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A favourite concept: The study of the distribution of primes using the analytic structure of Riemann's Zeta function. It is surprising that the "continuous" methods of analysis have any bearing on the "discrete" questions one may pose in number theory. Favourite techniques: Interchanging order...

Re: 2012 HSC MX2 Marathon |z\pm w|^2=(z\pm w)\overline{(z\pm w)}=(z\pm w)(\overline{z}\pm\overline{w})=(z\overline{z}\pm(z\overline{w}+\overline{z\overline{w}})+w\overline{w})\\=|z|^2\pm2\Re(z\overline{w})+|w|^2.\\ \\$Add the two equations together to arrive at c.$

You mean an angle of 0 right? |i|+|-i|\neq |i-i|

And the equality conditions stated are incorrect.

Re: 2012 HSC MX2 Marathon The point is an arbitrary point on one of the triangles sides, not a vertex.

Well I guess nothing is lost by trying :). Will look into it now.

Not really I guess, but its a pretty ambitious idea. Would need a lot of people working on it for it to be worthwhile.

A while ago I was thinking of making a wiki for the MX2 course, they work well with latex and would make it easy for many people to contribute...never got around to it though.

Re: 2012 HSC MX2 Marathon Answer to the argument question IS 7. \arg(z^n)=\arg(z)\Rightarrow \arg(z^{n-1})=0\Rightarrow \frac{-(n-1)\pi}{3}=2k\pi. for some integer k. Setting k=-1 gives us the simplest nontrivial solution of n=7.

Re: 2012 HSC MX2 Marathon To follow on from Carrotstick's question: \text{A theorem from analysis states that any decreasing sequence of positive real numbers} \\\text{ must converge to a limit. Use this theorem to prove that }\\ \sum_{k=1}^n \frac{1}{k}-\log(n) \\ \\ \text{converges as...

Re: 2012 HSC MX2 Marathon With perhaps a one marker in between asking to explain why the harmonic series diverges to justify our claim of the existence of such an n.

Re: 2012 HSC MX2 Marathon \text{Let }n\text{ be the least integer such that }\sum_{k=1}^n \frac{1}{k}>9000. \text{ Prove that: }\\e^{8999}<n<e^{9000}.

Re: 2012 HSC MX2 Marathon Ah okay cool, the phrase "need k=blah" is somewhat misleading then.

Re: 2012 HSC MX2 Marathon A couple of typos. i) Your first equation does not make sense as written. Swapping the role of k and n in the two sums fixes this. ii) We want only a partial sum in Part B. iii) We can never have n=e^9000 as the latter is not an integer. Solution: \text{By...

Went to St Aloysius college, we used Cambridge.

Re: 2012 HSC MX2 Marathon It can be made pretty fast...observe that arctan is an increasing function of the positive reals and that x^2+4=(x-2)^2+4x\geq 4x with equality attained at x=2. So \arctan(\theta)\geq \arctan(1)\Rightarrow \arg(z)\geq \pi/4. with equality attained for some z.

Re: 2012 HSC MX1 Marathon Will leave the first geometry question for someone else to do, it's fairly straightforward. Refer to my diagram for the second, it is not too difficult to show that there is no diagram dependence in the following working: The blue lines in the diagram represent...

Figured :).

The function you wrote down was e^{i*theta}

The period is just 2pi.