HSC 2013 MX2 Marathon (archive) (1 Viewer)

seanieg89 · Nov 24, 2012

Re: HSC 2013 4U Marathon

$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 $\sqrt{N}$.$

RealiseNothing · Nov 24, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$i) $Given that $ \ \ z=\cos \theta + i\sin \theta \ \ $Show that$ \ \ z^n + z^{-n} \ \ $ is real for positive integers n$$

$ii) $Hence or otherwise solve the equation$ \ \ \ 2z^4+3z^3+5z^2+3z+2=0$

For (i), after using De Moivre's Theorem we obtain:

$2cos(n\theta)$

Hence it is real etc.

For (ii):

$2z^4 + 3z^3+ 5z^2 + 3z + 2 = 0$

$2z^2(z^2 + z^{-2}) + 3z^2(z+z^{-1}) + 5z^2 = 0$

Taking $z^2 = 0$ as a solution.

$2(2cos2\theta) + 3(2cos\theta) + 5 = 0$

Using $cos2\theta = 2cos^2\theta - 1$

$4cos^2\theta - 2 + 6cos\theta + 5 = 0$

$4cos^2\theta + 6cos\theta + 3 = 0$

I cbf'd solving the quadratic, but you get the answer from it.

RealiseNothing · Nov 25, 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 $\langle z,w\rangle=\overline{\langle w,z\rangle}$

Going to attempt this in parts:

i) $\overline{(w,z)} = \overline{\sum_{j=1}^{n} w^j\overline{z_j}}$

$\overline{w_1\overline{z_1} + w_2\overline{z_2} + ... + w_n\overline{z_n}}$

$\overline{w_1\overline{z_1}} + \overline{w_2\overline{z_2}} + ... + \overline{w_n\overline{z_n}}$

$z_1\overline{w_1} + z_2\overline{w_2} + ... + z_n\overline{w_n}$

$\sum_{j=1}^{n} z_j\overline{w_j}$

$(z,w) = \overline{(w,z)}$

HeroicPandas · Nov 25, 2012

Re: HSC 2013 4U Marathon

$Solve for x: 2^x = \frac{1}{4}$

U cannot use LOGS

RealiseNothing · Nov 25, 2012

Re: HSC 2013 4U Marathon

HeroicPandas said:
$Solve for x: 2^x = \frac{1}{4}$

U cannot use LOGS

-2

this is a 2U question.

HeroicPandas · Nov 25, 2012

Re: HSC 2013 4U Marathon

Fus Ro Dah said:
Construct vectors A, B, B*, AB* and AB*-1. We immediately observe two congruent triangles, where matching sides include |A-B| and |AB*-1| and hence the result.

Show me

HeroicPandas · Nov 25, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
-2

this is a 2U question.

Yes i know, just a trick question...

twinklegal19 · Nov 25, 2012

Re: HSC 2013 4U Marathon

HeroicPandas said:
Yes i know, just a trick question...

no it isn't

Sy123 · Nov 25, 2012

Re: HSC 2013 4U Marathon

The Fibonacci sequence is a sequence of numbers whereby each term of the sequence is the sum of the previous two terms, the sequence starts off with two terms

1, 1

Then it goes on following the Fibonacci rule:

1, 1, 2, 3, 5, 8, 13, 21 ...

Notice how each term is the sum of the previous 2 terms.

$$If the kth term of the Fibonacci sequence is $ F_k$

$$Prove using mathematical induction or derive such otherwise that$ \\ \\ \sum_{k=1}^n F_k^2 = F_n F_{n+1}$

Carrotsticks · Nov 25, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
The Fibonacci sequence is a sequence of numbers whereby each term of the sequence is the sum of the previous two terms, the sequence starts off with two terms

1, 1

Then it goes on following the Fibonacci rule:

1, 1, 2, 3, 5, 8, 13, 21 ...

Notice how each term is the sum of the previous 2 terms.

$$If the kth term of the Fibonacci sequence is $ F_k$

$$Prove using mathematical induction or derive such otherwise that$ \\ \\ \sum_{k=1}^n F_k^2 = F_n F_{n+1}$

lol this question nearly made the BOS Trials.

Sy123 · Nov 25, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
lol this question nearly made the BOS Trials.

The proof by induction is quite easy, do you mean the derivation of the result?

Carrotsticks · Nov 25, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
The proof by induction is quite easy, do you mean the derivation of the result?

The induction method, then I extended the question to a limit problem.

seanieg89 · Nov 27, 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 $\sqrt{N}$.$

Bump.

Also:

Gamblor takes $10 on his visit to the casino. He plays a perfectly fair game: he has a 50% chance of winning a dollar and a 50% chance of losing a dollar in every round. He stops playing when he either reaches double or half of his starting bankroll. What is the probability of him ending up with $20?

Sy123 · Nov 27, 2012

Re: HSC 2013 4U Marathon

$$Define the function $ f(x)=(\cos x + i\sin x) \cdot e^{-ix}$

$$By considering $ f'(x) $ Prove the famous result $ \\ \\ e^{i\pi}=-1$

(Do not worry about whether it is actually possible to take exponentials of complex numbers)

Trebla · Nov 27, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$Define the function $ f(x)=(\cos x + i\sin x) \cdot e^{-ix}$

$$By considering $ f'(x) $ Prove the famous result $ \\ \\ e^{i\pi}=-1$

(Do not worry about whether it is actually possible to take exponentials of complex numbers)

This is going beyond syllabus now

Trebla · Nov 27, 2012

Re: HSC 2013 4U Marathon

Show that for some integer k and constant n:

$\displaystyle\int_0^{\pi}\left(\sin \dfrac{x}{2}\right)^n \cos kx\,dx = \displaystyle\int_0^{\pi}(\sin x)^n \cos 2kx\,dx$

cutemouse · Nov 28, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$Define the function $ f(x)=(\cos x + i\sin x) \cdot e^{-ix}$

$$By considering $ f'(x) $ Prove the famous result $ \\ \\ e^{i\pi}=-1$

(Do not worry about whether it is actually possible to take exponentials of complex numbers)

Something interesting to note that the complex exponential function is peroidic with period 2pi i

It does make the complex logarithm function a bit complicated though.

seanieg89 · Nov 28, 2012

Re: HSC 2013 4U Marathon

cutemouse said:
Something interesting to note that the complex exponential function is peroidic with period 2pi i

It does make the complex logarithm function a bit complicated though.

Technically, the logarithm MULTIfunction unless we choose a branch.

cutemouse · Nov 28, 2012

Re: HSC 2013 4U Marathon

Well, usually one chooses a branch cut according to what needs to be evaluated...

seanieg89 · Nov 28, 2012

Re: HSC 2013 4U Marathon

cutemouse said:
Well, usually one chooses a branch cut according to what needs to be evaluated...

I suppose, but I think it is somewhat limiting to only think of it as a function...much of complex analysis is based on the theory of multifunctions. Anyway, very off-topic.

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