HSC 2016 MX2 Marathon (archive) (1 Viewer)

Drsoccerball · Feb 6, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
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?

Paradoxica · Feb 8, 2016

Re: HSC 2016 4U Marathon

$\noindent 1. Given $P(z) = z^{n+1}-1$ and $ S_n = \frac{1}{1-r_1}+\frac{1}{1-r_2}+\cdots+\frac{1}{1-r_{n-1}}+\frac{1}{1-r_n}\\$where $r_k$ is the $k^{\text{th}}$ complex root of $P(z) \\$Note that $r_0 = 1$, which does not appear anywhere in $S_n.\\ $a) By transforming $P(z)$, evaluate $S_n \\ $b) By considering $\log(P(z))$, the fundamental theorem of algebra, differentiation, and using L'H$\hat{\text{o}}$pital's rule, evaluate $S_n \\ $c) By adding terms in pairs and considering the parity of $n$, evaluate $S_n \\ $Parity refers to whether something is odd or even.$

$\noindent 2. $I_n = \int_0^\pi \frac{1-\cos{nx}}{1-\cos{x}}$d$x \\ $a) Prove the recurrence relation: $ I_{n+2}-2I_{n+1}+I_n=0 \\ $b) Evaluate $I_0$ and $I_1. \\ $c) Prove $I_n = n\pi \,\,\,\, \forall n \in \mathbb{N} $ using mathematical induction.$

KingOfActing · Feb 8, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$\noindent 1. Given $P(z) = z^{n+1}-1$ and $ S_n = \frac{1}{1-r_1}+\frac{1}{1-r_2}+\cdots+\frac{1}{1-r_{n-1}}+\frac{1}{1-r_n}\\$where $r_k$ is the $k^{\text{th}}$ complex root of $P(z) \\$Note that $r_0 = 1$, which does not appear anywhere in $S_n.\\ $a) By transforming $P(z)$, evaluate $S_n \\ $b) By considering $\log(P(z))$ and using L'H$\hat{\text{o}}$pital's rule, evaluate $S_n \\ $c) By summing pairs of terms and considering the parity of $n$, evaluate $S_n \\ $Parity refers to whether something is odd or even.$

$\noindent 2. $I_n = \int_0^\pi \frac{1-\cos{nx}}{1-\cos{x}}$d$x \\ $a) Prove the recurrence relation: $ I_{n+2}-2I_{n+1}+I_n=0 \\ $b) Evaluate $I_0$ and $I_1. \\ $c) Prove $I_n = n\pi \,\,\,\, \forall n \in \mathbb{N} $ using mathematical induction.$

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 · Feb 8, 2016

Re: HSC 2016 4U Marathon

KingOfActing said:
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 · Feb 8, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
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 · Feb 8, 2016

Re: HSC 2016 4U Marathon

KingOfActing said:
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 · Feb 12, 2016

Re: HSC 2016 4U Marathon

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

$2\sqrt{2} ($ cos $ \frac{-7 \pi}{12} + i \ sin \frac{-7 \pi}{12})=(1-\sqrt{3})+(-1-\sqrt{3}) i$

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

KingOfActing · Feb 12, 2016

Re: HSC 2016 4U Marathon

davidgoes4wce said:
Are we expected to know the exact value of 75 degrees for this course?

No. If you look at the working, you see that the values are derived from the lines above.

davidgoes4wce · Feb 12, 2016

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 · Feb 12, 2016

Re: HSC 2016 4U Marathon

davidgoes4wce said:
This is from Fitzpatrick's book. I just wondering in going from

$2\sqrt{2} ($ cos $ \frac{-7 \pi}{12} + i \ sin \frac{-7 \pi}{12})=(1-\sqrt{3})+(-1-\sqrt{3}) i$

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 · Feb 12, 2016

Re: HSC 2016 4U Marathon

Is there a mistake here from Fitzpatrick's book?

$$I thought if we wanted to express a negative power it would be somewhere along the lines of$ \ z^{-10}=r^{-10} (cos (\frac{-10 \pi}{4}) + i \ sin (\frac{-10 \pi}{4}))$

InteGrand · Feb 12, 2016

Re: HSC 2016 4U Marathon

davidgoes4wce said:
Is there a mistake here from Fitzpatrick's book?

$$I thought if we wanted to express a negative power it would be somewhere along the lines of$ \ z^{-10}=r^{-10} (cos (\frac{-10 \pi}{4}) + i \ sin (\frac{-10 \pi}{4}))$

That's exactly what Fitzpatrick did. He then converted the argument to its principal argument value.

davidgoes4wce · Feb 12, 2016

Re: HSC 2016 4U Marathon

I had something along the lines of:

$z^{-10}=(\sqrt{2})^{-10} (cos (\frac{-10 \pi}{4}) + i \ sin (\frac{-10 \pi}{4}))$
$z^{-10}=2^{-5} (cos (\frac{-10 \pi}{4}) + i \ sin (\frac{-10 \pi}{4}))$
$z^{-10}=\frac{1}{32} (cos (\frac{-10 \pi}{4}) + i \ sin (\frac{-10 \pi}{4}))$

my 'r' value is a fraction

InteGrand · Feb 12, 2016

Re: HSC 2016 4U Marathon

davidgoes4wce said:
I had something along the lines of:

$z^{-10}=(\sqrt{2})^{-10} (cos (\frac{-10 \pi}{4}) + i \ sin (\frac{-10 \pi}{4}))$
$z^{-10}=2^{-5} (cos (\frac{-10 \pi}{4}) + i \ sin (\frac{-10 \pi}{4}))$
$z^{-10}=\frac{1}{32} (cos (\frac{-10 \pi}{4}) + i \ sin (\frac{-10 \pi}{4}))$

my 'r' value is a fraction

Ah, Fitzpatrick made a mistake with his 'r'. He did the reciprocal of what it should be.

davidgoes4wce · Feb 13, 2016

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 $(h \pm ae, k) Directrices \ x=h \pm \frac{a}{e}$

My thinking is if b>a,

$Foci (h, k \pm be) $and Directrices y$=k \pm \frac{b}{e}$

InteGrand · Feb 13, 2016

Re: HSC 2016 4U Marathon

davidgoes4wce said:
Curious to know how to get the last line involving the directrices?

From what I understand about the directrices , when a>b
Foci $(h \pm ae, k) Directrices \ x=h \pm \frac{a}{e}$

My thinking is if b>a,

$Foci (h, k \pm be) $and Directrices y$=k \pm \frac{b}{e}$

Correct!

davidgoes4wce · Feb 13, 2016

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 · Feb 13, 2016

Re: HSC 2016 4U Marathon

davidgoes4wce said:
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.

$\noindent For the ellipse, the equation is $B^2 = A^2 \left( 1- e^2 \right)$, where $B$ is the semi-minor axis and $A$ is the semi-major axis (i.e. $A \geq B$, so $A$ is the bigger number).$

Paradoxica · Feb 13, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$\noindent 1. Given $P(z) = z^{n+1}-1$ and $ S_n = \frac{1}{1-r_1}+\frac{1}{1-r_2}+\cdots+\frac{1}{1-r_{n-1}}+\frac{1}{1-r_n}\\$where $r_k$ is the $k^{\text{th}}$ complex root of $P(z) \\$Note that $r_0 = 1$, which does not appear anywhere in $S_n.\\ $a) By transforming $P(z)$, evaluate $S_n \\ $b) By considering $\log(P(z))$, the fundamental theorem of algebra, differentiation, and using L'H$\hat{\text{o}}$pital's rule, evaluate $S_n \\ $c) By summing pairs of terms and considering the parity of $n$, evaluate $S_n \\ $Parity refers to whether something is odd or even.$

$\noindent 2. $I_n = \int_0^\pi \frac{1-\cos{nx}}{1-\cos{x}}$d$x \\ $a) Prove the recurrence relation: $ I_{n+2}-2I_{n+1}+I_n=0 \\ $b) Evaluate $I_0$ and $I_1. \\ $c) Prove $I_n = n\pi \,\,\,\, \forall n \in \mathbb{N} $ using mathematical induction.$

Awaiting full responses...

Drsoccerball · Feb 13, 2016

Re: HSC 2016 4U Marathon

$\noindent 2. $I_n = \int_0^\pi \frac{1-\cos{nx}}{1-\cos{x}}$d$x \\ $a) Prove the recurrence relation: $ I_{n+2}-2I_{n+1}+I_n=0 \\ $b) Evaluate $I_0$ and $I_1. \\ $c) Prove $I_n = n\pi \,\,\,\, \forall n \in \mathbb{N} $ using mathematical induction.$

$a) After rigourous expansion and simplification the final integral is :$

$2 \int_0^{\pi}{\cos{(x(n+1))}}dx = 0$

$$ b) $ I_0 =0 , I_1=\pi$

$c) Statement is true for n=0,1 (since)$

$Assume n=k-1 and n=k is true$

$I_{(k-1)}=\pi (k-1) , I_k= \pi k$

$$ Via reduction formula: $ I_{k+1} = 2I_k -I_{k-1}$

$I_{k+1}= 2\pi k -\pi k + \pi$

$\therefore I_{k+1} = \pi{(k+1)}$

$If the statement is true for n=k-1 and n=k then it is also true for n=k+1. Since n=0 and n=1 are true the statement is true for all positive n integers and n=0 .$

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