HSC 2016 MX2 Marathon (archive) (2 Viewers)

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Drsoccerball

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Re: HSC 2016 4U Marathon

Can someone please give this a go? I have a complex/polynomial question waiting in line and this one is actually fairly easy as I have provided all the necessary guidance to solve the question.
I can do it tomorrow no time... but post your other question?
 

KingOfActing

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Re: HSC 2016 4U Marathon

Ugh I just typed it out in latex but didn't realise I wasn't logged in so when I press send it deleted. Too lazy to latex again.

Q(z) = z^(n+1) P(1-1/z)
The roots of Q are 1/(1-r) where P(r) = 0, for r =/= 1.
Hence S_n is the sum of the roots of Q(z)
Expanding Q(z) and using the binomial expansion for the first two terms, we get Q(z) = 2z^(n+1) - (n+1)z^n + [...]
Thus S_n = -b/a = (n+1)/2

Is that right?
 

Paradoxica

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Re: HSC 2016 4U Marathon

Ugh I just typed it out in latex but didn't realise I wasn't logged in so when I press send it deleted. Too lazy to latex again.

Q(z) = z^(n+1) P(1-1/z)
The roots of Q are 1/(1-r) where P(r) = 0, for r =/= 1.
Hence S_n is the sum of the roots of Q(z)
Expanding Q(z) and using the binomial expansion for the first two terms, we get Q(z) = 2z^(n+1) - (n+1)z^n + [...]
Thus S_n = -b/a = (n+1)/2

Is that right?
No. When you transform, one of the roots becomes impossible. Observe.
 

KingOfActing

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Re: HSC 2016 4U Marathon

No. When you transform, one of the roots becomes impossible. Observe.
Wait, I feel like I just expanded wrong.

Q(z) = -(n+1)z^n +(n+1)c2 z^(n-1) + [...]

-b/a = (n+1)c2 / (n+1) = (n+1)!/(2!(n-1)!) / (n+1) = n/2

Checking up to n = 4 suggests this is correct.
 

Drsoccerball

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Re: HSC 2016 4U Marathon

Wait, I feel like I just expanded wrong.

Q(z) = -(n+1)z^n +(n+1)c2 z^(n-1) + [...]

-b/a = (n+1)c2 / (n+1) = (n+1)!/(2!(n-1)!) / (n+1) = n/2

Checking up to n = 4 suggests this is correct.
n/2 is correct
 

davidgoes4wce

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Re: HSC 2016 4U Marathon



This is from Fitzpatrick's book. I just wondering in going from



Are we expected to know the exact value of 75 degrees for this course?
 

davidgoes4wce

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Re: HSC 2016 4U Marathon

I want to jump straight from the Polar form to the Cartesian form, I mean I can sort of visualize the numbers in going from Polar to Cartersian for instance -2.732050808. I know with some calculators that have the answer in exact form (i.e surd form). I have the HP 10 s+ Scientific calculator, if anyone knows the way to set it up in exact form that would be helpful as well.
 

InteGrand

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Re: HSC 2016 4U Marathon



This is from Fitzpatrick's book. I just wondering in going from



Are we expected to know the exact value of 75 degrees for this course?
No, they got the RHS of that line by expanding the Cartesian forms of the numbers. The LHS came from consideration of the polar (aka 'mod-arg' in HSC) forms.
 

davidgoes4wce

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Re: HSC 2016 4U Marathon



Curious to know how to get the last line involving the directrices?

From what I understand about the directrices , when a>b
Foci


My thinking is if b>a,

 

davidgoes4wce

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Re: HSC 2016 4U Marathon

Only for eccentricity , do we switch the 'a' and 'b' values around. But we must revert back to the original values in calculating the foci and directrices.

I made the mistake of using the 'swapped' values.
 

InteGrand

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Re: HSC 2016 4U Marathon

Only for eccentricity , do we switch the 'a' and 'b' values around. But we must revert back to the original values in calculating the foci and directrices.

I made the mistake of using the 'swapped' values.
 
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