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HSC 2013 MX2 Marathon (archive) (1 Viewer)

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RealiseNothing

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Re: HSC 2014 4U Marathon

so what does the z^3 do to it? im thinking via demoivres it might make it like a ray with a hollow circle except from 0 to pi instead of 0 to pi/3.
Remember

So consider what happens to the arguments when you multiply complex numbers together.
 

seanieg89

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Re: HSC 2014 4U Marathon

Ok so that gives me my lower bound:



Now I need to find an upper bound to use Squeeze Theorem on (well at least that is what I'm thinking will happen).
Could work. I did it a little differently, but the thing I used would have to be proven separately in an MX2 level solution.
 

RealiseNothing

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Re: HSC 2014 4U Marathon

Could work. I did it a little differently, but the thing I used would have to be proven separately in an MX2 level solution.
The only other thing I can think of is considering that and so which means from the AM-GM that the equality would actually hold. Though this seems very non-rigorous.
 

pHyRe

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Re: HSC 2014 4U Marathon

Remember

So consider what happens to the arguments when you multiply complex numbers together.

oh yeaa, so arguments will add together. that z^3 with complex numbers really sends you down the wrong path haha.

so wouldn't that have the same effect? pi/3 + pi/3 + pi/3 = pi?
 

RealiseNothing

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Re: HSC 2014 4U Marathon

oh yeaa, so arguments will add together. that z^3 with complex numbers really sends you down the wrong path haha.

so wouldn't that have the same effect? pi/3 + pi/3 + pi/3 = pi?
Not exactly. Try letting and finding
 

seanieg89

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Re: HSC 2014 4U Marathon

The only other thing I can think of is considering that and so which means from the AM-GM that the equality would actually hold. Though this seems very non-rigorous.
Well you are just dealing with a single variable problem atm (in trying to see what happens when p->0 while the xj are fixed). Think of what sort of tools you have for looking at limits and manipulate to bring the expression you are looking at to a form that is more appropriate for these tools.

This is the relatively easy part, the monotonicity of M_p in p is the tricky part (by mx2 standards).
 
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pHyRe

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Re: HSC 2014 4U Marathon

Not exactly. Try letting and finding
not really sure? cause if arg(z) = pi/3 and i'm trying to find arg(z*z*z) then you should be able to add the arguments of z together i.e. arg(z) + arg(z) + arg(z) = pi/3 + pi/3 + pi/3 = pi. i'm probably missing some fundamental theorem haha, i knew i shoulda asked to change teachers at the start of the year lol
 

RealiseNothing

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Re: HSC 2014 4U Marathon

not really sure? cause if arg(z) = pi/3 and i'm trying to find arg(z*z*z) then you should be able to add the arguments of z together i.e. arg(z) + arg(z) + arg(z) = pi/3 + pi/3 + pi/3 = pi. i'm probably missing some fundamental theorem haha, i knew i shoulda asked to change teachers at the start of the year lol
Yes but not
 

RealiseNothing

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Re: HSC 2014 4U Marathon

haha im stoopid :)

so it'd be 0 to pi/9 then?

also, how do you make the post look all mathsy if you know what i mean. LaTeX?
Yep that's correct now, which should be a lot easier to sketch.

To write in latex put your equations in [tex.] and [/tex.] (remove full stops) and use this site to get a feel of the inputs required http://www.codecogs.com/latex/eqneditor.php
 

Sy123

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Re: HSC 2014 4U Marathon



Note: This question should probably be split into several parts in order to make the difficulty more reasonable, but I would like to see what you guys can come up with off the bat.
For the first problem, I will attempt to prove it for integers p, q, hopefully then someone else can extend it to the reals somehow until I try to figure it out







Any ideas on how to generalise to rationals? I've seen it done for other results, where they'd make you prove integers, then rationals
 

seanieg89

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Re: HSC 2014 4U Marathon

For the first problem, I will attempt to prove it for integers p, q, hopefully then someone else can extend it to the reals somehow until I try to figure it out







Any ideas on how to generalise to rationals? I've seen it done for other results, where they'd make you prove integers, then rationals
Nice!

Yep, to get to rationals from here:

If a < b are rationals, we can choose a common denominator and write

a = p/r
b = q/r

where p < q.

We have M_p =< M_q, let x_k=y_k^{1/r} (and note that this is a bijection on the non-negative reals.)

Convert every 'x' in the statement M_p =< M_q to a 'y' and we have the desired statement M_a =< M_b.

To get to the reals, we would have to show that M_p is continuous in p > 0, which is pretty obvious.
 

Sy123

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Re: HSC 2014 4U Marathon

Ah ok I see the rational one, what do you mean proving M_p to be continuous? How would we do that?
 

seanieg89

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Re: HSC 2014 4U Marathon

Ah ok I see the rational one, what do you mean proving M_p to be continuous? How would we do that?
Using that the sum, product, and composition of continuous functions are continuous. (And that the exponential function a^p is continuous).

Actually "proving" these things is sort of overkill, and comes pretty easily from the epsilon-delta definitions of limits and continuity.

(And once we have continuity, the fact that we can find a sequence of rationals converging to any real number gives us what we want. If LHS(p_n)-RHS(p_n) =< 0 for all n, and p_n->p, then LHS(p)-RHS(p) =< 0.)

Nice idea on building up from the integers! Personally, I differentiated M_p w.r.t. p and showed this quantity was non-negative. (Very quickly by using Jensen's inequality, which is not too hard to prove.)

The other part I did using L'Hospital.
 
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Sy123

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Re: HSC 2014 4U Marathon

Using that the sum, product, and composition of continuous functions are continuous. (And that the exponential function a^p is continuous).

Actually "proving" these things is sort of overkill, and comes pretty easily from the epsilon-delta definitions of limits and continuity.

(And once we have continuity, the fact that we can find a sequence of rationals converging to any real number gives us what we want. If LHS(p_n)-RHS(p_n) =< 0 for all n, and p_n->p, then LHS(p)-RHS(p) =< 0.)

Nice idea on building up from the integers! Personally, I differentiated M_p w.r.t. p and showed this quantity was non-negative. (Very quickly by using Jensen's inequality, which is not too hard to prove.)

The other part I did using L'Hospital.
I guess I have some reading to do haha
 

RealiseNothing

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Re: HSC 2014 4U Marathon

My other idea which I don't think covers all of the reals was to consider the QM-AM inequality:



Letting we get:



If then we could choose arbitrarily small and apply this condition continuously and cover a lot of the reals. Not sure if this could be combined with Sy's induction on the integers to some how cover all reals.

Edit: I think the problem with this approach would be that the set of reals is uncountable whilst the set of integers is countable. So by using a recursive relation like the one above with the induction on the integers we won't be able to cover all reals.

Edit 2: Yer this would never work. Each "chain" would be independent of one another.

from induction

from relation

from relation

But each would be independent and would not imply things such as so :(
 
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RealiseNothing

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Re: HSC 2014 4U Marathon

Prove that between any two rational numbers, there exists at least one irrational number.
 

RealiseNothing

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Re: HSC 2014 4U Marathon

By lowest gap I'm assuming you mean smallest gap? I don't think is the smallest gap. Consider:



would be the smallest gap between all number of the form where the denominator is fixed and numerator varies. But the denominator isn't necessarily fixed.
 
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