Editorial, February 2018

Editorial, February 2018


Congratulations to Professor Alistair Fitt on his recent appointment as IMA President. He is a very active member of the mathematics community and will undoubtedly fulfil this role superbly over the next two years, as did his predecessor Professor Chris Linton. An interview with Alistair is transcribed on pages 5–7 and other IMA governance matters are considered in an article that summarises the 2017 Council Strategy Review on page 4.

Some time ago, when I was a final-year student at Loughborough University, our brilliant analysis lecturer Dr Peter Shiu set a problem to occupy our minds after the summer exams. He asked us to find a finite set of distinct natural numbers whose reciprocals sum to exactly four. This remains a delightful and stimulating challenge that is well worth investigating today. Although obvious solutions exist for sums of exactly one \{1\} and two \{1,2,3,6\}, finding a finite set of distinct natural numbers whose reciprocals sum to exactly three, let alone four, requires considerable effort. An even tougher problem involves finding the most efficient set as defined by having the fewest elements or least maximum. Exhaustive enumeration reveals that the above set for a sum to two is optimal in both respects.

editorial-february-2018-figure-1
Figure 1: 2/13 as an Egyptian fraction

This class of problems relates to Egyptian fractions, which are finite sums of unit fractions such as

    \[\frac{2}{13}=\frac{1}{8}+\frac{1}{52}+\frac{1}{104}\]

as illustrated in Figure 1. They date back to Ancient Egypt and are recorded on the Rhind Mathematical Papyrus in the British Museum. Various rules exist to determine Egyptian fractions, including Fibonacci’s greedy algorithm. This splits a fraction into the largest possible unit fraction and another fraction, which is then iteratively split in a similar way. This process leads to results such as

    \[\frac{14}{19}=\frac{1}{2}+\frac{1}{5}+\frac{1}{28}+\frac{1}{887}+\frac{1}{\np{2359420}}\]

but can introduce more terms and larger denominators than necessary. Indeed, the expansion

    \[\frac{14}{19}=\frac{1}{2}+\frac{1}{5}+\frac{1}{30}+\frac{1}{285}\]

is more efficient in both respects for this example.

A useful rule for expanding a unit fraction into a sum of unit fractions can be constructed as follows. First write

    \[\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}\]

based on a common denominator. For all (a,b)\in\mathbb{N}^2, the numerator and denominator of the right hand side are also natural numbers as the set \mathbb{N} is closed under addition and multiplication. Dividing through by a+b then leads to the generating equation

(1)   \begin{equation*} \frac{1}{ab}=\frac{1}{a+b}\left(\frac{1}{a}+\frac{1}{b}\right), \end{equation*}

which expresses a unit fraction as the sum of two unit fractions. Table 1 shows the results of evaluating Equation (1) for all a\le b\le 4, to illustrate this process while avoiding repetition due to symmetry. Although a=b generates identical unit fractions, subsequently expanding one of these generates distinct terms.

editorial-february-2018-table-1
Table 1: Unit fraction expansions

This algorithm expands a sixth in two ways

    \begin{align*} &(1,6)\rightarrow\frac{1}{6}=\frac{1}{7}+\frac{1}{42}\\ &(2,3)\rightarrow\frac{1}{6}=\frac{1}{10}+\frac{1}{15} \end{align*}

though we need to generalise it as other expansions exist. Multiplying or dividing both sides of Equation (1) by a specified natural number can generate expansions such as these:

    \begin{align*} &(1,1)\div 6\rightarrow\frac{1}{6}=\frac{1}{12}+\frac{1}{12}\\ &(1,2)\div 3\rightarrow\frac{1}{6}=\frac{1}{9}+\frac{1}{18}\\ &(1,3)\div 2\rightarrow\frac{1}{6}=\frac{1}{8}+\frac{1}{24}. \end{align*}

Having expressed a sixth as a sum of two unit fractions in several ways, we note that the smallest denominators include all integers in the interval (6,12] except for 11. This is because

    \[\frac{1}{6}=\frac{1}{11}+\frac{5}{66}\]

and the latter is not a unit fraction.

One way to resolve this matter involves applying the greedy algorithm, which generates

    \[\frac{1}{6}=\frac{1}{11}+\frac{1}{14}+\frac{1}{231}\]

for this example. Another approach involves generalising the rule in Equation (1) to allow for sums of three unit fractions

(2)   \begin{equation*} \frac{1}{abc}=\frac{1}{ab+ac+bc}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \end{equation*}

in order to generate expansions such as:

    \[(1,2,3)\rightarrow\frac{1}{6}=\frac{1}{11}+\frac{1}{22}+\frac{1}{33}.\]

Indeed, Equations (1) and (2) generalise to sums of n unit fractions

    \begin{equation*}\label{eqn:n} \left(\prod_{i=1}^{n}a_i\right)^{-1}=\left(\sum_{i=1}^{n}\prod_{j=1}^{n}a_j^{1-\delta_{ij}}\right)^{-1}\sum_{i=1}^{n}\frac{1}{a_i} \end{equation*}

where

    \[\delta_{ij}=\left\{\begin{array}{l}1;~i=j\\0;~i\ne j\end{array}\right.\]

is the Kronecker delta. Interested readers can easily prove this result by induction. First multiply both sides by the bracketed terms to simplify the expression. Next show trivially that equality holds for n=1. Then assume true for n=k, multiply both sides by a_{k+1} and add \prod_{i=1}^{k}a_i to both sides, so deducing that equality must also hold for n=k+1 as required.

Applying these methods to address the original problem, I generated the set \{1,\dots,9,18,20,24,42\} of 13 distinct natural numbers whose reciprocals sum to exactly three and the set

    \[\left\{\begin{array}{c}1,\dots,16,18,20,\dots,24,27,\dots,31,35,42,46,48,\\63,90,105,135,156,161,231,406,462,870,930\end{array}\right\}\]

of 42 distinct natural numbers whose reciprocals sum to exactly four. However, these are likely to be sub-optimal according to the efficiency criteria stated previously, so maybe you can construct sets with fewer elements or lesser maxima.

Perhaps Peter (Sir as I knew him) already had optimal solutions when he set this exercise, as he subsequently investigated other related problems [1]. Some useful sources of information to help with your quest include an inspirational website [2] maintained by the IMA’s North West Branch Chair Ron Knott, an introductory book [3, 4] on Egyptian arithmetic and an advanced book [5] that considers a broader variety of number-theoretic puzzles. Good luck!

February’s issue of Mathematics Today is printed using off-white paper with a silk finish, rather than with a gloss finish. This reduces the glare due to reflected light so that the content is easier to read and was requested by several readers in our recent survey.

Along with articles by our regular contributors to the Urban Maths, A Doctor Writes and Westward Ho! series, we have diversely informative and entertaining features on moments of inertia of rotating frames, how to estimate quantiles easily and reliably, and the barcode sequence.

The last of these relates to a problem posed by the United Kingdom Mathematics Trust’s Senior Kangaroo Challenge and the author is our former letters editor Graham Hoare. I am pleased to report that at a meeting in December, the MT Editorial Board decided to rename the annual Early Career Mathematicians Catherine Richards Prize as the Graham Hoare Prize in recognition of his substantial contributions to the IMA and mathematics education over many years. The main award will retain its name as the Catherine Richards Prize.

David F. Percy CMath CSci FIMA
University of Salford

References

  1. Shiu, P. (2009) Egyptian fraction representations of 1 with odd denominators. The Mathematical Gazette, vol. 93, pp. 271-276.
  2. Knott, R. (2017) Egyptian fractions, https://tinyurl.com/Knott-Egyptian (accessed 11 December 2017).
  3. Reimer, D. (2014) Count like an Egyptian, Princeton University Press.
  4. Hollings, C. (2015), Review of Count like an Egyptian, IMA website. Available at: https://tinyurl.com/IMA-review (accessed 11 December 2017).
  5. Guy, R.K. (2004) Unsolved problems in number theory, Springer-Verlag, New York.

Reproduced from Mathematics Today, February 2018

Download the article, Editorial, February 2018

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