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Re: HSC 2015 4U Marathon - Advanced Level You are a lot of the way there, these are most of the ideas I was looking for. But I don't think you have completed the f(1)=2 case. I don't think f(x+1)-f(x)=1 really implies that we are a linear polynomial. -What do you mean by "gradient" when...

Re: HSC 2015 4U Marathon - Advanced Level Find all functions f:\mathbb{Q}\rightarrow \mathbb{Q} such that f(xy)=f(x)f(y)-f(x+y)+1 for all rational x,y. (Note that \mathbb{Q} is the set of rational numbers.)

Re: HSC 2015 4U Marathon Key is the triangle inequality. For |z| <= 1, we have |p(z)| >= 3- |z+z^4| >= 3-|z|-|z|^4 >= 1 > 0. and |z^4q(1/z)| = |z^4+z^3+4| >= 4-|z|^3 - |z|^4 >=2 > 0. So p has no roots inside the closed unit disk and q has no roots outside the open unit disk (and so...

Re: HSC 2015 4U Marathon - Advanced Level Finishing off the cubic question (finally!): It remains to show that for a polynomial with three real roots, the inequalities: 1. |bd-c| < 1-d^2 and 2. |b+d| < |1+c| imply that all three roots live in the open unit interval. As discussed in my...

Re: HSC 2015 4U Marathon - Advanced Level (A almost identical argument establishes B & S(3) => A & S(3). p and q must have all roots in the unit interval so we get p(1) > 0, p(-1) < 0, q(1) > 0. Which gives us |b+d| < |1+c| = 1+c and 1-d^2 > c-bd. As with my previous post, the 1-d^2 > bd-c...

Re: HSC 2015 4U Marathon - Advanced Level For the cubic question: Let A be the statement that the pair of inequalities in the question hold. Let B be the statement that the polynomial has all roots in the open unit disk. Let S(n) be the statement that the polynomial has n real roots. It...

Re: HSC 2015 4U Marathon - Advanced Level Nope, not quite good enough for selection camp.

Re: HSC 2015 4U Marathon - Advanced Level I already proved this on the previous page, in fact there are at least three such intersection points. Yang's proof above is a much more elegant way of proving existence though, which is all we require for induction.

Re: HSC 2015 4U Marathon - Advanced Level What was the purpose of the post then?

Re: HSC 2015 4U Marathon - Advanced Level Sure. I think the only solutions are the trivial constant ones: 0, 1. First we note that there are no real roots, as if x is a real root, then (x+ 1)^2+1=x+(x+1/2)^2+7/4 is a larger real root. (And a non constant polynomial only has finitely many...

Re: HSC 2015 4U Marathon - Advanced Level Okay sure, sorry about the delay. For the record, I wouldn't expect many current MX2 students if any to come up with my proof, but I think there is probably some nicer way of doing it (its one of the earlier exercises in an olympiad training...

Re: HSC 2015 4U Marathon - Advanced Level Okay, I think I figured out a proof. I have to go out now but I can write it up later if people want. The idea: Use a stereographic projection (put a unit sphere on the origin and for any given point on the plane, draw a line in 3d space between this...

Re: HSC 2015 4U Marathon - Advanced Level I know that this is intuitively true, but what you have written isn't a proof. You would need to prove that "you would be over the n quota" before we add the line of zero contribution. I am not sure this is any easier than the initial problem really.

Re: HSC 2015 4U Marathon - Advanced Level Why is it not possible for a new line to meet all other lines over several previously existing int. points rather than "at once"? This possibility would also be the addition of a line without adding int. points.

Re: HSC 2015 4U Marathon - Advanced Level Could you explain in more detail why this is true please? Perhaps I am missing something basic, but I don't see how this quoted claim follows. It is clear that the (r+1)-th line must add an intersection: [Indeed for it to not do so, it would have to...

Re: MX2 2015 Integration Marathon Yep, yep, +. Cheers :).

Re: MX2 2015 Integration Marathon What do you think my third line is?

Re: MX2 2015 Integration Marathon Ps, you should use a "\" before commonly used trig functions in latex to make things look nicer.

Re: MX2 2015 Integration Marathon Too lazy to do the calculation at the end, but the reduction formula for n >= 2: \int \sec^n(x)\, dx \\ = \int \sec^{n-2}(x)\frac{d}{dx}(\tan(x))\, dx \\ = \sec^{n-2}(x)\tan(x)-(n-2)\int \tan^2(x)\sec^{n-2}(x)\, dx \\ \\ \Rightarrow...

Re: HSC 2015 4U Marathon - Advanced Level Yep, I had the same idea.