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Re: HSC 2013 4U Marathon Alright then: (i) Now, since the arguments of complex numbers are periodic by 2pi, in order for the sequence: cis(r\pi) , \ cis(2r\pi), \ cis(3r\pi), \ ... To have finitely many values, then we must observe that: 2\pi n + mr\pi = kr \pi That is, the kth term in...

Re: HSC 2013 4U Marathon Let: P(x) = (x-a)(x-(a+d))(x-(a+2d))(x-(a+3d)) $Let$ \ \ \alpha(x) = (x-a)(x-(a+d)) \ \ \ \ \beta (x) = (x-(a+2d))(x-(a+3d)) \therefore P'(x) = \alpha '(x) \beta (x) + \alpha (x) \beta '(x) \alpha '(x) = 2x - (2a+d) \beta '(x) = 2x - (2a+5d) \therefore \...

Re: HSC 2013 4U Marathon For (i) If r = pi then cis r = -1 \ \ cis 2r = 1 \ \ cis 3r = -1 ... Which is finitely many values (only 2 values), yet pi is irrational? Moreover if we let r=p/q, since there is a loop of 2pi, the only way I see if there are finitely many values is if r IS...

Re: HSC 2013 4U Marathon The first 3 parts of the question is ripped off of Moriah 2001, but the last part is my own which needed this, so I just copy pasted it in. For induction for 'Step 2' we are basically applying an assumption for n=k (which we can justify to be true for 1 value which...

Re: HSC 2013 4U Marathon I PMed you

I can''t find a really good substitution that will fit it,this is the closest I got: \int x e^{-x} \ dx x= -\ln u \ \ dx = \frac{-du}{u} \int -\ln u \cdot u \cdot -1/u \ du =\int \ln u \ du We don't need 4U to do this integral, this is a question very commonly asked in...

There comes a point where you can't just keep blaming silly mistakes anymore, if 'silly mistakes' are the sole reason for a mark from 100 raw drops to 80 (which achieves a 95 HSC mark I think), then there are different factors at play, maybe the student was really nervous for the exam - this is...

Re: HSC 2013 4U Marathon $Consider the set of polynomials that are defined by the 3 recurrence relations where$ 1) \ \ B_0(x)=1 2) \ \ B_n '(x) = nB_{n-1}(x) 3) \ \ \int_{0}^{1} B_n (x) \ dx = 0 $i) Find$ \ \ \ B_1 \ \ $in terms of$ \ \ x \ \ \ \ \fbox{2} $ii) If$ \ \ B_n...

'If you do 4U, 3U becomes a breeze' nearly every 4U student that I have met has said this. And that should be true as well.

Re: HSC 2013 4U Marathon A neat result that I proved: $Functions$ \ \ \ g(x) \ \ $and$ \ \ h(x) $are said to be within the same family if$ \ \ g(x) = k h(x) \ \ $ for some constant k value$ $i.e.$ \ \ \ g(x) = 2e^x \ \ \ \ h(x)=5e^x \ \ \ $are within the same family of functions$...

Yes the increased popularity of the subject from the scaling has made more people to choose it hence bringing the scaling down. However I am sure that the increased availability of tutors also has a role, some students could think that they are able to succeed in a course if they had a tutor...

Can we associate the the lower scaling of Mathematics Extension 2 to the increased availability of tutors? Discuss If not, what can we associate it to?

Re: HSC 2013 4U Marathon Very true, ignore the last part then.

Re: HSC 2013 4U Marathon Here is a nice result I proved: $Consider the polynomial$ \ \ \ \ P(z) = z^n + \sum_{k=0}^{n-1} b_{k}z^k $Where$ \ \ \ b_k \in \mathbb{R} \ \ \ $for$ \ \ k=0,1,2,...(n-1) $Prove that if ALL roots are real then$ \ \ \ \ \ b_{n-1}^2 > 2b_{n-2}

Well I was thinking of the situation of: \lim_{x \to \infty} \frac{x^n}{e^{x}} = 0 \ \ \ \ $for all n$ And I was thinking if we can do somethng similar here for the non-convergent integral divided by something that could have a faster convergence? (I'm not very familiar with the rigour of...

How about this then: \frac{-1}{2n}\int_{0}^{\pi} \left(\frac{1}{2\sin \alpha} - (-1)^n/2 - \int_{\pi/2}^{\alpha} \frac{\cos \theta}{2\sin^2 \theta} \ d\theta \right) \ d\alpha \leq \frac{1}{2n} \int_{0}^{\pi} \left (\frac{\cos 2n\alpha}{2\sin \alpha} - \frac{(-1)^n}{2} -...

Ah yes, sorry about that my 3rd line on the 1st page should be: = \frac{z-|z|^2z-z^{2n+1}+z^{2n-1}|z^2|^2}{|1-z^2|^2} The reason I didn't sub in cis theta earlier is due to it just being messier and longer to write for me. Its easier this way because I know |cis theta| = 1 so I have less...

Here we go: Part 1: Developing the identity Part 2: Integrating finding a form for Basel problem Part 3: Evaluating the integral That is the better way to do it lol, is my one sufficient though?

That would be greatly appreciated, thank you. ============== However I decided to tackle the problem from a different point, and I decided to change the limit of integration of alpha to 0, to alpha to pi/2. It has given some interesting results: \int_{\pi/2}^{\alpha} \cos \theta + \cos...

So its never a definite value? Just a range of values between csc(x)-1 and 1? (I guess this was predictable considering the functions are periodic) But oh well :( Any ideas on how to compute the integral?