## Friday, 9 March 2018

### On π Day

March 14th every year is $$\pi$$ day, because it is the 3rd month of the year, and 14th day of that month, and in the mm/dd format this is written as 3/14, or 3.14, which is $$\pi$$ correct to two decimal places (or three significant figures) — meaning, of course, that the day should be more accurately referred to as Pi-To-Two-Decimal-Places day, or Pi-To-Three-Significant-Figures day.  It is, nevertheless, a day for schools to unashamedly revel in a pure celebration of $$\pi$$ and, by implication and association, mathematics.

I offer in this post therefore a few little somethings that teachers may wish to share with their students, a few little somethings moreover that teachers may want to delve deeper into and thus take further with their students, marking $$\pi$$ day in the spirit of celebration that it promotes.  In addition, and in complement, schools and teachers may wish to carry out some Random Acts of Maths (download them here) over the course of the day, and/or hold a Favourite Number Election (find suggestions for use and download a ballot paper here).  You might also find some inspiration from the #PiDay2018 Twitter highlights here.

Archimedes' Constant

Archimedes (c. 287-212 BC) was the first to theoretically calculate $$\pi$$ — i.e. the ratio of a circle's circumference to its diameter (he never used the symbol) — rather than estimate it.  In proposition three of his ‘Measurement of the Circle’, Archimedes used a geometrical approach in which he repeatedly enclosed a circle by circumscribing it (i.e. constructing a regular polygon outside the circle) and by inscribing it (i.e. constructing a regular polygon inside the circle).

Starting with a regular hexagon, Archimedes progressively doubled the number of sides of the polygon in order to approximate a circle with increasing accuracy (from 6 to 96 sides).  He was thus able to prove that:

$\frac{{223}}{{71}} < \pi < \frac{{22}}{7}$
Or, in mixed numbers:
$3\frac{{10}}{{71}} < \pi < 3\frac{1}{7}$
Or, to six decimal places:
$3.140845 < \pi < 3.\dot 14285\dot 7$
This 'method of exhaustion' continued to be used for many years, until the invention of calculus and the use of arctangents allowed us to find $$\pi$$ to ever increasing numbers of decimal places ($$\pi$$ was found to 100 decimal places in such a way by John Machin, as published in William Jones' 1706 book Synopsis palmariorum matheseos).  The Archimedean 'method of exhaustion' was used most notably by German mathematician Ludoph van Ceulen in the 16th Century, who reached a regular polygon with 4,611,686,018,427,387,904 sides to arrive at a value of $$\pi$$ correct to 35 digits, calculated, of course, by hand.

Following Archimedes' approach with younger secondary students can influence their perception of mathematics, of what mathematics is, profoundly.  For more on how the Archimedes' upper and lower bounds for $$\pi$$ were found, see this video (17min).  With this interactive demonstration from the Wolfram Demonstrations Project, students can explore how the upper and lower bounds for $$\pi$$ change as the number of sides to the polygons change.

The origin of the symbol for $$\pi$$ as $$C/d$$

$$\pi$$ was first used to denote the ratio of a circle's circumference to its diameter by Welsh mathematician William Jones in his 1706 book Synopsis palmariorum matheseos.  In the same book, as mentioned above, $$\pi$$ was shown to 100 decimal places, as 'Computed', in Jones' words, 'by the Accurate and Ready Pen of the Truly Ingenious Mr. John Machin', using what has since become known as 'Machin's formula'.  (See this this post for a little more on Jones' notation.)

You can peruse the book electronically here, finding π (as C/d) for the first recorded time in history on p243, as extracted in the first image below, then more explicitly on p263, as extracted in the second image below.

$$\pi$$ formulae

Many awe-inspiring formulae have been found for $$\pi$$, in exact or approximate, iterative forms.  Below is a small selection of some classics that may be interesting to explore with students, possibly comparing the relative accuracy of respective formulae and their relative rates of convergence.

Viète's formula.  François Viète published in 1593 what is commonly accepted to be the first instance of an infinite product known in mathematics.  It is derived by comparing the areas of regular polygons with 2n and 2n + 1 sides inscribed in a circle, and converges to $$\pi$$ fairly rapidly.  (Play with Viète's formula with this interactive Wolfram demonstration.)

$\pi = 2 \times \frac{2}{{\sqrt 2 }} \times \frac{2}{{\sqrt {2 + \sqrt 2 } }} \times \frac{2}{{\sqrt {2 + \sqrt {2 + \sqrt 2 } } }} \times \frac{2}{{\sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt 2 } } } }} \times ...$

Wallis' product.  John Wallis published this formula in his 1656 Arithmetica Infinitorum.  Recently, Wallis' product was 'discovered,' quite incredibly, 'hidden in [the] hydrogen atom', as shared with the world through this paper — 'Quantum mechanical derivation of the Wallis formula for pi' — from the Journal of mathematical Physics, revealing a hitherto unknown and beautiful connection between mathematics and Physics.  (Visualise Wallis' sieve approximation for $$\pi$$ with this interactive Wolfram demonstration.)

$\frac{\pi }{2} = \prod\limits_{n = 1}^\infty {\left( {\frac{{2n}}{{2n - 1}} \times \frac{{2n}}{{2n + 1}}} \right)}$Giving
$\frac{\pi }{2} = \frac{2}{1} \times \frac{2}{3} \times \frac{4}{3} \times \frac{4}{5} \times \frac{6}{5} \times \frac{6}{7} \times ...$

Newton's approximation.  Sir Isaac Newton derived this formula in 1666, using it to calculate pi to 16 places 'using 22 terms of the series' (in Pickover, 2008).  In recognition of his obsessive calculations in the face of pi's irrationality, Newton wrote, 'I am ashamed to tell you to how many figures I carried these computations, having no other business at the time'.  (Play with Newton's approximation using this interactive Wolfram Demonstration.)
$\pi \approx \frac{3}{4}\sqrt 3 + 24\int\limits_0^{\frac{1}{4}} {\sqrt {x - {x^2}} } dx$giving
$\pi \approx \frac{3}{4}\sqrt 3 + 24\left( {\frac{1}{{12}} - \frac{1}{{5 \times {2^5}}} - \frac{1}{{28 \times {2^7}}} - \frac{1}{{72 \times {2^9}}} - ...} \right)$

Machin's formula.  Published in William Jones' 1706 book Synopsis palmariorum matheseos, the same book where $$\pi$$ was first used to denote the ratio of a circle's circumference to its diameter, Machin's rapidly converging series allowed $$\pi$$ to be computed to 100 decimal places.  (To see how Machin's formula converges to $$\pi$$, play with this interactive Wolfram Demonstration.)

$\frac{\pi }{4} = 4{\tan ^{ - 1}}\left( {\frac{1}{5}} \right) - {\tan ^{ - 1}}\left( {\frac{1}{{239}}} \right)$Or,
$\frac{\pi }{4} = 4{\cot ^{ - 1}}\left( 5 \right) - {\cot ^{ - 1}}\left( {239} \right)$

The Basel Problem.  A famous problem (as outlined here by Marianne Freiberger in Plus magazine) named after the home town of the renowned Bernoullis who worked on it, and famously solved by the great Leonarhd Euler in 1735, giving the surprising $$\pi$$-related result (see this accessible outline by Chris Sangwin).  Watch this superb video, from Grant Sanderson with Ben Hambrecht, giving a new take and beautiful proof of the problem, using light!
$\frac{{{\pi ^2}}}{6} = \frac{1}{{{1^2}}} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ...$

$$\pi$$ birthdays

At some point on $$\pi$$ day:
• A 3 year old born on 21 January 2015 will be $$\pi$$ years old.
• A 9 year old born on 30 April 2008 will be $${\pi ^2}$$ — or $$\pi$$$$\pi$$ — years old.
• A 12 year old born on 19 August 2005 will be 4$$\pi$$ — or $$\left\lceil \pi \right\rceil \pi$$ — years old.
• A 15 year old born on 28 June 2002 will be 5$$\pi$$ — or $$\left\lceil \pi \right\rceil \pi$$ + $$\pi$$ — years old.
• A 36 year old born on 26 September 1981 will be $${\pi ^\pi }$$ years old.
• Including Serena Williams, American tennis player with the most Grand Slam wins of any player in the Open Era, and Christina Milian, American singer, songwriter and actress.
If you teach in a primary, make sure to find students in your school born on 30 April 2008 (Y5) and make a fuss of their $$\pi$$$$\pi$$ birthday.  They may not know what $$\pi$$ is yet, but this is a lovely way to sew a seed of interest and give them something to take home and surprise their parents with.  If you teach in a secondary, make sure to find students in your school born on 19 August 2005 (Y8) and 28 June 2002 (Y11) and make a fuss of their 4$$\pi$$ and 5$$\pi$$ — or their $$\left\lceil \pi \right\rceil \pi$$ and $$\left\lceil \pi \right\rceil \pi$$ + $$\pi$$ — birthdays respectively.  And of course, if you have any 36 year old colleagues born on 26 September 1981, make sure you make a fuss of their $${\pi ^\pi }$$ birthdays!  (See here for my post with a calendar for the whole of 2018 outlining pi-related birthdays.)

$$\pi$$-day birthdays

The fact that Albert Einstein was born on $$\pi$$-day 1879 in Ulm, Germany, is well-known, and rightly celebrated.  Less known, however, and less celebrated, is the fact that the renowned Polish mathematician, Wacław Sierpiński, was born on $$\pi$$-day just $$\left\lceil \pi \right\rceil \pi$$ years later in 1882 in Warszawa, Poland.

Sierpiński made groundbreaking contributions to set theory, number theory, functions and topology, and has a number of well-known fractal objects named after him that are interesting to explore with students when exploring iterative procedures and nth term generalisations, as well as, of course, when playing around with the idea of infinity.  Three of these Sierpiński fractal objects are the Sierpiński carpet, the Sierpiński triangle (also referred to as the Sierpiński sieve or gasket), and the Sierpiński curve (including the Sierpiński arrowhead curve).

The first seven iterations of the Sierpiński carpet and Sierpiński triangle are shown in the images below.  An interesting problem to consider with students is to find the area of the carpet and triangle after 1, 2, 3, ...n iterations.

Taking the carpet above as an example, assuming the original square has a side length of 1 and thus an area of 1, after the 1st iteration the shaded area is $${\raise0.5ex\hbox{\scriptstyle 8} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 9}}$$ of 1.  Visualising this 1st iteration as being made up of 8 smaller shaded squares, i.e. each with a side length of $${\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 3}}$$, we can see that the area of each of these smaller squares after the subsequent 2nd iteration is $${\raise0.5ex\hbox{\scriptstyle 8} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 9}}$$ of $${\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 9}}$$, but there are 8 of these squares, thus meaning that after the second iteration the area of the original starting square that is shaded is $${\raise0.5ex\hbox{\scriptstyle 8} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 9}} \times {\raise0.5ex\hbox{\scriptstyle 8} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 9}} = {\left( {{\raise0.5ex\hbox{\scriptstyle 8} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 9}}} \right)^2}$$.  And so on, such that after iterations, the area of the original square $${A_n}$$ is

${A_n} = {\left( {\frac{8}{9}} \right)^n}$
Students could ponder what the area of the carpet is exactly, when $$n = \infty$$, i.e. after an infinite number of iterations (this problem was solved in 2012 by Yaroslav Sergeyev, as revealed in this paper).

The greatest $$\pi$$-day birthday in history

In 1592, it is possible (probable perhaps) that someone was $$\pi$$ years old, at $$\pi$$ o'clock precisely, on what we now know as $$\pi$$ day.  Or to put it another way, it is likely that at least one person was 3.141592653... years old on 3/14/1592 (mm/dd form) at 3:14am and 15.92653... seconds (or at 3.141592653... am).  So, when was this person born?

(Click here for the problem download, a solution think through, and the name of someone who would have been $$\pi$$ years old at some point on $$\pi$$ day in 1592.)

$$\pi$$ o'clock

You might like to try out this popular problem from my blog with students.  The clock face implies for students what is meant by $$\pi$$ o’clock in the problem, but perhaps it could mean something else, e.g. 3.141592... may suggest 3 hours and 0.141519... of an hour, rather than the 3:14 and 15.92 seconds the clock implies.  How accurate could students calculate the angle, i.e. will they use $$\pi$$ to 4, 6, 8, 10... decimal places?

$$\pi$$ is the 73rd day of the year

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 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.)  73 is also the number that marks when English-speaking children have learned the rules of counting sufficiently to overcome the cognitive need for memorisation, thus implying that once you can count to 73 in English, you can count forever (see this post for more detail about 73).

$$\pi$$ primes

A pi-prime (sequence A005042 in the OEIS) is a prime number made up of the initial digits of the decimal expansion of $$\pi$$.  To date we have found the first four pi-primes (sequence A060421), with another four found to be probable.  The first four pi-primes are:

$\begin{array}{l}3,\\31,\\314159,\\31415926535897932384626433832795028841\end{array}$
The fifth, a probable prime, 16,208 digits long, took four and a half months to compute:

3141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609433057270365759591953092186117381932611793105118548074462379962749567351885752724891227938183011949129833673362440656643086021394946395224737190702179860943702770539217176293176752384674818467669405132000568127145263560827785771342757789609173637178721468440901224953430146549585371050792279689258923542019956112129021960864034418159813629774771309960518707211349999998372978049951059731732816096318595024459455346908302642522308253344685035261931188171010003137838752886587533208381420617177669147303598253490428755468731159562863882353787593751957781857780532171226806613001927876611195909216420198938095257201065485863278865936153381827968230301952035301852968995773622599413891249721775283479131515574857242454150695950829533116861727855889075098381754637464939319255060400927701671139009848824012858361603563707660104710181942955596198946767837449448255379774726847104047534646208046684259069491293313677028989152104752162056966024058038150193511253382430035587640247496473263914199272604269922796782354781636009341721641219924586315030286182974555706749838505494588586926995690927210797509302955321165344987202755960236480665499119881834797753566369807426542527862551818417574672890977772793800081647060016145249192173217214772350141441973568548161361157352552133475741849468438523323907394143334547762416862518983569485562099219222184272550254256887671790494601653466804988627232791786085784383827967976681454100953883786360950680064225125205117392984896084128488626945604241965285022210661186306744278622039194945047123713786960956364371917287467764657573962413890865832645995813390478027590099465764078951269468398352595709825822620522489407726719478268482601476990902640136394437455305068203496252451749399651431429809190659250937221696461515709858387410597885959772975498930161753928468138268683868942774155991855925245953959431049972524680845987273644695848653836736222626099124608051243884390451244136549762780797715691435997700129616089441694868555848406353422072225828488648158456028506016842739452267467678895252138522549954666727823986456596116354886230577456498035593634568174324112515076069479451096596094025228879710893145669136867228748940560101503308617928680920874760917824938589009714909675985261365549781893129784821682998948722658804857564014270477555132379641451523746234364542858444795265867821051141354735739523113427166102135969536231442952484937187110145765403590279934403742007310578539062198387447808478489683321445713868751943506430218453191048481005370614680674919278191197939952061419663428754440643745123718192179998391015919561814675142691239748940907186494231961567945208095146550225231603881930142093762137855956638937787083039069792077346722182562599661501421503068038447734549202605414665925201497442850732518666002132434088190710486331734649651453905796268561005508106658796998163574736384052571459102897064140110971206280439039759515677157700420337869936007230558763176359421873125147120532928191826186125867321579198414848829164470609575270695722091756711672291098169091528017350671274858322287183520935396572512108357915136988209144421006751033467110314126711136990865851639831501970165151168517143765761835155650884909989859982387345528331635507647918535893226185489632132933089857064204675259070915481416549859461637180270981994309924488957571282890592323326097299712084433573265489382391193259746366730583604142813883032038249037589852437441702913276561809377344403070746921120191302033038019762110110044929321516084244485963766983895228684783123552658213144957685726243344189303968642624341077322697802807318915441101044682325271620105265227211166039666557309254711055785376346682065310989652691862056476931257058635662018558100729360659876486117910453348850346113657686753249441668039626579787718556084552965412665408530614344431858676975145661406800700237877659134401712749470420562230538994561314071127000407854733269939081454664645880797270826683063432858785698305235808933065757406795457163775254202114955761581400250126228594130216471550979259230990796547376125517656751357517829666454779174501129961489030463994713296210734043751895735961458901938971311179042978285647503203198691514028708085990480109412147221317947647772622414254854540332157185306142288137585043063321751829798662237172159160771669254748738986654949450114654062843366393790039769265672146385306736096571209180763832716641627488880078692560290228472104031721186082041900042296617119637792133757511495950156604963186294726547364252308177036751590673502350728354056704038674351362222477158915049530984448933309634087807693259939780541934144737744184263129860809988868741326047215695162396586457302163159819319516735381297416772947867242292465436680098067692823828068996400482435403701416314965897940924323789690706977942236250822168895738379862300159377647165122893578601588161755782973523344604281512627203734314653197777416031990665541876397929334419521541341899485444734567383162499341913181480927777103863877343177207545654532207770921201905166096280490926360197598828161332316663652861932668633606273567630354477628035045077723554710585954870279081435624014517180624643626794561275318134078330336254232783944975382437205835311477119926063813346776879695970309833913077109870408591337464144282277263465947047458784778720192771528073176790770715721344473060570073349243693113835049316312840425121925651798069411352801314701304781643788518529092854520116583934196562134914341595625865865570552690496520985803385072242648293972858478316305777756068887644624824685792603953527734803048029005876075825104747091643961362676044925627420420832085661190625454337213153595845068772460290161876679524061634252257719542916299193064553779914037340432875262888963995879475729174642635745525407909145135711136941091193932519107602082520261879853188770584297259167781314969900901921169717372784768472686084900337702424291651300500516832336435038951702989392233451722013812806965011784408745196012122859937162313017114448464090389064495444006198690754851602632750529834918740786680881833851022833450850486082503930213321971551843063545500766828294930413776552793975175461395398468339363830474611996653858153842056853386218672523340283087112328278921250771262946322956398989893582116745627010218356462201349671518819097303811980049734072396103685406643193950979019069963955245300545058068550195673022921913933918568034490398205955100226353536192041994745538593810234395544959778377902374216172711172364343543947822181852862408514006660443325888569867054315470696574745855033232334210730154594051655379068662733379958511562578432298827372319898757141595781119635833005940873068121602876496286744604774649159950549737425626901049037781986835938146574126804925648798556145372347867330390468838343634655379498641927056387293174872332083760112302991136793862708943879936201629515413371424892830722012690147546684765357616477379467520049075715552781965362132392640616013635815590742202020318727760527721900556148425551879253034351398442532234157623361064250639049750086562710953591946589751413103482276930624743536325691607815478181152843667957061108615331504452127473924544945423682886061340841486377670096120715124914043027253860764823634143346235189757664521641376796903149501910857598442391986291642193994907236234646844117394032659184044378051333894525742399508296591228508555821572503107125701266830240292952522011872676756220415420516184163484756516999811614101002996078386909291603028840026910414079288621507842451670908700069928212066041837180653556725253256753286129104248776182582976515795984703562226293486003415872298053498965022629174878820273420922224533985626476691490556284250391275771028402799806636582548892648802545661017296702664076559042909945681506526530537182941270336931378517860904070866711496558343434769338578171138645587367812301458768712660348913909562009939361031029161615288138437909904231747336394804575931493140529763475748119356709110137751721008031559024853090669203767192203322909433467685142214477379393751703443661991040337511173547191855046449026365512816228824462575916333039107225383742182140883508657391771509682887478265699599574490661758344137522397096834080053559849175417381883999446974867626551658276584835884531427756879002909517028352971634456212964043523117600665101241200659755851276178583829204197484423608007193045761893234922927965019875187212726750798125547095890455635792122103334669749923563025494780249011419521238281530911407907386025152274299581807247162591668545133312394804947079119153267343028244186041426363954800044800267049624820179289647669758318327131425170296923488962766844032326092752496035799646925650493681836090032380929345958897069536534940603402166544375589004563288225054525564056448246515187547119621844396582533754388569094113031509526179378002974120766514793942590298969594699556576121865619673378623625612521632086286922210327488921865436480229678070576561514463204692790682120738837781423356282360896320806822246801224826117718589638140918390367367222088832151375560037279839400415297002878307667094447456013455641725437090697939612257142989467154357846878861444581231459357198492252847160504922124247014121478057345510500801908699603302763478708108175450119307141223390866393833952942578690507643100638351983438934159613185434754649556978103829309716465143840700707360411237359984345225161050702705623526601276484830840761183013052793205427462865403603674532865105706587488225698157936789766974220575059683440869735020141020672358502007245225632651341055924019027421624843914035998953539459094407046912091409387001264560016237428802109276457931065792295524988727584610126483699989225695968815920560010165525637567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Play the $$\pi$$-day Lottery

This is (part of) the UK Lotto lottery ticket I have bought for the 2018 $$\pi$$-day draw:

As you can see, the first set of numbers are the first twelve digits of $$\pi$$ {31, 41, 59, 26, 53, 58}, and, moreover, the maximum number of digits from the start of $$\pi$$ that can be used in the UK Lotto (six numbers from 1 to 59 are drawn).  The second set of numbers are the numbers 1 to 6 {1, 2, 3, 4, 5, 6}.  The third set of numbers are a randomly generated set {18, 58, 41, 11, 10, 31}.

Suggested explorations/diversions with/for students:
• Find the probability of each separate set of numbers winning.
• Assuming one of the sets of numbers drawn in full, and thus wins the jackpot, which set would you expect to win the most money from?
• Whilst betting on the numbers 1, 2, 3, 4, 5 and 6 does not reduce your probability of winning, it does reduce the amount of money you are likely to win.  This is because, perhaps surprisingly, the numbers 1, 2, 3, 4, 5 and 6 are selected by around 10,000 people each week, thus meaning that the winning jackpot would be shared by 10,000.
• Similarly, maybe a few mathsy people will buy a ticket with the first set of numbers on $$\pi$$-day, therefore reducing the amount of money you would win if drawn because of the shared jackpot.
• Find the combinations of all 'wins' you could possibly make with all three sets of numbers.
• You 'win' —  relatively of course —  in the Lotto if you match 6 numbers (Jackpot), or if you match 5 and the 'Bonus Ball', or if you match 5, 4, 3, or 2.  (Visit the Lotto website for more detail.
• What is the least amount of initial digits from $$\pi$$ that can be used in the UK Lotto?
• Find all the combinations of initial $$\pi$$ digits that could buy a lottery ticket in the UK Lotto.
• Find the probability that a string of the initial digits of $$\pi$$ will win the Lottery.

$$\pi$$ as the Prime Counting Function

In 1909 Edmund Landau used the symbol $$\pi$$ to describe Prime Counting Function $$\pi \left( x \right)$$, which gives the number of primes not exceeding a given number $$x$$.  $$\pi$$ used in this sense has nothing to do with $$\pi$$ used in Jones' sense, i.e. $$\pi$$ as Archimedes' constant, $$\pi$$ as the ratio of a circle's circumference to its diameter.  By way of illustration, $$\pi \left( 7 \right) = 4$$ because the number of primes less than or equal to the number 7 is 4 (i.e. 2, 3, 5 and 7).

Suggested explorations/diversions with/for students:
• What is the largest n you can find the value of $$\pi \left( n \right)$$ for
• You may want to use the 'Sieve of Eratosthenes'
• The largest n for which a value of $$\pi \left( n \right)$$ has been computed (by Staple 2015, as part of his Masters' degree) is $$\pi \left( {{{10}^{25}}} \right)$$ = 1,699,246,750,872,437,141,327,603, which took 40,000 computing hours to find.
• Find the first five Ramanujan primes.
• A Ramanujan prime is the smallest number $${R_n}$$ such that $$\pi \left( x \right) - \pi \left( {x/2} \right) \ge n$$ for all $$x \ge {R_n}$$.

Fake proofs for $$\pi$$

These two now well-known fake 'proofs' are always interesting to share with students, in a 'surely this can't be right' sense.  Firstly, this 'proof' that $$\pi$$ = 4 always engages students visually, and is a nice conceit when thinking about gradients and derivatives (play with this Wolfram Demonstration for $$\pi$$ = 4).  Secondly, this following proof — which, on the face of it, seems fine to many — is a nice conceit when working with students on algebraic manipulation.

\begin{array}{c}\begin{align}x &= \frac{{\pi + 3}}{2}\\2x &= \pi + 3\\2x\left( {\pi - 3} \right) &= \left( {\pi + 3} \right)\left( {\pi - 3} \right)\\2\pi x - 6x &= {\pi ^2} - 3\pi + 3\pi - 9\\2\pi x - 6x &= {\pi ^2} - 9\\9 - 6x &= {\pi ^2} - 2\pi x\\9 - 6x + {x^2} &= {\pi ^2} - 2\pi x + {x^2}\\{\left( {3 - x} \right)^2} &= {\left( {\pi - x} \right)^2}\\3 - x &= \pi - x\\3 &= \pi \end{align}\end{array}
Suggested explorations/diversions with/for students:
• What is wrong with the 'proof' above?
• What happens when you 'fix' it?
• In this case, the problem comes from moving from $${\left( {3 - x} \right)^2} = {\left( {\pi - x} \right)^2}$$ to $$3 - x = \pi - x$$.  The 'proof' only provides the positive square root, which produces  the erroneous result.
• Make up you own fake $$\pi$$ proof.

Memorising $$\pi$$

Daniel Tammet, essayist and novelist, is described as an autistic savant.  Born with high-functioning autism, Daniel famously memorised and recited $$\pi$$ to 22,514 places on $$\pi$$-day in 2004, taking him just over five hours to do so.  This clip, lasting just over five minutes, is an excerpt from a longer documentary on Daniel, 'The Boy With The Incredible Brain'.

For more of Daniel:

Suresh Kumar Sharma of India recited 70,030 digits of pi on 21 October 2015, taking 17 hours and 14 minutes to do so.  See this article about the 'secret' to memorizing pi to such huge amounts of digits.  And see this article by @alexbellos in The Guardian about Akira Haraguch, who holds the unofficial World Record having recited pi to 100,000 digits in October 2006, over 16 hours.  It is fun to hold a class or school competition, linked to the 'self-referential stories' below perhaps, to see which student can recite $$\pi$$ to the greatest amount of decimal places.

Self-referential stories for $$\pi$$

Arguably the best known $$\pi$$ mnemonic, constructed by British astronomer Sir James Jeans, is: "How I want a drink, alcoholic of course, after the heavy chapters involving quantum mechanics", the number of letters in each word corresponding to the respective digit in the decimal expansion of  $$\pi$$.  This from Michael Keith in 1986 gives $$\pi$$ to 356 places.

The Music of $$\pi$$

This song, 'Pi', by Kate Bush from her 2005 album 'Aerial', is an ode to a 'sweet and gentle and sensitive man, with an obsessive nature and deep fascination for numbers, and a complete infatuation for the calculation of pi'.  Throughout the song Kate sings this number:

3.14159265358979323846264338327950288419716939937510582319749445923078164062862088214808651328230664709384460955058223

You will note, however, that whilst the 55th digit of π is 0, Kate sings 3 and then 1, before getting back on track and singing the next 24 digits correctly.  She then, however, completely misses out the next 22 digits of π before singing the next 37 digits correctly.

3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223

I wonder if she sings the song live?

(film, audio, websites, applets, articles, papers)

## Wednesday, 28 February 2018

### On A Mathsy St David's Day

Q: What links Bertrand Russell, the equals sign, the word in the English language with the most Zs, the use of π as the ratio of a circle's circumference to its diameter, the Online Encyclopedia of Integer Sequences, and the number 36?

A: Wales.

By way of explanation, and in mathsy celebration of St David's Day, I offer below a handful of mathematical somethings with a Welsh bent, peppered with one or two suggested explorations or diversions that teachers may wish to share and use with students on the day or, indeed, any other.

Dydd Gŵyl Dewi Hapus.

The Equals Sign

Welsh Doctor and mathematician Robert Recorde, born in 1510 in Tenby, Pembrokeshire, South Wales, was a popular author of a number of mathematical books — which he wrote, unusually for the time, in the English vernacular, thus making his writing more accessible than most scholarly books of the age, which were usually written in Latin.

In his 1557 book, The Whetstone of Witte, whiche is the seconde parte of Arithmetike: containyng thextraction of Rootes: The Cossike practise, with the rule of Equation: and the woorkes of Surde Nombers [1] (which you can peruse electronically here), Recorde 'invented' the '=' symbol, 'to avoide the tediouse repetition of these woordes : is equalle to :', which he had already used some 200 times in the book [2].

During Recorde's time, much of the mathematical notation we take for granted today was not yet in use.  In designing the symbol '=' — 'a paire of paralleles, or Gemowe [twin] lines of one length, thus: =====, bicause noe 2 thynges can be moare equalle' — Recorde's initial motivation to abbreviate was quickly overtaken by something more profound, more enduring.  As Joseph Mazur (2014) eloquently puts it, 'the concise character of the symbol came with an unintended benefit: it enabled an unadorned picture in the brain that could facilitate comprehension'.

Recorde was also, for example, and in the same book, the first to use the plus and minus signs in English: 'There be other 2 signes in often use of which the first is made thus + and betokeneth more: the other is thus made – and betokeneth lesse'.  And neither was there an easy way in the 16th century of denoting the powers of numbers, so Recorde coined the now unsurprisingly obsolete term ‘zenzizenzizenzic' [3] to ‘doeth represent the square of squares squaredly’, or in other words to denote the square of the square of a number's square:

${\left( {{{\left( {{n^2}} \right)}^2}} \right)^2} = {n^8}$
Recorde also used another word (which didn't quite catch on), the 'sursolid', meaning to be raised to a prime number greater than three.  So a power of five would be the first sursolid, a power of seven the second sursolid, a power of eleven the third, and so on.

Suggested explorations/diversions with/for students:
• Find the zenzizenzizenzic of n for 0 < n < 10.
• Devise questions in Recordian notation and answer them, for example: 'What is the fourth sursolid of two divided by the zenzizenzizenic of two?  Give your answer in modern and Recordian form'.
• Consider and explore the difference between:

${\left( {{{\left( {{n^2}} \right)}^2}} \right)^2}\;{\rm{and}}\;{n^{{2^{{2^2}}}}}$

The first use of π to denote C/d

Before being denoted π, the ratio of the circumference of a circle to its diameter was referred to typically in the Latin 'quantitas in quam cum multiflicetur diameter, proveniet circumferencia ('the quantity which, when the diameter is multiplied by it, yields the circumference)' (Rothman, 2009).

Welsh mathematician William Jones, born in 1675 in Llanfihangel Tre'r Beirdd, on the Isle of Anglesey, North Wales, was the first person to use π to denote the ratio of a circle's circumference to its diameter, doing so in his 1706 book Synopsis palmariorum matheseos: or, A new introduction to mathematics: containing the principles of arithmetic & geometry demonstrated, in a short and easie method; with their application to the most useful parts thereof ... Design'd for the benefit, and adapted to the capacities of beginners [4].  You can peruse the book electronically here [5].

This first ever appearance of π denoting the ratio of a circle's circumference to its diameter can be seen on p243, then more explicitly on p263, as excerpted below.  It can also be seen that Jones gave π correct to 100 decimal places, 'as Computed by the Accurate and Ready Pen of the Truly Ingenious Mr. John Machin', using an infinite series whose sum converged to π (see this on 'Machin's Formula' from Peter Rowlett in The Aperiodical).

Using π in this way was a significant philosophical step forwards; Jones was more than merely abbreviating.  Although unable to prove it, Jones recognised that the ratio of a circle's circumference to its diameter could not be expressed as a rational number — or in other words, that π was an irrational number — as can be seen in the p243 excerpt above: 'For as the exact Proportion between the Diameter and the Circumference can never be expres'sd in Numbers' [6].  Jones recognised, as such, that 'to represent an ideal that can be approached but never reached.... only a pure platonic symbol would suffice' (Rothman, 2009).

Jones' use of π as C/d was popularised when the great Swiss mathematician Leonhard Euler adopted it in his Introductio in analysin infinitorum (Introduction to the Analysis of the Infinite) in 1748 [7].

Suggested explorations/diversions with/for students:
• Share the excerpts above with students and try to make sense of them together.
• Have students work out their π-related birthdays, past or future, discussing precision (see this post).
• Note, for example, that at some point on St Davids' Day 2018:
• 3 year-olds born on 8 January 2015, will be π years old.
• 9 year-olds born on 17 April 2008, will be ππ years old.
• 36 year-olds born on 8 January 2015, will be ππ years old.

The OEIS

Mathematician Neil Sloane, born on 10 October 1939 in Beaumaris, North Wales — described by Erica Klarreich as 'The Connoisseur of Number Sequences' in this article in Quanta magazine, and as 'the Guy who Sorts All the World's Numbers in his Attic' in this reprint of the article in Wired — is considered by some to be one of the most influential mathematicians of our time, because in 1964 Sloane founded The Online Encyclopedia of Integer Sequences (OEIS).

The OEIS, as the name suggests — or Sloane, as it is often referred to by its users — is an online database of at the current count, over one quarter of a million integer sequences.  It is designed to be used by researchers in mathematics, but as John Conway and Tim Hsu put it in 2006, 'most Nerds should be able to get some enjoyment out of it'.

Enjoy this short selection of some gems that I first discovered through the OEIS:

And consider this sequence (sequence A168087): a(n) = the smallest number whose Welsh name (masculine or feminine versions) in the modern Decimal System contains n letters of the alphabet.  (For example, 224, dau gant dau ddeg pedwar, is the 20th number in the sequence, and is the smallest number with 20 letters.)

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
1 2 3 7 4 11 12 15 16 14 27 24 47 44 127 124 147 144 244

Suggested explorations/diversions with/for students:
• Try and ascertain the rule that describes each of my selection of 'gems' above.
• For sequence A168087
• Continue the sequence to n = 40.
• Find a(100).
• Do the same for sequence A168085 (using the traditional Vigesimal System).
• Generate the same sequence for numbers in English, and other languages.
• Describe this sequence 4, 2, 3, 3, 6, 4, 6, 5, 4, 3, 3, 8, 5... (sequence A140396), entered into the OEIS by Sloane himself in 2008, perhaps as a nod to his Welsh heritage.
• Generate a sequence of numbers that have the same amount of letters in Welsh as in English.  For example, a(1) = 2, because 'two' in English and 'dau' in Welsh has 3 letters.
• Maybe submit the sequence to the OEIS (it's not there; I've checked) on behalf of a student (with parental consent of course) who generates it and defines it best, according to the OEIS' format.
• Try this puzzle set by Sloane in Quanta Magazine:
• Can you figure out the 'simple' rule that describes this sequence 13, 26, 2, 4, 6, 3, 9, 12, 8, 10, 5, 15, 18, 14, 7, 21, 24, 16, … (click here for the solution when you're ready).
• The image below shows the 'zigzag triangle', via JohnConway and Tim Hsu (2006).  On the LHS of the triangle are the Zig (or secant or Euler) numbers, and on the RHS are the Zag (or tangent) numbers.
• Find the next Zig and Zag numbers.
• How far can you keep going?

The Number 36

Charles Yang, associate professor in the University of Pennsylvania Department of Linguistics, and not Welsh, has shown (2005, cf. 2015) that 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, that do not follow this rule and that must therefore be memorised, then

$e < {\theta _N}\;where\;{\theta _N}: = \frac{N}{{\ln \left( N \right)}}$
This model, dubbed the 'Tolerance Principle', can be applied to how we, as children, learn to count.  In short (see this post, 'On 73'for more detail), by using the Tolerance Principle, we can find the least amount of words that we need to learn ‘to overcome the exceptions we have to memorise’.  In Welsh, in the modern Decimal System of counting, the numbers 1 to 10 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.  Thus, the smallest value of N in Welsh such that θN = 10 is 36:

\begin{array}{l}\begin{align}N &= 36\;\\\because10 &= e < {\theta _N} = \frac{{36}}{{\ln \left( {36} \right)}}\\\;where\;{\rm{ }}\frac{{36}}{{\ln \left( {36} \right)}} &= 10.045991...\end{align}\end{array}
Or in other words, once a child has learned to count to 36 in Welsh, they have learned the rules of the game sufficiently to overcome the cognitive need for memorisation, and thus to keep going.

Bertrand Russell

Bertrand Russell, the mathematician, logician and humanist, was born on 18 May 1872, in Trellech, Monmouthshire.  Russell lived for most of his later years at Plas Penrhyn in Penrhyndeudraeth, Merioneth, North Wales, with a view south to Cardigan Bay and north to the mountains of Eryri (Snowdownia).  He died at Plas Penrhyn on February 2, 1970 (read his obituary in the New York Times here), was cremated at Colwyn Bay and had his ashes scattered over the Welsh hills.

Russell won a scholarship to read mathematics at Trinity College, Cambridge University, and with Alfred North Whitehead wrote his monumental three-volume work, Principia Mathematica, between 1910 and 1913.  ('Logicomix: An Epic Search for Truth', a wondrous graphic novel 'inspired by the epic story of the quest for the Foundations of Mathematics', described Russell and Whitehead's Principia as 'a heroic intellectual adventure.')   In 1950, Russell was made a Nobel Laureate in Literature “in recognition of his varied and significant writings in which he champions humanitarian ideals and freedom of thought”.

Read a short biography of Russell here, and watch his 1959 BBC 'Face to Face' interview with John Freeman here, and/or read the transcript here.

## Monday, 26 February 2018

### On Learning When You're Not Looking #1

Whether or not it is possible to be aware that we don't know something when we are not looking for it, we become only too aware when we find it — too aware that is of our not knowing, or of our neglect in having lost sight of the sense of wonder that comes with finding something out for the first time.

Finding things out when we are not looking has resonance for us, emotionally, intellectually, and formatively, and particularly so I think for teachers, because it is a rediscovery of sorts in ourselves of a sense of naïvety that we see —  rightly or wrongly — in our students, a naïvety moreover imbued with the pedagogical promise that it can be made something of, an inquisitiveness in other words and capacity to be surprised that we justly idealise and cultivate in our students.  The resonance of finding something out when we are not looking, of going places we weren't planning or perhaps prepared to go, is, in short, and to put it another way, a rediscovery of our own curiosity.  And the more we are subject to this resonance, the more we want more of it, the more we want to share it.

To this end, I offer here the first of an occasional series of posts in which I share a few little somethings that have made me wonder how I have got this far in life without either knowing, realising or appreciating more.  They may be titbits of trivia with a mathematical bent that I have stumbled across, or something whose resonance for others has vicariously re-awoken in me the fascination they once first instilled in me.  But most importantly, they have the pedagogical, educative and aspirational capital to exploit in support of our students' mathematical maturation, and to encourage students to begin to see themselves as young mathematicians in the making.

Blockbusters and the Game of Hex

The TV game show Blockbusters, created by prolific puzzle writer Steve Ryan [1], and still often used as a conceit to 'quiz' students in classrooms, was based on the Game of Hex invented by John Nash, the mathematician and winner of the Nobel Prize [2] for economics for his work on game theory, and whose life story was dramatised in the 2004 Oscar winning film 'A Beautiful Mind'.  [Heard from Tim Harford at the start of 'Tell Me Something I Don't Know' podcast episode 15.]

T
The game — Hex, that is, not Blockbusters — was actually first invented (or, arguably, discovered) in 1942 by Piet Hein, a Danish mathematician and polymath described by Martin Gardner as 'surely one of the most remarkable men in Denmark' (2008, p82), while he was working on the four-colour theorem.  The game 'became enormously popular in Denmark under the name of Polygon' (ibid), before John Nash subsequently and independently re-discovered the game in 1948, at Princeton (this deleted scene from 'A Beautiful Mind' is an interesting watch).  The game became popular amongst Princeton's mathematics graduate students who called it "Nash" or "John", the latter in affectionate homage to the hexagonal bathroom tiles that the students often played the game on.  (For more about the game, its history and its mathematics, see this from Wolfram Mathworld, and this from MIT.)

Hex is played on a diamond-shaped board made up of hexagons.  Two players take it in turns to mark a hexagon with their assumed colour, orange and white in the board illustrated above, and win by connecting their opposite sides of the board in an unbroken chain.  The game is a 'determined' game, i.e. a game that can never end in a draw with the first player always being able to force a win on a standard board of any size, as proved reductio ad absurdum by Nash in his 'strategy stealing' argument, as described by Gardner (2008, p85) and exquisitely here by Alexander Bogomolny.

In blocking your opponent's path across the board, you will of course connect your sides of the board, and if you go first you will always have that 'extra' first mark to facilitate the block and thus complete your unbroken chain — in theory.  But whilst Nash proved that on symmetrical boards of any size Hex is a 'first player' win (read an interesting discussion here), he did not show what the strategy is.  Indeed, the first explicit winning strategy (on a 7×7 board) was described in 2002, and as of 2016, by using brute force search computer algorithms, Hex boards up to size 9×9 have been completely solved — but no further.  As Bogomolny puts it, as 'simple as the game appears, no winning strategy has been discovered for bigger boards'.

This 4 minute video from Marc Chamberland of Tipping Point Math gives an excellent overview of the game and its significance.

To play the game  with students (on a seven-by-seven board) this spreadsheet, by James Robinson from @ATMMathematics, and this lovely geogebra file from Steve Phelps both work well — through projection, perhaps on an interactive whiteboard.  For more independent exploratory pencil and paper play, I provide an A4 version of the standard eleven-by-eleven board to download here.

As Gardner states, 'one of the best ways to learn the subtleties of Hex is to play the game on a field with a small number of hexagons' (2008, p84), building up gradually from a two-by-two board of four hexagons.  So after some initial exploratory play perhaps on the seven-by-seven and/or eleven-by-eleven boards provided above, by considering the game on boards of gradually increasing size (available to download in various formats here), students will quickly see that  and appreciate how — the first player wins on the two-by-two and three-by-three boards, but that things get more complicated quickly from then on (and with opportunities therein for deeper learning).  On the four-by-four board, for example, as the maestro explains, 'the first player is sure to win if he immediately occupies any one of the four cells [in the centre column of hexagons]. If he makes his opening play elsewhere, he can always be defeated. An opening play in [the two centre hexagons] insures a win on the fifth move; an opening play in [a hexagon above or below the two centre hexagons], a win on the sixth move.'  Gardner continued, 'the standard 11-by-11 board introduces such an astronomical number of complications that a complete analysis seems beyond the range of human computation', (ibid, p84).

It is interesting also to consider Nash's proof with students, and then to vary the rules of the game and explore with students what happens to their underlying strategy if, for example, the idea of winning is switched so that you win by avoiding winning (i.e. by not connecting your sides).  Or maybe by applying new rules, for example by having each player being allowed to swap positions once during each game.

The Isoperimetric Inequality

The Isoperimetric Inequality states, for the length L of a closed curve and the area A of the planar region that it encloses, that:
$A \le \frac{{{L^2}}}{{4\pi }}$
and that equality holds if and only if the curve is a circle.  [Heard (again) here thanks to @panlepan's nomination to @TeachFMaths 'Favourite Theorems and Conjectures' Twitter #MathOlympiad2018, and subsequently through perusing this by Robert Osserman in the November 1978 Bulletin of the American Mathematical Society]

F
Finding out some years ago that something I had independently and innocuously pondered upon as a youngster, that something I had intuitively understood as a schoolboy had a name, and that it had a name because it had a long history behind it [3], a history moreover of real, elegant and rigorous maths [4], had a profound effect on me — personally as a young learner, but more so I think professionally since I became a teacher.

So I was spiked by a desire to share the idea with students, or, moreover, to engineer a situation through which students may be able to pose and ponder the idea for themselves, without — to put it another way — giving it up up front.  As such I sketched out this problem [5], as shown below, linking the perimeter to the area of a curved 'fidget-spinnery' shape of sorts not typically considered in students' mathematics lessons, and that aims thus to nudge students (with careful questioning) towards wondering whether it is possible for a shape with the same perimeter as a circle's to have a larger area than the circle.

T
The idea is that working through this 'fidget-spinnery' problem with students would grant them the intellectual space to essentially pose the Isoperimetric Problem themselves, inducing the resonance of learning-when-you're-not-looking that in turn provides us as teachers with the opportunity to celebrate and validate students' own problematising (a key characteristic of the mathematically maturing student), by framing it within — and revealing the — history of the problem.

In other words, it is about using the problem to support the ongoing cultivation of a pedagogically empathetic classroom that tries to answer questions students think to ask, or questions we are asking as teachers that students if not love at least like, as Steven Strogatz elucidates beautifully in this article for the American Mathematical Society.

O
Once the problem is 'out there' [6], as it were, students may begin by exploring the area (A) of rectangles with a fixed perimeter (P) — e.g. for P = 20, {1,1,9,9} A = 9... {2,2,8,8} A = 16... {3,3,7,7} A = 21... {4,4,6,6} A = 24... {5,5,5,5} A = 25  before moving on to other shapes, ideally suggested by students themselves (validated or moderated by the teacher) and ultimately perhaps into how the area of an ellipse changes as we vary the lengths of its major (a) and minor (b) axes, whilst keeping the circumference constant and equal to the circumference of a circle (i.e. when a = b).

This provides the opportunity to reveal for students the perhaps surprising, exciting knowledge that whilst we know the area of an ellipse (πab), we have no exact formula for the cirumference of an ellipse, other, of course, than the sum of an infinite series (see this typically helpful consideration from mathsisfun.com).  This in turn would provide us with the 'excuse' to share some of the mathematics of Ramanujan with students, through the use of his approximations for an ellipse's circumference, perhaps:

$p \approx \pi \left[ {3\left( {a + b} \right) - \sqrt {\left( {3a + b} \right)\left( {a + 3b} \right)} } \right]$

Or this:

$p \approx \pi \left( {a + b} \right)\left[ {1 + \frac{{3{{\left( {\frac{{a - b}}{{a + b}}} \right)}^2}}}{{10 + \sqrt {4 - 3{{\left( {\frac{{a - b}}{{a + b}}} \right)}^2}} }}} \right]$

In exploring then how the area of an ellipse changes as a and b vary until a = b, students find respective values for a and b for a constant circumference, meaning that in terms thus of the Key Stage 4 National Curriculum in England — Algebra Programme of Study, 'additional mathematical content... taught to more highly attaining pupils' — students will be 'find[ing] approximate solutions to equations numerically using iteration'.  For example, for a circumference of 20, using Ramanujan's first formula above, when a = 1, b ≈ 4.739916655 and thus A ≈ 14.89088734.  When a = 2, b ≈ 4.173401125 and thus A ≈ 26.22225263.  When a = 3, b ≈ 3.361072807 and thus A ≈ 31.67736492.  When a = b = 3.183098862 and thus A = 31.83098862.

Wang Tiles (or dominoes)

I came across 'Wang Tiles' whilst absentmindedly browsing McSweeney's Internet Tendency (@mcsweeneys) and stumbled across Mark Amundsen's 'Tenth-Graders' Favorite Suggestive Terms from Geometry' from 2008.

The Chinese American mathematician, logician, and philosopher, Hao Wang first proposed a method for deciding an important case of David Hilbert's Entscheidungsproblem (the 'Decision Problem', a challenge posed by David Hilbert in 1928), in a 1961 paper.  This method is known as ‘the domino problem’, and was popularised in Wang’s November 1965 Scientific American article ‘Games, Logic and Computers’.

A Wang tile (or domino) is a unit square whose edges are coloured (or otherwise marked with specific symbols) in various ways, and Wang's problem was to find a procedure for deciding whether an infinite plane can be covered with a set of such tiles so that the colours on adjacent edges match, and with no tile being rotated or reflected.  Wang conjectured that if a set of tiles can cover the infinite plane, then there must be at least one arrangement in which they do so periodically (like wallpaper say), and thus that there would be a procedure for deciding whether a set of tiles can tile the infinite plane or not.  Robert Berger, however, one of Wang's students showed in 1966 that there exists sets of Wang tiles that cover the plane, but only aperiodically [7], and constructed such a set of  over 20,000 tiles (the illustration above shows such a set of 13 Wang tiles [8]).  In 2015 Emmanuel Jeandel and Michael Rao proved 'that there are no aperiodic tile set with less than 11 Wang tiles, and that there is an aperiodic tile set with 11 Wang tiles and 4 colors.'

Wang Tiles are an interesting conceit for exploring ideas of symmetry, combinations and permutations, logic, and algorithmic thinking with students.  One avenue of exploration could come from students creating their own tile sets and exploring how to tile grids of specific sizes.  They could use Jeandel and Rao's astonishing 2015 finding to see what periodic tilings they could make (albeit on a finite plane!) with a set of less than 11 tiles less using 4 colours or less.  They may want to start trivially and systematise their explorations.

Students may also find inspiration in the applications of Wang tiles, and particularly perhaps with reference to texturing in computer graphics.  This excellent introduction to the use of Wang tiles in computer graphics from Miguel Cepero, along with this on Wang Tiles and aperiodical tiling by Graham Shawcross, could provide some ideas and inspiration for work with students on combinations.  This from Alan Wolfe gives a succinct explanation about how to use Wang tiles to generate 'tile based graphics that don't look tiled at all', and this from Kevin Roberge is a fascinating account on using Wang Tiles for computation (e.g. 2 + 2 = 4).  Wolfram's Demonstrations Project has this demonstration, a variation of the idea of Wang Tiles, to generate tiling patterns.