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Re: HSC 2014 4U Marathon - Advanced Level $Let $P,Q,R$ be real polynomials and let $S(x)=P(x^3)+xQ(x^3)+x^2R(x^3)$.\\ \\If $S$ is of degree $n$ and has all roots distinct, find a polynomial of degree $n$ whose roots are the cubes of the roots of $S$. (In terms of $P,Q,R$).$

Re: HSC 2014 4U Marathon - Advanced Level The squeeze theorem isn't really necessary if we are working from MX2 knowledge, where e^x refers to a (not entirely rigorously defined) function which we know how to differentiate, and whose inverse we know how to differentiate. Here is my proof: Fix...

Yeah dude, probably mine too. Definitely my favourite two Australian bands in the metal/hardcore genre. Was stoked that too legit to quit did so well in the triple j hottest 100.

The entirety of Horizons is fantastic. Still my favourite PWD album. Anyway, I will list good songs by five different artists for variety. I think a question that might be easier is list your five favourite metal ALBUMS. It's so hard to just pick five songs lol.

Re: HSC 2014 4U Marathon - Advanced Level Yeah it tends to zero. Depends on whether m or n is larger. I would just write 0 though. The +/- notation is more commonly used when talking about one-sided limits.

Re: HSC 2014 4U Marathon - Advanced Level There is a bit of a logical flaw here, remember that we are trying to show that NO polynomial of the prescribed form has all roots positive and real. You have just provided one example of a polynomial of the prescribed form that does not have all...

Re: HSC 2014 4U Marathon I don't see how you get the product of beta_k as 1, since we don't know that |a_0|=1?

It's quite sloppily written question (unless you are paraphrasing). Are the tables circular? Did each of the 10 people come with a partner? Does "together" mean adjacent or "at the same table"? Are we looking for the probability of at least one couple sitting together, or of all couples...

Interesting, I will have a look later as that setting would probably be ideal for me too. I am guessing it will be an option in the settings page (check out the link that is hopefully in the top right hand corner of any page on BoS, right next to log out).

Yeah, they are glossing over that fact. In such situations it is usually clear when a factor will appear in both things you are counting (so you can ignore it), but for a worked solution that is a little sloppy. Their calculations are correct if you treat the three groups are ordered, eg you are...

Re: MX2 Integration Marathon bump.

Re: HSC 2014 4U Marathon bump.

Re: HSC 2014 4U Marathon - Advanced Level The inequality is not necessarily true if we don't also assume something like a,b,c >0. Eg a=-1, b=c=1.

Re: HSC 2014 4U Marathon - Advanced Level (You should state that a,b,c >0.)

Re: HSC 2014 4U Marathon - Advanced Level i) a = b = 2N is an example, for each N. ii) Are there meant to be additional assumptions or something? 1/3 = 1/6 + 1/6 = 1/4 + 1/12 is a counter-example to the claim.

Re: HSC 2014 4U Marathon - Advanced Level Well, evaluating Gamma(3/2) is the same integral as the Gaussian. Just from 0 to infinity, and with a possible constant factor floating around somewhere. This can be done (not rigorously of course, but as rigorous as the mx2 course gets.) I don't...

Re: HSC 2014 4U Marathon - Advanced Level We are just constructing a plausible definition for something that has been previously undefined (the factorials of half-integers), it is not a matter of using rigorous substitution to deduce the value of some previously defined and unknown quantity...

Re: HSC 2014 4U Marathon - Advanced Level $For $m,n\geq 1$, let:\\ \\ $B(m,n)=\int_0^1 x^m(1-x)^n \, dx=\int_0^1 \frac{d}{dx}\left(\frac{x^{m+1}}{m+1}\right)(1-x)^n\, dx\\ \\ = \frac{n}{m+1}\int_0^1 x^{m+1}(1-x)^{n-1}\, dx = \frac{m}{n+1}B(m+1,n-1).\\ \\ $So, for positive integers m,n: $\\ \\...

Re: HSC 2014 4U Marathon - Advanced Level No, it's http://en.wikipedia.org/wiki/Cross_product. It takes about as long to prove that it gives you area as it takes to prove Heron's.

Re: HSC 2014 4U Marathon - Advanced Level In an MX2 level proof I would prove the cross-product formula for a triangles area (using perp distance formula etc), but here is the idea: V=\frac{abc}{6}=\frac{d|\triangle ABC|}{3}\\ \\ \Rightarrow a^2b^2c^2 = d^2|(B-A)\times...