HSC 2013 MX2 Marathon (archive) (2 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.

HSC 2013 MX2 Marathon (archive) (2 Viewers)

what is that?It is Cowpea

Active Member

Well-Known Member

what is that?It is Cowpea

what is that?It is Cowpea

Well-Known Member

what is that?It is Cowpea

Well-Known Member

what is that?It is Cowpea

Active Member

what is that?It is Cowpea

Active Member

what is that?It is Cowpea

Active Member

what is that?It is Cowpea

Active Member

Active Member

Retired

Retired

Active Member

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