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Re: HSC 2016 MX2 Combinatorics Marathon That could well have been his intended question. This is why having rigorous notation for probabilistic statements and their proofs is important :).

Re: HSC 2016 MX2 Combinatorics Marathon Nope, just meant that others might argue differently to reach to same conclusion. That is the unique length.

Re: HSC 2016 MX2 Combinatorics Marathon I interpreted it as asking what length should we specify (instead of 1) for the whole interval, in order to have the probability of the middle interval having length at most 1/3 equal to 1/3? Paradoxica's is a correct answer to this question. (It's...

Re: HSC 2016 MX2 Combinatorics Marathon Those numbers look good to me :).

Re: First Year Uni Calculus Marathon \left|\frac{f(x)-f(0)}{x-0}\right|=\frac{|f(x)|}{|x|}\leq \frac{|x|^3}{|x|}=|x|^2\rightarrow 0\\ \\ \Rightarrow \frac{f(x)-f(0)}{x-0}\rightarrow 0. (The "=>" comes from the squeeze law. Generally if we suspect something might tend to zero, it is often...

Re: First Year Uni Calculus Marathon Yes it is possible. Eg: f(x)=x*sin(x). This is easy to make rigorous. For large x, x^3+1 < 2x^3 and x^2+1 < 2x^2. So log(x^3+1)/log(x^2+1) < log(2x^3)/log(x^2) = (log(2)+3log(x))/2log(x). and log(x^3+1)/log(x^2+1) >...

Re: First Year Uni Calculus Marathon (Sorry about delayed response, am wrestling with a problem of my own :p) Yep @leehuan, IVT, and EVT are key to rigorous argument here. You are on a nice train of thought and I think you have a good intuitive grasp of why k=2 is impossible, but making this...

Re: First Year Uni Calculus Marathon Indeed that is the essence of the question, I don't want to spoil it too quickly. It is very good exercise to reach a conclusion yourself and try to rigorously justify it. (The main tools at your disposal being properties of continuous functions like the...

Re: First Year Uni Calculus Marathon (A bit tricky) For which positive integers k is it possible to find a continuous function f:R->R that such that f(x)=y has exactly k solutions x for every real number y?

Notice you are applying a well known function (log) at x close to 1 (where log is quite nice). A first order Taylor approximation works: \log(1+x)=x+O(x^2) So \frac{\log(1-x^2)}{\log(1-x)\log(1+x)}=\frac{-x^2+O(x^4)}{-x^2+O(x^3)}=\frac{1+O(x^2)}{1+O(x)}\rightarrow 1. Don't worry if you...

Re: HSC 2016 MX2 Combinatorics Marathon A-5 is definitely a straight in the vast majority of poker variants.

Re: HSC 2016 Complex Numbers Marathon Here is a one way of proving the fundamental theorem of algebra (given an assumption from slightly outside syllabus). $Assumption: If $p(z)$ is a polynomial with complex coefficients, then for any positive real number $R$, we can find complex numbers...

Yeah, AU secondary maths education is pretty bad and is designed in a way that will inevitably make a lot of talented students dislike mathematics because of the petty dogmatic way things are taught. (As well as some of the material being straight up dry). Creativity and curiosity are very...

Re: HSC 2016 Complex Numbers Marathon By using complex numbers, show that for a quadrilateral ABCD, we have: AB.CD + AD.BC >= AC.BD. When does equality occur?

Well the fact it is continuous on an interval and does not vanish on this interval implies that it is of fixed sign.

Re: HSC 2016 MX2 Combinatorics Marathon Spoiler for second Monty Hall variant: (highlight to read) I think this is just an exercise in conditional probability. We let A(k) denote the event that the k-th door is correct and B(k) denote the event that Monty opens the k-th door. (Where we let...

Re: HSC 2016 MX2 Combinatorics Marathon Cool question! Did you think of it yourself braintic? Let's play this game with n doors. I believe the optimal strategy is to refuse to switch until the final offer, and switch then. This guarantees you a (n-1)/n chance of success. (It fails if and only...

All good, let me know if you want to play later.

Getting back into kerbal space program (pretty heavily modded) too. I am a beginner, just completed my first orbital rescue (some guy was stranded about 80km above kerbin). Matching up orbits and setting up the rendezvous was super fun actually. After doing some more missions to earn money /...

anyone for some chill games now?