HSC 2012 MX2 Marathon (archive) (1 Viewer)

seanieg89 · Apr 4, 2012

Re: 2012 HSC MX2 Marathon

$The three complex numbers $\alpha,\beta,\gamma$ form an equilateral triangle iff:\\ $\alpha^2+\beta^2+\gamma^2=\alpha\beta+\alpha\gamma+\beta\gamma$ (*).\\Using this, we get $(-a)^2-2b=b$ from which the result follows. We now prove (*).\\ By a straightforward computation, the polynomial\\ $P(\alpha,\beta,\gamma):=\alpha^2+\beta^2+\gamma^2-(\alpha\beta+\alpha\gamma+\beta\gamma)$ is invariant under the translation: $(\alpha,\beta,\gamma)\mapsto (\alpha+s,\beta+s,\gamma+s)$ for complex $s$. Hence by translation we may assume $\alpha=0$.\\ The condition $P=0$ then reduces to: $(\beta/\gamma)^2-(\beta/\gamma)+1=0$ which is equivalent to $\beta=\gamma\textrm{cis}(\pm\pi/3).$ This completes the proof of (*). \\(Note that throughout, we have ignored the degenerate case where all three roots are equal. For this situation, the claim is trivial.)$

seanieg89 · Apr 4, 2012

Re: 2012 HSC MX2 Marathon

lolcakes52 said:
I don't want to make a thread about this as i could just ask it here. How can a complex function that converges for Re(z)>s where s is a real number have a zero for Re(z)<s. I don't understand the logic and was wondering if the initial statement of convergence must be wrong as a imaginary part of high enough value might change the nature of the limit to infinity. I basically stumbled across this while reading about Reimanns zeta function which converges for Re(S)>1 but has zeroes at Re(s)=1/2 which seems like a contradiction.

Okay, I see what you are asking. This is a common misconception. The Riemann Zeta function is defined by the series:

$\sum_{n\in\mathbb{N}}n^{-s}$

ONLY for s with real part larger than one. It turns out there is only one "nice" function that is defined on the whole complex plane (excluding s=1) that agrees with this infinite series wherever the latter converges. THIS is the function that is conjectured to have all nontrivial zeros on the line Re(s)=1/2.

lolcakes52 · Apr 4, 2012

Re: 2012 HSC MX2 Marathon

So what your saying is that it only definitely converges for Re(s)>1 but can converge for values less than or equal to one depending on the imaginary part? In this case it can converge but only for specific values of Im(s)<1 ?

seanieg89 · Apr 4, 2012

Re: 2012 HSC MX2 Marathon

No, the series I wrote down converges if and ONLY if Re(s)>1. It diverges everywhere else. However, this fragment of a function can be extended to the whole complex plane in a very natural way. (Excluding the point s=1). This process is called analytic continuation and is a recurring theme in complex analysis.

cutemouse · Apr 5, 2012

Re: 2012 HSC MX2 Marathon

seanieg89 said:
This process is called analytic continuation and is a recurring theme in complex analysis.

Prove analytic continuation.

It actually is a VERY difficult theorem to prove.

AAEldar · Apr 5, 2012

Re: 2012 HSC MX2 Marathon

Come on guys, keep this stuff within the realms of MX2.

Carrotsticks · Apr 5, 2012

Re: 2012 HSC MX2 Marathon

Given that:

$e = \lim_{n \rightarrow \infty} \left ( 1 + \frac{1}{n} \right )^n$

Show that:

$e^x = \lim_{n \rightarrow \infty} \left ( 1 + \frac{x}{n} \right )^n$

math man · Apr 5, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Given that:

$e = \lim_{n \rightarrow \infty} \left ( 1 + \frac{1}{n} \right )^n$

Show that:

$e^x = \lim_{n \rightarrow \infty} \left ( 1 + \frac{x}{n} \right )^n$

damn you carrot

Carrotsticks · Apr 6, 2012

Re: 2012 HSC MX2 Marathon

math man said:
damn you carrot

What's wrong?

seanieg89 · Apr 6, 2012

Re: 2012 HSC MX2 Marathon

cutemouse said:
Prove analytic continuation.

It actually is a VERY difficult theorem to prove.

Which theorem are you referring to? The Monodromy theorem? It isn't too difficult to find the analytic continuation of the zeta function in particular...

deswa1 · Apr 6, 2012

Re: 2012 HSC MX2 Marathon

Going off Carrot's question, this is to prove the first result (from Cambridge):

<img src="http://latex.codecogs.com/gif.latex?\textup{Consider the area under }y=\frac{1}{x}\textup{ between x=n and x=n+1}\\ a) \textup{ Show that }\frac{1}{n+1}< \int_{n}^{n+1}\frac{1}{x}dx< \frac{1}{n}\\ b)\textup{ Hence show that }\frac{n}{n+1}< ln(1+\frac{1}{n})^n< 1\\ c)\textup{ Take the limit of this last result as n tends to infinity to}\\ \textup{show that }\lim_{n \to \infty }(1+\frac{1}{n})^n=e" title="\textup{Consider the area under }y=\frac{1}{x}\textup{ between x=n and x=n+1}\\ a) \textup{ Show that }\frac{1}{n+1}< \int_{n}^{n+1}\frac{1}{x}dx< \frac{1}{n}\\ b)\textup{ Hence show that }\frac{n}{n+1}< ln(1+\frac{1}{n})^n< 1\\ c)\textup{ Take the limit of this last result as n tends to infinity to}\\ \textup{show that }\lim_{n \to \infty }(1+\frac{1}{n})^n=e" />

seanieg89 · Apr 16, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Given that:

$e = \lim_{n \rightarrow \infty} \left ( 1 + \frac{1}{n} \right )^n$

Show that:

$e^x = \lim_{n \rightarrow \infty} \left ( 1 + \frac{x}{n} \right )^n$

Don't know what you are expecting us to assume, as arbitrary real powers are not defined in the HSC. But we can make sense of the question using the integral definition of the logarithm, and the log laws/differentiation laws that follow.

$Let $x_n:=(1+x/n)^n$. \\ \\We shall show $\log(x_n)\rightarrow x$ as $n\rightarrow\infty$.\\ \\ $\log(x_n)=x\cdot\frac{\log(1+x/n)}{x/n}$\\by rearrangement. \\ \\So as $n\rightarrow\infty$ we get:\\ \lim_{n\rightarrow\infty}\log(x_n)=x\cdot\lim_{y\rightarrow0}\frac{\log(1+y)-\log(1)}{y}=x\cdot\frac{d}{dt}(\log(t))|_{t=1}=x.$

Cyberbully · Apr 17, 2012

Re: 2012 HSC MX2 Marathon

How would you approach this algebraically (or any general explanation for that matter):

<img src="http://latex.codecogs.com/gif.latex?\textup{find the locus of } arg(z(z-(\sqrt{3} + i))) = \frac{\pi}{6}" title="\textup{find the locus of } arg(z(z-(\sqrt{3} + i))) = \frac{\pi}{6}" />

deswa1 · Apr 17, 2012

Re: 2012 HSC MX2 Marathon

Do it geometrically:
<img src="http://latex.codecogs.com/gif.latex?arg(z(z-(\sqrt{3}+i))=\frac{\pi}{6}\\ argz+arg(z-(\sqrt{3}+i))=\frac{\pi}{6}" title="arg(z(z-(\sqrt{3}+i))=\frac{\pi}{6}\\ argz+arg(z-(\sqrt{3}+i))=\frac{\pi}{6}" />

Try going from here. Tell us how you go (I haven't tried the question yet so I don't know the answer.

Carrotsticks · Apr 17, 2012

Re: 2012 HSC MX2 Marathon

Cyberbully said:
How would you approach this algebraically (or any general explanation for that matter):

find the locus of arg(z(z-(sqrt(3)+i))) = pi/6

$\\ \arg(z(z-(\sqrt{3}+i))) = \arg(z) + \arg(z-(\sqrt{3}+i)) \qquad $using a property of the argument.$$

Try it from here.

And beaten to it.

deswa1 · Apr 17, 2012

Re: 2012 HSC MX2 Marathon

Yes, I finally beat Carrotsticks to something

.

Cyberbully · Apr 17, 2012

Re: 2012 HSC MX2 Marathon

deswa1 said:
Do it geometrically:
<img src="http://latex.codecogs.com/gif.latex?arg(z(z-(\sqrt{3}+i))=\frac{\pi}{6}\\ argz+arg(z-(\sqrt{3}+i))=\frac{\pi}{6}" title="arg(z(z-(\sqrt{3}+i))=\frac{\pi}{6}\\ argz+arg(z-(\sqrt{3}+i))=\frac{\pi}{6}" />

Try going from here. Tell us how you go (I haven't tried the question yet so I don't know the answer.

tried that.

didn't get very far.

the "answer" is a straight line through origin and (sqrt(3) + i), excluding the point (sqrt(3) + i) on an argand diagram if it helps...

math man · Apr 29, 2012

Re: 2012 HSC MX2 Marathon

New question: Prove by induction that every natural number greater than 2 has a prime divisor.

Inference · May 1, 2012

Re: 2012 HSC MX2 Marathon

math man said:
New question: Prove by induction that every natural number greater than 2 has a prime divisor.

There is a stronger result, that every natural number greater than 1 has a prime divisor. Anyhow, several methods, first as per the question states:

Assume a set $Q = \{2, 3, \cdots, k-1\}$ where $\forall \ n \in Q$ , then n > 1 and n has a prime divisor. Now to show that k exists in Q is quite straightforward with the following partition:

If k has no other factors than k or 1, then k is trivially in Q.

If k is nonprime, then k will have a factor w such that $1<w<k$ , thus $w \in Q$ by the inductive hypothesis.

hence $k \in Q$

Next method is to show a even stronger result, in other words, the fundmental theorem of arithmetic.

Assume that there are numbers which can not be expressed as a product of primes. Let the smallest possible number of this kind be $m$ .

$m$ can not be $1$ since $1$ is neither composite nor prime. $m$ can not be prime since the PPF of a prime number is just itself. Thus $m$ must be a composite number.

Let the composition of $m = ab$ where $\{a,b\} < m$

Since $m$ was the smallest number that can not be expressed as a product of primes, this means $a$ and $b$ can be expressed as a product of primes and consequently we get $m = ab$ where $a$ and $b$ can be both expressed as primes. Contradiction!

Thus $m$ can also be expressed as a product of primes.

Lemma 1: If $p$ is a prime and $p|a_1a_2a_3 \cdots a_n$ then $p|a_i$ for some $i$ .

Lemma 2: If $p$ and $q$ are primes and $e$ is a natural number and $p|q^e$ then $p=q$ .

Let's assume that for some number $k$ that there are $2$ (at least) ways of expressing its PPF.

$\therefore k = p_1^{f_1}p_2^{f_2}p_3^{f_3} \cdots p_n^{f_n} = q_1^{e_1}q_2^{e_2}q_3^{e_3} \cdots q_m^{e_m}$

Clearly for all $\mbox{n}$ , $p_n|k$

$\implies p_n| q_1^{e_1}q_2^{e_2}q_3^{e_3} \cdots q_m^{e_m}$

By Lemma 1 $p_n|q_m^{e_m}$ for any $m$ .

By Lemma 2 $p_n=q_m$

This means that for all $\mbox{n}$ and all $m$ there are values of $p$ which equals to those of $q$ . For example, $p_3$ could equal to $q_5$ , or $p_6=q_3$ etc. This also means we have created a bijection between $p$ and $q$ such that $n = m$ .

Therefore if the number $k$ has $2$ PPF's then the prime number 'base' will be exactly the same, the only different would be in the powers, namely $f_n$ and $e_m$ .

Now since each $p_n$ has a corespondent equivalent $q_m$ we can rewrite $k$ as:

$p_1^{f_1}p_2^{f_2}p_3^{f_3} \cdots p_n^{f_n} = p_1^{e_1}p_2^{e_2}p_3^{e_3} \cdots p_n^{e_n}$

$p_1^{f_1-e_1}p_2^{f_2}p_3^{f_3} \cdots p_n^{f_n} = p_2^{e_2}p_3^{e_3} \cdots p_n^{e_n}$

$\mbox{LHS}|p_1$ however $\mbox{RHS}$ can not be divided by $p_1$ unless for some $2 \le j \le n$ such that $p_j = p_1$

But since $p_1 \neq p_2 \neq p_3 \neq \cdots \neq p_n$ we have a contradiction.

lolcakes52 · May 5, 2012

Re: 2012 HSC MX2 Marathon

Question 1: ß and ∂ are complex roots of the equation x^3 + 5x + 1=0. Show that the other root is -1/|ß|^2
Question 2: Show that ß∂ is a root of the equation x^3 -5x^2 -1=0.

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