HSC 2013 MX2 Marathon (archive) (1 Viewer)

Magical Kebab · Jan 5, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
There's a bit more to it than that =p

Firstly, we must have x lying on the unit circle as that is the maximal radius for which convergence occurs.

Secondly, you should probably add a part where they evaluate the integral of the n+1'th term, generally called the Error Term E(x), which you will then need to prove converges to 0 as n gets large.

Thirdly, it would be in your best interest to use say t as a dummy variable, and then integrate from 0 to x, where |x|<1, so then you can avoid issues with the constant term etc.

Sounds about right.

Sy123 · Jan 5, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
There's a bit more to it than that =p

Firstly, we must have x lying on the unit circle as that is the maximal radius for which convergence occurs.

Secondly, you should probably add a part where they evaluate the integral of the n+1'th term, generally called the Error Term E(x), which you will then need to prove converges to 0 as n gets large.

Thirdly, it would be in your best interest to use say t as a dummy variable, and then integrate from 0 to x, where |x|<1, so then you can avoid issues with the constant term etc.

Well, damn....

lol I give up trying to turn this into a question.

(For anyone else, the graph question is still unanswered)

Carrotsticks · Jan 5, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Well, damn....

lol I give up trying to turn this into a question.

(For anyone else, the graph question is still unanswered)

Don't give up, it's nearly formulated, just a few minor things!

Omed62 · Jan 5, 2013

Re: HSC 2013 4U Marathon

Hey all, good work guys keep up the good discussion

barbernator · Jan 5, 2013

Re: HSC 2013 4U Marathon

Omed62 said:
Hey all, good work guys keep up the good discussion

thank you for the good wishes

I appreciate the encouragement and look forward to your contribution to this thread

Sy123 · Jan 5, 2013

Re: HSC 2013 4U Marathon

Omed62 said:
Hey all, good work guys keep up the good discussion

Do you have anything difficult to contribute - it would be appreciated

barbernator · Jan 5, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Do you have anything difficult to contribute - it would be appreciated

his homework XD

Sy123 · Jan 5, 2013

Re: HSC 2013 4U Marathon

barbernator said:
his homework XD

lol anything but that

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

I made an easier one now....

$$Solve for $ \ \ z$

$4z^2-16z+30i=0$

AAEldar · Jan 5, 2013

Re: HSC 2013 4U Marathon

And while we're on the topic of easy ones:

$Find the perimeter of a regular $n$-gon inscribed in a circle of radius $r$.$

RealiseNothing · Jan 5, 2013

Re: HSC 2013 4U Marathon

Show that:

$\sum_{k=0}^{n-1}[\binom{n}{k}+2^k] = n+\sum_{k=0}^{n-1}\binom{n+1}{k}$

Sy123 · Jan 5, 2013

Re: HSC 2013 4U Marathon

$T_k \ \ \ $denotes the kth triangular number, which is defined by$$

$T_k=\sum_{m=0}^{k} m \equiv 1+2+3+...+k$

$Find the sum$

$\sum_{k=1}^{n} T_k \sin k \theta \equiv T_1\sin \theta + T_2 \sin 2\theta + T_3 \sin 3\theta + \dots +T_n \sin n\theta$

$HINT: You may want to consider the geometric sequence$

$z+z^2+z^3+...+z^n$

$You may also assume the formula$

$\sum_{k=1}^{n} k^2 = \frac{n}{6}(n+1)(2n+1)$

(you won't need the formula, but in case anyone has an 'otherwise' process, then that may be handy)

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

EDIT: I was initially trying to find the reciprocal times sin but I couldn't do anything with it, so as a little extension, if anyone else wants to help find:

$\sum_{k=1}^{n} \frac{\sin k \theta}{T_k}$

It seems the normal calculus approach to previous series will not work here.

EDIT2: It seems that I need to be able to evaluate: $\sum_{k=1}^{n} \sin k \theta / k$ which from experience in the derivation of pi thread shows to not be able to be expressed in elementary function -> however this is still open:

$\sum_{k=1}^{\infty} \frac{\sin k\theta}{T_k}$

RealiseNothing · Jan 5, 2013

Re: HSC 2013 4U Marathon

I edited it after, I had the "+n" on the wrong side, it was meant to be on the RHS not the LHS.

Sy123 · Jan 5, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
I edited it after, I had the "+n" on the wrong side, it was meant to be on the RHS not the LHS.

Alright then

Sy123 · Jan 5, 2013

Re: HSC 2013 4U Marathon

A short but not necessarily easy one:

$A prime number is a number where the only factors are 1 and itself, 1 is not prime$

$How many values of n is$ \ \ n^2-3n+2 \ \ $ a prime number?$

GoldyOrNugget · Jan 5, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
A short but not necessarily easy one:

$A prime number is a number where the only factors are 1 and itself, 1 is not prime$

$How many values of n is$ \ \ n^2-3n+2 \ \ $ a prime number?$

Hehe nice one. $n^2 - 3n + 2 = n(n-3) + 2.$ An even number * an odd number = an even number. n, n-3 are even, odd in some order. +2 leaves it even, and the only even prime is 2. So it's only true when the result is 2, that is for n=0 and n=2.

Sy123 · Jan 5, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
Hehe nice one. $n^2 - 3n + 2 = n(n-3) + 2.$ An even number * an odd number = an even number. n, n-3 are even, odd in some order. +2 leaves it even, and the only even prime is 2. So it's only true when the result is 2, that is for n=0 and n=2.

Alternatively, $n^2-3n+2=(n-2)(n-1)$

And notice that if it were to be prime, one bracket must be equal to 1, the other whatever it might be (as long as its positive)

$(3-2)(3-1) = 1 \cdot 2 = 2 = \ \ $prime$$

We could do the same for the other bracket where n=2, but that makes the whole thing zero which itself is not prime.

However we also observe that there is another way to get 2, which is by n=0

Either way, nice work =) (and your solution seems shorter anyway lol)

EDIT: Throughout reading your post I thought you were deswa1 heh

GoldyOrNugget · Jan 5, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
EDIT: Throughout reading your post I thought you were deswa1 heh

mission accomplished

RealiseNothing · Jan 6, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
Hehe nice one. $n^2 - 3n + 2 = n(n-3) + 2.$ An even number * an odd number = an even number. n, n-3 are even, odd in some order. +2 leaves it even, and the only even prime is 2. So it's only true when the result is 2, that is for n=0 and n=2.

$(n-1)(n-2)$

Apply the same logic now that an even times odd = even, and the only even prime is 2.

Sy123 · Jan 8, 2013

Re: HSC 2013 4U Marathon

$$i) Find the area underneath the curve$ \ \ \ \ f(x)=\frac{1}{x} \ \ \ \ \fbox{1}$

$$For$ \ \ \ 1\leq x \leq n$

$$ii) Find the volume created when rotating the area underneath$ \ \ f(x) \ \ $ by $ \ \ 2\pi \ \ $radians$ \ \ \ \ \fbox{1}$

$$iii) Compare the two results found in parts i and ii, when n approaches infinity. Describe what this means.$ \ \ \ \ \fbox{1}$

$: o$

asianese · Jan 8, 2013

Re: HSC 2013 4U Marathon

$\text{i)} A=\int_1^n \frac{1}{x} dx\\ = [\ln x]^n_1 \\ = \ln(n) u^2\\ \text{ii)} V= \lim_{\delta x \to 0}\sum_{x=1}^n \pi [f(x)]^2 \delta x\\ =\pi \lim_{\delta x \to 0} \sum_{x=1}^n \frac{1}{x^2} \delta x\\ = \pi \int_1^n x^{-2} dx \\ = -\pi \left[ \frac{1}{x} \right]^n_1\\=\pi \left( 1-\frac{1}{n} \right) u^3\\\\ \text{iii)} \text{As n approaches infinity, the area underneath the curve increases indefinitely, whilst the volume generated by a full revolution of this area converges to } \pi u^3.$

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