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Re: 2012 HSC MX2 Marathon Which theorem are you referring to? The Monodromy theorem? It isn't too difficult to find the analytic continuation of the zeta function in particular...

Re: 2012 HSC MX2 Marathon No, the series I wrote down converges if and ONLY if Re(s)>1. It diverges everywhere else. However, this fragment of a function can be extended to the whole complex plane in a very natural way. (Excluding the point s=1). This process is called analytic continuation and...

Re: 2012 HSC MX2 Marathon Okay, I see what you are asking. This is a common misconception. The Riemann Zeta function is defined by the series: \sum_{n\in\mathbb{N}}n^{-s} ONLY for s with real part larger than one. It turns out there is only one "nice" function that is defined on the whole...

Re: 2012 HSC MX2 Marathon $The three complex numbers $\alpha,\beta,\gamma$ form an equilateral triangle iff:\\ $\alpha^2+\beta^2+\gamma^2=\alpha\beta+\alpha\gamma+\beta\gamma$ (*).\\Using this, we get $(-a)^2-2b=b$ from which the result follows. We now prove (*).\\ By a straightforward...

I think I came third in an early subject test and came back...

Re: 2012 HSC MX2 Marathon Well that is how e^x is defined in higher mathematics, but at the school level that knowledge is not assumed. You can deduce the inequality pretty easily using properties of the exponential from the 3 unit course.

Re: 2012 HSC MX2 Marathon $Prove by induction that for any non-negative integer $n$, we have:\\ $e^x\geq \sum_{k=0}^n \frac{x^k}{k!}\quad\textrm{for }x\geq0.$\\ Hence deduce that $\lim_{x\rightarrow\infty} P(x)e^{-x}=0$ for any polynomial $P(x)$. \\ \\ (Extension) Let $f$ be the function...

You are adding false solutions when you square both sides, the answer is just the non-negative half of the imaginary axis.

Re: 2012 HSC MX1 Marathon For this question to make sense you also need to stipulate that the collection of rectangles you are looking at has fixed area or fixed perimeter (either will do). As stated, there is no rectangle with shortest diagonal length as one can simply draw a smaller...

Re: 2012 HSC MX1 Marathon $Prove that any function $f:\mathbb{R}\rightarrow\mathbb{R}$ can be written uniquely as the sum of an odd function and an even function.$

I didn't mean real vs nonreal I meant real vs indeterminate. Eg the role x plays in polynomials.

The infinite expression makes sense if you are letting a,b,c be real numbers. But then this is saying nothing other than that a real number can be written as a product of arbitrary length. Nothing amazing... Anyway, I'm done here. I think enough has been said.

"Stating that a(b-c) is not fully factorised, nor is a(sqrtb-sqrtc). It is proving that there are an infinite amount of factors to the expression a(b-c). " Having an 'infinite amount of factors' is a vague property unless you specify what these factors must be. Eg we factorise polynomials into...

If "right" means writing down an infinite product that is equal to a/(b-c) for any a,b,c>0, then no you are not...sorry! :(

Well I haven't understood your explanations of what you are TRYING to achieve, or what you THINK you have achieved, but all I see is an infinite product over some (still ambiguous) set which is equal to zero.

You mean you missed the condition b,c>0. In any case, read my previous post. This infinite product will always converge to 0...

So you are taking the product of b^(1/n)-c^(1/n) over all even squares n in addition to n=1? For starters, that doesn't make sense when either of b,c is negative. Secondly for positive b,c this should converge to 0...since b^1/n and c^1/n both approach 1. Again, being a member of n^2 doesnt...

The factor contains n-th roots now but my other complaints remain...

still makes no sense, you are using n to denote both a variable and a set? And the factor inside the product you latexed is independent of n...

Be a little more explicit and we might be able to tell you...write it up?