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Re: MX2 Integration Marathon \int_0^{\pi/2}x\cot(x)\, dx its been done earlier in this thread, but if you resist the temptation to look at the solution it is a very good question.

Re: MX2 Integration Marathon We don't even need substitution: I=\int \frac{1}{x^n+x} \, dx=\int \frac{1}{x}-\frac{x^{n-2}}{x^{n-1}+1} \, dx=\log |x|-\frac{1}{n-1}\log |x^{n-1}+1|+C

Re: HSC 2013 4U Marathon No heavy machinery required here, the simplest proof of this fact that I can think of is essentially identical to the standard proof of a course theorem.

https://www.youtube.com/watch?v=eAjhG09X9YA

Re: HSC 2013 4U Marathon After completing Sy123's question on conjugate roots, it is a nice challenge exercise to consider the analogous problem for cube roots: $If $P(x)$ is a polynomial with rational coefficients and $\alpha+\sqrt[3]{\beta}$ is a root of $P$ for some rational $\alpha$ and...

A quick note on the intuition behind the solution: This sort of problem screams out binary decomposition, which led to finding a likely candidate for a tight bound by solving a recurrence relation. The rest was just verification.

I think that: (n*log(n))/(2*log(2)) is an upper bound which is tight. (For n >= 2). I shall check my working tonight and write up a proof if I am happy with it.

Nice question, will have a crack in a few hours after my supervisor meeting.

Re: HSC 2013 4U Marathon Huh? The result is still true for monic polynomials...and having a root of the form p+q^{1/2} is perfectly possible for monic polynomials.

Re: HSC 2013 4U Marathon Am interested in seeing peoples solutions actually. This is a v.good question because it is not very technically demanding at all, but if you don't properly understand the relevant part of the MX2 course you probably won't have any chance of solving it. ie it slay rote...

Re: HSC 2013 4U Marathon (You must also assume that q is not the square of a rational number.)

Re: MX2 Integration Marathon Not really, its just one branch.

Physically I dig babies who have got back. Slim is good too but generally prefer curves. Also more into brunettes that other hair colours on average. Other than that if a girl ticks some of the following boxes she becomes a lot more attractive to me: -Intelligence. -Is an interesting person...

Re: HSC 2013 4U Marathon Good, although your answer to 3. is more of a plausibility argument than a proof. The point is to do it in a similar way to to 2, using that a continuous function can only change sign on an interval if it possesses a root on this interval.

Re: HSC 2013 4U Marathon Let a < b be two consecutive roots of a polynomial P(x), (so P is nonzero between a and b). Suppose these roots are each of multiplicity 1, so that: P(x)=(x-a)(x-b)g(x) for some polynomial g(x) with g(a),g(b) != 0 by the factor theorem. 1. Prove that g(a) and g(b)...

am glad someone said it.

Cryptography is probably their most important "real world" application as Goldy said. From the perspective of pure mathematics they are important simply because they are the multiplicative "building blocks" of the natural numbers, and although they have been studied for thousands of years...

Re: HSC 2013 4U Marathon Well not entirely, he should probably change the log to something else. Logs explode at 0 so its not at all clear that the integral is well defined ;).

Of course, but I figured that physical attraction was implied as being of universal importance. I think my weighting on the two things I mentioned is less so.

Haha well obviously. And attraction is often because of really intangible factors...just thought I'd chuck out two of the more important things that a girl I would want to date would have to have for the sake of discussion, they are two of the less obvious deal-breakers for me. I wouldn't trust...