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I'm pretty sure spherical coordinates are not required... cylindrical coordinates make the problem much nicer, but they're not required to solve the problem "conceptually". Anyway you can do it also like, by using spherical coordinates but in terms of like, cones etc. Not too hard, no...

Re: 2012 HSC MX2 Marathon No its not... \int_0^d \frac{1+\sin x}{\cos x} dx = \int_0^d (\tan x + \sec x) dx = \left[-\ln |\cos x| + \ln |\tan x + \sec x| \right]_0^d = \left[ \ln \left(\frac{\tan x + \sec x}{\cos x} \right)\right]_0^d = \ln \left|\frac{\tan d + \sec d}{\cos d}\right| Then...

Re: 2012 HSC MX2 Marathon uhhhhh... how can this be true? the right hand series (the harmonic series) diverges

Re: 2012 HSC MX2 Marathon Neat question ^^ First we make the substitution u = x^2. u = x^2 $ implies $ du = 2x dx and thus I = \int\frac{x^3 e^{x^2}}{(x^2+1)^2}dx = \frac{1}{2}\int\frac{u e^u}{(u+1)^2)}du And now we note that u = (u+1) - 1 so we can split up the integral: I =...

Re: 2012 HSC MX1 Marathon You do polynomial division. x = (e^y)/(3+e^y) = 1 - 3/(3+e^y) so 3/(3 + e^y) = 1 - x so 3/(1-x) = 3+e^y so y = ln[3/(1-x) - 3]

Re: 2012 HSC MX2 Marathon It's also true because of http://en.wikipedia.org/wiki/Circles_of_Apollonius

Re: 2012 HSC MX1 Marathon Or go to sleep :P

Re: 2012 HSC MX1 Marathon Timske's answer is correct

There's a lot of total bullcrap that goes around the place about how to get into overseas unis. First, academic results do matter from school (ATAR above 99 is great) but not an enormous amount, and SAT scores really don't matter if they're above a certain point. So as long as you don't...

The difficulty of various subjects depends a lot not just on the subject but on what sort of student you are and in particular how hard you work. Now I did both Latin and Classical Greek (extensions in both as well) for my HSC so I can say something about these subjects (and hopefully languages...

Ive got another way which I think is much faster, using real/imaginary parts and modulus/argument and de moivres: Im\left(z + \frac{1}{z}\right) = Im\left(rcis(\theta) + \frac{1}{r}cis(-\theta)\right) = rsin(\theta) + \frac{1}{r}sin(-\theta) = sin(\theta)\left(r - \frac{1}{r}\right) And thus...

This is dubious and quite hard to prove. If anyone is interested, you need to basically know about minimal polynomials and algebraic numbers (its pretty cool). Basically, each complex (and real!) number is either "algebraic" or "transcendental". What that means is this: an algebraic number is...

Re: 2012 HSC MX1 Marathon You can do this sort of thing and make it be perfectly valid - its called "reverse reconstruction". You need to be a bit careful though about the uniqueness of points and such (nooblet94's proof isnt actually right because of things lying on different sides of lines).

Re: 2012 HSC MX1 Marathon wahhhh... A, B and D are always going to be concyclic so long as <ADB is not zero (i.e. A,D,B collinear) since any three noncollinear points are concyclic. If you meant A, B, D are on a circle centre C, thats not true; although A, B, C and the centre of the circle...

Re: 2012 HSC MX1 Marathon You dont even need to use the centre. here's a pretty simple proof: Note that the combined length of arc PAQ is the same as arc PBQ (since arcs AP = PB, BQ = QA). Hence the angles they subtend are equal, so /_PAQ = /_PBQ, but by cyclic quadrilateral interior angle sum...

Can be pretty pure too; c.f. http://en.wikipedia.org/wiki/Hall's_marriage_theorem

I really haven't done enough applied maths to know, but I really like discrete mathematics (graph theory!) and number theory (and what I've seen of its university extension in abstract algebra). I did a kinda no rigorous course in topology a while ago that was fun but occasionally kinda...

Anyway so results of enrolment day: MATH1901, MATH1902, MATH2961, MATH2962. And wrarrr Im going to through my hands into the pure camp. But as much as I understand it, some aspects of cosmology/theoretical physics are actually much closer to pure maths (topology etc) than they are to anything...

Re: 2012 HSC MX2 Marathon Consider R(x) = P(x) - Q(x): note that it is of degree at most n. Now suppose R(x) is not the zero polynomial; in this case it has at most n roots (from unique factorisation, which is provable from the division algorithm). However it clearly has n+1 roots x1, x2...

Oh I really like discrete maths, but thats sorta why Im not going to take it straight away :P And Im doing physics anyway as one of the compulsory non-maths science subjects. Yeah looking at my level of overload doing both is probably too much... so Ill probably do metric spaces and not galois...