Friday, 9 March 2018

On π Day

March 14th every year is \(\pi \) day[1], 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.

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Archimedes' Constant
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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 \(C/d\)
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\(\pi \) 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
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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 and watch this typically wonderful video from Grant Sanderson providing a 'new proof of the Wallis formula'.)

\[\frac{\pi }{2} = \prod\limits_{n = 1}^\infty  {\left( {\frac{{2n}}{{2n - 1}} \times \frac{{2n}}{{2n + 1}}} \right)} \]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
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At some point on \(\pi \) day 2019:
  • A 3 year old born on 21 January 2016 will be \(\pi \) years old.  
  • A 9 year old born on 30 April 2009 will be \({\pi ^2}\) — or \(\pi \)\(\pi \) — years old.  
  • A 12 year old born on 19 August 2006 will be 4\(\pi \) — or \(\left\lceil \pi  \right\rceil \pi \) — years old.  
  • A 15 year old born on 28 June 2003 will be 5\(\pi \) — or \(\left\lceil \pi  \right\rceil \pi \) + \(\pi \) — years old.  
  • A 36 year old born on 26 September 1982 will be \({\pi ^\pi }\) years old.

    If you teach in a primary, make sure to find students in your school born on 30 April 2009 (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 2006 (Y8) and 28 June 2003 (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 1982, 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
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    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 \({\raise0.5ex\hbox{$\scriptstyle 8$}
    \lower0.25ex\hbox{$\scriptstyle 9$}}\) of 1.  Visualising this 1st iteration as being made up of 8 smaller shaded squares, i.e. each with a side length of \({\raise0.5ex\hbox{$\scriptstyle 1$}
    \lower0.25ex\hbox{$\scriptstyle 3$}}\), we can see that the area of each of these smaller squares after the subsequent 2nd iteration is \({\raise0.5ex\hbox{$\scriptstyle 8$}
    \lower0.25ex\hbox{$\scriptstyle 9$}}\) of \({\raise0.5ex\hbox{$\scriptstyle 1$}
    \lower0.25ex\hbox{$\scriptstyle 9$}}\), 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 \({\raise0.5ex\hbox{$\scriptstyle 8$}
    \lower0.25ex\hbox{$\scriptstyle 9$}} \times {\raise0.5ex\hbox{$\scriptstyle 8$}
    \lower0.25ex\hbox{$\scriptstyle 9$}} = {\left( {{\raise0.5ex\hbox{$\scriptstyle 8$}
    \lower0.25ex\hbox{$\scriptstyle 9$}}} \right)^2}\).  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
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    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
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    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
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    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
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    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:

    The fifth, a probable prime, 16,208 digits long, took four and a half months to compute:


    Play the \(\pi \)-day Lottery
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    This is (part of) the UK Lotto lottery ticket I 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
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    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 \(\pi \) 
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    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 \(\pi \)
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      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 under 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.

      Continued Fractions \(\pi \) 
      Milü — 密率 — the 4th convergent of the continued fraction for π, found by Tsu Ch'ung-Chihin, is the best rational approximation of π with a denominator of 4 digits or fewer, accurate to 6 decimal@places, within 0.000009%.  For more on the beauty of continued fractions and π, the 'most romantic way to represent numbers', see this from Evelyn Lamb.

      Self-referential stories for \(\pi \)
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      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 \(\pi \)
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      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:


      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.


      I wonder if she sings the song live?

      Other links to explore 
      (film, audio, websites, applets, articles, papers)
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        Notes, References & Links: