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Yeah you are right and the book is wrong, unless you typoed the question.

Err, that's not true. Any real z that isn't 1 will give z/(z-1) real.

Re: HSC 2013 4U Marathon 0. The only constant solutions are p=0,1. Let us now assume that p is nonconstant. By comparing leading coefficients we must have p monic. 1. For any such p, the identity p(x)p(x+1)=p(x^2) must hold for all complex numbers, as nonzero polynomials have only finitely...

Re: HSC 2013 4U Marathon Your answer to iii) is correct, but I was looking for some justification as to why "v1->v2".

Re: HSC 2013 4U Marathon Volumes/Integration: $Consider the surface $S$ above the $xy$-plane with height given by $h(x,y)=e^{-(x^2+y^2)}$.\\ \\ i) Find the volume $V_1$ of the region bounded by the $xy$-plane, the surface $S$, and the infinitely long rectangular prism $\max\{|x|,|y|\}=R$ for...

Re: HSC 2013 4U Marathon Harder 3U (Combinatorics): In the board game Risk, if there are at least three attacking units and at least two defending units, then the attacker rolls three dice and the defender rolls two dice. The top two dice of the attacker are paired up with the two dice of the...

Re: HSC 2013 4U Marathon This is easy enough: If X+1/X=r rational, then X is a nonzero irrational root of a rational quadratic. So we can write X=a+b*sqrt(c), with a,b,c rational, c not a perfect square (from the definition of X). By cubing this implies that we must have 3a^2b+cb^3=0. As c is...

Re: HSC 2013 4U Marathon Exactly, well done. Are you a student or a teacher?

Sure, we would all need to go into an internet chat room or something and from there I would be able to announce brackets and stuff. I doubt we would get enough people interested though...

Re: Chess thread? I know. Another?

Re: Chess thread? It was a forced mate in about 4 anyway, but good game...the second one was especially quite fun...am a bit rusty dropping material so easily.

Re: Chess thread? I'm up for a game after :).

Re: HSC 2013 4U Marathon So close, but why "must" we have Q=pR? Why must any Q be divisible by p? This needs proof but otherwise good.

Re: HSC 2013 4U Marathon Cool, so you are mostly done. What you have done is constructed a particular rational polynomial (your cubic) with alpha as a root. You have then found the two other roots of this poly (which look like the two roots I am looking for, although I haven't double-checked)...

Re: HSC 2013 4U Marathon With MX2 a week away, I will post one or two difficult questions each day, hopefully covering all topics (polynomials, integration, etc). As usual, this is more to stretch good students and hone their problemsolving skill for Q16s than to provide exercises at doing the...

Ceylon black tea and various green teas mostly.

Want some games later?

Bump. Anyone want to resurrect this thread and get some wifi going? It would be cool if we could get enough people to start up an online league or something.

Re: HSC 2013 4U Marathon \log(S_n)=\frac{1}{n}\sum_{k=1}^n\log\left(1+\frac{k}{n}\right).\\ \textrm{ which is a Riemann sum.} \\ \\ \Rightarrow \lim_{n\rightarrow\infty} \log(S_n) = \int_1^2 \log(t)\, dt = 2\log(2)-1.\\ \\ \textrm{Exponentiate both sides to end up with }4e^{-1}.

Yeah this. Or read a book on it after taking 2962.