HSC 2013 MX2 Marathon (archive) (1 Viewer)

Lieutenant_21 · Mar 21, 2013

Re: HSC 2013 4U Marathon

$Evaluate the sum:$
$\sum_{n=1}^{\infty }\frac{1}{2^{n}}tan\frac{a}{2^{n}}$

Homo Sapien · Mar 21, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
It's MX1.

So seanieg meant the question is too easy by "troll question" haha
I didn't know it was MX1 though

Sy123 · Mar 22, 2013

Re: HSC 2013 4U Marathon

Lieutenant_21 said:
$Evaluate the sum:$
$\sum_{n=1}^{\infty }\frac{1}{2^{n}}tan\frac{a}{2^{n}}$

Great question:

First lets establish the identity

$\sin x = 2^r \cos(x/2) \cos (x/4) \cos(x/8) \dots \cos(x/2^r) \sin(x/2^r)$

Take the natural logarithm of both sides, ensuring that cos(x/2^k) is always positive (absolute value), and use logarithm laws

$\ln|\sin x| - \ln|\sin x/2^r| = \ln(2^r) + \ln |\cos x/2| + \ln |\cos x/4| + \dots + \ln |\cos x/2^r|$

Differentiate both sides

$-\cot x + \frac{1}{2^r} \cot (x/2^r) = \frac{1}{2} \tan (x/2) + \frac{1}{4} \tan (x/4) + \dots + \frac{1}{2^r} \tan (x/2^r) \equiv \sum_{n=1}^{r} \frac{1}{2^n} \tan \frac{x}{2^n}$

Take r to infinity

$\lim_{r to \infty} \frac{\cos (x/2^r)}{2^r \sin (x/2^r)} - \cot x = \sum_{n=1}^{\infty} \frac{1}{2^n} \tan (x/2^n) = S$

$\alpha = 1/2^r$

$\frac{\lim_{r \to infty} \cos(x/2^r)}{\lim_{\alpha \to 0} \frac{\sin \alpha x}{\alpha}} - \cot x = S$

$S= \frac{1}{x} - \cot x$

$\therefore \ \sum_{n}^{\infty} \frac{1}{2^n} \tan \left(\frac{a}{2^n} \right) = \frac{1}{a} - \cot a$

$\square$

Lieutenant_21 · Mar 22, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Great question:

First lets establish the identity

$\sin x = 2^r \cos(x/2) \cos (x/4) \cos(x/8) \dots \cos(x/2^r) \sin(x/2^r)$

Take the natural logarithm of both sides, ensuring that cos(x/2^k) is always positive (absolute value), and use logarithm laws

$\ln|\sin x| - \ln|\sin x/2^r| = \ln(2^r) + \ln |\cos x/2| + \ln |\cos x/4| + \dots + \ln |\cos x/2^r|$

Differentiate both sides

$-\cot x + \frac{1}{2^r} \cot (x/2^r) = \frac{1}{2} \tan (x/2) + \frac{1}{4} \tan (x/4) + \dots + \frac{1}{2^r} \tan (x/2^r) \equiv \sum_{n=1}^{r} \frac{1}{2^n} \tan \frac{x}{2^n}$

Take r to infinity

$\lim_{r to \infty} \frac{\cos (x/2^r)}{2^r \sin (x/2^r)} - \cot x = \sum_{n=1}^{\infty} \frac{1}{2^n} \tan (x/2^n) = S$

$\alpha = 1/2^r$

$\frac{\lim_{r \to infty} \cos(x/2^r)}{\lim_{\alpha \to 0} \frac{\sin \alpha x}{\alpha}} - \cot x = S$

$S= \frac{1}{x} - \cot x$

$\therefore \ \sum_{n}^{\infty} \frac{1}{2^n} \tan \left(\frac{a}{2^n} \right) = \frac{1}{a} - \cot a$

$\square$

Great solution! There is a little shorter way:

$Note that:$

$tanx=\frac{1}{tanx}-\frac{1-tan^{2}x}{tanx}=$

$\frac{1}{tanx}-2\frac{1}{tan2x}=cotx-2cot2x$

$This implies that:$

$\frac{1}{2^{n}}tan\frac{a}{2^{n}}=\frac{1}{2^{n}}-\frac{1}{2^{n-1}}cot\frac{a}{2^{n-1}}$

$The sum telescopes:$

$\lim_{n\rightarrow \infty }\frac{1}{2^{n}}cot\frac{a}{2^{n}}-cota= \cdot \cdot \cdot =\frac{1}{a}-cota$

================================================================================

Have a go at this integral:

$\int\frac{xsinx}{1+(cosx)^{2}}dx$

seanieg89 · Mar 22, 2013

Re: HSC 2013 4U Marathon

Pretty sure that integral can't be expressed in terms of elementary functions.

Lieutenant_21 · Mar 22, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Pretty sure that integral can't be expressed in terms of elementary functions.

IBP

PS: Never mind, I think arctan(cosx) can't be expressed in terms of elementary functions.

seanieg89 · Mar 22, 2013

Re: HSC 2013 4U Marathon

Arctan and cos are both elementary functions themselves lol. I still don't think it can be done though.

Lieutenant_21 · Mar 22, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Arctan and cos are both elementary functions themselves lol. I still don't think it can be done though.

Yes, I know arctan and cos are elementary functions but you need poly-logarithm functions to do it.

seanieg89 · Mar 22, 2013

Re: HSC 2013 4U Marathon

Lieutenant_21 said:
Yes, I know arctan and cos are elementary functions but you need poly-logarithm functions to do it.

Yeah I know, I was just responding to you saying you think arctan(cos x) can't be expressed in terms of elementary functions.

Anyway, all good

.

Lieutenant_21 · Mar 22, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Yeah I know, I was just responding to you saying you think arctan(cos x) can't be expressed in terms of elementary functions.

Anyway, all good .

Should have said integral of arctan(cosx)

My bad.

Sy123 · Mar 22, 2013

Re: HSC 2013 4U Marathon

$f_0(x)= \frac{1}{1-x}$

$$let$ \ \ f_r (x) = f(f(f(\dots (f(x))\dots ))$

$Where there are r iterations of the function$

$$So$ \ \ f_2(x) = f(f(f(x)))$

$$Evaluate$ \ \ \ f_{2013} (\pi)$

Lieutenant_21 · Mar 23, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$f_0(x)= \frac{1}{1-x}$

$$let$ \ \ f_r (x) = f(f(f(\dots (f(x))\dots ))$

$Where there are r iterations of the function$

$$So$ \ \ f_2(x) = f(f(f(x)))$

$$Evaluate$ \ \ \ f_{2013} (\pi)$

$f_{0}(x)= \frac{1}{1-x}$

$f_{1}(x)=\frac{1}{1-\left ( \frac{1}{1-x} \right )}=\frac{x-1}{x}$

$f_{2}(x)=\frac{1}{1-\left ( \frac{1}{1-\left ( \frac{1}{1-x} \right )} \right )}=x$

$f_{3}(x)=\frac{1}{1-\left ( \frac{1}{1-\left ( \frac{1}{1-\left ( \frac{1}{1-x} \right )} \right )} \right )}=\frac{1}{1-x}$

$We can see that:$

$f_{2013}(x)=f_{0}(x)= \frac{1}{1-x} \therefore f_{2013}(x)=\frac{1}{1-\pi }$

$\square$

Lieutenant_21 · Mar 23, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
From the early 70s. Those problems are often doable by good mx2 students with no prior olympiad training and this one is no exception.

Could you please elaborate on that?

What do people do to prepare for Olympiads? They clearly need excellent problem solving skills which is something I want to improve.

seanieg89 · Mar 23, 2013

Re: HSC 2013 4U Marathon

Solve problems haha. And you can find sort of unofficial IMO syllabi various places online to get an idea of useful pieces of theory that are still considered to be at a high school level. (As opposed to advanced university uses of calculus, which are not tested as calculus is not taught properly in high school.)

The point is more to answer things very rigorously with elementary methods. There are many books/articles on problem solving strategies out there, but honestly practicing and reviewing your though processes is probably the most beneficial "training".

Lieutenant_21 · Mar 23, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Solve problems haha. And you can find sort of unofficial IMO syllabi various places online to get an idea of useful pieces of theory that are still considered to be at a high school level. (As opposed to advanced university uses of calculus, which are not tested as calculus is not taught properly in high school.)

The point is more to answer things very rigorously with elementary methods. There are many books/articles on problem solving strategies out there, but honestly practicing and reviewing your though processes is probably the most beneficial "training".

Thank you!
So I will just keep doing questions.

hayabusaboston · Mar 23, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Assume it is rational, so $\sqrt{2}$ can be expressed as a fraction where the numerator and denominator have no common factors:

$\sqrt{2} = \frac{p}{q}$

$q\sqrt{2} = p$

$2q^2 = p^2$

So now we know that $p^2$ is even and hence $p$ is even. So now we let $p=2k$ :

$p^2 = 4k^2$

$2q^2 = 4k^2$

$q^2 = 2k^2$

So now we know that $q^2$ is even, and so $q$ is even.

However if both $p$ and $q$ are even, then they have a common factor of $2$ . This is a contradiction and so our assumption is wrong, that is, $\sqrt{2}$ is not rational - it is irrational.

Is this not directly quoted from cambridge?

hayabusaboston · Mar 23, 2013

Re: HSC 2013 4U Marathon

Lieutenant_21 said:
I am pretty sure that was the only mark I lost (hopefully, as I haven't gotten the whole paper back yet). The test wan't really that hard, was very similar to questions found in Cambridge.
I never used Cambridge before but I used it to study for this test because my teacher uses it and I think Cambridge and Terry Lee complement each other very well for 4U.

According to a lady from Dymocks on george street (Is it george street? lol I think so), cambridge 4u is "for students who are really struggling to grasp the concepts of extension 2 maths, and need help at a basic level"

Lol.

If ur test similar to cambridge, must have been friggin easy hahaha.

Lieutenant_21 · Mar 23, 2013

Re: HSC 2013 4U Marathon

hayabusaboston said:
Is this not directly quoted from cambridge?

Why do you think so?

hayabusaboston · Mar 23, 2013

Re: HSC 2013 4U Marathon

Lieutenant_21 said:
Why do you think so?

iirc ive read those words verbatim out of either cambridge or some other place. I had a flashback of the exact moment I read them haha.

Lieutenant_21 · Mar 23, 2013

Re: HSC 2013 4U Marathon

$Cool complex numbers question. Can be tedious but definitely MX2 level, may be question 16 difficulty.$

$a) By equating imaginary parts in DeMoivre's formula prove that:$

$sin(n\theta )=sin^{n}\theta \left \{ \binom{n}{1}cot^{n-1}\theta -\binom{n}{3}cot^{n-3}\theta +\binom{n}{5}cot^{n-5}\theta -+ \cdots \right \}$

$b) If$ $0<\theta<\frac{\pi }{2}$ $$, prove that:$

$sin(2m+1)\theta =sin^{2m+1}\theta P_{m}(cot^{2}\theta )$

$where$ $P_{m}$ $is the polynomial of degree m given by:$

$P_{m}=\binom{2m+1}{1}x^{m}-\binom{2m+1}{3}x^{m-1}+\binom{2m+1}{5}x^{m-2}-+\cdots$

$Use this to show that$ $P_{m}$ $has zeros at the m distinct points$ $x_{k}=cot^{2}(\frac{\pi k}{2m+1}) for k=1, 2, 3, \dots, m$

$c) Show that the sum of the zeros of$ $P_{m}$ $is given by$ $\sum_{k=1}^{m}cot^{2}\frac{\pi k}{2m+1}=\frac{m(2m-1)}{3},$

$and that the sum of their squares is given by:$

$\sum_{k=1}^{m}cot^{4}\frac{\pi k}{2m+1}=\frac{m(2m-1)(4m^{2}+10m-9)}{45}$

$d) Hence, prove that$ $\sum_{n=1}^{\infty }n^{-2}=\frac{\pi ^{2}}{6}$ $and$ $\sum_{n=1}^{\infty }n^{-4}=\frac{\pi ^{4}}{90}$

HSC 2013 MX2 Marathon (archive) (1 Viewer)

Member

Member

This too shall pass

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Member

This too shall pass

Member

Member

Well-Known Member

Member

Well-Known Member

Well-Known Member

Member

Well-Known Member

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)