HSC 2015 MX2 Marathon (archive) (4 Viewers)

VBN2470 · Feb 9, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
New question:
$$The equation$ y=x^3 +2x^2 +3x +3 $has the roots$ \alpha, \beta $and$ \gamma$

$$Find the equation with roots$ \alpha,-\alpha,\beta,-\beta,\gamma,-\gamma$

$Using squares of the sum and products of roots we can find expressions for the coefficients of our new polynomial to get $x^6+2x^4-3x^2-9$.$

NEXT QUESTION

$$ If $\alpha$ is a non-real $n$th root of 1, show that $(1-\alpha)(1-\alpha^2)...(1-\alpha^{n-1})=n$ $.$

Sy123 · Feb 9, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
$Using squares of the sum and products of roots we can find expressions for the coefficients of our new polynomial to get $x^6-10x^4+21x^2-9$.$

NEXT QUESTION

$$ If $\alpha$ is a non-real $n$th root of 1, show that $(1-\alpha)(1-\alpha^2)...(1-\alpha^{n-1})=1$ $.$

Easier way:

$p(x) = x^3 + 2x^2 + 3x + 3$

$p(-x) = -x^3 + 2x^2 - 3x + 3$

$p(x) \ $has roots$ \ \alpha, \beta , \gamma$

$p(-x) \ $has roots$ \ -\alpha, -\beta, -\gamma$

$\\ $Therefore$ \ p(x) \cdot p(-x) = q(x) \ $has all these roots$$

braintic · Feb 9, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
New question:
$$The equation$ y=x^3 +2x^2 +3x +3 $has the roots$ \alpha, \beta $and$ \gamma$

$$Find the equation with roots$ \alpha,-\alpha,\beta,-\beta,\gamma,-\gamma$

Since you didn't specifically ask for a polynomial equation, couldn't we just say:

y = | x |³ + 2x² + 3 | x | + 3

FrankXie · Feb 9, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
$Using squares of the sum and products of roots we can find expressions for the coefficients of our new polynomial to get $x^6-10x^4+21x^2-9$.$

NEXT QUESTION

$$ If $\alpha$ is a non-real $n$th root of 1, show that $(1-\alpha)(1-\alpha^2)...(1-\alpha^{n-1})=1$ $.$

wrong question, RHS should be n.

Drsoccerball · Feb 9, 2015

Re: HSC 2015 4U Marathon

braintic said:
Since you didn't specifically ask for a polynomial equation, couldn't we just say:

y = | x |³ + 2x² + 3 | x | + 3

Would you write that in the exam

Drsoccerball · Feb 9, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
Easier way:

$p(x) = x^3 + 2x^2 + 3x + 3$

$p(-x) = -x^3 + 2x^2 - 3x + 3$

$p(x) \ $has roots$ \ \alpha, \beta , \gamma$

$p(-x) \ $has roots$ \ -\alpha, -\beta, -\gamma$

$\\ $Therefore$ \ p(x) \cdot p(-x) = q(x) \ $has all these roots$$

Wouldnt that be long to expand? you would have 16 terms

FrankXie · Feb 9, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Wouldnt that be long to expand? you would have 16 terms

you do not really expand term by term, you can expand by inspection, say, you know that the term x^4 comes from products of x^3 and x, x^2 and x^2, and x and x^3. meanwhile, for p(x)p(-x), if you group even power terms and odd power terms, you will have sum times difference which is difference of squares

p.s., by inspection, I actually find the answer should be $x^6+2x^4-3x^2-9$

Drsoccerball · Feb 9, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
$Using squares of the sum and products of roots we can find expressions for the coefficients of our new polynomial to get $x^6-10x^4+21x^2-9$.$

NEXT QUESTION

$$ If $\alpha$ is a non-real $n$th root of 1, show that $(1-\alpha)(1-\alpha^2)...(1-\alpha^{n-1})=1$ $.$

$z^n=1$

$\theta= \frac{2\pi k}{n}$

$\therefore z= cis\frac{2\pi}{n}, cis\frac{4\pi}{n}....,cis\frac{2\pi(n-1)}{n} $Where a$ =cis\frac{2\pi}{n}$

$(z-a)(z-a^2)....(z-a^{n-1})=1$

$$since$ z^n =1$

$(1-a)(1-a^2)....(1-a^{n-1})=1$

FrankXie · Feb 9, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$z^n=1$

$\theta= \frac{2\pi k}{n}$

$\therefore z= cis\frac{2\pi}{n}, cis\frac{4\pi}{n}....,cis\frac{2\pi(n-1)}{n} $Where a$ =cis\frac{2\pi}{n}$

$(z-a)(z-a^2)....(z-a^{n-1})=1$

$$since$ z^n =1$

$(1-a)(1-a^2)....(1-a^{n-1})=1$

the third last line is wrong. refer to my previous post, the RHS of the question should be n.

Sy123 · Feb 9, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Wouldnt that be long to expand? you would have 16 terms

FrankXie's method is good but another way that I can see is to:

$\\ (x^3+2x^2+3x+3)(-x^3+2x^2-3x+3) \\ = ((2x^2+3) + (x^3+3x))((2x^2+3) - (x^3+3x)) \\ = (2x^2+3)^2 - (x^3+3x)^2 = \dots$

Drsoccerball · Feb 9, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
the third last line is wrong. refer to my previous post, the RHS of the question should be n.

but the RHS = z^n ?

VBN2470 · Feb 9, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
$Using squares of the sum and products of roots we can find expressions for the coefficients of our new polynomial to get $x^6-10x^4+21x^2-9$.$

NEXT QUESTION

$$ If $\alpha$ is a non-real $n$th root of 1, show that $(1-\alpha)(1-\alpha^2)...(1-\alpha^{n-1})=n$ $.$

Yeah sorry, the RHS should be 'n' not '1', just fixed it, thanks Frank.

Drsoccerball · Feb 9, 2015

Re: HSC 2015 4U Marathon

Me and ekman attempting to make a question:
$Let z=$ cis(\frac{\pi+2\pi k}{n})$

$$For n greater than or equal to $ 3, $ and k is an integer greater than or equal to $ 1$

$Find what $ (-2i)^{\frac{n}{2}}\xs cos\frac{\pi}{n} \xs cos\frac{3\pi}{n}......\xs cos\frac{\pi+2\pi k}{n}$ Is equal to?$

Sy123 · Feb 9, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Me and ekman attempting to make a question:
$Let z=$ cis(\frac{\pi+2\pi k}{n})$

$$For n greater than or equal to $ 3, $ and k is an integer greater than or equal to $ 1$

$Find what $ (-2i)^{\frac{n}{2}}\xs cos\frac{\pi}{n} \xs cos\frac{3\pi}{n}......\xs cos\frac{\pi+2\pi}{n}$ Is equal to?$

Is that last fraction $\cos \frac{2k\pi+\pi}{n}$ ?

FrankXie · Feb 9, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
NEXT QUESTION

$$ If $\alpha$ is a non-real $n$th root of 1, show that $(1-\alpha)(1-\alpha^2)...(1-\alpha^{n-1})=n$ $.$

There's still a serious problem in the question: say, if $n=6, \alpha=cis\frac{2\pi}{3},$ then the LHS will be zero. So I modify this question into

$$ If $ z_1, z_2, \cdots, z_{n-1} $ are distinct $ n $-th root of $ 1 $ other than $ 1, $ show that $ (1-z_1)(1-z_2)...(1-z_{n-1})=n.$

Drsoccerball · Feb 9, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
Is that last fraction $\cos \frac{2k\pi+\pi}{n}$ ?

yes

Ekman · Feb 9, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Me and ekman attempting to make a question:
$Let z=$ cis(\frac{\pi+2\pi k}{n})$

$$For n greater than or equal to $ 3, $ and k is an integer greater than or equal to $ 1$

$Find what $ (-2i)^{\frac{n}{2}}\xs cos\frac{\pi}{n} \xs cos\frac{3\pi}{n}......\xs cos\frac{\pi+2\pi k}{n}$ Is equal to?$

BUMP. I really want to see what people get for our question.

FrankXie · Feb 9, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Me and ekman attempting to make a question:
$Let z=$ cis(\frac{\pi+2\pi k}{n})$

$$For n greater than or equal to $ 3, $ and k is an integer greater than or equal to $ 1$

$Find what $ (-2i)^{\frac{n}{2}}\xs cos\frac{\pi}{n} \xs cos\frac{3\pi}{n}......\xs cos\frac{\pi+2\pi k}{n}$ Is equal to?$

is k any integer>=1, like can it be greater than n ?

Drsoccerball · Feb 9, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
is k any integer>=1, like can it be greater than n ?

Sorry 1 greater than or equal to k which is greater than or equal to n-1

seanieg89 · Feb 9, 2015

Re: HSC 2015 4U Marathon

Ekman said:
BUMP. I really want to see what people get for our question.

Er, are you really sure you have a nice expression for this?

I am a bit sceptical. You should probably phrase it as a "prove that ..." question so people can easily check that the question works before spending time on it.

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