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I decided to focus on instead: $Prove$ \ \ \ \ \lim_{n \to \infty} \frac{\sin 2n \theta}{\sin \theta} = 1 Wolfram Alpha outputs: What does this even mean? csc(x) -1 to 1.....

badump bump

Re: HSC 2013 4U Marathon Here is my solution to the question. ============== Ok, so we need to model the moving marker, and I know I can simplify the problem by finding the parametric equations of the red marker. So I need to find the parametric equations of the marker, but since this red...

I might also add that doing this problem is equivalent to proving: \lim_{n \to \infty} \int_{0}^{\alpha} \frac{\sin (2n\theta)}{\sin \theta} \ d\theta = \alpha If I can prove this, then I end up with the value \frac{\pi^2}{8} Which is indeed the correct value of \sum_{k=1}^{\infty}...

There is a typo in that question, I need sin(2ntheta)/2sin(theta) Oh and I need one of the limits to be variable in order to then integrate once more :/ (in order to attempt to solve the Basel problem)

Re: HSC 2013 4U Marathon $Prove that if$ \ \ \cos n\theta \ \ \ $would be expressed in terms of$ \ \ \cos \theta \ \ $ then $ \ \ \cos n\theta \ \ $would be a polynomial of degree n in$ \ \ \cos \theta

Q8a is just a proof of the result that I initially used, the background on this problem is: I can prove that odd cosine thing result easily with geometric series (no need for induction as the HSC specifies) I simply integrate twice after that: \sin \theta + \frac{\sin 3\theta}{3} + \frac{\sin...

Is there anyway to find: \lim_{n \to \infty} \int_{0}^{\alpha} \frac{\sin n\theta}{2\sin \theta} \ d\theta In terms of alpha preferably, and using HSC methods if possible (the most advanced I will accept is taylor series but even then I can't justify integration of infinite terms) Thanks

Re: HSC 2013 4U Marathon Yeah, my answer was wrong since I messed up the initial conditions, it should be correct now.

Re: HSC 2013 4U Marathon This is true, I gave it to sort of make the elimination of variables easier. Because its easier to imagine it if you are given a time But reason why this is true is: l=r \theta \rightarrow l=\theta

Re: HSC 2013 4U Marathon HINT: Parametric equations

Re: HSC 2013 4U Marathon Not quite heh, I made it a show that to compare solutions

Re: HSC 2013 4U Marathon $A circle of unit radius is rolled along the x-axis in the positive direction initially at the origin. There is a red marker on circle that is fixed on the circle as it rolls along the x-axis. The marker is originally at the origin, and as the ball moves, the marker...

Have you made any goals in general for this HSC year? i.e. Study for X hours per day, Improve in Y subject Have you stuck to/fulfilled such goals?

Re: HSC 2013 4U Marathon Apologies, it should be the other way around. Fixed now.

Re: HSC 2013 4U Marathon This may or may not bypass the rigour issues that happened last time I posted this idea as a question: $i) Show that$ \binom{n}{j} \frac{1}{n^j} = \frac{1}{j!} \left ( 1 \cdot (1-\frac{1}{n}) \cdot (1-\frac{2}{n}) \cdot \dots \cdot (1-\frac{j-1}{n}))\right ) $ii)...

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Re: HSC 2013 4U Marathon Correct - good job. Alternatively consider: \frac{1}{2} \int_{0}^{\infty} \frac{1}{1+x^2} \ dx < \sum_{k=1}^{\infty} \frac{1}{1+k^2} < \int_{0}^{\infty}\frac{1}{1+x^2} \ dx Where we can justify the lower bound because the height of the biggest rectangle when we...

Re: HSC 2013 4U Marathon Lower bound is pi/4 from the question, upper bound is pi/2. The lower limit of the integral is 0 however, and that integral that I just posted = pi/2

Re: HSC 2013 4U Marathon Not sure what you mean by the first sentence. You are correct in that it involves the integral of 1/(1+x^2) HINT: Notice how the upper bound is \lim_{n \to \infty}\int_{0}^{n} \frac{1}{1+x^2} \ dx