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To justify the "equating like parts", use the following result. If a,b,c,d are rationals, c,d >= 0 and c is not the square of a rational, then a+sqrt(c)=b+sqrt(d) implies a=b and c=d. Proof: Note that if we either have a=b or c=d that the other equality falls out immediately, so let us...

I dual boot a very fast and lightweight Arch Linux installation (Openbox WM) with windows 8.1. I mostly use Arch for study and day-to-day usage whilst windows is there for games and supporting some closed source drivers better.

It just means z is an integer that is not divisible by any squares. (And this condition uniquely determines x,y,z.)

Re: HSC 2016 4U Marathon Created a new marathon thread for undergrad stuff: http://community.boredofstudies.org/1003/maths/346314/undergraduate-mathematics-marathon.html Hopefully some of the uni students here will make use of it!

There was a marathon thread in this subforum that died a long time ago, so I am going to try to start a new one for people to post, solve, and discuss undergraduate mathematics problems. (Of course things like the stackexchange already exist that are fantastic for this purpose, but it would be...

Re: HSC 2016 4U Marathon Finding an expression for this quantity is precisely a derivation of one form of the Leibniz rule. (And so assuming a weaker version of the Leibniz rule as a starting point is cheating a little.) What I was looking for was such a derivation based on first principles...

Re: HSC 2016 4U Marathon Having thought about this a little more, indeed the question requires a little more than can be expected of HS students. If any want to still give it a crack, the following tools from outside curriculum are the basic things you might need: -The epsilon-delta...

Re: HSC 2016 4U Marathon Here is a good exercise to test understanding of the connection between integration and differentiation: $Let $F(x):=\int_{a(x)}^{b(x)} f(x,t)\, dt$\\ \\where $a$, $b$, and $f$ are smooth (infinitely differentiable) functions.\\ \\ Prove that $F$ is differentiable, and...

Not sure what is leading him to believe that...

Its not as fancy as it might look. If you learn a little about functional equations then the tricks I used were very standard. :)

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No it does not, but we can easily find all smooth solutions to the functional equation by putting in slightly more work. Note that if we find all solutions f defined on the domain of positive reals, then it is trivial to find all solutions on the full domain of nonzero reals. (Let f,g be...

There are LOTS of solutions. Let g(x) be an arbitrary nonvanishing continuous function on the set [-1,1]\{0}, such that g(1)^2=g(-1)^2=1. Then g extends uniquely to a continuous solution of the functional equation on the nonzero reals. (Define it outside of [-1,1]\{0} using g(x):=1/g(1/x).)

Indeed, and I had this in mind when making my original post. Although the tricks you use and the problems you tackle in olympiad are a lot more varied so I find them more fun. It's still pretty different to what real maths is like though. T. Tao had a good article about this on his blog.

Sure, there definitely are a class of functions that have elementary symbolic primitives that we can find that certain software cannot. When it comes to something like engineering though, this skill is completely moot, as we need a numerical output. And doing symbolic gymnastics to reach...

Re: HSC 2016 4U Marathon - Advanced Level From the definition of g, we have g(x)=b_0\prod_{i=1}^n (x-\alpha_i). So upon multiplying both sides of the equation \frac{f(x)}{g(x)}=\sum_{j=1}^n A_j(x-\alpha_j)^{-1} by \prod_{i=1}^n (x-\alpha_i), we are left with \frac{f(x)}{b_0}=\sum_{j=1}^n...

Re: HSC 2016 4U Marathon - Advanced Level For the polynomial question without assuming C-S: P(x)P(x^{-1})=\sum_{k=0}^n \sum_{j=0}^n a_ka_jx^{k-j}=\sum_{k=0}^n a_k^2 +\sum_{k<j}a_ka_j(x^{j-k}+x^{k-j})\\ \\ \geq \sum_{k=0}^n a_k^2 + 2\sum_{k<j} a_ka_j = \left(\sum_{k=0}^n a_k\right)^2=P(1)^2...

Re: HSC 2016 4U Marathon - Advanced Level ai) We are solving the quadratic p(z)=z^2-2wz+1=0, where w is in the open upper half unit circle. The complex numbers u+i and v+i then must be roots of q(z)=p(z-i)=z^2-2(i+w)z+2wi. It suffices then to prove that the roots of q lie on the same open...

Re: HSC 2016 4U Marathon - Advanced Level This seems close to the right idea, but not quite I think. To start with, P(x_1,\ldots,x_n)-P(x_1,\ldots,x_{n-1,0}) is NOT necessarily divisible by x_1\ldots x_n (the error in your argument being that we cannot write this difference using the set \chi...

Re: HSC 2016 4U Marathon - Advanced Level Not too sure what you are asking here? I assume by AM-GM polynomials, you mean: P_n(x_1,\ldots,x_n)=\frac{1}{n}\sum_{k=1}^n x_k^n-\prod_{k=1}^n x_k. (The AM-GM inequality then being the statement that P_n is non-negative if each x_k is...