HSC 2014 MX2 Marathon ADVANCED (archive) (4 Viewers)

hit patel · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
View attachment 30261

Here is my attempt

edit: damn just realized there is an error on like 3rd last line, how do i prove l is the positive case?

Nice Davo/

seanieg89 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
Hmm what do you mean by "distinct mod 3"? and do you mind showing me your method, sounds interesting

Have different remainder upon division by 3. (So when you add P(x^3), xQ(x^3) and x^2R(x^3), the leading terms can't cancel each other out. Otherwise we could not conclude that the degree of the sum is the max of the individual degrees.)

Will write out my solution later today, have to dash to uni now.

jyu · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Find all polynomials$ \ f \ $such that$ \ \ f(x) + f(1-x) = 1 \ $for all$ \ x$

f(x)=1/2 + linear combination of odd power of (x-1/2)

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Find all polynomials$ \ f \ $such that$ \ \ f(x) + f(1-x) = 1 \ $for all$ \ x$

$f(x)=x-(1-x)^2+x^2-(1-x)^3+x^3-...-(1-x)^n+x^n$

Here is one, should be some more, how many are you after?

hit patel · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
$f(x)=x-(1-x)^2+x^2-(1-x)^3+x^3-...-(1-x)^n+x^n$

Here is one, should be some more, how many are you after?

Are you doing the HSC this year? Just curious.

braintic · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Find all polynomials$ \ f \ $such that$ \ \ f(x) + f(1-x) = 1 \ $for all$ \ x$

It will work for any polynomial which has line symmetry in the line x=1/2 [so P(x+1/2) must be even], and for which P(1/2) = 1/2.

So P(x) = (a_n) (x-1/2)^(2n) + (a_(n-1)) (x-1/2)^(2n-2) + (a_(n-2)) (x-1/2)^(2n-4) + ... + (a_1) (x-1/2)^2 + c

where c is chosen to make P(1/2)=1/2

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

hit patel said:
Are you doing the HSC this year? Just curious.

Year 10 just started 4u

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

hit patel said:
Are you doing the HSC this year? Just curious.

Hahaha I'm kidding I was meant to do it this year but was accelerated so last year

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Prove that $\pi=lim_{n\to\infty} \; 2^n\sqrt{2-\sqrt{2+\sqrt{2+...\sqrt{2}}}}$ (Where there are a total of n square root brackets)

Sy123 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

For the polynomials one what I did was:

$\\ f(x) + f(1-x) = 1 \\ f'(x) = f'(1-x) \\ \therefore \ f'(x) = \sum_{k=0}^n c_k x^{a_k} (1-x)^{b_k} \\ $Then$ \ f(x) = \int \sum_{k=0}^n c_k x^{a_k}(1-x)^{b_k} \ dx + c \\ \\ $where$ \ \ c = \frac{1}{2} - \int_0^{1/2} \sum_{k=0}^n c_k x^{a_k}(1-x)^{b_k} \ dx \\ $Then find a formula for beta function then apply$$

I thought this covers all polynomials but perhaps not

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Something I made up

seanieg89 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

braintic said:
It will work for any polynomial which has line symmetry in the line x=1/2 [so P(x+1/2) must be even], and for which P(1/2) = 1/2.

So P(x) = (a_n) (x-1/2)^(2n) + (a_(n-1)) (x-1/2)^(2n-2) + (a_(n-2)) (x-1/2)^(2n-4) + ... + (a_1) (x-1/2)^2 + c

where c is chosen to make P(1/2)=1/2

I think you mean (1/2 + odd) rather than even.

seanieg89 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
Have different remainder upon division by 3. (So when you add P(x^3), xQ(x^3) and x^2R(x^3), the leading terms can't cancel each other out. Otherwise we could not conclude that the degree of the sum is the max of the individual degrees.)

Will write out my solution later today, have to dash to uni now.

$S=P(x^3)+xQ(x^3)+x^2R(x^3)=c\prod_j (x-\alpha_j).\\ \\ $If $T(x^3):=c^3\prod_j (x^3-\alpha_j^3)=c^3\prod_j (x-\alpha_j)(\omega x-\alpha_j)(\omega^2 x-\alpha_j) \\ \\ = S(x)S(\omega x)S(\omega^2 x)=P(x^3)^3+x^3Q(x^3)^3+x^6R(x^3)^3-3P(x^3)Q(x^3)R(x^3)\\ \\ \Rightarrow T=P^3+xQ^3+x^2R^3-3PQR.\\ \\ T$ has roots $\alpha_j^3$ by construction, and has degree $n$ by your argument and the comment on degree mod 3. $$

braintic · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
I think you mean (1/2 + odd) rather than even.

I didn't mean that at the time ... but yes, that was dumb.

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
$S=P(x^3)+xQ(x^3)+x^2R(x^3)=c\prod_j (x-\alpha_j).\\ \\ $If $T(x^3):=c^3\prod_j (x^3-\alpha_j^3)=c^3\prod_j (x-\alpha_j)(\omega x-\alpha_j)(\omega^2 x-\alpha_j) \\ \\ = S(x)S(\omega x)S(\omega^2 x)=P(x^3)^3+x^3Q(x^3)^3+x^6R(x^3)^3-3P(x^3)Q(x^3)R(x^3)\\ \\ \Rightarrow T=P^3+xQ^3+x^2R^3-3PQR.\\ \\ T$ has roots $\alpha_j^3$ by construction, and has degree $n$ by your argument and the comment on degree mod 3. $$

Woah! Thats a crazy smart solution! Had no clue how you would use cube roots of unity.
one thing though, i think there should be an x in front of PQR?

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Prove the following expression:
$(2+\sqrt{2})^n\cos{\frac{n\pi}{4}}=\sum\limits_{k=0}^{2n} \binom {2n} {k}\cos{\frac{k\pi}{4}}$

and similarly prove:
$(2+\sqrt{2})^n(\cos{\frac{n\pi}{4}} +\sin{\frac{n\pi}{4}})=\sum\limits_{k=0}^{2n} \binom {2n} {k}(\cos{\frac{k\pi}{4}} +\sin{\frac{k\pi}{4}})$

VBN2470 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

How many pairs $(a, b)$ of positive integers are there such that $a$ and $b$ are factors of $6^6$ and $a$ is a factor of $b$ ?

hit patel · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
Hahaha I'm kidding I was meant to do it this year but was accelerated so last year

I was surprised when you said you are in year 10. Then I realised the you had a post right after it hahahahah. You are intelligent for your age probably in atleast the 2nd percentile in your age group of NSW from what I can see. If you dont mind me asking how much did you get for hsc cos thats what to uni's matter more than intelligence and the actual ability.

VBN2470 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

How many functions $f(x)=ax^2+bx+c$ are there with the property that, for all $x$ , $f(x)=f(-x)=f(x^2)$

VBN2470 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

In how many ways can you choose four positive integers $a, b, c$ and $d$ with $a<b<c<d$ , so that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$ is an integer?

HSC 2014 MX2 Marathon ADVANCED (archive) (4 Viewers)

New Member

Well-Known Member

Member

Active Member

New Member

Well-Known Member

Active Member

Active Member

Active Member

This too shall pass

Active Member

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Active Member

Well-Known Member

New Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 4)