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Re: HSC 2013 4U Marathon Cool, good question by the way.

Re: HSC 2013 4U Marathon If and only if? I don't see how it follows from your working that an f which does not satisfy f''=kff' cannot give rise to an I which has a primitive in terms of f,f',k and n...

Re: HSC 2013 4U Marathon A hint for my earlier question: The corresponding result holds true for differentiable functions and is probably easier to understand for an mx2 student. The ideas behind the proofs are the same though. $Let $f$ be a positive, continuously differentiable, decreasing...

Re: HSC 2013 4U Marathon Your first step in II doesn't work unfortunately, remember that cos(blah) isn't always positive.

Re: HSC 2013 4U Marathon Essentially correct although your notation hurts my head... (a_n) typically refers to the sequence (a_0,a_1,...) rather than the corresponding series. For things like that it suffices to provide an example, in this case the harmonic series would be adequate.

Re: HSC 2013 4U Marathon Ah yep, a more general phrasing of the problem would use cis instead of cos, in which case we would need the assumption for all complex sequences.

Re: HSC 2013 4U Marathon |a|=a, remember this sequence is positive. 2 because they walk you through everything, although I will admit that there are many places for the average MX2 student to go wrong. The scale is more the calibrated to the questions I post here than the HSC questions...

If your tutor wrote that, get one who can write questions in proper English. It makes no difference here (you are not asked to and not expected to find roots), but in more difficult questions it can easily lead to confusion.

Re: HSC 2013 4U Marathon $In this question you may assume:\\ Let $(a_j)$ be a sequence of complex numbers. If there exists an $M>0$ such that:\\ \\ $\sum_{j=0}^n |a_j|\leq M$\\ \\ for all $n$, then the series\\ $\sum_{j=0}^\infty a_j$\\ converges. \\ \\ \\ Let $(a_n)$ be a positive and...

Its a pretty common view for people in science and especially mathematics I think.

Re: HSC 2013 4U Marathon I_n+I_{n-1}=\int_0^{\pi/4} \tan^{2n}(x)+\tan^{2n-2}(x)\, dx \\=\int_0^{\pi/4} \tan^{2n-2}(x)\sec^2(x)\, dx=\frac{1}{2n-1}\\ \\ $so$\\ \\ \left|\frac{\pi}{4}-\sum_{j=1}^n \frac{(-1)^j}{2j-1}\right|=\left|\frac{\pi}{4}-\sum_{j=1}^n (-1)^j(I_j+I_{j-1})\right|=I_n <...

I am an agnostic atheist: I do not personally believe in the existence of a god, nor do I think that satisfactory proof or disproof of his existence can be found. If gods existence was proven tomorrow, I do not think it would change the way I live my life. I do not judge people on their...

Honeymoon period on average 4-6 months for me. That doesn't mean it has to get shit after, unless you are dating someone boring or you discover something you can't stand about them that wasn't obvious at first.

Re: HSC 2013 4U Marathon abc=\frac{1}{6}((a+b+c)^3-3(a^2+b^2+c^2)(a+b+c)+2(a^3+b^3+c^3)) too lazy to put the numbers in :P. edit: -4.

Re: HSC 2013 4U Marathon Any rational function of trig functions has elementary primitive. t substitutions -> partial fractions -> integrating rational functions with at most quadratic denoms, all of which are elementary.

A positive integer is written on a blackboard. Players A and B take turns subtracting a positive square integer from it such that the result is still positive. The person who plays the last legal move wins. There are ~180,000 out of the first 40,000,000 positive integers that result in a win...

Re: HSC 2013 4U Marathon Yep. A fast way of generalising this is proving that you can differentiate term-by-term in the interior of a power series disc of convergence, and then differentiating the infinite geometric series with leading term 1 and ratio x a bunch of times.

Re: HSC 2013 3U Marathon Thread It doesn't, as AB and CD are arbitrary the claim is false.

Fair enough then. Yeah sashimi especially salmon/kingfish is amazeballs.

Re: HSC 2013 3U Marathon Thread As you have currently phrased the question, X and Y are both the point M itself...