – 4xy\\

&= (x+y)^2 – 4xy \text{.}

\end{align*} \]

Hence \[ \begin{align*}

4xy &\le (x+y)^2\\

\sqrt{xy} &\le \frac{x+y}{2} \text{,}

\end{align*} \]

with equality iff $x=y$. So $\mathrm{GM}(a,b)\le\mathrm{AM}(a,b)$, but are they necessarily close?

By exhaustion, it is straightforward to show (for single-digit integers, given the rule to round numbers where possible) that the largest error occurs when $a=1$ and $b=9$. Then \[ \sqrt{1 \times 9} = 3 \ne 5 = \frac{1+9}{2} \] and the error is $2$ which, relative to the biggest number $9$ might be seen as quite significant.

I’d say you are not likely to use this method if the numbers are of the same order of magnitude, because the idea is to come up with fairly wild approximations and if they were quite close it might be sensible to think of them as not really different. Then the error is going to be at least one order of magnitude smaller than the upper bound, i.e. $10^\frac{x+y}{2} \ll 10^y$. For example, if your numbers were $1$ and $900$ (as a pretty bad case), then: \[ \mathrm{GM}(1,900)=\sqrt{900}=30 \ne 50=\mathrm{AGM}(1,900) \] and a difference of $20$ on a top value of $900$ is not as significant as a difference of $2$ was on a top value of $9$.

So I suppose I would argue that this makes the error relatively insignificant. However, this thinking left me somewhat unsatisfied. I felt there ought to be a nicer way to demonstrate why the approximate geometric mean works as an approximation for the geometric mean. Following my talk at Maths Jam, Philipp Reinhard has been thinking about this, and he will share his thoughts in a post here in a few days (the post is now online).

I didn’t have time to fit into my talk what I would recommend if the two numbers differed by an odd number of orders of magnitude. For example, $\mathrm{AGM}(1,1000)$ generates another square root in $1 \times 10^{\frac{3}{2}}$ – precisely what we were trying to avoid! What I have recommended to students is to simply rewrite one of the numbers so that the difference in exponents is even. For example, writing $1=1 \times 10^0$ and $1000 = 10 \times 10^2$ gives \[\mathrm{AGM}(1,1000)=5.5 \times 10^{1} \text{.}\]

Following Maths Jam, the esteemed Colin Beveridge made the sensible suggestion of just treating $10^{\frac{1}{2}}$ as $3$, making \[

\begin{align*}

&\mathrm{AGM}(1,1000)\\

&= 1 \times 10^{\frac{3}{2}}\\

&\approx 1 \times 3^3 = 27\text{.}

\end{align*}

\]

This increases our problems, though, because we have the potential to deal with larger differences (hence larger errors) than when dealing with single-digit numbers. Actually, it was wondering why this increased error happens that got me thinking seriously on this topic in the first place. I’ll stop now to let Philipp share what he has been thinking on this.

]]>I just noticed that last Wednesday was ten years since that lecture. It was basic maths for forensic science students. I was given a booklet of notes and told to either use it or write my own (I used it), had a short chat about how the module might work with another lecturer, and there I was in front of the students. That was spring in the academic year 2007/8 and this is the 21st teaching semester since then. This one is the 15th semester during which I have taught — the last 12 in a row, during which I got a full-time contract and ended ten years of part-time working.

I have this awful feeling this might lead people to imagine I’m one of the people who knows what they are doing.

P.S. The other thing that I started when I started working for the IMA was blogging – yesterday marks ten years since my first post. So this post represents the start of my second ten years of blogging.

]]>The first will reward a well-made, delicious item; the second will reward the item which has been decorated the most beautifully and looks most like what it’s supposed to be; and the third will reward the most ingenious mathematical theming.

You can view the entries from this year on the MathsJam website.

The other regular competition is the Competition Competition. This invites attendees to submit a competition, which other attendees can enter. There are some rules, including minimum font size, paper size and maximum value of prize. To be clear, the rules say “any type of competition is permitted as long as it can be judged by the setter (or a winner randomly chosen from the correct entries)”.

Prizes are awarded for best competition, popular vote winner (the competition with the most entries) and “best attempt at circumvention of the rules while still strictly sticking to the rules”. Seeing people attempt the latter is quite delightful.

Chatting to people at MathsJam this year, I was reminded of my entry into the Competition Competition when it first ran in 2014. I invited entrants to write down an integer between 0 and 100, then I said that I would run a Shapiro-Wilk test of normality on the numbers people had written down. This tests the null hypothesis that the data come from a normally distributed population. The competitive element of the competition asked people to guess the p-value obtained from that test.

While we were chatting about this, The Aperiodical’s own Paul Taylor asked me what p-value came out as. I couldn’t remember, but I’ve looked it up. The numbers entered were as follows:

Number entered |
Number of people entering it |
---|---|

2 | 1 |

6 | 1 |

12 | 1 |

16 | 1 |

50 | 1 |

71 | 1 |

72 | 2 |

73 | 2 |

86 | 2 |

97 | 2 |

99 | 7 |

In a sample of 21 integers from 1-99 where only four numbers are below 50 and one third are precisely 99, it may not surprise you to learn that the statistical test gave strong evidence to reject the hypothesis that these data were from a normally distributed population. The p-value (from R) was 0.0002211865 and the winner was Francis Hunt, who guessed 0.0001.

You can find out about the MathsJam competitions and other side activities that take place on the MathsJam Extras page.

]]>The MTaP carnival traditionally starts with a puzzle, game or piece of trivia. As this edition marks the start of a new year, I thought I’d share a new year puzzle I played with my students in the last session before the Christmas break.

Good news everyone – the 2017 game works (though it's easier than previous years, I think). pic.twitter.com/VRyfQ1bWym

— Peter Rowlett (@peterrowlett) December 12, 2016

In fact, this bring me straight onto the submissions this month. Let’s start with some puzzles and games. Denise Gaskins (@letsplaymath) submitted a post about the 2017 Mathematics Game, providing a set of rules similar to the game above that open it up to 1-100. Denise points out that “the goal is adjustable: Young children can start with looking for 1-10, middle grades with 1-25.”

Manan Shah (@shahlock) offers Alphanumeric Puzzle #1, saying “I thought readers might get a kick out of this”. It’s a type of puzzle known as a cryptarithmetic puhezzle.

Megan Schmidt (@veganmathbeagle) submitted Math Game: Draw 10 by Annie Forest. She says “Annie always has thought provoking ideas to share and this one is no different. She brings the math and the excitement for those days before Christmas when engagement is tough to come by.”

Next up, Measurement Games and Activities, a list of measurement/estimation games submitted by Crystal (@Tri_Learning), who says “The ability to estimate and measure are important skills for every day life and many professions. Find measurement games and activities for all ages.”

And if you still haven’t had enough games, Denise Gaskins (@letsplaymath) has it covered in My Favorite Math Games. Denise says “I like to use games as a warm-up with my co-op math circle. Here’s a collection of my favorites.”

Next, there are several posts offering stories and mathematical investigations.

This entertaining post by James Propp tells a tale centred around the (Prouhet-)Thue-Morse Sequence which covers religious law, fussy eaters, fractals, music, poetry, chess, rowing and more: Avoiding chazakah with the Prouhet-Thue-Morse sequence.

1226221 Is A Fascinating Number was submitted by Manan Shah (@shahlock), who says “What’s so fascinating about 1226221? It’s a palindrome, but that’s not all!”

I saw this fun post by Ben Orlin which asks the question: how many apparently different looking but actually the same sudoku puzzles can be generated from a single sudoku: 1.2 Trillion Ways to Play the Same Sudoku.

This being a mathematics teacher blog carnival, it may not come as a surprise that most submissions were tips and approaches to bringing a little fun to the classroom.

First, Tips for Teaching Students How to Identify Functions was submitted by Mrs. E (@mrseteachesmath), who says “This is a fun lesson idea for teaching students how to identify functions. It keeps kids engaged and a song helps them remember the definition of a function.”

This second post submitted by Mrs. E is about using paper folding to help students discover and think about geometry constructions: Altitudes and Angle Bisectors Paper Folding Activity.

And this third from Mrs. E is about using a puzzle/pattern-spotting approach to introducing logarithms: How I Teach Intro to Logs.

Crystal (@Tri_Learning) submitted Our Favorite Math Manipulatives, about using physical objects to help students learn, for students at various levels.

Next, the submitter didn’t leave their name, but we have Tell me everything you know about… by Jen McAleer, which offers some ideas and in-class activities to improve engagement, first by asking a loosely-defined question and second by holding a quick competition.

Amber Thomas submitted Playing with Our Food, Nutrition, and Fraction Line Plots, saying:

I found this fun nutrition website that I fondly refer to as “Smash My Food.” The kids see how much oil, salt, and sugar is in food (by watching it get squeezed out). It’s good, gross fun! Where the math comes in, is I take those amounts and create fractions on a line plot. You know, that pesky common core standard that math textbooks don’t cover at all (CCSS Math 4.MD.B.4)? Yeah, that one. Kids find the information relevant and relate-able, and I’ve gotten good feedback from others who have tried teaching fraction line plots this way.

Rupesh Gesota submitted Understanding v/s Answer-getting, a post about trying to teach understanding, even if it doesn’t use the standard set of rules. Rupesh says “This article shows some interesting (non-standard) ways of approaching and solving couple of mathematical problems. These are the ways developed by the students themselves based on understanding rather than procedures/ rules instructed by the teacher…”

Finally, in one last burst of teaching-related fun, Jo Morgan (@mathsjem) submitted the results of her World Cup of Maths, a Twitter competition to find UK maths teachers’ favourite topic to teach.

To round off this carnival, although I should disclaim that I am mentioned in it, this post by mathsbyagirl offers a guide to getting started listening to mathematical podcasts in 5 Maths Podcasts.

If you enjoyed this and the MTaP carnival in general, consider also reading the Carnival of Mathematics blog carnival which is coordinated by Katie Steckles here at The Aperiodical. Both Carnivals are always open for submissions and always happy to hear from potential future hosts. Submit a post you’ve enjoyed (whether you wrote it or not), and consider hosting a carnival at your blog!

]]>I’m so far from understanding the mind of a mathematical genius that it’s simply inconceivable that you could tell a person an apparently random number and he could intuit or deduce the kind of fact that he deduced about that taxi license number. I mean, I can’t run a four-minute mile, but I once ran a five-minute mile, and I can extrapolate from my own experience, in a way understand how someone might just be a lot better than me at something that, in an inferior way, I can also do. But Ramanujan isn’t like that. It’s as though this man were a different species, not just a superior example of the same species. Can you learn to do this kind of thing? Could I, if I had applied myself? Or is it that goddess again, is it really just genius?

Answers on a postcard!

]]>The Math Teachers at Play (MTaP) blog carnival is a monthly collection of tips, tidbits, games, and activities for students and teachers of preschool through pre-college mathematics. We welcome entries from parents, students, teachers, homeschoolers, and just plain folks. If you like to learn new things and play around with ideas, you are sure to find something of interest.

I’ll be hosting the January 2017 edition of MTaP here at Travels in a Mathematical World. Of course, a blog carnival is only as good as its submissions, so if you join me in aspiring to the claim “you are sure to find something of interest” then please keep your eyes open for interesting blog posts and submit them to MTaP. Please submit posts you’ve enjoyed by others or yourself. Posts you wrote that are appropriate to the theme are strongly encouraged. Submit through the MTaP submission form, leave a comment here or tweet me. Thank you!

Submissions are open now, and anything received by Friday 20th January 2017 will be considered for the edition hosted here.

]]>This space was renovated for mathematics a little before I arrived. It was designed to enhance student engagement and to create this sense of community, to allow collaborative learning and encourage inter-year interactions.

Over the last year, we conducted a study of use of the space. This included observations of use of the space as well as questionnaires and interviews with students about their use of the space, including students who had studied in the department in the old and new locations.

The results have just been published as ‘The role of informal learning spaces in enhancing student engagement with mathematical sciences‘ by Jeff Waldock, Peter Rowlett, Claire Cornock, Mike Robinson & Hannah Bartholomew, which is online now and will appear in a future issue of *International Journal of Mathematical Education in Science and Technology* (doi:10.1080/0020739X.2016.1262470).

Three married couples want to cross a river in a boat that is capable of holding only two people at a time, with the constraint that no woman can be in the presence of another man unless her (jealous) husband is also present. How should they cross the river with the least amount of rowing?

I’m planning to use this again next week. It’s a nice puzzle, good for exercises in problem-solving, particularly for Pólya’s “introduce suitable notation”. I wondered if there could be a better way to formulate the puzzle – one that isn’t so poorly stated in terms of gender equality and sexuality.

There’s a related, but not identical, problem – but this doesn’t help as it has its own, different issues. Here’s the version of the missionaries and cannibals problem given by Wikipedia:

Three missionaries and three cannibals must cross a river using a boat which can carry at most two people, under the constraint that, for both banks, if there are missionaries present on the bank, they cannot be outnumbered by cannibals (if they were, the cannibals would eat the missionaries). The boat cannot cross the river by itself with no people on board.

Wikipedia says the jealous husbands problem is older, dating back in some forms in Europe to the 800s, with the ‘husbands and wives’ formation coming between the 13th and 15th centuries.

Anyway, absent of a clever revelation I asked Twitter. There are minor spoilers below, so you might want to have a go at the puzzle first if you haven’t seen it before.

First, Christian Lawson-Perfect suggested simply to replace each wife with a heavy, inanimate object that belongs to one person and is coveted by the others. The object must be heavy, or at least bulky, in order that the boat can only hold one person and one object on each journey. I pointed out that whatever those coveting the object want to do with it must be done during a boat ride. In the classic formulation, I suppose each husband fears his wife would be charmed during time alone with another man. Christian suggested unlocked suitcases and Colin Beveridge suggested that these could contain top-secret information. Matthew Arbo pointed out what I had missed: at some point in the solution, we’d require one of these suitcases to row the boat.

Christian suggested replacing the wives with people who know TV spoilers. It’s a nice thought, but I think this would be very complicated to state because of the pairing of characters in the puzzle. We’d need each person who knew spoilers to know different spoilers and be paired with one of those who don’t know spoilers known by the others.

Ian Preston suggested a formulation that I wrote up like this:

Three children, each accompanied by one of their parents, each want to cross a river in a boat that is capable of holding only two people at a time. Children behave very well with each other and with their parent, but misbehave in the presence of other adults when their parent is not present. Everyone must therefore cross the river with the constraint that no child can be in the presence of an adult who is not their parent unless their parent is also present. How should they cross the river with the least amount of rowing?

This is longer than the classic statement and more convoluted. The requirement that children behave together is necessary so that we don’t think they need to stay with the parent at all times, but it’s a big hint that at some point some children are going to be left alone. Even so, there is a further problem. James Grime was confused about whether the children could row the boat, suggesting I replace children and their parents with dogs and their owners. Since at some point we require children to row the boat, perhaps I should say they can do this in the statement – yet another hint.

James Grime also suggested prison wardens and prisoners on a boat to Alcatraz. This is a creative idea, but at some point in the solution we have all the guards at Alcatraz and the prisoners, with the boat, on the shore at San Francisco. Plus, I think this is closer to the missionaries and cannibals than the jealous husbands because of the lack of pairing.

Alison Kiddle suggested a formulation in which we have three mods and three rockers, with each mod having a rocker sibling. People tolerate their own clique or their own sibling, and in a mixed group they won’t kick off if their sibling is present. I think this is a good statement of the problem and I like it quite a lot, though the cultural reference might need updating and its a bit more complicated to explain what will happen if the two groups are allowed to mix.

out of the norm said he’d heard it with Harry, Ron and Hermione with three ogres, or three nuns and three ogres, since overpowering is equivalent to jealousy. Karen Hancock suggested the allergies puzzle at the bottom of this list of interesting river-crossing problems. Nice statements, but I don’t think either is equivalent to the jealous husbands.

Then we came to the suggestion I think I am happiest with. James Sumner made a suggestion that I’ve written up as the following:

Three actors and their three agents want to cross a river in a boat that is capable of holding only two people at a time, with the constraint that no actor can be in the presence of another agent unless their own agent is also present, because each agent is worried their rivals will poach their client. How should they cross the river with the least amount of rowing?

This maintains the jealousy, so is hopefully easy to understand and should minimise the need for additional explanation. As James pointed out, we might wonder why on earth they’re crossing a river in a boat made for two, but I think that’s a minor quibble.

]]>Today my phone told me that the app Photomath has an update and now supports handwriting recognition. This means I can write something like this:

and Photomath does this with it:

Well. My immediate reaction is to be quite terrified. Clearly this is a fantastic technical achievement and a wonderful resource, but my thoughts go straight to assessment. I remember when I heard Wolfram Alpha was released, I was working to input questions a lecturer had written into an e-assessment system and realised that all the questions on the assessment I was inputting could be answered, with zero understanding, by typing them into Wolfram Alpha. Actually, not quite zero understanding, because at least you had to be able to reliably type the question. Now Photomath closes that gap (or will do soon – of course, it’s not yet perfect).

However, a lot of water has passed under the bridge since I was inputting questions into an e-assessment system. I’m a lecturer at Sheffield Hallam University now, where students who don’t arrive knowing about Wolfram Alpha are told about it, because students are encouraged to learn to use any technology available to them. Indeed, this year I was involved with marking a piece of coursework where engineering students were asked to show by hand how they had worked out their solutions and provide evidence that they checked their answer by an alternative method, usually by Wolfram Alpha screenshot.

It if often the case that lecturers use computers when setting assessments (beyond typesetting, I mean), even when they don’t expect students to use them in answering. I asked this question in a survey for my PhD and even about half of people who don’t use e-assessment with their students still use computers when setting questions (to check their answers are correct, perhaps). (Link to PhD thesis, see section 3.4.5 on p. 60.) Perhaps we should encourage our students to embrace technology in the same way.

In the academic year that is about to start, I am to teach on the first year modelling module. This is where our first year mathematics degree students get their teeth into some basic mathematical models, ahead of more advanced modelling modules in the second and final year. If you accept that a lot of mathematics is a process of: understand and formulate the problem, solve it, then translate and understand that solution – then this sort of technology only helps with the ‘solve it’ step. In the case of modelling, taking a real world situation, interpreting that as a mathematical model and extracting meaning from your solution are difficult tasks of understanding which these technologies do not help with, even as they help you get quickly and easily to a solution.

So, should I view Photomath as a terrifying assault on our ability to test students’ ability to apply mathematical techniques? Probably I should view it instead as a powerful tool to add to the mathematician’s toolkit, which hints at a world where handwritten mathematics can be solved or converted to nicely typeset documents, and so allow my students to gravitate from the tedious mechanics of the subject to greater ability to apply and show off their understanding. Probably.

]]>Oh blimey pic.twitter.com/OdKS1MmY1N

— Peter Rowlett (@peterrowlett) September 4, 2016