Search results

Re: MX2 2015 Integration Marathon I guess it can be made quicker. I_R=\int_R^R \arctan(x+1/2)-\arctan(x-1/2)\, dx = 2\int_R^{R+1/2}\arctan(x)\, dx.\\ \\ $So $2\arctan(R)\leq I_R \leq 2\arctan(R+1/2)\\ \\ $from which the squeeze law completes the proof.$

Re: MX2 2015 Integration Marathon Is it a consequential difference? Either way we get an arctan integral, which we perform by integrating by parts.

Re: MX2 2015 Integration Marathon I_R=\int_{-R}^R \frac{d}{dx}(x)\cdot\arctan\left(\frac{1}{x^2+3/4}\right)\, dx\\ \\ = \int_{-R}^R \frac{4x^2\, dx}{1+(x^2+3/4)^2}\, dx\\ \\ = \int_{-R}^R \left(\frac{x}{1+(x-1/2)^2}-\frac{x}{1+(x+1/2)^2}\right)\, dx\\ \\ = \int_{-R-1/2}^{R-1/2}...

Re: MX2 2015 Integration Marathon I think they are both equal to \pi. For the sum one: \sum_{n\in\mathbb{Z}}\arctan\left(\frac{1}{n^2+3/4}\right)\\ \\= \lim_{N\rightarrow\infty}\sum_{|n|\leq N}(\arctan(n+1/2)-\arctan(n-1/2)\right))\\ \\=...

Re: 2015 permutation X2 marathon And for comparison to your solutions, I believe braintic's answer for the tetrahedron problem is correct (for all n).

Re: 2015 permutation X2 marathon Okay, I will illustrate the idea for a simpler problem. Q/ How many ways are there to colour the 4 sides of a square using n colours? A/ Consider strings of the form (ABCD). In this problem we have 4 symmetries (that correspond to rotating the square...

Re: 2015 permutation X2 marathon I haven't done the calculation yet myself, but the answer will look something like braintics. I won't completely give away the method yet, because I think it is a fantastic exercise in counting things with symmetry (although a decent bit harder than syllabus...

Re: 2015 permutation X2 marathon How many ways can we paint the 4 faces of a tetrahedron using n colours? (colours can be repeated.) Answer the same question for the cube.

Re: 2015 permutation X2 marathon Given any four distinct points, there is exactly one way of dividing the four points into two pairs and joining each pair with a line such that the intersection is interior to the convex polygon defined by the initial four points. (The lines are just the...

Re: 2015 permutation X2 marathon Ah cool, I feel like I've heard that name before. Have definitely seen these objects before.

Re: 2015 permutation X2 marathon You can visualise this problem as counting the number of paths (consisting of moves right and moves up) along the edges of a 6x6 grid from the bottom left corner to the top right corner that always stay below the diagonal. We will solve the generalisation...

Re: HSC 2015 4U Marathon - Advanced Level $Take $y=\frac{u+v}{2},x=\frac{u-v}{2}$ (it is easy to see this is a bijection on $\mathbb{R}^2$). \\ \\ The rearranged inequality then becomes\\ \\ $f(v)\geq f(u)-f\left(\frac{u^2-v^2}{4}\right)\quad \textrm{ for all }u,v\in\mathbb{R}.$\\ \\ But for...

Re: HSC 2015 4U Marathon \frac{(\sqrt{a}-\sqrt{b})^2}{2}\geq 0 \Rightarrow \frac{a+b}{2}\geq \sqrt{ab}. \frac{a+b+c+d}{4}=\frac{\frac{a+b}{2}+\frac{c+d}{2}}{2}\geq \frac{\sqrt{ab}+\sqrt{cd}}{2}\geq \sqrt[4]{abcd.} LHS=\frac{a+b+c}{4}+\frac{a+b+c}{12}=\frac{a+b+c}{3}...

Re: HSC 2015 4U Marathon Also, an application for the above extension problem in geometry: Find all triangles with sides lengths that are consecutive positive integers that also have their area as an integer. You may find it helpful to first find an expression for a triangles area purely in...

Re: HSC 2015 4U Marathon We will prove existence and uniqueness separately. Existence we do inductively. It is trivial for n=1. Now suppose that there exist positive integers p_n,q_n such that (1+\sqrt{2})^n=p_n+q_n\sqrt{2}. Then...

Re: HSC 2015 4U Marathon The sum diverges, because the absolute value of the n-th term tends to infinity.

Re: HSC 2015 4U Marathon - Advanced Level It's been a few days, so I will finish off your solution so we can move to something new :). You were very close, and my method started out essentially the same as yours. Since we have f(x+1)=f(x)+1, we have f(x+m)=f(x)+m for all integers m. Since we...

Re: HSC 2015 4U Marathon I might as well just prove the inductive step of your extension question, as the same method is how we answer the earlier parts of the question. Consider the polynomial p(x)=(x+a)^n-bx. By considering its first two derivatives, it has at most two stationary points...

Re: HSC 2015 4U Marathon Another approach that is nice and works is to consider the recurrence relation u_n-u_{n-1}=p(n),u(0)=0. for polynomials p of degree d. (When p(x) is the monomial x^d, finding a closed solution to the recurrence is the same thing as finding a closed form for S_d(n).)...

Re: HSC 2015 4U Marathon I don't think this works, we need a fixed polynomial that evaluates the n-th partial sum. Here is a way of proving that it is a degree d+1 polynomial (where d is the power we are raising the terms to). Suppose by inductive hypothesis that...