HSC 2012 MX2 Marathon (archive) (1 Viewer)
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Nooblet94
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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.
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5 minutes to work out. 1 hour to type up. 30 minutes to post. I need to get better at latex.
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deswa1
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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...
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bleakarcher
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Re: 2012 HSC MX2 Marathon
Dude how did you do the first part of the question?
5 minutes to work out. 1 hour to type up. 30 minutes to post. I need to get better at latex.
nightweaver066
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Re: 2012 HSC MX2 Marathon
By expanding it in the first part, you get ^^^ + nw^n which becomes n.
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
Who isn't?Lol you are way too obsessed with complex analysis
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.
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
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{ 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.}})
A couple of typos.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.
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.
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
seanieg89
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Re: 2012 HSC MX2 Marathon
Ah okay cool, the phrase "need k=blah" is somewhat misleading then.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
Carrotsticks
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seanieg89
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Carrotsticks
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Re: 2012 HSC MX2 Marathon
Question has been updated with amendments.
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)
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)
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
Also:
Determine the values of k for which the simultaneous equations |z-2i| = k and |z - (3+2i)|=2 have exactly 2 solutions.
thanks