HSC 2013 MX2 Marathon (archive) (7 Viewers)

SpiralFlex · Nov 16, 2012

Re: HSC 2013 4U Marathon

Hey math all nighter? Oh wai- you guys have school tomorrow

whilst I go shopping.

Carrotsticks · Nov 16, 2012

Re: HSC 2013 4U Marathon

Writing BOS 2013 Trials now haha. Just made a really sweet Q15.

As for a problem...

$\\ $Let $ m $ be a positive odd integer $ > 3. $ Prove that $ \frac{2^{2m}+1}{5} $ is a composite integer.$$

cutemouse · Nov 16, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
Writing BOS 2013 Trials now haha. Just made a really sweet Q15.

Are you using LaTeX?

seanieg89 · Nov 16, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
Writing BOS 2013 Trials now haha. Just made a really sweet Q15.

As for a problem...

$\\ $Let $ m $ be a positive odd integer $ > 3. $ Prove that $ \frac{2^{2m}+1}{5} $ is a composite integer.$$

$Suppose to the contrary that $1+4\cdot 4^{2k}=5p$ for some positive integer $k > 1$ where $p$ is prime. Then\\ \\ $5p=(1+2^{k+1}+2^{2k+1})(1-2^{k+1}+2^{2k+1})$ by difference of squares. \\ \\But for $k>1$ each of these factors is greater than $5$. As the only way to express $5p$ as the product of two nontrivial factors is $5\cdot p$, this is a contradiction. $\box$$

Sy123 · Nov 16, 2012

Re: HSC 2013 4U Marathon

SpiralFlex said:
$$(a) Prove$\;\frac{1+\cos \theta+i\sin \theta}{1+\cos \theta-i\sin \theta}=i\cot \frac{\theta}{2}$

$\frac{1+\cos \theta +i\sin \theta}{1+\cos\theta -i\sin\theta}=\frac{(1+\cos \theta + i\sin \theta)^2}{(1+\cos \theta)^2+\sin^2 \theta } \\ \\ =\frac{1+2\cos \theta +\cos^2 \theta -\sin^2\theta+2i(\sin \theta(1+\cos \theta)}{2+2\cos \theta}=\frac{2\cos \theta(1+\cos \theta)+i2\sin \theta(1+\cos \theta)}{2(1+\cos \theta} \\ \\ =\cos \theta + i \sin \theta???$

bobmcbob365 · Nov 16, 2012

Re: HSC 2013 4U Marathon

I'd assume it's meant to be 1-cos(theta)-isin(theta) for the denominator. In which case, I think using double angle formulae will work out.

Trebla · Nov 16, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
Writing BOS 2013 Trials now haha. Just made a really sweet Q15.

As for a problem...

$\\ $Let $ m $ be a positive odd integer $ > 3. $ Prove that $ \frac{2^{2m}+1}{5} $ is a composite integer.$$

lol already?

twinklegal19 · Nov 16, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$\frac{1+\cos \theta +i\sin \theta}{1+\cos\theta -i\sin\theta}=\frac{(1+\cos \theta + i\sin \theta)^2}{(1+\cos \theta)^2+\sin^2 \theta } \\ \\ =\frac{1+2\cos \theta +\cos^2 \theta -\sin^2\theta+2i(\sin \theta(1+\cos \theta)}{2+2\cos \theta}=\frac{2\cos \theta(1+\cos \theta)+i2\sin \theta(1+\cos \theta)}{2(1+\cos \theta} \\ \\ =\cos \theta + i \sin \theta???$

yeah I got the same

http://imgur.com/OxbXq

twinklegal19 · Nov 16, 2012

Re: HSC 2013 4U Marathon

bobmcbob365 said:
I'd assume it's meant to be 1-cos(theta)-isin(theta) for the denominator. In which case, I think using double angle formulae will work out.

yay it does

http://i.imgur.com/cyIfH.jpg

twinklegal19 · Nov 16, 2012

Re: HSC 2013 4U Marathon

Strong triple post I know, but

$$Let $z=\cos{\frac{\pi}{n}}+i\sin{\frac{\pi}{n}},$ where $n$ is a positive integer. Show that$\\$i) $1+z+z^2+...+z^{2n-1}=0\\$ii) $1+z+z^2+...+z^{n-1}=1+i\cot{\frac{\pi}{n}}$

SpiralFlex · Nov 16, 2012

Re: HSC 2013 4U Marathon

Sorry guys, change the bottom fraction to all minuses. I mis-typed it.

RealiseNothing · Nov 16, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
Strong triple post I know, but

$$Let $z=\cos{\frac{\pi}{n}}+i\sin{\frac{\pi}{n}},$ where $n$ is a positive integer. Show that$\\$i) $1+z+z^2+...+z^{2n-1}=0\\$ii) $1+z+z^2+...+z^{n-1}=1+i\cot{\frac{\pi}{n}}$

i) $1 + z + z^2 + ... + z^{2n-1} = \frac{z^{2n}-1}{z-1}$

Let:

$cos\frac{\pi}{n} + isin\frac{\pi}{n} = cis\frac{\pi}{n}$

Substituting this in gives:

$\frac{(cis\frac{\pi}{n})^{2n} - 1}{cis\frac{\pi}{n} - 1}$

By De Moivre's Theorem:

$\frac{cis\frac{2n\pi}{n} - 1}{cis\frac{\pi}{n} - 1}$

Now note that:

$cis\frac{2n\pi}{n} = cis2\pi = 1$

Hence:

$\frac{cis\frac{2n\pi}{n} - 1}{cis\frac{\pi}{n} - 1} = \frac{1-1}{cis\frac{\pi}{n} - 1} = 0$

As required.

deswa · Nov 16, 2012

Re: HSC 2013 4U Marathon

^Oh I did it a different way. Recognise that z^2n=1 (By De Moivres).
So z^2n-1=0 and therefore:
(z-1)(z^2n-1+...+z^2+z+1)=0 upon factorising and the result follows

Sy123 · Nov 17, 2012

Re: HSC 2013 4U Marathon

SpiralFlex said:
$(b) Let$\;n \;$be a positive integer. Show that the roots of the equation$

$(z-1)^n+(z+1)^n=0......(*)$

$$can be written as$\;ia_{k}, \;$where$\; a_{k}\in \mathbb{R} , k=0,1....,n-1$

$(z-1)^n+(z+1)^n=0 \\ \\ 1+(\frac{z+1}{z-1})^n=0 \\ \\ 1+(-i\cot \frac{\theta}{2})^n=0?? $ (not purely imaginary?$)$

RealiseNothing · Nov 18, 2012

Re: HSC 2013 4U Marathon

SpiralFlex said:
$(b) Let$\;n \;$be a positive integer. Show that the roots of the equation$

$(z-1)^n+(z+1)^n=0......(*)$

$$can be written as$\;ia_{k}, \;$where$\; a_{k}\in \mathbb{R} , k=0,1....,n-1$

$(z+1)^5 + (z-1)^5 =0$

$z^5 + 5z^4 + 10z^3 + 10z^2 + 5z + 1 + z^5 - 5z^4 + 10z^3 - 10z^2 + 5z - 1 = 0$

$2z^5 + 20z^3 +10z = 0$

$z^4 + 10z^2 + 5 = 0$

$\sqrt{10^2 - 4(5)} = \sqrt{100 - 20} > 0$

???

SpiralFlex · Nov 18, 2012

Re: HSC 2013 4U Marathon

I'll check what's wrong tomorrow. Sorry about that.

SpiralFlex · Nov 18, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
$(z+1)^5 + (z-1)^5 =0$

$z^5 + 5z^4 + 10z^3 + 10z^2 + 5z + 1 + z^5 - 5z^4 + 10z^3 - 10z^2 + 5z - 1 = 0$

$2z^5 + 20z^3 +10z = 0$

$z^4 + 10z^2 + 5 = 0$

$\sqrt{10^2 - 4(5)} = \sqrt{100 - 20} > 0$

???

What? It's a "n"

RealiseNothing · Nov 18, 2012

Re: HSC 2013 4U Marathon

SpiralFlex said:
What? It's a "n"

"Let 'n' be a positive integer."

It should work for 5 then, but it doesn't.

seanieg89 · Nov 18, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
"Let 'n' be a positive integer."

It should work for 5 then, but it doesn't.

It does work for 5. Elaborate on your working if you think otherwise, I don't see how you have disproven anything.

RealiseNothing · Nov 18, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
It does work for 5. Elaborate on your working if you think otherwise, I don't see how you have disproven anything.

Ok yes, I see it now, I assumed something when I probably shouldn't have.

HSC 2013 MX2 Marathon (archive) (7 Viewers)

Well-Known Member

Retired

Account Closed

Well-Known Member

This too shall pass

Member

Administrator

.

.

.

Well-Known Member

what is that?It is Cowpea

Member

This too shall pass

what is that?It is Cowpea

Well-Known Member

Well-Known Member

what is that?It is Cowpea

Well-Known Member

what is that?It is Cowpea

Users Who Are Viewing This Thread (Users: 0, Guests: 7)