HSC 2013 MX2 Marathon (archive) (10 Viewers)

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

The way I did it was to use:

$\sum_{k=1}^{n} kcis(k\theta)$

Then equate imaginary parts to get the result:

$\frac{sin(n\theta) + nsin(\theta)[1-sin(n\theta)]}{2-2cos\theta} = \sum_{k=1}^{n} ksin(k\theta)$

Which is wrong, so idk.

Carrotsticks · Dec 27, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
The way I did it was to use:

$\sum_{k=1}^{n} kcis(k\theta)$

Then equate imaginary parts to get the result:

$\frac{sin(n\theta) + nsin(\theta)[1-sin(n\theta)]}{2-2cos\theta} = \sum_{k=1}^{n} ksin(k\theta)$

Which is wrong, so idk.

Can you tell us how exactly you acquired

$\sum_{k=1}^{n} k cis (k\theta)$

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
Can you tell us how exactly you acquired

$\sum_{k=1}^{n} k cis (k\theta)$

$cis\theta + 2cis2\theta + 3cis3\theta + ... + ncisn\theta$

$cis\theta + 2cis^2\theta + 3cis^3\theta + ... + ncis^n\theta$ by De Moivre's Theorem

$(cis\theta + cis^2\theta + cis^3\theta + ...) + (cis^2\theta + cis^3\theta + ...) + (cis^3\theta + ...)$

Sum each one individually (a neat pattern arises). Then add them together and equate imaginary components.

Carrotsticks · Dec 27, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
$cis\theta + 2cis2\theta + 3cis3\theta + ... + ncisn\theta$

$cis\theta + 2cis^2\theta + 3cis^3\theta + ... + ncis^n\theta$ by De Moivre's Theorem

$(cis\theta + cis^2\theta + cis^3\theta + ...) + (cis^2\theta + cis^3\theta + ...) + (cis^3\theta + ...)$

Sum each one individually (a neat pattern arises). Then add them together and equate imaginary components.

Yep, so quite like the first method I mentioned. That method SHOULD work, but there is much larger room for errors to be made. Try finding a closed form again, but using the second method I provided.

Sy123 · Dec 27, 2012

Re: HSC 2013 4U Marathon

$$Denote$ \ \ \ \ S_k(n) \equiv \sum_{r=0}^n r^k$

$$You may assume$ \ \ \ S_1(n)=\frac{n}{2}(n+1) \ \ \ \ \ S_2 (n) = \frac{n}{6}(n+1)(2n+1)$

$$i) By considering $ \sum_{r=0}^n (r+1)^k - r^k$

$$Show that$ \\ \\ kS_{k-1}(n) = (n+1)^k - (n+1) - \binom{k}{2}S_{k-2}(n) - \binom{k}{3}S_{k-3}(n) - ... - \binom{k}{k-1}S_1 (n)$

$ii) Hence, explain why$ \ \ S_k(n) \ \ $ is a polynomial of degree $ \ \ k+1 \ \ $ in $ \ \ n \ \ $ Show that this polynomial is divisible by n, and the sum of the coefficients is 1$

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

$$i)$ \sum_{r=0}^{n}(r+1)^k - r^k = (n+1)^k$ by a telescoping sum.

Now expanding the binomial first instead of using a telescoping sum gives:

$(r+1)^k = \binom{k}{0} + \binom{k}{1}r + \binom{k}{2}r^2 + ... + \binom{k}{k}r^k$

Now if we apply the summation, we sum each one individually, that is:

$\sum_{r=0}^{n}(r+1)^k = \sum_{r=0}^{n}\binom{k}{0} + \sum_{r=0}^{n}\binom{k}{1}r + ... + \sum_{r=0}^{n}\binom{k}{k}r^k$

Now using our definition of $S_k(n)$ we obtain:

$\sum_{r=0}^{n}(r+1)^k = (n+1) + \binom{k}{1}S_1(n) + \binom{k}{2}S_2(n) + ... + \binom{k}{k-1}S_{k-1}(n) + \binom{k}{k}S_k(n)$

Using the symmetry property of binomial expansions gives:

$\sum_{r=0}^{n}(r+1)^k = (n+1) +\binom{k}{k-1}S_1(n) + \binom{k}{k-2}S_2(n) + ... + \binom{k}{1}S_{k-1}(n) + \binom{k}{0}S_k(n)$

Now since $\binom{k}{1} = k$ and $\binom{k}{0} = 1$ we get:

$\sum_{r=0}^{n}(r+1)^k = (n+1) +\binom{k}{k-1}S_1(n) + \binom{k}{k-2}S_2(n) + ... + kS_{k-1}(n) + S_k(n)$

Now consider the second part of the equation:

$\sum_{r=0}^{n} -r^k = -\sum_{r=0}^{n} r^k = -S_k(n)$ by definition.

Hence if we add the two parts together we obtain:

$\sum_{r=0}^{n}(r+1)^k - r^k = (n+1) +\binom{k}{k-1}S_1(n) + \binom{k}{k-2}S_2(n) + ... + kS_{k-1}(n) + S_k(n) - S_k(n)$

So the two terms on the end cancel out and give:

$\sum_{r=0}^{n}(r+1)^k - r^k = (n+1) +\binom{k}{k-1}S_1(n) + \binom{k}{k-2}S_2(n) + ... + kS_{k-1}(n)$

But this was equal to the value of the telescoping sum, so we can equate the two and get:

$(n+1)^k = \sum_{r=0}^{n}(r+1)^k - r^k = (n+1) +\binom{k}{k-1}S_1(n) + \binom{k}{k-2}S_2(n) + ... + kS_{k-1}(n)$

Taking everything from the RHS to the LHS except for the last term gives us the required result of:

$kS_{k-1}(n) = (n+1)^k - (n+1) - \binom{k}{k-1}S_1(n) - \binom{k}{k-2}S_2(n) - ... - \binom{k}{2}S_{k-2}(n)$

Which can be re-written in the form asked:

$kS_{k-1}(n) = (n+1)^k - (n+1) - \binom{k}{2}S_{k-2}(n) - \binom{k}{3}S_{k-3}(n) - ... - \binom{k}{k-1}S_1(n)$

Quod Erat Demonstrandum.

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

I'll do part (ii) when my hands can be bothered latexing again lol.

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

How do I make it so that it is all on one line and doesn't go to the next?

Carrotsticks · Dec 27, 2012

Re: HSC 2013 4U Marathon

Don't think you can do that. It auto formats and goes into the next line when too long.

bleakarcher · Dec 27, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
Don't think you can do that. It auto formats and goes into the next line when too long.

Carrot, mind sharing some of the rejected BOS trial questions?

Carrotsticks · Dec 27, 2012

Re: HSC 2013 4U Marathon

bleakarcher said:
Carrot, mind sharing some of the rejected BOS trial questions?

2013 ones?

bleakarcher · Dec 27, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
2013 ones?

yeah, whichever ones you have to spare. I don't mind if it's a 2012 one.

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$ii) Hence, explain why$ \ \ S_k(n) \ \ $ is a polynomial of degree $ \ \ k+1 \ \ $ in $ \ \ n \ \ $ Show that this polynomial is divisible by n, and the sum of the coefficients is 1$

Given:

$kS_{k-1}(n) = \sum_{r=0}^{n} kr^{k-1}$

Integrating the RHS gives:

$\sum_{r=0}^{n} r^k$

$=S_k(n)$

Therefore:

$\int kS_{k-1}(n) = S_k(n)$

Thus we can integrate the following to obtain $S_k(n)$ on the LHS:

$kS_{k-1}(n) = (n+1)^k - (n+1) - \binom{k}{2}S_{k-2}(n) - \binom{k}{3}S_{k-3}(n) - ... - \binom{k}{k-1}S_1 (n)$

Since the RHS is a polynomial in 'n' itself with highest degree 'k', it follows that after integration, the highest power will become 'k+1', and hence $S_k(n)$ is a polynomial of degree 'k+1' in 'n'.

bleakarcher · Dec 27, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Given:

$kS_{k-1}(n) = \sum_{r=0}^{n} kr^{k-1}$

Integrating the RHS gives:

$\sum_{r=0}^{n} r^k$

$=S_k(n)$

Therefore:

$\int kS_{k-1}(n) = S_k(n)$

Thus we can integrate the following to obtain $S_k(n)$ on the LHS:

$kS_{k-1}(n) = (n+1)^k - (n+1) - \binom{k}{2}S_{k-2}(n) - \binom{k}{3}S_{k-3}(n) - ... - \binom{k}{k-1}S_1 (n)$

Since the RHS is a polynomial in 'n' itself with highest degree 'k', it follows that after integration, the highest power will become 'k+1', and hence $S_k(n)$ is a polynomial of degree 'k+1' in 'n'.

You could have also replaced k with k+1.

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

bleakarcher said:
You could have also replaced k with k+1.

lol that is so much easier.

Sy123 · Dec 27, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Given:

$kS_{k-1}(n) = \sum_{r=0}^{n} kr^{k-1}$

Integrating the RHS gives:

$\sum_{r=0}^{n} r^k$

$=S_k(n)$

Therefore:

$\int kS_{k-1}(n) = S_k(n)$

Thus we can integrate the following to obtain $S_k(n)$ on the LHS:

$kS_{k-1}(n) = (n+1)^k - (n+1) - \binom{k}{2}S_{k-2}(n) - \binom{k}{3}S_{k-3}(n) - ... - \binom{k}{k-1}S_1 (n)$

Since the RHS is a polynomial in 'n' itself with highest degree 'k', it follows that after integration, the highest power will become 'k+1', and hence $S_k(n)$ is a polynomial of degree 'k+1' in 'n'.

Nice work! Your approach to the first and second one was different to mine, I actually ripped this question off of STEP 2008 III Q2

Carrotsticks · Dec 27, 2012

Re: HSC 2013 4U Marathon

bleakarcher said:
yeah, whichever ones you have to spare. I don't mind if it's a 2012 one.

Not bothered atm to dig up the old BOS Trials thread, and can't find my spare question file for 2013 actually.

bleakarcher · Dec 27, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
Not bothered atm to dig up the old BOS Trials thread, and can't find my spare question file for 2013 actually.

it's okay dude.

Sy123 · Dec 28, 2012

Re: HSC 2013 4U Marathon

I made this question just now:

$$If $ \ \ \ z= \cos \theta + i \sin \theta$

$$Consider $ \ \ \ (z+\frac{1}{z})^{2n}$

$$Show that$ \int_{0}^{\frac{\pi}{2}} \cos^{2n} \theta \ d\theta = \frac{\pi (2n)!}{2^{2n+1}(n!)^2}$

Carrotsticks · Dec 28, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
I made this question just now:

$$If $ \ \ \ z= \cos \theta + i \sin \theta$

$$Consider $ \ \ \ (z+\frac{1}{z})^{2n}$

$$Show that$ \int_{0}^{\frac{\pi}{2}} \cos^{2n} \theta \ d\theta = \frac{\pi (2n)!}{2^{2n+1}(n!)^2}$

That feel when questions you made (legitimately) have already been asked in the HSC ='(

2009 Q7

HSC 2013 MX2 Marathon (archive) (10 Viewers)

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