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HSC 2013 MX2 Marathon (archive) (2 Viewers)

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seanieg89

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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:

 

Sy123

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Re: HSC 2013 4U Marathon

Let integral be I
let x^2 = tanθ

2x.dx = sec^2 θ .dθ

When x= 0, θ = 0
When x = INFTY, θ = pi/2

therefore, int(θ=0 to θ=pi/2) = 0.5 dθ

therefore, I = pi/4?
You have made a mistake when subbing dx back into the integral
 
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Re: HSC 2013 4U Marathon

Let integral be I
let x^2 = tanθ

2x.dx = sec^2 θ .dθ

When x= 0, θ = 0
When x = INFTY, θ = pi/2

therefore, int(θ=0 to θ=pi/2) = 0.5 dθ

therefore, I = pi/4?
You can't use that substitution because you can't get rid off the x after you sub dx back.
The only substitution you can use is x=tanx; which doesn't really help.
 
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Re: HSC 2013 4U Marathon

Hint: fugly completion of squares :)
 

Sy123

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Re: HSC 2013 4U Marathon

Finding roots of unity then expressing in quadratic factors then applying partial fractions, then manipulating to get it into a suitable integral form:



By integrating both sides, we yield:

First and second fractions are logarithms, both elements inside the logarithm functions are monic quadratics, at infinity they cancel to 1, making the first 2 terms together 0, at x=0, the same happens and we get the logarithm of 1 which is again zero.
The second 2 terms are in the form for the atan function. After completing the square and manipulations, we evaluate them at infinity keeping in mind that at infinity atan is pi/2.

In the end we should arrive at



Also it would be awesome if people started posting these integrals in the integration marathon please =)
 
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Re: HSC 2013 4U Marathon

Finding roots of unity then expressing in quadratic factors then applying partial fractions, then manipulating to get it into a suitable integral form:



By integrating both sides, we yield:

First and second fractions are logarithms, both elements inside the logarithm functions are monic quadratics, at infinity they cancel to 1, making the first 2 terms together 0, at x=0, the same happens and we get the logarithm of 1 which is again zero.
The second 2 terms are in the form for the atan function. After completing the square and manipulations, we evaluate them at infinity keeping in mind that at infinity atan is pi/2.

In the end we should arrive at



Also it would be awesome if people started posting these integrals in the integration marathon please =)
Good job!

By the way,

No problem, I will post there next time.
 

Sy123

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Re: HSC 2013 4U Marathon

Even though its integration I think it should be asked here because its not really a 'Q1 style' integral.

=============



















 
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seanieg89

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Re: HSC 2013 4U Marathon

This can be so easily done using abstract algebra but I don't think that is HSC level.
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.
 

Sy123

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Re: HSC 2013 4U Marathon





(there is a reason I asked this in this 4U thread)
 

RealiseNothing

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Re: HSC 2013 4U Marathon

In this question you may assume that:















 
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seanieg89

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Re: HSC 2013 4U Marathon

Lol rather large assumptions there Realise.
 

seanieg89

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Re: HSC 2013 4U Marathon

He eventually found a rigorous proof. You actually need quite a lot of machinery to make that particular proof work (much more than you have stated I am pretty sure).

Edit: To be a little more concrete in my criticism, my main objection is that it is not at all clear that the symmetric polynomials in the roots of a power series have to be related to the coefficients of the power series in the same way that is true of polynomials. (In fact this is not true in general.)

To make this argument work we need to assume something along the lines of the Weierstrass factorisation theorem, as well as some facts about convergence and uniform convergence. In short, I think that this isn't a good choice for an MX2 question because I think it is better for the assumptions to be milder and more precise. That way, the student is at least engaging in (close-to) rigorous mathematical reasoning just skipping the theory leading up to the assumption rather than doing a whole series of hand-wavy steps.
 
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