HSC 2012 MX2 Marathon (archive) (1 Viewer)

deswa1

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Re: 2012 HSC MX2 Marathon

The answer to my question, a few posts back
 

deswa1

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Re: 2012 HSC MX2 Marathon

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

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Re: 2012 HSC MX2 Marathon

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...

 
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nightweaver066

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Re: 2012 HSC MX2 Marathon

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:
 
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Carrotsticks

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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

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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.

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:

 
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Carrotsticks

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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 , then the partial sum would fail to be above 9000.
 

seanieg89

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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 , then the partial sum would fail to be above 9000.
Ah okay cool, the phrase "need k=blah" is somewhat misleading then.
 

seanieg89

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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

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Re: 2012 HSC MX2 Marathon

Question has been updated with amendments.
 

kingkong123

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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

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Re: 2012 HSC MX2 Marathon

To follow on from Carrotstick's question:

 

kingkong123

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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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