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Re: MX2 Integration Marathon \int{\frac{(1+x^{2})dx}{(1-x^{2})\sqrt{1+x^{4}}}} \int \frac{dx}{(x^2+a^2)^n}

I would remove volumes, graphs, and harder 3U (probably put this in the 3U course?). I would replace them with real analysis, differential equations and linear algebra. I would make polynomials harder.

Finished everything. Now I really need to get my self to doing more english and science...

Re: HSC 2013 4U Marathon Didn't think of that. Seems like an easier method. Did you make up this question?

Re: MX2 Integration Marathon Method 1: Let\ I=\int \sqrt{\tan(x)}dx Let\ u=\tan(x), du=(1+\tan^{2}(x))dx I=\int \frac{\sqrt{u}}{u^{2}+1} Let\ v=\sqrt{u}, dv=\frac{du}{2\sqrt{u}} I=2\int \frac{v^{2}}{v^{4}+1} \int_0^\infty\frac{x^2}{1+x^4}dx Let\ t=\frac{1}{v} \therefore...

Re: HSC 2013 4U Marathon It is definitely correct.

LOL

Re: MX2 Integration Marathon \int \frac{1+sinx}{cosx}dx

Thank you! This explains why they call it 'nuking mosquitos' haha

Switch to medicine or engineering/Law :)

So I was looking at Terry Tao's (one of the best mathematicians alive) profile (http://mathoverflow.net/users/766/terry-tao) and I went through some of his questions and answers just out of curiosity and I came across this: Every person who saw is was amazed by it. Since I am a noob can...

Re: HSC 2013 4U Marathon In this proof I will assume that x -\frac{x^2}{2}+\frac{x^3}{3}\cdots converges to \ln (1+x) (Taylor Series). $Let,$ S_1 = \sum_{n=1}^{\infty}\frac{\cos n\theta}{n} S_2 = \sum_{n=1}^{\infty}\frac{\sin n\theta}{n} $Then$ S_1 + iS_2 =...

Re: HSC 2013 4U Marathon That is correct. Good job.

Re: HSC 2013 4U Marathon A rectangle is drawn where the lengths of the sides are chosen randomly from [0, 10] and independently of one another. Find the probability that the length of its diagonal is smaller than or equal to 10.

May be two? I too agree that most people exaggerate its difficulty, ESPECIALLY teachers.

Do it and see how you go. If you got 70% in MX1 then I think you will find MX2 difficult because MX2 is very similar to MX1 but I'd say 4 times harder.

Re: HSC 2013 4U Marathon \sum_{k=0}^n\binom{2n}{2k}x^{2k} =\frac12\left(\sum_{k=0}^{2n}\binom{2n}{k}x^k+\sum_{k=0}^{2n}\binom{2n}{k}(-x)^k\right) =\frac{(1+x)^{2n}+(1-x)^{2n}}{2} $Plug in$ x=i $to get$ \sum_{k=0}^n(-1)^k\binom{2n}{2k} =\frac{(1+i)^{2n}+(1-i)^{2n}}{2}...

Re: HSC 2013 4U Marathon hmm I will try to do Apéry's constant then :P PS: this can't be done, right? Since it doesn't converge to a nice value.

Re: HSC 2013 4U Marathon Seen a question slightly similar to this in 2000 paper. Don't know if any one in the state managed to get it lol

Re: HSC 2013 4U Marathon I like the one using Fourier series but I still don't completely understand it haha I want to make a 4U level question to compute \zeta (4), don't know if that is possible though.