Jekyll2021-09-27T20:17:06+01:00https://www.flyingcoloursmaths.co.uk/feed.xmlFlying Colours MathsFlying Colours Maths helps make sense of maths at A-level and beyond.A Fermi Estimation Problem: How Many People Have A Maths A-Level?2021-09-27T00:00:00+01:002021-09-27T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/fermi-estimation-a-levels<p>In an idle moment, I wondered how many people in the UK ((Rather, England and Wales)) have an A-level in maths.</p>
<p>This strikes me as a great Fermi estimation question – you can come up with any number of answers and bounds on it and probably nobody knows the answer even roughly. (You might be able to get the number of A-levels <em>awarded</em>, but some recipients will no longer be with us.)</p>
<h3 id="some-wide-bounds">Some wide bounds</h3>
<p>Somewhere in the back of my head, I know that about 100,000 people took A-level maths last year. And that feels about right – if there are 40,000,000 in England and Wales, that works out as about 500,000 18 year-olds (give or take), and 20% of them taking maths doesn’t seem like a huge stretch. So: at least 100,000 A-levels are almost certainly out in the wild, and that’s a (very low) lower bound. We could say the numbers have been pretty stable for the last half-decade or so and bump it up, but let’s stick with it.</p>
<p>A-level numbers have been going up, and it’s likely that 100,000 is about the biggest cohort ever. A-levels were introduced in 1951, so there’s about 70 years available – making 7,000,000 a (very generous) upper bound.</p>
<h3 id="a-good-start">A good start</h3>
<p>That’s already a good start: more than 100,000 and less than 7,000,000 suggests “about a million” as a reasonable ballpark answer straight away. But can we do better?</p>
<p>Probably.</p>
<p>I would probably model the A-level uptake as a straight line, rising steadily from “not very many” in the 1950s to today’s numbers – that would suggest a number around 3,500,000. I suspect that’s an overestimate and that “two or three million” is a reasonable number, about 5% of the population.</p>
<hr />
<p>How would you approach it? Do you get a different answer? I’d love to hear about it!</p>In an idle moment, I wondered how many people in the UK ((Rather, England and Wales)) have an A-level in maths.Ask Uncle Colin: Why does this have repeated roots?2021-09-22T00:00:00+01:002021-09-22T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-repeated-roots<blockquote>
<p>Dear Uncle Colin,</p>
<p>I have solved $16x^5 - 20x^3 + 5x + 1 = 0$ by letting $x = \cos(\theta)$ and considering $\cos(5\theta)$ – but how do I explain that there are only three roots?</p>
<p>My Understanding Limited To Interesting Problems Like Integration; Can’t Interpret Trigonometry Yet</p>
</blockquote>
<p>Hi, MULTIPLICITY ((That’s a good one!)) and thanks for your message!</p>
<p>For the benefit of those playing along at home, let’s go through the steps fully.</p>
<p>The method for this relies on De Moivre’s theorem:</p>
<ul>
<li>Start with the fact that $\br{\cos(5\theta) + \i \sin(5\theta)} = \br{\cos(\theta) + \i \sin(\theta)}^5$</li>
<li>Expand the bracket on the right using the binomial theorem: $\dots = \cos^5(\theta) + 5\i \cos^4(\theta) \sin(\theta) - 10 \cos^3(\theta) \sin^2(\theta) - 10\i\cos^2(\theta)\sin^3(\theta) + 5\cos(\theta)\sin^4(\theta) + \i \sin^5(\theta)$</li>
<li>We only care about the real part:
<ul>
<li>$\cos(5\theta) = \cos^5(\theta) - 10 \cos^3(\theta)\sin^2(theta) + 5\cos(\theta)\sin^4(\theta)$</li>
</ul>
</li>
<li>Let $c = \cos(\theta)$ to save on typing, and note that $\sin^2(\theta) = 1- c^2$</li>
<li>… so we have $\cos(5\theta) = c^5 - 10c^3(1-c^2) + 5c\br{1-c^2}^2$</li>
<li>… or $\cos(5\theta) = c^5 - 10c^3 + 10c^5 + 5c - 10c^3 + 5c^5$</li>
<li>… or $\cos(5\theta) = 16c^5 - 20c^3 + 5c$</li>
</ul>
<p>Hey! That looks familiar!</p>
<p>If we change our original variable to $c$, we now need to solve $\cos(5\theta) = -1$ for $\theta$, and find all of the values of $\cos(\theta)$ that result.</p>
<h3 id="solving-for-theta">Solving for $\theta$</h3>
<p>The equation $\cos(5x) = k$ typically has ten solutions between $0$ and $2\pi$, but our quintic can only have, at most, five ((Because it’s a quintic.)). We can reconcile this by remembering that we’re ignoring the imaginary part: we could have picked $\cos(\theta) - \i \sin(theta)$ as the basis for our De Moivre expansion and got just the same expression for $\cos(5\theta)$. Each solution for $\theta$ “works” for both $\cos(\theta) + \i \sin(\theta)$ and $\cos(\theta) - \i \sin(\theta)$, so we can discard the lower half of the unit circle without losing any solutions.</p>
<p>If we let $\alpha = 5\theta$, and solve $\cos(\alpha) = -1$ for $0 \le \alpha \le 5\pi$ ((The upper limit needs to be inclusive; we’re discarding only the <em>negative</em>-imaginary parts.)), we get $\alpha = \pi, 3\pi, 5\pi$.</p>
<p>So $\theta = \frac{1}{5}\pi, \frac{3}{5}\pi$ and $\pi$.</p>
<p>But that’s only three solutions, when we wanted five!</p>
<p>“Ah, the line $y = -1$ is tangent to the curve $y=\cos(5\theta)$ there!”.</p>
<p>But that implies <em>six</em> solutions, doesn’t it?</p>
<p>It’s a bit subtle. If we consider $\cos(5\theta)=-1 + \epsilon$, for some small value of $\epsilon$, we get two solutions near the “internal” values of $\theta$ – but only one at $\theta = \pi$, because there’s no “back up” part of the curve. So while it <em>feels</em> like there should be six, there are only five!</p>
<h3 id="solving-for-c">Solving for $c$</h3>
<p>After that, it’s straightforward: $\cos\br{\frac{1}{5}\pi} = \frac{\phi}{2}$, $\cos\br{\frac{3}{5}\pi} = -\frac{1}{2\phi}$, and $\cos(\pi) = -1$.</p>
<p>So the solutions for $x$ are $\frac{\phi}{2}$ (twice), $-\frac{1}{2\phi}$ (twice) and -1.</p>
<h3 id="other-methods">Other methods?</h3>
<p>The solution of $x = -1$ is a fairly obvious one to the original problem.</p>
<p>Dividing out the factor of $(x+1)$ gives $16x^4 - 16x^3 - 4x^2 + 4x + 1 = 0$. Let’s call that left-hand side $f(x)$.</p>
<p>Now, generally, factorising quartics is possible-but-messy. This one <em>looks</em> like it should be nice, though! If I let $y = 2x-\frac{1}{2}$, I know that $y^4$ starts $16x^4 - 16x^3$ and it might be that things drop out cleanly from there.</p>
<p>So, $y^4 = 16x^4 - 16x^3 + 6x^2 - x + \frac{1}{16}$, so $y^4 = f(x) + 10x^2 - 5x - \frac{15}{16}$.</p>
<p>And we are lucky! $y^2 = 4x^2 - 2x + \frac{1}{4}$, so $\frac{5}{2}y^2 = 10x^2 - 5x + \frac{5}{8}$.</p>
<p>That means $y^4 - \frac{5}{2}y^2 = f(x) - \frac{25}{16}$… or $f(x) = y^4 - \frac{5}{2}y^2 + \frac{25}{16}$, which is a quadratic we can solve!</p>
<ul>
<li>Multiply by 16: $16y^4 - 40y^2 + 25 = 0$</li>
<li>Complete the square: $(4y^2-5)^2 = 0$</li>
</ul>
<p>Gosh! It’s a perfect square! We get a pair of double roots: $y = \frac{\sqrt 5}{2}$ and $y =-\frac{\sqrt 5}{2}$</p>
<p>But we want $x$, which is $\frac{2y+1}{4}$, giving $x = \frac{1 \pm \sqrt{5}}{4}$, each twice – and those correspond to the roots we got the other way.</p>
<p>(For clarity: I wouldn’t have attempted this if I hadn’t known it came out clearly, and don’t recommend it as a general method.)</p>
<hr />
<p>Hope that helps!</p>
<p>- Uncle Colin</p>
<ul>
<li>Many thanks to @mrsouthernamths for the prompt, and for setting me straight when I made an error.</li>
</ul>Dear Uncle Colin, I have solved $16x^5 - 20x^3 + 5x + 1 = 0$ by letting $x = \cos(\theta)$ and considering $\cos(5\theta)$ – but how do I explain that there are only three roots? My Understanding Limited To Interesting Problems Like Integration; Can’t Interpret Trigonometry YetAn infinite sum2021-09-20T01:00:00+01:002021-09-20T01:00:00+01:00https://www.flyingcoloursmaths.co.uk/an-infinite-sum<p>A puzzle from @dmarain:</p>
<blockquote class="twitter-tweet"><p lang="en" dir="ltr">Explain using at least 2 methods:<br />(1+x²)(1-x²+x⁴-x^6+…)=1, -1<x<1<br />And why that domain?<a href="https://twitter.com/johnjoy1966?ref_src=twsrc%5Etfw"><span class="citation" data-cites="johnjoy1966">@johnjoy1966</span></a> <a href="https://twitter.com/aranglancy?ref_src=twsrc%5Etfw"><span class="citation" data-cites="aranglancy">@aranglancy</span></a> <a href="https://twitter.com/HarMath?ref_src=twsrc%5Etfw"><span class="citation" data-cites="HarMath">@HarMath</span></a> <a href="https://twitter.com/mrdardy?ref_src=twsrc%5Etfw"><span class="citation" data-cites="mrdardy">@mrdardy</span></a> <a href="https://twitter.com/hashtag/MTBoS?src=hash&ref_src=twsrc%5Etfw">#MTBoS</a></p>— David Marain (<span class="citation" data-cites="dmarain">@dmarain</span>) <a href="https://twitter.com/dmarain/status/766398857978740736?ref_src=twsrc%5Etfw">August 18, 2016</a></blockquote>
<script async="" src="https://platform.twitter.com/widgets.js" charset="utf-8"></script>
<p>As usual, have a crack if you want to. Spoilers below the line.</p>
<hr />
<p>I bet there are dozens of methods here.</p>
<h3 id="approach-number-1">Approach Number 1</h3>
<p>My first approach is to just multiply out the brackets: one multiplied by the second bracket is $(1 -x^2 + x^4 - x^6 + \dots)$, and $x^2$ multiplied by the second bracket is $(x^2 - x^4 + x^6 - \dots)$, and all of the terms cancel out except for the 1. We’re allowed to rearrange because the sequence converges absolutely for $|x|<1$.</p>
<h3 id="approach-number-2">Approach Number 2</h3>
<p>A cheeky binomial expansion: $(1+x^2)^{-1} = 1 - x^2 + x^4 - x^6 + \dots$ for $|x|<1$, so multiplying this by $(1+x^2)$ gives 1.</p>
<h3 id="approach-number-3">Approach Number 3</h3>
<p>The second bracket is a geometric series with a first term of $a=1$ and a common difference of $r= -x^2$. Its sum to infinity is $\frac{a}{1-r}$ (as long as $|r|<1$), which works out to be $\frac{1}{1+x^2}$. Multiplying this by $x^2$ gives 1.</p>
<h3 id="approach-number-4">Approach Number 4</h3>
<p>Boringly, divide 1 by $(1+x^2)$ in whatever way you fancy. You’ll get the infinite series in the second bracket.</p>
<hr />
<p>Part of me says “there has to be a trigonometric way to do this!”, but I can’t immediately see it. There’s probably a calculus approach, too. Any other ideas? I’d love to hear them.</p>A puzzle from @dmarain:Ask Uncle Colin: A rotated hyperbola2021-09-15T01:00:00+01:002021-09-15T01:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-a-rotated-hyperbola<blockquote>
<p>Dear Uncle Colin,</p>
<p>I have the curve $y = \frac{1}{x}$ and need to rotate it by 45 degrees clockwise. I can’t figure out what curve I get! Any clues?</p>
<p>Where Asymptotes Lie, Like I Said</p>
</blockquote>
<p>Hi, WALLIS, and thanks for your message!</p>
<p>I can think of several ways to do this. Let’s look at some!</p>
<h3 id="polar-coordinates">Polar coordinates</h3>
<p>One way is to convert it into polar coordinates, adjust the angle, and convert back.</p>
<p>If $y = \frac{1}{x}$, then $xy = 1$; since $x = r\cos(\theta)$ and $y = r\sin(\theta)$, that gives $r^2 \sin(\theta)\cos(\theta) = 1$.</p>
<p>Better yet, we can make that $r^2 \sin(2\theta) = 2$.</p>
<p>Rotating it by $-\piby 4$ radians ((as Euler intended)) turns $\theta$ into $\theta+\piby 4$, so we have - for the rotated curve - $r^2 \sin\left(2\theta + \piby 2\right)=2$.</p>
<p>Now, $\sin(2\theta + \piby2) \equiv -\cos(2\theta)$, so our curve is $-r^2 \cos(2\theta) = 2$</p>
<p>Let’s get that back into the $x-y$ plane. $\cos(2\theta) \equiv 2\cos^2(\theta) - 1$, and $\cos(\theta) = \frac{x}{r}$, while $r^2 \equiv x^2 + y^2$ so the curve is $-(x^2+y^2)\br{2\frac{x^2}{x^2+y^2}- 1} = 2$.</p>
<p>Rearranging: $-\br{x^2 + y^2}\br{2x^2 - (x^2 + y^2)} = 2\br{x^2 + y^2}$, or $\br{x^2+y^2}\br{y^2-x^2} = 2\br{x^2 + y^2}$.</p>
<p>We also know that $x^2 + y^2 > 0$ because $x\ne0$, so we can divide by it to get $y^2 = x^2 - 2$.</p>
<p>Does that make sense? As $x$ gets large, $y$ approaches $\pm x$, as we would like; the curve also passes through $\br{\sqrt{2},0}$, which is where we want it to be.</p>
<h3 id="matrices">Matrices</h3>
<p>If we represent $y = \frac{1}{x}$ as $\colvectwo{t}{\frac{1}{t}}$ for $t \ne 0$ and rotation by $-\piby4$ as $\frac{1}{\sqrt{2}}\mattwotwo{1}{1}{-1}{1}$, then the transformed curve is $\frac{1}{\sqrt{2}}\colvectwo{t + \frac{1}{t}}{\frac{1}{t}-t}$.</p>
<p>So, in parametric form, we have $y = \frac{1}{\sqrt{2}} \br{\frac{t^2+1}{t}}$ and $x = \frac{1}{\sqrt{2}}\br{\frac{1-t^2}{t}}$</p>
<p>Then we need to isolate $t$, which is a bit messy.</p>
<ul>
<li>$xt\sqrt{2} = 1 + t^2$</li>
<li>$yt\sqrt{2} = 1 - t^2$</li>
</ul>
<p>If we square both of those:</p>
<ul>
<li>$2 x^2 t^2 = 1 + 2t^2 + t^4$</li>
<li>$2 y^2 t^2 = 1 - 2t^2 + t^4$</li>
</ul>
<p>Subtract:</p>
<ul>
<li>$2t^2 \br{x^2 - y^2} = 4t^2$</li>
</ul>
<p>Since $t^2 \ne 0$, this reduces to $x^2 - y^2 = 2$, as before.</p>
<h3 id="whoosh">Whoosh</h3>
<p>“Should have known I’d find you here, sensei.”</p>
<p>A better substitution is $y = e^t$ and $x = e^{-t}$. Then your bullet points are $xe^t \sqrt{2} = 1 + e^{2t}$ and $ye^t\sqrt{2} = 1 - e^{2t}$.</p>
<p>“And that’s better? You’re right, your bullets are - as always - better than my bullets.”</p>
<p>Alternatively, $x = \sqrt{2}\cosh(x)$ and $y = \sqrt{2}\sinh(x)$. Thus $x^2 - y^2 = 2$.</p>
<h3 id="whoosh-1">Whoosh</h3>
<p>Hope that helps!</p>
<p>- Uncle Colin</p>Dear Uncle Colin, I have the curve $y = \frac{1}{x}$ and need to rotate it by 45 degrees clockwise. I can’t figure out what curve I get! Any clues? Where Asymptotes Lie, Like I SaidMorrie’s Law2021-09-13T00:00:00+01:002021-09-13T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/morries-law<p>I suppose this ought to be a Dictionary of Mathematical Eponymy post, but it isn’t. So there ((for now, at least)).</p>
<p>The legend is that Richard Feynman, as a child ((History does not record whether he was a jerk at that point.)), was discussing trigonometry with his friend Morrie Jacobs in the Jacobs family leatherworks, when Morrie divulged that $\cos(20^o) \cos(40^o) \cos(80^o) = \frac{1}{8}$. This surprised and delighted Feynman, who eventually referred to it as Morrie’s law.</p>
<h3 id="why-does-it-work">Why does it work?</h3>
<p>It works for very neat trigonometric reasons, related to the fact that $\sin(2x) = 2\sin(x)\cos(x)$.</p>
<p>In particular, $\sin(40^o)\sin(80^o)\sin(160^o) = \left(2\sin(20^o)\cos(20^o)\right)\times$ $\left(2\sin(40^o)\cos(40^o)\right)\times$ $\left(2 \sin(80^o)\cos(80^o)\right)$.</p>
<p>Most of those sines cancel!</p>
<p>We’re left with $\sin(160^o) = 8\sin(20^o)\cos(20^o)\cos(40^o)\cos(80^o)$.</p>
<p>However, $\sin(160^o) = \sin(20^o)$, so <em>those</em> cancel, as well! We’re left with $1 = 8\cos(20^o)\cos(40^o)\cos(80^o)$, which is Morrie’s law.</p>
<h3 id="can-it-be-extended">Can it be extended?</h3>
<p>Of course it can! You can do generalise the idea to $2^{n+1}\cos(x)\cos(2x)\cos(4x)\dots\cos\left(2^nx\right) = \frac{\sin(2^{n+1} x)}{\sin(x)}$. If you pick your value of $x$ so that the right hand side takes on specific values (for example, 1 is a handy one; I imagine one could contrive various fractions and square roots with a bit of thinking), you get an identity as delightful as Morrie’s.</p>
<p>Let’s try one that goes one step further: $16\cos(x)\cos(2x)\cos(4x)\cos(8x) = \frac{\sin(16x)}{\sin(x)}$.</p>
<p>Now I want $\sin(16x)$ and $\sin(x)$ to be the same - I could pick $\frac{180^o}{17}$, which is of course a lovely number. Isn’t it great that 360 has so many factors? It makes trigonometry so much simpler. In fact, I believe Morrie’s law is the furthest this can be taken with integer-degree angles ((Of course, in radians, it all works so much more nicely.))</p>I suppose this ought to be a Dictionary of Mathematical Eponymy post, but it isn’t. So there ((for now, at least)).Ask Uncle Colin - An Implicit Curve2021-09-08T01:00:00+01:002021-09-08T01:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-an-implicit-curve<blockquote>
<p>Dear Uncle Colin,</p>
<p>I’m told that $x^2 \tan(y) = 9$, for $0<y< \piby 2$. I have to show that $\dydx = \frac{-18x}{x^4+81}$, and that there’s a point of inflection at $x = 27^{1/4}$. Where do I even start!?</p>
<p>- I Might Plot Loci… I Can’t, It’s Tricky</p>
</blockquote>
<p>Hi, IMPLICIT, and thanks for your message!</p>
<p>Plotting this does seem a bit over the top, now you mention it! You could probably figure out a few points – when $x = \pm 1$, $y = \piby 4$, for example; as $y$ approaches 0, $x^2$ increases without bound (so $x$ goes to infinity in both directions), and when $y$ approaches $\piby 2$, $x^2$ approaches 0$. I can picture that! It looks a bit like the Gaussian.</p>
<p>But that’s beside the point! We need to differentiate this, and doing that implicitly is probably for the best.</p>
<h3 id="first-part">First part</h3>
<p>The left hand side is a product, so we need the product rule: we get $(2x)(\tan(y)) + \br{x^2}\br{\sec^2(y) \dydx} = 0$.</p>
<p>Now, how to manipulate that into what they have? It makes sense to eliminate the $y$s, since there are none of those in the final answer: $\tan(y) = \frac{9}{x^2}$ and we can use the fact that $\sec^2(y) = 1 + \tan^2(y)$ to say $\sec^2(y) = \frac{x^4 + 81}{x^4}$. That has things in common with our answer!</p>
<p>Where are we now, then? $(2x)\br{\frac{9}{x^2}} + \br{x^2}\frac{x^4 + 81}{x^4}\dydx = 0$.</p>
<p>We know that $x \ne 0$, because $y$ never reaches $\piby 4$, so we can mulitply through by $x^2$, cancel and consolidate a little: $18x + \frac{x^4 + 81}\dydx = 0$.</p>
<p>Aha! We’re pretty much there – just shuffle and divide to get $\dydx = \frac{-18x}{x^4 + 81}$ as required.</p>
<h3 id="second-part">Second part</h3>
<p>For a point of inflection, the second derivative needs to be zero. We can differentiate using the quotient rule (for a moment I wondered about differentiating implicitly again, but haha nope.)</p>
<p>We’ve got $u = -18x$ and $v = x^4 + 81$, so $u’ = -18$ and $v’ = 4x^3$.</p>
<p>$\diffn{2}{y}{x} = \frac{vu’ - uv’}{v^2} = \frac{-18\br{x^4+81} + 72x^4}{x^8}$.</p>
<p>We’re interested in where the top vanishes, i.e., where $-18\br{x^4+81} + 72x^4 =0$.</p>
<p>Taking out $18$ gives $18\br{3x^4 - 81}=0$, so $x^4 = 27$ as required.</p>
<p>Hope that helps!</p>
<p>- Uncle Colin</p>Dear Uncle Colin, I’m told that $x^2 \tan(y) = 9$, for $0<y< \piby 2$. I have to show that $\dydx = \frac{-18x}{x^4+81}$, and that there’s a point of inflection at $x = 27^{1/4}$. Where do I even start!? - I Might Plot Loci… I Can’t, It’s TrickyDictionary of Mathematical Eponymy: The Haynsworth Inertia Additivity Formula2021-09-06T01:00:00+01:002021-09-06T01:00:00+01:00https://www.flyingcoloursmaths.co.uk/dictionary-of-mathematical-eponymy-haynsworth-inertia-additivity-formula<h3 id="what-is-the-haynsworth-inertia-additivity-formula">What is the Haynsworth Inertia Additivity Formula?</h3>
<p>Start with a <em>Hermitian</em> matrix, $\bb{H}$ – which is one that’s equal to its conjugate transpose $\bb{H^*}$ ((That is to say: switch the rows and columns, then take the complex conjugate of every entry – if this gives you the original matrix, it’s Hermitian.)).</p>
<p>Haynsworth defined the <em>inertia</em> of a Hermitian matrix to be a triple, $\text{In}(\bb{H}) = \br{ \pi(\bb{H}), \nu(\bb{H}), \delta(\bb{H})}$, where:</p>
<ul>
<li>$\pi(\bb{H})$ is the number of positive eigenvalues of $\bb{H}$;</li>
<li>$\nu(\bb{H})$ is the number of negative eigenvalues of $\bb{H}$;</li>
<li>$\delta(\bb{H})$ is the number of zero eigenvalues of $\bb{H}$.</li>
</ul>
<p>If you split a Hermitian matrix into parts, $\bb{H} = \mattwotwo{\bb H_{11}}{\bb H_{12}}{\bb H^*_{12}}{\bb H_{22}}$ (where $\bb{H_{11}}$ is non-singular) then there’s a relationship between the inertias of the parts:</p>
<p>$\text{In}(\bb{H}) = \text{In}\br{\bb{H_{11}}} + \text{In}\br{\bb{H}/ bb{H_{11}}}$</p>
<p>What’s the $/$ here? It’s not division, but a <em>Schur complement</em> – $\bb{H}/\bb{H_{11}} = \bb{H_{22}} - \bb{H^*_{12}} \bb{H^{-1}_{11}} \bb{H_{12}}$.</p>
<p>This means, if we’re just interested in the signs of the eigenvalues of a Hermitian matrix, we can break it into smaller parts and work with those instead.</p>
<h3 id="who-was-emilie-virginia-haynsworth">Who was Emilie Virginia Haynsworth</h3>
<p>Haynsworth was born in Sumter, South Carolina, in 1916. After studying at Coker College and Columbia University in New York, she taught mathematics in high school, worked at the Aberdeen Proving Ground during WW2 and then lectured at the University of Illinois.</p>
<p>She completed her doctorate at UNC Chapel Hill in 1952, then worked at Wilson College in Pennsylvania, the National Bureau of Statistics, and finally at Auburn University. Her research into the eigenvalues of matrices was renowned for its originality and elegance.</p>
<p>Haynsworth retired in 1983 and died in South Carolina in 1985.</p>What is the Haynsworth Inertia Additivity Formula?Ask Uncle Colin: Comparing Irrational Exponents2021-09-01T01:00:00+01:002021-09-01T01:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-comparing-irrational-exponents<blockquote>
<p>Dear Uncle Colin,</p>
<p>How do I compare two numbers with irrational exponents? I want to know whether $2^{\sqrt{3}}$ is larger or smaller than $3^{\sqrt{2}}$.</p>
<p>- How’d You Power A Thing Irrationally? Aaargh!</p>
</blockquote>
<p>Hi, HYPATIA, and thanks for your message! There’s a sensible way to do this and an alternative that I rather like.</p>
<h3 id="the-sensible-way">The sensible way</h3>
<p>Let $x = 2^{\sqrt{3}}$ and $y = 3^{\sqrt{2}}$.</p>
<p>Then $\ln(x) = \sqrt{3} \ln(2)$ and $\ln(y) = \sqrt{2} \ln(3)$.</p>
<p>These are both positive, so squaring them will keep them in the same order.</p>
<p>$(\ln(x))^2 = 3 (\ln(2))^2$ and $(\ln(y))^2 = 2(\ln(3))^2$.</p>
<p>Now, $\ln(3)> 1$, so the second of these is greater than 2.</p>
<p>Also, $\ln(2) < \sqrt{\frac{1}{2}}$, so the second of these is smaller than $\frac{3}{2}$. Therefore $y$ is larger.</p>
<h3 id="another-way">Another way</h3>
<p>$\sqrt{2} > 1.4$ so $3^{\sqrt{2}} > 3^{7/5}$.</p>
<p>$\sqrt{3} < 1.75$, so $2^{\sqrt{3}} < 2^{7/4}$.</p>
<p>Now, $3^4 > 2^5$, so $3^{7/5} > 2^{7/4}$</p>
<p>Putting it all together, $3^{\sqrt{2}} > 3^{7/5} > 2^{7/4} > 2^{\sqrt{3}}$</p>
<h4 id="woosh">Woosh</h4>
<p>“Ahem. $2^{\sqrt{3}} < 4$. $3^{\sqrt{2}} > 3^{4/3} > 4$.”</p>
<p>“How do y…”</p>
<p>“$81 > 64$ so $3^4 > 4^3$ and $3^{4/3} > 4$. As you were.”</p>
<p>“Thank you, sensei.”</p>
<hr />
<p>Hope that helps!</p>
<p>- Uncle Colin</p>Dear Uncle Colin, How do I compare two numbers with irrational exponents? I want to know whether $2^{\sqrt{3}}$ is larger or smaller than $3^{\sqrt{2}}$. - How’d You Power A Thing Irrationally? Aaargh!Reversible numbers2021-08-30T01:00:00+01:002021-08-30T01:00:00+01:00https://www.flyingcoloursmaths.co.uk/reversible-numbers<blockquote>
<p>“Find all of the five-digit numbers that are reversed when multiplied by (a) 9 and (b) 4.”</p>
</blockquote>
<p>I can’t say I’m a fan of this type of puzzle, but a student asked if I’d help them prep some number theory, so here we are.</p>
<h3 id="how-i-went-about-it">How I went about it</h3>
<p>Well, with a great deal of bad language and worse grace, frankly. But here’s my eventual thinking for (a):</p>
<ul>
<li>Let $n = 10^4 a + 10^3 b + 10^2 c + 10 d + e$, such that $N = 10^4 e + 10^3 d + 10^2 c + 10b + a$.</li>
<li>Considering $9n - N = 0$ gives $89,999a + 990b + 800c - 910d - 9,991e = 0$</li>
<li>$a = 1$ and $e = 9$, or else $n$ would have six digits.</li>
<li>So $89,999 + 990b + 800 c- 910d - 89,919 = 0$, or $80 + 990b + 800c + 910d = 0$ - or even $8 + 99b + 80c - 91d = 0$</li>
<li>Working modulo 10: if $b =0$, $d=8$; if $b=1$, then $d=7$.</li>
<li>Two options:
<ul>
<li>if $b=0$ and $d=8$, then $8 + 80c - 728 = 0$ and $c = 9$</li>
<li>if $b=1$ and $d=7$, then $8 + 99 + 80c - 637 = 0$ and $80c = -530$, which doesn’t work.</li>
</ul>
</li>
<li>So $n = 10,989$ and $N = 98,901$.</li>
</ul>
<p>In a similar vein, I got that $n = 21,978$ and $N = 87,912$ for part (b).</p>
<h3 id="but-hang-on-a-minute">But hang on a minute!</h3>
<p>I noticed a few things here:</p>
<ul>
<li>The $n$ for part (b) is double that of part (a)</li>
<li>In part (a), $n = 11,000 - 11$ and $N = 99,000 - 99$; in part (b), $n = 22,000 - 22$ and $4n = 88,000 - 88$.</li>
</ul>
<p>And I wondered a few things:</p>
<ul>
<li>What about other five-digit numbers of the same form?
<ul>
<li>It turns out $(33,000 - 33) \times \frac{7}{3} = 77,000, - 77$, $(44,000 -44)\times\frac{3}{2}$ works, and so does $(55,000 - 55)\times 1$ - although that one is hardly unique.</li>
</ul>
</li>
<li>What about numbers of different lengths? Is there a pattern?</li>
<li>Is there significance to the fact that $\frac{1}{91} = 0.\dot 01098\dot9$?</li>
</ul>
<p>What do you find when you play with it?</p>“Find all of the five-digit numbers that are reversed when multiplied by (a) 9 and (b) 4.”What’s My Plot, Episode 32021-08-27T00:00:00+01:002021-08-27T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/whats-my-plot-3<p>A couple of months ago, I asked about this plot, and promptly forgot about it.</p>
<blockquote class="twitter-tweet" data-partner="tweetdeck"><p lang="en" dir="ltr">OK, twitter: you know the drill: I plotted something, and you’ve got to guess what it is. For the third time: what’s the plot? (Postcards for responses I deem to be the winners.) <a href="https://t.co/nwiKK3pCfT">pic.twitter.com/nwiKK3pCfT</a></p>— Colin Beveridge (@icecolbeveridge) <a href="https://twitter.com/icecolbeveridge/status/1407694057007091720?ref_src=twsrc%5Etfw">June 23, 2021</a></blockquote>
<script async="" src="https://platform.twitter.com/widgets.js" charset="utf-8"></script>
<p>This wasn’t a great What’s My Plot, in honesty. The broad class of thing was far too easy to recognise, but the specific thing far too hard.</p>
<p>I think the best response was by @realityminus3:</p>
<blockquote>
<p>Astrophotography of distant galaxy on nine occasions (three of which feature lens cap mishaps)</p>
</blockquote>
<p>A postcard is on its way.</p>
<p>Honourable mentions to @tpfto (“Those look awfully like plots of two-dimensional distributions”) and @ashwinvis (“Correlation between two 2nd order symmetric tensors?”), both on Mathstodon.</p>
<h3 id="what-it-was">What it was</h3>
<ul>
<li>I split the text of <em>Moby-Dick</em> up into roughly-tweet-sized messages</li>
<li>I counted how many times each bigram, trigram and quadgram of letters occurred</li>
<li>I calculated the implied log-probability of the bigrams, trigrams and quadgrams in each message</li>
<li>I binned the results from each and made a contour plot of how the various n-gram probabilities correlate.</li>
</ul>
<p>Can’t believe nobody got it, tbh.</p>
<p>Stay tuned for another episode of <em>What’s My Plot?</em> at some future point!</p>
<p>* Thanks to my good friend Matt for reminding me I hadn’t given an answer to this!</p>A couple of months ago, I asked about this plot, and promptly forgot about it.