HSC 2013 MX2 Marathon (archive) (2 Viewers)

Lieutenant_21 · Mar 23, 2013

Re: HSC 2013 4U Marathon

hayabusaboston said:
iirc ive read those words verbatim out of either cambridge or some other place. I had a flashback of the exact moment I read them haha.

This is the standard proof. You get like 2 million results if you search "prove sqrt2 is irrational" on google.

Lieutenant_21 · Mar 23, 2013

Re: HSC 2013 4U Marathon

I am not going to be able to use BOS that much because the HSC is very close :$ Keep this thread alive guys! See you in uni

Sy123 · Mar 23, 2013

Re: HSC 2013 4U Marathon

Lieutenant_21 said:
$Cool complex numbers question. Can be tedious but definitely MX2 level, may be question 16 difficulty.$

$a) By equating imaginary parts in DeMoivre's formula prove that:$

$sin(n\theta )=sin^{n}\theta \left \{ \binom{n}{1}cot^{n-1}\theta -\binom{n}{3}cot^{n-3}\theta +\binom{n}{5}cot^{n-5}\theta -+ \cdots \right \}$

$b) If$ $0<\theta<\frac{\pi }{2}$ $$, prove that:$

$sin(2m+1)\theta =sin^{2m+1}\theta P_{m}(cot^{2}\theta )$

$where$ $P_{m}$ $is the polynomial of degree m given by:$

$P_{m}=\binom{2m+1}{1}x^{m}-\binom{2m+1}{3}x^{m-1}+\binom{2m+1}{5}x^{m-2}-+\cdots$

$Use this to show that$ $P_{m}$ $has zeros at the m distinct points$ $x_{k}=cot^{2}(\frac{\pi k}{2m+1}) for k=1, 2, 3, \dots, m$

$c) Show that the sum of the zeros of$ $P_{m}$ $is given by$ $\sum_{k=1}^{m}cot^{2}\frac{\pi k}{2m+1}=\frac{m(2m-1)}{3},$

$and that the sum of their squares is given by:$

$\sum_{k=1}^{m}cot^{4}\frac{\pi k}{2m+1}=\frac{m(2m-1)(4m^{2}+10m-9)}{45}$

$d) Hence, prove that$ $\sum_{n=1}^{\infty }n^{-2}=\frac{\pi ^{2}}{6}$ $and$ $\sum_{n=1}^{\infty }n^{-4}=\frac{\pi ^{4}}{90}$

It's a bit harsh to make people try and make their own inequality and bounds AND squeeze theorem twice isn't it? :s

Sy123 · Mar 23, 2013

Re: HSC 2013 4U Marathon

$Define a polynomial $ \ \ C_n(x) \ \ $such that it behaves that$

$C_n(\cos \theta) = \cos(n\theta)$

$$i) Prove that the Polynomial satisfies the recurrence relation with the initial conditions$ \ \ \ \ \ \ \fbox{3}$

$\begin{cases} C_0(x) = 1 \\ C_1(x)=x \\ C_n(x) = 2xC_{n-1}(x) - C_{n-2}(x) \end{cases}$

$$ii) Define the power series$ \ \ C(z) \ \ $in$ \ \ z \ \ $with co-efficient as the polynomials$ \ \ C$

$C(z) = \sum_{k=0}^{\infty} C_k(x) z^k$

$Prove using the recurrence relation proven in part (i)$

$C(z) = \frac{1-xz}{1-2xz+z^2} \ \ \ \ \ \ \ \fbox{3}$

$$iii) Hence prove that the sum of the co-efficients of the polynomial$ \ \ C_n \ \ $for all integers$ \ \ n \ \ $ is 1$ \ \ \ \ \ \ \ \fbox{2}$

$iv) Using de Moivre's theorem prove that$

$\cos (n\theta) = \cos^n \theta - \binom{n}{2} \cos^{n-2} \theta \sin^2 \theta + \binom{n}{4} \cos^{n-4} \theta \sin^4 \theta - \dots \ \ \ \ \ \ \ \fbox{1}$

$v) By using part (iv) verify the result proven in part (iii)$

$$ by showing that the sum of the co-efficients of$ \ \ C_n \ \ $is 1$ \ \ \ \ \ \ \ \fbox{2}$

HeroicPandas · Mar 23, 2013

Re: HSC 2013 4U Marathon

Lieutenant_21 said:
$Cool complex numbers question. Can be tedious but definitely MX2 level, may be question 16 difficulty.$

I think i posted up a similar question a long time ago and Sy answered it

Trebla · Mar 23, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Define a polynomial $ \ \ C_n(x) \ \ $such that it behaves that$

$C_n(\cos \theta) = \cos(n\theta)$

$$i) Prove that the Polynomial satisfies the recurrence relation with the initial conditions$ \ \ \ \ \ \ \fbox{3}$

$\begin{cases} C_0(x) = 1 \\ C_1(x)=x \\ C_n(x) = 2xC_{n-1}(x) - C_{n-2}(x) \end{cases}$

$$ii) Define the power series$ \ \ C(z) \ \ $in$ \ \ z \ \ $with co-efficient as the polynomials$ \ \ C$

$C(z) = \sum_{k=0}^{\infty} C_k(x) z^k$

$Prove using the recurrence relation proven in part (i)$

$C(z) = \frac{1-xz}{1-2xz+z^2} \ \ \ \ \ \ \ \fbox{3}$

$$iii) Hence prove that the sum of the co-efficients of the polynomial$ \ \ C_n \ \ $for all integers$ \ \ n \ \ $ is 1$ \ \ \ \ \ \ \ \fbox{2}$

$iv) Using de Moivre's theorem prove that$

$\cos (n\theta) = \cos^n \theta - \binom{n}{2} \cos^{n-2} \theta \sin^2 \theta + \binom{n}{4} \cos^{n-4} \theta \sin^4 \theta - \dots \ \ \ \ \ \ \ \fbox{1}$

$v) By using part (iv) verify the result proven in part (iii)$

$$ by showing that the sum of the co-efficients of$ \ \ C_n \ \ $is 1$ \ \ \ \ \ \ \ \fbox{2}$

Chebyshev polynomials of the first kind

RealiseNothing · Mar 23, 2013

Re: HSC 2013 4U Marathon

hayabusaboston said:
iirc ive read those words verbatim out of either cambridge or some other place. I had a flashback of the exact moment I read them haha.

Cos that's how you prove it???

Sy123 · Mar 24, 2013

Re: HSC 2013 4U Marathon

Trebla said:
Chebyshev polynomials of the first kind

They're pretty cool aren't they?

psychotropic · Mar 26, 2013

Re: HSC 2013 4U Marathon

so cool!! =)

seanieg89 · Mar 27, 2013

Re: HSC 2013 4U Marathon

$Given two polynomials $p(z)$, $d(z)$ with complex coefficients prove that there exist unique polynomials $q(z)$ and $r(z)$ such that: \\ \\$p(z)=q(z)d(z)+r(z)$ \\and $\deg(r)<\deg(d)$. \\ \\(Note that by convention, $\deg(0):=-\infty$.)$

Sy123 · Mar 27, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Define a polynomial $ \ \ C_n(x) \ \ $such that it behaves that$

$C_n(\cos \theta) = \cos(n\theta)$

$$i) Prove that the Polynomial satisfies the recurrence relation with the initial conditions$ \ \ \ \ \ \ \fbox{3}$

$\begin{cases} C_0(x) = 1 \\ C_1(x)=x \\ C_n(x) = 2xC_{n-1}(x) - C_{n-2}(x) \end{cases}$

$$ii) Define the power series$ \ \ C(z) \ \ $in$ \ \ z \ \ $with co-efficient as the polynomials$ \ \ C$

$C(z) = \sum_{k=0}^{\infty} C_k(x) z^k$

$Prove using the recurrence relation proven in part (i)$

$C(z) = \frac{1-xz}{1-2xz+z^2} \ \ \ \ \ \ \ \fbox{3}$

$$iii) Hence prove that the sum of the co-efficients of the polynomial$ \ \ C_n \ \ $for all integers$ \ \ n \ \ $ is 1$ \ \ \ \ \ \ \ \fbox{2}$

$iv) Using de Moivre's theorem prove that$

$\cos (n\theta) = \cos^n \theta - \binom{n}{2} \cos^{n-2} \theta \sin^2 \theta + \binom{n}{4} \cos^{n-4} \theta \sin^4 \theta - \dots \ \ \ \ \ \ \ \fbox{1}$

$v) By using part (iv) verify the result proven in part (iii)$

$$ by showing that the sum of the co-efficients of$ \ \ C_n \ \ $is 1$ \ \ \ \ \ \ \ \fbox{2}$

i) First let x=cos theta, substitute it into the expression they ask to prove. We get:

Prove:

$\cos n\theta = 2\cos \theta \cos(n-1) \theta - \cos(n-2)\theta$

With simple algebra and compound angle formula we arrive that LHS = RHS, also see that C_1(cos theta) = cos theta
C_1(x) = x
C_0(cos theta) = cos 0*theta = 1
C_0(x)=1

ii) $C(z) = C_0(x) + C_1(x) z + \sum_{k=2}^{\infty} C_{k}(x) z^k$

$C(z) = 1+ xz + \sum_{k=2}^{\infty} 2xC_{k-1}z^k - C_{k-2}z^k$

$C(z)=1+xz+2xz(C(z)-1) - z^2C(z)$

$C(z) = $required result$$

iii) Sum of co-efficients is simply C_n(1), so into C(z) sub in x=1

$C(z)= \frac{1-z}{1-2z+z^2} = \frac{1}{1-z} = 1 + z + z^2 + z^3 + \dots + z^n + \dots = \sum_{k=0}^{\infty} C_k(1) z^k$

So equating in general the co-efficient of z^n, we get

$C_n(1) = 1$

So for all polynomials sum of co-efficients is 1.

iv) Straightforward

v) Let $\theta = \cos^{-1}x$

By using the fact that:

$\sin(\cos^{-1}x) = \sqrt{1-x^2}$

Substitute everything in, note that the LHS becomes simply C_n(x)

$C_n(x) = x^n - \binom{n}{2} x^{n-2} (1-x^2) + \dots$

So indeed every sine term is even, now when we sub in x=1 (to get sum of co-efficients,) all the terms after the first cancel to 0

Leaving with C_n(1) = 1^n = 1

Sy123 · Mar 28, 2013

Re: HSC 2013 4U Marathon

$$Find$

$1+3x+6x^2+10x^3+ \dots$

funnytomato · Mar 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Find$

$1+3x+6x^2+10x^3+ \dots$

$S=1+3x+6x^2+10x^3+15x^4+\dots$

$Sx=x+3x^2+6x^3+10x^4+15x^5+\dots$

$S(1-x)=1+2x+3x^2+4x^3+5x^4+\dots$

$Sx(1-x)=x+2x^2+3x^3+4x^4+5x^5+\dots$

$S(1-x)(1-x)=1+x+x^2+x^3+x^4+x^5+\dots$

$S(1-x)^2)=1+x+x^2+x^3+x^4+x^5+\dots=\frac{1}{1-x}$, if $|x|<1$

$S=\frac{1}{(1-x)^3}$ , if $|x|<1$

RealiseNothing · Mar 28, 2013

Re: HSC 2013 4U Marathon

If you split it up into individual series, you sum them and get:

$\frac{1}{1-x} + \frac{2x}{1-x} + \frac{3x^2}{1-x} + \frac{4x^3}{1-x} + ...$

$\frac{1+2x+3x^2+4x^3+...}{1-x}$

Now consider the series in the numerator:

$1+2x+3x^2+4x^3+...$

Summing these individually again gives:

$\frac{1}{1-x} + \frac{x}{1-x} + \frac{x^2}{1-x} + ...$

$\frac{1+x+x^2+...}{1-x}$

$\frac{\frac{1}{1-x}}{1-x}$

$\frac{1}{(1-x)^2}$

Substitute this back in for the first numerator:

$\frac{\frac{1}{(1-x)^2}}{1-x}{$

$\frac{1}{(1-x)^3}$

RealiseNothing · Mar 28, 2013

Re: HSC 2013 4U Marathon

I'm pretty sure you could continue this pattern and show that:

$1+4x+10x^2+20x^3+35x^4+... = \frac{1}{(1-x)^4}$

funnytomato · Mar 28, 2013

Re: HSC 2013 4U Marathon

Let G be the centroid of triangle ABC which is inscribed in a circle
Produce AG,BG and CG so that they intersect the circle again at points X,Y and Z respectively

Show that $\frac{AG}{GX}+\frac{BG}{GY}+\frac{CG}{GZ}=3$

seanieg89 · Mar 28, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
I'm pretty sure you could continue this pattern and show that:

$1+4x+10x^2+20x^3+35x^4+... = \frac{1}{(1-x)^4}$

Yep. A fast way of generalising this is proving that you can differentiate term-by-term in the interior of a power series disc of convergence, and then differentiating the infinite geometric series with leading term 1 and ratio x a bunch of times.

RealiseNothing · Mar 28, 2013

Re: HSC 2013 4U Marathon

Or in other terms:

$\binom{3}{0} + \binom{4}{1}x + \binom{5}{2}x^2 + \binom{6}{3}x^3 + ... = \frac{1}{(1-x)^4}$

RealiseNothing · Mar 28, 2013

Re: HSC 2013 4U Marathon

In general:

$\binom{n}{0} + \binom{n+1}{1}x + \binom{n+2}{2}x^2 + \binom{n+3}{3}x^3 + ... = \frac{1}{(1-x)^{n+1}}$

I'm pretty sure this works.

Sy123 · Mar 29, 2013

Re: HSC 2013 4U Marathon

Nice work guys,

$Simplify$

$\sin \theta \sin(2\theta) + \sin (2\theta)\sin(3\theta) + \dots + \sin(n\theta) \sin((n+1)\theta)$

HSC 2013 MX2 Marathon (archive) (2 Viewers)

Member

Member

This too shall pass

This too shall pass

Heroic!

Administrator

what is that?It is Cowpea

This too shall pass

Member

Well-Known Member

This too shall pass

This too shall pass

Active Member

what is that?It is Cowpea

what is that?It is Cowpea

Active Member

Well-Known Member

what is that?It is Cowpea

what is that?It is Cowpea

This too shall pass

Users Who Are Viewing This Thread (Users: 0, Guests: 2)