HSC 2013 MX2 Marathon (archive) (1 Viewer)

asianese · Feb 12, 2013

Re: HSC 2013 4U Marathon

I got the same result at Sy. You get $(1-y)^n-1=0$ . Expand the binomial out, leaving C(n,0) which cancels with the 1. Divide by y since we eliminate a solution. This leaves -n as the constant.

hayabusaboston · Feb 12, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
$$ The equation z^n - 1 = 0$ $ has roots: 1, z_1, z_2, \dots , z_{n-1}. $ $ Show that the constant term of the polynomial with roots: (1 - z_1), (1 - z_2), \dots , (1 - z_{n-1})$ $ is n ($ $ where n is even)$

divide z^n-1 by z-1

you get 1+z+z^2+z^3+...+z^n-1

so 1+z+z^2+z^3+...+z^n-1=(z-a1)(z-a2)...(z-asubscript n-1)

let z=1

(1-a1)(1-a2)...(1-asubscript n-1)=n

EDIT: I should probably replace a1, a2 etc with z1, z2 etc lol, to not confuse. But hey, I like realise's method

asianese · Feb 12, 2013

Re: HSC 2013 4U Marathon

$z^2-1=0 $ has roots: $ 1, z_1. $The polynomial with roots $ 1-1, 1-z_1 $ is: $\\$Let $x=1-z \Rightarrow z=1-x\\(1-x)^2-1=0 $ has roots $ 1-1, 1-z_1.\\ (1-x^2)-1\\=1-2x+x^2-1\\=x(x-2) \Rightarrow $The constant is $ -2 $ upon division by $ x\neq0.$

I think the answer is -n.

@Realise: what exactly are you saying about the roots and what does it have to do with the new equation with roots 1-z? (Confused)

@Haya: Just because a polynomial can be written as (1-z_1)(1-z_2)=k, some constant, doesn't mean the constant term of the polynomial is k.

Another example:

$(1-X)^3-1\\=1-3x+3x^2-x^3-1\\=x(x^2+3x-3)\\ $The constant in the polynomial with roots $(1-z_1), (1-z_2) $ is $-3.$

Sy123 · Feb 12, 2013

Re: HSC 2013 4U Marathon

asianese said:
$z^2-1=0 $ has roots: $ 1, z_1. $The polynomial with roots $ 1-1, 1-z_1 $ is: $\\$Let $X=1-z \Rightarrow z=1-X\\(1-X)^2-1=0 $ has roots $ 1-1, 1-z_1.\\ (1-x^2)-1\\=1-2x+x^2-1\\=x(x-2) \Rightarrow $The constant is $ -2 $ upon division by $ x\neq0.$

I think the answer is -n.

@Realise: what exactly are you saying about the roots and what does it have to do with the new equation with roots 1-z? (Confused)

@Haya: Just because a polynomial can be written as (1-z_1)(1-z_2)=k, some constant, doesn't mean the constant term of the polynomial is k.

Another example:

$(1-X)^3-1\\=1-3x+3x^2-x^3-1\\=x(x^2+3x-3)\\ $The constant in the polynomial with roots $(1-z_1), (1-z_2) $ is $-3.$

Verified for n=4

$x^4-1 = 0$

Roots $1, i, -i, -1$

Polynomial with roots:

$1-i, 1+i, 1--1$ as specified by the question

$P(x)=(x-2)(x-(1+i))(x-(1-i)) = (x-2)(x^2-4x+2) = x^3-6xx^2+10x-4$

asianese · Feb 12, 2013

Re: HSC 2013 4U Marathon

Either heroic had a typo or he was thinking of the (1-z1)(1-z2)...(1-zn-1)=n which doesn't qualify for what he's asking.

Sy123 · Feb 13, 2013

Re: HSC 2013 4U Marathon

$The point A lies on the y axis and the point B lies on the x axis$

$The point C is such that the points, A, B and C are collinear$

$$The distance between B and C is$ \ \ q$

$$The distance between A and B is$ \ \ p$

$The line is ABC makes an angle of$ \ \ \theta \ \ $with the x-axis$

$$Prove that the locus of C is an ellipse$

$Find the eccentricity of this ellipse$

asianese · Feb 13, 2013

Re: HSC 2013 4U Marathon

Does ABC make an angle of theta with the positive x-axis or negative x-axis?

Sy123 · Feb 13, 2013

Re: HSC 2013 4U Marathon

asianese said:
Does ABC make an angle of theta with the positive x-axis or negative x-axis?

HeroicPandas · Feb 13, 2013

Re: HSC 2013 4U Marathon

hayabusaboston said:
divide z^n-1 by z-1

you get 1+z+z^2+z^3+...+z^n-1

so 1+z+z^2+z^3+...+z^n-1=(z-a1)(z-a2)...(z-asubscript n-1)

let z=1

(1-a1)(1-a2)...(1-asubscript n-1)=n

EDIT: I should probably replace a1, a2 etc with z1, z2 etc lol, to not confuse. But hey, I like realise's method

yes

asianese said:
Either heroic had a typo or he was thinking of the (1-z1)(1-z2)...(1-zn-1)=n which doesn't qualify for what he's asking.

why cant (1-z1)(1-z2)...(1-zn-1)=n?

If u have cambridge, go to polynomials further questions, the last few ones and u will see "show that (1-z1)(1-z2)...(1-zn-1)=n"

doesnt (1-z1)(1-z2)...(1-zn-1)=n show that the polynomial with roots 1-z1, 1-z2, ...,1-zn has a constant term of n (using product of roots)

Sy123 · Feb 13, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
yes

why cant (1-z1)(1-z2)...(1-zn-1)=n?

If u have cambridge, go to polynomials further questions, the last few ones and u will see "show that (1-z1)(1-z2)...(1-zn-1)=n"

Ah yes but here is the thing, they are not the same thing in this case.

See for every odd degree polynomials:

$P(x)=x^{n-1}+ a_{n-2}x^{n-2}+ \dots + a_{0}$

Now, in this case n is even, the above polynomial is in the form of mine and asianese's polynomial.

If the above polynomial has roots alpha_1 alpha_2 alpha_3 .....

$\alpha_1 \alpha_2 ...\alpha_{n-1} = -1 \cdot a_0 \neq \ $constant term of the polynomial$$

HeroicPandas · Feb 14, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Ah yes but here is the thing, they are not the same thing in this case.

See for every odd degree polynomials:

$P(x)=x^{n-1}+ a_{n-2}x^{n-2}+ \dots + a_{0}$

Now, in this case n is even, the above polynomial is in the form of mine and asianese's polynomial.

If the above polynomial has roots alpha_1 alpha_2 alpha_3 .....

$\alpha_1 \alpha_2 ...\alpha_{n-1} = -1 \cdot a_0 \neq \ $constant term of the polynomial$$

oh right... i understand xD

asianese · Feb 16, 2013

Re: HSC 2013 4U Marathon

Sy123 said:

$e=\sqrt{\frac{p}{p+q}}$ ?

Kipling · Feb 16, 2013

Re: HSC 2013 4U Marathon

$P(x) is a polynomial of degree at least 2 such that P'(a) = 0. Show that when P(x) is divided by$ (x-a)^2 $the remainder is P(a)$

Sy123 · Feb 16, 2013

Re: HSC 2013 4U Marathon

asianese said:
$e=\sqrt{\frac{p}{p+q}}$ ?

Not quite, what did you get as the ellipse that satisfies C?

asianese · Feb 16, 2013

Re: HSC 2013 4U Marathon

Awks should be

$\begin{align*}e&=\sqrt{1-\left(\frac{q}{p+q}\right)^2}\\&=\frac{\sqrt{p(p+2q)}}{p+q} \end{align*}$

$\frac{x^2}{(p+q)^2}+\frac{y^2}{q^2}=1$ .

Sy123 · Feb 16, 2013

Re: HSC 2013 4U Marathon

asianese said:
Awks should be

$\begin{align*}e&=\sqrt{1-\left(\frac{q}{p+q}\right)^2}\\&=\frac{\sqrt{p(p+2q)}}{p+q} \end{align*}$

$\frac{x^2}{(p+q)^2}+\frac{y^2}{q^2}=1$ .

Correct.

==================

This is an easier one but interesting. Also I will post a solution to Kipling's question by 12:00 if no-one has posted theirs by then.

$$Express in modulus-argument form$ \ \ \ 1+\cos \theta + i\sin \theta$

hayabusaboston · Feb 16, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Correct.

==================

This is an easier one but interesting. Also I will post a solution to Kipling's question by 12:00 if no-one has posted theirs by then.

$$Express in modulus-argument form$ \ \ \ 1+\cos \theta + i\sin \theta$

dafuq sy, what happened to your characteristically hard questions?

hayabusaboston · Feb 16, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
yes

What do you mean "yes", what is that referring to? lol

Sy123 · Feb 16, 2013

Re: HSC 2013 4U Marathon

hayabusaboston said:
dafuq sy, what happened to your characteristically hard questions?

They are getting more and more non-rigorous so I decided to not make as many questions

hayabusaboston · Feb 16, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
They are getting more and more non-rigorous so I decided to not make as many questions

and how come you havent replied to my messaage?
lol
I am curious

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