HSC 2013 MX2 Marathon (archive) (1 Viewer)

Sy123 · Dec 28, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
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.

The smallest gap yes, I mean that the smallest gap between any 2 rational numbers is 1/(common denominator).

Take arbitrary rationals with integer variables

$\frac{a}{b} - \frac{c}{d}$

$\frac{ad}{bd} - \frac{bc}{bd} = \frac{ad-bc}{bd}$

In order to minimize the gap between two arbitrary numbers $ad - bc = 1$

$ad = 1 + bc$

But using the p,q,r notation is easier I think it says the same thing however

$\frac{bc}{bd} < \frac{bc}{bd} + \frac{1}{\sqrt{2} bd} < \frac{bc}{bd} + \frac{1}{bd} = \frac{ad}{bd}$

Meaning the middle bound is still rational

ganji · Dec 28, 2013

Re: HSC 2014 4U Marathon

Show that the points z_1,z_2,z_3 are collinear if and only if αz_1+βz_2+γz_3=0, where α+β+γ=0 (α,β,γ are real and non zero)

P.s. How do you use Latex here?

RealiseNothing · Dec 28, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
The smallest gap yes, I mean that the smallest gap between any 2 rational numbers is 1/(common denominator).

Take arbitrary rationals with integer variables

$\frac{a}{b} - \frac{c}{d}$

$\frac{ad}{bd} - \frac{bc}{bd} = \frac{ad-bc}{bd}$

In order to minimize the gap between two arbitrary numbers $ad - bc = 1$

$ad = 1 + bc$

But using the p,q,r notation is easier I think it says the same thing however

$\frac{bc}{bd} < \frac{bc}{bd} + \frac{1}{\sqrt{2} bd} < \frac{bc}{bd} + \frac{1}{bd} = \frac{ad}{bd}$

Meaning the middle bound is still rational

Ok I see what you mean. Using p, q, r notation though there is no guarantee that $\frac{p}{r}$ and $\frac{p+1}{r}$ immediately follow each other. So I think instead of using the smallest gap as $\frac{1}{r}$ I think you should let the two rational numbers be say $A$ and $B$ and assume that they immediately follow each other (that is there is no rational number between A and B). From here the smallest gap is $A-B$ by assumption, which can be used in your inequality to get instead:

$A < A + \frac{A-B}{C} < B$

Where $C$ is some irrational number greater than 1, and $A-B$ is rational as it is the difference of two rational numbers.

seanieg89 · Dec 29, 2013

Re: HSC 2014 4U Marathon

$If $a<b$ are the two rationals, then $a+\frac{1}{\sqrt{2}}(b-a)$ is an irrational between them.\\ \\ It is easy to check that it lies between $a$ and $b$, and if it were rational, we could deduce that $\sqrt{2}$ was rational, which is absurd.$

Sy123 · Dec 29, 2013

Re: HSC 2014 4U Marathon

seanieg89 said:
$If $a<b$ are the two rationals, then $a+\frac{1}{\sqrt{2}}(b-a)$ is an irrational between them.\\ \\ It is easy to check that it lies between $a$ and $b$, and if it were rational, we could deduce that $\sqrt{2}$ was rational, which is absurd.$

This is a much more better version of my proof lol (I too went for a 3 inequality)

seanieg89 · Dec 29, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
This is a much more better version of my proof lol (I too went for a 3 inequality)

Haha yep, right idea just a bit of clutter.

Sy123 · Dec 29, 2013

Re: HSC 2014 4U Marathon

This is a good inequality:

$\\ $Consider the positive real numbers$ \ a_1 , a_2 , \dots , a_n \ $such that$ \\ a_1 + a_2 + \dots + a_n = 1$

$\\ $Show that$ \\ \\ \sum_{k=1}^n\frac{a_k}{\sqrt{1-a_k}} \geq \frac{1}{\sqrt{n-1}} \sum_{k=1}^n \sqrt{a_k}$

seanieg89 · Dec 29, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
This is a good inequality:

$\\ $Consider the positive real numbers$ \ a_1 , a_2 , \dots , a_n \ $such that$ \\ a_1 + a_2 + \dots + a_n = 1$

$\\ $Show that$ \\ \\ \sum_{k=1}^n\frac{a_k}{\sqrt{1-a_k}} \geq \frac{1}{\sqrt{n-1}} \sum_{k=1}^n \sqrt{a_k}$

Another opportunity to illustrate the power of Jensen's inequality!

$As $f(x):=1/\sqrt{1-x}$ is convex on $(0,1)$, we have:\\ \\ $LHS=\sum_j f(x_j) \geq n f\left(\frac{1}{n}\sum_j x_j\right)=nf(1/n)\\ \\=\frac{\sqrt{n}}{\sqrt{n-1}}\geq \frac{\sum_j \sqrt{a_j}}{\sqrt{n-1}}.$\\ \\ (The last inequality just popping out from Cauchy--Schwartz) .$

Sy123 · Dec 29, 2013

Re: HSC 2014 4U Marathon

seanieg89 said:
Another opportunity to illustrate the power of Jensen's inequality!

$As $f(x):=1/\sqrt{1-x}$ is convex on $(0,1)$, we have:\\ \\ $LHS=\sum_j f(x_j) \geq n f\left(\frac{1}{n}\sum_j x_j\right)=nf(1/n)\\ \\=\frac{\sqrt{n}}{\sqrt{n-1}}\geq \frac{\sum_j \sqrt{a_j}}{\sqrt{n-1}}.$\\ \\ (The last inequality just popping out from Cauchy--Schwartz) .$

Wow haha, my way is much much longer but uses more basic inequalities

seanieg89 · Dec 29, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
Wow haha, my way is much much longer but uses more basic inequalities

Haha the gap in lengths would probably be at least partially closed by me having to prove Jensen's before using it. I will post a statement and proof of it here later today, just out of interest to see how compact and elementary I can make it.

seanieg89 · Dec 29, 2013

Re: HSC 2014 4U Marathon

View attachment jensen.pdf

RealiseNothing · Dec 30, 2013

Re: HSC 2014 4U Marathon

Show that $\sum_{k=1}^{\infty} \sin(\frac{1}{k})$ diverges to infinity.

Note the angle is in radians.

Sy123 · Dec 30, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
Show that $\sum_{k=1}^{\infty} \sin(\frac{1}{k})$ diverges to infinity.

Note the angle is in radians.

Here is my attempt:

$\sum_{k=1}^{\infty} \sin \left(\frac{1}{k} \right) = \sum_{k=1}^{\infty} \cos \left(\frac{\pi}{2} - \frac{1}{k} \right)$

$\\ $Consider the graph$ \ y= \cos x \ \ \ \ 0 \leq x \leq \frac{\pi}{2} \ $and the straight line through$ \ (0,1) \ \left(\frac{\pi}{2} , 0 \right) \\ \\ $We can observe that$ \ \cos x \geq 1-\frac{2}{\pi}x \\ \\ $Hence$ \ \sum_{k=1}^{\infty} \cos \left(\frac{\pi}{2} - \frac{1}{k} \right) > \lim_{N \to \infty} \sum_{k=1}^N 1 - \frac{2}{\pi} \left(\frac{\pi}{2} - \frac{1}{k}\right) = \frac{2}{\pi} \lim_{N \to \infty} \left(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{N} \right)$

$\\ $We see that$ \ \sum_{k=1}^{\infty} \frac{1}{k} \rightarrow \infty \ $as$ \ 1+\int_1^{\infty} \frac{1}{x} \ dx < \sum_{k=1}^{\infty}\frac{1}{k} \ \ $and, the lower bound tends to infinity, therefore the Harmonic sum must tend to infinity$$

$\\ $Since the Harmonic sum is in the lower bound of the previous inequality, it must follow that$ \ \ \sum_{k=1}^{\infty} \cos \left(\frac{\pi}{2} - \frac{1}{k} \right) \ \ $tends to infinity$ \\ \\ $Therefore implying that$ \ \ \sum_{k=1}^{\infty} \sin \left(\frac{1}{k} \right) \rightarrow \infty$

Sy123 · Dec 30, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
This is a good inequality:

$\\ $Consider the positive real numbers$ \ a_1 , a_2 , \dots , a_n \ $such that$ \\ a_1 + a_2 + \dots + a_n = 1$

$\\ $Show that$ \\ \\ \sum_{k=1}^n\frac{a_k}{\sqrt{1-a_k}} \geq \frac{1}{\sqrt{n-1}} \sum_{k=1}^n \sqrt{a_k}$

Here is my long solution;

$\\ (\sqrt{a_1} + \sqrt{a_2} + \dots + \sqrt{a_n})^2 = (a_1 + \dots + a_n) + 2(\sqrt{a_1a_2} + \dots) \\ \\ $Use$ \ a_k + a_m \geq 2\sqrt{a_k a_m} \\ \\ \therefore \ 1 + (n-1)(a_1 + \dots + a_n) \geq (\sqrt{a_1}+ \dots + \sqrt{a_n})^2 \\ \\ $Hence$ \ \sqrt{\frac{n}{n-1}} \geq \frac{1}{\sqrt{n-1}} (\sqrt{a_1} + \dots + \sqrt{a_n}) \\ \\ $Now we must prove that$ \ \sum_{k=1}^n \frac{a_k}{\sqrt{1-a_k}} \geq \sqrt{\frac{n}{n-1}} \\ \\ $let$ \ b_k^2 = 1-a_k \ \ \therefore \ \sum b_k^2 = (n-1) \\ \\ \sum \frac{a_k}{\sqrt{1-a_k}} = \sum \frac{1-b_k^2}{b_k} = \sum \frac{1}{b_k} - \sum b_k \\ \\ $By Cauchy$ \ \ \left(\sum b_k^2 \right) \left(\sum \frac{1}{b_k} \right) \geq \left(\sum b_k \right)^2 \\ \\ \therefore \ \sum \frac{1}{b_k} \geq \frac{1}{n-1} \left(\sum b_k \right)^2$

$\\ $let$ \ S = \sum b_k \\ \\ \therefore \ \sum \frac{1-b_k^2}{b_k} \geq \left(\frac{S^2 - (n-1)S}{n-1} \right) \\ \\ $Since$ \ b_1^2 + \dots + b_n^2 \leq S^2 \\ \\ $Therfore$ \ S \geq \sqrt{n-1} \\ \\ $Also$ \ \ S^2 = (n-1) + 2(b_1b_2+ \dots)$

$\\ $Therefore$ \ (n-1)^2 + (n-1) \geq S^2 \\ \\ $Therefore$ \ \sqrt{n-1} \leq S \leq \sqrt{n(n-1)} \\ \\ \left(\sum \frac{1}{b_k} \right) \left(\sum b_k \right) \geq n^2 \\ \\ \therefore \ \ \sum \frac{1}{b_k} \geq \frac{n^2}{S} \\ \\ $Since$ \ S \leq \sqrt{n(n-1)} \\ \\ \sum \frac{1}{b_k} \geq \frac{n^2}{\sqrt{n(n-1)}} \\ \\ $And$ \ -S \geq -\sqrt{n(n-1)}$

$\\ $Therefore$ \ \ \ \sum \frac{1}{b_k} - \sum b_k \geq \frac{n^2}{\sqrt{n(n-1)}} - \sqrt{n(n-1)} = \sqrt{\frac{n}{n-1}} \\ \\ $Sub back in$ \ b_k^2 = 1-a_k \\ \\ $and the final result is immediate$ \\ \\ \sum_{k=1}^n \frac{a_k}{\sqrt{1-a_k}} \geq \sqrt{\frac{n}{n-1}} \geq \frac{1}{\sqrt{n-1}} (\sqrt{a_1} + \dots + \sqrt{a_n})$

So yeaa this might be longer than your proof including the proof of Jensen's inequality

seanieg89 · Dec 30, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
Here is my long solution;

$\\ (\sqrt{a_1} + \sqrt{a_2} + \dots + \sqrt{a_n})^2 = (a_1 + \dots + a_n) + 2(\sqrt{a_1a_2} + \dots) \\ \\ $Use$ \ a_k + a_m \geq 2\sqrt{a_k a_m} \\ \\ \therefore \ 1 + (n-1)(a_1 + \dots + a_n) \geq (\sqrt{a_1}+ \dots + \sqrt{a_n})^2 \\ \\ $Hence$ \ \sqrt{\frac{n}{n-1}} \geq \frac{1}{\sqrt{n-1}} (\sqrt{a_1} + \dots + \sqrt{a_n}) \\ \\ $Now we must prove that$ \ \sum_{k=1}^n \frac{a_k}{\sqrt{1-a_k}} \geq \sqrt{\frac{n}{n-1}} \\ \\ $let$ \ b_k^2 = 1-a_k \ \ \therefore \ \sum b_k^2 = (n-1) \\ \\ \sum \frac{a_k}{\sqrt{1-a_k}} = \sum \frac{1-b_k^2}{b_k} = \sum \frac{1}{b_k} - \sum b_k \\ \\ $By Cauchy$ \ \ \left(\sum b_k^2 \right) \left(\sum \frac{1}{b_k} \right) \geq \left(\sum b_k \right)^2 \\ \\ \therefore \ \sum \frac{1}{b_k} \geq \frac{1}{n-1} \left(\sum b_k \right)^2$

$\\ $let$ \ S = \sum b_k \\ \\ \therefore \ \sum \frac{1-b_k^2}{b_k} \geq \left(\frac{S^2 - (n-1)S}{n-1} \right) \\ \\ $Since$ \ b_1^2 + \dots + b_n^2 \leq S^2 \\ \\ $Therfore$ \ S \geq \sqrt{n-1} \\ \\ $Also$ \ \ S^2 = (n-1) + 2(b_1b_2+ \dots)$

$\\ $Therefore$ \ (n-1)^2 + (n-1) \geq S^2 \\ \\ $Therefore$ \ \sqrt{n-1} \leq S \leq \sqrt{n(n-1)} \\ \\ \left(\sum \frac{1}{b_k} \right) \left(\sum b_k \right) \geq n^2 \\ \\ \therefore \ \ \sum \frac{1}{b_k} \geq \frac{n^2}{S} \\ \\ $Since$ \ S \leq \sqrt{n(n-1)} \\ \\ \sum \frac{1}{b_k} \geq \frac{n^2}{\sqrt{n(n-1)}} \\ \\ $And$ \ -S \geq -\sqrt{n(n-1)}$

$\\ $Therefore$ \ \ \ \sum \frac{1}{b_k} - \sum b_k \geq \frac{n^2}{\sqrt{n(n-1)}} - \sqrt{n(n-1)} = \sqrt{\frac{n}{n-1}} \\ \\ $Sub back in$ \ b_k^2 = 1-a_k \\ \\ $and the final result is immediate$ \\ \\ \sum_{k=1}^n \frac{a_k}{\sqrt{1-a_k}} \geq \sqrt{\frac{n}{n-1}} \geq \frac{1}{\sqrt{n-1}} (\sqrt{a_1} + \dots + \sqrt{a_n})$

So yeaa this might be longer than your proof including the proof of Jensen's inequality

Nice!

Yeah imo Jensen's is extremely powerful for how easy it is to prove.

Fizzy_Cyst · Dec 30, 2013

Re: HSC 2014 4U Marathon

seanieg89 said:
$we could deduce that $\sqrt{2}$ was rational, which is absurd.$

Absurd or a surd?

Lmfao.

RealiseNothing · Dec 30, 2013

Re: HSC 2014 4U Marathon

nvm

RealiseNothing · Dec 30, 2013

Re: HSC 2014 4U Marathon

Carrotsticks said:
You gave your sum of a_k to go from 1 to n, but your product makes no mention of n.

I presume you want the limit as n-> infinity, in which case the a_k series you provided converges to 1?

Yep that's it, I initially wanted the question to be in terms of 'n' then changed to letting n -> infinity but forgot to change the sum of a_k.

Sy123 · Jan 1, 2014

Re: HSC 2014 4U Marathon

$\\ $Help needed$ \\ \\ $Prove$ \\ \\ \left(\frac{1-a_1}{a_1} \right) \left(\frac{1-a_2}{a_2} \right) \dots \left(\frac{1-a_n}{a_n} \right) \geq (n-1)^n \\ \\ $for positive real$ \ a_k \ $and$ \ \ a_1 + a_2 + \dots +a_n = 1$

AM-GM will lead to an inequality that is not tight enough to cover the bound in question

seanieg89 · Jan 1, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Help needed$ \\ \\ $Prove$ \\ \\ \left(\frac{1-a_1}{a_1} \right) \left(\frac{1-a_2}{a_2} \right) \dots \left(\frac{1-a_n}{a_n} \right) \geq (n-1)^n \\ \\ $for positive real$ \ a_k \ $and$ \ \ a_1 + a_2 + \dots +a_n = 1$

If you take the log of both sides, it is a straightforward application of Jensen's inequality.

HSC 2013 MX2 Marathon (archive) (1 Viewer)

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