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Re: MX2 Integration Marathon It can be resolved to a cosine recurrence integral with the substitution x = atan u Or by parts we can do: I_n = \int (a^2+x^2)^{-n} \ dx I_n = \frac{x}{(x^2+a^2)^n} + 2n \int \frac{x^2}{(x^2+a^2)^{n+1}} \ dx I_n = \frac{x}{(x^2+a^2)^n} + 2n (I_n - a^2...

Re: MX2 Integration Marathon Great Question By dividing numerator and denominator by x^2, we arrive at: I = \int \frac{\frac{1}{x^2} + 1}{\left(\frac{1}{x}- x \right ) \sqrt{x^2 + \frac{1}{x^2}}} \ dx y = x - \frac{1}{x} dy = 1 + \frac{1}{x^2} \ dx I = \int \frac{dy}{-y \sqrt{y^2+2}}...

Re: HSC 2013 4U Marathon IF there does exist a multiple of 2013 like that, then for some pair of integers (n,m) The following equality must hold: 2013n = m And for some value k, if m has last four digits 2012 then k is integer. k= \frac{m - 2012}{10 000} On the k-m plane it is a linear...

Re: HSC 2013 4U Marathon Using the result from the previous question: \sum_{k=1}^{n} \frac{\sin(k\theta)}{k} = 0.5(\pi -\theta) + \int_{\pi}^{\theta} \frac{\cos(n\alpha) + \cos(n+1)\alpha}{4\sin^2 \alpha/2} \ d\alpha Integrating both sides from pi to 2, and using the double angle rule, the...

Re: HSC 2013 4U Marathon I made up the solution (with help from sean in its rigour), the problem is a discarded one from STEP.

Graphs because it belongs in the 2U course.

Mathematics

b) Assuming u, v, w are functions of x. y=uvw y'= (uv)(w) = (uv)'w + (uv) w' y'= w(u'v+v'u) + w'(uv) y' = u'vw + uv'w + uvw' $Where$ \ \ f' = \frac{df}{dx} ii) Simply let u = x^5, v = (x-1)^4, w=(x-2)^3 Then substitute it in the above equation. c) f(x) = x^r (1-x)^s We can...

Re: HSC 2013 4U Marathon Yes that is true. === Also Registered User, are you sure your series is correct? \sum_{k=-\infty}^{\infty} \frac{\sin^2 k}{k^2} Doesn't look very nice because of the sines of integers, is it a typo?

Re: HSC 2013 4U Marathon Nice work, it can also be done by considering the geometric series: z+z^2+ \dots +z^n = \frac{z(z^n-1)}{z-1} Substituting cis theta, equating real parts for a series in cosine. Then integrating with bounds alpha and pi yields the LHS and the RHS plus a limit...

Re: HSC 2013 4U Marathon Can you please criticise this proof? I am starting to have doubts about it. === Start with the polynomial, x=p+\sqrt{q} We need to make the lowest reducible rational polynomial that has that as a root, so we simply: (x-p)^2 - q = 0 The above polynomial has the...

Re: HSC 2013 4U Marathon Hopefully this is correct. Varying the sides of lengths x and y, then the curve represented by the length of the diagonal is x^2 + y^2 \leq 100 x > 0 y > 0 Therefore the probability is that area that satisfies the above equation, divided by the square 10 x 10 since...

Yes silly me, thanks.

Re: HSC 2013 4U Marathon That is true.

Re: HSC 2013 4U Marathon The problem simply boils down to computing: P = \sum_{k=0}^{m} \binom{n+k}{n} (P(T))^n (P(H))^k \frac{1-P(H)}{P(H)} = q \ \ P(H) = \frac{1}{q+1} \ \ P(T) = \frac{q}{q+1} Is it possible to compute \sum_{k=0}^{m} \binom{n+k}{k} x^k ? Its easy if x is 1 but....

Re: MX2 Integration Marathon Yep, so all you need to do is pretty much is if I_n = \int_0^{\pi/2} \cos^{2n} x \ dx $Find$ \ \ \sum_{k=0}^{m} I_{k} Which is longer than you might think

Re: HSC 2013 4U Marathon yes that is also true. Yeah sure if you want, just make sure you can prove everything that you need to, and justify things rigorously (such as radius of convergence etc.) I would prefer a HSC solution though. EDIT: Also Heroic, I don't like the step 'I'll assume R(x)...

y = \frac{e^x}{3+e^x} 3y + y \cdot e^x = e^x 3y = e^x(1-y) e^x = \frac{1-y}{3y} \therefore \ f^{-1}(x) = \ln \left |\frac{1-x}{3x} \right |

Re: MX2 Integration Marathon J_m = \int_0^{\pi / 2} \frac{1-\cos^{2m} x}{\sin^2 x} \ dx $Prove that$ J_m = \frac{\pi}{2} \left (\frac{(2m+2)! - m! (m+1)! 2^{2m+1}}{m! (m+1)! 2^{2m+1}} \right ) \ \ \ \ \ \ \ \fbox{6} I apologise if there is a slight algebraic mistake (it should only be a...

Re: HSC 2013 4U Marathon z=k+it for some constant k, and variable t. z^2 = k^2 - t^2 +i 2kt Then taking the parametric equations of them: x = k^2 - t^2 y = 2kt x= k^2 - \frac{y^2}{4k^2} which is a parabola since k is constant and x is linear while y is quadratic. =================...