HSC 2012 MX2 Marathon (archive) (2 Viewers)

math man

Member
Joined
Sep 19, 2009
Messages
503
Location
Sydney
Gender
Male
HSC
N/A
Re: 2012 HSC MX2 Marathon

He explained his method to me through PM and I don't think it is valid, but we will see if someone can post a valid solution using MX2 only techniques (I don't think this is possible).
to prove my method i had to construct a quarter circle contour to show the integrals were the same, so the method works, but yeh it does need proof why it works using contour integration.
 

math man

Member
Joined
Sep 19, 2009
Messages
503
Location
Sydney
Gender
Male
HSC
N/A
Re: 2012 HSC MX2 Marathon

Here is a 4u question from cambridge, but done in a different way:

regions in plane.png
 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Re: 2012 HSC MX2 Marathon

Here is a decent question:

The three roots of the complex polynomial:



all lie on the unit circle in the complex plane. Prove that the three roots of:



also lie on the unit circle.
 
Joined
Sep 20, 2010
Messages
2,225
Gender
Undisclosed
HSC
2012
Re: 2012 HSC MX2 Marathon

I'm assuming a, b, c are real?
 

lolcakes52

Member
Joined
Oct 31, 2011
Messages
286
Gender
Undisclosed
HSC
2012
Re: 2012 HSC MX2 Marathon

Second part is done with Newton's method for x=k to find the approximation. I'm having trouble with the first part, possibly because I'm not sure about whether the next root is between k=1 and 1, greater than k, or less than zero.
 

barbernator

Active Member
Joined
Sep 13, 2010
Messages
1,439
Gender
Male
HSC
2012
Re: 2012 HSC MX2 Marathon

Second part is done with Newton's method for x=k to find the approximation. I'm having trouble with the first part, possibly because I'm not sure about whether the next root is between k=1 and 1, greater than k, or less than zero.
Well the sum of the roots is k, so if one root is between 0 and 1, one is between k-1 and k, then the next must be between 0 and 1 as well to satisfy k
 

math man

Member
Joined
Sep 19, 2009
Messages
503
Location
Sydney
Gender
Male
HSC
N/A
Re: 2012 HSC MX2 Marathon

a better question would be to deduce the smallest positive root plus the smallest negative root is positve
 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Re: 2012 HSC MX2 Marathon

Here is a decent question:

The three roots of the complex polynomial:



all lie on the unit circle in the complex plane. Prove that the three roots of:



also lie on the unit circle.
.
 

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
Re: 2012 HSC MX2 Marathon

Here is a decent question:

The three roots of the complex polynomial:



all lie on the unit circle in the complex plane. Prove that the three roots of:



also lie on the unit circle.
So far I've got:



y/n?
 

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
Re: 2012 HSC MX2 Marathon

I don't think q can possibly be that, that would imply that the product of roots of p has modulus 3.
Then what have I done wrong? Have I assumed something that isn't true maybe?

Let the roots of p(x) be alpha, beta, and gamma. Then:









Since the roots lie on the unit circle and hence .

I then do this for 'b' and 'c' as well to obtain my expression for q(x).
 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Re: 2012 HSC MX2 Marathon

From your second line to your third line of displayed equations.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top