HSC 2012 MX2 Marathon (archive) (3 Viewers)

IamBread · Jan 15, 2012

Re: 2012 HSC MX2 Marathon

bleakarcher said:
yeh its 9.

What's 9?

deswa1 · Jan 15, 2012

Re: 2012 HSC MX2 Marathon

The answer to my question, a few posts back

Nooblet94 · Jan 15, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
Question

5 minutes to work out. 1 hour to type up. 30 minutes to post. I need to get better at latex.

deswa1 · Jan 15, 2012

Re: 2012 HSC MX2 Marathon

Nooblet94 said:
5 minutes to work out. 1 hour to type up. 30 minutes to post. I need to get better at latex.

1 hour's good by my standards for that length. It took me 25 minutes to type up a solution to the fourth roots of unity (about 4 lines of working). I've quit latex now...

IamBread · Jan 15, 2012

Re: 2012 HSC MX2 Marathon

deswa1 said:
1 hour's good by my standards for that length. It took me 25 minutes to type up a solution to the fourth roots of unity (about 4 lines of working). I've quit latex now...

Haha yeah latex takes a bit to get used too...

$$If $ sinx = \frac{e^{ix}-e^{-ix}}{2i} $, show sin^{-1}x = -i\ln(ix+\sqrt{1-x^2})$

bleakarcher · Jan 15, 2012

Re: 2012 HSC MX2 Marathon

Nooblet94 said:
5 minutes to work out. 1 hour to type up. 30 minutes to post. I need to get better at latex.

Dude how did you do the first part of the question?

nightweaver066 · Jan 15, 2012

Re: 2012 HSC MX2 Marathon

bleakarcher said:
Dude how did you do the first part of the question?

As w is the nth root of unity of z^n = 1, 1 + w + w^2 + ... + w^(n - 1) = 0

By expanding it in the first part, you get ^^^ + nw^n which becomes n.

Question:

cutemouse · Jan 16, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
$$If $ sinx = \frac{e^{ix}-e^{-ix}}{2i} $, show sin^{-1}x = -i\ln(ix+\sqrt{1-x^2})$

Lol you are way too obsessed with complex analysis

Carrotsticks · Jan 16, 2012

Re: 2012 HSC MX2 Marathon

cutemouse said:
Lol you are way too obsessed with complex analysis

Who isn't?

Carrotsticks · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

Here's a bit of a fun question. It really shows how slowly the Harmonic Series diverges and how it can be modelled by the natural logarithmic function for large values of n.

seanieg89 · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Here's a bit of a fun question. It really shows how slowly the Harmonic Series diverges and how it can be modelled by the natural logarithmic function for large values of n.

A couple of typos.

i) Your first equation does not make sense as written. Swapping the role of k and n in the two sums fixes this.

ii) We want only a partial sum in Part B.

iii) We can never have n=e^9000 as the latter is not an integer.

Solution:

$\text{By considering the upper and lower Riemann sums of }\log(x)\text{ with respect to}\\ \text{the partition: }\{1,2,\ldots,n\}, \text{ we get }\\ \\ \sum_{k=2}^n\frac{1}{k}<\int_1^n \frac{1}{x}\,dx=\log(n)<\sum_{k=1}^{n-1}\frac{1}{k}.\\ \\ \text{Can you clarify Part B? If we set }n:=\lfloor e^{9000}\rfloor \text{ we have:}\\h_n>\log{(n+1)}>\log{(e^{9000})}=9000.\\ \text{However I see no guarantee that this is the least such n.}\\ \text{In fact the least such n will be roughly }e^{9000-\gamma},\\ \text{ where }\gamma\text{ is the Euler-Mascheroni constant.}}$

Carrotsticks · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

i) My fault for getting the letters mixed up. Thank you for pointing that out.

ii) The top limit should have said 'k' and it should have read 'Partial sum of Harmonic Series'

iii) I think Microsoft Word didn't recognise my floor function from Mathtype so it didn't come out.

It is most surely not the least such n but it is an approximation for the magnitude of k required for such a sum to exist.

If we were to use $e^{8999}$ , then the partial sum would fail to be above 9000.

seanieg89 · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
i) My fault for getting the letters mixed up. Thank you for pointing that out.

ii) The top limit should have said 'k' and it should have read 'Partial sum of Harmonic Series'

iii) I think Microsoft Word didn't recognise my floor function from Mathtype so it didn't come out.

It is most surely not the least such n but it is an approximation for the magnitude of k required for such a sum to exist.

If we were to use $e^{8999}$ , then the partial sum would fail to be above 9000.

Ah okay cool, the phrase "need k=blah" is somewhat misleading then.

Carrotsticks · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

seanieg89 said:
Ah okay cool, the phrase "need k=blah" is somewhat misleading then.

Couldn't think of a better way to express it. Any ideas?

seanieg89 · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Couldn't think of a better way to express it. Any ideas?

$\text{Let }n\text{ be the least integer such that }\sum_{k=1}^n \frac{1}{k}>9000. \text{ Prove that: }\\e^{8999}<n<e^{9000}.$

seanieg89 · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

With perhaps a one marker in between asking to explain why the harmonic series diverges to justify our claim of the existence of such an n.

Carrotsticks · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

Question has been updated with amendments.

kingkong123 · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

Sorry guys, simple question but i dont really understand it

If z = 1-√3i
Find a possible value for n (n>1) such that arg(z) = arg(zⁿ)

seanieg89 · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

To follow on from Carrotstick's question:

$\text{A theorem from analysis states that any decreasing sequence of positive real numbers} \\\text{ must converge to a limit. Use this theorem to prove that }\\ \sum_{k=1}^n \frac{1}{k}-\log(n) \\ \\ \text{converges as }n\rightarrow\infty.$

kingkong123 · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

Also:

Determine the values of k for which the simultaneous equations |z-2i| = k and |z - (3+2i)|=2 have exactly 2 solutions.

thanks

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