HSC 2013 MX2 Marathon (archive) (12 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?)

HSC 2013 MX2 Marathon (archive) (12 Viewers)

This too shall pass

This too shall pass

This too shall pass

same

This too shall pass

same

Member

This too shall pass

This too shall pass

same

.

Cult of Personality

This too shall pass

Cult of Personality

Well-Known Member

This too shall pass

Cult of Personality

This too shall pass

Account Closed

same

Users Who Are Viewing This Thread (Users: 0, Guests: 12)