HSC 2013 MX2 Marathon (archive) (1 Viewer)

Lieutenant_21 · Feb 4, 2013

Re: HSC 2013 4U Marathon

Hey Sy123, many of your questions are unsolved so is it possible for you to post a solution when no one can solve the question?
I guess Latex is too slow and inefficient for many people (including myself) so they just use pen and paper. Nice questions though! Keep posting

Sy123 · Feb 4, 2013

Re: HSC 2013 4U Marathon

William Sidis said:
Hey Sy123, many of your questions are unsolved so is it possible for you to post a solution when no one can solve the question?
I guess Latex is too slow and inefficient for many people (including myself) so they just use pen and paper. Nice questions though! Keep posting

If there is enough demand for it sure, definitely

twinklegal19 · Feb 4, 2013

Sy123 said:
If there is enough demand for it sure, definitely

*adds to demand*

Sy123 · Feb 4, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
*adds to demand*

Haha, I will post solutions maybe 1-2 nights after I post them then.
So I will do the one in the 3U thread now

twinklegal19 · Feb 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Haha, I will post solutions maybe 1-2 nights after I post them then.
So I will do the one in the 3U thread now

yay

asianese · Feb 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$A sequence is defined recursively as follows$

$F_0 = 0 \ \ F_1=1$

$F_n=F_{n-1}+F_{n-2} \ \ \ n \geq 2$

$$i) Write down the first 7 numbers of this sequence$ \ \ \ \fbox{1}$

$ii) Consider a power series in$ \ z \ $whereby the co-efficients are the numbers of the defined sequence. That is$

$F(z)=\sum_{k=0}^{\infty} F_k z^k \equiv z+z^2+2z^3+3z^4+5z^5+8z^6+ \dots$

$$Show that$ \ \ \ F(z)=z+zF(z)+z^2F(z) \ \ \ \fbox{2}$

$$iii) Hence show that$ \ \ \ F(z) = \frac{-z}{z^2+z+1} \ \ \ \ \fbox{1}$

$iv) Find$ \ \ A, \ B, \ \alpha, \ \beta \ \ $within the below expression$

$F(z)=\frac{A}{1-\frac{z}{\alpha}} + \frac{B}{1-\frac{z}{\beta}}$

$$Where$ \ \ A \ \ $and$ \ \ B \ \ $are independent of$ \ \ z \\ \\ $ and$ \ \ \alpha, \ \beta \ \ $are roots of$ \ \ \ z^2+z+1 \ \ \ \ \fbox{2}$

$v) Hence prove that$

$F_k = \frac{1}{\sqrt{5}} \left ( \left(\frac{1+\sqrt{5}}{2} \right )^k - \left (\frac{1-\sqrt{5}}{2} \right )^k \right ) \ \ \ \fbox{2}$

Lol Syd grammar. It's a good question

Sy123 · Feb 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$i) Show that$

$1+2z+3z^2+\dots+nz^{n-1} = \frac{nz^{n+1}-(n+1)z^n +1}{(1-z)^2} \ \ \ |z|<1 \ \ \ \fbox{2}$

$$ii) Hence find in terms of $ \ z \ \ \ \ \sum_{k=1}^{\infty} kz^k \ \ \ \fbox{1}$

$iii) The Triangular numbers are a sequence of numbers defined by the recursive relations$

$T_1=1 \ \ \ T_k=k+T_{k-1}$

$Define the power series$ \ \ T(z) \ \ $, where the co-efficents are the triangular numbers$

$T(z) = \sum_{k=1}^{\infty} T_k z^k \equiv z + 3z^2+6z^4+10z^3+...$

$By using the recursive definition of$ \ \ T_k \ \ $ and using part ii, prove that$

$T(z) = \frac{z}{(1-z)^3} \ \ \ \ \fbox{4}$

$i) Consider the geometric series$

$1+z+z^2+ \dots + z^{n} = \frac{1-z^{n+1}}{1-z}$

Differentiate: (skipping the grunt work)

$1+2z+3z^2+ \dots + nz^{n-1} = \frac{nz^{n+1}-(n+1)z^n+1}{(1-z)^2}$

$$ii) Let n go to infinity, we are already given that$ \ |z| < 1$

So therefore anything z^n or something will converge to zero (even with the (n+1) coefficient)
After we do that we also multiply by z, so to get:

$\therefore \ \ \sum_{k=1}^{\infty} = \frac{z}{(1-z)^2}$

$$iii)$ \ \ T(z) = z + \sum_{k=2}^{\infty} T_k z^k = z+ \sum_{k=2}^{\infty} kz^k + \sum_{k=2}^{\infty} T_{k-1}z^k$

via the recurrence relation, then we do this:

$T(z) = \sum_{k=1}^{\infty} kz^k + z \sum_{k=2}^{\infty} T_{k-1}z^{k-1}$

$T(z) = \frac{z}{(1-z)^2} + zT(z)$

$\therefore \ \ T(z) = \frac{z}{(1-z)^3}$

twinklegal19 said:
yay

hehehe not sure how they correlate but ok

Sy123 · Feb 4, 2013

Re: HSC 2013 4U Marathon

asianese said:
Lol Syd grammar. It's a good question

The Sydney Grammar one was about the Chebyshev polynomials

I modified their method to find an expression for Fibonacci.

EDIT: I would do the same for the Triangular number question but there is no point considering we already know that T_k=k/2 (k+1)

asianese · Feb 4, 2013

Re: HSC 2013 4U Marathon

Oh haha. I couldn't remember. Good job on modifying it!

Sy123 · Feb 4, 2013

Re: HSC 2013 4U Marathon

There is a typo in my question, where it used to be:

$\frac{-z}{z^2+z+1}$

should be:

$\frac{-z}{z^2+z-1}$

sorry for inconveniences.

cutemouse · Feb 4, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
yay

Lol, a bit of 1st year micro economics...

Lieutenant_21 · Feb 4, 2013

Re: HSC 2013 4U Marathon

$Prove that for any integer$ $n>1, \frac{cos2\pi }{2^{n}-1}.\frac{cos4\pi }{2^{n}-1}...cos\frac{2^{n}\pi }{2^{n}-1}=\frac{1}{2^{n}}$

Sy123 · Feb 4, 2013

Re: HSC 2013 4U Marathon

William Sidis said:
$Prove that for any integer$ $n>1, \frac{cos2\pi }{2^{n}-1}.\frac{cos4\pi }{2^{n}-1}...\frac{2^{n}\pi }{2^{n}-1}=\frac{1}{2^{n}}$

Are you sure the result is correct?

(Im assuming the last term is cos(2^n pi)/(2^n-1)

Lieutenant_21 · Feb 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Are you sure the result is correct?

(Im assuming the last term is cos(2^n pi)/(2^n-1)

Yes the result is correct and I edited the post.
Hint: multiply the left side by: $sin\frac{2\pi }{2^{n}-1}$

Sy123 · Feb 4, 2013

Re: HSC 2013 4U Marathon

William Sidis said:
Yes the result is correct and I edited the post.
Hint: multiply the left side by: $sin\frac{2\pi }{2^{n}-1}$

Wait so is the result:

$\cos{\frac{2\pi}{2^n-1}} \cdot \cos \frac{4\pi}{2^n-1} \cdot \dots \cos \frac{2^n \pi}{2^n-1} = \frac{1}{2^n}$

Sy123 · Feb 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Wait so is the result:

$\cos{\frac{2\pi}{2^n-1}} \cdot \cos \frac{4\pi}{2^n-1} \cdot \dots \cos \frac{2^n \pi}{2^n-1} = \frac{1}{2^n}$

Assuming its the above result:

Establish:

$\sin \theta = 2\sin \theta/2 \cos \theta /2$

$\sin \theta = 4\sin \theta / 4 \cos \theta /2 \cos \theta/4$

And so on:

$\sin \theta = 2^n \cos \frac{\theta}{2} \cos \frac{\theta}{4} \dots \cos \frac{\theta}{2^n} \sin \frac{\theta}{2^n}$

Rearrange it, and sub in:

$\theta = \frac{2^{n+1}}{2^n-1}$

$\frac{1}{2^n} = \cos \frac{2\pi}{2^n-1} \cdot \cos \frac{4\pi}{2^n-1} \cdot \dots \cdot \cos \frac{2^n}{2^n-1} \cdot \frac{\sin \frac{2\pi}{2^n-1}}{\sin \frac{2^{n+1}\pi}{2^n-1}}$

Now:

$\frac{2\pi}{2^n-1} = \frac{2^{n+1}\pi}{2^n-1} -2\pi$

Sub this into the last sine part:

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

$=\frac{\sin 2^{n+1}/(2^-1) \cos 2\pi - \sin 2\pi \cos 2^{n+1}/(2^n-1)}{\sin 2^{n+1}/(2^n-1)} = 1$

$\therefore \frac{1}{2^n} = \cos \frac{2\pi}{2^n-1} \cdot \cos \frac{4\pi}{2^n-1} \cdot \dots \cdot \cos \frac{2^n}{2^n-1}$

Lieutenant_21 · Feb 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Assuming its the above result:

Establish:

$\sin \theta = 2\sin \theta/2 \cos \theta /2$

$\sin \theta = 4\sin \theta / 4 \cos \theta /2 \cos \theta/4$

And so on:

$\sin \theta = 2^n \cos \frac{\theta}{2} \cos \frac{\theta}{4} \dots \cos \frac{\theta}{2^n} \sin \frac{\theta}{2^n}$

Rearrange it, and sub in:

$\theta = \frac{2^{n+1}}{2^n-1}$

$\frac{1}{2^n} = \cos \frac{2\pi}{2^n-1} \cdot \cos \frac{4\pi}{2^n-1} \cdot \dots \cdot \cos \frac{2^n}{2^n-1} \cdot \frac{\sin \frac{2\pi}{2^n-1}}{\sin \frac{2^{n+1}\pi}{2^n-1}}$

Now:

$\frac{2\pi}{2^n-1} = \frac{2^{n+1}\pi}{2^n-1} -2\pi$

Sub this into the last sine part:

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

$=\frac{\sin 2^{n+1}/(2^-1) \cos 2\pi - \sin 2\pi \cos 2^{n+1}/(2^n-1)}{\sin 2^{n+1}/(2^n-1)} = 1$

$\therefore \frac{1}{2^n} = \cos \frac{2\pi}{2^n-1} \cdot \cos \frac{4\pi}{2^n-1} \cdot \dots \cdot \cos \frac{2^n}{2^n-1}$

Nice proof but it can be done in 3 lines

Sy123 · Feb 4, 2013

Re: HSC 2013 4U Marathon

William Sidis said:
Nice proof but it can be done in 3 lines

Ah yes I see it, my proof was just a roundabout way to essentially the same thing

(except yours goes backwards and mine forward)

Lieutenant_21 · Feb 5, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Ah yes I see it, my proof was just a roundabout way to essentially the same thing (except yours goes backwards and mine forward)

This question is from a Romanian textbook IIRC. One can find very cool questions from around the world.

asianese · Feb 6, 2013

Re: HSC 2013 4U Marathon

$\text{Find }\int \sin^{-1} \left(\sqrt{x}\right) \; dx.$

HSC 2013 MX2 Marathon (archive) (1 Viewer)

Member

This too shall pass

.

This too shall pass

.

Σ

This too shall pass

This too shall pass

Σ

This too shall pass

Account Closed

Member

This too shall pass

Member

This too shall pass

This too shall pass

Member

This too shall pass

Member

Σ

Users Who Are Viewing This Thread (Users: 0, Guests: 1)