HSC 2013 MX2 Marathon (archive) (5 Viewers)

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Here's what I did which ended up wrong (I think?)

Since one root is $\alpha$ , the other must be $\frac{1}{\alpha}$

So we get:

$\alpha^2 + A\alpha + 1 = 0$

$\alpha^{-2} + A\alpha^{-1} + 1 = 0$

Using:

$z^n + z^-n = 2cos(n\theta)$ and adding the two equation together we obtain:

$2cos2\theta + 2Acos\theta + 1 = 0$

Using the identity:

$cos2\theta =2cos^2\theta -1$

We get a quadratic in $cos\theta$ which when solved using the quadratic formula gives:

$\theta = cos^{-1}(\frac{A}{2})$ or $\theta = cos^{-1}(\frac{-3A}{4})$

Now the first solution works for all $|A| < 2$ , but the second solution doesn't as $\frac{-3A}{4} > 1$ for certain values of A within the domain of $|A| < 2$

Hence solving for A:

$-1 \leq \frac{-3A}{4} \leq 1$

$\frac{4}{3} \geq A \geq \frac{-4}{3}$

Which is a different result to $|A| < 2$

(note I forgot to add the fact that $|\alpha| = 1$ )

What's wrong with this solution?

2nd last line.

bleakarcher · Dec 27, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
I stated that they were lol (well it was a late edit lol).

Well, by conjugate root theorem, the modulus has to be 1 given the roots are non-real anyway. Say some non-real number a is a root of x^2+Ax+1=0 then conjugate(a) is a root. But 1/a is also a root. => conjugate(a)=1/a <=> lal=1

seanieg89 · Dec 27, 2012

Re: HSC 2013 4U Marathon

oh okay, that makes the question trivial then, roots must be conjugates and $A=2\cos(\theta)$ for some theta in (0,pi). The range of A follows.

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
oh okay, that makes the question trivial then, roots must be conjugates and $A=2\cos(\theta)$ for some theta in (0,pi). The range of A follows.

But how come my solution gives a range of A that is wrong?

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

bleakarcher said:
Well, by conjugate root theorem, the modulus has to be 1 given the roots are non-real anyway. Say some non-real number a is a root of x^2+Ax+1=0 then conjugate(a) is a root. But 1/a is also a root. => conjugate(a)=1/a <=> lal=1

True, I don't even have to state it has a modulus of 1 I guess.

seanieg89 · Dec 27, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
But how come my solution gives a range of A that is wrong?

I think you made the mistake when solving the quadratic.

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
I think you made the mistake when solving the quadratic.

It's probably some small calculation error along the way, because that method should work right?

seanieg89 · Dec 27, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
It's probably some small calculation error along the way, because that method should work right?

Well yeah, but its just a roundabout way of getting to the same A=-2cos(theta).

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Well yeah, but its just a roundabout way of getting to the same A=-2cos(theta).

I wanted to try and make the question so that the polynomial was to the n'th degree, but it seemed to be a lot messier and harder, so I stuck with a quadratic.

bleakarcher · Dec 27, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
oh okay, that makes the question trivial then, roots must be conjugates and $A=2\cos(\theta)$ for some theta in (0,pi). The range of A follows.

sean, one question bro, how do you know (0,pi) is open and not [0,pi]?

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

bleakarcher said:
sean, one question bro, how do you know (0,pi) is open and not [0,pi]?

It can't be 0 because then it would be real, same for pi.

bleakarcher · Dec 27, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
It can't be 0 because then it would be real, same for pi.

ah right, shit.

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

Show that:

$\frac{sin(n\theta) + nsin(\theta)[1-sin(n\theta)]}{2-2cos\theta} = \sum_{k=1}^{n} ksin(k\theta)$

bleakarcher · Dec 27, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Show that:

$\frac{sin(n\theta) + nsin(\theta)[1-sin(n\theta)]}{2-2cos\theta} = \sum_{k=1}^{n} ksin(k\theta)$

The only way I can think of is mathematical induction. Also, when n=1,
LHS=[sin(θ)+sin(θ)[1-sin(θ)]]/[2-2cos(θ)]=sin(θ)[2-sin(θ)]/[2-2cos(θ)]
RHS=sin(θ)
LHS=/=RHS

Maybe your formula is wrong?

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

bleakarcher said:
The only way I can think of is mathematical induction. Also, when n=1,
LHS=[sin(θ)+sin(θ)[1-sin(θ)]]/[2-2cos(θ)]=sin(θ)[2-sin(θ)]/[2-2cos(θ)]
RHS=sin(θ)
LHS=/=RHS

Maybe your formula is wrong?

Probably, it worked when I put in $n=5$ and $\theta =30$ so i assumed it was right for any values (cos what are the odds of it working for a random value I choose lol).

edit: ok I have no idea where I went wrong, and the working out is too annoying to check, basically what I tried to do was find a closed form for:

$\sum_{k=1}^{n} ksin(k\theta)$

Some one else might be able to get it.

bleakarcher · Dec 27, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Probably, it worked when I put in $n=5$ and $\theta =30$ so i assumed it was right for any values (cos what are the odds of it working for a random value I choose lol).

edit: ok I have no idea where I went wrong, and the workin out is too annoying to check, basically what I tried to do was find a closed form for:

$\sum_{k=1}^{n} ksin(k\theta)$

Some one else might be able to get it.

lol, it always comes to probability with you, doesn't it haha..

bleakarcher · Dec 27, 2012

Re: HSC 2013 4U Marathon

I'm going to keep thinking about it. Got nothing so far.

Carrotsticks · Dec 27, 2012

Re: HSC 2013 4U Marathon

$\\ $A way of finding $ \sum_{k=0}^{n} k \cos (k \theta) $ is to consider the sum $ \sum_{k=0}^{n} \cos (k \theta) + \sum_{k=1}^{n} \cos (k \theta) + ... + \sum_{k=n-1}^{n} \cos (k \theta ) + \sum_{k=n}^{n} \cos (k \theta) = \sum_{j=0}^{n} \sum_{k=j}^{n} \cos (k \theta) \\\\ $And each of these individual sums can be evaluated by considering the real part of $ 1+z+z^2+...+z^n = \frac{z^{n+1}-1}{z-1} $ then $ z+z^2+z^3+...+z^n = z \times \frac{z^{n}-1}{z-1} $ or more generally $ z^k + z^{k+1}+ ... +z^n = z^k \times \frac{z^{n-k+1}}{z-1} $ where $ k \leq n$

Carrotsticks · Dec 27, 2012

Re: HSC 2013 4U Marathon

$\\ $Another way is to consider the series $ 1+z+z^2+z^3+...+z^n = \frac{z^{n+1}-1}{z-1} $ and then differentiate both sides, yielding $ 1+2z+3z^2+4z^3+...+nz^{n-1} = ... $ and then multiply both sides by $ z $ to give $ z+2z^2+3z^3+...+nz^n = ... $ and then let $ z=cos \theta + i \sin \theta$

bleakarcher · Dec 27, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
$\\ $A way of finding $ \sum_{k=0}^{n} k \cos (k \theta) $ is to consider the sum $ \sum_{k=0}^{n} \cos (k \theta) + \sum_{k=1}^{n} \cos (k \theta) + ... + \sum_{k=n-1}^{n} \cos (k \theta ) + \sum_{k=n}^{n} \cos (k \theta) = \sum_{j=0}^{n} \sum_{k=j}^{n} \cos (k \theta) \\\\ $And each of these individual sums can be evaluated by considering the real part of $ 1+z+z^2+...+z^n = \frac{z^{n+1}-1}{z-1} $ then $ z+z^2+z^3+...+z^n = z \times \frac{z^{n}-1}{z-1} $ or more generally $ z^k + z^{k+1}+ ... +z^n = z^k \times \frac{z^{n-k+1}}{z-1} $ where $ k \leq n$

Nice, looks like it's going to be pretty tedious though.

Second method looks much better to me.

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