Jekyll2021-04-21T01:56:20+01:00https://www.flyingcoloursmaths.co.uk/feed.xmlFlying Colours MathsFlying Colours Maths helps make sense of maths at A-level and beyond.Ask Uncle Colin: A Geometric Subset2021-04-21T00:00:00+01:002021-04-21T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-a-geometric-subset<blockquote>
<p>Dear Uncle Colin,</p>
<p>I was working on a MAT question that asked about finding a subset, $S$, of the 2D-plane, and a point $P$ such that no point in $S$ was the closest to $P$. I had no idea where to start!</p>
<p>Omitted Point, Euclidean Norm</p>
</blockquote>
<p>Hello, OPEN, and thanks for your message!</p>
<p>My first answer wasn’t the one the MAT answer suggested, but I stand by it: if $S$ is the empty set, it’s definitely a subset of the 2D-plane ((every point in the empty set is also a member of the set of points in the real plane, vacuously)). And since there are no points in $S$ to be nearest to $P$, this satisfies the condition.</p>
<p>But they had something a bit trickier in mind.</p>
<p>According to the answer sheet, a valid answer would be something like:</p>
<blockquote>
<p>$S$ is the set of points such that $x^2+y^2 < 1$, and $P$ is a distance greater than 1 from the origin.</p>
</blockquote>
<p>Let’s unpack that a bit. The set $S$ is an open disc, a filled circle centred on the origin, with a radius of 1, but such that the circumference is not part of the set. $P$ is outside of the circle. So why is there not a point in $S$ that’s closest to $P$?</p>
<p>Suppose (without loss of generality) that $P$ is on the positive $x$-axis. Any candidate closest point would also have to be on the positive $x$-axis, with an $x$-coordinate smaller than 1 to be a member of $S$.</p>
<p>If someone claims “The point $(X,0)$, is the closest member of $S$ to $P$”, you can immediately say “No, it isn’t – the point $\left(\frac{1+X}{2},0\right)$, midway between $P$ and the circumference, is inside of $S$ and closer to $P$ than your point.” Contradiction. Boom.</p>
<p>It’s a bit subtle (which is why I prefer my empty set method), and I’m not sure it’s a fair thing to ask an A-level student to come up with on the fly in an exam, but it’s interesting enough to write up.</p>
<p>I hope that helps!</p>
<p>- Uncle Colin</p>Dear Uncle Colin, I was working on a MAT question that asked about finding a subset, $S$, of the 2D-plane, and a point $P$ such that no point in $S$ was the closest to $P$. I had no idea where to start! Omitted Point, Euclidean NormA pretty puzzle2021-04-19T00:00:00+01:002021-04-19T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/a-pretty-puzzle<p>I heard it from @benjamin_leis, and he says he heard it from @CMonMattTHINK, and I love it:</p>
<blockquote>
<p>The number of integer solutions to $x^2 + xy + y^2 = a$ appears to be a multiple of six for all $a \in \mathbb{Z_+}$ . Why?</p>
</blockquote>
<p>How good a puzzle is this? I started my swimming class on Thursday and decided to swim non-stop until I solved it. Forty lengths later they kicked me out of the pool.</p>
<p>As always, spoilers below the line.</p>
<hr />
<p>This is a very pretty puzzle. Some observations to start with:</p>
<ul>
<li>The curve describes an ellipse</li>
<li>There are two lines of symmetry to the ellipse, $x=y$ and $x=-y$.</li>
<li>The extreme points of the ellipse are on the lines $x+2y=0$ and $2x+y=0$.</li>
</ul>
<h3 id="how-i-did-it">How I did it</h3>
<p>I think it’s useful to break my solution into two bits, and acknowledge that the “correct”, underlying answer came after a lot of head-scratching and missteps. The key bit of reasoning was:</p>
<blockquote>
<p>If $(X,Y)$ is an integer solution, fixing $y=Y$ gives a quadratic $x^2 + Yx + Y^2-a=0$. > I could solve that explicitly, but I don’t need to; $Y^2-a$ is an integer, so the second solution is also an integer; the solutions sum to $-Y$, so the other solution is $x=-X-Y$, which (for convenience) I’ll call $Z$.</p>
</blockquote>
<p>That means, if $(X,Y)$ is an integer solution, so is $(Z,Y)$. Applying similar reasoning to the resulting point, fixing $x=Z$ gives a solution at $(Z,X)$.</p>
<p>We can carry on around to get $(Y,X)$, $(Y,Z)$ and $(X,Z)$ before returning to $(X,Y)$, giving six algebraically distinct solutions.</p>
<p>BUT! Rotational symmetry also gives a solution at $(-X, -Y)$ - so changing the signs on all of the points we found before gives a further six solutions, a total of 12.</p>
<p>However, they aren’t all necessarily distinct! If any pair of $X$, $Y$ and $Z$ are equal, then each point we’ve found appears twice, leaving us with six solutions (in fact, two lie on the minor axis and the remaining four at the extreme points).</p>
<h3 id="underlying-group-structure">Underlying group structure</h3>
<p>I wasn’t <em>happy</em> with my answer until I managed to shoehorn a group structure onto the solutions. The points can be classified by three pieces of information:</p>
<ul>
<li>Which letter is missing (isomorphic to $\mathbb{Z}_3$, with $X \sim 0$, $Y \sim 1$ and $Z \sim 2$)</li>
<li>Whether the signs are + or - (isomorphic to $\mathbb{Z}_2$)</li>
<li>Whether the second letter immediately follows the first (isomorphic to $\mathbb{Z}_2$ as well). By convention, $Y$ follows $X$, $Z$ follows $Y$ and $X$ follows $Z$ in a pleasing cycle.</li>
</ul>
<p>And this gives a natural way to combine any pair of integer solutions!</p>
<p>The point $(X,Z)$ has $Y$ missing, positive signs and is out of order.</p>
<p>The point $(-Y,-X)$ has letter $Z$, negative signs and is out of order.</p>
<p>Combining them together, $Y + Z \sim (1 + 2)$, which is $0 \sim X$; the sign is negative, and the letters must be in order, giving $(-Y, -Z)$.</p>
<p>This is a group, because it’s isomorphic to $\mathbb{Z}_3 \times \mathbb{Z}_2 \times \mathbb{Z}_2$, and has order 12.</p>
<p>(In the degenerate case when a pair of variables is equal, each point coincides with exactly one other.)</p>
<h3 id="even-more-underlying">Even more underlying</h3>
<p>After reading around a bit on Diophanitine equations - by which I mean scanning a few paragraphs in a huff - I realised that the ellipse equation can be rewritten as:</p>
<ul>
<li>$2x^2 + 2xy + y^2 = 2a$</li>
<li>$x^2 + x^2 + 2xy + y^2 + y^2 = 2a$</li>
<li>$x^2 + (x+y)^2 + y^2 = 2a$</li>
</ul>
<p>Or, equivalently, $x^2 + (-x-y)^2 + y^2 = 2a$. Why that way?</p>
<p>To tell the truth, that’s still a bit nebulous. However, it has the lovely pair of properties:</p>
<ul>
<li>by symmetry, if $(X,Y)$ is a solution, any permutation of two elements of ${ X, Y, -X-Y }$ is also a solution</li>
<li>by a different symmetry, if $(X,Y)$ is a solution, any permutation of two elements of ${-X, -Y, X+Y}$ is also a solution.</li>
</ul>
<p>If $X$, $Y$ and $-X-Y$ are all different, then there are six solutions from the first set and six from the second; if two are the same, then there are three from each.</p>
<p>In either case, the number of solutions is a multiple of six.</p>
<hr />
<p>Can I stop swimming now? I need to put my computer in rice.</p>I heard it from @benjamin_leis, and he says he heard it from @CMonMattTHINK, and I love it:Ask Uncle Colin: A classical mixture2021-04-14T00:00:00+01:002021-04-14T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-a-classical-mixture<blockquote>
<p>Dear Uncle Colin,</p>
<p>The puzzle asks: you have two glasses with equal volumes of water and juice. You take a tablespoon of water and mix it into the juice; you take a tablespoon of the juice-water mix and mix it into the water. Is there more water in the juice, or juice in the water? Apparently the amounts are the same, but I don’t see how they can be!</p>
<p>Combining Liquids And Stirring Strangely - It’s Confusing!</p>
</blockquote>
<p>Hi, CLASSIC, and thanks for your message!</p>
<p>There are many ways to approach this puzzle, and I’m going to leave what I think is the simplest one until last.</p>
<h3 id="algebra">Algebra</h3>
<p>Initially, one glass contains $V$ of water; the other contains $V$ of juice.</p>
<p>Take a volume $v$ of water and add it to the juice; there is now $V - v$ of water in one glass; in the other, there is $V$ of juice and $v$ of water.</p>
<p>Now the tricky bit: we’re going to move a total of $v$ of the mixture back into the first glass, but it could be all juice, all water, or any mixture in between. So, let’s say the $v$ is made up of $t$ of juice and $v-t$ of water.</p>
<p>After moving it, the first glass has $(V - v) + (v-t)$, or $V-t$ of water and $t$ of juice.</p>
<p>The second glass has $V-t$, of juice, and $v - (v-t)$ or $t$ of water.</p>
<p>There’s as much juice in the water as there is water in the juice.</p>
<h3 id="cards">Cards</h3>
<p>The algebra is all well and good, but it’s not exactly intuitive. Instead, I like to think about things you can count, like cards of different colours.</p>
<p>Imagine you have a number of red cards and a number of blue cards. You take a certain number of red cards and mix them with the blue ones; you take the same number from the mixed pack and put them back in the red pack.</p>
<p>Now, there’s the same number of cards in each pack, and there’s still the same number of red and blue cards split between them. For every red card in the blue pack, a blue card must have gone to the red pack, and vice versa: there are as many red cards in the blue pack as vice versa.</p>
<h3 id="intuitively">Intuitively</h3>
<p>Everything in the juice glass that isn’t juice, is water.</p>
<p>Everything in the water glass that isn’t water, is juice.</p>
<p>There’s the same amount of water and juice altogether, so the “isn’ts” must be the same.</p>
<hr />
<p>Hope that helps!</p>
<p>- Uncle Colin</p>Dear Uncle Colin, The puzzle asks: you have two glasses with equal volumes of water and juice. You take a tablespoon of water and mix it into the juice; you take a tablespoon of the juice-water mix and mix it into the water. Is there more water in the juice, or juice in the water? Apparently the amounts are the same, but I don’t see how they can be! Combining Liquids And Stirring Strangely - It’s Confusing!Circles in Seattle2021-04-12T00:00:00+01:002021-04-12T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/circles-in-seattle<p>A day off ill, and the chance to look muddle-headedly at a tweet from @trianglemanscd long ago:</p>
<blockquote class="twitter-tweet"><p lang="en" dir="ltr">OK Geometers, I'm gonna need you to problem solve for me. <a href="https://t.co/00njFhiBEw">pic.twitter.com/00njFhiBEw</a></p>— Christopher Danielson (<span class="citation" data-cites="Trianglemancsd">@Trianglemancsd</span>) <a href="https://twitter.com/Trianglemancsd/status/1056955508995883010?ref_src=twsrc%5Etfw">October 29, 2018</a></blockquote>
<script async="" src="https://platform.twitter.com/widgets.js" charset="utf-8"></script>
<p>I’m a geometer! And I like problem-solving!</p>
<p>There are <em>loads</em> of valid approaches here, and I’m going to talk of three.</p>
<h3 id="a-really-boring-approach">A really boring approach</h3>
<p>Step 1: go to Google maps Step 2: find the scale of the map Step 3: use that to calibrate the distances.</p>
<p>Of course I <em>could</em> do that. But that’s geography, not maths.</p>
<h3 id="slightly-nicer">Slightly nicer</h3>
<p>A more mathematical approach would be to centre a circle on each of the the landmarks with a radius proportional to the given distance to the mystery hotel. Slide the scale until the circles all intersect, or close enough. <a href="/images/Screenshot-2020-09-17-at-12.11.05.png"><img src="/images/Screenshot-2020-09-17-at-12.11.05.png" alt="" /></a></p>
<p><a href="/images/Screenshot-2020-09-17-at-12.11.23.png"><img src="/images/Screenshot-2020-09-17-at-12.11.23.png" alt="" /></a></p>
<p>This is <em>becoming</em> more mathematical, but still a bit boring: my best effort gives a location somewhere between Madison and Union, where there are half a dozen hotels.</p>
<h3 id="the-proper-way">The proper way</h3>
<p>But as a mathematician, fudging $k$ to get an answer feels a bit dirty. The circles don’t intersect properly because the distances are rounded. Surely we can do it another way?</p>
<p>Of course we can.</p>
<p>For example, the mystery hotel is marked as 0.3m from both Pike Place Market and Columbia Center. Or rather, both of those landmarks are between 0.25m and 0.35m away.</p>
<p>Or rather rather, the <em>ratio</em> of the distances from our hotel to those landmarks is between $\frac{5}{7}$ and $\frac{7}{5}$. What does that look like?</p>
<p>Well, like this:</p>
<p><a href="/images/Screenshot-2020-09-17-at-12.17.41.png"><img src="/images/Screenshot-2020-09-17-at-12.17.41.png" alt="" /></a></p>
<p>A large circle around each landmark is excluded - and I can do the same with any pair I choose: <a href="/images/Screenshot-2020-09-19-at-08.52.02.png"><img src="/images/Screenshot-2020-09-19-at-08.52.02.png" alt="" /></a></p>
<p>Several observations here: first, the orange region, centred on A and G, seems to disagree with the other pairs I picked. I don’t know why that is - possibly the distance is measured from a different part of CenturyLink Field. Secondly, the others converge on a point 2.5 blocks from both Union and I-5 ((Oh good grief, sudden flashback to the one time I drove on that)) - which could reasonably <a href="https://www.google.com/maps/@47.6077527,-122.3361587,17.12z">be the Executive Hotel Pacific or the West Seattle</a>. (There are several other plausible options nearby).</p>
<h3 id="where-was-it">Where was it?</h3>
<p>It was, it turns out, neither of those, but the Kimpton Alexis, several blocks further from the freeway – astonishingly, bang on where the bulk of the circles coincide in the “slightly nicer way”!</p>
<p>I don’t know what the take-home message here is, only that it was a nice thing to play with, and I’d be curious to hear what other methods you might try!</p>
<p>* Accessibility features on images are currently unavailable owing to a glitch on the website. I hope to find and address this before it goes live, but who can predict?</p>A day off ill, and the chance to look muddle-headedly at a tweet from @trianglemanscd long ago:Ask Uncle Colin: A Couple of Tangents2021-04-07T00:00:00+01:002021-04-07T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-a-couple-of-tangents<blockquote>
<p>Dear Uncle Colin,</p>
<p>I’ve got two circles ($x^2 + y^2 = 3^2$ and $(x-10)^2 + y^2 =5^2$) and I want to find the equations of the common tangents. I’ve been stuck for ages!</p>
<p>- Tackling A New Geometry Exercise, Need Tuition</p>
</blockquote>
<p>Hi, TANGENT, and thanks for your message!</p>
<p>I can see two fairly nice ways to approach this.</p>
<h3 id="with-algebra">With algebra!</h3>
<p>Suppose the equation of a tangent line is $y = m(x-x_0)$ for some fixed values of $m$ and $x_0$. ((Other equations are available.))</p>
<p>This line meets the first circle when $x^2 + m^2(x-x_0)^2 = 3^2$; if the line is tangent, the equation has a repeated root.</p>
<p>Rearrange to get $x^2 (1+m^2) - 2m^2x_0x + (m^2x_0^2 - 9) = 0$; this has equal roots if $4m^4x_0^2 = 4(1+m^2)(m^2 x_0^2 - 9)$.</p>
<p>Expanding that and dividing by 4 gives $m^4 x_0^2 = m^2x_0^2 - 9 + m^4x_0^2 - 9m^2$</p>
<p>There’s an $m^2 x_0^2$ on each side, so $m^2(x_0^2 - 9) = 9$, or $m^2 = \frac{9}{x_0^2 - 9}$.</p>
<p>So, this is the relationship between $m$ and $x_0$ for any line that touches the first circle. What about the second one?</p>
<p>We can do a similar analysis. Substitute to get $(x-10)^2 + m^2(x-x_0)^2 = 5^2$, and rearrange to get $x^2 (1+m^2) - (2m^2 x_0 + 20)x + (75 + m^2x_0^2)= 0$.</p>
<p>Next, this needs to have repeated roots, so $(2m^2x_0 + 20)^2 = 4(1+m^2)(75 + m^2x_0^2)$.</p>
<p>Expanding and dividing by 4: $m^4 x_0^2 + 20m^2 x_0 x + 100 = 75 + m^2x_0^2 + 75m^2 + m^4x_0^2$.</p>
<p>Again, the $m^4$ term vanishes to leave us, this time, with $m^2(x_0^2-20x_0 + 75) = 25$.</p>
<p>We have two relationships now:</p>
<ul>
<li>$m^2 = \frac{9}{x_0^2 - 9}$</li>
<li>$m^2 = \frac{25}{x_0^2 - 20x_0 + 75}$</li>
</ul>
<p>We can combine those to give $\frac{9}{x_0^2 - 9} = \frac{25}{x_0^2 - 20x_0 + 75}$, or $9(x_0^2 - 20x_0 + 75) = 25(x_0^2 - 9)$.</p>
<p>This rearranges to $0 = 16x_0^2 + 180x_0 - 900$, or $4x_0^2 + 45x_0 - 225 = 0$…</p>
<p><strong>whoosh</strong></p>
<p>“$(4x_0-15)(x_0+15)$. You’re welcome.”</p>
<p><strong>whoosh</strong></p>
<p>So $x_0 = \frac{15}{4}$ or $-15$, and $m^2 = \frac{16}{9}$ or $\frac{1}{24}$.</p>
<p>The tangent lines are $y = \pm \frac{4}{3}\left( x - \frac{15}{4}\right)$ and $y = \pm \frac{1}{\sqrt{24}}\left( x + 15 \right)$.</p>
<p>The Mathematical Ninja, I couldn’t help noticing, wrinkled their face in passing.</p>
<h3 id="with-geometry">With geometry:</h3>
<p>We have a circle with radius 3, centred at the origin.</p>
<p>We have a circle with radius 5, centred at point P, (-10, 0).</p>
<p>Suppose we have a line that’s tangent to the first circle at T and the second circle at U, and that crosses the X-axis at X.</p>
<p>Triangle OTX and triangle PUX are similar, with a scale factor of 3:5.</p>
<p>Now, there are two classes of tangent: inner tangents (that cross between the circles) and outer tangents (that cross outside). Let’s look at those in turn.</p>
<p>If the tangents cross the axis outside the circles, we know that the difference between the hypotenuses of the triangles is 10. That means the longer hypotenuse is 25 units long and the shorter one 15 – and these must meet the axis at (-15,0).</p>
<p>If the shorter hypotenuse is 15 and its shorter leg is 3, its longer leg is $\sqrt{225-9} = 6\sqrt{6}$. The gradient of the tangent is then $\pm \frac{3}{6\sqrt{6}} = \pm \frac{1}{\sqrt{24}}$, so the equation of the line is $y = \pm \frac{1}{\sqrt{24}}(x+15)$. (The $\pm$ is because the tangent could be above or below the axis).</p>
<p>Looking at the inner tangents, the sum of the hypotenuses is 10, so their lengths are $\frac{15}{4}$ and $\frac{25}{4}$, and they cross at $\left(\frac{15}{4},0\right)$.</p>
<p>The shorter hypotenuse is $\frac{15}{4}$ long, and one of the legs is 3, so the other is $\frac{9}{4}$. The gradient of the tangent is $\pm\frac{9}{3/4} = \pm\frac{4}{3}$, and the equation of the line is $y = \frac{4}{3}\left(x - \frac{15}{4}\right)$.</p>
<hr />
<p>Hope that helps!</p>
<p>- Uncle Colin</p>Dear Uncle Colin, I’ve got two circles ($x^2 + y^2 = 3^2$ and $(x-10)^2 + y^2 =5^2$) and I want to find the equations of the common tangents. I’ve been stuck for ages! - Tackling A New Geometry Exercise, Need TuitionThe Dictionary of Mathematical Eponymy: Borromean rings and the Clélie curve2021-04-05T00:00:00+01:002021-04-05T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/borromean-rings-and-the-clelie-curve<p><em>I’m cheating a bit with this one; the Borromean rings are named after the whole of the Borromeo family, and the Clélie curve is (unusually) named after a mathematician’s first name. But this is my dictionary, and I’m going to claim a twofer here.</em></p>
<h3 id="who-was-clelia-grillo-borromeo">Who was Clelia Grillo Borromeo?</h3>
<p>Clelia (or Celia) Grillo Borromeo (1684-1777) was a Genovese scientist and mathematician. The daughter of a duke, she spoke eight languages and was described as an ‘independent’ person, which was an eccentric attribute for a woman of her time.</p>
<p>She died in Milan in 1777.</p>
<h3 id="what-are-the-borromean-rings">What are the Borromean rings?</h3>
<p>The Borromean rings are one of my favourite mathematical objects: three rings that interlock in such a way that removing any one of them causes the other two to fall apart. They’re named for the aristocratic Borromeo family, but have a much longer history: Norse image stones show a topologically equivalent <em>valknut</em>, made of three equilateral triangles, as early as the 7th century.</p>
<h3 id="what-is-a-clélie-curve">What is a Clélie curve?</h3>
<p>A <em>Clélie</em> is a curve on a sphere whose longitude ($\phi$) and colatitude ($\theta$) are connected by the equation $\phi = c \theta$ for $c > 0$.</p>
<p>The curve traces out a spiral on the spherical surface. It’s neat!</p>
<h3 id="why-is-it-important">Why is it important?</h3>
<p>A Clélie describes the motion of a satellite that passes over the Earth’s poles. Plus, they’re pretty neat!</p>I’m cheating a bit with this one; the Borromean rings are named after the whole of the Borromeo family, and the Clélie curve is (unusually) named after a mathematician’s first name. But this is my dictionary, and I’m going to claim a twofer here.Ask Uncle Colin: Remembering the Unit Circle2021-03-31T00:00:00+01:002021-03-31T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-remembering-the-unit-circle<blockquote>
<p>Dear Uncle Colin,</p>
<p>How do you remember the values for the trig functions in the unit circle?</p>
<p>- Can’t Abide Stupid Trigonometry</p>
</blockquote>
<p>Hi, CAST, and thanks for your message!</p>
<p>The short answer is, by knowing what sine and cosine <em>mean</em> on a unit circle, knowing a couple of special triangles, and knowing a bit about symmetry.</p>
<h3 id="what-the-functions-mean">What the functions mean</h3>
<p>Mentally draw a radius to any point on a unit circle, and mentally measure an angle from the x-axis, anticlockwise, until you hit the radius. Call that angle $\theta$.</p>
<p>$\sin(\theta)$ is, <em>by definition</em>, the $y$-coordinate of that point.</p>
<p>$\cos(\theta)$ is, <em>by definition</em>, the $x$-coordinate.</p>
<p>Try a few (mentally). If you pick the point at the top of the circle, $\theta$ is a right angle; $\sin(90º) = 1$ and $\cos(90º) = 0$. ((I’ve picked degrees here because I’m in that sort of a mood. Lay off, OK, sensei?))</p>
<p>Pick the rightmost point, at $(-1,0)$. The angle there is 180º, so $\cos(180º)=-1$ and $\sin(180º) = 0$. The corners of the circle ((you heard me)) are all straightforward to work out.</p>
<p><strong>The sine and cosine functions pick out the co-ordinates of a point on a unit circle, as a function of its angle from the $x$-axis.</strong> This is important.</p>
<h3 id="special-triangle-number-1">Special triangle number 1</h3>
<p>There are two especially useful triangles, the first of which is the isosceles right triangle, with two angles of 45º. Knowing this allows you to work out the values for any angle that’s a multiple of 45º.</p>
<p>The isosceles right-angled triangle has side lengths in the ratio of $1:1:\sqrt{2}$, which you can verify with Pythagoras. That means $\sin(45º) = \cos(45º) = \frac{1}{\sqrt{2}}$, using basic trigonometry.</p>
<p>So, if you put a point on the circle at 45º anticlockwise from the x-axis, its coordinates are $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ ((No, I’m not going to rationalise. Rationalising is irrational.))</p>
<p>If you think about reflecting the circle across the $y$-axis, this point will become the point at 135º, with coordinates of $\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$, which gives you $\cos(135º) = -\frac{1}{\sqrt{2}}$ and $\sin(135º)= \frac{1}{\sqrt{2}}$.</p>
<p>You can continue around the circle to find the values for 225º and 315º.</p>
<h3 id="special-triangle-number-2">Special triangle number 2</h3>
<p>The second special triangle is an equilateral triangle cut in half (I usually imagine the base horizontal, and slice straight up the middle vertically). Then the angles are 60º (at the bottom), 30º (at the top) and 90º (in the corner). The side from the 60º corner to the right angle is half the distance from the 60º to the top (because we’ve cut an equilateral triangle in half), and with a bit of Pythagoras, it becomes clear that the side lengths are in the ratio $1 : \sqrt{3} : 2$, with the longer leg opposite 60º.</p>
<p>That means that $\cos(60º) = \frac{1}{2}$ and $\sin(60º) = \frac{\sqrt{3}}{2}$ - I find that remembering $\sqrt{3} > 1$ is helpful for not muddling these up. Whenever I have to figure out (say) $\cos(300º)$, I imagine where I’d put the triangle to put my point in the right place. For 300º, the short base goes on the positive $x$-axis and the ‘top’ is directly below it. That means $\cos(300º) = \frac{1}{2}$ and $\sin(300º) = -\frac{\sqrt{3}}{2}$.</p>
<p>If I wanted 330º instead, I’d need to put the base vertically on the negative $y$-axis and point the top to the right, so $\cos(330º) = \frac{\sqrt{3}}{2}$ and $\sin(330º) = -\frac{1}{2}$.</p>
<hr />
<p>I hope that helps you start to get the ideas, CAST!</p>
<p>- Uncle Colin</p>Dear Uncle Colin, How do you remember the values for the trig functions in the unit circle? - Can’t Abide Stupid TrigonometryWhy $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$2021-03-29T00:00:00+01:002021-03-29T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/why-gammaleftfrac12right-sqrtpi<p><em>This is based on a <a href="https://www.cambridge.org/core/journals/mathematical-gazette/article/mathematical-note/C00CCBEA36A704A24EE954F48442016E">Mathematical Note</a> written by George Osborn: link to DOME. ((I have made some modifications to the notation, treated limits with a little more care, and introduced a minor change in the line of reasoning, but otherwise this is entirely a gloss on Osborn’s work.))</em></p>
<p>Start by defining a function, $G$, such that $G(n, 1+x) = \frac{n^x}{\Pi_{i=1}^{n}\left(1+\frac{x}{i}\right)}$.</p>
<p>For example, $G(3, 1 + 1) = \frac{3^1}{\left(1+ \frac{1}{1}\right)\left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{3}\right)}$, which is $\frac{3}{4)}$. In fact, there’s a lot of cancelling in the denominator, and in general $G(n, 1+1) = \frac{n}{n+1}$.</p>
<p>We’ll also let $\Gamma(1+x)$ be the limit of $G(n,1+x)$ as $n \to \infty$. This is the gamma function you’re possibly used to, but we’ve not proved that yet; you might have spotted that $\Gamma(1+1) = 1$.</p>
<h3 id="is-this-gamma-our-gamma">Is this $\Gamma$ our $\Gamma$?</h3>
<p>We can look also at $G(n, x)$, which is $\frac{n^{x-1}}{\Pi_{i=1}^n 1 + \frac{x-1}{i}}$, and consider $\frac{G(n,1+x))}{xG(n,x)}$. That’s ((I’m taking the limits as read because I’m lazy)) $\frac{n^x \Pi 1 + \frac{x-1}{i}}{xn^{x-1} \Pi 1 + \frac{x}{i}}$; almost all of the $n$s cancel to give $\frac{n \Pi 1 + \frac{x-1}{i}}{x \Pi 1 + \frac{x}{i}}$.</p>
<p>Almost all of the terms in the products cancel, too! The product on top is $\frac{x}{1} \cdot \frac{1+x}{2} \dots \frac{n+x-1}{n}$; on the bottom, it’s $\frac{1+x}{1} \cdot \frac{2+x}{2} \dots \frac{n+x}{n}$. The denominators cancel, and all of the factors in the numerators vanish except for $x$ on top and $n+x$ on the bottom.</p>
<p>That leaves us with $\frac{G(n, 1+x)}{xG(n, x)} = \frac{nx}{x(n+x)} = \frac{n}{n+x}$.</p>
<p>As a result, when $n \to \infty$, $\Gamma(1+x) = x \Gamma(x)$. Also, $G(n, 1+0)$ is the ratio of two empty products, so $\Gamma(1) = 1$; by induction, you can show that $\Gamma(1+x) = x!$ ((Osborn’s note uses a sort of “L” shape around the $x$ to denote a factorial. Apparently that’s an old-fashioned thing.))</p>
<h3 id="gammaleftfrac12right">$\Gamma\left(\frac{1}{2}\right)$</h3>
<p>Now let’s consider $G(n, 1+x) \cdot G(n,1-x)$. Clearly the $n^x$s cancel, leaving just $\frac{1}{\left(\Pi 1 + \frac{x}{i}\right)\left(\Pi 1 - \frac{x}{i}\right)}$, so $\Gamma(1+x) \cdot \Gamma(1-x) = \Pi_{i=1}^{\infty} \left( 1 - \left(\frac{x}{i}\right)^2\right)^{-1}$.</p>
<p>What’s that product? It is undefined whenever $x$ is a positive integer, it’s 1 when $x=0$, and a little messing about will hopefully convince you that it’s equivalent to $\pi x \cosec(\pi x)$. ((Seriously, how neat is that?))</p>
<p>In particular, if $x= \frac{1}{2}$, then $\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right) = \piby 2$ (because $\cosec\left(\piby 2\right) = 1)$.</p>
<p>Also, $\Gamma\left(\frac{3}{2}\right) = \frac{1}{2}\Gamma\left(\frac{1}{2}\right)$, so $\left[\Gamma\left(\frac{1}{2}\right)\right]^2 = \pi$.</p>
<p>$\blacksquare$.</p>This is based on a Mathematical Note written by George Osborn: link to DOME. ((I have made some modifications to the notation, treated limits with a little more care, and introduced a minor change in the line of reasoning, but otherwise this is entirely a gloss on Osborn’s work.))Ask Uncle Colin: A restricted determinant2021-03-24T00:00:00+00:002021-03-24T00:00:00+00:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-a-restricted-determinant<blockquote>
<p>Dear Uncle Colin,</p>
<p>Can you show that 4 is the largest determinant of a 3 by 3 matrix made of 1s and -1s?</p>
<p>- Just A Crisis Of Brainy Ideas</p>
</blockquote>
<p>Hi, JACOBI, and thanks for your message!</p>
<h3 id="pluggety-and-chuggety">Pluggety and chuggety</h3>
<p>Let’s let the matrix be $\matthreethree{a}{&b}{&c}{d}{&e}{&f}{g}{&h}{&i}$.</p>
<p>Its determinant is $a\left| \mattwotwo{e}{f}{h}{i} \right| + b \left| \mattwotwo{f}{d}{i}{g} \right| + c \left|\mattwotwo{d}{e}{g}{h}\right|$, and the largest each of those two-by-two determinants could possibly be is 2 - which gives us an immediate upper bound of 6 for the full determinant.</p>
<p>How would we end up with one of the submatrices having a determinant of 2? We’d need a two-by-two matrix to be something like $\mattwotwo{1}{1}{-1}{1}$ - i.e., any three elements matching in sign and the other one opposite.</p>
<p>Suppose (without loss of generality), $\mattwotwo{e}{f}{h}{i} = \mattwotwo{1}{1}{-1}{1}$, leaving $d$ and $g$ as variables for now. What happens to the other two matrices? Those are now $\mattwotwo{f}{d}{i}{g} = \mattwotwo{1}{d}{1}{g}$ and $\mattwotwo{d}{e}{g}{h} = \mattwotwo{d}{1}{g}{-1}$.</p>
<p>The first of those has a determinant of $\pm 2$ if $d \ne g$ and 0 if they’re equal; with the second, it’s the other way around - if they’re equal, the determinant is $\pm 2$ and it’s zero if they’re different. In either case, the determinant of one of the matrices is $\pm 2$ and the other is 0, so the largest determinant for the 3 by 3 matrix is 4.</p>
<p>But that feels a bit unsatisfactory.</p>
<h3 id="geometry-and-symmetry">Geometry and symmetry</h3>
<p>What does a 3-by-3 matrix do? It maps three vertices of the unit cube ( $\mathbf i = (1,0,0)$, $\mathbf j = (0,1,0)$ and $\mathbf k = (0,0,1)$ ) to three new vertices, and drags the rest of cube into a parallelepiped. The volume of the parallelepiped is the determinant of the matrix.</p>
<p>An arbitrary 3-by-3 matrix can send the vertices anywhere in 3D space, but we’re restricted to sending them to the eight vertices of a cube, which are $(\pm_1 1, \pm_2 1, \pm_3 1)$. However, we’re stuck with keeping one of the vertices at the origin, the centre of the cube.</p>
<p>So what’s the biggest we can make our parallelepiped under this restriction?</p>
<p>Suppose (again, wlog) I send $(1,0,0)$ to $(1,1,1)$. A healthy start, with an edge of length $\sqrt{3}$. Where can I send the other vertices to maximise the parallelepiped’s volume?</p>
<p>There’s no point in sending a vertex to $(-1,-1,-1)$ because the resulting parallelepiped would have zero volume, so we have six vertices available to us. Three are ‘neighbours’ to $(1,1,1)$ (i.e., you can reach them by flipping only one sign), and three require two flips.</p>
<p>If we pick two neighbouring vertices (and exclude their opposites), the third vertex must lie on a face of the big cube. If we pick two non-neighbouring vertices (and exclude their opposites), the third vertex is either a neighbour of both (and we’ve covered that case), or a neighbour of neither (and forms an equilateral triangle).</p>
<p>So we only have two possible shapes!</p>
<p>Rather than the parallelepiped ((I’ve asked @alisonkiddle to reset the counter, don’t worry)), let’s think about the tetrahedron formed by $\mathbf 0, \mathbf i, \mathbf j$ and $\mathbf k$. The scale factor of the transformation that takes it to a new tetrahedron will be the same as that of the cube, because linear magic. (The original tetrahedron has a volume of $\frac{1}{6}$, which you might like to verify.)</p>
<p>The first shape has the three free vertices going to three points on the same face, which I’ll think of as the base. The base has an area of 2 units, and is one unit from the origin, so the new tetrahedron has an volume of $\frac{2}{3}$ units, four times that of the original tetrahedron.</p>
<p>The second shape is (for me at least) a bit harder to reason about. In fact, I’m going to map $\mathbf i, \mathbf j$ and $\mathbf k$ to $(-1,1,1)$, $(1,-1,1)$ and $(1,1,-1)$, respectively, because it’s neater in my head.</p>
<p>The edges of the equilateral triangle they form have length of $2\sqrt{2}$. That means the area of the base can be worked out using $\frac{1}{2}ab \sin(C)$ to be $2{\sqrt{3}}$. But what about the height?</p>
<p>The centroid of the base lies at $\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$, and by symmetry has to be the foot of the perpendicular from the origin. That distance is $\frac{1}{\sqrt{3}}$.</p>
<p>The volume of this tetrahedron is then $\frac{1}{3}(2\sqrt{3})\left(\frac{1}{\sqrt{3}}\right) = \frac{2}{3}$ as well! The scale factor is again 4.</p>
<h3 id="jiggery-and-pokery">Jiggery and pokery</h3>
<p>I think this shows that the determinant of a matrix made of 1s and -1s is either 4, 0 or -4. There are $2^9=512$ possible such matrices, although if you restrict the first column to be all 1s (wlog), there are only 64. Looking at the next column, two of the eight possibilities give a determinant of zero (they’re multiples of ‘all 1s’); given we’re not among those two, a random final column has a 50-50 chance of being a multiple of one of the first two. So, of the original 64, 24 have determinant of $\pm 4$ (12 of each), and the remaining 40 have determinant zero.</p>
<hr />
<p>That’s probably more than you wanted to know, isn’t it? Never mind; I still hope it helps.</p>
<p>- Uncle Colin</p>Dear Uncle Colin, Can you show that 4 is the largest determinant of a 3 by 3 matrix made of 1s and -1s? - Just A Crisis Of Brainy IdeasA Continued Fraction for $e$2021-03-22T00:00:00+00:002021-03-22T00:00:00+00:00https://www.flyingcoloursmaths.co.uk/a-continued-fraction-for-e<p>I was a little surprised to see a continued fractions conjecture on <a href="https://en.wikipedia.org/wiki/List_of_representations_of_e#Conjectures">Wikipedia</a>, stating:</p>
<p>$e = { 3 + \frac{-1}{4 + \frac{-2}{5 + \frac{-3}{6 + \dots}}}}$</p>
<p>Obviously, my first thought was, “that doesn’t look too hard to prove”. My second thought was “I imagine it already has been”, and such is the case: an <a href="https://arxiv.org/pdf/1907.05563.pdf">arXiv paper</a> and a <a href="https://mathoverflow.net/a/336399">MathOverflow comment</a> both beat me to the punch.</p>
<p>But just because something has been proved doesn’t mean it’s not worth proving again!</p>
<h3 id="playing-with-the-convergents">Playing with the convergents</h3>
<p>I am not a native continuedfractionologist. I didn’t touch them during my degree, and I’ve only recently started meddling with them. I tend to approach them with a bit of trepidation, and gingerly work out the first few <em>convergents</em> - the fractions you get from truncating the expansion as you go along.</p>
<p>Here, the first approximation is 3. That’s not a bad approximation to $e$, as these things go.</p>
<p>The second approximation is $3 + -1/4$, or $\frac{11}{4}$. 2.75: definitely in the right ballpark.</p>
<p>Then we have $3 + -1/(4 + -2 / 5)$, which works out to be $\frac{49}{18}$, 2.7222…. It looks plausible at the very least!</p>
<p>In fact, as we work it through, we notice that the $n$th convergent seems to correspond to the $n$th partial sum in $3 - \frac{1}{4} - \frac{1}{36} - \dots - \frac{1}{(n+1)!(n+1)n} - \dots$.</p>
<p>And that’s <em>also</em> listed in the representations of $e$ article as a valid series. If we can prove that the two are the same, we’ve proved it!</p>
<h3 id="a-way-in">A way in!</h3>
<p>I know I’ve told the joke repeatedly that in maths, discoveries are typically named after the <em>second</em> person to discover it, because otherwise everything would be named after Euler.</p>
<p>$e$ is named after Euler ((He didn’t discover it)). <a href="https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula">Euler’s continued fraction formula</a> is also named after him ((He <em>did</em> discover this.))</p>
<p>It states that, if you can write a number as a convergent sum $a_0(1 + a_1(1+ a_2(1 + \dots)))$, then its continued fraction can be written as:</p>
<p>$\frac{a_0}{1 + \frac{a_1}{(1+a_1) - \frac{a_2}{(1+a_2 - \frac{a_3}{(1+a_4 - \dots)}}}}$</p>
<p>As in the example we’re looking at, the $n$th convergent matches up with the $n$th partial sum.</p>
<p>To make things easier for my simple mind, I’m going to start from the equivalent statement $3-e = \frac{1}{4} + \frac{1}{36} + \frac{1}{288} + \dots + \frac{1}{(n+2)!(n+2)(n+1)}$.</p>
<p>Alternatively, this is $3-e = \frac{1}{4}\left(1 + \frac{1}{9}\left(1 + \frac{2}{16}\left(1 + \dots \right)\right)\right)$. The $a_i$s here are $a_0 = \frac{1}{4}$, $a_1 = \frac{1}{9}$, $a_2 = \frac{2}{16}$, and generally $a_k \frac{k}{(k+2)^2}$ for $k \ge 1$.</p>
<p>And we can apply Euler directly to write down a continued fraction representation:</p>
<p>$3-e = \frac{ 1/4 }{1 - \frac{1/9}{10/9 - \frac{2/16}{18/16 - \frac{\dots}{\dots k/(k+2)^2}{1 + k/(k+2)^2 - \frac{(k+1)/(k+3)^2}{\dots}}}}}$</p>
<h3 id="ooft">Ooft.</h3>
<p>Now we’re going to repeatedly flush out the fractions. In general, $\frac{a}{b + \frac{c}{d + \dots}} = \frac{pa}{pb + \frac{pc}{d + \dots}}$, so if we multiply the top of a fraction, the lead term of the bottom, and the top of the next fraction by the same constant, we don’t change the value of the continued fraction. In particular, if I want to turn the 1/4 on top into 1, I need to multiply the 1 and the 1/9 on the next line by 4 as well:</p>
<p>$3-e = \frac{1}{4 - \frac{4/9}{10/9 - \frac{2/16}{18/16 - \dots}}}$</p>
<p>I can repeat the process, multiplying the next bit down by 9/2:</p>
<p>$3 - e = \frac{1}{4 - \frac{2}{5 - \frac{9/16}{18/16 - \frac{3/25}{\dots}}}}$</p>
<p>On the next line, the multiplier is 16/3:</p>
<p>$3 - e = \frac{1}{4 - \frac{2}{5 - \frac{3}{6 - \frac{16/25}{\dots}}}}$</p>
<p>That’s looking pretty good so far – and the basis of an induction proof!</p>
<h3 id="induction-sketch">Induction (sketch)</h3>
<p>If I have $\frac{(k+1)^2/(k+2)^2}{(1 + k/(k+2)^2) - \frac{(k+1)/(k+3)^2}{\dots}}$ at some point in the fraction structure, I can rewrite the lead term in the denominator as $((k+2)^2+ k)/(k+2)^2$. The top of that expands as $k^2 + 5k + 4$, or $(k+1)(k+4)$.</p>
<p>We can therefore multiply all three of the relevant terms by $(k+2)^2 / (k+1)$ to leave us with $\frac{k+1}{k+4 - \frac{(k+2)^2/(k+3)^2}{\dots}}$</p>
<p>Using $k=1$ as the base case (we deal with the fraction starting with 4/9 in the previous section), this will show that following the flushing out will always give us the correct form: the top will become $k+1$, and the bottom line $k+4 - \frac{(k+2)^2/(k+3)^2}{\dots}$. The integer parts match up with the continued fraction we’re trying to reach, and the next fraction to deal with is the ratio of (the correct) consecutive squares so we can apply the same trick to the next line of the fraction.</p>
<p>I’ve left some details for the reader, so I won’t throw out a $\blacksquare$ here, but I believe I’ve given you enough scaffolding to make it rigorous if you’d like to!</p>
<hr />
<p>I thought that was a neat thing, and I’m always keen to learn a little more about continued fractions. What did you make of it?</p>I was a little surprised to see a continued fractions conjecture on Wikipedia, stating: