HSC 2013 MX2 Marathon (archive) (1 Viewer)

Sy123 · Jan 23, 2013

Re: HSC 2013 4U Marathon

$$Let$ \ \ I_{n} = \int_{0}^{\pi} \sin^n x \ dx$

$$i) Prove that$ \ \ \ \frac{I_{n}}{I_{n-1}} = \frac{n-1}{n} \ \ \ \ \fbox{2}$

$ii) By using the above result twice at least, or otherwise, prove that$

$\prod_{n=1}^{\infty} \left (\frac{2n}{2n-1} \cdot \frac{2n}{2n+1} \right ) \equiv \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \dots = \frac{\pi}{2} \ \ \ \ \ \fbox{4}$

vbzxwgy · Jan 23, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
I fixed the wording for my answer to part (i)

(ii)

$$for$ \ \ c>0$

Establish the fact that there will always exist some complex u |u|=1 , such that:

$|u-z| < c \ \ \ $for$ \ \ c>0$

There are an infinitely many u such that this condition is true, for ANY z.

Now, if we look at the complex number:

$cis nr\pi$

For $n=1,2,3,4 ...$ That complex number

$cis nr\pi$ is a 'region' of the WHOLE unit circle, that is, every and any complex number will satisfy $cis nr\pi$ if and only if r is irrational, this is the converse from the result proven in part (i)

Hence there must exist infinitely many cis nrpi such that the inequality holds true for any z.

Q.E.D

EDIT: Fixed the proof.

probably shouldnt put qed unless u are sure mate.

"cis(nr*pi) is a 'region' of the WHOLE unit circle, that is, every and any complex number will satisfy if and only if r is irrational, this is the converse from the result proven in part (i)"

its not true that every complex number on the circle will be cis(nr*pi) for some n whenever r is irrational if that is what you are trying to say. part (i) says nothing of this sort.

Sy123 · Jan 23, 2013

Re: HSC 2013 4U Marathon

vbzxwgy said:
probably shouldnt put qed unless u are sure mate.

"cis(nr*pi) is a 'region' of the WHOLE unit circle, that is, every and any complex number will satisfy if and only if r is irrational, this is the converse from the result proven in part (i)"

its not true that every complex number on the circle will be cis(nr*pi) for some n whenever r is irrational if that is what you are trying to say. part (i) says nothing of this sort.

Well I was quite sure so I put it there.

It is the converse argument from the first part.
The first part says that there are finitely many values if and only if r is rational
Conversely, there are infinitely man values if and only if r is irrational. This is the condition given in the question.

If there are infinitely many values for cis nr pi for all n=1,2,3.. and so on, for infinitely many values of n, cis n pi r will satisfy the every complex number of modulus 1.

There exists infinitely many values of u such that |u-z| < c for any c.
Of these infinitely many values of u, cis npi r covers infinitely many of them for infinitely many n. That should finish it shouldn't it?

vbzxwgy · Jan 23, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Well I was quite sure so I put it there.

It is the converse argument from the first part.
The first part says that there are finitely many values if and only if r is rational
Conversely, there are infinitely man values if and only if r is irrational. This is the condition given in the question.

If there are infinitely many values for cis nr pi for all n=1,2,3.. and so on, for infinitely many values of n, cis n pi r will satisfy the every complex number of modulus 1.

There exists infinitely many values of u such that |u-z| < c for any c.
Of these infinitely many values of u, cis npi r covers infinitely many of them for infinitely many n. That should finish it shouldn't it?

"Of these infinitely many values of u, cis npi r covers infinitely many of them for infinitely many n"

no

infinitely many integers are even
infinitely many integers are odd
therefore there must be infinitely many integers that are both even and odd at the same time

qed

Sy123 · Jan 23, 2013

Re: HSC 2013 4U Marathon

vbzxwgy said:
"Of these infinitely many values of u, cis npi r covers infinitely many of them for infinitely many n"

no

infinitely many integers are even
infinitely many integers are odd
therefore there must be infinitely many integers that are both even and odd at the same time

qed

cis npi r can cover the whole unit circle, since it satisfies infinitely many values for integers n.

That is, u= cis npi r for some irrational r and number n.

Now there are infinitely many u, such that |u-z| < c

So it must follow that there is an infinite number of cis n pi r which satisfy |cis npi r - z| < c

vbzxwgy · Jan 23, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
cis npi r can cover the whole unit circle, since it satisfies infinitely many values for integers n.

That is, u= cis npi r for some irrational r and number n.

Now there are infinitely many u, such that |u-z| < c

So it must follow that there is an infinite number of cis n pi r which satisfy |cis npi r - z| < c

did you not read my example of why that logic doesn't work? you are doing exactly the same thing as in my obviously erroneous argument. think about it.

braintic · Jan 23, 2013

Re: HSC 2013 4U Marathon

I have not read the original question. But regarding this most recent point, Sy123, I don't think you understand the concept of there being different infinities. You are trying to equate two different infinities.

Sy123 · Feb 2, 2013

Re: HSC 2013 4U Marathon

Here is something I have been working on:

$$(i) Prove that the parabola$ \ \ x^2=4ay \ \ $is the locus of equi-distance from a straight line and a fixed point$ \ \ \fbox{2}$

$$Consider the circle$ \ \ x^2+y^2=a^2$

$$Consider the points$ \ \ D(a\cos \theta, a\sin \theta \ \ \ F(-a\cos \theta, -a\sin \theta)$

$A tangent to the circle at the point$ \ \ D \ \ $ is constructed$

$Using the definition of the parabola proven in part (i)$

$$(ii) Prove that a parabola rotated by$ \ \ \theta \ \ $ radians, with the vertex at the origin has the equation$ \ \ \ \ \fbox{4}$

$x^2 \sin^2 \theta + y^2 \cos^2 \theta -xy\sin 2\theta + 4a(x\cos \theta + y\sin \theta)$

EDITED

EDITED Mk2

vbzxwgy · Feb 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Here is something I have been working on:

$$(i) Prove that the parabola$ \ \ x^2=4ay \ \ $is the locus of equi-distance from a straight line and a fixed point$ \ \ \fbox{2}$

$$Consider the circle$ \ \ x^2+y^2=a^2$

$$Consider the points$ \ \ D(a\cos \theta, a\sin \theta \ \ \ F(-a\cos \theta, -a\sin \theta)$

$A tangent to the circle at the point$ \ \ D \ \ $ is constructed$

$Using the definition of the parabola proven in part (i)$

$$(ii) Prove that a parabola rotated by$ \ \ \theta \ \ $ radians, with the vertex at the origin has the equation$ \ \ \ \ \fbox{4}$

$x^2\left (1-\frac{\cos^2 \theta}{a^2}\right ) + y^2 \left (1-\frac{\sin^2 \theta}{a^2} \right ) - xy \frac{\sin 2\theta}{a^2} + 2x\cos \theta \left (a+ \frac{1}{a} \right ) + 2y\sin \theta \left (a+\frac{1}{a} \right ) + a^2 = 1$

wrong. let theta be zero for example and it doesnt work.

heres a better way of rotating parabolas.

if we identify complex numbers and the points in the argand plane that represent them, the standard form parabola is given parametrically by:

z(t)=2at+iat^2.

to rotate by @ counterclockwise, multiply by cis(@).

gives (2atcos(@)-at^2sin(@))+i(2atsin(@)+at^2cos(@))

so x=2atcos(@)-at^2sin(@)
y=2atsin(@)+at^2cos(@)

for our rotated parabola.

so ycos(@)-xsin(@)=at^2=(2at)^2/4a=(xcos(@)+ysin(@))^2/4a.

finally leading to x^2cos^2(@)+y^2sin^2(@)+2xysin(@)cos(@)+4a(xsin(@)-ycos(@))=0

the correct cartesian equation.

Sy123 · Feb 2, 2013

Re: HSC 2013 4U Marathon

vbzxwgy said:
wrong. let theta be zero for example and it doesnt work.

heres a better way of rotating parabolas.

if we identify complex numbers and the points in the argand plane that represent them, the standard form parabola is given parametrically by:

z(t)=2at+iat^2.

to rotate by @ counterclockwise, multiply by cis(@).

gives (2atcos(@)-at^2sin(@))+i(2atsin(@)+at^2cos(@))

so x=2atcos(@)-at^2sin(@)
y=2atsin(@)+at^2cos(@)

for our rotated parabola.

so ycos(@)-xsin(@)=at^2=(2at)^2/4a=(xcos(@)+ysin(@))^2/4a.

finally leading to x^2cos^2(@)+y^2sin^2(@)+2xysin(@)cos(@)+4a(xsin(@)-ycos(@))=0

the correct cartesian equation.

If you put in theta=0 and a=1

y^2=-4x

Which is correct because the focus is at (-1,0) and directrix at x=1....
It is also correct for theta = pi/2, pi, 3pi/2

vbzxwgy · Feb 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
If you put in theta=0 and a=1

y^2=-4x

Which is correct because the focus is at (-1,0) and directrix at x=1....
It is also correct for theta = pi/2, pi, 3pi/2

lol convenient that you only checked situations with a=1, this is the only time your equation is correct.

Sy123 · Feb 2, 2013

Re: HSC 2013 4U Marathon

vbzxwgy said:
lol convenient that you only checked situations with a=1, this is the only time your equation is correct.

Looking over my working for the third time, I found that when I should of multiplied by 1/a I instead multiplied by 1/a^2 which is why it only works for a=1

The result of:

$(X+a\cos \theta)^2+(Y+a\sin \theta)^2 = \frac{1}{a}(X\cos \theta + Y\sin \theta - a)^2$

Should be the answer.

vbzxwgy · Feb 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Looking over my working for the third time, I found that when I should of multiplied by 1/a I instead multiplied by 1/a^2 which is why it only works for a=1

The result of:

$(X+a\cos \theta)^2+(Y+a\sin \theta)^2 = \frac{1}{a}(X\cos \theta + Y\sin \theta - a)^2$

Should be the answer.

no, that's still wrong.

Sy123 · Feb 2, 2013

Re: HSC 2013 4U Marathon

vbzxwgy said:
no, that's still wrong.

With the conditions above, create the variable point P(X,Y)

$PF=\sqrt{(X+a\cos \theta)^2+(Y+a\sin \theta)^2}$

The line of the tangent at D is:

$y-a\sin \theta = \frac{-\cos \theta}{\sin \theta} (x-a\cos \theta)$

$y\sin \theta + x\cos \theta -a$

Use perpendicular distance formula for PD' (D' being the point on the line)

$PD' =\frac{|X\cos \theta + Y\sin \theta - a|}{\sqrt{(\cos^2 \theta + \sin^2 \theta)}}$

$PD' = |X\cos \theta + Y\sin \theta -a|$

Square both and equate (my working was right the first time, it is supposed to be 1/a^2)

What is wrong here?

EDIT: Nevermind, I saw what I did wrong and I now fixed it, a should not have been in the denominator, it was a silly mistake:

$(x+a\cos \theta)^2+(y+a\sin \theta)^2=(x\cos \theta + y\sin \theta -a)^2$

Should be correct

vbzxwgy · Feb 2, 2013

Re: HSC 2013 4U Marathon

cbf fixing your working for you. i posted a correct answer with proof above.

Sy123 · Feb 2, 2013

Re: HSC 2013 4U Marathon

$P(x) $is a polynomial of degree$ \ \ n \ \ $with rational co-efficients$$

$$The leading co-efficient is$ \ \ a_0 \ \ $and the polynomial has roots$ \ \ \ \alpha_1, \alpha_2, .... , \alpha_n$

$Prove$

$P'(x)=\frac{P(x)}{x-\alpha_1} + \frac{P(x)}{x-\alpha_2}+\frac{P(x)}{x-\alpha_3}+ \dots + \frac{P(x)}{x-\alpha_n}$

HeroicPandas · Feb 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$P(x) $is a polynomial of degree$ \ \ n \ \ $with rational co-efficients$$

$$The leading co-efficient is$ \ \ a_0 \ \ $and the polynomial has roots$ \ \ \ \alpha_1, \alpha_2, .... , \alpha_n$

$Prove$

$P'(x)=\frac{P(x)}{x-\alpha_1} + \frac{P(x)}{x-\alpha_2}+\frac{P(x)}{x-\alpha_3}+ \dots + \frac{P(x)}{x-\alpha_n}$

1997 4U HSC Sydney boys? I remmeber doing this!!

Sy123 · Feb 2, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
1997 4U HSC Sydney boys? I remmeber doing this!!

Correct, it was a really good question

HeroicPandas · Feb 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Correct, it was a really good question

Yeh man!

$$ Since P(x) has roots \alpha_1, \alpha_2, ..., \alpha_n \therefore P(x) = (x - \alpha_1)(x - \alpha_2)\dots(x - \alpha_n) \\ \\$ $ Log both sides and applying the useful rule: log_e ab = log_e a + log_e b\\ \\ ln(P(x)) = ln(x-\alpha_1) + \dots +ln(x-\alpha_n) \\ \\ $ $Differentiating both sides with respect to x \\ \\ \therefore \frac{P'(x)}{P(x)} = \frac{1}{x-\alpha_1} + \frac{1}{x-\alpha_2} +\dots +\frac{1}{x-\alpha_n} \\ \\$ $ Times both sides by P(x) and zu end up with ze answer!$

Helvetica · Feb 2, 2013

Re: HSC 2013 4U Marathon

Let m and n be the roots of the equation $t^{2}+bt+c=0$ . Show that b and c are the roots of the equation $t^{2}+(m+n-mn)t-mn(m+n)=0$ .

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