Recent content by Fus Ro Dah

Looking at solutions, I think I dropped 1 mark overall, in Question 13 (a) (ii). I used the formula for half of a circle instead of a quarter circle =(

Was nice meeting you too!

BTW I know what the last question is in the 4U exam. It's a really awesome result and it is just so unexpected! jk, but it's probably going to be a neat result anyway, knowing Carrot

Hardly think there's much to be a fan of really... but alright ^^

But I didn't score gold! Wouldn't you want the autograph of somebody who at least got gold?

Well this is awkward.

Great guy. lol. Good job Team Aus. Congratulations to everybody.

Re: HSC 2013 4U Marathon It is a good method, but why do you say this as if it is alien to you? You can imagine that the proof for this would be very similar to the proof of the existence of complex conjugate pairs in real polynomials, since the complex conjugate is analogous to the irrational...

For what reason(s) is it the most sensible choice?

Re: MX2 Integration Marathon \int \sqrt{\tan x}dx

Re: HSC 2013 3U Marathon Thread There is a shorter way. x^2+y^2+z^2 = (x+y+z)^2-2(xy+yz+xz) =1-2(xy+yz+xz) \geq 1-2(x^2+y^2+z^2) \Rightarrow x^2+y^2+z^2 \geq \frac{1}{3}

Re: HSC 2013 4U Marathon \\ k(k+1)(k+2)...(k+n) = \frac{(k+n)!}{(k-1)!}, where k>1 and we want to prove that \frac{(k+n)!}{(k-1)!n!} is an integer. Consider \binom{k+n}{k-1}, which we know is an integer K. \binom{k+n}{k-1} = \frac{(k+n)!}{(k-1)!(n+1)!}=K so \frac{(k+n)!}{(k-1)!n!}=K(n+1)...

Re: HSC 2013 4U Marathon Normally would have taken the usual z' \mapsto \frac{1+z}{1-z}$ approach to observe that z = \frac{z'-1}{z'+1}, and then substitute into P(z). We acquire a polynomial that is easy to prove has all complex roots but proving that they are purely imaginary is so long...

Finish anything to do with Belonging. Frankly, I am tired of that word now.

It depends on the individual and their abilities and accuracy.