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A fun MX2 exercise could be to calculate the radius and centre of the generalised circle that the unit circle gets mapped to by T in terms of the coefficients of T.

More generally, suppose you are solving for w^n=1 (1), where w=(az+b)/(cz+d)=T(z) for some complex a,b,c,d with ad-bc not equal to zero (*). Such maps T are called "fractional linear transformations" on the complex plane, and have lots of nice properties. One of these properties that is...

Re: HSC 2015 4U Marathon - Advanced Level What is the purpose of that graph? It is neither convex or sexy. And why are you bringing up continuity? I never mentioned it...

You are confusing cos and cis! Your last line is invalid for this reason. (z+z^{-1})^4\\=(z^4+z^{-4})+4(z^2+z^{-2})+6\\=2\cos(4\theta)+8\cos(2\theta)+6.

Re: HSC 2015 4U Marathon - Advanced Level What do you mean? I can choose these conditions how I want. All I am saying is that functions that satisfy these conditions are sexy, I am not saying that all sexy functions satisfy these conditions. All that is needed is a construction that shows we...

Re: HSC 2015 4U Marathon - Advanced Level It has been a while, so I will post my solution. I think the answer is \alpha=2. First, we show that f(x)\leq 2x for all x and all f. (1) This inequality is certainly true in the interval (1/2,1] by monotonicity and the fact that f(1)=1. We now...

Yep, by all means. Cheers, will amend a little later.

Re: MX2 2015 Integration Marathon I know how they work and what kind of answer I am looking for. I am asking students to justify the accuracy of Simpsons rule being greater than that of the trapezoidal rule, and quantify to what extent it is better. Eg, something like a statement about...

Ah yes, neither for the hyperbola. We require k > |z1-z2| for the ellipse and k < |z1-z2| for the hyperbola to avoid degeneracy.

Re: MX2 2015 Integration Marathon bump.

The ones involving just distances to two points that are good to recognise: |z-z_1|=|z-z_2| (the line that bisects z1z2). |z-z_1|=k|z-z_2| \quad (k> 0) (the circle with diameter given by the line segment joining the internal and external points of k:1 division). |z-z_1|+|z-z_2| =...

Re: HSC 2015 4U Marathon - Advanced Level Re-read the original question. The condition is: f(x+y) >= f(x) + f(y) (often called super-additivity) not convexity.

Re: HSC 2015 4U Marathon - Advanced Level Nope, alpha is not 1. Try to find an upper bound first :).

Re: 2015 permutation X2 marathon You might find it easier to think about if the committees are indistinguishable, in which case the bracelets are monochromatic and are just splitting the people up into groups separated by the the partnership relation a finite number of times, adjacency meaning...

Re: 2015 permutation X2 marathon Yeah, sorry about the wording and explanation being a bit terse. Bracelets because this thing has a looped structure. Say I am in two committees, choose one of them arbitrarily. This committee has one other member, this member is in one further...

Re: HSC 2015 4U Marathon - Advanced Level Here is a further hint. If a function f satisfies f(0)=0 and is convex, then f(x+y) >= f(x) + f(y) for all x and y, as: f(x)+f(y) \leq \frac{x}{x+y}f(x+y)+\frac{y}{x+y}f(x+y)=f(x+y).

Re: 2015 permutation X2 marathon A recursive formula is easy enough. Picture the n people as black beads, numbered 1 through n, and the n committees as white beads, numbered 1 through n. How many ways can we create a finite collection of bracelets, where the beads on each bracelet alternate...

Re: 2015 permutation X2 marathon Will double check this solution / look for a nicer way of computing it (ideally a computation generalised to n people/committees) when sober but: If A has the same partner in both of his committees, then so must the other two people. So the number of such...

Re: HSC 2015 4U Marathon - Advanced Level Uh, I think you are getting a bit confused, this quote makes very little sense. Any sexy function f satisfies f(x+y) >= f(x) + f(y) for all non-negative x and y such that x+y =< 1. x and y are just variables used for the sake of defining what it...

Re: HSC 2015 4U Marathon - Advanced Level The statement is required to hold true for any pair of non-negative numbers x and y such that x+y =< 1. (So their sum is still in the domain of the function.) This condition was already included in the initial and revised wording of the question.