HSC 2014 MX2 Marathon (archive) (1 Viewer)

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glittergal96

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Re: HSC 2014 4U Marathon

We have



so at least one term on the LHS must be non-positive. As each of these polynomials is concave up, at least one must be concave up with vertex on below the x-axis. This polynomial must then have at least one real root.
 

glittergal96

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Re: HSC 2014 4U Marathon

We have



so at least one term on the LHS must be non-positive. As each of these polynomials is concave up, at least one must be concave up with vertex on below the x-axis. This polynomial must then have at least one real root.
Here is a perhaps cleaner way of presenting this same argument:

Suppose each of these polynomials has no real roots. As they are concave up, this means that they are positive definite.

Then the sum of the three polynomials must also be positive definite.

But the sum of the three quadratics by direct calculation is , which is NOT positive definite. Contradiction.
 

Tugga

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Re: HSC 2014 4U Marathon

Could you also do this:

Assume all of the eqns have non-real complex roots. Then by conjugate root theorem and summing product of roots you get a sum of real squares = 0.

Therefore, all the roots must be zero, a contradiction of the assumption.
 

Axio

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Re: HSC 2014 4U Marathon

Could you also do this:

Assume all of the eqns have non-real complex roots. Then by conjugate root theorem and summing product of roots you get a sum of real squares = 0.

Therefore, all the roots must be zero, a contradiction of the assumption.
Do all the roots have to be zero just because the sum of their products is zero?
__________
And could somebody please post another complex number question, thanks.
 

Tugga

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Re: HSC 2014 4U Marathon

Do all the roots have to be zero just because the sum of their products is zero?
__________
And could somebody please post another complex number question, thanks.
When you sum the product of roots you get this: (gonna try use fancy latex)

 

Axio

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HSC 2015 4U Marathon

New Question (Cambridge):

If |z| = r and arg z = theta, show that z/[(z^2)+(r^2)] is real and give its value.
 

Chlee1998

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Re: HSC 2014 4U Marathon

You change r to mod z. Then factor z out of denominator which cancels with the top. Then u get 1/ 2rcos theta


EDIT: did u type the question wrongly or did Realisenothing read it wrong?
 
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Chlee1998

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Re: HSC 2015 4U Marathon

The inverse of a nonzero real number is real.
Yes, but what's the point of doing it Realise's way? like in the end he still has to find the real value. I don't think there's anything wrong with my solution? (in the post after the question)
 

glittergal96

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Re: HSC 2014 4U Marathon

No-one said there was anything wrong with your solution, he just wrote up what is logically the same thing in a different notation. (With the typo of having the reciprocal of what we actually want, and he didn't finish up by writing .)

I just meant that its pretty obvious what he meant, so picking on the typo seemed a bit petty.
 

RealiseNothing

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Re: HSC 2014 4U Marathon

No-one said there was anything wrong with your solution, he just wrote up what is logically the same thing in a different notation. (With the typo of having the reciprocal of what we actually want, and he didn't finish up by writing .)

I just meant that its pretty obvious what he meant, so picking on the typo seemed a bit petty.
Wasn't a typo, I just flipped it since like you said the inverse of a real number is still real, and I thought it was easier to manipulate when it was flipped compared to the original question.
 

RealiseNothing

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Re: HSC 2014 4U Marathon

Wasn't a typo, I just flipped it since like you said the inverse of a real number is still real, and I thought it was easier to manipulate when it was flipped compared to the original question.
Though looking back on it, there really isn't much difference in flipping it or not lol.
 
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