HSC 2013 MX2 Marathon (archive) (1 Viewer)

Registered User · May 6, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Prove that$

$\sum_{r=0}^{n} \frac{(-1)^{r}}{m+r+1} \binom{n}{r} = \frac{m! n!}{(m+n+1)!}$

The RHS is the beta function https://en.wikipedia.org/wiki/Beta_function, I'm I allowed to use it to prove this?

Sy123 · May 6, 2013

Re: HSC 2013 4U Marathon

Registered User said:
The RHS is the beta function https://en.wikipedia.org/wiki/Beta_function, I'm I allowed to use it to prove this?

Definitely not

Though I am curious to see how you would do so.

EDIT: Yes that is the method that I had in mind for this question (though I didn't use the Beta function specifically), you need to 'think up of' a function such that you integrate you get the LHS, the RHS comes through.....

Sy123 · May 6, 2013

Re: HSC 2013 4U Marathon

Registered User said:
Do you do this questions using the conjugate root theorem?

Well I didn't, you may find a way using it, this question is more about argument than algebraic skill (like my series question just now)

Registered User · May 6, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Prove that$

$\sum_{r=0}^{n} \frac{(-1)^{r}}{m+r+1} \binom{n}{r} = \frac{m! n!}{(m+n+1)!}$

Good question but 'thinking up of' a function etc. is very tricky and I think this question would be more 4U level if you said consider $\sum_{r=0}^{n} (-1)^{r} \binom{n}{r} x^{m+r}$ hence prove etc.

$Consider the sum:$

$\sum_{r=0}^{n} (-1)^{r} \binom{n}{r} x^{m+r} = x^m \sum_{r=0}^{n} (-1)^{r} \binom{n}{r} x^{r} = x^m(1-x)^{n}$

$Integrating both sides of the above equation with respect to x from 0 to 1, we have$

$\sum_{r=0}^{n} (-1)^{r} \binom{n}{r}\frac{1}{m+r+1} x^{m+r+1}\Big|_{0}^{1}=\int_{0}^{1}x^m(1-x)^{n} dx$

$\sum_{r=0}^{n} (-1)^{r} \binom{n}{r}\frac{1}{m+r+1}=\int_{0}^{1}x^m(1-x)^{n} dx$

$Now find$ $I_{m,n}=\int_{0}^{1}x^m(1-x)^{n} dx$

$Let$ $u=(1-x)^n, du=-n(1-x)^{n-1}dx, dv=x^{m}dx, v=\frac{x^{m+1}}{m+1}$

$I_{m,n}=\frac{n}{m+1}\int_{0}^{1}x^{m+1}(1-x)^{n-1}$

$\therefore I_{m,n}=\frac{n}{m+1}I_{m+1,n-1}$

$I_{m,n}=\frac{n}{m+1}I_{m+1,n-1}$

$I_{m+1,n-1}=\frac{n-1}{m+2}I_{m+2,n-2}$

$I_{m+2,n-2}=\frac{n-2}{m+3}I_{m+3,n-3}$

$I_{m+3,n-3}=\frac{n-3}{m+4}I_{m+4,n-4}$

$\cdot$

$\cdot$

$\cdot$

$I_{m+n-1,1}=\frac{1}{m+n}I_{m+n,0}$

$I_{m+n,0}=\int_{0}^{1}x^{m+n}dx=\frac{x^{m+n+1}}{m+n+1}\Big|_{0}^{1}=\frac{1}{m+n+1}$

$I_{m,n}=\frac{n}{m+1}\cdot \frac{n-1}{m+2}\cdot \frac{n-2}{m+3}\cdot \frac{n-3}{m+4} \cdots \frac{1}{m+n} \cdot \frac{1}{m+n+1}=\frac{m!n!}{(m+n+1)!}$

$\therefore \sum_{r=0}^{n} (-1)^{r} \binom{n}{r}\frac{1}{m+r+1}= I_{m,n} = \beta(m+1,n+1) = \frac{m!n!}{(m+n+1)!}$

$\square$

Sy123 · May 6, 2013

Re: HSC 2013 4U Marathon

Registered User said:
Good question but 'thinking up of' a function etc. is very tricky and I think this question would be more 4U level if you said consider $\sum_{r=0}^{n} (-1)^{r} \binom{n}{r} x^{m+r}$ hence prove etc.

$Consider the sum:$

$\sum_{r=0}^{n} (-1)^{r} \binom{n}{r} x^{m+r} = x^m \sum_{r=0}^{n} (-1)^{r} \binom{n}{r} x^{r} = x^m(1-x)^{n}$

$Integrating both sides of the above equation with respect to x from 0 to 1, we have$

$\sum_{r=0}^{n} (-1)^{r} \binom{n}{r}\frac{1}{m+r+1} x^{m+r+1}\Big|_{0}^{1}=\int_{0}^{1}x^m(1-x)^{n} dx$

$\sum_{r=0}^{n} (-1)^{r} \binom{n}{r}\frac{1}{m+r+1}=\int_{0}^{1}x^m(1-x)^{n} dx$

$Now integrate$ $I_{m,n}=\int_{0}^{1}x^m(1-x)^{n} dx$

$$Let$[tex] [tex]u=(1-x)^n, du=-n(1-x)^{n-1}dx, dv=x^{m}dx, v=\frac{x^{m+1}}{m+1}$

$I_{m,n}=\frac{n}{m+1}\int_{0}^{1}x^{m+1}(1-x)^{n-1}$

$\therefore I_{m,n}=\frac{n}{m+1}I_{m+1,n-1}$

$I_{m,n}=\frac{n}{m+1}I_{m+1,n-1}$

$I_{m+1,n-1}=\frac{n-1}{m+2}I_{m+2,n-2}$

$I_{m+2,n-2}=\frac{n-2}{m+3}I_{m+3,n-3}$

$I_{m+3,n-3}=\frac{n-3}{m+4}I_{m+4,n-4}$

$\cdot$

$\cdot$

$\cdot$

$I_{m+n-1,1}=\frac{1}{m+n}I_{m+n,0}$

$I_{m+n,0}=\int_{0}^{1}x^{m+n}dx=\frac{x^{m+n+1}}{m+n+1}\Big|_{0}^{1}=\frac{1}{m+n+1}$

$I_{m,n}=\frac{n}{m+1}\cdot \frac{n-1}{m+2}\cdot \frac{n-2}{m+3}\cdot \frac{n-3}{m+4} \cdots \frac{1}{m+n} \cdot \frac{1}{m+n+1}=\frac{m!n!}{(m+n+1)!}$

$\therefore \sum_{r=0}^{n} (-1)^{r} \binom{n}{r}\frac{1}{m+r+1}= \beta(m+1,n+1)=\frac{m!n!}{(m+n+1)!}$

$\square$

Good job.

Ah yes but I've seen 3U questions (like in 2012 Independent) where they simply replace the m with 2. Because that is easier to integrate at 3U level (substitution).

But looking at the LHS 'reminds' you with something to do with integration, which may lead to thinking of $\int_0^1 x^m(1-x)^n \ dx$

===========

$Prove that$

$\sum_{n=1}^{\infty} \frac{\sin(n\theta)}{n} = \frac{\pi}{2} - \frac{\theta}{2} \ \ \ \ \ \fbox{7}$

$0 \leq \theta \leq 2\pi$

You must also prove the domain of theta.

You must be rigorous (as possible) in your approach to the operations with infinite sequences.
The mark count is an estimated length and value of the question if it appeared in a HSC exam.

seanieg89 · May 7, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Good job.

Ah yes but I've seen 3U questions (like in 2012 Independent) where they simply replace the m with 2. Because that is easier to integrate at 3U level (substitution).

But looking at the LHS 'reminds' you with something to do with integration, which may lead to thinking of $\int_0^1 x^m(1-x)^n \ dx$

===========

$Prove that$

$\sum_{n=1}^{\infty} \frac{\sin(n\theta)}{n} = \frac{\pi}{2} - \frac{\theta}{2} \ \ \ \ \ \fbox{6}$

You must be rigorous (as possible) in your approach to the operations with infinite sequences.
The mark count is an estimated length and value of the question if it appeared in a HSC exam.

Good question, I hope some students can make progress on it as it certainly can be done using syllabus methods.

(Just a quick note, the desired equality will only hold in the interval (0,2*pi). Using the 2pi-periodicity of the LHS and the fact that it is trivially zero at even multiples of pi this tells us what the LHS is equal to everywhere.)

Sy123 · May 7, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Good question, I hope some students can make progress on it as it certainly can be done using syllabus methods.

(Just a quick note, the desired equality will only hold in the interval (0,2*pi). Using the 2pi-periodicity of the LHS and the fact that it is trivially zero at even multiples of pi this tells us what the LHS is equal to everywhere.)

Yep that's true, edited.

seanieg89 · May 7, 2013

Re: HSC 2013 4U Marathon

Registered User said:
Do you do this questions using the conjugate root theorem?

Keeping the ideas behind the complex conjugate root theorem at the back of your mind leads to a pretty slick proof.

Registered User · May 7, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Prove that$

$\sum_{n=1}^{\infty} \frac{\sin(n\theta)}{n} = \frac{\pi}{2} - \frac{\theta}{2} \ \ \ \ \ \fbox{7}$

$0 \leq \theta \leq 2\pi$

This is a simple Fourier series called sawtooth function.
I still can't find a way to do it using MX2 methods.
Can you give a hint?

Sy123 · May 7, 2013

Re: HSC 2013 4U Marathon

Registered User said:
This is a simple Fourier series called sawtooth function.
I still can't find a way to do it using MX2 methods.
Can you give a hint?

HINT: Integration

Registered User · May 8, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
In this question you may assume that:

$cos(x) = \sum_0^{\infty} (-1)^k \frac{x^{2k}}{(2k)!}$

$i) Explain why the polynomial$

$P(x) = \frac{(x-a)(x-b)(x-c)(x-d)}{abcd}$

$$has roots a,b,c,d and$~P(0)=1$

$ii) Show that$

$P(x) = (1-\frac{x}{a})(1-\frac{x}{b})(1-\frac{x}{c})(1-\frac{x}{d})$

$iii) Assuming the result in part (ii) holds for polynomials of infinite degree, prove that$

$\sum_{k=1}^{\infty} \frac{1}{(2k-1)^2} = \frac{\pi^2}{8}$

I tried to sub the roots of cosx into part (ii) then differentiate but I can't get the answer yet

I know that I have to follow the steps blah blah blah but it can be proved much more rigorously using $\zeta (2)$

There are many proofs of $\zeta (2)$ here http://www.uam.es/personal_pdi/ciencias/cillerue/Curso/zeta2.pdf

$So if$ $\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}=\dfrac{\pi ^{2}}{6}$

$Then$

$\sum_{n=1}^\infty \frac{1}{(2n)^2}=\frac{\pi^2}{24}$

$And, finally:$

$\sum_{n=1}^\infty \frac{1}{(2n-1)^2} = \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty \frac{1}{(2n)^2} = \frac{3\pi^2}{24} = \frac{\pi^2}{8}$

RealiseNothing · May 8, 2013

Re: HSC 2013 4U Marathon

Registered User said:
I tried to sub the roots of cosx into part (ii) then differentiate but I can't get the answer yet
I know that I have to follow the steps blah blah blah but it can be proved much more rigorously using $\zeta (2)$

There are many proofs of $\zeta (2)$ here http://www.uam.es/personal_pdi/ciencias/cillerue/Curso/zeta2.pdf

$So if$ $\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}=\dfrac{\pi ^{2}}{6}$

$Then$

$\sum_{n=1}^\infty \frac{1}{(2n)^2}=\frac{\pi^2}{24}$

$And, finally:$

$\sum_{n=1}^\infty \frac{1}{(2n-1)^2} = \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty \frac{1}{(2n)^2} = \frac{3\pi^2}{24} = \frac{\pi^2}{8}$

You prove $\zeta (2)$ basically the same way as my question lol, you just use sine instead of cosine. But of course without the assumptions and all. This was actually the original method Euler used to prove it.

RealiseNothing · May 8, 2013

Re: HSC 2013 4U Marathon

Registered User said:
There are many proofs of $\zeta (2)$ here http://www.uam.es/personal_pdi/ciencias/cillerue/Curso/zeta2.pdf

Proof 11, shades of 2010 HSC.

study1234 · May 8, 2013

Re: HSC 2013 4U Marathon

(From Patel - Ex 4O Q17)

The point P (which represents z = x +iy) moves in a straight line parallel to the imaginary axis. Prove that the point Q which represents z^2 moves in a certain parabola. Find the focus. Also describe the locus of Q when P moves on the imaginary axis.

Sy123 · May 8, 2013

Re: HSC 2013 4U Marathon

study1234 said:
(From Patel - Ex 4O Q17)

The point P (which represents z = x +iy) moves in a straight line parallel to the imaginary axis. Prove that the point Q which represents z^2 moves in a certain parabola. Find the focus. Also describe the locus of Q when P moves on the imaginary axis.

$z=k+it$ for some constant k, and variable t.

$z^2 = k^2 - t^2 +i 2kt$

Then taking the parametric equations of them:

$x = k^2 - t^2$

$y = 2kt$

$x= k^2 - \frac{y^2}{4k^2}$

which is a parabola since k is constant and x is linear while y is quadratic.

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

$Prove$

$2^n \cos \left (\frac{n \pi}{2} \right ) = \sum_{k=0}^{n} (-1)^k \binom{2n}{2k}$

asianese · May 8, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Proof 11, shades of 2010 HSC.

I believe Daniel Daners of USYD contributed to the basel problem for the 2010 HSC:

http://www.maths.usyd.edu.au/u/daners/publ/abstracts/zeta2/zeta2.pdf

He notes:

Acknowledgement This work is derived from work done for the Board of
Studies of NSW, Australia.

Registered User · May 8, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Prove that$

$\sum_{n=1}^{\infty} \frac{\sin(n\theta)}{n} = \frac{\pi}{2} - \frac{\theta}{2} \ \ \ \ \ \fbox{7}$

$0 \leq \theta \leq 2\pi$

You must also prove the domain of theta.

You must be rigorous (as possible) in your approach to the operations with infinite sequences.
The mark count is an estimated length and value of the question if it appeared in a HSC exam.

I think it should be $0 < \theta < 2\pi$

Can I do this using Fourier or Taylor series since they use elementary functions and some integration?

Registered User · May 8, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
You prove $\zeta (2)$ basically the same way as my question lol, you just use sine instead of cosine. But of course without the assumptions and all. This was actually the original method Euler used to prove it.

haha nice, I didn't know this was the original method. I heard that his original method wasn't very rigorous so now I know why.

Which one is your favorite method to compute $\zeta (2)$ ?

RealiseNothing · May 8, 2013

Re: HSC 2013 4U Marathon

Registered User said:
haha nice, I didn't know this was the original method. I heard that his original method wasn't very rigorous so now I know why.

Which one is your favorite method to compute $\zeta (2)$ ?

I quite like the original, mostly because a lot of the others are too advanced for me to understand right now haha.

HeroicPandas · May 8, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$The polynomial$ \ \ P(x) \ \ $has integer co-efficients$

$p+\sqrt{q} \ \ $is a root of the polynomial, where$ \ \ p, q\ \ $are rational$$

$Prove that$ \ \ p-\sqrt{q} \ \ $must also be a root of the polynomial$

hm...

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