Tuesday, 28 November 2017

Problem... Countdown

#MentalArithmetic, #MentalCalculation, #OrderOfOperations,
#AlgorithmicThinking, #ComputationalThinking

Monday, 27 November 2017

Problem... It's π O'Clock

#Fractions, #Time, #Angle, #Turn, #Conversions, #Pi #π, #PiDay, #πDay

On the little things in teaching #1

That beautifully judgemental silence from a class faced with your meticulously engineered pun, which, as a function of their understanding of the concept underpinning it, can only mean that they really get it...

M
ME [during a discussion about correlation]:
Yes, the more overweight men with white beards wearing red suits we see, the greater the chance it is that it's nearing Christmas.  That's a good example.  So, could we say this in a more statty way?

STUDENT:
There is a strong positive correlation between the number of overweight men with white beards and red suits and when Christmas is.

ME:
I'm not sure how we could judge the strength, but other than that yes, we can say that there is a positive correlation between the number of overweight men we observe with white beards wearing red suits and how close Christmas is.  Or maybe it's a negative correlation?  I don't know.  Does it matter?

STUDENT:
Yes, because the more men you see dressed up as Father Christmas, the less time it is until Christmas.

ANOTHER STUDENT:
But you could say that the closer Christmas is the more people we see dressed up as Santa.  And that's more positive... But I suppose we're saying the same thing.

ANOTHER STUDENT:
But if you drew a graph with the date on the x axis and the number of people dressed up as Santa on the y, it would go up.

ME [slightly panicky; the chance for the pun slipping away]:
It's a bit like the chocolate consumption and Nobel Prize data we looked at [see here].  Before I told you what the data was, you assumed that more of x would make more of y.  But when you saw what the data was...

STUDENT:
...Oh that thing where the more chocolate people ate in a country the more Nobel prizes they won...

ME:

STUDENT:
It would mean that you have more chance of winning a Nobel Prize if you eat a lot of chocolate.

ANOTHER STUDENT:
Which is just dumb...

ME [quickly]:
Which is why describing it in more of a statty way is safer.  So, if we think about the men dressing up as Santa example in the same way, what would we be saying?  What would it mean if we thought about it in the wrong way?

STUDENT:
If we wanted Christmas to come quicker we would just need to see more people dressed up as Father Christmas...

ANOTHER STUDENT:
Or that we could make Christmas come quicker by getting more people to dress up as Santa.

ME [relieved]:
Exactly.  But the fact that we see an increase in overweight men with white beards wearing red suits as we get closer to Christmas, does not cause Christmas to come quicker does it?  Do you get the distinction?  If we wanted Christmas to be tomorrow would we just get millions of people to dress up as Santa?  No, of course not. The important thing to get here is that correlation may suggest a relationship between two things, but it does not imply Clausality.

AyThangYaw.

(See here for a selection of correlation and causation conflations.)

Friday, 24 November 2017

On the Crescent Area Problem

Problem... A circle is inscribed in a square that has a side length of one unit, which is also the radius of a quarter-circle.  Find the area of as many of the different shapes and regions that you can see.  (Which shapes, or regions, are you finding difficult?  What is it that is stopping you find their areas?)

Students typically define the square, the inscribed circle, the quarter circle and often a quarter of the inscribed circle as shapes and regions to find the areas of, and proceed to do so without issue.  Students also note the dark orange 'crescent' region as an area to find, and, indeed, the lighter orange regions around it, and these delineate the aspects of the problem that have the potential to develop into the most formative discussions.

Students, with the teacher's guidance / encouragement of course, can outline a strategy of how to find these more complex areas, ignoring whether or not they yet have the knowledge, skills or understanding to follow such a strategy through.  Depending on where students are in their mathematical careers, this can then lead into more direct teaching, or, moreover, can function as a segue into a body of work that will provide students with the knowledge and skills they need to apply their strategy.

To see a 'thinking through' of a typical solution / approach with students to the problem of the 'crescent' area, click on 'Read more' below.

Suggested Solution:

Thursday, 23 November 2017

Problem... The Greatest π-day Birthday in History

#Time, #TimeConversion, #Conversions
#ComutationalThinking, #AlgorithmicThinking
#Pi #π, #PiDay, #πDay

SPOILER:

Whether you take π o'clock to mean between 03:14:15 and 03:14:16, or as 3.141592653... am, people born on January 21, 1589 in the Gregorian calendar would have been π years old on π day 1592, i.e. on 3/14/1592.  One of these people was William Strode Jr, or William Strode of Barington, philanthropist and Roundhead in the English Civil War, who was born on 11 January 1589 according to the Julian calendar.  In 1589, however, the Gregorian calendar was 10 days ahead of the Julian calendar, which was still in use in Britain.

Wednesday, 22 November 2017

Problem... Crescent Area

#Area, #Circle, #Arc, #Sector, #Segment, #Trigonometry, #CosineRule

Tuesday, 21 November 2017

On 73

73 is a permutable (or anagrammatic) prime, a prime that can have its digits' rearranged (in base 10) in any permutation and still be a prime number. 73 is also an emirp, a number whose reverse, 37, is also prime — a property also evident in terms of the respective ordinal positions of 73 and its emirp partner, 73 being the 21st prime [1] while 37 is the 12th prime.  73 is a Sophie Germain prime and is palindromic in binary 1001001 (interestingly, all Fermat primes and Mersenne primes are subsets of the binary palindromic primes).  73 is also an octal palindrome 111 and the only octal prime repunit.  73 is, moreover, and for some of the reasons given here, Sheldon Cooper's favourite integer in The Big Bang Theory — as was first referenced in the show’s 73rd episode.  (Jim Parsons, incidentally, the actor who plays Sheldon Cooper, was born in 1973.)

However, after listening to Charles Yang, associate professor in the University of Pennsylvania Department of Linguistics, in this episode of the superb Tell Me Something I Don't Know podcast, hosted by Stephen Dubner of Freakonomics fame, it is another little number-73-centred-nugget-of-information that has prompted this post.  Namely, that once an English speaking child has learned to count to 73, they realise they can keep going, and thus ‘understand’ the concept of infinity (NB the 1st two comments at the end of this blog).

Having delved into Yang's fascinating research, let me try and elaborate:  If there is a linguistic rule, a generalisation in other words that can be applied to a set of N words, but within this set of N words there is a subset of words, e [2], that do not follow this rule and that must therefore be memorised, then Yang (2005, cf. 2015) has shown that:

$e < {\theta _N}\;where\;{\theta _N}: = \frac{N}{{\ln \left( N \right)}}$

This model, 'dubbed' the Tolerance Principle by Yang (2005, cf. 2015 pp16-17), asserts that for the generalisation to be ‘productive’, i.e. for the rule that generates words to tolerate the exceptions and thus be applied instead of the pure memorisation of all N, e must not exceed / ln(N).  It delineates, in other words, a threshold value for e (as a function of N) at which a word-forming rule ceases to be a productive mechanism for the learner to apply, and thus provides a ‘sufficiency measure’ for such linguistic generalisation.

By way of illustration, Yang (2015, pp948-949) gave this example of regular and irregular verbs: ‘If a typical English speaker knows e = 120 irregular verbs, the productivity of the -ed rule is guaranteed [my emphasis] only if there are many more regular verbs. Specifically, there must be N verbs, including both regulars and irregulars, such that θN = / ln(N) ≥ 120.  The minimum value of N is 800.  In other words, if there are at least 680 regular verbs in English, the -ed rule can tolerate 120 irregular verbs’.

In his Tell Me Something I Don’t Know appearance, Yang described applying this Tolerance Principle to how we as children learn to count.  We know how many of our counting words are exceptions to the rule, those numbers that is that ‘bear no relation to their quantity, [that are] are completely arbitrary’, those numbers to put it another way that ‘won’t tell you anything about how numbers are forming' (Yang, in Berger, 2017) — in the case of children counting in English, there are 17 of these exceptions, e = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 20, 30, 50} [3].

Once we have memorised, ‘by brute force,’ the number words one to ten, we start to observe and learn rules for continuing.  For example, we start to see that fourteen is formed from ‘four and ten’, sixteen from ‘six and ten’, twenty-four from ‘two tens and four’, 'twenty-six from 'two tens and six', etc.  But there remain of course those exceptions to the rule — why is eleven not one-teen, or twelve not two-teen, etc. —  where ‘memorisation [is] the sole educational mechanism’ (Yang, in Berger 2016).  To this end, using the Tolerance Principle, we can find the least amount of words that we need to learn ‘to overcome the [17] exceptions we have to memorise’.

The smallest value of N such that θN = 17 is 73:

\begin{array}{l}\begin{align}N &= 73\;\\\because17 &= e < {\theta _N} = \frac{{73}}{{\ln \left( {73} \right)}}\\\;where\;{\rm{ }}\frac{{73}}{{\ln \left( {73} \right)}} &= 17.014...\end{align}\end{array}

Or in other words, once a child has learned to count to 73, they have learned the rules of the game sufficiently to overcome the cognitive need for memorisation [4].  Once children have passed this 'tipping point', when they come to a number that they are unaware has a word outside of the system they have subconsciously generalised, a thousand for example, they make up a word that fits their generalised system.  Thus, once an English speaking child can count to 73, they can count forever.

[1]  Incidentally, also the product of multiplying 7 and 3.

[2]  Not to be confused with Euler's number, e.

[3]  Yang noted that in Chinese, e = 11, meaning that a Chinese speaking child need only count up to 42 before they are able to continue and thus appreciate the concept of infinity, some one whole year before English speaking children.

[4]  Welsh may be one of the languages with the smallest threshold at θ= 36.  The numbers 1-10 in Welsh are the only 'exceptions': Un (one), Dau (two), Tri (three), Pedwar (four), Pump (five), Chwech (six), Saith (seven), Wyth (eight), Naw (nine), and Deg (ten).  All numbers beyond this are generalised from them, for example eleven is un deg un (one ten one), twelve is un deg dau (one ten two), seventy three is saith deg tri (seven ten three), etc.

Friday, 17 November 2017

On Santamaths (An xmaths conversation with a 5 year old)

OK, let's get straight to the point: it is a fallacy, or to put it another way, a gross oversimplification to think that Santa will deem you, in binary fashion, to have been either naughty or nice, discretely, categorically, one way or the other, and that this will in turn delineate whether you receive presents or not.  This is — plainly, to put it politely, and as I will show in what follows — utter horse nonsense.  There are gradations of naughtiness or niceness: one person can be naughtier than another, or nicer, and it is evident to boot that naughtiness is a function of niceness, and vice-versa.  To put it another way, an increase in naughtiness (say η) and a decrease in niceness (say η) are biconditional:
$\rlap{-} \eta \;\; \Leftrightarrow \;\;\eta$
To elaborate: Sometimes you will be naughty and sometimes you will be nice. And of course the more you are naughty the less time you have to be nice — and vice-versa: the more you are nice the less time you have to be naughty.  So, and in other words, the naughtier you are the less likely it is that you will be nice, and the nicer you are the less likely it is that you will be naughty.  Yes?  The truth is that two naughty people will be naughty to different degrees: one will be more or less naughty than the other.  (The same is obviously true for two nice people.)  And of course, one naughty act might not be as naughty as another, so there are gradations of naughtiness and, indeed, gradations of niceness.  It is nice to say please and thank you, for example, but it is nicer to slot a pound into a Noddy car ride in the shopping centre and leave it to be discovered by an unsuspecting — but soon to be overjoyed — toddler.  Similarly, it is naughty to take your sister's cuddly Pooh bear when she doesn't want you to, but it is naughtier to hide Pooh and tell your sister that he’s run away because he doesn’t love her her anymore.

Santa is not stupid; he knows all of this.  And he uses Santamaths to deal with the inherent variability in η and hence η.  Put simply, he uses relative frequencies to assign a weighted 'niceness index' (known as the ηi pronounced /niː/) to each person (𝓍), and the presents each person receives are thus dependent on such.  In short:

$\begin{array}{l}\forall x,\;0 \le {\eta _i} = \frac{\eta }{{\eta + \rlap{-} \eta }} \le 1\\\\and\;as\;{\eta _i} \to 1,\;\forall x\left( {\rlap{-} \eta \Rightarrow \eta } \right),\\and\;P\left( {x\;gets\;what\;s/he\;wants} \right) \to 1\\\\Similarly,\;as\;{\eta _i} \to 0,\;\forall x\left( {\eta \Rightarrow \rlap{-} \eta } \right),\\and\;P\left( {x\;gets\;what\;s/he\;wants} \right) \to 0\end{array}$
Santa then uses probability theory and combinatorics to calculate what and how many presents you get.  As such, this year, as you know, I have been keeping a tally of all the times you have done something naughty (η) and all of the times you have done something nice (η).  You get one scratch for a naughty act and one for a nice act, an act of kindness if you will, and dependent of course on the act.  Remember that time you fooled your sister  your type 1 hypersensitive to tomatoes sister  into thinking that ketchup you had mixed in her ice-cream was actually raspberry juice, well that got you ten scratches in the naughty column, for example, whilst tidying up your Lego without being asked that one time you did it in the whole year got you five scratches in the nice column.

Here's a little section of the tallies I've been keeping for this year:
Overall you had 740 naughty scratches this year and 623 nice scratches, giving ηi = 0.46:
\begin{array}{l}\begin{align}\frac{{623}}{{740 + 623}} &= \frac{{623}}{{1363}}\\\\ &= 0.46\end{align}\end{array}
Now, this doesn't mean that you have a 46% chance of getting your presents this year.  Oh no.  You wrote to Santa to ask for the following five presents:
• An umbrella in the shape of an Octopus,
• A Lego Batman car [I presume you mean the Batmobile]
• A green and orange ukulele,
• A dinosaur that can really move and roar and scare people like a monkey on a horse,
• A helicopter (remote-controlled will do if Santa can't source a real one for you).

Hence the following Santamaths model applies:

$\begin{array}{l}{\left[ {{\eta _i} + \left( {1 - {\eta _i}} \right)} \right]^5} = {}^5{C_5}{\left( {{\eta _i}} \right)^5}{\left( {1 - {\eta _i}} \right)^0} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad {}^5{C_4}{\left( {{\eta _i}} \right)^4}{\left( {1 - {\eta _i}} \right)^1} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {}^5{C_3}{\left( {{\eta _i}} \right)^3}{\left( {1 - {\eta _i}} \right)^2} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {}^5{C_2}{\left( {{\eta _i}} \right)^2}{\left( {1 - {\eta _i}} \right)^3} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {}^5{C_1}{\left( {{\eta _i}} \right)^1}{\left( {1 - {\eta _i}} \right)^4} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {}^5{C_0}{\left( {{\eta _i}} \right)^0}{\left( {1 - {\eta _i}} \right)^5}\end{array}$
and inputting the respective values for ηand 1 – ηi...

$\begin{array}{l}{\left[ {{\eta _i} + \left( {1 - {\eta _i}} \right)} \right]^5} = 1{\left( {0.46} \right)^5}{\left( {0.54} \right)^0} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad 5{\left( {0.46} \right)^4}{\left( {0.54} \right)^1} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 10{\left( {0.46} \right)^3}{\left( {0.54} \right)^2} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 10{\left( {0.46} \right)^2}{\left( {0.54} \right)^3} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 5{\left( {0.46} \right)^1}{\left( {0.54} \right)^4} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 1{\left( {0.46} \right)^0}{\left( {0.54} \right)^5}\\\quad \quad \quad \quad \quad \quad \quad = 0.0205... + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad 0.1208... + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0.2838... + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0.3332... + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0.1955... + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0.0459...\end{array}$
So as you can see, there is therefore only a 2.06% chance that you will receive all the presents you have asked Santa for this year — indeed, it is more than twice as likely that you will receive nothing.  The most likely thing to happen is that you will receive 2 of the 5 presents you have asked Santa for, but there is more than a 50% chance that you will receive 2 or 3 of the presents you have asked for.

I am sharing this analysis in the detail I am with you so as to prepare you for any possible disappointment on Christmas morning.  I must of course emphasise that this is a purely theoretical model and thus that the experimental error that is life may get in the way, and result in you having all of the presents you have asked for, or, indeed, even more.  (Your sister, incidentally, has asked for 4 presents.  I wonder what proportion of her behaviour would need to have been deemed nice in order for her to have a 90% or more chance of receiving at least 3 of them?)

Night-night.

Thursday, 16 November 2017

Problem... The Sophie Germain Identity

#AlgebraicManipulation,  #ExpandingBrackets, #Proof

Wednesday, 15 November 2017

On Proving the Sophie Germain Identity

Marie-Sophie Germain (1776-1831) was, in short, an extraordinary mathematician.  Noted for her work across a range of disciplines, from number theory to physics and astronomy, she is widely recognised as one of the first women to make significant and original contributions to mathematical research.

Germain's story is, in more ways than one, a remarkable one, and it is not hyperbole to describe her achievements, whether framed within the context of time and place or not, as inspirational. Forbidden to attend University and thus unable to make a career from her mathematics, Germain, with the financial and moral support of her father, famously taught herself.  She was inspired by Jean-Étienne Montucla's Histoire des Mathématiques, read Newton and Euler, and, under the nom de plume of a Monsieur LeBlanc, corresponded with preeminent mathematicians of the time, most notably Adrien-Marie LegendreJoseph-Louis Lagrange and Carl Friedrich Gauss.  It was through her persistence, resilience, ingenuity, inquisitiveness and most importantly the sheer, outrageous brilliance of her enduring work, that she overcame — to an arguable exten— the prejudices of 19th-century French society that threatened to impede her way in mathematics.

Germain was inspired by Ernst Chladni's beautiful, awe-inspiring 'figures' to formulate a mathematical theory of elastic surfaces that described Chladni's experimental observations, eventually publishing Récherches sur la théorie des surfaces élastiques in 1821 (see Gardner's and Kowk's fascinating paper), and in doing so winning the prix extraordinaire from the esteemed Académie des sciences.  She was also one of a select number of mathematicians to have worked on Fermat's Last theorem, including 'Euler, Legendre, Gauss, Abel, Dirichlet, Kummer and Cauchy' (Dickson, 1917), resulting in arguably her most notable achievement, what is now known as the Sophie Germain Theorem (for more detail see herethis typically insightful account from Simon Singhand this from Robin Whitty of theoremoftheday.org, who notes that Germain provided 'the first real breakthrough on Fermat's Last Theorem since Fermat’s death in 1665').

Whilst this post is not about Germain's life or accomplishments per se, it is an advocation — it is about sharing Germain's story with our students, but in a way moreover that achieves more than mere reach, or appreciation, or admiration.  This post is about sharing Germain's story with our students in a way that supports their learning and enhances their mathematical experience.  It is about using narrative to personify, if you like, Germain's mathematics — in an effort to make it more accessible, to make the seemingly difficult be more 'learn-able', to undermine in turn the pervasive (and often persuasive) fear of 'hard maths'.  For all of our students, boy or girl, for the sake of their own mathematical maturation and in the perspectivist sense of the messages we inadvertently transmit as teachers, it is the way in which we tell Germain's story that matters more than the story itself [1].

Germain, along with Emmy Noether, Hypatia of AlexandriaSofia Kovaleskaya and Ada Lovelace, are often the women schools refer — or should that be, if you will forgive me a degree of provocativeness here, defer — to when celebrating the women of mathematics, usually on a perfunctory once or twice a year, such as on the albeit superb Ada Lovelace Day[2].  But of course every day is (or should be) Ada Lovelace Day, and of course the stories of these amazing mathematicians should be told, celebrated, and learned from.  However, to encourage the greatest effect (i.e. deep and lasting learning for our students), we must take care to share these stories at appropriate — i.e. not overly contrived — points in the curriculum, and that when doing so, in the face of such overwhelmingly compelling socio-cultural narratives, we do not lose or dilute the underlying story of their astounding mathematics and mathematical achievements.

As I have elaborated upon in more detail in an earlier post, we must do more as teachers than just not deny (inadvertently or otherwise) the girls we teach access to the positive gender role models they need (see, for example, Dennehy and Dasgupta, 2017Lockwood, 2006Allen and Eby, 2004).  And this is about more than just telling their stories, it is also about considering what it is that makes them positive role models, considering what it was that drove their intellectual achievements.  To this end, we should want all of our students, boy or girl, to be inspired by Germain's mathematics, to be inspired by Germain because of her mathematics and the manner in which she did it, not just by the albeit inspirational fact that her mathematics was done in the face of almost insurmountable odds.

I am suggesting, therefore, that we look for genuine ways of exposing our students to the accessible aspects of Germain's mathematics (and the mathematics of her ilk) as early as possible in their mathematical careers  wherever that is possible, wherever that is doable.  It is a deliberate effort, in other words, to expose students to what they will perceive to be ‘hard maths’ earlier than is perhaps typical, when they may not ordinarily be considered 'ready', to increase therefore the opportunities for ‘inspiration’ and support our students' mathematical maturation, to sail against the prevailing cultural and pedagogical winds and make students begin to see themselves as mathematicians in the making.  Exploring Sophie Germain primes when working on properties of number, for example, or, indeed, when working on algebraic manipulation, this:

Prove the Sophie Germain Identity
${a^4} + 4{b^4} \equiv \left( {{a^2} + 2{b^2} + 2ab} \right)\left( {{a^2} + 2{b^2} - 2ab} \right)$

The outline presented as follows is a ‘thinking through’ of how this problem could be approached in a way to make it accessible to secondary students yet to have formally learned how to factorise expressions.  The problem is intended, thus, to be shared and explored with students before they typically learn the relatively more advanced techniques that would normally (and more efficiently) be used, i.e. completing the square and factorising by finding the difference between two squares.  It is intended in other words for use and exploration with secondary students who have some knowledge of simplifying and manipulating algebraic expressions by collecting like terms, multiplying a single term over a bracket, expanding products of two binomials, and the multiplication index law.

The idea behind this exercise is to consider with students how they could prove the identity, using their existing knowledge, without the direct verification of the identity that they are likely to suggest — without, in other words, ‘going that way’, multiplying out the right-hand side and simplifying (which should still of course be applauded and reinforced as a sound proof).  This then situates students at the limits of their existing knowledge and frames their approaches within a more ‘problem solving’ context.  Students’ understanding of the techniques they have already learned — or are in the process of learning — should deepen as a result of working with this problem.

The accompanying text is intended purely to help teachers consider how to approach the problem with students, structurally [3] and pedagogically, and consider what are the questions that can/should be asked of their students, should they not yet be able to ask them for themselves. (This from Brilliant.org could be used to illustrate how the identity is often used in modern mathematics, and  thus perhaps lead onto other areas of learning.)