HSC 2012 MX2 Marathon (archive) (1 Viewer)

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

Carrotsticks · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

Second Question

Observe the diagram. The two loci share the same y value for their centres. The green circle and the blue circle are the largest possible values of k for it to intersect with the red circle whilst having two solutions (well, strictly 1 solution, but we will be using an inequality).

Looking at the radius of the red circle (which is obviously 2 as defined by the locus), the green and blue circle can only intersect on either side of the red circle, in order to have 1 solution.

This occurs when x=1 and x=5.

However since the centre of the circle lies on the line x=0, we can safely say that these x co-ordinates will also be the radius of the blue/green circle.

But the radius of the blue/green circle is the value of k.

Hence $1 < k < 5$

kingkong123 · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

for q2, thanks heaps man; perfect diagram understood everything

you're a king. i was drawing |z-2i| = k as a circle with centre (0,-2) instead of (0,2) ie |z+2i|=k. i think it makes the question a lot harder? true?

anyway thanks a lot!

Carrotsticks · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

kingkong123 said:
for the first question, i did it the same way;
i solved -npi/3 = 2kpi. therefore n =-6k where (k C Z) let k = -1 then n =6 but the answer in the solutions say n=7 :S

for q2, thanks heaps man; perfect diagram understood everything you're a king. i was drawing |z-2i| = k as a circle with centre (0,-2) instead of (0,2) ie |z+2i|=k. i think it makes the question a lot harder? true?

anyway thanks a lot!

Suppose the locus was |z+2i|=k.

You will need to use an algebraic proof as opposed to a geometric proof like I did. Here is the method if you were to receive such a question:

1. Find the Cartesian equation of both circles.

2. Attempt to solve them simultaneously. You will acquire a quadratic in terms of constants, x and k.

3. Let the discriminant of this quadratic be greater than 0 (for 2 roots)

4. Simplify and write down the inequality.

kingkong123 · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
The answers must be incorrect then. If we were to have n=7, this means that we are rotating backwards 7 times. If we do so, we will not go back to our original vector (again, ignoring modulus). We will in fact overshoot it.

Suppose the locus was |z+2i|=k.

You will need to use an algebraic proof as opposed to a geometric proof like I did. Here is the method if you were to receive such a question:

1. Find the Cartesian equation of both circles.

2. Attempt to solve them simultaneously. You will acquire a quadratic in terms of constants, x and k.

3. Let the discriminant of this quadratic be greater than 0 (for 2 roots)

4. Simplify and write down the inequality.

oh okay, thanks for the method! ill use it if that type of question comes up

kingkong123 · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

lol "You must spread some Reputation around before giving it to Carrotsticks again."

seanieg89 · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

Answer to the argument question IS 7.

$\arg(z^n)=\arg(z)\Rightarrow \arg(z^{n-1})=0\Rightarrow \frac{-(n-1)\pi}{3}=2k\pi.$

for some integer k.

Setting k=-1 gives us the simplest nontrivial solution of n=7.

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