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Re: HSC 2015 4U Marathon - Advanced Level No mate, don't worry too much about it...especially if you are at the start of your HSC year. This thread is mostly just quite high level students / uni students pushing each other by asking interesting questions outside syllabus. Doing well in these...

Re: MX2 2016 Integration Marathon Oh lol, yeah. Got a bit complacent there :p. The primitive of arctan(x) is x*arctan(x)-log(1+x^2)/2 + C (a straightforward IBP) so use that instead. Cheers!

Re: MX2 2016 Integration Marathon tan(A-B)=(tan(A)-tan(B))/(1+tan(A)tan(B)) So if A-B is in (-pi/2,pi/2), we have: A-B=arctan((tan(A)-tan(B))/(1+tan(A)tan(B))) Let A=arctan(a) and B=arctan(b) and this simplifies to arctan(a)-arctan(b)=arctan((a-b)/(1+ab))=arccot((1+ab)/(a-b))...

Haha yes, spot on with it being a crucial ingredient in Nash's work. This is one of the applications in abstract maths I referred to in my last line which I added after you took your quote. And nope, not an intentional reference. Russia was actually an arbitrary choice lol.

I have always found fixed point theorems quite pretty. Probably the most well-known one is the Brouwer fixed point theorem. One version of this states that if you have a continuous function f from a closed n-ball (eg the set of points of distance =< 1 from the origin in n-dimensional Euclidean...

Ah okay cool. Yeah, of course using words like "most" is vague and imprecise. Talking about measure theory will go over most high schoolers heads though, so waving hands and being a bit colloquial in this thread seems more appropriate.

It's still unrelated to countability though. "Most" meant almost all with respect to a certain measure. Both the set and it's complement are still uncountable.

I more meant that countability is kind of an irrelevant notion to the thing I mentioned so it seemed weird to bring it up as a reply to my post. (And yeah, meant first para. Second para is good for this thread :). Countability is such a nice and low-tech example of higher maths to show HS...

I know lol, why are you reminding me? In any case, none of the objects I mentioned is countably infinite.

Being continuous is a property that says that a function is in some sense "nice" near a point. Being differentiable is also a "niceness" property at a point. In fact it is a much stronger property, in the sense that differentiability at a point implies continuity at a point. It is clearly...

Well if I wanted to describe/study that union of two curves I would literally just use x^y+y^x-1.3844... . The branches should be symmetric, so finding a way of representing a single one of them but not the other probably wouldn't be particularly useful. As for what this looks like globally, I...

Nothing all that special is going on here, you will get this kind of behaviour with a lot of functions. I will just give a brief partial answer illustrating where some of the properties come from. Noting that f(x)=x^x has a minimum at x=1/e, we have x^x+y^y >= 2f(1/e) = 1.3844... So for k_1...

Yeah, sorry about that. Misinterpreted the original post and what kind of bias was entailed. (overall students vs acceptance rates)

I love intermediate value theorem consequences like this. You can get quite a lot out of such a seemingly common sense result. Another one is the one that colloquially states that you can always rotate a square four legged table on wobbly ground into a position in which it balances.

It can't be that EACH of the individual faculties has a bias towards women, just that most of them do. At least one faculty must have male bias if the overall bias is male.

For 3, it is useful to consider the function d(f(x),f(y)) defined on the product space K x K.

For 2, as Paradoxica mentioned, we can find n linearly independent exponential solutions. (Let's assume the polynomial only has simple roots for now, although the case of roots with multiplicity > 1 is not much harder). To show then that these are the only solutions, it suffices to show that an...

I will get people started on problem 1: $Let $G(x)=\int_{0}^{b(x)} f(x,t)\, dt.$\\ \\ Clearly, knowledge of how to differentiate such a $G$ will tell us how to solve the original problem. We then have\\ \\ $\frac{G(x+h)-G(x)}{h}=\int_{0}^{b(x+h)} f(x+h,t)\, dt-\int_{0}^{b(x)} f(x,t)\, dt\\ \\...

Past papers are mostly for when you understand the material and want to optimise your exam technique. (Especially past HSC papers.) First step is to make sure you understand the theory of the course, why things are true, how the proofs of things in the course work etc. (Try to prove some of...

Radius of convergence will just be 1.