CSSA Trials: Maths (2 Viewers)

TrentsUnicorn · Jul 31, 2014

So who else had their maths exam today, and what did everyone think???

4U was a pretty good paper, but for someone like me who sucks at the subject.... I'm gonna fail

Carrotsticks · Jul 31, 2014

What was the very last question of the Extension two paper?

TrentsUnicorn · Jul 31, 2014

They were a bunch of harder 3U proof questions. Don't remember exactly what it was

wagig · Jul 31, 2014

Carrotsticks said:
What was the very last question of the Extension two paper?

I didn't get time to look at it properly but it had infinite sums and powers of 2 to the powers of n, etc

Carrotsticks · Jul 31, 2014

The only proof I know over the top of my head involving powers of two to the power of N is proving that a sum (too hard to type on phone) converges to pi squared on 6 or or pi squared on eight. A similar proof demonstrates the Leibniz Series.

This one involves the limits being powers of two, was that yours? Or was it the actual summand involving 2^n?

TrentsUnicorn · Jul 31, 2014

I don't remember seeing pi
But it involved x^(2^n) all I can tell you, sorry. I rushed that question in the last 10 minutes

Davo_01 · Jul 31, 2014

Carrotsticks said:
What was the very last question of the Extension two paper?

$\; $Evaluate$ \sum_{n=0}^{\infty} \dfrac{1}{2014^{2^n}-2014^{-2^n}}$

Which i believe is $\dfrac{1}{2013}$

hecticated · Jul 31, 2014

yeah i had no idea how to do it because i didn't study for harder 3U hahaha

integral95 · Jul 31, 2014

Davo_01 said:
$\; $Evaluate$ \sum_{n=0}^{\infty} \dfrac{1}{2014^{2^n}-2014^{-2^n}}$

Which i believe is $\dfrac{1}{2013}$

Are there any previous parts to it?

Davo_01 · Jul 31, 2014

hecticated said:
yeah i had no idea how to do it because i didn't study for harder 3U hahaha

You use the previous part by substituting $x=2014^{-1}$

Davo_01 · Jul 31, 2014

integral95 said:
Are there any previous parts to it?

yes il post it from memory

panakap · Jul 31, 2014

I have a thread for the CSSA paper on a discussion of all questions

Davo_01 · Jul 31, 2014

integral95 said:
Are there any previous parts to it?

$$i) Show that$ \dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{1}{1- x^{2^n}}-\dfrac{1}{1-x^{2^{n+1}}}$

$$ii) Hence show that$ \sum_{n=0}^{N} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{1}{1-x}-\dfrac{1}{1-x^{2^{N+1}}}$

$$iii)Given that$ \lim_{n\rightarrow \infty}\sum_{n=0}^{N} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\sum_{n=0}^{\infty} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}\; $and$\; -1<x<1$

$$Hence show that $ \sum_{n=0}^{\infty} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{x}{1-x}$

$$iv)Evaluate$ \sum_{n=0}^{\infty} \dfrac{1}{2014^{2^n}-2014^{-2^n}}$

hecticated · Jul 31, 2014

the previous part was pretty long, I don't remember the question though.

EDIT: HOW DO YOU GUYS REMEMBER THE QUESTIONS? lol

Carrotsticks · Jul 31, 2014

From an initial glance, it looks like you just construct a telescoping sum and then you'll get a simple expression at the end. I'm going to take a stab and guess that they led you into it by making you prove some sort of identity at first and then take the sum, that's usually how they lead you to telescoping sums ie: 2011 HSC 2 unit Q9 and 2010 HSC 4 unit Q8.

EDIT: Haha there you go (two posts above mine). If nobody writes solutions by 8pm, I'll do so when I finish teaching. I'm sure there are many able-bodied students here who can do it.

Davo_01 · Jul 31, 2014

Carrotsticks said:
From an initial glance, it looks like you just construct a telescoping sum and then you'll get a simple expression at the end. I'm going to take a stab and guess that they led you into it by making you prove some sort of identity at first and then take the sum, that's usually how they lead you to telescoping sums ie: 2011 HSC 2 unit Q9 and 2010 HSC 4 unit Q8.

Yeh it was exactly that, a telescoping sum

Davo_01 · Jul 31, 2014

hecticated said:
the previous part was pretty long, I don't remember the question though.

EDIT: HOW DO YOU GUYS REMEMBER THE QUESTIONS? lol

Lol i got a good memory when it comes to math

Davo_01 · Jul 31, 2014

That polynomial with the roots of unity i have posted on the 4u marathon a while ago

panakap · Jul 31, 2014

Post all the specific question on a CSSA thread I got.

Poll there as well :thumbsup:

Davo_01 · Jul 31, 2014

Davo_01 said:
$$i) Show that$ \dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{1}{1- x^{2^n}}-\dfrac{1}{1-x^{2^{n+1}}}$

$$ii) Hence show that$ \sum_{n=0}^{N} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{1}{1-x}-\dfrac{1}{1-x^{2^{N+1}}}$

$$iii)Given that$ \lim_{n\rightarrow \infty}\sum_{n=0}^{N} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\sum_{n=0}^{\infty} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}\; $and$\; -1<x<1$

$$Hence show that $ \sum_{n=0}^{\infty} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{x}{1-x}$

$$iv)Evaluate$ \sum_{n=0}^{\infty} \dfrac{1}{2014^{2^n}-2014^{-2^n}}$

Solution:
$$i)\dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{x^{2^n} (1-x^{2^{n}})}{(1-x^{2^{n+1}})(1-x^{2^{n}})}=\dfrac{(1-x^{2^{n+1}})-(1-x^{2^{n}})}{(1-x^{2^{n+1}})(1-x^{2^{n}})}$
$=\dfrac{1}{1- x^{2^n}}-\dfrac{1}{1-x^{2^{n+1}}}$

$$ii) Hence show that$ \sum_{n=0}^{N} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{1}{1-x}-\dfrac{1}{1-x^{2^{N+1}}}$
$$Telescoping sum:$ \sum_{n=0}^{N}(a_n-a_{n+1})=a_0-a_{N+1}$

$$Substituting$\; a_n=\dfrac{1}{1- x^{2^n}}\; $gives the required result$

$$iii) Since$\; -1<x<1\; $then$ \lim_{n\rightarrow \infty} x^{2^n}=0$

$\sum_{n=0}^{\infty} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{1}{1-x}-1=\dfrac{x}{1-x}$

$$iv)$ \sum_{n=0}^{\infty} \dfrac{1}{2014^{2^n}-2014^{-2^n}}=\dfrac{2014^{-2^n}}{1-2014^{-2^{n+1}}}=\dfrac{2014^{-1}}{1-2014^{-1}}=\dfrac{1}{2013}$

CSSA Trials: Maths (2 Viewers)

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