HSC 2012 MX2 Marathon (archive) (7 Viewers)

[deswa1]

Re: 2012 HSC MX2 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" title="U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" /></a>

 

[IamBread]

Re: 2012 HSC MX2 Marathon

I am serious. >_>

I came near bottom out of 117 hahaha. I thought I got everything correct...I skipped questions (Oops).

That's what happens when you go penguin

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" title="U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" /></a>

Oh I love recursive relations!

 

[IamBread]

Re: 2012 HSC MX2 Marathon

Oh I love recursive relations!

Me too! <3

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" title="U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" /></a>

Never seen anything like this before. Can someone post a worked solution? I'm curious.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Never seen anything like this before. Can someone post a worked solution? I'm curious.

There are a couple of ways to do the question.

The most 'mindless' of which being induction.

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

There are a couple of ways to do the question.

The most 'mindless' of which being induction.

That's what I immediately thought of, but I figured there must be a nicer way.

 

[deswa1]

Re: 2012 HSC MX2 Marathon

There are a couple of ways to do the question.

The most 'mindless' of which being induction.

About that. The question said to use induction but you can do it however you want...

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" title="U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" /></a>

The characteristic polynomial for this equation is:



Hence the solution to this recursive relation is in the form:



We are given the conditions that:



Substituting this into our general expression for U_n yields a system of equations (PS: this is one application of solving systems of linear equations, if anybody is interested)



Solving these two equations simultaneously for alpha and beta yields:



Hence the solution is:

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

PS: This technique isn't taught at schools, but it is MUCH faster than using Induction (plus I find this proof to be more elegant than one by Induction, which is more of a means of verification).

Here's a bit of practise. Follow my steps above to try to find a closed form for the Nth Fibbonaci Number.

I'll summarise my method for the previous question in case I was ambiguous:

1. Make the characteristic polynomial.

2. Solve the characteristic polynomial and find the roots.

3. Make the general solution.

4. Substitute the conditions (They will always give you the minimalistic conditions) and make a system of linear equations (usually in Extension 2, the system usually only involves two variables)

5. Solve this system by any method (substitution/elimination is probably easiest for you guys) and acquire alpha and beta.

6. Substitute back into your general solution to acquire answer.

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

PS: This technique isn't taught at schools, but it is MUCH faster than using Induction (plus I find this proof to be more elegant than one by Induction, which is more of a means of verification).

Here's a bit of practise. Follow my steps above to try to find a closed form for the Nth Fibbonaci Number.

I'll summarise my method for the previous question in case I was ambiguous:

1. Make the characteristic polynomial.

2. Solve the characteristic polynomial and find the roots.

3. Make the general solution.

4. Substitute the conditions (They will always give you the minimalistic conditions) and make a system of linear equations (usually in Extension 2, the system usually only involves two variables)

5. Solve this system by any method (substitution/elimination is probably easiest for you guys) and acquire alpha and beta.

6. Substitute back into your general solution to acquire answer.

It certainly is a lot faster and more elegant, but when you say it isn't taught in schools, does that mean we can't actually use it in the HSC?

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

It certainly is a lot faster and more elegant, but when you say it isn't taught in schools, does that mean we can't actually use it in the HSC?

I am actually not 100% sure. My gut says it will be accepted, but I will need to speak to a senior marker to verify this.

 

[Drongoski]

Re: 2012 HSC MX2 Marathon

Oh I love recursive relations!

recurrence

 

[deswa1]

Re: 2012 HSC MX2 Marathon

PS: This technique isn't taught at schools, but it is MUCH faster than using Induction (plus I find this proof to be more elegant than one by Induction, which is more of a means of verification).

Here's a bit of practise. Follow my steps above to try to find a closed form for the Nth Fibbonaci Number.

I'll summarise my method for the previous question in case I was ambiguous:

1. Make the characteristic polynomial.

2. Solve the characteristic polynomial and find the roots.

3. Make the general solution.

4. Substitute the conditions (They will always give you the minimalistic conditions) and make a system of linear equations (usually in Extension 2, the system usually only involves two variables)

5. Solve this system by any method (substitution/elimination is probably easiest for you guys) and acquire alpha and beta.

6. Substitute back into your general solution to acquire answer.

That's epic . I just did the Fibonnaci sequence with it. Thanks.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

recurrence

Thank you for that.

There is also a way to prove it using matrices, but I don't know how to type (nor can I be bothered lol) matrices.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

And thanks to Math Man who reminded me that the proper term for the 'general solution' was the 'auxiliary equation'. For some reason as I was typing it out, I just could NOT remember that word for my life.

 

[math man]

Re: 2012 HSC MX2 Marathon

And thanks to Math Man who reminded me that the proper term for the 'general solution' was the 'auxiliary equation'. For some reason as I was typing it out, I just could NOT remember that word for my life.

it happens to the best of us sometimes

 

[IamBread]

Re: 2012 HSC MX2 Marathon

PS: This technique isn't taught at schools, but it is MUCH faster than using Induction (plus I find this proof to be more elegant than one by Induction, which is more of a means of verification).

Here's a bit of practise. Follow my steps above to try to find a closed form for the Nth Fibbonaci Number.

I'll summarise my method for the previous question in case I was ambiguous:

1. Make the characteristic polynomial.

2. Solve the characteristic polynomial and find the roots.

3. Make the general solution.

4. Substitute the conditions (They will always give you the minimalistic conditions) and make a system of linear equations (usually in Extension 2, the system usually only involves two variables)

5. Solve this system by any method (substitution/elimination is probably easiest for you guys) and acquire alpha and beta.

6. Substitute back into your general solution to acquire answer.

That's so much better!

 

[cutemouse]

Re: 2012 HSC MX2 Marathon

What maths is that taught in? Haven't come across it yet...

PS: This technique isn't taught at schools, but it is MUCH faster than using Induction (plus I find this proof to be more elegant than one by Induction, which is more of a means of verification).

Here's a bit of practise. Follow my steps above to try to find a closed form for the Nth Fibbonaci Number.

I'll summarise my method for the previous question in case I was ambiguous:

1. Make the characteristic polynomial.

2. Solve the characteristic polynomial and find the roots.

3. Make the general solution.

4. Substitute the conditions (They will always give you the minimalistic conditions) and make a system of linear equations (usually in Extension 2, the system usually only involves two variables)

5. Solve this system by any method (substitution/elimination is probably easiest for you guys) and acquire alpha and beta.

6. Substitute back into your general solution to acquire answer.

 

[gurmies]

Re: 2012 HSC MX2 Marathon

What maths is that taught in? Haven't come across it yet...

It's somewhat reminiscent of DEs.