HSC 2012 MX2 Marathon (archive) (3 Viewers)

cutemouse · Mar 22, 2012

Re: 2012 HSC MX2 Marathon

Everyone else wrote that it equals zero (odd function etc.) but I thought that you can only integrate something if it is defined in the domain so I wrote that it diverges as x can't equal 1 or -1. Is that right?

Indeed the integral doesn't converge absolutely.

What everyone else wrote is the Cauchy Principal Value. This assigns a value to such integrals. But of course that was not asked here.

seanieg89 · Mar 29, 2012

Re: 2012 HSC MX2 Marathon

$Prove by induction that for any non-negative integer $n$, we have:\\ $e^x\geq \sum_{k=0}^n \frac{x^k}{k!}\quad\textrm{for }x\geq0.$\\ Hence deduce that $\lim_{x\rightarrow\infty} P(x)e^{-x}=0$ for any polynomial $P(x)$. \\ \\ (Extension) Let $f$ be the function defined by:\\$f(x):=e^{-1/x}\quad(\textrm{for }x>0),$\\ $f(x):=0 \quad(\textrm{for }x\leq 0)$. \\ \\Using the previous part of this question or otherwise, prove that $f$ is differentiable of all orders, with every derivative vanishing at $x=0$. (You may assume without proof that exponential functions and ratios of polynomials have derivatives of all orders on their domain.)$

lolcakes52 · Mar 30, 2012

Re: 2012 HSC MX2 Marathon

Couldn't you just use the taylor series of e^x?

seanieg89 · Mar 30, 2012

Re: 2012 HSC MX2 Marathon

Well that is how e^x is defined in higher mathematics, but at the school level that knowledge is not assumed. You can deduce the inequality pretty easily using properties of the exponential from the 3 unit course.

lolcakes52 · Apr 4, 2012

Re: 2012 HSC MX2 Marathon

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.

Carrotsticks · 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.

BOS is stuffing up a bit. I quoted your statement to point out something which didn't make sense, then once it appeared in the quote tags, some text appeared *which made the sentence complete*, which was not originally there. I'm betting that if I were to reply, the quote would come off missing parts as well.

You said the Riemann Zeta Function has zeros for Re(s) = 0.5, but this is actually one of the greatest unsolved problems in Mathematics, which you may have heard of.

It is called the Riemann Hypothesis: http://en.wikipedia.org/wiki/Riemann_hypothesis

It essentially says that all non-trivial zeros for this function have real component 0.5, but is yet to be proved.

Carrotsticks · Apr 4, 2012

Re: 2012 HSC MX2 Marathon

Here is a very nice question. If you have done many geometrical applications of Complex Numbers questions, then this shouldn't be too difficult for you, as it draws from a very well known condition. If you don't know this condition... the question may be quite difficult.

Consider the cubic non-zero polynomial with complex coefficients:

$P(x)=x^3+ax^2+bx+c$

Show that the 3 roots form an equilateral triangle if and only if:

$a^2=3b$

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}" />

HSC 2012 MX2 Marathon (archive) (3 Viewers)

Account Closed

Well-Known Member

Member

Well-Known Member

Member

Retired

Retired

Well-Known Member

Well-Known Member

Member

Well-Known Member

Account Closed

Premium Member

Retired

Member

Retired

Well-Known Member

Well-Known Member

Well-Known Member

Where do you live?

Users Who Are Viewing This Thread (Users: 0, Guests: 3)