Jekyll2021-12-06T03:00:02+00:00https://www.flyingcoloursmaths.co.uk/feed.xmlFlying Colours MathsFlying Colours Maths helps make sense of maths at A-level and beyond.The Dictionary of Mathematical Eponymy: Kawasaki’s Theorem2021-12-06T00:00:00+00:002021-12-06T00:00:00+00:00https://www.flyingcoloursmaths.co.uk/dictionary-of-mathematical-eponymy-kawasakis-theorem<p>I am not, by nature, an origamist. While I’m all for it in principle, I get frustrated at my inability to fold straight lines, and when I do succeed at making something, I never quite know what to do with it afterwards.</p> <p>But the <em>maths</em> behind it is fascinating.</p> <h3 id="what-is-kawasakis-theorem">What is Kawasaki’s Theorem?</h3> <p>Kawasaki’s theorem explains when it’s possible (in principle) to fold a pattern flat at a given vertex, assuming no other folds in the paper.</p> <p>To do this, there must be a difference of two between the number of valley folds and mountain folds (that’s Maetana’s theorem), and the alternating sum of the angles between adjacent folds must be zero.</p> <p>For example, if your pattern calls for angles of 30º, 45º, 60º, 75º, 90º and 60º, you would work out $30 - 45 + 60 - 75 + 90 - 60$ and conclude that the pattern can be folded flat at this vertex.</p> <p>In this example, the angles sum to 360º, but that’s not a condition of the theorem! Kawasaki’s theorem works on a curved surface where the angles at a point way sum to more or less than a full circle.</p> <h3 id="why-is-it-interesting">Why is it interesting?</h3> <p>Origami has mathematical applications I wouldn’t have expected before reading up about them – for example, trisecting an angle with straight-edge and compass is impossible, but with a sheet of paper and a few folds? It’s child’s play. More practically, origami informs how to package up (say) a solar sail so it can be transported into space without taking up much room, before unfolding elegantly and efficiently later in the voyage.</p> <p>Understanding how origami works mathematically opens up new opportunities – both for art and for science.</p> <h3 id="who-is-toshikazu-kawasaki">Who is Toshikazu Kawasaki?</h3> <p>Toshikazu Kawasaki was born in Kurume, Fukuoka, Japan in 1955. He’s better known for his origami than for his maths, but teaches maths at Sasebo Technical Junior College in Nagasaki, Japan.</p>I am not, by nature, an origamist. While I’m all for it in principle, I get frustrated at my inability to fold straight lines, and when I do succeed at making something, I never quite know what to do with it afterwards.Ask Uncle Colin: A Cursed Integral2021-12-01T00:00:00+00:002021-12-01T00:00:00+00:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-a-cursed-integral<blockquote> <p>Dear Uncle Colin,</p> <p>I tried to work out $\int \frac{1}{\sqrt{3 - 4x - 4x^2}}\dx$ going via the complex numbers and end up with a different answer compared to the ‘proper’ way – I get an $\arcosh$ instead of an $\arcsin$. Can you sort out this witchcraft?</p> <p>Am Genuinely Not Expecting Strange Integrals</p> </blockquote> <p>Hi, AGNESI, and thanks for your message!</p> <p>Let’s run through this both ways and see where the problem lies.</p> <h3 id="the-proper-way">The “proper” way</h3> <p>That’s an awkward quadratic inside the square root, but it turns out that $3 - 4x - 4x^2 \equiv 2^2 - (2x+1)^2$.</p> <p>Then applying the rule $\int \frac{1}{\sqrt{a^2 - u^2}} du = \arcsin\br{\frac{u}{a}} + C$, you get $\frac{1}{2} \arcsin \br{x + \frac{1}{2}} + C$.</p> <p>So far so tasty.</p> <h3 id="the-complex-way">The complex way</h3> <p>Factoring -1 out of the bottom turns the integral into $\frac{1}{i} \int \frac{1}{\sqrt{4x^2 + 4x-3}} \dx$ – and the quadratic in the square root is now $(2x+1)^2 - 2^2$.</p> <p>Using the rule $\int \frac{1}{\sqrt{u^2 - a^2}} = \arcosh\br{\frac{u}{a}}$, the integral gives $\frac{1}{2i} \arcosh \br{x + \frac{1}{2}} + c$.</p> <p>Those don’t look the same! What gives?</p> <h3 id="are-they-really-different">Are they really different?</h3> <p>I’m going to try to work out under what circumstances these two answers <em>would</em> be the same.</p> <p>Let the first result be $I_1$ and the second $I_2$.</p> <p>Rearranging the first gives $\sin\br{2I_1 - C} = x + \frac{1}{2}$.</p> <p>Rearranging the second gives $\cosh\br{2i I_2 - ic} = x + \frac{1}{2}$.</p> <p>However, $\cosh(ti) \equiv \cos(t)$, so this second left-hand side is $\cos\br{2I_2 - c}$.</p> <p>However, $\cos\br{t-\piby2} \equiv \sin(t)$, so if I let $c = C + \piby 2$, the second left-hand side is $\sin\br{2I_2 - C}$ – which is the same as the first answer!</p> <p>So, $I_1$ and $I_2$ differ by a constant – so they’re both valid solutions to the integral!</p> <p>(I’d go with the arcsine one, myself. Don’t want to be unleashing forces beyond our control, now, do we?)</p> <p>Hope that helps!</p> <p>- Uncle Colin</p>Dear Uncle Colin, I tried to work out $\int \frac{1}{\sqrt{3 - 4x - 4x^2}}\dx$ going via the complex numbers and end up with a different answer compared to the ‘proper’ way – I get an $\arcosh$ instead of an $\arcsin$. Can you sort out this witchcraft? Am Genuinely Not Expecting Strange IntegralsAn Interesting Integral2021-11-29T00:00:00+00:002021-11-29T00:00:00+00:00https://www.flyingcoloursmaths.co.uk/an-interesting-integral<p>This cropped up on Reddit:</p> <blockquote> <p>$I = \int_0^1 \br { \Pi_{r=1}^{10}(x+r)}\br{\sum_{r=1}^{10}\frac{1}{x+r}} \dx$</p> <p>Show that $I=(a)(b!)$, where $a$ and $b$ are positive integers to be found.</p> </blockquote> <p>Hm. Tricky. I have some <em>ideas</em>, but I’m going to tackle it a bit more brutally at first.</p> <p>What would happen if the top limit was something other than 10, like, say, 1?</p> <p>$I_1 = \int_0^1 (x+1) \frac{1}{x+1} \dx = 1$. So that’s not much to write home about.</p> <p>What about 2?</p> <p>$I_2 = \int_0^1 (x+1)(x+2) \br{\frac{1}{x+1} + \frac{1}{x+2} } \dx$</p> <p>That’s more interesting. It’s $\int_0^1 \br{2x+3}\dx$, or $[x^2+3x]_0^1$, which is 4.</p> <p>Is there going to be some sort of pattern when the sum of fractions gets added? I fear $I_3$ is going to be messy, but I can see what the pattern looks like:</p> <p>$I_3 = \int_0^1 (x+1)(x+2) + (x+2)(x+3) + (x+3)(x+1) \dx$</p> <p>$\dots = \int_0^1 3x^2 + 12x + 11 \dx$, or $[x^3 + 6x^2 + 11x]_0^1$, which is 18.</p> <p>Is there a pattern forming? I can see $I_n = (n)(n!)$ as a possibility… but I don’t see a way to prove it straight away.</p> <h3 id="how-about-a-substitution">How about a substitution?</h3> <p>What if I write the original integral as $\int_0^1 P S \dx$, where $P$ is the product and $S$ the sum?</p> <p>Then $\diff{P}{x} = S \br{\frac{1}{x+1} + \frac{1}{x+2} + \dots + \frac{1}{x+10}} = SP$.</p> <p>Oo! That’s accidentally done the integration for us! $\int SP dx = P$ ((Plus a constant.))</p> <p>So, $I = [\Pi_{r=1}^{10}(x+r)]_0^1$, which is $11! - 10!$, or $(10)(10!)$ as we suspected.</p> <p>I feel like there may be other ways to do it. What would you have tried?</p>This cropped up on Reddit:Ask Uncle Colin: Which Way Does Friction Go?2021-11-24T00:00:00+00:002021-11-24T00:00:00+00:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-which-way-does-friction-go<blockquote> <p>Dear Uncle Colin,</p> <p>I keep getting tripped up in my mechanics questions because I can never figure out which way the frictional force has to go. Do you have any advice?</p> <p>Mechanics Understanding</p> </blockquote> <p>Hi, MU, and thanks for your message!</p> <p>This is a situation where I have One Big Trick and not much else to say. The thing to do is to ask “what would happen if friction wasn’t there?” and reason that friction has to oppose that motion.</p> <p>For example, if you have a ladder leaning against a wall, you can argue that the ladder would slide down a smooth wall, so the friction force at that end must act upwards. Similarly, the ladder would slide away from the wall on a smooth floor, so friction must act towards the wall.</p> <p>It takes a little practice to get used to visualising this, but once you have the idea in place and it clicks, you find yourself wondering how you ever used to get it wrong!</p> <p>Hope that helps,</p> <p>- Uncle Colin</p>Dear Uncle Colin, I keep getting tripped up in my mechanics questions because I can never figure out which way the frictional force has to go. Do you have any advice? Mechanics UnderstandingA Facet of Factoring2021-11-22T00:00:00+00:002021-11-22T00:00:00+00:00https://www.flyingcoloursmaths.co.uk/a-facet-of-factoring<p>Someone on reddit asked how to show that $2^{15}-1$ was not a prime number, and I suddenly understood something I’d previously just ‘become used to’.</p> <p>In binary, $2^{15}-1$ is 111,111,111,111,111. You can break that down in groups of three (or five) – but it’s fairly obviously $111 \times 1,001,001,001,001$.</p> <p>Previously, I could certainly have proved that $a^{bc} - 1$ was a multiple of $a^b - 1$, probably using a geometric series formula, but I don’t think I had a visual feel that it was true – even if the geometric series is formally just the same as splitting the number up.</p>Someone on reddit asked how to show that $2^{15}-1$ was not a prime number, and I suddenly understood something I’d previously just ‘become used to’.Ask Uncle Colin: A Nonlinear Graph Transformation2021-11-17T00:00:00+00:002021-11-17T00:00:00+00:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-a-nonlinear-graph-transformation<blockquote> <p>Dear Uncle Colin,</p> <p>I’ve figured out the graph of $y = e^x\br{2x^2 -5x + 2}$ using algebra and calculus; the second part of the question asks about $y = e^{x^2}\br{2x^4 - 5x^2 + 2}$. I feel like there must be a quicker way than redoing all of that work!</p> <p>Some Heuristic Or Rule To Circumvent Unnecessary Toil?</p> </blockquote> <p>Hi, SHORTCUT, and thanks for your message!</p> <p>You can indeed circumvent unnecessary toil here – the trick is to notice that the right hand side of the second expression is the same as the first, but with all of the $x$s replaced by $x^2$s. Alternatively said, if the first equation is $y=f(x)$, then the second is $y = f\br{x^2}$.</p> <p>We’re looking at graph transformations.</p> <h3 id="recap">Recap</h3> <p>Now, you’re likely familiar with linear graph transformations “inside the bracket”: given something like $y=f(2x)$, you need to “undo” the multiplication and halve the width of the curve. Given something like $y = f(x-3)$, you need to “undo” the subtraction and shift the curve three units to the right.</p> <p>The reason for this is that when you want the $y$ value corresponding to a given $x$, you evaluate the function at the $x$-value after you’ve transformed it – so for the $y$ value of the curve where $x=-3$, you have to work out the value of the function with $x=-6$.</p> <h3 id="what-about-the-square">What about the square?</h3> <p>Well, you need to undo it. Each of the points on the curve where $x&gt;=0$ will correspond to two points on the new curve, where the new $x$ value is a square root of the old one.</p> <p>So, your $x$-intercepts of $\frac{1}{2}$ and $2$ become $\pm \frac{1}{\sqrt{2}}$ and $\pm \sqrt{2}$, and your minimum at $\frac{3}{2}$ becomes $\pm \sqrt{\frac{3}{2}}$.</p> <p>The $y$-intercept at $(0,2)$ stays put (and becomes a maximum).</p> <p>Hope that helps!</p> <p>- Uncle Colin</p>Dear Uncle Colin, I’ve figured out the graph of $y = e^x\br{2x^2 -5x + 2}$ using algebra and calculus; the second part of the question asks about $y = e^{x^2}\br{2x^4 - 5x^2 + 2}$. I feel like there must be a quicker way than redoing all of that work! Some Heuristic Or Rule To Circumvent Unnecessary Toil?Some Smart STEP Trigonometry2021-11-15T00:00:00+00:002021-11-15T00:00:00+00:00https://www.flyingcoloursmaths.co.uk/some-smart-step-trigonometry<p>A question we looked at in class:</p> <blockquote> <p>If $f(x) = \arctan(x) + \arctan\br{\frac{1-x}{1+x}}$, find $f’(x)$. Hence, or otherwise, find a simple expression for $f(x)$.</p> </blockquote> <p>I’m not sure if I <em>like</em> this question, but there’s a good deal of depth to it.</p> <p>Have a go if you want to – I’ll tackle it below the line, where there be spoilers.</p> <hr /> <h3 id="option-1-brutal-calculus">Option 1: brutal calculus</h3> <p>I don’t know the derivative of $\arctan(x)$ off the top of my head ((I should note that the student did!)), but it’s not too hard to work out.</p> <p>If $y = \arctan(x)$ then $\tan(y) = x$; differentiating implicitly, $\sec^2(y) \dydx = 1$, but $\sec^2(y) = 1 + \tan^2(y)$, or $1+x^2$. Therefore $\dydx = \frac{1}{1+x^2}$.</p> <p>So the first term isn’t too bad. The second term… well, yuk. The obvious options are the chain rule and some monstrous rearrangement/implicit differentiation dance that I don’t fancy. Let’s chain rule it.</p> <p>We’ve got $y = \arctan(u)$, where $u = \frac{1-x}{1+x}$.</p> <p>$\diff{y}{u} = \frac{1}{1+u^2}$. We could use quotient rule on $u$, but I’d sooner spot that it’s $\frac{2}{1+x} - 1$, which differentiates easily to $\diff{u}{x} = \frac{-2}{(1+x)^2}$.</p> <p>So, $\dydx = \frac{1}{1+u^2}\times \frac{-2}{(1+x)^2}$, or $\frac{-2}{\br{1+u^2}\br{1+x}^2}$.</p> <p>I’ve written it like that because it makes the bottom turn out nicely: $\br{1+u^2}\br{1+x}^2 = (1+x)^2 + (1-x)^2$, which is $2 + 2x^2$.</p> <p>So that leads us to $\dydx = \frac{-2}{2+2x^2}$, or $\frac{-1}{1+x^2}$ for the second term.</p> <p>Adding the first and second terms together gives a derivative of zero, so the function is <em>clearly</em> ((“Gosh! He’s using italics there! I wonder why?”)) constant, and sticking in $x=0$ gives $f(x) = \frac{\pi}{4}$.</p> <h3 id="option-2-moderate-trigonometry">Option 2: moderate trigonometry</h3> <p>Let $A=\arctan(x)$ and $B = \arctan\br{\frac{1-x}{1+x}}$, so $f(x) = A+B$.</p> <p>Let $y = f(x)$, so that $\tan(y) = \tan(A+B)$.</p> <p>$\tan(A+B) = \frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$</p> <p>But we know what $\tan(A)$ and $\tan(B)$ are!</p> <p>$\tan(A+B) = \frac{x + \frac{1-x}{1+x}}{1 - x \frac{1-x}{1+x}}$. Ugh! We can simplify that.</p> <p>$\dots = \frac{x(1+x) + (1-x)}{(1+x) - x(1+x)}$</p> <p>$\dots = \frac{x^2 + 1}{1+x^2} = 1$.</p> <p>So, $\tan(f(x)) = 1$ and $f(x)$ is <em>clearly</em> $\piby4$.</p> <h3 id="option-3-simple-trigonometry">Option 3: simple trigonometry</h3> <p>$\tan\br{\piby4-t} = \frac{1-\tan(t)}{1+\tan(t)}$, so $\arctan\br{\frac{1-x}{1+x}} = \piby4 - \arctan(x)$.</p> <p>$f(x) = \arctan(x) + \piby4 - \arctan(x)$, or $\piby4$.</p> <p>How lovely! All three of the methods give the same answer.</p> <h3 id="unfortunately-its-the-wrong-answer">Unfortunately, it’s the wrong answer</h3> <p>Again, have a little think and see if you can spot anywhere I’ve gone wrong. It’s a little subtle.</p> <p>…</p> <p>Did you spot it? The problem is that I’ve implicitly said that $f’(x) = 0$ <strong>everywhere</strong> – and there’s one exception to that.</p> <p>When $x=-1$, $f(x)$ is undefined – the bottom of the fraction in the second term is zero, and we can’t be having that. There’s a discontinuity there.</p> <p>In fact, if we put a huge negative number ($-X$) in for $x$, we get $\arctan(-X) + \arctan\br{\frac{1+X}{1-X}}$, which approaches $-\piby2 - \piby4$, or $-\frac{3}{4}\pi$ as $X \to \infty$.</p> <p>The correct answer is:</p> $f(x) = \begin{cases} -\frac{3}{4}\pi, &amp; x &lt; -1 \\\\ \text{undefined}, &amp; x=-1 \\\\ \piby 4, &amp; x&gt;1 \end{cases}$ <p>Just when I thought it had come out nicely, it turned out gnarly after all.</p> <p>Did you get it? Did you use a different approach? I’d love to hear about it in either case.</p>A question we looked at in class:Ask Uncle Colin: A Dungeon Decision2021-11-10T00:00:00+00:002021-11-10T00:00:00+00:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-a-dungeon-decision<blockquote> <p>Dear Uncle Colin,</p> <p>I’m playing a solo D&amp;D game and want some strategy advice for the final battle, which consists of a series of rounds, alternating between monster attacks and my attacks. In a monster attack, it rolls nine dice and I roll eight; if it rolls more sixes than I do, I lose the difference in health points. In my attack, I roll three dice and it rolls nine; if I roll more sixes than it does, I win. How many health points to I need to have a 50-50 chance of winning?</p> <p>- Many Orcs Need Swords To Exact Revenge</p> </blockquote> <p>Hi, MONSTER, and thanks for your message!</p> <p>My first thought is, yuk. Those probabilities are going to be a bit hairy, especially for the monster attacks. Let’s look at your attacks first.</p> <h3 id="your-attacks">Your attacks</h3> <p>You win if:</p> <ul> <li>You roll three sixes and the monster rolls two or fewer</li> <li>You roll two sixes and the monster rolls one or none</li> <li>You roll one six and the monster rolls none.</li> </ul> <p>The probability of the monster rolling:</p> <ul> <li><strong>two sixes</strong> is $\nCr{9}{2} \br{\frac{1}{6}}^2 \br{\frac{5}{6}}^7 \approx 0.279$ (call that $S_2$).</li> <li><strong>one six</strong> is $\nCr{9}{1} \br{\frac{1}{6}} \br{\frac{5}{6}}^8 \approx 0.349$ (call that $S_1$).</li> <li><strong>no sixes</strong> is $\br{\frac{5}{6}}^9 \approx 0.194$ (call that $S_0$).</li> </ul> <p>The probability of you rolling:</p> <ul> <li><strong>three sixes</strong> is $\frac{1}{216} \approx 0.00463$</li> <li><strong>two sixes</strong> is $\nCr{3}{2} \br{\frac{1}{6}}^2\br{\frac{5}{6}} \approx 0.0694$</li> <li><strong>one six</strong> is $\nCr{3}{1} \br{\frac{1}{6}}\br{\frac{5}{6}}^2 \approx 0.347$.</li> </ul> <p>So the probability of you winning a round is $\frac{1}{216}(S_0+S_1+S_2) + \frac{15}{216}(S_0+S_1) + \frac{75}{216}(S_0)$, which works out to be about 0.1088 – you’ll win about one round in 10.</p> <h3 id="the-monsters-attacks">The monster’s attacks</h3> <p>Honestly, we <em>could</em> do the same analysis, but we’re not going to. Instead, we’re going to work it out with a generating function.</p> <p>Skipping a lot of details, we can model the number of sixes the monster rolls as $\br{\frac{5}{6} + \frac{1}{6}x}^9$, and the number of sixes you roll as $\br{\frac{5}{6} + \frac{1}{6}y}^8$.</p> <p>And in fact, if we let $y = {x^{-1}}$, then multiplying those two expressions together gives us magic ((Not a spell you can cast, sadly)): the coefficient of the $x^k$ term is exactly the probability of the monster rolling $k$ more sixes than you.</p> <p>Unfortunately, it’s - so to speak - monstrous. Neglecting the terms with non-positive powers of $x$ gives a generating function of $\frac{390625}{16926659444736} x^9 + \frac{18203125}{16926659444736}x^8 + \frac{47515625}{2115832430592}x^7 + \frac{585490625}{2115832430592}x^6 + \frac{9431471875}{4231664861184}x^5 + \frac{51945142375}{4231664861184}x^4 + \frac{99389922275}{2115832430592}x^3 + \frac{263182236755}{2115832430592}x^2 + \frac{1888873979963}{8463329722368}x$</p> <p>Frankly, we’re in simulation territory here. There <em>is</em> an analytic solution, but just no.</p> <p>Running a few thousand simulations (which we could, of course, have done straight away) shows that with 4 health points, you win 44% of the time and with 5, you win 52% – so five health points will be enough about half of the time. With nine, you win about three-quarters of the simulations, and with 16 you have a 90% success rate.</p> <p>Hope that helps!</p> <p>- Uncle Colin</p>Dear Uncle Colin, I’m playing a solo D&amp;D game and want some strategy advice for the final battle, which consists of a series of rounds, alternating between monster attacks and my attacks. In a monster attack, it rolls nine dice and I roll eight; if it rolls more sixes than I do, I lose the difference in health points. In my attack, I roll three dice and it rolls nine; if I roll more sixes than it does, I win. How many health points to I need to have a 50-50 chance of winning? - Many Orcs Need Swords To Exact RevengeA Cubic Conundrum2021-11-08T00:00:00+00:002021-11-08T00:00:00+00:00https://www.flyingcoloursmaths.co.uk/a-cubic-conundrum<p>In class, we tackled a Senior Maths Challenge problem that goes something like this:</p> <p>“Given that $y^3 f(x) = x^3 f(y)$ for all real $x$ and $y$, and that $f(3) \ne 0$, find the value of $\frac{f(20)-f(2)}{f(3)}$.</p> <p>A: 6; B: 20; C: 216; D: 296; E:2023”</p> <p>Have a play with it. Spoilers below the line. The interesting bits for me come after the problem is solved.</p> <hr /> <p>The question itself is more intimidating than it is difficult. In particular, $27 f(x) = x^3 f(3)$ for all $x$, so we can multiply by 27 and replace all of the $f$ evaluations like this ((This is simply the neatest way I can see; other options are available.)):</p> <p>$\frac{27 f(20)- 27f(2)}{27f(3)} = \frac{20^3 f(3) - 2^3 f(3)}{27 f(3)}$.</p> <p>All of the $f(3)$s cancel, leaving $\frac{8000 - 8}{27}$.</p> <p>A quick sense check says that’s between 100 and 1000, so the answer must be C or D; use whatever arithmetic you like to find that it’s 296. (Under pressure, I’d probably divide by 9 and by 3; alternatively, $27\times 250 = 6750$, so the answer is greater than 250). The answer is D.</p> <h3 id="a-wry-twist">A wry twist</h3> <p>The answer can’t be 216, because 216 is a cube. If the answer were 216 then $20^3 + (-2)^3 = \br{3^3}{6^3}$, which would be a counterexample to Fermat’s Last Theorem.</p> <h3 id="but-whats-the-function">But what’s the function?</h3> <p>I also wondered what functions might satisfy the given relationship, which I rewrote as $\frac{f(y)}{f(x)} = \frac{y^3}{x^3}$. It’s fairly obvious that $f(x) = x^3$ satisfies this, but are there others?</p> <p>I don’t much like the division. I prefer subtraction. I’m going to let $F(x) = \ln(f(x))$, so that $F(y) - F(x) = 3\ln(y) - 3\ln(x)$.</p> <p>Now let $y = x + h$: $F(x+h) - F(x) = 3\ln(x+h) - 3\ln(x)$</p> <p>Divide by $h$: $\frac{F(x+h)-F(x)}{h} = 3\frac{ \ln(x+h) - \ln(x) }{h}$</p> <p>So the derivative of $F(x)$ is equal to the derivative of $3\ln(x)$ – or $\diff{}{x} \ln(f(x)) = \diff{}{x}\ln(x^3)$.</p> <p>Integrate both sides with respect to $x$: $\ln(f(x)) = \ln(x^3)+ C$</p> <p>So $f(x) = Ax^3$ for any constant $A$ ((Except for 0, which is disallowed in the question.))</p> <hr /> <p>I thought that was a great question, as much for where it led as for the puzzle it posed. What did you make of it?</p>In class, we tackled a Senior Maths Challenge problem that goes something like this:Ask Uncle Colin: A Jigsaw Puzzle2021-11-03T00:00:00+00:002021-11-03T00:00:00+00:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-a-jigsaw-puzzle<blockquote> <p>Dear Uncle Colin,</p> <p>I bought a 750-piece jigsaw from a charity shop and found it had only 740 pieces! If I bought the same jigsaw again, and found it was missing ten random pieces, how likely is it I’d be able to complete the puzzle?</p> <p>Just Imagine Going Shopping As Well</p> </blockquote> <p>Hi, JIGSAW, and thanks for your message!</p> <p>This looks to me like a birthday problem. I can solve it exactly, but I’m going to ballpark it first, then solve it exactly.</p> <h3 id="ballpark">Ballpark</h3> <p>The probability of the first missing piece being present in the second puzzle is $\frac{740}{750}$, or $\frac{74}{75}$. If that were constant, the probability of all ten pieces being present would be $\br{\frac{74}{75}}^{10}$.</p> <p>This looks like Ninja work. Taking logs gives $10\ln\br{1 - \frac{1}{75}}$, which is about $-\frac{2}{15}$. Unlogging gives about $\frac{13}{15}$, or roughly 87%.</p> <p>This is an overestimate – the presence probability of the later pieces varies depends on the presence of the earlier one – given all of the previous ones are there, the probability drops, and my guess would be closer to 80%.</p> <h3 id="doing-it-properly">Doing it properly</h3> <p>The true probability is $\frac{740}{750}\times \frac{739}{749} \times \dots \frac{731}{741}$.</p> <p>That’s $\frac{740!}{730!}\times\frac{740!}{750!}$. Gosh! To get a exact answer, I’d stick that directly into Wolfram|Alpha, or the original fraction into a calculator.</p> <p>It works out to be 0.874 – so my first answer was considerably closer!</p> <p>Hope that helps. Enjoy the jigsaw!</p> <p>- Uncle Colin</p>Dear Uncle Colin, I bought a 750-piece jigsaw from a charity shop and found it had only 740 pieces! If I bought the same jigsaw again, and found it was missing ten random pieces, how likely is it I’d be able to complete the puzzle? Just Imagine Going Shopping As Well