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Well, sure, the formula for a geometric series can be applied...but I think its just to confirm it using another method.

Very nice paper.

Will attempt when I'm back in Sydney lol

Great questions.probing some deep theory. Will have an in depth look and a try later.

So excited! Carret can you pm me the papers ;) fankyew.

No at trial means after trial.

Played so shit at uni games. Cannot believe it. I don't need words - I just need some time.

Illin at the gold coast. So sad I can't come.

Just try to speak English all the time. You will improve much more than if you speak Chinglish/Cantonese and English. Good luck.

ermagrd i wanna come ~_~

It sometimes also depends on how you write the function. For example, y= \dfrac{1}{\frac{1}{x}} = x but in the first case since I wrote it like that then the natural domain would be x \text{ in the reals, with }x\ne 0. Or perhaps y = e^{\ln x} has a natural domain of x>0 even though you can...

note that conics that are not centred at the origin are not in the syllabus.

That's not dumb. THAT'S CORRECT!

What is the relationship between the gradient of the tangent and the gradient of the normal?

It doesn't matter in this question - if you differentiate wrt y you get the same answer.

Yes.

Yes. When you enrol just specify the 'stream' to be B Sc (Adv Math).

IDK about his maths tutoring abilities, but this guy is shredded as fuck. Top eyecandy 2k14. Get in kwik peeps. #ddtuition4lyfe

\ln\left(\dfrac{4}{3}\right) is more simplified - this is only one term as opposed to 2.

Re: HSC 2013 3U Marathon Thread Just because I am in a LaTexy mood: \frac{d}{dx}\sin x = \lim_{h\to 0} \dfrac{\sin(x+h)-\sin x}{h} \\ = \lim_{h\to 0} \dfrac{\sin x \cos h + \sin h \cos x -\sin x}{h} \\ =\lim_{h\to 0} \dfrac{\sin x(\cos h-1)+\sin h\cos x}{h}\\=\sin x \lim_{h\to 0}\frac{\cos...