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Re: HSC 2016 4U Marathon - Advanced Level Very close. Why must p(z)-p(t) have exactly q roots in the rational multiple of pi case? Also are you saying this is true for all t? Or some particular choice of t? I agree that if t is nonzero, then p(z)-p(t) must have at LEAST q roots.

Re: HSC 2016 4U Marathon - Advanced Level Related: Find all polynomials p(z) such that p(cis(t))=p(cis(t+a)) for all t, where a is a given positive number. You may need to consider cases.

Some linear algebra and matrices could definitely slot into the HSC courses in terms of level of difficulty. It is just a matter of what one would replace with them. And there are plenty of basic applications that would seem interesting to some students (and provide motivation): -The first...

It's roughly 2nd year of advanced maths courses when you learn enough about analysis (a branch of maths concerning things like limits) that I would consider your knowledge of HSC level calculus to be "rigorous". First year is usually only as rigorous as is possible without going into any of the...

The "assumed knowledge" doesn't really refer to stuff you need to know otherwise you won't be able to do the course. You will have books/course notes and stuff that are pretty self-contained. It's just that the average student who has only taken 2U might struggle in some first year uni courses...

From scratch as in from the definitions of a function, continuity, differentiability, etc etc. Basically "jumping in" right after the real numbers are constructed and their elementary properties understood. In later years if you take further maths courses, you can then look at the construction...

Yep, these are good concerns to have. In HS they sweep a bunch of details under the carpet and you just have to trust them (not specifically your HS teachers, many of whom don't know better themselves lol, more the people who write the syllabus and books). If they tried to do things more...

Oh I didn't read that you wanted to do it at an actual place, rather than just learning from books. In that case your options are more limited, and I am less knowledgable about them...sorry!

If you are doing it purely for self-study there is absolutely no reason to handcuff yourself to the actual MX2 syllabus. You can actually learn the base topics like calculus and complex numbers properly, and then pick a bunch of topics of a similar level that sound interesting to you personally...

Let P(n)= 10^n - 7^n - 5^n. Then P(2)=100 - 49 - 25 > 0. Now suppose P(n) > 0. Then P(n+1)=10^(n+1)-7^(n+1)-5^(n+1) >10^(n+1)-10*7^n-10*5^n =10*P(n) >0. Hence P(n) > 0 for all positive integers n >= 2 by induction.

Yeah, sure. There's no way they would mark that wrong lol. As for "simplification", that is debatable,as it requires more writing rather than less. This fraction will usually be far from simplest form anyway.

Yep, this is good and valid. :)

I guess we will have to wait till the exam notes come out from BOS to know for sure, but usually the questions at this point in the paper are written by people who know well enough that the answers they are designed to elicit are actually quite rigorous. (At least rigorous as deductions from...

Would be wary about writing this solution in an MX2 exam. Finding the limit of what the probability is as n gets big is not necessarily the same as just treating the problem as taking place on an infinity x q grid. Furthermore what does it even mean to choose a square randomly from an infinite...

Haha there are almost always some technical problems in Terrys solutions.

Which parts are confusing you? For the first one you want to change that real integral into a contour integral with circular contour, by taking z=e^(i*theta), and this reduces the problem to integrating a rational function over the counterclockwise unit circle. The denominator can be...

Re: HSC 2015 4U Marathon Hence? These facts are pretty unrelated, I would be surprised if there was a proof of divergence that used in an essential way that the H_n are non-integers. An elementary proof of divergence can be obtained by writing: 1+1/2+1/3+...

Re: HSC 2015 4U Marathon An alternate solution to iii): Let m be the largest integer such that 2^m =< n. We must then have 2^m =< n < 2^{m+1}, which implies that 2^m > n/2. Hence k.2^m > n for any positive integer k larger than 1. Another way of saying this is that 2^m is the...

Re: HSC 2015 4U Marathon Yep :).

Re: HSC 2015 4U Marathon v=\frac{g}{k}(1-e^{-kt})=\frac{g}{k}(1-(1-kt+O(k^2t^2)))\\=\frac{g}{k}(kt+O(k^2t^2))=gt+o(1).\\ \\x=\frac{g}{k}\int_0^t 1-e^{-ks}\, ds=\frac{g}{k^2}(-1+kt+e^{-kt})\\=\frac{g}{k^2}(-1+kt+(1-kt+k^2t^2/2+O(k^3t^3)))=\frac{gt^2}{2}+o(1).