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Don't know. It's just a very vague and non-rigorous thing to talk about "long-term cycles" as opposed to the limit of a concrete sequence of probabilities. If you tried to write up a rigorous solution based on this heuristic it would probably be a lot more clear as to what's going wrong.

Nah this isn't valid. Interesting that it is so close numerically though.

Yep. Your notation is a little different to mine but the answer is correct unless I have messed up. 4a) Either we have {biased coin & 3 consec heads} or {unbiased coin & 3 consec heads}. These have probability (1/2)*(1) and (1/2)*(1/8) respectively. So the odds that the first happens given...

Ah okay, that is exactly what I suggested to Sy. It comes out v. fast.

What do you mean?

He stays in town B on the n-th night iff he stays in town A on the (n-1)-th night, so p_n(A)=p_{n-1}(B). Letting n-> infty gives a = b. Do something like this for a,b,c,d,e, you will get a linear equation for each one. Just solve these. Also, I think your answers to 4 are incorrect, I got 8/9...

Judging by the wording of the question, I think you are allowed to assume that there IS a long-term probability associated to each town. ie lim P(person stays in town X on n-th night) exists for X=A,B,C,D,E. (Even if you just assume it for X=E, this gives you convergence for the other towns.)...

me too :).

Slight correction, the inequality in (i) only necessarily holds for x >= -1.

Re: HSC 2013 4U Marathon A bit flawed, the radius of convergence of a Taylor series can be 0. Eg consider the function defined piecewise as e^{-1/x} for x > 0 and 0 for x < 0. sine just happens to have an infinite radius of convergence.

How do you think you would go? I am pretty confident I could do every question (and would be quicker at the harder stuff than yr12 sean) but would probably be slower at the routine stuff than I used to be. Without practicing, 3 hours would probably be quite tight.

I actually probably can't make it up to Syd (as it is a weekday), but if Carrot emails me the paper I could do it in the evening and scan my solutions. Of course, you guys would have to trust me that I limited myself to three hours.

lrn2maths.

Re: HSC 2013 3U Marathon Thread How many terms are you adding to the LHS every time you increase n by one? What are these terms all bigger than?

I would probably use the words: "raise 10 to the power of both sides" rather than "raise both sides by 10", but these are of course just words. The idea is much more important. Would rep for the latter paragraph on the idea of "inversion" but have not spread it around enough.

The domain of interest is obvious from the forum the question was posted in. If I asked you to solve some exponential equation without stating which field/ring to quantify over, and then called you a liar for not including quaternionic solutions, that would be equally ridiculous. Of course a...

Re: HSC 2013 4U Marathon Well we only need EVENTUAL monotonicity to use the theorem you are thinking of, but that theorem isn't really MX2 assumed knowledge. But yeah, syllabus aside...if your working is all correct then the method is valid. Well done for bashing it out :).

Re: HSC 2013 4U Marathon Hmm yeah, cannot think of any immediate way to show that the limit is finite. My method was to assume n was large, (which is all we care about anyway) and use that binomial coefficients get bigger towards the middle. So: 0\leq \sum_{k=0}^n {n\choose k}^{-1} -...

Re: HSC 2013 4U Marathon But there are an increasing number of terms, so that isn't quite rigorous. Eg. \lim_{n\rightarrow\infty}\sum_{j=0}^n \left(\frac{1}{\sqrt{n+j}}\right)=\infty You can check this counterexample yourself by mx2 level methods, but heuristically it is obvious...we are...

Forgot about this question. Got around to looking at it and it turned out to be rather straightforward: If we let the prize be 1 for simplicity, the expectation for a given (p,q,r) is E=1/3 + r(2q/3 - p/3) by straightforward calculation. If 2q > p then the optimal r is 1, which improves...