HSC 2013 MX2 Marathon (archive) (1 Viewer)

seanieg89 · Dec 15, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Hope you don't mind me expanding on this:

$iii) $Deduce that$~\sum_{k=1}^{\infty} \frac{1}{k} - \frac{1}{k+1} = \sum_{k=1}^{\infty} \frac{1}{2^k}$

$iv) $Find the first term of each series$$

$v) $Find a value of 'k' such that$~k(k+1) \neq 2^k$

$vi) $Hence deduce that there is a positive real solution to$~ k(k+1)=2^k ~$other than$~ k=1$

Lol such a roundabout way of showing that equation has another real sol.

kazemagic · Dec 15, 2012

Re: HSC 2013 4U Marathon

hard questions are hard D: or im just stupid

Sy123 · Dec 15, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Hope you don't mind me expanding on this:

$iii) $Deduce that$~\sum_{k=1}^{\infty} \frac{1}{k} - \frac{1}{k+1} = \sum_{k=1}^{\infty} \frac{1}{2^k}$

$iv) $Find the first term of each series$$

$v) $Find a value of 'k' such that$~k(k+1) \neq 2^k$

$vi) $Hence deduce that there is a positive real solution to$~ k(k+1)=2^k ~$other than$~ k=1$

I have written up a solution, but I will do what you are doing and wait until night and post my solution after yours.

mraznboy · Dec 15, 2012

Re: HSC 2013 4U Marathon

Looooooooooooooooooooool

jeffwu95 · Dec 15, 2012

Re: HSC 2013 4U Marathon

its amazing how much you can forget in a month

Sy123 · Dec 15, 2012

Re: HSC 2013 4U Marathon

Here is another question:

$Question Fibonacci - 15 Marks$

$$The Fibonacci sequence is a sequence of natural numbers based on the following recurrence relation$ \\ \\ $Where $F_n$denotes the nth Fibonacci number$ \ \\ \\ \ F_0=1 \ \ \ F_1=1 \\ F_n=F_{n-1}+F_{n-2} \ \ ...... \ \ (*)$

$$The beginning of the Fibonacci sequence is:$\ \ \ 1 , 1 , 2, 3, 5, 8, 13, 21, 34 ...$

$$a) Define $ \varphi = \frac{1+\sqrt{5}}{2} \ \ \ \tau=\frac{1-\sqrt{5}}{2} \\ \\ 1, \varphi, \varphi^2, \varphi^3, \varphi^4 , ... \ \ \ \ \ $AND$ \ \ \ \ 1, \tau, \tau^2, \tau^3, \tau^4, ... \ \ \ \ \ $Show that both these sequences satisfy the recurrence relation $ a_n=a_{n-1}+a_{n-2} \ \ \ \ $Where$ \ \ a_n \ \ $is the nth term$ \ \ \ \ \fbox{2}$

$$It can be shown that$\ \ \ F_n= a \varphi^n + b\tau^n \\ $For a constant value of a and b for all natural numbers n (DO NOT prove this)$

$$b) Show that$ \ \ \ F_n=\frac{1}{\sqrt{5}}(\varphi^{n+1}-\tau^{n+1}) \ \ \ \ \fbox{3}$

$$c) Using part (c) or via utilising the definition of the limit, or otherwise.$ \\ $Show that$ \ \ \ \lim_{n \to \infty} \frac{F_{n+1}}{F_n}=\varphi \ \ \ \ \fbox{3}$

$$d) Define the power series$ \\ \\ S(x)=\sum_{k=0}^{\infty}F_k x^k \\ \\ $Show that$ \ \ \ S(x)=1+x+\sum_{k=2}^{\infty}F_{k-1}x^k + \sum_{k=2}^{\infty}F_{k-2}x^k \ \ \ \fbox{2}$

$$e) Show that$ \\ \\ S(x)= \frac{1}{1-x-x^2} \ \ \ \fbox{3}$

$$f) Show that$ \sum_{k=0}^{\infty} \frac{F_k}{n^k}=\frac{n^2}{n^2-n-1} \ \ \ \fbox{2}$

seanieg89 · Dec 15, 2012

Re: HSC 2013 4U Marathon

Just a point on nomenclature: the object in d) is called a formal power series, not a polynomial.

Sy123 · Dec 15, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Just a point on nomenclature: the object in d) is called a formal power series, not a polynomial.

Apologies hehe, I'll fix it now.

bleakarcher · Dec 15, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Here is another question:

$Question Fibonacci - 15 Marks$

$$The Fibonacci sequence is a sequence of natural numbers based on the following recurrence relation$ \\ \\ $Where $F_n$denotes the nth Fibonacci number$ \ \\ \\ \ F_0=1 \ \ \ F_1=1 \\ F_n=F_{n-1}+F_{n+2} \ \ ...... \ \ (*)$

$$The beginning of the Fibonacci sequence is:$\ \ \ 1 , 1 , 2, 3, 5, 8, 13, 21, 34 ...$

$$a) Define $ \varphi = \frac{1+\sqrt{5}}{2} \ \ \ \tau=\frac{1-\sqrt{5}}{2} \\ \\ 1, \varphi, \varphi^2, \varphi^3, \varphi^4 , ... \ \ \ \ \ $AND$ \ \ \ \ 1, \tau, \tau^2, \tau^3, \tau^4, ... \ \ \ \ \ $Show that both these sequences satisfy the recurrence relation $ a_n=a_{n-1}+a_{n-2} \ \ \ \ \fbox{2}$

$$It can be shown that$\ \ \ F_n= a \varphi^n + b\tau^n \\ $For a constant value of a and b for all natural numbers n (DO NOT prove this)$

$$b) Show that$ \ \ \ F_n=\frac{1}{\sqrt{5}}(\varphi^{n+1}-\tau^{n+1}) \ \ \ \ \fbox{3}$

$$c) Using part (c) or via utilising the definition of the limit, or otherwise.$ \\ $Show that$ \ \ \ \lim_{n \to \infty} \frac{F_{n+1}}{F_n}=\varphi \ \ \ \ \fbox{3}$

$$d) Define the power series$ \\ \\ S(x)=\sum_{k=0}^{\infty}F_k x^k \\ \\ $Show that$ \ \ \ S(x)=1+x+\sum_{k=2}^{\infty}F_{k-1}x^k + \sum_{k=2}^{\infty}F_{k-2}x^k \ \ \ \fbox{2}$

$$e) Show that$ \\ \\ S(x)= \frac{x+1}{1-x-x^2} \ \ \ \fbox{3}$

$$f) Show that$ \sum_{k=0}^{\infty} \frac{F_k}{n^k}=\frac{n^2+n}{n^2-n-1} \ \ \ \fbox{2}$

Sy, can you check e)? I got 1/[1-x-x^2].

Sy123 · Dec 15, 2012

Re: HSC 2013 4U Marathon

bleakarcher said:
Sy, can you check e)? I got 1/[1-x-x^2].

Yep you are correct, I made a silly error, I will change e and f accordingly.

RealiseNothing · Dec 16, 2012

Re: HSC 2013 4U Marathon

Also Sy, you made a typo when you defined the sequence. No biggie, just people unfamiliar with the Fibonacci sequence might define it using the typo.

The $F_n = F_{n-1} + F_{n+2}$ part, should be "n-2".

Sy123 · Dec 16, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Also Sy, you made a typo when you defined the sequence. No biggie, just people unfamiliar with the Fibonacci sequence might define it using the typo.

The $F_n = F_{n-1} + F_{n+2}$ part, should be "n-2".

Oops, well I'll fix that now. Thanks

RealiseNothing · Dec 16, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$Define the nth Triangular number as$ \ \ T_n=\sum_{k=1}^{n}k = 1+2+3+...+n$

$$i) Show that $ \ \ \sum_{k=1}^{\infty} \frac{1}{T_k} = 2\sum_{k=1}^{\infty} \frac{1}{k}-\frac{1}{k+1} \ \ \ \ \ \fbox{2}$

$$ii) Hence evaluate $ \ \ \sum_{k=1}^{\infty} \frac{1}{T_k} \ \ \ \ \ \fbox{2}$

i) $\sum_{k=1}^{\infty} \frac{1}{T_k}$

Note $t_k = \frac{k}{2}(k+1)$

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

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

$2 \sum_{k=1}^{\infty} \frac{1}{k} - \frac{1}{k+1}$

ii) $\sum_{k=1}^{\infty} \frac{1}{T_k} = 2 \sum_{k=1}^{\infty} \frac{1}{k} - \frac{1}{k+1}$

This makes a telescoping sum, which leaves us with:

$\lim_{k \to \infty} 2(1-\frac{1}{k+1})$

$= 2$

Sy123 · Dec 16, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Hope you don't mind me expanding on this:

$iii) $Deduce that$~\sum_{k=1}^{\infty} \frac{1}{k} - \frac{1}{k+1} = \sum_{k=1}^{\infty} \frac{1}{2^k}$

$iv) $Find the first term of each series$$

$v) $Find a value of 'k' such that$~k(k+1) \neq 2^k$

$vi) $Hence deduce that there is a positive real solution to$~ k(k+1)=2^k ~$other than$~ k=1$

iii) Not sure how to actually deduce it from previous results but I can easily evaluate both sides.

LHS is a telescoping sum from which we get:

$\lim_{k \to \infty}1-\frac{1}{k+1}) = 1$

RHS is a limiting sum of a geometric series:

$RHS=\frac{1/2}{1-1/2}=1=LHS$

iv)

First term of LHS -> 1/2
First term of RHS -> 1/2

v) k=5 -> 5(5+1)=30 =/=2^5=32
Therefore for k=5 the non-equality holds

vi) Note how k=1 is a solution, when we see k=2
For k(k+1)=2^k

2(3) > 2^2
6 > 4

However we observed before that k=5

30 < 32

Hence there must exist a solution in between k=2 and k=5 whereby the transition from LHS > RHS and LHS < RHS takes place.

Not sure if this is the intended solution. Also I will post solution to Fibonacci question tomorrow night if no-one posts a solution by then.

Sy123 · Dec 16, 2012

Re: HSC 2013 4U Marathon

Something a little easier:

$$Given the polynomial$ \ \ \ S(x)=Ax^3+Bx^2+Bx+A$

$The roots are in arithmetic sequence$

$Hence find the roots$

RealiseNothing · Dec 16, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
iii) Not sure how to actually deduce it from previous results but I can easily evaluate both sides.

LHS is a telescoping sum from which we get:

$\lim_{k \to \infty}1-\frac{1}{k+1}) = 1$

RHS is a limiting sum of a geometric series:

$RHS=\frac{1/2}{1-1/2}=1=LHS$

iv)

First term of LHS -> 1/2
First term of RHS -> 1/2

v) k=5 -> 5(5+1)=30 =/=2^5=32
Therefore for k=5 the non-equality holds

vi) Note how k=1 is a solution, when we see k=2
For k(k+1)=2^k

2(3) > 2^2
6 > 4

However we observed before that k=5

30 < 32

Hence there must exist a solution in between k=2 and k=5 whereby the transition from LHS > RHS and LHS < RHS takes place.

Not sure if this is the intended solution. Also I will post solution to Fibonacci question tomorrow night if no-one posts a solution by then.

I should have put something like "without substitution of further k values, deduce that there is a positive real solution besides k=1".

The intended solution was to see that for some value k (ie you did k=5), the sums are not equal and hence the ratios are not equal (one is larger).

But as you approach infinity the sums become equal, and hence the magnitudes of the ratios of the series must switch so that the other is larger.

If the ratios cross over, then there must be a positive real solution.

Not sure if this makes sense since it's 3am, but w/e.

Sy123 · Dec 16, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
I should have put something like "without substitution of further k values, deduce that there is a positive real solution besides k=1".

The intended solution was to see that for some value k (ie you did k=5), the sums are not equal and hence the ratios are not equal (one is larger).

But as you approach infinity the sums become equal, and hence the magnitudes of the ratios of the series must switch so that the other is larger.

If the ratios cross over, then there must be a positive real solution.

Not sure if this makes sense since it's 3am, but w/e.

Oh I see, nice solution.

Sy123 · Dec 16, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Something a little easier:

$$Given the polynomial$ \ \ \ S(x)=Ax^3+Bx^2+Bx+A$

$The roots are in arithmetic sequence$

$Hence find the roots$

Just in case other people did not see it.

HeroicPandas · Dec 17, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Just in case other people did not see it.

Test:
S(-1) = 0

Therefore x=-1 is a root

Since roots are in an arithmetic sequence, let roots be $-d-1 ,-1, d-1$

$\sum \alpha:3=\frac{B}{A}$

$\prod \alpha:$ (d+1)(d-1)=-1
d^2=0
So d=0....

Therefore triple root at x = -1? (or roots are x = -1, -1, -1)

Sy123 · Dec 17, 2012

Re: HSC 2013 4U Marathon

HeroicPandas said:
Test:
S(-1) = 0

Therefore x=-1 is a root

Since roots are in an arithmetic sequence, let roots be $-d-1 ,-1, d-1$

$\sum \alpha:3=\frac{B}{A}$

$\prod \alpha:$ (d+1)(d-1)=-1
d^2=0
So d=0....

Therefore triple root at x = -1? (or roots are x = -1, -1, -1)

That is one case of it, yes.

I should of extended the question to:

$Find the possible roots of S(x)$

So yeah consider if -1 is the first or the last term etc.

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