Jekyll2022-05-22T08:00:00+01:00https://www.flyingcoloursmaths.co.uk/feed.xmlFlying Colours MathsFlying Colours Maths helps make sense of maths at A-level and beyond.A Red Rag To A Bull (The Duckworth-Lewis-Stern Method)2022-05-16T00:00:00+01:002022-05-16T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/red-rag-duckworth-lewis-stern<blockquote> <p>The original D/L model started by assuming that the number of runs that can still be scored (called $Z$), for a given number of overs remaining (called $u$) and wickets lost (called $w$), takes the following exponential decay relationship:</p> <p>$Z(u,w)=Z_{0}(w)\left({1-e^{-b(w)u}}\right)$ where the constant $Z_{0}$ is the asymptotic average total score in unlimited overs (under one-day rules), and $b$ is the exponential decay constant. Both vary with $w$ (only). The values of these two parameters for each $w$ from 0 to 9 were estimated from scores from ‘hundreds of one-day internationals’ and ‘extensive research and experimentation’, <strong>though were not disclosed due to ‘commercial confidentiality’</strong>.</p> <p><a href="https://en.wikipedia.org/wiki/Duckworth%E2%80%93Lewis%E2%80%93Stern_method#Mathematical_theory">Wikipedia: Duckworth-Lewis-Stern Method; Mathematical Theory</a></p> </blockquote> <p>Commercial confidentiality, you say? Pah! You can’t keep something as important as the cricket score secret! It just isn’t… well.</p> <p>(I’m not going to go into the ins and outs of DLS here - it’s a way of deciding who won a rain-interrupted cricket match. Given that I live in England, that happens a lot.)</p> <p>Anyway. Since my dander was well and truly up, I figured I could probably work out the constants $Z_0$ and $b$ for any given number of wickets, given that the <a href="http://www.ccua.ca/Duckworth-Lewis_Tables._Table_of_ball_by_ball_.pdf">tables are easy enough to find</a> <a href="https://www.easycalculation.com/sports/dl-method-table.php">in several places</a>.</p> <p>What’s more, it’s easy enough to tidy up the tables in the spreadsheet program of your choice, then copy-and-paste into <a href="https://www.desmos.com/calculator/ke78btdcj5">Desmos</a> – although it seems to have a five-column limit. ((That’s not as big a problem as it might be. I can just do it in two blocks.))</p> <p>And, using Desmos’s handy $\simeq$ functionality, I can fit the given model to the data and come up with very good estimates of $Z_0$ and $b$ for every $w$:</p> <table> <tbody> <tr> <td>W</td> <td>Z_0</td> <td>b</td> </tr> <tr> <td>0</td> <td>134.1022939</td> <td>0.027393558</td> </tr> <tr> <td>1</td> <td>118.5253606</td> <td>0.030996527</td> </tr> <tr> <td>2</td> <td>101.9143572</td> <td>0.036038757</td> </tr> <tr> <td>3</td> <td>84.45285677</td> <td>0.043504654</td> </tr> <tr> <td>4</td> <td>66.99557009</td> <td>0.054869035</td> </tr> <tr> <td>5</td> <td>50.28100499</td> <td>0.073076761</td> </tr> <tr> <td>6</td> <td>35.11610741</td> <td>0.104616858</td> </tr> <tr> <td>7</td> <td>21.98992361</td> <td>0.167213062</td> </tr> <tr> <td>8</td> <td>11.90743731</td> <td>0.309870559</td> </tr> <tr> <td>9</td> <td>4.700112328</td> <td>0.763221136</td> </tr> </tbody> </table>The original D/L model started by assuming that the number of runs that can still be scored (called $Z$), for a given number of overs remaining (called $u$) and wickets lost (called $w$), takes the following exponential decay relationship: $Z(u,w)=Z_{0}(w)\left({1-e^{-b(w)u}}\right)$ where the constant $Z_{0}$ is the asymptotic average total score in unlimited overs (under one-day rules), and $b$ is the exponential decay constant. Both vary with $w$ (only). The values of these two parameters for each $w$ from 0 to 9 were estimated from scores from ‘hundreds of one-day internationals’ and ‘extensive research and experimentation’, though were not disclosed due to ‘commercial confidentiality’. Wikipedia: Duckworth-Lewis-Stern Method; Mathematical TheoryAsk Uncle Colin: What’s the Bisector of a Tesseract2022-05-09T00:00:00+01:002022-05-09T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-the-bisector-of-a-tesseract<blockquote> <p>Dear Uncle Colin</p> <p>If you hang a square up by one corner and cut it in the plane perpendicular to the vertical diagonal, you get a regular hexagon. What’s the corresponding result for a tesseract?</p> <p>- Have You Proved Everything, Really?</p> </blockquote> <p>Hi, HYPER, and thanks for your message! Let’s start in 3D.</p> <p>If you dangle a 3D cube from one vertex (call it O), you can mark each other vertex with a height – those that share an edge with O have a height of 1, the three vertices in the next layer have a height of 2, and the final opposite vertex has a height of 3.</p> <p>The mid-plane is therefore at height 1.5, and cuts all of the edges connecting 1-vertices to 2-vertices. There are six of those, and by symmetry they have to form a regular hexagon.</p> <p>When you move to 4D, the first thing you get is a headache.</p> <p>However, once you get past that, a similar argument holds sway: pick a vertex and call it at ‘height’ 0. It’s not really a height, it’s whatever you want to call the fourth dimension. Time, maybe.</p> <p>The four vertices connected to the initial vertex have time 1; the six vertices connected to those have time 2; then four of the remaining vertices have time 3 and the final one has time 4.</p> <p>The mid-hyperplane passes through the six vertices with height 2. So it’s a hexagon as well, right? Wrong.</p> <p>In a unit tesseract((the best kind)), the six points have coordinates:</p> <ul> <li>A: (1,1,0,0)</li> <li>B: (1,0,1,0)</li> <li>C: (1,0,0,1)</li> <li>D: (0,1,0,1)</li> <li>E: (0,1,1,0)</li> <li>F: (0,0,1,1)</li> </ul> <p>(Hey look! Their coordinates each sum to 2. That’s interesting.)</p> <p>Point A is $\sqrt{2}$ units away from B, C, D and E, and two units away from F. It’s similar for each of the points: they’re $\sqrt{2}$ units away from four others, and two from the last. Also, the mid-hyperplane of a 4D shape should give us a 3D object.</p> <p>Six vertices, each with four neighbours? We’re looking at a <em>regular octahedron</em>.</p> <p>Hope that helps!</p> <p>- Uncle Colin</p>Dear Uncle Colin If you hang a square up by one corner and cut it in the plane perpendicular to the vertical diagonal, you get a regular hexagon. What’s the corresponding result for a tesseract? - Have You Proved Everything, Really?The Dictionary of Mathematical Eponymy: The Lute of Pythagoras2022-05-02T01:00:00+01:002022-05-02T01:00:00+01:00https://www.flyingcoloursmaths.co.uk/dictionary-of-mathematical-eponymy-lute-of-pythagoras<h3 id="what-is-the-lute-of-pythagoras">What is the Lute of Pythagoras?</h3> <ul> <li>Draw a regular pentagon (lightly).</li> <li>Draw its diagonals (darkly)</li> </ul> <p>The outline of the dark shape is a concave, equilateral decagon. Pick one of the concave ((“Inny”)) corners and its two neighbours; the edges linking these can be two edges of a regular pentagon.</p> <ul> <li>Draw this regular pentagon (lightly)</li> <li>Draw its diagonals (darkly)</li> </ul> <p>Repeat these two steps for as long as you want to. You’ll end up (in the limit) with a lightly-drawn kite with pretty diagonals drawn.</p> <h3 id="why-is-it-interesting">Why is it interesting?</h3> <p>Pretty!</p> <p>Also, it’s just <em>fun</em> to mess around with it. What if you go off in random directions? What if you start with an irregular pentagon, can you make that work? Can you make it tessellate? Why does it fit together nicely? There’s all sorts you can do.</p> <h3 id="what-has-it-to-do-with-pythagoras">What has it to do with Pythagoras?</h3> <p>Bog all, so far as anyone knows. The earliest known mention of it is from 1990, which is generally classified as “post-Pythagorean.”</p> <h3 id="who-was-pythagoras-of-samos">Who was Pythagoras of Samos?</h3> <p>He’s the one mathematician almost everyone has heard of. If you stop a random person on the street and ask them to name a theorem, they’ll probably avoid eye contact and walk on. But failing that, they’ll probably say Pythagoras.</p> <p>He was born is Samos, in the eastern Aegean Sea, about 570BCE; he died around 495BCE. Little is known for sure of his life, and many of the stories surrounding him are almost certainly legendary (his approach to Hippasus’s proof that $\sqrt{2}$ is irrational may have inspired generations of Reviewer 2s, but almost certainly didn’t happen). He’s reputed to have done good work on triangles, astronomy and music (among other things).</p> <p>Oh, and Pythagoras’s Theorem? He didn’t discover that, the Babylonians did, likely a millennium before the bearded wonder made his entrance.</p>What is the Lute of Pythagoras?IRAC and Mathematical Communication2022-04-25T01:00:00+01:002022-04-25T01:00:00+01:00https://www.flyingcoloursmaths.co.uk/irac-and-mathematical-communication<p>If you <a href="https://twitter.com/icecolbeveridge">follow me on Twitter</a>, you might have noticed that I’m a fan of legal blogger <a href="https://twitter.com/davidallengreen">David Allen Green</a> – not because I always agree with him, but because even when I don’t, he sets out a clear and compelling case for why I should.</p> <p>He’s famed for writing in one-sentence paragraphs, because there’s “no place to hide” – it forces him to distil his argument down to its absolute key points, avoiding Johnsonesque word salad.</p> <p>So, I wondered, what can the world of legal writing tell us about mathematical communication?</p> <h3 id="irac">IRAC</h3> <p>I stumbled on an <a href="https://media.cheatography.com/storage/thumb/theheartbeating_legal-problem-solving-technique.750.jpg?last=1478727197">interesting cheat-sheet</a> by Izzy (theheartbeating), about a “problem-solving technique for legal issues((I would say it’s as much a presentation technique, but tomatoes, edible berries of the plant Solanum lycopersicum.))” called IRAC.</p> <p>So what is it ((For those of you who don’t want to read the sheet, for some reason))?</p> <p>It stands for:</p> <ul> <li><strong>I</strong>ssue</li> <li><strong>R</strong>ule</li> <li><strong>A</strong>nalysis</li> <li><strong>C</strong>onclusion</li> </ul> <p>That is:</p> <ul> <li>State the problem you’re faced with</li> <li>State the rule or rules you rely on in your answer</li> <li>Write down how the rule applies to the problem</li> <li>Clearly state your answer.</li> </ul> <h3 id="an-example">An example</h3> <p>Let’s try it, based on a Madas IYGB A-level question (it’s question 1 on <a href="https://madasmaths.com/archive/iygb_practice_papers/mp2_practice_papers/mp2_s.pdf">this paper</a>). I’ve <em>already solved</em> the question, and am now trying to communicate my solution clearly.</p> <h4 id="issue">Issue</h4> <p><em>Here, I’m just going to paraphrase the question. A diagram wouldn’t go amiss.</em></p> <p>Square ABCD has sides of length 2. Point M is the midpoint of CD. Points A, B and M lie on a circle. What is the radius of the circle?</p> <h4 id="rule">Rule</h4> <p><em>I’m going to state the main rule I rely on in my solution.</em></p> <p>The <strong>intersecting chord theorem</strong> states that if two chords of a circle, PQ and RS, meet at a point X inside the circle, then $|PX||XQ| = |RX||XQ|$.</p> <h3 id="analysis">Analysis</h3> <p><em>A run down of my reasoning, in one-word paragraphs.</em></p> <ul> <li>Let MN be the diameter of the circle perpendicular to CD.</li> <li>Let X be the point where MN meets AB.</li> <li>We know that: <ul> <li>$|AX| = |XB| = 1$, since the diameter bisects the chord</li> <li>$|MX| = 2$, since MX is the shortest distance between AB and CD</li> </ul> </li> <li>Applying the rule as $|AX||XB| = |MX||XN|$, we get $(1)(1) = (2)|XN|$</li> <li>Thus $|XN| = \frac{1}{2}$</li> <li>The length of the diameter is therefore $2 + \frac{1}{2} = \frac{5}{2}$.</li> </ul> <h3 id="conclusion">Conclusion</h3> <p><em>And finally…</em></p> <p><strong>The radius of the circle is half of the diameter, or $\frac{5}{4}$</strong></p> <hr /> <p>I think that’s a clear and tidy way to present the solution to a question! In an exam, you may not have the time or inclination to go through the rigmarole, but exams aren’t proper maths. If you’re trying to communicate your thinking – to someone else, or to future-you – then taking the time to refine and express your thoughts clearly will pay off.</p>If you follow me on Twitter, you might have noticed that I’m a fan of legal blogger David Allen Green – not because I always agree with him, but because even when I don’t, he sets out a clear and compelling case for why I should.Ask Uncle Colin: the ff(n) puzzle2022-04-18T00:00:00+01:002022-04-18T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-the-ffn-puzzle<blockquote> <p>Dear Uncle Colin,</p> <p>Someone gave me the puzzle:</p> <ul> <li>$ff(n) = 3n$ for all real $n$,</li> <li>$f(n)$ is a positive integer for all positive integer $n$</li> <li>$f(n+1) &gt; f(n)$ for all positive integer $n$</li> <li>What is $f(13)$?</li> </ul> <p>Any ideas?</p> <p>Functions? Fun? No.</p> </blockquote> <p>Hi, FFN,</p> <p>This is a fun puzzle! Here’s how I tackled it:</p> <ul> <li>$ff(1) = 3$, so I reckon $f(1) = 2$ and $f(2)=3$</li> </ul> <p>This takes a little unpicking; $f(1)$ can’t be 1, or else $ff(1) = 1$. Could it be 3, and $f(3)=3$? again, no: $f(3) &gt; f(2) &gt; f(1)$. So, it has to be 2, and $f(2) = 3$.</p> <ul> <li> <p>We know that $ff(2) = 6$ and $f(2) = 3$, so $f(3) = 6$.</p> </li> <li> <p>Now $ff(3) = 9$, so $f(6) = 9$</p> </li> </ul> <p>Let’s take stock:</p> <ul> <li>$f(1) = 2$</li> <li>$f(2) = 3$</li> <li>$f(3) = 6$</li> <li>$f(6) = 9$</li> </ul> <p>That means $f(4) = 7$ and $f(5) = 8$, if we want to satisfy the “increasing” condition.</p> <ul> <li>We know $ff(4) = 12$ and $f(4) = 7$, so $f(7) = 12$.</li> <li>Similarly, $ff(5) = 15$ and $f(5) = 8$, so $f(8) = 15$.</li> </ul> <p>There’s a pattern emerging!</p> <ul> <li>$f(1) = 2$</li> <li>$f(2) = 3$</li> <li>$f(3) = 6$</li> <li>$f(4) = 7$</li> <li>$f(5) = 8$</li> <li>$f(6) = 9$</li> <li>$f(7) = 12$</li> <li>$f(8) = 15$</li> </ul> <p>I reckon, loosely, it goes up in threes up to a power of 3, then up in ones until reaching double a power of three.</p> <p>If that’s true, we expect $f(9) = 18$, then increases of one until we reach $n=18$.</p> <p>Indeed, $ff(6)=18$, so $f(9) = 18$; $ff(7)=21$, so $f(12)=21$ (filling in 10 and 11 automatically); $ff(8) = 24$, so $f(15)=24$, which makes $f(13) = 22$.</p> <hr /> <p>Is there an explicit form for this? Yes, but you’re not going to like it.</p> <p>Notice that $f\br{3^k} = 2\br{3^k}$ and $f\br{2\br{3^k}} = 3^{k+1}$ for integer $k$; the function is linear between those points.</p> <p>If we divide $n$ by the preceding power of 3 and truncate the result, we get either 1 or 2; if it’s a 1, we’re in the part with gradient 1 and if it’s a 3, we’re in the part with gradient 3. Equivalently, we can look at the fractional part of $\log_3(n)$: the truncated result being 1 is the same as that fractional part being less than $\log_3(2)$.</p> <p>So! If the fractional part of the base-three log is less than $\log_3(2)$, and the integer part of it is $k$, we’re on a segment with gradient 1 passing through $\br{3^k, 2\br{3^k}}$.</p> <p>If it’s larger than $\log_3(2)$, then our segment has gradient 3 and passes through $\br{3^{k+1}, 2\br{3^{k+1}}}$.</p> <p>Putting this all together gives:</p> <p>$f(n) = \begin{cases} n + 3^k &amp; {\log_3(n)-k \le \log_3(2)} \ 3n - 3^{k+1} &amp; \floor{\log_3(n)-k&gt; \log_3(2)}\end{cases}$,</p> <p>where $k = \floor{\log_3(n)}$.</p> <p>I told you you wouldn’t like it, but I hope it helps!</p> <p>- Uncle Colin</p> <p>* Thanks to @benjamin_leis, who first sent me the puzzle</p>Dear Uncle Colin, Someone gave me the puzzle: $ff(n) = 3n$ for all real $n$, $f(n)$ is a positive integer for all positive integer $n$ $f(n+1) &gt; f(n)$ for all positive integer $n$ What is $f(13)$? Any ideas? Functions? Fun? No.The Legendary Question 62022-04-11T00:00:00+01:002022-04-11T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/the-legendary-question-6<p>I say it’s legendary – I hadn’t heard of it before stumbling across this Numberphile video:</p> <iframe width="560" height="315" src="https://www.youtube.com/embed/L0Vj_7Y2-xY" title="YouTube video player" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture" allowfullscreen=""></iframe> <p>That said, it’s a question with a good story behind it: it’s from the 1988 International Mathematics Olympiad (which immediately says “this is a hard question”); it was marked with two asterisks to say “this is especially hard”.</p> <p>So hard, in fact, that Terry Tao – who would go on to earn a Fields Medal and general renown as an incredible mathematician and role model – “only” scored one point out of a possible seven. (He still got a gold medal. It was around the time of his 13th birthday. “Only.” Pah.)</p> <p>Anyway, here’s the question:</p> <blockquote> <p>Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 + b^2$.</p> <p>Show that $\frac{a^2 + b^2}{ab+1}$ is the square of an integer.</p> </blockquote> <p>If you’d like to have a go at it, please do! Obviously, don’t get disheartened if you struggle. It’s a two-asterisk IMO question. After the break, I’ll take you through my solution (which relied on a big hint from the video). And in case you missed the implication, there are spoilers below the line.</p> <hr /> <h3 id="my-solution">My solution</h3> <p>Here’s how far I got on my own.</p> <p>I know that $\frac{a^2 + b^2}{ab+1}$ is a positive integer, so I can write down that $\frac{a^2 + b^2}{ab+1} = k$ – and rearrange to make it $a^2 + b^2 = k(ab + 1)$.</p> <p>I think it looks nicer if it’s all on one side:</p> <ul> <li>$a^2 - kab + b^2 - k = 0$</li> </ul> <p>And I’m trying to show that $k$ has to be a square.</p> <p>I tried many things here, not limited to falling down a rabbit hole of discriminants. However, the hint of “look for other solutions” was the key one.</p> <p>That is, what I’ve got here is a quadratic in either $a$ or $b$. (I’ll pick $a$, but it doesn’t really matter.) And the thing about the roots of quadratics is, you know what they sum to – the roots of $x^2 + Bx + C = 0$ sum to $-B$.</p> <p>So here, given $k$ and $b$, there are (typically) two possible values for $a$, and they sum to $kb$.</p> <p>Put another way, if you know $a$, $b$ and $k$ are a solution, then so are $kb-a$, $b$ and $k$.</p> <p>For example, if you spot that $(a,b,k) = (0,2,4)$ is a solution, you immediately know that $(8,2,4)$ is also a solution. Because it’s symmetrical in $a$ and $b$, $(2,8,4)$ must also work. Then you can apply the trick again to get $(30,8,4)$, and so on – you can generate an infinite series of possible values for $a$; let’s call it ${ a_{i} }$. The sequence is different for each given value of $k$. (In general, $a = a_i$, $b=a_{i\pm 1}$ is a solution for the given value of $k$.)</p> <p>That sequence comes from the recurrence relation $a_{i+2} - ka_{i+1} + a_{i} = 0$ – and recurrence relations like this are simple enough to solve!</p> <p>You solve the characteristic equation, $\lambda^2 - k\lambda + 1 = 0$, to get $\lambda = \frac{k \pm \sqrt{k^2-4}}{2}$, and write down the solution: $a_n = A\br{k+\sqrt{k^2-4}}^n + B\br{k - \sqrt{k^2-4}}^n$ (subsuming the half into the constants).</p> <p>There’s a clever bit here, too.</p> <p>Thinking about the binomial expansion of each of the pieces, the even terms will be nice integers, but the odd terms will contain some multiple of the square root. So, any integer solutions for $a_n$ will need to account for that. There are two options:</p> <ol> <li>Make the square root vanish - in the positive integers, $k^2 - 4$ is a square only for $k=2$.</li> <li>Cancel the square root with an astute selection of constants – if $B = A$, the square roots will simply disappear.</li> </ol> <p>Case (1) can be shown not to work: going back to $a^2 - kab + b^2 -k = 0$, if $k=2$ then we have $\br{a-b}^2=2$, which has no integer solution ((splosh)).</p> <p>In case (2), it’s clear that $a_0 = 0$. Going back to the original setup, if $a=0$, we have $\frac{b^2}{1} = k$, which can only work if $k$ is a square number!</p> <p>And there you have it! $k$ has to be a square.</p> <hr /> <h3 id="analysis">Analysis</h3> <p>I’m not sure how convincingly I’ve shown that this is the only way to account for the square roots – it might be better to say that $B = A + C$ in general and say:</p> <ul> <li>$a_n = A\br{\br{ k + \sqrt{k^2-4}}^n + \br{k - \sqrt{k^2-4}}^n} + C\br{k - \sqrt{k^2-4}}^n$</li> </ul> <p>The $A$ term has no square roots in it, but the $C$ one certainly does – unless $C=0$.</p> <p>I’m curious to hear about different solutions! As always, if you’ve got something to add, <a href="mailto:colin@flyingcoloursmaths.co.uk">drop me an email</a>.</p>I say it’s legendary – I hadn’t heard of it before stumbling across this Numberphile video:Dictionary of Mathematical Eponymy: The Ollerenshaw-Brée Theorem2022-04-04T01:00:00+01:002022-04-04T01:00:00+01:00https://www.flyingcoloursmaths.co.uk/dictionary-of-mathematical-eponymy-ollerenshaw-bree-theorem<p>I’m currently reading (on @drmackiver’s recommendation) <em>Across the Board</em> by John J Watkins, a study of the mathematical side of some chessboard-related puzzles. Among them is using knight’s tours to generate magic squares, which is the kind of useless but pleasing trick I am absolutely here for.</p> <p>So I’m just as pleased that today’s entry in the DOME is also about magic squares – in particular, <em>most-perfect</em> magic squares.</p> <h3 id="what-is-the-ollerenshaw-brée-theorem">What is the Ollerenshaw-Brée Theorem</h3> <p>I quote <a href="https://www.theoremoftheday.org/CombinatorialTheory/MostPerfect/TotDMostPerfect.pdf">Robin Whitty of Theorem of the Day</a>:</p> <blockquote> <p>Let $n$, a positive integer, have prime factorisation $n = p_1^{s_1} p_2^{s_2}\dots p_r^{s_r}$, in which $p_1 = 2$ and $s_1 \ge 2$, so that $n$ is doubly even ((“A multiple of four,” as I would have it)). The number of $n \times n$ most-perfect magic squares, up to horizontal, vertical and diagonal symmetry, is given by:</p> <p>$N(n) = M(n) \sum_{v=0}^{\tau(n)} W(v) \left[ W(v) + W(v+1)\right]$,</p> <p>where $\tau(n)$ is the number of divisors of $n$; $M(n)$ is given by $M(n) = 2^{n−2} \left[(n/2)!\right]^2$; and $W(v) = \sum_{i=0}^{v} (-1)^{v+i} \nCr{v+1}{i+1} \Product_{j=1}^{r} \nCr{s_j+i}{i}$</p> </blockquote> <h3 id="whats-a-most-perfect-magic-square">What’s a most-perfect magic square?</h3> <p>A most-perfect magic square is one that satisfies two conditions:</p> <ul> <li>Any cell in the grid has one natural ‘opposite’ cell, the one that’s $n/2$ units away diagonally. In a most-perfect magic square, the numbers in a cell and its opposite sum to $n^2-1$.</li> <li>The numbers in any 2-by-2 block in the grid sum to $2\left(n^2-1\right)$</li> </ul> <p>For example, Whitty gives the example (and how to construct it):</p> <table> <tbody> <tr> <td>0</td> <td>14</td> <td>5</td> <td>11</td> </tr> <tr> <td>7</td> <td>9</td> <td>2</td> <td>12</td> </tr> <tr> <td>10</td> <td>4</td> <td>15</td> <td>1</td> </tr> <tr> <td>13</td> <td>3</td> <td>8</td> <td>6</td> </tr> </tbody> </table> <p>Every row, column, diagonal, split diagonal (such as 14-2-1-13 or 14-7-1-8) and 2-by-2 block in the grid sums to 30.</p> <h3 id="why-is-it-interesting">Why is it interesting?</h3> <p>Well, because magic squares are kind of cool. But beyond that, the Ollerenshaw-Brée formula was the first counting formula for any class of magic squares.</p> <h3 id="who-was-kathleen-ollerenshaw">Who was Kathleen Ollerenshaw?</h3> <p>Kathleen Timpson was born in Withington, Manchester in 1912. She became both deaf ((She wouldn’t have an effective hearing aid until 1949)) and fascinated by mathematics at school, studying at Lady Barn House School in Cheadle and St Leonards in St Andrews. She studied maths at Somerville College, Oxford and remained there to earn her DPhil in 1945.</p> <p>After the Second World War, Ollerenshaw ((She married Robert in 1939)) lectured part-time at the University of Manchester and took an active part in local and national politics. She died in 2014, aged 101.</p>I’m currently reading (on @drmackiver’s recommendation) Across the Board by John J Watkins, a study of the mathematical side of some chessboard-related puzzles. Among them is using knight’s tours to generate magic squares, which is the kind of useless but pleasing trick I am absolutely here for.Ask Uncle Colin: The RMS Line2022-03-28T00:00:00+01:002022-03-28T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-rms-line<blockquote> <p>Dear Uncle Colin,</p> <p>My textbook claims that the RMS voltage of a sine wave is $\sqrt{\frac{1}{2}}$ because between $x=0$ and $x=\pi$, the signed area between the curves $y=\sin(x)$ and $y=\sqrt{2}$ is 0 – but I checked and it isn’t. What’s going on?</p> <p>Really A Disappointing Instructional Overview</p> </blockquote> <p>Hi, RADIO, and thanks for your message!</p> <p>You’re quite right – we can use the symmetry of the graph to restrict ourselves to $0 \le x \le \piby 2$, and then we just need to look at the integral $\int_0^{\piby2} \left(\sin(x) - \sqrt{\frac{1}{2}}\right)\dx$.</p> <p>That’s $\left[ -\cos(x) - x\sqrt{\frac{1}{2}} \right]_0^{\piby2}$, or $1-\piby2 \sqrt{\frac{1}{2}}$. That’s clearly not 0 (it’s about -0.11).</p> <p>So, two questions: one, where should the equal-area line lie; and two, what should the RMS explanation have said?</p> <p>We’ve done most of the work for the first one already: we just need to replace $\sqrt{\frac{1}{2}}$ with $k$ and solve for it.</p> <p>The integral is $\left[ -\cos(x) - kx \right]_0^{\piby2}$, or $1 - k\piby 2 = 0$ – which means $k = \frac{2}{\pi}$. But that’s not the RMS.</p> <hr /> <p>The RMS of a function $f(x)$ over an interval $a \lt x \lt b$ is defined as $y_{RMS} = \sqrt {\frac{1}{b-a}\int_a^b [f(x)]^2 \dx}$.</p> <p>I find it slightly nicer to square and cross-multiply to give $(b-a)y^2_{RMS} = \int_a^b [f(x)]^2 \dx$, which at least looks like you’ve got the same kind of thing on both sides.</p> <p>In any case, we need to work out $\int_0^{\piby2} \sin^2(x) \dx$. Using the identity $\sin^2(x) \equiv \frac{1-\cos(2x)}{2}$, this integrates to $\left[ \frac{x}{2} - \frac{\sin(2x)}{4} \right]_0^{\piby2}$, which is $\piby 4$.</p> <p>So we have $\left(\piby 2\right) y_{RMS}^2 = \left( \piby 4\right)^2$, which simplifies down to $y_{RMS} = \frac{1}{\sqrt{2}}$, which is a relief.</p> <p>It turns out that it’s not that the curves have the same area beneath them – it’s that the volumes of revolution they define (about the $x$-axis) have the same volume.</p> <p>Hope that helps!</p> <p>- Uncle Colin</p>Dear Uncle Colin, My textbook claims that the RMS voltage of a sine wave is $\sqrt{\frac{1}{2}}$ because between $x=0$ and $x=\pi$, the signed area between the curves $y=\sin(x)$ and $y=\sqrt{2}$ is 0 – but I checked and it isn’t. What’s going on? Really A Disappointing Instructional OverviewAn Argand Diagram Puzzler2022-03-21T00:00:00+00:002022-03-21T00:00:00+00:00https://www.flyingcoloursmaths.co.uk/argand-puzzler<p>On Twitter, @whitehughes posted a nice complex numbers problem:</p> <blockquote class="twitter-tweet" data-partner="tweetdeck"><p lang="en" dir="ltr">I really like this complex number geometry question that I came across this morning. There are so many different ways to tackle it but some are definitely more efficient than others! <a href="https://t.co/yrt1n9TqBU">pic.twitter.com/yrt1n9TqBU</a></p>&mdash; Susan Whitehouse (@Whitehughes) <a href="https://twitter.com/Whitehughes/status/1413822950868979712?ref_src=twsrc%5Etfw">July 10, 2021</a></blockquote> <script async="" src="https://platform.twitter.com/widgets.js" charset="utf-8"></script> <p>Have a go yourself, if you’d like to; below the line are spoilers.</p> <hr /> <p>As Susan says, there are several ways to tackle this. Off the top of my head, I can see:</p> <ul> <li>An algebraic method, finding the equation of the line BC and a circle of the right radius, then solving</li> <li>A vectors method, finding the vector AB, a perpendicular vector, and hence BC</li> <li>A trigonometric method, solving triangle ABC</li> </ul> <p>There are probably multiple variations on these; I went directly for manipulating complex numbers.</p> <h3 id="my-method">My method</h3> <p>Let $z = u-v$, which corresponds to the vector $\vec{BA}$. This works out to be $-2 - (2\sqrt{3}-2)\i$.</p> <p>The point C is in the first quadrant, so we need to rotate AB by 90º <em>clockwise</em> – that corresponds to multiplying $z$ by $-i$. (I think this is the nugget of my solution).</p> <p>But that won’t give us C – it gives us a complex number corresponding to a multiple of a vector between B and C. To get our final answer, we’ll need to double it, and add on $v$.</p> <p>So:</p> <ul> <li>$-2zi = (4\sqrt{3} - 4) + 4\i$</li> <li>$v-2zi = (4\sqrt{3} - 1) + 6\i$</li> </ul> <p>And we’re done!</p> <h3 id="or-nearly">Or nearly</h3> <p>What do we do when we’ve finished a problem, children?</p> <p>That’s right! We check our answer, and we look for other solutions.</p> <p>And naturally, I leave those as an exercise; if you come up with anything interesting, let me know!</p>On Twitter, @whitehughes posted a nice complex numbers problem:Ask Uncle Colin: How can I tell if my quartic is a square?2022-03-14T00:00:00+00:002022-03-14T00:00:00+00:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-quartic-square<blockquote> <p>Dear Uncle Colin,</p> <p>I have a quartic expression and I want to know whether it can be expressed as a perfect square. How would you do it?</p> <p>Quite Ugly Algebra – Rooting Turns Into Clean Squares?</p> </blockquote> <p>Hi, QUARTIC, and thanks for your message!</p> <p>Like quadratics, quartics have a discriminant. Unfortunately, the discriminant is too unwieldy to be much use here, and doesn’t actually answer your question – so we won’t be looking at it.</p> <p>I can think of two methods for this: an algebraic one (matching coefficients) and a calculus one (finding turning points).</p> <p>For the purposes of this post, I’ll consider two expressions:</p> <ul> <li>$9x^4 - 30x^3 + 37x^2 - 20x + 4$, which is a square</li> <li>$9x^4 - 30x^3 + 37x^2 - 20x + 9$, which is not a square</li> </ul> <h3 id="matching-coefficients">Matching coefficients</h3> <p>If the quartic is a perfect square, its square root is an arbitrary quadratic that can be expressed as $ax^2 + bx + c$. So the quartic is $\br{ax^2 + bx + c}^2$, or $a^2x^4 + 2abx^3 + (b^2 + 2ac)x^2 + 2bcx + c^2$.</p> <p>That’s a bit of a mess, but it does allow us to pattern match quite easily.</p> <p><em>If</em> the quartic is a square, then its $x^4$ coefficient must be a square ($a^2$) and its unit term must be a square $c^2$. Both of our candidates satisfy this test. Without loss of generality, we can say that $a$ is positive.</p> <ul> <li>For the first expression, $a=3$ and $c = \pm 2$</li> <li>For the second, $a=3$ and $c = \pm 3$</li> </ul> <p>We can work out a candidate $b$ from the $x^3$ coefficient, which must equal $2ab$. In both cases, that gives $b = -5$.</p> <p>Then we can check whether it works using the $x$ coefficient, which must equal $2bc$.</p> <p>In both cases, the $x$ coefficient is $-20$, so $c = 2$ is the only possible value – which means the first expression is $\br{3x^2 - 5x + 2}^2$ and the other is not a square.</p> <h3 id="calculus">Calculus</h3> <p>If the quartic is a perfect square, all of its zeros – including the complex ones – must also be turning points.</p> <p>Let’s look at the derivative of $(ax^2 + bx + c)^2$, which is $(2ax + b)(ax^2 + bx + c)$.</p> <p>That has (generally) three solutions: one at $x = -\frac{b}{2a}$ and two more at $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.</p> <p>That is to say, the two ‘outer’ turning points are evenly spaced around the inner one – the curve is symmetric about its inner turning point. (The turning points might involve complex numbers. That’s ok.) This is a necessary but not sufficient condition for the quartic to be a square – as we’ll see immediately: in both of the expressions, we have the same cubic derivative: $36x^3 - 90x^2 + 74x - 20$.</p> <p>If the expression has a line of symmetry, then it’s midway between the points of inflexion! So we can differentiate again: $108x^2 - 180x + 74$; we don’t even need to solve it, the zeros are evenly spaced around $\frac{180}{2\times108}$, which is $\frac{5}{6}$.</p> <p>This tells us that $6x -5$ is a factor of $36x^3 - 90x^2 + 74x - 20$. We can do the division and get $6x^2 - 10x + 4$ back out; this factorises as $2(3x-2)(x-1)$.</p> <p>The outer turning points are therefore at $x = \frac{2}{3}$ and $x=1$, which are indeed equally spaced about $x=\frac{5}{6}$.</p> <p>But are they zeros? You have two options. You may plonk them back into the original expressions and see whether you get 0; or you may work out $(3x-2)^2(x-1)^2$ and see that you match exactly one of the options.</p> <hr /> <p>Do you know of a better way? Let me know!</p> <p>- Uncle Colin</p>Dear Uncle Colin, I have a quartic expression and I want to know whether it can be expressed as a perfect square. How would you do it? Quite Ugly Algebra – Rooting Turns Into Clean Squares?