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Hikorski Triples By Jonny Griffiths UEA, May 2010

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The mathematician's patterns, like the painter's or the poet's must be beautiful; the ideas, like the colors or the words must fit together in a harmonious way. Beauty is the first test: there is no permanent place in this world for ugly mathematics. G. H. Hardy (1877 - 1947), A Mathematician's Apology

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Mathematics

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What does mean to you?

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GCSE Resit Worksheet, 2002 How many different equations can you make by putting the numbers into the circles? Solve them!

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Suppose a, b, c, and d are in the bag. If ax + b = cx + d, then the solution to this equation is x = There are 24 possible equations, but they occur in pairs, for example: ax + b = cx + d and cx + d = ax + b will have the same solution. So there are a maximum of twelve distinct solutions.

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This maximum is possible: for example, if 7, -2, 3 and 4 are in the bag, then the solutions are:

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If x is a solution, then –x, 1/x and -1/x will also be solutions. ax + b = cx + d a + b(1/x) = c + d(1/x) c(-x) + b = a(-x) + d a + d(-1/x) = c + b(-1/x)

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The solutions in general will be: {p, -p, 1/p, -1/p} {q, -q, 1/q, -1/q} and {r, -r, 1/r, -1/r} where p, q and r are all ≥ 1

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It is possible for p, q and r to be positive integers. For example, 1, 2, 3 and 8 in the bag give (p, q, r) = (7, 5, 3). In this case, they form a Hikorski Triple.

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Are (7, 5, 3) linked in any way? Will this always work?

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a, b, c, d in the bag gives the same as b, c, d, a in the bag, gives the same as … Permutation Law

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a, b, c, d in the bag gives the same as a + k, b + k, c + k, d + k in the bag. Translation Law

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a, b, c, d in the bag gives the same as ka, kb, kc, kd in the bag. Dilation Law

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So we can start with 0, 1, a and b (a, b rational numbers with 0 < 1 < a < b) in the bag without loss of generality.

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a, b, c, d in the bag gives the same as -a, -b, -c, -d in the bag. Reflection Law

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Suppose we have 0, 1, a, b in the bag, with 0 < 1 < a < b and with b – a < 1 then this gives the same as –b, -a, -1, 0 which gives the same as 0, b - a, b - 1, b which gives the same as 0, 1, (b -1)/(b - a), b/(b - a) Now b/(b - a) - (b -1)/(b - a) = 1/(b - a) > 1

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If the four numbers in the bag are given as {0, 1, a, b} with 1 1, then we can say the bag is in Standard Form. So our four-numbers-in-a-bag situation obeys four laws: the Permutation Law, the Translation Law, the Reflection Law and the Dilation Law.

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Given a bag of numbers in Standard Form, where might the whole numbers for our HT come from?

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The only possible whole numbers here are:

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(b-1)/a must be the smallest here.

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Pythagorean Triples (√(x 2 +y 2 ), x, y) Hikorski Triples (p, q, (pq+1)/(p+q))

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How many HTs are there? Plenty...

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Is abc unique?

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Twelve solutions to bag problem are:

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What do mean to you?

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Adding Speeds Relativistically Suppose we say the speed of light is 1. How do we add two speeds?

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Try the recurrence relation: x, y, (xy+1)/(x+y)… Not much to report... Try the recurrence relation: x, y, (x+y)/(xy+1)… still nothing to report...

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But try the recurrence relation: x, y, -(xy+1)/(x+y)… Periodic, period 3

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Now try the recurrence relation: x, y, -(x+y)/(xy+1)… Also periodic, period 3 Are both periodic, period 6

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Parametrisation for Pythagorean Triples: (r(p 2 +q 2 ), 2rpq, r(p 2 -q 2 )) For Hikorski Triples?

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The Cross-ratio

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If a, b, c and d are complex, when is the cross-ratio real?

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Takes six values as A, B and C permute: Form a group isomorphic to S 3 under composition

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So the cross-ratio and these cross-ratio-type functions all obey the four laws: the Permutation Law, the Translation Law, the Reflection Law and the Dilation Law.

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Elliptic Curve Connection Rewrite this as Y 2 = X(X -1)(X - D) Transformation to be used is: Y = ky, X = (x-a)/(b-a), or... D runs through the values

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So we have six isomorphic elliptic curves. The j-invariant for each will be the same.

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is an elliptic curve

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has integral points (5,3), (3,-2), (-2,5) If the uniqueness conjecture is true... and (30,1), (1,-1), (-1,30) and (1,30), (30,-1), (-1,1) and (3,5), (-2,3), (5,-2)

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A periodic recurrence relation with period 5. A Lyness sequence: a ‘cycle’. (R. C. Lyness, once mathematics teacher at Bristol Grammar School.)

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x y Find a in terms of x and e in terms of y and then substitute... Cross-ratio-type functions and Lyness Cycles

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What if we try the same trick here? x y z ?

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And here? x y ? So this works with the other cross-ratio type functions too...

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Why the name?

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www.jonny-griffiths.net

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