HSC 2013 MX2 Marathon (archive) (2 Viewers)

Status
Not open for further replies.

deswa1

Well-Known Member
Joined
Jul 12, 2011
Messages
2,256
Gender
Male
HSC
2012
Re: HSC 2013 4U Marathon

0, since if all the coefficients are real, the roots must occur in conjugate pairs, therefore the arguments will be in opposite in sign?

edit: oh right it can also be pi as well if both roots are real but one's negative and other's positive.
but he said in terms of a, b, c
0(a+b+c-3ab +2(a^2)b - c^3)=0 (that's in terms of a,b,c)
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Re: HSC 2013 4U Marathon

0, since if all the coefficients are real, the roots must occur in conjugate pairs, therefore the arguments will be in opposite in sign?

edit: oh right it can also be pi as well
How can it be pi as well?
 

SpiralFlex

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
Re: HSC 2013 4U Marathon

I'm impressed that the advertisement of this thread worked to a degree.
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon

damn I think I missed all the shortcuts :(
Nice work, alternatively while we can do it your way, do you see how you needed to find

And sum of them two at a time, then three at a time -> why not find a polynomial with roots 1/alpha 1/beta 1/gamma
And it actually isn't time consuming at all, in order to find a polynomial with those roots, you must 'reverse' the co-efficients
So a poly with roots 1/a 1/b 1/c
would be:

2x^3+3x^2+0x+1 = 0

So then you could find sum of them one at a time, two at a time and three by just using co-efficients instead of finding common denominator and going through that torturous process :p

And you could do the same for sum of squares and stuff as well.
 
Last edited:

Immortalp00n

Member
Joined
Dec 27, 2012
Messages
272
Gender
Undisclosed
HSC
2009
Re: HSC 2013 4U Marathon

i came to this thread for d babes as per sy's sig
where them bad bitches at?
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon

i came to this thread for d babes as per sy's sig
where them bad bitches at?
Err, not sure what you're talking about :p

Only maths here, if that's what you're talking about.
 

GoldyOrNugget

Señor Member
Joined
Jul 14, 2012
Messages
583
Gender
Male
HSC
2012
Re: HSC 2013 4U Marathon

I was playing around with these sorts of problems on the bus the other day, but I ended up confusing myself and didn't get very far.

Say you had a polynomial with roots and you want to find a polynomial with roots (Sy's question would be a special case of this where ).

You can find an equation with these roots by replacing every instance of in the polynomial with (this is an interesting property to investigate/prove). In some cases, that equation can be algebraically manipulated into a polynomial without creating any new roots. This is why the trick of inverting coefficients works for -- it's a shortcut to replacing all terms with and multiplying out the resulting fractions.

When isn't monotonic (meaning is not a function), it still works; I'm pretty sure you can choose a maximal interval on its domain over which it's monotonic (probably the wrong terminology) and use that for the inverse, because the only necessity is that . Also, I'm not convinced that all equations generated in this manner can be manipulated into polynomials without creating new roots (for example, I had some trouble doing it with Sy's question, but I could have just messed up the algebra). Perhaps someone else will have more luck investigating this?
 
Status
Not open for further replies.

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

Top