HSC 2013 MX2 Marathon (archive) (10 Viewers)

Sy123 · Dec 4, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Cbb checking haha but that is the correct idea.

As an addendum to your question, which I think I have asked before:

Prove that E has NO real roots if n is even.

I'll add it in the latex so people don't miss it

Sy123 · Dec 5, 2012

Re: HSC 2013 4U Marathon

$$Given $ \\ \\ e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+... \equiv \sum_{r=0}^{\infty} \frac{1}{r!}$

$$Define the nth partial sum $ S_n=\sum_{r=0}^{n} \frac{1}{r!}$

$$i) Explain why $ \ \ \ \frac{1}{k}+\frac{1}{k(k+1)}+ \frac{1}{k(k+1)(k+2)}+...<\frac{1}{k}+\frac{1}{k^2}+\frac{1}{k^3}+...$

$$ii) By constructing an infinite geometric series or otherwise, show that $ \\ \\ 0 < n!(e-S_n) < \frac{1}{n} \ \ \ \ \ \ \fbox{2}$

$$iii) Explain why $ e-S_n $ is NOT an integer$ \ \ \ \ \ \ \fbox{1}$

$iv) Assume that $ e=\frac{p}{q} $ whereby p and q are integers which have no common divisor other than -1 and 1$

$$Prove that e is irrational$ \ \ \ \ \ \ \ \ \fbox{3}$

EDIT: I added another first part to make the question easier. It should be doable now.

Sy123 · Dec 8, 2012

Re: HSC 2013 4U Marathon

$$i) Show that $ z^5-1=(z-1)(z^2-2z\cos \frac{2\pi}{5}+1)(z^2-2z\cos \frac{4\pi}{5}+1) \ \ \ \ \ \fbox{2}$

$$ii) Hence form a quadratic equation with roots $ \ \ \ \cos \frac{2\pi}{5} \ \ \ $ AND $ \ \ \ \cos \frac{4\pi}{5} \ \ \ \ \ \ \fbox{3}$

Difficulty: Moderate

It would also be great if people could rate the difficulty of their own question as I have just done.

Examine · Dec 8, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$i) Factorise among the irrational numbers $ z^5-1 \ \ \ \ \ \fbox{2}$

$$ii) Form a quadratic equation with roots $ \ \ \ \cos \frac{2\pi}{5} \ \ \ $ AND $ \ \ \ \cos \frac{4\pi}{5} \ \ \ \ \ \ \fbox{3}$

Difficulty: Moderate

It would also be great if people could rate the difficulty of their own question as I have just done.

What does it mean by irrational numbers? (like (z-1)(z^4+z^3+z^2+z+1)) ?

Sy123 · Dec 8, 2012

Re: HSC 2013 4U Marathon

Examine said:
What does it mean by irrational numbers? (like (z-1)(z^4+z^3+z^2+z+1)) ?

HINT: Factorise into 1 linear factor, and 2 quadratic factors. The factorisation you have is among the integers, the one you should have among the irrationals which will be in the form of 1 linear and 2 quadratic factors will have irrational co-efficient

EDIT: I changed first one to show that.

Examine · Dec 8, 2012

Re: HSC 2013 4U Marathon

oops pls wait

jyu · Dec 8, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
It would also be great if people could rate the difficulty of their own question as I have just done.

isn't it subjective?

Sy123 · Dec 8, 2012

Re: HSC 2013 4U Marathon

jyu said:
isn't it subjective?

Yeah it is subjective but ask yourself where you would see it in an exam.

If Q1-3 then easy

Q4-6 moderate

Q7-8 hard.

Beyond HSC yet still in syllabus = Extreme (I guess?)

Sy123 · Dec 8, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
$Let $z=(z_1,\ldots,z_n)$ and $w=(w_1,\ldots,w_n)$ be two $n$-tuples of complex numbers. We define:\\ $\langle z,w\rangle:=\sum_{j=1}^n z_j \overline{w_j}.$ \\ I) Prove that $\langle z,w\rangle=\overline{\langle w,z\rangle}$ and $t\langle z,w\rangle=\langle tz,w\rangle=\langle z,\overline{t}w\rangle$ for all complex $t$.\\ II) Explain why $\langle z-tw,z-tw\rangle$ is real and non-negative for all complex $t$.\\ III) By applying the above with \\ \\ $t:=\frac{\langle z,w \rangle}{\langle w,w\rangle}$\\ \\prove that\\ $|\langle z,w\rangle|\leq \left(\sum_{j=1}^n |z_j|^2\right)^{1/2}\left(\sum_{j=1}^n |w_j|^2\right)^{1/2}.$\\ \\IV) A mathematically gifted grasshopper is at the centre of an Argand diagram. Due to a tragic birth defect, he can only make hops of length one, and can only hop in the directions of the $p$-th roots of unity for some fixed positive integer $p>1$.$

$He chooses his hopping direction at random. Prove that his average distance from the origin after $N$ hops is smaller that $\sqrt{N}$.$

Got up to iii, faced a problem:

$\langle z,w \rangle = t \langle w, w \rangle \\ \\ |\langle z,w \rangle|^2=t \langle w, z \rangle \sum_{j=1}^{n} |w_j|^2 \\ \\ \therefore \langle tw,z \rangle \ \ \in \mathbb{R} \\ \\ \therefore tw=z \\ \therefore $EQUALITY, not inequality for the expression we are trying to prove$$

Am I missing something here?

(Also I am very close to solving the last question)

Examine · Dec 8, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$i) Show that $ z^5-1=(z-1)(z^2-2z\cos \frac{2\pi}{5}+1)(z^2-2z\cos \frac{4\pi}{5}+1) \ \ \ \ \ \fbox{2}$

$$ii) Hence form a quadratic equation with roots $ \ \ \ \cos \frac{2\pi}{5} \ \ \ $ AND $ \ \ \ \cos \frac{4\pi}{5} \ \ \ \ \ \ \fbox{3}$

Difficulty: Moderate

It would also be great if people could rate the difficulty of their own question as I have just done.

for the first part

$z^5=1 \\ \\ cis^5\theta=1\\ \\ 5\theta=cis2k\pi \\ \\ \theta=cis\frac{2k\pi}{5}\\ \\ \theta=cis0, cis\frac{2\pi}{5}, cis\frac{-2\pi}{5}, cis\frac{4\pi}{5}, cis\frac{-4\pi}{5},\\ \\ (x-1)(x-cis\frac{2\pi}{5})(x-cis\frac{-2\pi}{5})(x-cis\frac{4\pi}{5})(x-cis\frac{-4\pi}{5})\\ \\ (x-1)(x^2-2cos\frac{2\pi}{5}+1)(x^2-2cos\frac{2\pi}{5}+1)$

twinklegal19 · Dec 8, 2012

Re: HSC 2013 4U Marathon

for your fourth and fifth lines, you can drop off the "cis" on RHS because you are talking about the arguments. Also you used z in the first part and suddenly you switched to x when factorising

Shadowdude · Dec 8, 2012

Re: HSC 2013 4U Marathon

So I was reading some discussion on how this thread was just full of ridiculous questions 0.5% of the population can do, so how about we try something like...

$\textup{Factorise } 1 + z + z^{2} + z^{3} + z^{4} + z^{5} \textup{ into linear and quadratic factors over }\mathbb{C}$

(This is an easy-ish question... if you know "the trick to it")

Sy123 · Dec 8, 2012

Re: HSC 2013 4U Marathon

Shadowdude said:
So I was reading some discussion on how this thread was just full of ridiculous questions 0.5% of the population can do, so how about we try something like...

$\textup{Factorise } 1 + z + z^{2} + z^{3} + z^{4} + z^{5} \textup{ into linear and quadratic factors over }\mathbb{C}$

(This is an easy-ish question... if you know "the trick to it")

$\frac{z^6-1}{z-1}=1+z+z^2+z^3+z^4+z^5 \\ \\ = \frac{(z-1)(z-cis \frac{2\pi}{6})(z-cis\frac{4\pi}{6})(z-cis \frac{6\pi}{6})(z-cis\frac{8\pi}{6})(z-cis\frac{10\pi}{6})}{(z-1)}=(z+1)(z^2-2z\cos \frac{2\pi}{6}+1)(z^2-2z\cos \frac{4\pi}{6}+1)$

Keep in mind my question is still unanswered (the second part) also people should be able to do the e is irrational one for those who know series, and the one on the previous page about E(x)

Shadowdude · Dec 8, 2012

Re: HSC 2013 4U Marathon

Nice.

$\textup{Expand }(4+i)(5+3i)\textup{ and hence show that }\frac{\pi}{4} = \tan^{-1}\left ( \frac{1}{4} \right ) + \tan^{-1}\left (\frac{3}{5} \right )$

bit harder.

deswa1 · Dec 8, 2012

Re: HSC 2013 4U Marathon

Shadowdude said:
Nice.

$\textup{Expand }(4+i)(5+3i)\textup{ and hence show that }\frac{\pi}{4} = \tan^{-1}\left ( \frac{1}{4} \right ) + \tan^{-1}\left (\frac{3}{5} \right )$

bit harder.

Ahh I like these sort of questions because you are using imaginary numbers to prove a 'real' result- I've always found that cool. I'll leave it for a current 4U student to solve it though

Sy123 · Dec 8, 2012

Re: HSC 2013 4U Marathon

Shadowdude said:
Nice.

$\textup{Expand }(4+i)(5+3i)\textup{ and hence show that }\frac{\pi}{4} = \tan^{-1}\left ( \frac{1}{4} \right ) + \tan^{-1}\left (\frac{3}{5} \right )$

bit harder.

$(4+i)(5+3i)=17+17i \\ \\ arg(17+17i)=arg((4+i)(5+3i)) \\ \\ \frac{\pi}{4}=arg(4+i)+arg(5+3i) \\ \\ \therefore \frac{\pi}{4}=\tan^{-1}\frac{1}{4}+\tan^{-1} \frac{3}{5}$

Shadowdude · Dec 8, 2012

Re: HSC 2013 4U Marathon

deswa1 said:
Ahh I like these sort of questions because you are using imaginary numbers to prove a 'real' result- I've always found that cool. I'll leave it for a current 4U student to solve it though

True that. Especially for those who think "ugh complex numbers, why are we using imaginary numbers anyway - what's the point" - it's pretty awesome.

My lecturer for Complex Analysis shared the same feelings you have - and as an avid number theorist - and he gave us a question to do in class - you might want to try this one because it's not strictly in the syllabus, it just uses complex number ideas:

$\textup{Write 20 and 29 as the sum of two squares and hence write their product as the sum of two squares, in two different ways}$

(hint: use complex numbers)

And the difficulty is... quite easy, you just have to know how to use the hint properly.

Sy123 · Dec 8, 2012

Re: HSC 2013 4U Marathon

Shadowdude said:
True that. Especially for those who think "ugh complex numbers, why are we using imaginary numbers anyway - what's the point" - it's pretty awesome.

My lecturer for Complex Analysis shared the same feelings you have - and as an avid number theorist - and he gave us a question to do in class - you might want to try this one because it's not strictly in the syllabus, it just uses complex number ideas:

$\textup{Write 20 and 29 as the sum of two squares and hence write their product as the sum of two squares, in two different ways}$

(hint: use complex numbers)

And the difficulty is... quite easy, you just have to know how to use the hint properly.

$20=4^2+2^2=(4-2i)(4+2i) \ \ \ 29=5^2+2^2=(5-2i)(5+2i)$

$20 \cdot 29 = (4-2i)(5-2i) (4+2i)(5+2i) = (16-18i)(16+18i) = 16^2+18^2 \\ \\ 20 \cdot 29 = (4-2i)(5+2i) (4+2i)(5-2i) = (24-2i) (24+2i)=24^2+2^2$

(still a question unanswered above)

cutemouse · Dec 9, 2012

Re: HSC 2013 4U Marathon

Shadowdude said:
My lecturer for Complex Analysis

Lol P Brown... Did you do it this year?

Examine · Dec 9, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
for your fourth and fifth lines, you can drop off the "cis" on RHS because you are talking about the arguments. Also you used z in the first part and suddenly you switched to x when factorising

~oops~ and ty. (so you don't have to write cis for the 4th and 5th lines?)

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