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Re: HSC 2013 3U Marathon Thread \int \frac{\cos \left(\frac{1}{x} \right)}{x^2} \ dx

Re: HSC 2013 4U Marathon i) 0, 1, 1, 2, 3, 5, 8 ii) and iii) F(z)= 0z^0 + 1z + \sum_{k=2}^{\infty} F_k z^k = z + z\sum_{k=2}^{\infty} F_{k-1}z^{k-1} + z^2 \sum_{k=2}^{\infty} F_{k-2}z^{k-2} F(z) = z+z(F(z)-0z^0) + z^2(F(z)) -z=F(z)(z^2+z-1) \therefore F(z)=\frac{-z}{z^2+z-1} iv) To...

I don't see how that justifies the proof though, the only reason why torque is Nm is because t=Fd. So its like proving a definition using its own definition. Doesn't make sense

Re: HSC 2013 4U Marathon $let$ \ \ x=\cos \theta \cos^2 \theta + \cos^2 3\theta = 1 \cos 3\theta = \pm \sin \theta Now notice that the function: f(x)=x^2+(4x^3-3x)^2 = f(-x) So its an even function, hence the roots are opposites of each other. So I will only take one of the above...

I am very familiar with the HSC concept of torque since the questions that utilise it are all very similar and simple. I am looking for a proper definition of torque, not the washed up HSC version.

I see, thank you for the clarification

I am guessing r is radius and p momentum right? A=r \cdot p p=mv \\ \ \frac{dp}{dt} = m \frac{dv}{dt} = ma = F \frac{d}{dt}A = r \frac{dp}{dt} = F \cdot r = \tau This is a correct proof yeah? Thanks man

So there really isn't a simple answer to this? That page has a lot of vector calculus in it hah I'll look at it in my free time I think

It is given that the torque is: \tau = F \cdot d (disregarding the angle by which the force is applied, which just results in a trivial trigonometry problem) How do you prove this formula? I have been told that torque by definition is the distance multiplied by the force. But I don't...

If people rote learned less, they would have more time on their hands hence they would not need to deprive themselves of sleep.

What are mock trials?

Re: HSC 2013 4U Marathon Ah yes I see it, my proof was just a roundabout way to essentially the same thing :P (except yours goes backwards and mine forward)

Re: HSC 2013 4U Marathon Assuming its the above result: Establish: \sin \theta = 2\sin \theta/2 \cos \theta /2 \sin \theta = 4\sin \theta / 4 \cos \theta /2 \cos \theta/4 And so on: \sin \theta = 2^n \cos \frac{\theta}{2} \cos \frac{\theta}{4} \dots \cos \frac{\theta}{2^n} \sin...

Re: HSC 2013 4U Marathon Wait so is the result: \cos{\frac{2\pi}{2^n-1}} \cdot \cos \frac{4\pi}{2^n-1} \cdot \dots \cos \frac{2^n \pi}{2^n-1} = \frac{1}{2^n}

Re: HSC 2013 4U Marathon Are you sure the result is correct? (Im assuming the last term is cos(2^n pi)/(2^n-1)

First establish: p \neq \frac{3}{5} \frac{3p+2}{5p-3} = 4 \ \ \ 3p+2 = 20p-12 \ \ \ p=\frac{14}{17} So we have found 2 critical points p=3/5 p=14/17 Now, we need to test points on either side of these 2 critical points and test whether they satisfy the condition p > 14/17 p=1...

Re: HSC 2013 4U Marathon There is a typo in my question, where it used to be: \frac{-z}{z^2+z+1} should be: \frac{-z}{z^2+z-1} sorry for inconveniences.

Re: HSC 2013 4U Marathon The Sydney Grammar one was about the Chebyshev polynomials :P I modified their method to find an expression for Fibonacci. EDIT: I would do the same for the Triangular number question but there is no point considering we already know that T_k=k/2 (k+1)

Re: HSC 2013 4U Marathon $i) Consider the geometric series$ 1+z+z^2+ \dots + z^{n} = \frac{1-z^{n+1}}{1-z} Differentiate: (skipping the grunt work) 1+2z+3z^2+ \dots + nz^{n-1} = \frac{nz^{n+1}-(n+1)z^n+1}{(1-z)^2} $ii) Let n go to infinity, we are already given that$ \ |z| < 1 So...

Re: HSC 2013 3U Marathon Thread Solution I was looking for: $Consider the expression$ (1+x)^m + (1+x)^{m+1}+ (1+x)^{m+2}+ \dots + (1+x)^{n} =(1+x)^m (1+(1+x)+(1+x)^2+ \dots + (1+x)^{n-m}) =(1+x)^m \frac{1-(1+x)^{n-m+1}}{1-(1+x)} =(1+x)^m \frac{(1+x)^{n-m+1} -1}{x} =\frac{1}{x}...