HSC 2014 MX2 Marathon ADVANCED (archive) (2 Viewers)

seanieg89 · Jun 1, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
I'm getting:

$\frac{1}{4}(1+\frac{1}{3^n})$

I got:

$\frac{1}{4}\left(1+3\left(-\frac{1}{3}\right)^n\right)$

Notice that the answer should be 0 for n=1.

RealiseNothing · Jun 1, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
I got:

$\frac{1}{4}\left(1+3\left(-\frac{1}{3}\right)^n\right)$

Notice that the answer should be 0 for n=1.

Ah right, I probs summed up the series I obtained slightly wrong. This may be influenced by the 18th I was at tonight.

My logic was:

$P(n) = \frac{1}{3}[1-P(n-1)]$

Where $P(k)$ is the probability that after flipping it $k$ times we are back to the original side.

I then just summed this using a recursive relation.

seanieg89 · Jun 1, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
Ah right, I probs summed up the series I obtained slightly wrong. This may be influenced by the 18th I was at tonight.

My logic was:

$P(n) = \frac{1}{3}[1-P(n-1)]$

Where $P(k)$ is the probability that after flipping it $k$ times we are back to the original side.

I then just summed this using a recursive relation.

Your recurrence is right, but your solution to it wasn't.

Don't drink and derive! (But actually do, it is the best.)

Sy123 · Jun 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Prove that if an integer polynomial is reducible over the integers, then the factor polynomials can not out put an infinite amount of primes (when given an integer input)$$

EDIT: disregard

seanieg89 · Jun 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Prove that if an integer polynomial is reducible over the integers, then the factor polynomials can not out put an infinite amount of primes (when given an integer input)$$

I don't think this is actually known to be true or false.

Eg (x^2+1)^2 is reducible over the integers, but it is not known (and regarded to be a rather difficult problem) whether or not there are infinitely many primes of the form n^2+1.

We only have this sort of result for linear polynomials atm (Dirichlet's theorem on primes in arithmetic progression), and even this proof is not easy.

Am I misinterpreting?

Sy123 · Jun 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
I don't think this is actually known to be true or false.

Eg (x^2+1)^2 is reducible over the integers, but it is not known (and regarded to be a rather difficult problem) whether or not there are infinitely many primes of the form n^2+1.

We only have this sort of result for linear polynomials atm (Dirichlet's theorem on primes in arithmetic progression), and even this proof is not easy.

Am I misinterpreting?

My bad my wording is wrong

$\\ $Prove that if an integer polynomial$ \ P \ $is reducible over the integers, then$ \ P \ $cannot output an infinite amount of primes, given integral input$$

seanieg89 · Jun 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
My bad my wording is wrong

$\\ $Prove that if an integer polynomial$ \ P \ $is reducible over the integers, then$ \ P \ $cannot output an infinite amount of primes, given integral input$$

Ah right, all good then.

aDimitri · Jun 5, 2014

Re: HSC 2014 4U Marathon - Advanced Level

By subbing x = sqrt(x) you get a polynomial with roots r1^2, r2^2... etc.
By expanding Q(r1)*Q(r2)*...*Q(r5). You get an expansion in terms of r1^2 r2^2 etc.
This is in terms of the coefficients of the new polynomial. But I dont have paper here so I cant get the number atm because I cbb without paper. Questions is probs a bit easy for the advanced thread!!

edit:
answer = 9

seanieg89 · Jun 5, 2014

Re: HSC 2014 4U Marathon - Advanced Level

aDimitri said:
By subbing x = sqrt(x) you get a polynomial with roots r1^2, r2^2... etc.
By expanding Q(r1)*Q(r2)*...*Q(r5). You get an expansion in terms of r1^2 r2^2 etc.
This is in terms of the coefficients of the new polynomial. But I dont have paper here so I cant get the number atm because I cbb without paper. Questions is probs a bit easy for the advanced thread!!

edit:
answer = 9

You have constructed one polynomial to make your life easier, why not another? Make the substitution x->x+2 after your first construction and you have a polynomial with roots rj^2-2. You don't even need to expand it to compute what the question is asking, just find it's constant coefficient giving you the product of roots.

aDimitri · Jun 5, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
You have constructed one polynomial to make your life easier, why not another? Make the substitution x->x+2 after your first construction and you have a polynomial with roots rj^2-2. You don't even need to expand it to compute what the question is asking, just find it's constant coefficient giving you the product of roots.

wow. i can't believe i didn't see that...

aDimitri · Jun 5, 2014

Re: HSC 2014 4U Marathon - Advanced Level

probs made a mistake expanding, no way i'm going to look where hahaha

Sy123 · Jun 6, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
My bad my wording is wrong

$\\ $Prove that if an integer polynomial$ \ P \ $is reducible over the integers, then$ \ P \ $cannot output an infinite amount of primes, given integral input$$

$\\ $If a polynomial$ \ P \ $is reducible, then$ \ P = R \cdot S \ $for integer polynomials$ \ R,S \\ $If$ \ P \ $is prime, then$ \ R=1, S=P \ $or$ \ R=P , S=1 \\ \\ $Then if there are infinite inputs to get an infinite amount of primes for output, then either/or$ \ R = 1 \ $or$ \ S=1 \ $infinitely many times$ \\ \\ $Meaning the polynomial$ \ R-1 \ $has infinite amount of roots, but by the Fundamental Theorem of Algebra, this cannot be the case, leading to a contradiction$ \\ \\ $Therefore if$ \ P \ $is reducible over the integers, then it cannot output an infinite amount of primes, given input$$

Sy123 · Jun 6, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $A polynomial$ \ P \ $is irreducible over the integers if there exists no$ \ R, S \ $integer polynomials so that$ \ P = R \cdot S \\ \\ $Consider the polynomial$ \\ \\ f(x) = (x-a_1)(x-a_2) \dots (x-a_n) - 1 \\ \\ $Where$ \ n \ $is odd and$ \ a_1,a_2,\dots,a_n \ $are distinct$ \\ \\ $Show that$ \ f \ $is irreducible over the integers$$

Sy123 · Jun 7, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Find, with proof (without using L'Hopitals Rule)$ \\ \\ \lim_{x \to 0} \frac{\ln(\tan \left(x + \frac{\pi}{4} \right))}{\tan x}$

seanieg89 · Jun 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

TL1998 said:
ABCD is a quadrilateral. A line AF parallel to BC meets BD at F. A line BE parallel to AD meets AC at E. Prove that EF is parallel to CD.

A mathematics Extension 2 state ranker from last year couldn't complete this question (well I guess with enough he could have) so i can guarantee this one's difficulty

It's just similar triangles:

Alternate angles tells us:

$\triangle AXF \sim \triangle CXB$

and

$\triangle AXD \sim \triangle EXB$

This implies:

$\frac{EX}{CX}=\frac{EX}{BX}\cdot\frac{BX}{CX}= \frac{AX}{DX} \cdot\frac{FX}{AX}=\frac{FX}{DX}$

Therefore:

$\triangle XFE \sim \triangle XDC$

(equal angle and adjacent sides have a common ratio.)

Finally, this similarity gives us by corresponding angles that $FE\parallel DC$ .

mathing · Jun 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Nice Sean! Please post the alternative solution to continuing f for x < 0.

seanieg89 said:
It's just similar triangles:

Alternate angles tells us:

$\triangle AXF \sim \triangle CXB$

and

$\triangle AXD \sim \triangle EXB$

This implies:

$\frac{EX}{CX}=\frac{EX}{BX}\cdot\frac{BX}{CX}= \frac{AX}{DX} \cdot\frac{FX}{AX}=\frac{FX}{DX}$

Therefore:

$\triangle XFE \sim \triangle XDC$

(equal angle and adjacent sides have a common ratio.)

Finally, this similarity gives us by corresponding angles that $FE\parallel DC$ .

mathing · Jun 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Not sure if this is what you are looking for but we can apply the tan addition formula to get

$\lim_{x \to 0} \frac{\ln(\tan(x) + \tan(\pi/4)) - \ln(1 - \tan(x)\tan(pi/4))}{\tan(x)} \\ $Since $ \tan $ is continuous at$ x=0 $ we have:$ \\ \lim_{u = 0} \frac{\ln(1 + u) - \ln(1 - u)}{u}. $The derivatives of the numerators are $ 1/(1+u) - 1/(1-u) $which is are 1 and -1 when u = 0 so we know that they are approaching at the same gradient as the denominator at u = 0, so the limit is 2.$

You can also see the last fraction is the definition of the two derivatives at u = 0.

Sy123 said:
$\\ $Find, with proof (without using L'Hopitals Rule)$ \\ \\ \lim_{x \to 0} \frac{\ln(\tan \left(x + \frac{\pi}{4} \right))}{\tan x}$

mathing · Jun 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $A polynomial$ \ P \ $is irreducible over the integers if there exists no$ \ R, S \ $integer polynomials so that$ \ P = R \cdot S \\ \\ $Consider the polynomial$ \\ \\ f(x) = (x-a_1)(x-a_2) \dots (x-a_n) - 1 \\ \\ $Where$ \ n \ $is odd and$ \ a_1,a_2,\dots,a_n \ $are distinct$ \\ \\ $Show that$ \ f \ $is irreducible over the integers$$

Please tell us the solution!

seanieg89 · Jun 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

mathing said:
Nice Sean! Please post the alternative solution to continuing f for x < 0.

For x < 0, define f(x) = af(-x) + bf(-2x)+cf(-3x) (*) for some undetermined constants a,b,c.

It is clear that this will satisfy the differentiability and boundedness assumptions of f for positive x, just by the laws of how differentiation behaves with respect to addition, constant multiplication, and composition. This means that the only thing to worry about is how the definitions of f for positive x and negative x "match up" at 0.

If we let x -> 0- in (*) and use continuity, we get f(0) = (a+b+c)f(0).

Similarly, since we have differentiability, we can differentiate (*) (at least twice) and then use continuity of derivatives to get:

f'(0) = (-a-2b-3c)f'(0).

f''(0) = (a+4b+9c)f''(0).

Now if we choose our a,b,c to satisfy the simultaneous equations:

a+b+c = 1
-a-2b-3c = 1
a+4b+9c = 1

(this is a short exercise in simultaneous equations, but I cbb doing it again right now) then from the earlier discussion the function and its first two derivatives will match up continuously, and we are done.

seanieg89 · Jun 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

mathing said:
Please tell us the solution!

Suppose to the contrary that f IS reducible over Z and f = gh is one such expression with deg(g) > n/2 > deg(h). Note that we can assume strict inequality of degrees because their sum is deg(f) = n which is odd. This turns out to be key.

g(aj)h(aj) = -1 for each j, but since both of these polynomials has integer coefficients, both of these factors must be integers. That implies that for each j we either have:

I) g(aj) = 1 h(aj) = -1

or

II) g(aj) = -1 h(aj) = 1.

One of these must occur strictly more often than the other (again as n is odd), and so we have h(x) = c for some c (either 1 or -1, we don't really care) and at least n/2 > deg(h) different values of x. But this is a contradiction, as the definition of reducibility (Sy forgot to mention this ) requires that the two factor polynomials are non-constant, and no non-constant polynomial can assume the same value as many times as (its degree + 1) for the same reason that no non-constant polynomial can have more roots than its degree.

So we are done.

HSC 2014 MX2 Marathon ADVANCED (archive) (2 Viewers)

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