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If you enjoy maths (and are topping it) you are almost certainly not COMPLETELY roting it. Roting in its rawest form is just imitation of things you have seen before (changing numbers here and there). someone who completely rotes definitely can't answer the hardest of mx2 questions and hence...

Okay, so your initial expression for \sum_{k=1}^n \cos(k\theta) is incorrect. To check this chuck in \theta=\pi, n=1. Re-do this part and simplify and you will end up with my expression for S_n(\phi).

Will respond to this post in a second but first a response to your edit: your manipulation with I_n and I_{n+1} is only valid if I_n tends to some finite limit. I will go through your other working carefully soon.

You want to be taking definite integrals for these sorts of questions. The output of an indefinite integral is an equivalence class of functions which differ by a constant, the output of a definite integral is a number. We want to show that our integral is small, this notion is well defined for...

Less than or equal to. As the second term's absolute value is less than some positive thing that tends to zero, the second term must itself tend to zero as n-> inf.

$Okay, so let $S_n(\phi):=\sum_{k=1}^n \cos(k\phi)$ for $\phi\in(0,2\pi)$. We have by working similar to Spiral's that:\\$S_n(\phi)=-\frac{1}{2}+\frac{\sin(n+1/2)\theta}{2\sin(\theta/2)}.$\\ Now integrate both sides of this expression from $\pi$ to $\theta\in(0,2\pi)$. We obtain:\\$\sum_{k=1}^n...

My method was the same as spiralflex's with k1=pi k2=theta in (0,2pi) (my original identity was in terms of phi to avoid confusion). Regarding the integrand: The -1/2 pops out to give you what you want, you can show the remaining term tends to zero as n->inf by using IBP once. If no-one posts...

Re: HSC 2013 4U Marathon Lies! p1p2...pn doesn't have to be prime! But it does have to contain a prime not previously listed as a factor.

I studied mathematics in uni as it is what I always loved and wanted to do with my life. Upon graduating there were pretty abundant job opportunities in finance, trading, applied mathematics etc but I chose to continue on with what I enjoyed above all...I find financial mathematics rather dull...

Re: HSC 2013 4U Marathon My solution to 1. $Let $u=\sin^2\theta,v=\cos^2\theta.$\\Then $\lim_{\theta\rightarrow 0} \frac{u}{v}\cdot\frac{1-u^n}{1-v^n}=\lim_{\theta\rightarrow 0}\frac{u}{v}\cdot\frac{1-u}{1-v}\cdot\frac{1+u+\ldots+u^{n-1}}{1+v+\ldots+v^{n-1}}\\=\lim_{\theta\rightarrow...

Re: HSC 2013 4U Marathon Yep, so then the first question is fine and pretty straightforward using mx2 methods: 1/n. The latter question turns out to be true taking limits in either order, ie: \lim_{n\rightarrow \infty}\lim_{\theta\rightarrow 0}...

Re: HSC 2013 4U Marathon Well firstly you asked for its value at theta=0 rather than its limit as theta->0. But assuming you meant the latter then your second question might not work out the way you want, as the two limiting processes (theta->0,n->inf) don't have to commute. I will think about...

Re: HSC 2013 4U Marathon Isn't the first thing undefined at theta=0?

People seem to like game theory on here. Here is a nice and simple game that shows some of the mathematics present in poker (beyond mere probability calculations). Suppose there are three cards in a deck, A > K > Q. Players 1 and 2 are each dealt a card out of this three card deck. Each player...

Re: HSC 2013 4U Marathon Well yeah, but its just a roundabout way of getting to the same A=-2cos(theta).

Re: HSC 2013 4U Marathon I think you made the mistake when solving the quadratic.

Re: HSC 2013 4U Marathon oh okay, that makes the question trivial then, roots must be conjugates and A=2\cos(\theta) for some theta in (0,pi). The range of A follows.

Re: HSC 2013 4U Marathon How do we know the roots are of unit modulus?

Re: HSC 2013 4U Marathon Shortest way I can think of is: 1. Whatever we choose A to be the poly has two complex roots counting multiplicity. 2. One of them being non-real is the opposite of both of them being real. 3. Both of them are real <=> A=-(r+1/r) for some real r. 4. The range of...

Re: HSC 2013 4U Marathon z^n+z^{-n}=2\cos(n\theta) ? Assuming theta is meant to be the argument of z, this is not true.