Jekyll2021-08-05T04:07:27+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 fiddly Area2021-08-04T01:00:00+01:002021-08-04T01:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-a-fiddly-area<blockquote>
<p>Dear Uncle Colin,</p>
<p>How would I find the area between the curve $y = \frac{1}{x^2}$, the coordinate axes, and the lines $y=4$ and $x=2$?</p>
<p>Don’t Really Appreciate What It Needs Graphically</p>
</blockquote>
<p>Hi, DRAWING, and thanks for your message!</p>
<p>The sketch is the important thing here. We’ve got a shape with five boundary pieces: the $x$-axis from 0 to 2, the line $x=2$ from 0 to where it meets the curve at $\left(2, \frac{1}{4}\right)$, the curve from that point to where it meets the upper line at $\left(\frac{1}{2},4\right)$, the line from $\left(\frac{1}{2},4\right)$ to $(0,4)$ and the $y$-axis from $(0,4)$ to $(0,0)$.</p>
<p>The area of that regionsplits nicely into two sections: a rectangle with a base of $\frac{1}{2}$ and height 4 (so area 2), and an area under the curve we’ll need to integrate.</p>
<p>Using the limits we worked out (and writing $\frac{1}{x^2}$ as $x^{-2}$), that’s going to be $\int_{1/2}^{2} x^{-2} \dx$.</p>
<p>We get $\left[ -x^{-1}\right]_{1/2}^{2}$, which is $-\frac{1}{2}+2$, or $\frac{3}{2}$.</p>
<p>Overall, the area is $2 + \frac{3}{2} = \frac{7}{2}$.</p>
<p>Hope that helps!</p>
<p>- Uncle Colin</p>Dear Uncle Colin, How would I find the area between the curve $y = \frac{1}{x^2}$, the coordinate axes, and the lines $y=4$ and $x=2$? Don’t Really Appreciate What It Needs GraphicallyDictionary of Mathematical Eponymy: The Gray Graph2021-08-02T01:00:00+01:002021-08-02T01:00:00+01:00https://www.flyingcoloursmaths.co.uk/dictionary-of-mathematical-eponymy-the-gray-graph<h3 id="what-is-the-gray-graph">What is the Gray Graph?</h3>
<p>Imagine a Rubik’s cube – just for visualisation purposes, we’re not solving it or anything. It’s got 27 cubelets (including the middle). It also has 27 lines of three cubelets parallel to the axes (nine in each direction).</p>
<p>Here comes the graph: represent each of the lines and each of the cubelets by a node in the graph. Two nodes are connected if and only if the cubelet represented by one node lies on the line represented by the other.</p>
<p>Here are two pictures of it:</p>
<p><img src="/images/Gray_graph_1.png" alt="" /> <img src="/images/Gray_graph_2.png" alt="" /></p>
<p>Since each cubelet lies on three lines (and each line passes through three cubelets), the graph has 81 edges linking the 54 nodes.</p>
<h3 id="why-is-it-important">Why is it important?</h3>
<p>The Gray graph is the smallest possible cubic semi-symmetric graph: any pair of <em>edges</em> can be transformed one into the other, but that’s not true of the <em>nodes</em> – you can’t transform a node representing a line into a node representing a cubelet or vice versa.</p>
<p>I’m not certain <em>why</em> this is significant – but it has a neat picture. Gray apparently thought similarly; she thought it a curiosity and didn’t bother to publish it. It was studied by Bouwer in the 1970s and found to answer some questions about symmetries.</p>
<h3 id="who-was-marion-cameron-gray">Who was Marion Cameron Gray?</h3>
<p>Marion Gray was born in Ayr, Scotland in 1902 and studied mathematics and natural philosophy at Edinburgh, graduating with a first-class degree. In the 1920s, she gained her PhD at Bryn Mawr College, Pennsylvania.</p>
<p>In 1930, she joined AT&T as an engineer, where she came up with the Gray graph in 1932. From 1934 until her retirement, she worked at Bell Labs, and served on all sorts of committees (including the one responsible for the goldmine Handbook of Mathematical Functions).</p>
<p>Gray died in Edinburgh, Scotland in 1977.</p>What is the Gray Graph?Ask Uncle Colin: Runs of numbers2021-07-28T01:00:00+01:002021-07-28T01:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-runs-of-numbers<blockquote>
<p>Dear Uncle Colin,</p>
<p>I noticed that $\frac{987654321}{123456789} \approx 8.0000000729$ in base 10. On further investigation, $\frac{321}{123}\approx 2$ in base 4, $\frac{654321}{123456}\approx 5$ in base seven and a similar pattern recurs in any base: $\frac{[b-1][b-2]\dots 54321}{12345\dots[b-2][b-1]}$ (where $[a][b]$ means the concatenation of digits $a$ and $b$) is very close to $b-2$. Any idea why?</p>
<p>Runs Are Tremendously Interesting, Oooya!</p>
</blockquote>
<p>Hi, RATIO, and thanks for your message! The answer lies in the binomial expansion.</p>
<h3 id="in-base-10">In base 10</h3>
<p>Let’s look at the base 10 version first, and then generalise it.</p>
<p>The nugget here is that $\frac{1}{81} = 0.\dot 01234567\dot 9$ - the nice pattern is slightly spoilt by a carry from the 10. This comes from $(1-x)^{-2} = 1 + 2x + 3x^2 + \dots$; evaluated at $x=\frac{1}{10}$, you get $\frac{100}{81} = 1.2345\dots$, and you can get to $\frac{1}{81}$ - or, more pertinently, something very close to 123456790 - from there. In fact, $\frac{10^{10}}{81} = 123,456,790.\dot 123456789\dot0$, which is only one-and-a-bit more than the denominator.</p>
<p>Let’s also think about (for reasons that will become clear) the sum of the numerator and denominator, which is 1,111,111,110. You might recognise an almost-pattern there, too; it looks very much like a ninth. In fact, it’s one and a smidge smaller than $\frac{10^{10}}{9}$.</p>
<p>Recap: 123,456,789 is about $\frac{x}{81}$ and 123,456,789 + 987,654,321 is about $\frac{x}{9}$, with $x =10^{10}$. That means 987,654,321 is roughly $\frac{8x}{81}$.</p>
<p>And voila: $\frac{987,654,321}{123,456,789} \approx \frac{8x/81}{x/81} = 8$.</p>
<h2 id="in-base-b">In base $b$</h2>
<p>A similar pattern holds in general, only replacing 81 with $(b-1)^2$, 9 with $(b-1)$ and $10^{10}$ with $b^b$.</p>
<p>More concretely:</p>
<ul>
<li>$123\dots[b-2][b-1] \approx \frac{b^b}{(b-1)^2}$</li>
<li>$[b-2][b-1]\dots321 + 123\dots[b-2][b-1] = 111\dots10 \approx \frac{b^b}{b-1}$</li>
<li>So $[b-2][b-1]\dots321 \approx \frac{(b-2)b^b}{(b-1)^2}$</li>
<li>And $\frac{[b-2][b-1]\dots321}{123\dots[b-2][b-1]} \approx b-2$.</li>
</ul>
<p>Hope that helps!</p>
<p>- Uncle Colin</p>Dear Uncle Colin, I noticed that $\frac{987654321}{123456789} \approx 8.0000000729$ in base 10. On further investigation, $\frac{321}{123}\approx 2$ in base 4, $\frac{654321}{123456}\approx 5$ in base seven and a similar pattern recurs in any base: $\frac{[b-1][b-2]\dots 54321}{12345\dots[b-2][b-1]}$ (where $[a][b]$ means the concatenation of digits $a$ and $b$) is very close to $b-2$. Any idea why? Runs Are Tremendously Interesting, Oooya!Five reasons your trapezium rule answer might be off2021-07-26T01:00:00+01:002021-07-26T01:00:00+01:00https://www.flyingcoloursmaths.co.uk/five-reasons-your-trapezium-rule-answer-might-be-off<p>Despite “appropriate use of technology” being a key part of the A-level syllabus, it is apparently Still A Thing to have to perform the trapezium rule by hand (rather than, say, asking Wolfram|Alpha like a normal mathematician would).</p>
<p>But still. If it’s in the syllabus, I suppose you can’t simply state a moral objection to the question and expect any marks, so you’re stuck with learning it. And you’re probably stuck with making the same errors as everyone else. Here are some questions to ask yourself if your answer is off:</p>
<h3 id="1-are-you-using-radians">1. Are you using radians?</h3>
<p>I think this is probably the biggest source of errors in trapezium rule questions involving trig functions. Calculus, even numerical calculus, is <em>always</em> in radians. If your calculator isn’t, you will get the wrong answer.</p>
<h3 id="2-have-you-got-the-right-number-of-strips-and-ordinates">2. Have you got the right number of strips and ordinates?</h3>
<p>The question might ask for a particular number of <em>strips</em> or a particular number of <em>ordinates</em> (the values either side of each strip). If you have $n$ strips, you have $n+1$ ordinates. You always divide the width of the domain you’re interested by the number of <em>strips</em>, not the number of ordinates – imagine you just had one strip; then you’d have to divide the domain by one to get the height of the trapezium.</p>
<h3 id="3-have-you-got-the-multipliers-right">3. Have you got the multipliers right?</h3>
<p>The $y$-value of the first and last ordinates are multiplied by 1; the others are multiplied by 2, and the whole lot added up then multiplied by half the strip-width. It’s easy to mess this up; I recommend using a table of values to keep track.</p>
<h3 id="4-have-you-multiplied-by-half-the-strip-width">4. Have you multiplied by half the strip width?</h3>
<p>Again, this step is very easy to miss. I suggest coming up with a ballpark answer before you do the calculation – just eyeball it to see what range of answers you’d be prepared to believe.</p>
<h3 id="5-have-you-accounted-for-positive-and-negative-y-values">5. Have you accounted for positive and negative $y$-values?</h3>
<p>It’s unusual for them to pull a sneaky trick like this, but not beyond the scope: make sure you’re looking out for it. If your curve drops below the $x$-axis, some or all of the result of the integration will be negative. If you’re looking for the area, you will need to account for this.</p>
<p>Anything I’ve missed? <a href="mailto:colin@flyingcoloursmaths.co.uk">Drop me an email!</a></p>Despite “appropriate use of technology” being a key part of the A-level syllabus, it is apparently Still A Thing to have to perform the trapezium rule by hand (rather than, say, asking Wolfram|Alpha like a normal mathematician would).Ask Uncle Colin: Messy Geometry2021-07-21T01:00:00+01:002021-07-21T01:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-messy-geometry<blockquote>
<p>Dear Uncle Colin,</p>
<p>I’m told that the normal to the curve $y=ax^2 + bx + 1$ through the point $(3,1)$ is parallel to the line $3y - x + 4 = 0$ and I need to find $a$ and $b$. There’s a lot going on there and I’m a bit lost!</p>
<p>Bit Unhappy Solving, Yuck!</p>
</blockquote>
<p>Hi, BUSY, and thank for your message! There <em>is</em> a lot going on there.</p>
<p>There are two key pieces of information to pull out there:</p>
<ul>
<li>The curve passes through $(3,1)$</li>
<li>The normal to the curve at that point is parallel to a given line.</li>
</ul>
<p>These are going to give us a pair of equations to solve.</p>
<p>Let’s start with the first piece of information: when $x=3$, $y=1$, so $1 = 9a + 3b + 1$. Therefore, $9a + 3b = 0$ - or $b = -3a$ after a bit of shuffling.</p>
<p>The second piece is more involved. I would start by finding the gradient of the line: rewriting it as $y = \frac{1}{3}x + \frac{4}{3}$ tells you that the gradient is $\frac{1}{3}$.</p>
<p>This is perpendicular to the gradient of the curve, so the curve must have a gradient of $-3$ when $x=3$.</p>
<p>What is the gradient of the curve there? Let’s differentiate: it’s $\dydx = 2ax + b$, so when $x=3$, $-3 = 6a + b$.</p>
<p>We know that $b= -3a$, so $-3 = 3a$, which gives $a=-1$ and $b=3$.</p>
<p>A quick check: if $y = -x^2 + 3x + 1$, it does indeed go through $(3,1)$; its derivative is $\dydx = -2x + 3$, which is indeed perpendicular to the line.</p>
<p>Hope that helps!</p>
<p>- Uncle Colin</p>Dear Uncle Colin, I’m told that the normal to the curve $y=ax^2 + bx + 1$ through the point $(3,1)$ is parallel to the line $3y - x + 4 = 0$ and I need to find $a$ and $b$. There’s a lot going on there and I’m a bit lost! Bit Unhappy Solving, Yuck!A STEP question that just drops out2021-07-19T01:00:00+01:002021-07-19T01:00:00+01:00https://www.flyingcoloursmaths.co.uk/a-step-question-that-just-drops-out<p>If there’s one thing I like better than a STEP mechanics question, it’s a STEP mechanics question that just drops out when you get the diagram right.</p>
<p>It’s question 11 from the <a href="https://stepdatabase.maths.org/database/db/17/17-S1.pdf">2017 STEP I</a>: a ladder in a box on a slope.</p>
<p>I’m going to answer it twice, once using the “routine” diagram-forces-moments-algebra method, and then the really nice way. But first, let’s get the diagram right.</p>
<h3 id="the-mandatory-diagram-improvement">The mandatory diagram improvement</h3>
<p>One of the best hacks I know for awkward inclined plane diagrams is to tilt the page so the floor is flat and the horizontal and vertical are on an incline. Sure, this means you have to account for the weight not acting vertically, but (in this case) it means all of the other forces are vertical and horizontal.</p>
<h3 id="the-three-equations">The three equations</h3>
<p>Coming up with three equations is fairly straightforward:</p>
<ul>
<li>Resolving parallel to the plane: $R_B = \mu R_A + W \sin(\alpha)$ (1)</li>
<li>Resolving perpendicular: $R_A + \mu R_B = W \cos(\alpha)$ (2)</li>
<li>Taking moments about the midpoint: $L\mu R_B \sin(\beta) + L R_B \cos(\beta) + L \mu R_A \sin(\beta) = L R_A \cos(\beta)$ (3)</li>
</ul>
<p>Now we can work on eliminating as many variables as possible. One is very easy to remove: there’s a common factor of $L$ through the last equation, so we can divide that out:</p>
<ul>
<li>$\mu R_B \sin(\beta) + R_B \cos(\beta) + \mu R_A \cos(\beta) = R_A \sin(\beta)$ (3’)</li>
</ul>
<p>I’m also going to multiply equation (2) by $\tan(A)$ so I can eliminate $W\sin(\alpha)$:</p>
<ul>
<li>$R_A \tan(\alpha) + \mu R_B \tan(\alpha) = W \sin(\alpha)$ (2’)</li>
</ul>
<p>So</p>
<ul>
<li>$R_B = \mu R_A + R_A \tan(\alpha) + \mu R_B \tan(\alpha)$ (1’)</li>
</ul>
<p>Or, if $t = \tan(\alpha)$:</p>
<ul>
<li>$R_B(1 - t\mu) = R_A(\mu + t)$ (1’’)</li>
</ul>
<p>We can treat equation (3’) similarly, letting $T = \tan(\beta)$:</p>
<ul>
<li>$R_B(1 + \mu T) = R_A(T-\mu)$ (3’’)</li>
</ul>
<p>Dividing these last two gives:</p>
<ul>
<li>$\frac{1-t\mu}{1+\mu T} = \frac{\mu + t}{T-\mu}$</li>
</ul>
<p>Which becomes:</p>
<ul>
<li>$(1 - t\mu)(T - \mu) = (\mu + t)(1 + \mu T)$</li>
</ul>
<p>This algebra is already tedious, isn’t it?</p>
<ul>
<li>$T - \mu - tT\mu + t\mu^2 = \mu + t + \mu^2 T + \mu Tt$</li>
<li>$(T-t) - (T-t)\mu^2 = 2\mu + 2\mu Tt$</li>
<li>$(T-t)(1-\mu^2) = 2\mu(1 + Tt)$</li>
</ul>
<p>Now cross-divide:</p>
<ul>
<li>$\frac{T-t}{1+Tt} = \frac{2\mu}{1-\mu^2}$</li>
</ul>
<p>Of course, $T$, $t$ and $\mu$ are all $tan$s in disguise:</p>
<ul>
<li>$\frac{\tan(\beta) - \tan(\alpha)}{1 + \tan(\alpha)\tan{\beta}} = \frac{2\tan(\gamma)}{1 - \tan^2(\gamma)}$</li>
</ul>
<p>And those are compound angle identities!</p>
<p>$\tan(\beta - \alpha) = \tan(2\gamma)$.</p>
<p>We know that $\piby 2 > \beta - \alpha > 0$ and that $0 < 2\gamma < \pi$, so the only option is for $\beta-\alpha = 2\gamma$ as required.</p>
<p>Can you see why I looked for a nicer solution?</p>
<h3 id="a-much-nicer-method">A much nicer method</h3>
<p>I learned this trick from @JusSumChick, and am always eager to apply it!</p>
<p>First, I’m going to do something with the contact forces: I can combine the normal reaction force and friction force at $A$ into a single force $C_A$ which acts at an angle of $\gamma$ to the left of the ‘vertical’. Similarly at $B$, the forces can combine into a single force $C_B$ at the same angle $\gamma$ above the ‘horizontal’.</p>
<p>Including $W$, we now have three forces. According to the Jus Sum Trick, the rod can only be in equilibrium if lines of action of these three forces pass through the same point ((You can take moments about anywhere. Take moments about a point where two lines of actions intersect. The moment of the other force can only be zero if it passes through that point.)).</p>
<p>The lines of action of $C_A$ and $C_B$ meet at a right angle at point $G$, so the lines of action and the rod form a right-angled triangle. $W$ acts through the midpoint $M$ of the rod, which is the circumcentre of the triangle. Therefore, point $G$ is a distance of $L$ from $M$ and - in particular - triangle $MGA$ is isosceles, and angles $M\hat G A$ and $G \hat A M$ are equal.</p>
<p>Angle $M\hat G A$ is $\alpha + \gamma$. Angle $G \hat A M$ is $\beta - \gamma$. Therefore $\beta - \alpha = 2\gamma$, as required.</p>
<p>* Edited 2021-07-19 to fix LaTeX</p>If there’s one thing I like better than a STEP mechanics question, it’s a STEP mechanics question that just drops out when you get the diagram right.Ask Uncle Colin: An Algebraic Puzzler2021-07-14T01:00:00+01:002021-07-14T01:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-algebraic-puzzler<blockquote>
<p>Dear Uncle Colin,</p>
<p>If you know that $a$ is an algebraic number, how do you prove that $ai$ is also algebraic?</p>
<p>Am Lost! Get Equations But Real Analysis Is Confusing</p>
</blockquote>
<p>Hi, ALGEBRAIC, and thanks for your message!</p>
<p>I’m in the same boat – my real analysis lectures were at the other end of town from my French classes, so I always pitched up ten minutes late, out of puff and missing the context of whatever Professor Falconer was talking about.</p>
<h3 id="exploration">Exploration</h3>
<p>So, with that in mind, I typically have to work stuff like this up from the ground using examples, and try to generalise from them.</p>
<p>In this case, I went for the simplest thing I could think of (although I’ve since come up with a simpler example - of which more later): I know that 2 and 3 are algebraic because $(x-2)(x-3) = x^2 - 5x + 6$ – and an algebraic number is one that’s a zero of any polynomial with real, rational coefficients.</p>
<p>How can I generate a satisfactory polynomial that has $2i$ and $3i$ as zeros? $(x-2i)(x-3i)$ doesn’t work, because $x^2 - 5ix - 6$ doesn’t have real coefficients.</p>
<p>However, there’s a lemma I can use: if $p + iq$ is a root of a real polynomial, so is $p - qi$ – so I need to include $-2i$ and $-3i$ as roots as well. That give $(x-2i)(x-3i)(x+2i)(x+3i) = x^4 + 13x^2 + 36$, which is the kind of polynomial we want.</p>
<p>And what I’ve done there is take a polynomial $p(x)$ with roots $A$ and $B$, then work out $p(ix)p(-ix)$ – which has roots $\pm Ai$ and $\pm Bi$.</p>
<p>So much for an example. Does it always work?</p>
<h3 id="a-sketchy-proof">A sketchy proof</h3>
<p>Recap: we’re trying to prove that if $a$ is algebraic, then so is $ai$.</p>
<p>We know (by definition) that there’s a polynomial with real, rational coefficients, $p(x)$, such that $p(a)=0$.</p>
<p>We want to show that $q(x) = p(ix)p(-ix)$ has two properties:</p>
<ul>
<li>It has $ai$ as a root</li>
<li>It has rational, real coefficients.</li>
</ul>
<p>The first of those is trivial: $q(ai) = p(-a)p(a)$, and $p(a)=0$.</p>
<p>The second is a little trickier, but not by much. What do the coefficients of $p(ix)$ look like? The coefficients of even powers of $x$ remain rational, but the odd-power coefficients become imaginary (but still rational). We can write $p(ix)$ as $s(x) + it(x)$, where $s$ and $t$ both have real, rational coefficients.</p>
<p>The same is true of $p(-ix)$, only with a change of sign in the odd terms: $p(-ix) = s(x) - it(x)$.</p>
<p>Their product is $\left[s(x)\right]^2 + \left[t(x)\right]^2$, which is a polynomial with real, rational coefficients – and with $ai$ as a root.</p>
<p>I hope that helps!</p>
<p>- Uncle Colin</p>Dear Uncle Colin, If you know that $a$ is an algebraic number, how do you prove that $ai$ is also algebraic? Am Lost! Get Equations But Real Analysis Is ConfusingThe Angle Windmill2021-07-12T01:00:00+01:002021-07-12T01:00:00+01:00https://www.flyingcoloursmaths.co.uk/the-angle-windmill<p>One of the places I would always fall down in A-level mechanics was on inclined plane force diagrams: I was chronically incapable of putting the correct angle in the correct place.</p>
<p>Until, that is, I figured out…</p>
<h3 id="the-angle-windmill">The Angle Windmill</h3>
<p>It looks like this: <img src="/images/angle_windmill.png" alt="The Angle Windmill" /></p>
<p>The process goes as follows:</p>
<ul>
<li>Draw a big diagram without any forces on
<ul>
<li>Mark the angle between the plane and the horizontal as $\alpha$</li>
</ul>
</li>
<li>Draw a line perpendicular to the inclined plane (in grey on the diagram) through your object</li>
<li>Draw a horizontal and vertical line (in green) through the object</li>
<li>(You now have eight rays coming out of the object</li>
<li>Find and mark a pair of rays separated by $\alpha$</li>
<li>Mark alternate spaces between rays with $\alpha$</li>
</ul>
<p>Boom! Job done. Now you know which angle goes where. You might alsonote that the angles in the gaps are all $90º - \alpha$.</p>
<h3 id="but-wait-theres-more">But wait. There’s more.</h3>
<p><img src="/images/angle_windmill2.png" alt="The Angle Windmill, revisited" /></p>
<p>Suppose you need to resolve a force that lies on one perpendicular grid to the other.</p>
<p>If you draw the <strong>s</strong>ails on the windmill like this, it tells you which component of the force you’re resolving requires a <strong>s</strong>ine.</p>
<p>This is, of course, a lazy shortcut – working out the trig isn’t <em>that</em> hard. It’s just easy to get wrong in a hurry, and I was always in a hurry.</p>
<p>(You may like, as an exercise, to convince yourself that all of the assertions in this post are true.)</p>One of the places I would always fall down in A-level mechanics was on inclined plane force diagrams: I was chronically incapable of putting the correct angle in the correct place.Ask Uncle Colin: A Tournament2021-07-07T00:00:00+01:002021-07-07T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-tournament<blockquote>
<p>Dear Uncle Colin,</p>
<p>I’m told that two teams play a five-game series. The first team wins the series, but the second team wins the second game. What’s the probability that the second team also won the first game? (Assume that the probability of winning each match is 50% and the matches are independent.)</p>
<p>Mulling A Tournament’s Combinarical Hurdles</p>
</blockquote>
<p>Hi, MATCH, and thanks for your message!</p>
<p>I would imagine there are many ways to approach this, but I’m going to go at it sledgehammer fashion.</p>
<p>If we know that the second team (let’s call them B) won the second match, we can just ignore that game and consider the remaining four, of which the first team (A) won at least three.</p>
<p>Either team A won all four of the remaining games (one way), or they won three and B won one (four ways).</p>
<p>There are five possibilities, and B won the first game in only one of them – so the probability is 1/5.</p>
<p>Hope that helps!</p>
<p>- Uncle Colin</p>Dear Uncle Colin, I’m told that two teams play a five-game series. The first team wins the series, but the second team wins the second game. What’s the probability that the second team also won the first game? (Assume that the probability of winning each match is 50% and the matches are independent.) Mulling A Tournament’s Combinarical HurdlesDictionary of Mathematical Eponymy: The Fischer 1960 Ellipsoid2021-07-05T01:00:00+01:002021-07-05T01:00:00+01:00https://www.flyingcoloursmaths.co.uk/dictionary-of-mathematical-eponymy-the-fischer-1960-ellipsoid<h3 id="what-is-the-fischer-1960-ellipsoid">What is the Fischer 1960 Ellipsoid?</h3>
<p>The world is round, right? If you’re not on board with that, I suggest you take a long swim.</p>
<p>It’s round all right, but it’s not a sphere: in the 1600s, models of the planet as an <strong>ellipsoid</strong> began to be taken seriously – the Earth is slightly flattened at the poles. (The difference is about 20km out of 6,400km or so, roughly 0.3%. If the Earth was the size of a football, the difference would be about a millimetre.) This amount of flattening – usually expressed a ratio – is the important thing in this story.</p>
<p>In 1924, the International Union of Geodesy and Geophysics adopted the Hayford ellipsoid (proposed by John Fillmore Hayford in 1910), with a flattening of 1:297. Hayford’s was one of the first multicontinental efforts (Bessel’s value, based only on European measurements, was 1:299.15).</p>
<p>Hayford was unchallenged for many years – until Fischer calculated a better value of 1:298.3 ((The value in use today is 1:298.257223563, about 20m different from Fischer’s.)). However, she was forbidden from publishing papers based on this value because it contradicted the one accepted in the literature. Of course, when satellites came along, they established that Fischer’s value was indeed correct and she was allowed to go back and correct her earlier papers.</p>
<h3 id="why-is-it-important">Why is it important?</h3>
<p>At the time, the USA was trying to recover from a shocking start to the space race. As part of its efforts in the Mercury mission, it needed accurate orbital calculations – and for that, it needed accurate three-dimensional vectors relating the locations of its tracking stations.</p>
<p>While the difference between the Fischer model and the Hayford one was only 90m or so at the poles, that was enough to make a significant difference in triangulation, and the one adopted by NASA for the mission.</p>
<h3 id="who-was-irene-k-fischer">Who was Irene K Fischer?</h3>
<p>Irene Kaminka Fischer was born in Vienna, Austria in 1907, and studied descriptive and projective geometry and the Technical University of Vienna.</p>
<p>She and her family fled Austria for Boston in 1939, teaching herself geodesy while working for the Army Map Service, where she would eventually become chief. She wrote a high school geometry book in the 1960s, more than 120 papers and a memoir of her career - the brilliantly-titled <strong>Geodesy? What’s That? My Personal Involvement in the Age-Old Quest for the Size and Shape of the Earth, With a Running Commentary on Life in a Government Research Office.</strong></p>
<p>She died in Boston in 2009.</p>What is the Fischer 1960 Ellipsoid?