(Text as at 18/12/2010 19:58:05)

__Background__

A reunion was arranged for 24th January 2009 for mathematics graduates of King’s College Cambridge, of which I am one (just about). See the programme (Web Link (http://www.kings.cam.ac.uk/assets/d/d2/240109_programme.pdf) Defunct). Though not announced as such, it appears to have been in celebration of Ken Moody, the former Director of Studies in Mathematics and Computer Science. Ken has just been made a Life Fellow of King’s, according to the King’s Annual Report, and has recently retired, according to his own website (Web Link). Some time or other I intend to write to Ken about my Web-technology^{1} project, but the time is not quite yet. I don’t want to be laughed at.

There were lectures and a dinner, and to add a bit of spice, some questions were set and a prize offered for the best solutions. They are of the “challenging but elementary” category, presumably on the assumption that the more antiquated of us outside of academia will have forgotten everything we ever learnt beyond ‘A’ level. See the questions (Web Link (http://www.kings.cam.ac.uk/assets/6/6e/MathsDinnerQuestions.pdf) Defunct). Despite being (I imagine) one of the worst mathematicians ever to be allowed to study at King’s (would that I had known what philosophy was at the time!), I decided to have a bash at them, with the results below. Initially, I thought I’d got them out, but closer inspection revealed some mistakes and arm-waving in my rough notes. The end result of all this – and the fact that we’d recently had a reunion for my matriculation year – was that I didn’t actually attend the event, which was a shame as it seems to have been rather a jolly affair, at least if the photos^{2} are anything to go by.__Attempted Solutions to Questions__

My website generator isn’t geared up for diagrams and mathematical formulae, so some tedious long hand involving MS WordArt and Formula Editor, saved as PDF files, has been required. The reader’s forbearance is requested.

**The Lazy Librarian**:**Question**: In the 30th century the Encyclopedia Britannica consists of 3000 volumes. They are laid out on a very long single shelf in the new King's College Library, but unfortunately not in the right order. Every day, the Librarian approaches the shelf, and in a completely unsystematic fashion finds a volume k in the wrong place and replaces it in the correct (that is kth) place. Does his task ever end?**Answer**: Well, yes the process does terminate – but it will take the librarian his whole career, and some, most likely. I’ve written a computer simulation of this problem, which includes (what I take to be) the key insight – that once Volume 1 is selected and put in the right place, it stays there. Then, when Volume 2 is selected ….; and similarly with volumes 3,000, 2,999 and so on. The initial worry is that volumes in mid-range don’t “stick”, because if a volume is removed from above and placed below, or vice-versa, any intervening volumes that__had__been correctly filed get displaced. However, there’s continual partial ordering going on and – as noted – there’s gradual “freezing” from the outsides. I ran my simulation for Volumes ranging from 10 to 30,000 (in general, 100 trials for each: see Raw Data and Summarised Data). The number of moves seems to increase approximately linearly with number of volumes (see Graph, Extended Graph). The time taken by the algorithm increases geometrically, because of the effort involved in selecting a misfiled volume randomly, and all the shuffling along the shelves required (see Graph, Extended Graph). No doubt there are more efficient algorithms to simulate the Lazy Librarian. Here’s my Code^{3}, (in MS Access Basic; unfortunately, your browser will probably tidy away all the indentations making the code less legible). A question remains whether the Lazy Librarian is__certain__to complete her task, or whether this happy result is just__overwhelmingly likely__. I address the possibility that (say) Volume 1 is never selected in a further Note^{4}.**All Triangles are Isosceles**:**Question**: see the question sheet.**Answer**: this is a “dodgy diagram” problem. The diagram can never be drawn with point O not below the baseline, except in the case where the triangle is isosceles, when it lies on the baseline. This requires proof … so follow the link^{5}.**In Favour of the Little Guy**:**Question**: Two scorpions chase an ant on the edges of a regular tetrahedron. The scorpions start at the midpoints of two opposite edges while the ant is at a midpoint of one of the remaining edges. The ant can move twice as fast as the scorpions. How should the ant move to escape its pursuers?**Answer**: basically, the ant legs it to one of the vertices that one of the scorpions moves away from. The ant can only be caught if the two scorpions are both less than half a side away from the two vertices terminating the side the ant is on … this needs more explanation and a diagram^{6}.**Chain Gang (Part 1)**:**Question**: The chain is coiled on a smooth horizontal table and one end is falling over the edge. Let the length which has fallen at time t be x(t). (i) Show that d^{2}x/dt^{2}= g/3 where g is the vertical acceleration due to gravity, so that the free end drops more slowly than a freely falling particle. (ii). Is energy conserved?**Answer**: It’s a little difficult to set this problem up. The reference to the chain being “coiled up” on the table is a little obscure. Basically, the set-up that gives the right answer is for the chain to be continually teetering on the edge so that it doesn’t need to be dragged. Models that give the wrong answer are a long spread-out chain and a coiled chain that would act like a rotating plate. For some sums, see the following note^{7}. Energy is, of course conserved, but it appears not to be because the kinetic energy gained by the chain that has fallen off the table is less than the loss of potential energy by the same x-length of chain. The difference is made up by kinetic energy in the bit of chain that has not yet fallen off the table. Follow the link^{8}for the proof.**Chain Gang (Part 2)**:**Question**: The two ends of the chain are held at a single point with the chain hanging, folded in half, under gravity. One end is held firmly and the other end is released. Let the length which the free end has fallen at time t be x(t). (i) Is energy conserved? (ii) Show that (dx/dt)^{2}= gx(2L – x)/(L-x) and comment on the limits as x tends to 0 and as x tends to L.**Answer**: This appears to be a straightforward question using conservation of energy. It looks as though (in contrast to part 1) that we can assume conservation of energy, and then net off increasing kinetic energy against decreasing potential energy, as in the attached Proof^{9}. As for the limiting cases:-- As x tends to 0, dx/dt tends to 0, indicative of a sluggish start to the chain’s fall as the first link is poked over the edge.
- As x tends to L, dx/dt tends to infinity. The literal infinity is an artefact of the idealised conditions, but is exemplified approximately in nature by the crack of a whip, where the end of the whip moves faster than sound, causing a sonic boom.

**Chain Gang (Conclusion)**:**Question**: Why are these problems so different? (You can make a reasonable approximation to the chain by daisy-chaining wire paperclips.)**Answer**: I’ve no idea at the moment; nor do I know why paperclips are supposed to help. I suppose the answer is that Part 2 involves the whole chain (well, half of it really) all the time, whereas in Part 1 the chain only gets involved link by link.

- Follow (this link) for level 5.

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