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Re: HSC 2016 MX2 Combinatorics Marathon I will not spoil (1), because it is the closest to MX2 level. (The same method works there though). Remark: The standard way of dealing with counting problems involving symmetry is Burnside's lemma. Unfortunately this is far enough out of syllabus to...

(If you understand the above, then it would be good exercise for your algebra skills to generalise the result, both by replacing 100 by an arbitrary even number, and then by allowing for arbitrary parity of A,B,n.)

Yep, the computations work out. Suppose A > B are odd integers. If you follow the above guidelines, you should obtain: J_0=J_{100}=3333 \Rightarrow J_n=2n^2-200n+3333 \Rightarrow \sum_{k=1}^{99}\frac{\sin(Ak\pi/100)\sin(Bk\pi/100)}{1+\cos(k\pi/100)}\\ \\...

Okay, I actually really can't be bothered to write everything up, but I will sketch each step of my computation. (This will evaluate the sum for any pair of odd integers in the numerator. Extending this to treat non-odd integers is not difficult, but it clutters the computations with (-1)^n...

Thought of a method that should work (and give us a more general formula), but it takes some doing. Will work through the details later today and post them when I do.

Yeah. The composition of two finite-degree field extensions is finite (degrees being multiplicative), and finite-degree extensions are algebraic.

https://www.ocf.berkeley.edu/~abhishek/chicmath.htm These reviews are quite well written (don't agree with everything but there is subjectivity involved obviously) and I found this list fairly useful in undergrad.

For 2) it is just sin(1)/3 because k=3 is the first time its positive, and for k>3, the thing is bounded by 1/4 which is smaller than sin(1)/3.

$1. $L(xy)=\int_1^{xy}\,\frac{dt}{t}=\int_{y^{-1}}^x \frac{1}{xt}\cdot x\, dt\\=L(x)-\int_1^{y^{-1}}\,\frac{dt}{t}=L(x)-\int_1^{y}t\,\left(-\frac{dt}{t^2}\right)=L(x)+L(y).

Re: HSC 2016 4U Marathon - Advanced Level I got an equivalent series and the same numerics as Integrand. Will spend a little time trying to evaluate it before posting.

Okay, had another look at it after my meeting today. My parametric form looks correct, and leads to a cartesian equation pretty straightforwardly. $As the angle $\theta$ varies along the director circle, let $(x,y)$ be the point we are trying to find the locus of.\\ \\...

a-b,a-c,b-c \geq 0 \\ \\ \Rightarrow (a-b)(a-c)(b-c)\geq 0 \\ \\ \Rightarrow \textrm{the inequality you want.}

But do you know whether or not there is a known cartesian equation for this thing or are you just exploring?

Equally useful would be a visual confirmation of the above using Geogebra. I will do this tomorrow if no-one else does.

Do you know for a fact that a nice cartesian form exists? Or will a parametric description suffice? It is not hard to obtain a parametric formula. (MX2 level conics.) My method: 1. Show that the director circle has radius sqrt(a^2+b^2) for an ellipse in the standard form. 2. For a point X on...

Yep exactly. Because we want our domain to be an interval, the turning points of the cubic are "bad" in the sense that f is not injective in any inteval containing either of these turning points in their interior. Elsewhere we don't have a problem, because f is piecewise monotonic, so the...

And the relationship between these conditions and invertibility: A function is f is surjective <=> it has a right inverse. A function f is injective <=> it has a left inverse. And a function f is bijective iff it is invertible.

When you specify a function f: X->Y you are also specifying the domain X and the co-domain Y. With this in mind: f is said to be injective if any two distinct elements of X get sent to distinct elements of Y by f. f is said to be surjective if for every y in the codomain Y there exists an x...

Alternatively alternatively: P(x)=(x^5-1)/(x-1) for x =/= 1. But x^5 is increasing, so the numerator has the same sign as the denominator and P(x) > 0 apart from possibly at x=1. (And P(1)=5 so we are positive here too.) Edit: Haha, Integrand beat me to it.

I+2\pi\mathbb{Z}$ is pretty standard. More generally $A+B$ to denote the set of all sums of elements in $A$ and elements in $B$. This is pretty unambiguous and common notation, although the corresponding product notation sometimes has different meanings (eg the product of two ideals in a ring.) $