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  1. G

    HSC 2015 MX2 Marathon ADVANCED (archive)

    Re: HSC 2015 4U Marathon - Advanced Level Bump with clearer wording.
  2. G

    HSC 2015 MX2 Marathon ADVANCED (archive)

    Re: HSC 2015 4U Marathon - Advanced Level All good, looking back on the initial wording I can understand the misinterpretation. I will edit accordingly.
  3. G

    HSC 2015 MX2 Marathon ADVANCED (archive)

    Re: HSC 2015 4U Marathon - Advanced Level I think you might have some quantifiers the wrong way around? We want to find a real number a such that f(x) =< ax for all f satisfying f(x+y) >= f(x) + f(y). In particular, we want to find the least a that works for every such f. We know a = 1...
  4. G

    HSC 2015 MX2 Marathon ADVANCED (archive)

    Re: HSC 2015 4U Marathon - Advanced Level A useful fact that has not yet been mentioned/proven is that any such f must be monotonic.
  5. G

    HSC 2015 MX2 Marathon ADVANCED (archive)

    Re: HSC 2015 4U Marathon - Advanced Level Yep, \alpha \geq 1 follows from considering the function f(x)=x. You are also right that this isn't the only function obeying the inequality in the question, so there might be solutions which are not bounded by x. See if you can find an alpha...
  6. G

    HSC 2015 MX2 Marathon ADVANCED (archive)

    Re: HSC 2015 4U Marathon - Advanced Level It has been a while, so here are some hints. (I don't want to give it away just yet because I think this is a great/fun question!) 1. What must f(0) be? 2. Can you think of any relationships between convexity and the property f(x+y) >= f(x)+f(y)...
  7. G

    HSC 2015 MX2 Integration Marathon (archive)

    Re: MX2 2015 Integration Marathon Your statement is equivalent to the statement of continuity. Ie f(x) -> 0 as x -> 0. What is important though is that we can also differentiate f everywhere on this interval (this is easy away from the point 0, but we can also find f'(0) by first...
  8. G

    HSC 2015 MX2 Integration Marathon (archive)

    Re: MX2 2015 Integration Marathon To be precise, the things I was lazy about can be summarised in the following exercise: Let f(x)=\frac{1}{\sin(x)}-\frac{1}{x} for x\in [-\pi/2,\pi/2]\setminus \{0\} and f(0)=0. Prove that f is continuous and differentiable on the interval [-\pi/2,\pi/2]...
  9. G

    HSC 2015 MX2 Integration Marathon (archive)

    Re: MX2 2015 Integration Marathon Okay, I will continue. Note in the below that R ranges over non-negative integers. \int_\mathbb{R} \frac{\sin(x)}{x}\, dx \\ \\ = \lim_{R \rightarrow \infty}\int_{-(2R+1)\pi/2}^{(2R+1)\pi/2}\frac{\sin(x)}{x}\, dx\\ \\...
  10. G

    HSC 2015 MX2 Integration Marathon (archive)

    Re: MX2 2015 Integration Marathon To get people started on Paradoxica's integral: \int_\mathbb{R} \frac{\sin^2(ax)}{x^2}\, dx\\ \\ = \int_\mathbb{R} \sin^2(ax)\frac{d}{dx}(-x^{-1})\, dx\\ \\ = \int_\mathbb{R} \frac{2a\sin(ax)\cos(ax)}{x}\, dx\\ \\ =\int_\mathbb{R} \frac{a\sin(2ax)}{x}\, dx\\...
  11. G

    HSC 2015 MX2 Integration Marathon (archive)

    Re: MX2 2015 Integration Marathon Lol it's positive everywhere, its integral over the real line is definitely NOT 0.
  12. G

    HSC 2015 MX2 Marathon ADVANCED (archive)

    Re: HSC 2015 4U Marathon - Advanced Level If f:[0,1]\rightarrow \mathbb{R} is a non-negative function such that f(1)=1 and f(x+y)\geq f(x)+f(y)\quad \textrm{for }0\leq x,y\textrm{ and }x+y\leq 1 we say that f is a sexy function. Find the smallest real \alpha such that...
  13. G

    Harder HSC Papers for MX2

    Re: Harder HSC Papers And "by convention" means it is a matter of convention what domain to define P on (whether or not to include (0,0)), given that main interest is for positive integers clearly.
  14. G

    Harder HSC Papers for MX2

    Re: Harder HSC Papers My point is that the give you a physical problem from which you can deduce both a recurrence relation and initial values. In passing to the problem of solving a recurrence, we need to obtain two pieces of initial data, in this case the values of P(1,0) and P(0,1). As you...
  15. G

    Harder HSC Papers for MX2

    Re: Harder HSC Papers I would say so yes, but this is of course a matter of convention. And no, I think that that recurrence formula requires two specified values to give you a unique solution. A general formula does not follow just from knowledge of P(0,0).
  16. G

    Harder HSC Papers for MX2

    Re: Harder HSC Papers Yep.
  17. G

    Harder HSC Papers for MX2

    Re: Harder HSC Papers I don't think a single value like that is enough to uniquely specify a function using the recurrence.
  18. G

    Harder HSC Papers for MX2

    Re: Harder HSC Papers Sorry, yep it should be "exactly".
  19. G

    Harder HSC Papers for MX2

    Re: Harder HSC Papers Another way of thinking about it: Each possible string "HTTHHTTHT..." can be given two running counts of the heads and tails in the first n letters. Eg for the above example these counts are: 0,1,1,1,2,3,... 0,0,1,2,2,2,... P(m,n) is the probability that the first...
  20. G

    Harder HSC Papers for MX2

    Re: Harder HSC Papers P(1,0) is the probability of throwing one head before throwing zero tails. But you have thrown zero tails to start with, without throwing any heads. So this event can never occur.
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