HSC 2013 MX2 Marathon (archive) (1 Viewer)

pHyRe · Dec 26, 2013

Re: HSC 2014 4U Marathon

nightweaver066 said:
Sketch the locus of z if $0 \leq arg(z^3) \leq \frac{\pi}{3}$

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.

RealiseNothing · Dec 26, 2013

Re: HSC 2014 4U Marathon

pHyRe said:
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 $z^3 = z \times z \times z$

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

seanieg89 · Dec 26, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
Ok so that gives me my lower bound:

$M_p \geq (\prod_{j=1}^{n} x_j)^\frac{1}{n}$

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 · Dec 26, 2013

Re: HSC 2014 4U Marathon

seanieg89 said:
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 $\lim_{p \to 0^+} x_j^p = 1$ and so $x_1=x_2=...=x_n$ which means from the AM-GM that the equality would actually hold. Though this seems very non-rigorous.

pHyRe · Dec 26, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
Remember $z^3 = z \times z \times z$

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 · Dec 26, 2013

Re: HSC 2014 4U Marathon

pHyRe said:
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 $arg(z) = \theta$ and finding $arg(z^3)$

seanieg89 · Dec 26, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
The only other thing I can think of is considering that $\lim_{p \to 0^+} x_j^p = 1$ and so $x_1=x_2=...=x_n$ 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).

pHyRe · Dec 26, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
Not exactly. Try letting $arg(z) = \theta$ and finding $arg(z^3)$

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 · Dec 26, 2013

Re: HSC 2014 4U Marathon

pHyRe said:
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 $arg(z^3) \leq \frac{\pi}{3}$ not $arg(z) \leq \frac{\pi}{3}$

pHyRe · Dec 26, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
Yes but $arg(z^3) \leq \frac{\pi}{3}$ not $arg(z) \leq \frac{\pi}{3}$

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?

RealiseNothing · Dec 26, 2013

Re: HSC 2014 4U Marathon

pHyRe said:
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 · Dec 27, 2013

Re: HSC 2014 4U Marathon

seanieg89 said:
$Let $x_1,x_2,\ldots,x_n$ be positive real numbers. Let \\ \\$M_p:=\left(\frac{1}{n}\sum_{j=1}^n x_j^p\right)^{1/p}$\\ \\ for positive real $p$.\\ \\ Prove that $M_p\leq M_q$ whenever $p\leq q$.\\ \\ Prove that $\lim_{p\rightarrow 0^+} M_p = \left(\prod_{j=1}^n x_j\right)^{1/n}.$$

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

$\\ $We must prove that$ \ M_p \leq M_q \ $for now we assume that$ \ p,q \in \mathbb{N} \ $We will attempt to prove by induction that$ \ \ M_k \leq M_{k+1} \ \ $for positive integers$ \ k \\ \\ $Step 1$ \ \ $Prove$ \ M_1 \leq M_2 \\ \\ $If$ \ M_1 \leq M_2 \ $then$ \ \frac{1}{n}(x_1+\dots+x_n)^2 \leq (x_1^2 + \dots + x_n^2)^2 \\ \\ $Expanding we see that it comes down to proving$ \\ \\ \ 2x_1 x_2 + 2x_1 x_3 + \dots +2x_1x_n + 2x_2 x_3 + \dots \leq (n-1)(x_1^2 + x_2^2 + \dots +x_n^2)$

$\\ $This is easily proven by the AM-GM inequality of 2 variables, utilising it$ \ (n-1) \ $times for each selected 2 variables$ \\ \\ $Step 2$ \ \ $Assume$ \ M_m \leq M_{m+1} \\ \\ $This assumption leads to the inequality that$ \\ \\ S(m) =\frac{(x_1^m+ \dots + x_n^m)^{m+1}}{(x_1^{m+1} + \dots + x_n^{m+1})^{m}} \leq n \\ \\ \\ $Step 3$ \ $We try to prove$ \ M_{m+1} \leq M_{m+2} \ \ $meaning we must prove using$ \ S(m) \ $that$ \ S(m+1) \leq n \\ \\ $This is done by multiplying$ \ S(m) \ $by$ \\ \\ \\ \frac{S(m+1)}{S(m)} \\ $And showing this is less than$ 1 \\ \\ $Taking$ \ \ \frac{S(m+1)}{S(m)} \ \ \ $it is observed that we only need to prove that (by re-arranging)$ \\ \\ (x_1^{m+1} + \dots + x_n^{m+1})^2 \leq (x_1^{m+2} + \dots + x_n^{m+2})(x_1^m+ \dots +x_n^m)$

$\\ $Taking Cauchy$ \ \ (a_1b_1 + \dots + a_nb_n)^2 \leq (a_1^2 + \dots + a_n^2)(b_1^2 + \dots + b_n^2) \\ \\ $Let$ \ a_k = x_k^{\frac{m+2}{2}} , \ b_k = x_k^{\frac{m}{2}} \\ \\ $The desired inequality falls out straight away$ \\ \\ $Proof is complete, therefore$ \ \ M_k \leq M_{k+1} \ \ $meaning it is montonic increasing therefore allowing us to generalise to$ \ \ M_p \leq M_q \ \ $for integers$ \ p,q$

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 · Dec 27, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
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

$\\ $We must prove that$ \ M_p \leq M_q \ $for now we assume that$ \ p,q \in \mathbb{N} \ $We will attempt to prove by induction that$ \ \ M_k \leq M_{k+1} \ \ $for positive integers$ \ k \\ \\ $Step 1$ \ \ $Prove$ \ M_1 \leq M_2 \\ \\ $If$ \ M_1 \leq M_2 \ $then$ \ \frac{1}{n}(x_1+\dots+x_n)^2 \leq (x_1^2 + \dots + x_n^2)^2 \\ \\ $Expanding we see that it comes down to proving$ \\ \\ \ 2x_1 x_2 + 2x_1 x_3 + \dots +2x_1x_n + 2x_2 x_3 + \dots \leq (n-1)(x_1^2 + x_2^2 + \dots +x_n^2)$

$\\ $This is easily proven by the AM-GM inequality of 2 variables, utilising it$ \ (n-1) \ $times for each selected 2 variables$ \\ \\ $Step 2$ \ \ $Assume$ \ M_m \leq M_{m+1} \\ \\ $This assumption leads to the inequality that$ \\ \\ S(m) =\frac{(x_1^m+ \dots + x_n^m)^{m+1}}{(x_1^{m+1} + \dots + x_n^{m+1})^{m}} \leq n \\ \\ \\ $Step 3$ \ $We try to prove$ \ M_{m+1} \leq M_{m+2} \ \ $meaning we must prove using$ \ S(m) \ $that$ \ S(m+1) \leq n \\ \\ $This is done by multiplying$ \ S(m) \ $by$ \\ \\ \\ \frac{S(m+1)}{S(m)} \\ $And showing this is less than$ 1 \\ \\ $Taking$ \ \ \frac{S(m+1)}{S(m)} \ \ \ $it is observed that we only need to prove that (by re-arranging)$ \\ \\ (x_1^{m+1} + \dots + x_n^{m+1})^2 \leq (x_1^{m+2} + \dots + x_n^{m+2})(x_1^m+ \dots +x_n^m)$

$\\ $Taking Cauchy$ \ \ (a_1b_1 + \dots + a_nb_n)^2 \leq (a_1^2 + \dots + a_n^2)(b_1^2 + \dots + b_n^2) \\ \\ $Let$ \ a_k = x_k^{\frac{m+2}{2}} , \ b_k = x_k^{\frac{m}{2}} \\ \\ $The desired inequality falls out straight away$ \\ \\ $Proof is complete, therefore$ \ \ M_k \leq M_{k+1} \ \ $meaning it is montonic increasing therefore allowing us to generalise to$ \ \ M_p \leq M_q \ \ $for integers$ \ p,q$

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 · Dec 27, 2013

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 · Dec 27, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
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.

Sy123 · Dec 27, 2013

Re: HSC 2014 4U Marathon

seanieg89 said:
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 · Dec 28, 2013

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:

$\sqrt{\frac{a_1^2+a_2^2+...+a_n^2}{n}} \geq \frac{a_1+a_2+...+a_n}{n}$

Letting $a_j=x_j^p$ we get:

$\sqrt{M_{2p}} \geq M_p$

If $M_p > 1$ then we could choose arbitrarily small $p$ 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.

$M_1 < M_2 < M_3 < ...$ from induction

$M_{0.75} < M_{1.5} < M_3 < ...$ from relation

$M_{0.5} < M_1 < M_2 < ...$ from relation

But each would be independent and would not imply things such as $M_{1.5} < M_2$ so

RealiseNothing · Dec 28, 2013

Re: HSC 2014 4U Marathon

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

Sy123 · Dec 28, 2013

Re: HSC 2014 4U Marathon

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

Hopefully this is valid,

$Consider 2 rational numbers with the same denominator, in the form$ \ \ \frac{p}{r} \ $and$ \ \frac{q}{r} \ \ $which can be done with any 2 rational numbers$

$Let$ \ q > p \ $be integers$

$$And so the lowest gap between the two distinct numbers is$ \ \frac{p+1}{r} - \frac{p}{r} = \frac{1}{r}$

$\frac{1}{r} \ $is the lowest amount of gap between the two numbers, assuming$ \ \sqrt{2} \ $is irrational and$ \ \sqrt{2} > 1$

$\frac{p}{r} < \frac{p}{r} + \frac{1}{r \sqrt{2}} < \frac{p+1}{r} \ $the middle number is irrational, hence there is at least 1 irrational number between every 2 rationals$$

RealiseNothing · Dec 28, 2013

Re: HSC 2014 4U Marathon

By lowest gap I'm assuming you mean smallest gap? I don't think $\frac{1}{r}$ is the smallest gap. Consider:

$\frac{3}{7} < \frac{1}{2} < \frac{4}{7}$

$\frac{1}{r}$ would be the smallest gap between all number of the form $\frac{m}{r}$ where the denominator is fixed and numerator varies. But the denominator isn't necessarily fixed.

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