HSC 2015 MX2 Marathon (archive) (1 Viewer)

RealiseNothing · Jan 3, 2015

Re: HSC 2015 4U Marathon

My latex is stuffing up in two places above, not really sure why.

Carrotsticks · Jan 3, 2015

Re: HSC 2015 4U Marathon

RealiseNothing said:
My latex is stuffing up in two places above, not really sure why.

Fixed.

You just needed some spaces.

Also, although not necessary, sin should be \sin to avoid it being italicised.

t_davis9 · Jan 3, 2015

Re: HSC 2015 4U Marathon

Nice work all,
Next Question
http://imgur.com/ZLWDxr8
$1, \, \omega \,\,and \,\, \omega^2 \, are \, the \, cube \, roots \, of \, unity.\\ \, State \, the\,values\,of\,\omega^3\,and\,\,1+\omega+ \omega^2 \, \\ Hence\,show\,that:\\(1+\omega^2)^1^2\,=\,1\\and\\(1-\omega)(1-\omega^2)(1-\omega^4)(1-\omega^5)(1-\omega^7)(1-\omega^8)\,=\,27$
It's quite simple, try to do part 2 with as little expanding as possible

porcupinetree · Jan 3, 2015

Re: HSC 2015 4U Marathon

t_davis9 said:
Nice work all,
Next Question
http://imgur.com/ZLWDxr8
$1, \, \omega \,\,and \,\, \omega^2 \, are \, the \, cube \, roots \, of \, unit. \, State \, the\,values\,of\,\omega^3\,and\,1\,+\,\omega\,+\,\omega^2\,\\ Hence\,show\,that:\\(1+\omega^2)^1^2\,=\,1\\and\\(1-\omega)(1-\omega^2)(1-\omega^4)(1-\omega^5)(1-\omega^7)(1-\omega^8)\,=\,27$
It's quite simple, try to do part 2 with as little expanding as possible

http://imgur.com/FZnq83p

Kaido · Jan 3, 2015

Re: HSC 2015 4U Marathon

t_davis9 said:
It's quite simple, try to do part 2 with as little expanding as possible

Just...LOL

When i first did these types of question, I have to say, I was a bit retarded...

Sy123 · Jan 3, 2015

Re: HSC 2015 4U Marathon

$\\ $Find the polynomial of lowest degree, with rational coefficients, that has the root$ \\ \\ x = \sqrt[3]{\frac{-1 + \sqrt{5}}{2}} - \sqrt[3]{\frac{1 + \sqrt{5}}{2}}$

t_davis9 · Jan 3, 2015

Re: HSC 2015 4U Marathon

Kaido said:
Just...LOL

When i first did these types of question, I have to say, I was a bit retarded...

Haha, you're not alone - I've spent way too long expanding some of those

porcupinetree said:
http://imgur.com/FZnq83p

Nice work, you too a different approach to my attempt for both parts.

PhysicsMaths · Jan 6, 2015

Re: HSC 2015 4U Marathon

t_davis9 said:
Haha, you're not alone - I've spent way too long expanding some of those

Nice work, you too a different approach to my attempt for both parts.

I had the same method as porcupine for the 2nd part. Mind sharing your approach?

t_davis9 · Jan 7, 2015

Re: HSC 2015 4U Marathon

I don't actually have my original approach as our test paper is still at school.

Although, looking back at porcupine's answer, mine was somewhat similar, but I expanded at one stage.
I'll have a look when school goes back

t_davis9 · Jan 7, 2015

Re: HSC 2015 4U Marathon

I don't actually have my original approach as our test paper is still at school.

Although, looking back at porcupine's answer, mine was somewhat similar, but I expanded at one stage.
I'll have a look when school goes back

Kaido · Jan 8, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Find the polynomial of lowest degree, with rational coefficients, that has the root$ \\ \\ x = \sqrt[3]{\frac{-1 + \sqrt{5}}{2}} - \sqrt[3]{\frac{1 + \sqrt{5}}{2}}$

Lol just realised there's an unanswered question.
Btw Sy, is it "THE ROOT" or should it be "A ROOT"
Because THE root would not have rational coefficients (because it would be linear)

InteGrand · Jan 8, 2015

Re: HSC 2015 4U Marathon

Kaido said:
Lol just realised there's an unanswered question.
Btw Sy, is it "THE ROOT" or should it be "A ROOT"
Because THE root would not have rational coefficients (because it would be linear)

The polynomial just needs to have that as a zero (it can have others too).

Kaido · Jan 8, 2015

Re: HSC 2015 4U Marathon

Right...
From observation, something like x^3+3x+1=0

Holy crap, that question was from 5 days ago, how come noone noticed

InteGrand · Jan 8, 2015

Re: HSC 2015 4U Marathon

Kaido said:
Right...
From observation, something like x^3+3x+1=0

Holy crap, that question was from 5 days ago, how come noone noticed

Proof that it's the lowest degree one?

PhysicsMaths · Jan 8, 2015

Re: HSC 2015 4U Marathon

Kaido said:
Right...
From observation, something like x^3+3x+1=0

Holy crap, that question was from 5 days ago, how come noone noticed

Some insight into how you got to there? o.o

InteGrand · Jan 8, 2015

Re: HSC 2015 4U Marathon

PhysicsMaths said:
Some insight into how you got to there? o.o

Sy123 said:
$\\ $Find the polynomial of lowest degree, with rational coefficients, that has the root$ \\ \\ x = \sqrt[3]{\frac{-1 + \sqrt{5}}{2}} - \sqrt[3]{\frac{1 + \sqrt{5}}{2}}$

For convenience, we let:

$\bullet \text{ } \alpha =\sqrt[3]{\frac{-1+\sqrt{5}}{2}}-\sqrt[3]{\frac{1+\sqrt{5}}{2}}$

$\bullet \text{ } \rho = \frac{-1+\sqrt{5}}{2}$

$\bullet \text{ } \varphi = \frac{1+\sqrt{5}}{2}$

so that $\alpha = \sqrt[3]{\rho}-\sqrt[3]{\varphi}$ .

Note that $\rho - \varphi=\frac{-1+\sqrt{5}-1-\sqrt{5}}{2}=\frac{-2}{2}=-1$

and $\rho \varphi = \frac{-1+\sqrt{5}}{2}\cdot\frac{1+\sqrt{5}}{2}=\frac{5-1}{4}=1$ .

We first find a cubic polynomial with rational coefficients that has $\alpha$ as a zero. We then show that no polynomial with rational coefficients of degree 1 or 2 can have $\alpha$ as a zero.

Using the identity $(A-B)^3 = A^3 - 3AB(A-B)-B^3$ , with $A=\sqrt[3]{\rho}, B=\sqrt[3]{\varphi}$ (so $A - B = \alpha$ ), we have

$\alpha ^3 = (\sqrt[3]{\rho}-\sqrt[3]{\varphi})^3$

$= \rho - 3\sqrt[3]{\rho}\sqrt[3]{\varphi}\alpha -\varphi$

$=-1-3\sqrt[3]{\rho \varphi}\cdot \alpha$

$=-1-3\sqrt[3]{1}\cdot \alpha$

$=-1-3\sqrt[3]{\rho \varphi}\cdot \alpha$

$=-1-3 \alpha$ ,

$\Rightarrow \alpha \text{ satisfies the equation } x^3 = -1-3x$

$\Leftrightarrow x^3 +3x +1 = 0$ .

So the required polynomial is $P(x)=x^3 +3x + 1$ , assuming that no polynomial with rational coefficients of degree 1 or 2 can have $\alpha$ as a zero, which we show below.

Clearly, $\alpha$ cannot be the root of a degree 1 polynomial equation $ax+b=0, \text{ }a,b\in \mathbb{Q}, \text{ }a \neq 0$ , because otherwise we would have

$a\alpha+b=0, \text{ }a,b\in \mathbb{Q}$

$\Rightarrow \alpha = -\frac{b}{a}$ , a contradiction as the LHS is irrational whilst the RHS is rational.

Now we show that we cannot have $\alpha$ be the root of a degree 2 polynomial equation with rational coefficients. Assume by way of contradiction that we could, so that $\alpha$ satisfies the equation $x^2+ax+b=0, \text{ for some }a,b\in \mathbb{Q}$ . Let $\beta$ be the other root of this equation.

Then by product of roots, $\alpha \beta = b \Rightarrow \beta = \frac{b}{\alpha}$ .

We now rationalise the denominator:

$\beta = \frac{b}{\alpha}$

$=b\cdot\frac{1}{\sqrt[3]{\rho}-\sqrt[3]{\varphi}}\cdot\frac{\sqrt[3]{\rho ^2}+\sqrt[3]{\rho}\cdot\sqrt[3]{\varphi}+\sqrt[3]{\varphi ^2}}{\sqrt[3]{\rho ^2}+\sqrt[3]{\rho}\cdot\sqrt[3]{\varphi}+\sqrt[3]{\varphi ^2}}$

$=b\cdot\frac{\sqrt[3]{\rho^2}+1+\sqrt[3]{\varphi^2}}{\rho - \varphi}$ , since $X^3 - Y^3 = (X-Y)(X^2 + XY + Y^2)$ and $\rho \varphi = 1$

$\Rightarrow \beta = -b(\sqrt[3]{\rho^2}+1+\sqrt[3]{\varphi^2})$ , as $\rho - \varphi = -1$

$\Rightarrow \beta = -b(\sqrt[3]{\rho^2}+1+\sqrt[3]{\varphi^2})$

$\Rightarrow \beta = -b\left ( \sqrt[3]{\frac{3-\sqrt{5}}{2}}+1+\sqrt[3]{\frac{3+\sqrt{5}}{2}} \right )$ .

Now, by sum of roots, $\alpha + \beta = -a$

$\Rightarrow \sqrt[3]{\frac{-1+\sqrt{5}}{2}} - \sqrt[3]{\frac{1+\sqrt{5}}{2}} -b\left ( \sqrt[3]{\frac{3-\sqrt{5}}{2}}+1+\sqrt[3]{\frac{3+\sqrt{5}}{2}} \right )=-a$ , a contradiction, as the RHS is rational but the LHS is irrational for any rational b.

(I'm not entirely sure how to prove that last line though)

Kaido · Jan 8, 2015

Re: HSC 2015 4U Marathon

I question whether this guy is really a 2014er. ^

Kaido · Jan 9, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
Proof that it's the lowest degree one?

Sadly, I can't do these types of proofs yet. But after reading your solution, I think I get the general concept of contradiction

Btw Sy, could you show us your solution.

Sy123 · Jan 9, 2015

Re: HSC 2015 4U Marathon

Kaido said:
Sadly, I can't do these types of proofs yet. But after reading your solution, I think I get the general concept of contradiction

Btw Sy, could you show us your solution.

Pretty much exactly the same as InteGrand's one

dan964 · Jan 9, 2015

Re: HSC 2015 4U Marathon

@kaido proof by contradiction is
to take an assumption and take it to its logical conclusion (mathematically) and then show how the original assumption cannot be true, because you end up with a contradiction. You can use it to prove sqrt 2 is irrational.

then this is coming from someone whose strength includes mathematical induction (which we learnt in second lesson in yr 11 - because our teacher taught it then)

HSC 2015 MX2 Marathon (archive) (1 Viewer)

what is that?It is Cowpea

Retired

New Member

not actually a porcupine

be.

This too shall pass

New Member

Active Member

New Member

New Member

be.

Well-Known Member

be.

Well-Known Member

Active Member

Well-Known Member

be.

be.

This too shall pass

what

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