HSC 2013 MX2 Marathon (archive) (4 Viewers)

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Sy123

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Re: HSC 2013 4U Marathon

Do we have to use Euler's method and compare real part etc. or is there a MX2 way to do it ?
What do you mean by euler's method?

there is a MX2 way of doing it
 

JJ345

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Re: HSC 2013 4U Marathon

What do you mean by euler's method?

there is a MX2 way of doing it
Euler's Formula sorry, applying
e^{iy}=cos y+isin y
But if you didn't think of it, I probably did something retarded...
 
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Sy123

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Re: HSC 2013 4U Marathon

Euler's Formula sorry
e^{iy}=cos y+sin y
But if you didn't think of it, I probably did something retarded...
Well the HSC method would be very similar to it, in your method, just change every e^(iy) into cos y + i sin y, and it will still work out with De Moivere's
 

gustavo28

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Re: HSC 2013 4U Marathon

Multiply it by a sufficiently large integer to make it an integer polynomial Q(x) which is also good. Let the leading coefficient of Q be A and the constant be C. Now, considering the graph of Q(x)-C, we can make a prime p large enough so that Q(x)+p-c is above the x-axis for -1<=x<=1. We can also make p a prime that does not divide A. So, let such a prime be called q. Now, Q(x)+q-C where q is prime is also good and hence has only rational roots. Its constant term is q. Consider a root of this polynomial c/d. So, d divides A and, since q does not divide A, q doesn't divide d. However, c divides q. So, c=1 or c=q. If c=1, then -1<=c/d<=1 which is a contradiction since Q(x)+q-c is above the x-axis for such values of x. So, c=q.

Hence, the product of the roots is equal to q^n/stuff where stuff is an integer with no factors of q in it and n is the degree of Q. This is equal to q/A or -q/A. So, q^(n-1)*A=stuff but since n>1, q divides an integer with no factors of q in it.

Contradiction. SO the only good polynomials are linear.
 

Sy123

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Re: HSC 2013 4U Marathon























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Not sure if I 'proved' the method correctly lol
 

Sy123

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Re: HSC 2013 4U Marathon

Here is a good one:





 

JJ345

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Re: HSC 2013 4U Marathon

Here is a good one:





Let p(x) have its minimum value point at Q.
So p(x)>= p(Q) (1)
p(Q)>=p''(Q) ...given (2)
p''(Q)>=0 (because its the concavity at the min point of even degree positive coeff. polynomial)...(3)
So combining (1), (2) & (3)
We get that p(x)>=0

I think my (3) is pretty dodgy...
 

Sy123

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Re: HSC 2013 4U Marathon

Let p(x) have its minimum value point at Q.
So p(x)>= p(Q) (1)
p(Q)>=p''(Q) ...given (2)
p''(Q)>=0 (because its the concavity at the min point of even degree positive coeff. polynomial)...(3)
So combining (1), (2) & (3)
We get that p(x)>=0

I think my (3) is pretty dodgy...
That is the sketch of the proof, but can you explain why we get the conclusion from (1) (2) and (3)?
 

JJ345

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Re: HSC 2013 4U Marathon

That is the sketch of the proof, but can you explain why we get the conclusion from (1) (2) and (3)?
I don't get what you want me to explain.
Like actually define a polynomial and then do some random algebra or just explain what I did in more words?
 

Sy123

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Re: HSC 2013 4U Marathon

I don't get what you want me to explain.
Like actually define a polynomial and then do some random algebra or just explain what I did in more words?
Wait yea nevermind my method was a little different to yours, it involved a proof by contradiction (kinda).
Well done
---





 

Sy123

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Re: HSC 2013 4U Marathon





(can definitely be done within syllabus)
 
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