Tuesday, 14 August 2018

On Descartes Numbers

A number is said to be Perfect if it is equal to the sum of its proper positive factors, i.e. the sum of its factors excluding the number itself.  Summing all of the factors of a number \(n\) — i.e. including the number \(n\) itself — is known as the sum-of-divisors function:

\[\sigma \left( n \right) = \sum\limits_{d\left| n \right.} d \]
A number \(n\) is thus said to be Perfect when \(\sigma \left( n \right)\) = 2n.  For example, the factors of 28 — 1, 2, 4, 7, 14 and 28 — sum to 28 + 28 = 56, thus \(\sigma \left( {28} \right)\) = 2 \(\times \) 28 and so 28 is a Perfect number.  In contrast, and by way of illustration, the factors of 42 — 1, 2, 3, 6, 7, 14, 21 and 42 — sum to 96, or 2.2857... \(\times \) 42, thus \(\sigma \left( {42} \right) \ne\) 2 \(\times \) 42 and so 42 is not a Perfect number.  In fact, 42 is said to be an Abundant number because the sum of its proper positive divisors is greater than the number itself, i.e. \(\sigma \left( n \right) > 2n\).  When the sum of a number's proper positive divisors is less than the number itself, i.e. \(\sigma \left( n \right) < 2n\), the number is said to be Deficient.  For example, the factors of 50 — 1, 2, 5, 10, 25 and 50 — sum to 93, or 1.86 \(\times \) 50, and thus \(\sigma \left( {50} \right)\) < 2 \(\times \) 42.  As such:

\[\begin{array}{l}\sigma \left( n \right) < 2n \Rightarrow n\;{\rm{is}}\;{\rm{deficient}}\\\sigma \left( n \right) = 2n \Rightarrow n\;{\rm{is}}\;{\rm{perfect}}\\\sigma \left( n \right) > 2n \Rightarrow n\;{\rm{is}}\;{\rm{abundant}}\end{array}\]
Perfect numbers have intrigued us since Euclid, who observed via Proposition 36 in Book IX of his Elements that a number of the form (2\(^{p - 1}\))(2\(^p\) − 1) is a perfect number whenever 2\(^p\) − 1 is prime (i.e. what subsequently became known as a Mersenne prime) [1].  The fact that every even perfect number is of this type, i.e. that every even perfect number can be written in the form (2\(^{p - 1}\))(2\(^p\) − 1) whenever 2\(^p\) − 1 is prime, was proposed by René Descartes in his famed correspondence with Marin Mersenne (specifically in his letter to Mersenne of 15 November 1638) [2], and later proven in 1849 by Leonhard Euler [3].  Descartes was, indeed, 'among the first to consider the existence of odd perfect numbers' (Greathouse and Weisstein, 2012).

As of writing, of the fifty perfect numbers that have been found, all are even, and it is not known if there are infinitely many or, indeed, whether any odd perfect numbers exist — although Pascal Ochem and Michaël Rao showed (2012) that no number up to 10\(^{1500}\) is an odd perfect number.

One number has been found, however, that would have been an odd perfect number if only one of its factors was prime rather than a 'spoof prime' (i.e. a composite number wrongly assumed to be prime).  This number — 198,585,576,189 — was found in 1638 by Descartes, documented in his 15 November letter to Mersenne, and is known as a Descartes Number, or, as an 'odd spoof perfect number' (note that the spoof prime in question is 22021, i.e. 19\(^2\) \(\times \) 61).

\[198,585,576,189 = {3^2} \times {7^2} \times {11^2} \times {13^2} \times 22021\]
Descartes showed that if 22021 was prime, the proper factors of 198,585,576,189 [4] would sum to 198,585,576,189 — making it the only odd perfect number ever found.  As it is, because 22021 is not prime, or because 22021 is a spoof prime, the proper factors of 198,585,576,189 do not sum to 198,585,576,189 (they actually sum to 227,441,894,589).  This makes 198,585,576,189 an odd spoof perfect number, which in itself is the only such number ever found!

To elaborate: Descartes showed that an equivalence to the sum-of-divisors function using the prime factorisation of a number is the following [5]:

\[\sigma \left( n \right) = \prod\limits_{{p^a}\left\|{\;n} \right.} {\frac{{{p^{a + 1}} - 1}}{{p - 1}}} {\rm{ }}\]
where \(p\) is a distinct prime factor.  This means that a number's factors can be summed using only the number's prime factors.  Taking 28 again as an example, we have seen that it is a perfect number because

\[\sigma \left( 28 \right) = \sum\limits_{d\left| 28 \right.} d  = 1 + 2 + 4 + 7 + 14 + 28 = 56 = 2n\]
We can also observe this result using the prime factors of 28, i.e. 2\(^2 \times \) 7:

\[\prod\limits_{{p^a}\left\|{\;n} \right.} {\frac{{{p^{a + 1}} - 1}}{{p - 1}}}  = \frac{{{2^{2 + 1}} - 1}}{{2 - 1}} \times \frac{{{7^{1 + 1}} - 1}}{{7 - 1}} = \frac{7}{1} \times \frac{{48}}{6} = 56 = 2n\]
And so, similarly, the sum of the divisors of Descartes' number:

\[\begin{array}{c}\begin{align}\sigma \left( 198,585,576,189 \right) &= \prod\limits_{{p^a}\left\| {\;n} \right.} {\frac{{{p^{a + 1}} - 1}}{{p - 1}}} \\ \\&= \frac{{{3^{2 + 1}} - 1}}{{3 - 1}} \times \frac{{{7^{2 + 1}} - 1}}{{7 - 1}} \times \frac{{{{11}^{2 + 1}} - 1}}{{11 - 1}} \times \frac{{{{13}^{2 + 1}} - 1}}{{13 - 1}} \times \frac{{{{22021}^{1 + 1}} - 1}}{{22021 - 1}}\\  \\&= 13 \times 57 \times 133 \times 183 \times 22022\\ &= 13 \times \left( {3 \times 19} \right) \times \left( {7 \times 19} \right) \times \left( {3 \times 61} \right) \times \left( {2 \times 7 \times 11 \times 11 \times 13} \right)\\ &=2 \times {3^2} \times {7^2} \times {11^2} \times {13^2} \times \left( {{{19}^2} \times 61} \right)\\ &= 2 \times {3^2} \times {7^2} \times {11^2} \times {13^2} \times 22021\\ &= 2 \times 198,585,576,189\end{align}\end{array}\]
Hence, if 22021 were prime rather than 19\(^2\) \(\times \) 61, Descartes' number 198,585,576,189 would  be an odd perfect number because \(\sigma \left( n \right)\) = 2n.

Notes, References & Links:

Wednesday, 11 July 2018

Calendar of Mathsy Moments

The calendar of mathsy 'moments' (downloads provided below) is a calendar marking at least one mathsy 'moment' (or event) of interest for each day from 1 August 2018 to 31 December 2019.  It is primarily intended for schools and teachers to use in gentle cultivation of a respectful yet intellectually playful approach to mathematics in their students, in support moreover of the development of students' deeper and perhaps wider appreciation and framing of mathematics in the historical, cultural and intellectual sense.

The calendar is shared to help teachers establish and cultivate an ethos of mathematics as progress in the classroom, to help teachers in other words help their students develop and sustain happy and fulfilling relationships with their mathematics, encouraging the desire to explore further, deeper, and more independently — fuelling students' curiosity and their capacity for such.  It is hoped that in sharing these brief moments of mathematics with students, in unabashedly celebrating the joy that is to be found in mathematics, teachers will fulfil their roles as enthusiasts, advocates and celebrators of mathematics, both as an intellectual discipline and cultural artefact, and students will in turn stand a greater chance that the subject will inspire — rather than beguile — them.

The calendar is provided below for access from this page.  Downloads in a range of formats are also provided for easy import.  Each 'moment' is outlined in more detail within the description / notes section of each calendar entry, which may serve as a prompt for the teacher with respect to how s/he may use / share it, if at all.  Where applicable, links to sources, and other sources of information — including online articles, historical artefacts, videos, podcasts, etc. — are provided within each description as a start for further exploration, as desired.

The moments are categorised within the calendar as follows:
  • Moments / Events occurring at various times throughout the year (note that a 'Day' relates to the day of the year, e.g. 14 March 2019 is the 73rd day of the year, and so a curio about 73 may be given.  A 'Date' relates to the date as a concatenated number, in dd/mm/yyyy format unless otherise indicated, e.g. 14 March 2019, the 14th day of the 3rd month in the 2019th year, 14/3/2019, which concatenates to 1432019).  These moments include:
    • Prime Number Dates, Mathematicians' Birthdays, Historical Dates to note in Mathematics, Circular Prime Number Dates, Emirp Dates, Frustrating Prime Number Dates, Days for Curious Calculations, Days for Number Facts, Palindromic Number Dates, Palindromic Prime Number Days, Pi Dates, et al.
    • The 'Day for a Number Fact' days aims to feed our natural love of numbers (or perhaps revitalise it), exposing students to — or maybe tantalising them with — a wide range of more exotic sounding mathematics from number theory that they may not ordinarily come across in their day to day curriculum.  For example, in sharing such moments students will know of: good primes, home primes, long primes, Sophie Germain primes, Ramanujan primes, Fibonacci primes, near-repdigit primes, quadruprimes, prime triplets, prime sums, primes that remain prime when added to their reverse, circular primes, emirps, permutable primes, near primorial primes, palindromic primes, dihedral primes, twin primes, sexy primes, numbers commonly assumed incorrectly to be prime, semiprimes, sphenic numbers, other prime products, unprimeable numbers, happy numbers, sad numbers, distinctly happy numbers, perfect numbers, superperfect numbers, pseudoperfect numbers, weird numbers, abundant numbers, deficient numbers, almost integers, duck-duck-goose numbers, Mersenne numbers, lucky numbers, amicable numbers, polite numbers, sociable numbers, automorphic numbers, narcissistic numbers, odious numbers, pernicious numbers, evil numbers, practical numbers, arithmetic numbers, polygonal numbers, undulating numbers, telephone numbers, antisigma numbers, highly composite numbers (or anti-primes), refactorable numbers, taxicab numbers, tetranacci numbers, vampire numbers, untouchable numbers, Bell numbers, Euclid numbers, Franel numbers, Frobenius numbers, mountain numbers, star numbers, Smith / Joke numbers, Friedman numbers, the divisor (factor) function, the partition function, Jacobi's formula, Aronson's sequence, Fibonacci factorials, Euler's 6n+1 theorem, repdigits, primorials, subfactorials, double factorials, superfactorials, hyperfactorials, trees, Golomb's sequence, numbers in different bases, etc.
  • Moments / Events occurring once in the year (including some national or globally recognised awareness days with particular reference to mathematics).  These moments include:
    • A Perfect Day, An Anti-Perfectt Day, Ada Lovelace Day, e day (Europe and US), Fibonacci Day (US), Half Year Day, Halloween (Vampire Numbers), Mathematics and Statistics Awareness month, McNugget Numbers Day, National Numeracy Day, Phi Day, Phi Day (US), Pi Approximation Day, Pi Day, Root-2 Date, Spreadsheet Day, The Golden Ratio Moment, The Oddest Prime Day, et al.
    • Other Moments / Events not directly related to mathematics (that schools / teachers may nonetheless wish to mark, either because of the mathematics latent within them, or for other more school-centred reasons around values, community and relationships).  These moments include: 
      • Ask A Stupid Question Day, Autumnal Equinox, Black History Month, Book Lovers' Day, European Day of Languages, International Joke Day, International Women's Day, Make Up Your Own Holiday Day, National Coding week, National Poetry Day, National Read a Book Day, National School Nurse Day, National Teaching Assistant's Day, Origami Day, Random Act of Kindness Day, Roald Dahl Day, Space Day, Summer Solstice, Vernal Equinox, Winter Solstice, World Book Day, World Kindness Day, World Music Day, World Teachers' Day.

    Mathsy Moments calendar
    (Click here to go back to the top)
    There are and will be, of course, many other 'moments' that could and maybe should be included in such a calendar.  I make no claim that the calendar offers anything comprehensive.  If there are other 'moments' that you think other teachers may enjoy sharing with students, please let me know in the comments and I will update in subsequent versions.  Equally, whilst every effort has been taken to eliminate errors, if users notice any error in the content of the calendar, I would be very grateful if you could again let me know in the comments.

    GCSE Maths 2019 Countdown calendar
    (Click here to go back to the top)
    This calendar gives a daily countdown to GCSE mathematics exams, both in terms of days left and school days left.  It is intended as a support for teachers to help students (and their families) manage their own study time.

    Further sources
    Sources of information (or explanation), where applicable, have been provided in link form in the description of each moment / event.  If you wish to play further, explore the following sites:

      Tuesday, 1 May 2018

      On Mountain Numbers

      Cairn Toul, a beautifully wild and remote mountain in the Cairngorms massif of the eastern Scottish highlands, is notable not only for being the fourth highest mountain in the UK, but also for being the only mountain in the UK whose elevation in metres — 1291m above sea level — is also a 'mountain'.  Großglockner, by way of a further example, the highest mountain in Austria, is a mountain whose elevation in feet — 12461ft above sea level — is also a 'mountain'.

      To elaborate: A number is a 'mountain' if its decimal digits start with 1, i.e. 'base camp', ascend continuously to a unique summit, i.e. to one largest digit, then descend continuously back to 1, i.e. back to base camp.  Such mountain numbers are recorded as sequence A134941 (itself a mountain) in the On-line Encyclopedia of Integer Sequences (OEIS), from which you can also find this table of all possible 21,846 mountains (including 1 itself).

      Mountain primes, as the name suggests, are prime mountain numbers, prime numbers in other words whose decimal digits start with 1, ascend continuously to the summit of a single largest digit, then descend continuously back to 1.  (The area chart image above shows a mountain range of the first 17 such numbers.)  Mountain primes are recorded in the OEIS as sequence A134951 (yes, also a mountain prime), from which you can find this table of all 2620 such primes.

      When climbing a mountain, it is typical to descend back to base camp, i.e. back to where you started your ascent, but it is not always the case.  Mountain numbers that ascend from one location (as an elevation) but descend to another — e.g. 3,598,432 — are considered to be generalised mountain numbers, recorded as sequence A134853 in the OEIS (there are 173,247 such mountains).

      Suggested explorations, diversions & links:
      • Explain why there must be a finite number of mountain numbers.
      • Construct a mountain range diagram for mountains of your choice.
      • What is the Everest of mountain numbers?
        • How will you define 'Everest'; how will you define 'elevation'?  (In the image above, the 'height' of mountain 1291, for example, is greater than the 'height' of mountain 1571.)  
      • Explore the distribution of mountain number digits, i.e. how many mountains have 1 digit, 2 digits, 3 digits, etc. (OEIS sequence A135417).
      • If we regard the number of digits in a mountain number as the horizontal distance travelled when climbing such a mountain:
        • What mountain(s) has (have) the shallowest ascent?
        • What mountain has the steepest ascent and descent?
        • What about generalised mountains?
      • Many people find the most beautiful mountains to be those most pyramidal in shape, such as for example the Matterhorn in the Alps and Machapuchare in the Himalaya.
        • Define and find your most beautiful mountain numbers: 
          • Palindromic mountain numbers perhaps (OEIS sequence A173070),
          • Or palindromic mountain prime numbers (the largest is 123467898764321).
        • Find all possible — i.e. 45 — Giza numbers.
          • Giza numbers are so-called because they represent the pyramids of Giza in the sense that their first digits increase in consecutive order to a largest central digit, and their last digits decrease in the same consecutive order as they increased.  
          • The largest Giza number is 12,345,678,987,654,321 (OEIS sequence A134810).
      • Of two mountain numbers chosen at random, what is the probability that their numerical heights — i.e. the magnitude of the number — will be in alignment with the elevation of actual mountains?  
        • For example, mountain number 1291 is 'higher' than mountain number 1571, but as actual elevations, 1291m < 1591m, so these two mountains are not in 'alignment'. 
      • Find as many actual mountains whose heights are mountain numbers, or mountain primes.
        • What is the highest mountain in the world whose height is also a mountain (use metres and/or feet for the height)?

      Friday, 27 April 2018

      On Overheard Conversations About Maths #1

      The floor.  Just before bedtime.
      Three siblings are islands in a sea of Lego.
      The eldest is trying to make a Lego Penrose triangle.

      3 YEAR OLD:

      (Silently counting the studs on a red 2 × 4 Lego brick,
      mouthing each number until, to no-one in particular...)


      (He opens up his thumb and fingers successively
      from the fist of one hand as he counts, this time out loud.) 
      One, two, three, four, five...

      (Now to the other hand.)

      ...six, seven, eight.

      (Some time passes as he collects all the 2 × 4 bricks
      he can find, piles them together, and pretends to count them all.)

      A hundred!  A hundred is bigger than four.  Or five.  Or eight.

      6 YEAR OLD:

      It’s not as big as a million though.

      YEAR OLD:

      (With a nod to his Numberphile education...)

      Or Graham.

      YEAR OLD:

      Or ten.  Ten's bigger.

      YEAR OLD:

      No, a million's bigger than ten.  A hundred is bigger than ten.  A million is really big.

      YEAR OLD:

      Who's Graham?

      YEAR OLD:

      Infinity is bigger than Graham.  

      YEAR OLD:


      YEAR OLD:

      No, infinity.  In-fin-ity.

      YEAR OLD:

      (Straining to speak; arms outstretched as wide as he can make them go.)  

      Is Nifity big like this?

      YEAR OLD:

      That’s not even a number.

      YEAR OLD:


      YEAR OLD:

      Infinity.  In-fi-nity.  In-finity.  No, it’s not a number it just means the biggest thing.  People say it's a number but it's not because you could just say infinity plus one but that's still just infinity because...

      YEAR OLD:

      Infinity plus a hundred then.

      YEAR OLD:


      Infinity plus infinity.

      YEAR OLD:

      Two times infinity.

      YEAR OLD:

      Duh, that's the same.  And it's still just infinity.

      YEAR OLD:

      (Quietly, in the ear of his 6 year old sibling.)

      Is Graham friendly?

      YEAR OLD:

      Infinity times infinity then.

      YEAR OLD:

      Do you even know what that means?  OK then, infinity times infinity infinity times.  It's still just infinity!  It's still...

      (Establishing the correct perspective
      from which to see the Lego Penrose triangle
      doesn't seem to matter quite as much right now.)

      Isn't it...?  Erm...

      YEAR OLD:



      Suggested explorations, diversions and links:

      What is infinity?
      • Share this typically accessible introduction to the idea of infinity from MathsIsFun.com.  (Play, for example, with the provocation 'An infinite series of A's followed by a B will NEVER have a B'.)  The page gently introduces infinity as 'the idea that something has no end', that doesn't grow, and subsequently moves into arithmetic properties and the exciting idea that there are different sizes of infinity.  

        Why does \(\infty  + 1 = \infty  + 100 = \infty  + \infty  = \infty  \times \infty  = \infty \)?

        We can (sometimes) add an infinity of numbers.

        Play around with infinite sums, starting perhaps with this article by Luciano Rila from Plus Magazine.   And consider, for example:
        \[\underbrace {1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...}_\infty \]Or, to put it another way:
        \[\begin{array}{c}\begin{align}\sum\limits_{k = 0}^\infty  {\left( {\frac{1}{{{2^k}}}} \right)}  &= \frac{1}{{{2^0}}} + \frac{1}{{{2^1}}} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ...\\ &= \frac{1}{1} + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...\end{align}\end{array}\]
        If we add up the first few terms (i.e. find the first few partial sums):
        \[\begin{array}{c}\begin{align}\frac{1}{1} + \frac{1}{2} &= 1\frac{1}{2}\\\frac{1}{1} + \frac{1}{2} + \frac{1}{4} &= 1\frac{3}{4}\\\frac{1}{1} + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} &= 1\frac{7}{8}\\\frac{1}{1} + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} &= 1\frac{{15}}{{16}}\\etc.\end{align}\end{array}\]
        We can see that each iteration produces a number that gets closer and closer to 2.  We produce another infinite sequence in others words with a limit of 2, thus:
        \[\left\{ {1,\;1\frac{1}{2},\;1\frac{3}{4},\;1\frac{7}{8},\;1\frac{5}{{16}},\;...} \right\}\]
        And we can therefore show that the sum of infinite series is 2.  Say:
        \[\sum\limits_{k = 0}^\infty  {\left( {\frac{1}{{{2^k}}}} \right)}  = s = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...\]
        \[\begin{array}{c}\begin{align}\ 2s &= 2\left( 1 \right) + 2\left( {\frac{1}{2}} \right) + 2\left( {\frac{1}{4}} \right) + 2\left( {\frac{1}{8}} \right) + ...\\2s &= 2 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...\end{align}\end{array}\]
        And so:
        \[\begin{array}{c}\begin{align}\ 2s - s &= \left( {2 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...} \right) - \left( {1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...} \right)\\s &= 2\end{align}\end{array}\]
        \[\sum\limits_{k = 0}^\infty  {\left( {\frac{1}{{{2^k}}}} \right)}  = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... = 2\]

        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.

        Click to jump to:

        Archimedes' Constant
        (Click here to go back to the menu)

        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 \(C/d\)
        (Click here to go back to the menu)

        \(\pi \) 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
        (Click here to go back to the menu)

        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: