HSC 2013 MX2 Marathon (archive) (1 Viewer)

Sy123 · Jan 4, 2013

Re: HSC 2013 4U Marathon

bleakarcher said:
It's as if this thread is invisible to everyone besides the usual bosers. Usually you'd have some other randoms participate in marathon threads.

- its probably the questions. And that I answer them too fast not giving a chance for others.

Ok, next time someone posts a normal type question I will leave it for others to answer it so the participation rate is too high.
So yeah if someone could post a question that would be great, if I answer it I won't post the answer straight away (unless it is a difficult one and I somehow manage to find it)

twinklegal19 · Jan 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
A question that is a little easier to other people don't get scared off or anything:

$x^3+3x+2=0 \ \ \ \ \rightarrow $roots$ \ \ \ \ \alpha , \beta, \gamma$

$$Find the polynomial equation with roots$ \ \ \alpha + \frac{1}{\alpha} , \beta + \frac{1}{\beta} , \gamma + \frac{1}{\gamma}$

(it isn't that long if you can create shortcuts for yourself.)

did you get something like 2x^3+3x^2+8=0? just wanted to check, probably made heaps of sillies along the way.

GoldyOrNugget · Jan 4, 2013

Re: HSC 2013 4U Marathon

Sy posted that easy polynomial roots question at the bottom of pg 27, that's definitely approachable by any 4u student, and no one else's tried it yet... weird

Sy123 · Jan 4, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
did you get something like 2x^3+3x^2+8=0? just wanted to check, probably made heaps of sillies along the way.

Yeah something like that, I only remember that the first 2 terms were 2, 3 and co-efficient of x was 0. So yeah good enough lol.

How did you do it?

RealiseNothing · Jan 4, 2013

Re: HSC 2013 4U Marathon

The quadratic $az^2+bz+c$ has roots $\alpha ~and~\beta$

$Find$~ arg(\alpha)+arg(\beta)$

seanieg89 · Jan 4, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
The quadratic $az^2+bz+c$ has roots $\alpha ~and~\beta$

$Find$~ arg(\alpha)+arg(\beta)$

Err, it could be anything? You probably want to say something about a,b,c first.

Sy123 · Jan 4, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
The quadratic $az^2+bz+c$ has roots $\alpha ~and~\beta$

$Find$~ arg(\alpha)+arg(\beta)$

Are a, b, c real?
I won't answer it in this thread if I get the answer.

RealiseNothing · Jan 4, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Err, it could be anything? You probably want to say something about a,b,c first.

Well I meant find the argument in terms of a, b, c.

RealiseNothing · Jan 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Are a, b, c real?
I won't answer it in this thread if I get the answer.

Oh yes, they are real.

seanieg89 · Jan 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yeah, I rephrased it:

Still not what I meant, the lim should be on the side of the cosines.

Carrotsticks · Jan 4, 2013

Re: HSC 2013 4U Marathon

$\\ $Let $ p_n = \cos \frac{\theta}{2} \times \cos \frac{\theta}{2^2 } \times ... \times \cos \frac{\theta}{2^n} \qquad ,\forall n\geq 1. \\\\ $Since $ 0 < \frac{\theta}{2^n} < \frac{\pi}{2} \qquad , \forall n \geq 1 $ and $ \forall 0<\theta<\frac{\pi}{2} $ we can conclude that for all $ n \geq 1, p_{n}>0 $ and $ \frac{p_{n+1}}{p_n}= \cos \frac{\theta}{2^n} < 1 $ and hence by the Monotone Convergence Theorem, $ p_n $ converges as $ n \rightarrow \infty. \\\\ $Multiply both sides by $ \sin \frac{\theta}{2^n} $ which initiates an inductive process, and so we acquire $ p_n = \frac{\frac{\sin \theta}{2^n}}{\sin \frac{\theta}{2^n}} $ and then work it from there.$$

Sy123 · Jan 4, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Still not what I meant, the lim should be on the side of the cosines.

Well I did have the 3 dots implying that it continued to infinity after the nth term, but I will change it again (even though no-one else will answer it)

twinklegal19 · Jan 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yeah something like that, I only remember that the first 2 terms were 2, 3 and co-efficient of x was 0. So yeah good enough lol.

How did you do it?

Edit: cbb to latex, will scan solution

RivalryofTroll · Jan 4, 2013

Re: HSC 2013 4U Marathon

Polynomials Question:

If alpha, beta and gamma are roots of the equation:

x^3 - px - q = 0

Find the equation whose roots are alpha/(beta.gamma), beta/(alpha.gamma), gamma/(alpha.beta).

NOTE: Sorry, I don't use latex.

twinklegal19 · Jan 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
A question that is a little easier to other people don't get scared off or anything:

$x^3+3x+2=0 \ \ \ \ \rightarrow $roots$ \ \ \ \ \alpha , \beta, \gamma$

$$Find the polynomial equation with roots$ \ \ \alpha + \frac{1}{\alpha} , \beta + \frac{1}{\beta} , \gamma + \frac{1}{\gamma}$

(it isn't that long if you can create shortcuts for yourself.)

damn I think I missed all the shortcuts

HeroicPandas · Jan 4, 2013

Re: HSC 2013 4U Marathon

RivalryofTroll said:
Polynomials Question:

If alpha, beta and gamma are roots of the equation:

x^3 - px - q = 0

Find the equation whose roots are alpha/(beta.gamma), beta/(alpha.gamma), gamma/(alpha.beta).

NOTE: Sorry, I don't use latex.

x^3 - px - q = 0

Roots: $\alpha, \beta, \gamma$

$\prod \alpha: \alpha\beta\gamma=q$

We want equation with: alpha/(beta.gamma), beta/(alpha.gamma), gamma/(alpha.beta).

Re-arranging product of roots and subbing into ^

New aim is to find eqn with roots: $\alpha^2/q, \beta^2/q, \gamma^2/q$

Let y= x^2/q

x= sqrt(yq) -> sub this into the original equation: x^3 - px - q = 0

THEREFORE: $(\sqrt{yq})^{\frac{3}{2}} - p(\sqrt{yq})^{\frac{1}{2}} - q = 0$

Chuck the q over and square both sides

$[(\sqrt{yq})^{\frac{3}{2}} - p(\sqrt{yq})^{\frac{1}{2}}]^2 =q^2$

$q^3y^3-2pq^2y^2+p^2qy-q^2=0$

HeroicPandas · Jan 4, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
The quadratic $az^2+bz+c$ has roots $\alpha ~and~\beta$

$Find$~ arg(\alpha)+arg(\beta)$

nvm

Carrotsticks · Jan 4, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
3arg(z) + arg(-b-az) - arg(-bz-c)??

Hint: Think about the conjugate root theorem.

RealiseNothing · Jan 4, 2013

Re: HSC 2013 4U Marathon

Let me change the question to make the value a bit more specific:

$z^2 + (1+i)z + (1-i)$

Find $arg(\alpha)+arg(\beta)$

HeroicPandas · Jan 4, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Let me change the question to make the value a bit more specific:

$z^2 + (1+i)z + (1-i)$

Find $arg(\alpha)+arg(\beta)$

-pi/4?

HSC 2013 MX2 Marathon (archive) (1 Viewer)

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