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HSC 2013 MX2 Marathon (archive) (2 Viewers)

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Sy123

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Re: HSC 2013 4U Marathon

I have stated that they are non-zero real numbers.
I just solved it:

http://i981.photobucket.com/albums/ae294/sy08071996/Image4-2.jpg


However I believe that extends it to non-zero complex numbers as well.

Brief explanation of solution:

- Let polynomial have roots x, y, z
Find co-efficients of b, c using 1/x + 1/y + 1/x = x+y+z=0

Leave d alone for now. Sub in x, y, z to create equality with the zero. Then add all 3 equations to get expression for x^3+y^3+z^3
Similarly multiply x^3/y^3/z^3 to each of the 3 previous equations. Find expression for x^6+y^6+z^6

Then simplify and express d=-xyz
 
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Sy123

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Re: HSC 2013 4U Marathon

Well, there is something called "analytic continuation".
Well yes, if I prove it to be true for complex numbers, then it must be true for the real numbers. But I was just pointing out that the question can be extended to non-zero complex numbers.
 

jyu

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Re: HSC 2013 4U Marathon

I have stated that they are non-zero real numbers.
I don't think there are non-zero real numbers x, y and z satisfying x + y + z = 1/x + 1/y + 1/z = 0. If there are such real numbers, what are they?
 

Rezen

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Re: HSC 2013 4U Marathon

There isn't any. You can prove it by rearranging the conditions to get it in terms of z;

.

Then form the difference function


which can be shown with some basic multivariate calculus to not have any zeros among the reals. (well, atleast that's what my quick working shows. I might be wrong.)
 

Sy123

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Re: HSC 2013 4U Marathon

There isn't any. You can prove it by rearranging the conditions to get it in terms of z;

.

Then form the difference function

which can be shown with some basic multivariate calculus to not have any zeros among the reals. (well, atleast that's what my quick working shows. I might be wrong.)
Yes I think there are no all three real solutions. This comes from the fact that, if we have a polynomial roots x, y, z with the restrictions in the question.
Then the polynomial equation will be:



Where it is a polynomial in alpha, and d is any constant, from this equation we can automatically deduce that there can only be 1 real root (and non-zero).
 

Rezen

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Re: HSC 2013 4U Marathon

Yes I think there are no all three real solutions. This comes from the fact that, if we have a polynomial roots x, y, z with the restrictions in the question.
Then the polynomial equation will be:



Where it is a polynomial in alpha, and d is any constant, from this equation we can automatically deduce that there can only be 1 real root (and non-zero).
That's true. It also gives us that the only possible form of x,y and z are which after a quick substitution prove to be solutions.
 

cutemouse

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Re: HSC 2013 4U Marathon

Well yes, if I prove it to be true for complex numbers, then it must be true for the real numbers.
No actually I THINK it's the other way around... It's been a while since I've done complex analysis though (which is VERY beautiful).
 

Carrotsticks

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Re: HSC 2013 4U Marathon

No actually I THINK it's the other way around... It's been a while since I've done complex analysis though (which is VERY beautiful).
The reals are a subset of C, so if proven true for all C, then it is true for all R. The converse isn't necessarily the same.
 

RealiseNothing

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Re: HSC 2013 4U Marathon

No actually I THINK it's the other way around... It's been a while since I've done complex analysis though (which is VERY beautiful).
The real numbers are apart of the complex numbers though, so proving true for all complex numbers would imply it's true for all real numbers. Like all real numbers can be expressed as a complex number

ie 3 = 3 + 0i
 

Shadowdude

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Re: HSC 2013 4U Marathon

No actually I THINK it's the other way around... It's been a while since I've done complex analysis though (which is VERY beautiful).
you're thinking about analytic continuity, right?
 

Shadowdude

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Re: HSC 2013 4U Marathon

I think they just got confused for a bit, personally.
 

seanieg89

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Re: HSC 2013 4U Marathon

Straightforward exercise to flex algebra muscles:

Find a polynomial with integer coefficients that has:



as one of its roots.
 

Sy123

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Re: HSC 2013 4U Marathon

DONE:

I'm not bothered to post working out, basically let x = that root, then find x^6, x^4 and x^2 using binomial theorem, then manipulation then:


I better not of made a silly mistake :/

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



 
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seanieg89

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Re: HSC 2013 4U Marathon

DONE:

I'm not bothered to post working out, basically let x = that root, then find x^6, x^4 and x^2 using binomial theorem, then manipulation then:


I better not of made a silly mistake :/

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



Cbb checking haha but that is the correct idea.

As an addendum to your question, which I think I have asked before:

Prove that E has NO real roots if n is even.
 
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