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Past Paper help - HSN forum

# Past Paper help

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### #1Kenny

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Posted 18 April 2005 - 11:25 AM

I was just trying some old CSYS papers, and found a rather nasty quesiton (only worth 2 marks mind). I don't know if this is simple and I have forgotton, or it is just hard (or we don't need to know how to do it anymore).

it was in the 2001 CSYS paper, quesiton 4b)

find the EXACT VALUE of the fixed point of the recurrence relation:

x(n+1) = 1/4(xn + 7/xn^2)

(sorry if that is hard to understand, tricky to do equations on this )

Now, I know the answer is the cube root of 7 (from some intelligent guessing) but I have no idea how to get to that mathematically.

I remember being given an example on something similar (gettgin a recurrence realtion for square root of 7) but we were told we would never be asked it so I didn't copy it down .

Does anyone know how to show this, or if we even don't have to do this?.

### #2sparky

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Posted 18 April 2005 - 07:02 PM

I have never heard of fixed points of recurrence relations!

The one thing we have done related to recurrence relations is iterative processes.
Mark

### #3George

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Posted 18 April 2005 - 08:36 PM

We've just been doing fixed points at uni, though more in the context of functions.

Basically, at fixed points the function has no effect. ie f(x) = x

Now, in the recurrence relation notation, xn+1 = xn

Just let me know if you didn't follow that

### #4Kenny

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Posted 19 April 2005 - 07:00 AM

Thanks George! I remember that now

Since others haven't heard of it (and you are doing it at uni) I assume we don't have to know this for the exam? I am going to ask my teacher next time I see them. Willl post back here with what he says.

Thanks again,

Kenny

### #5sparky

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Posted 10 May 2005 - 07:27 PM

It actually is in the 2003 Paper but we never done any questions remotely like it. I assume our teacher will go over it in class this week when we get onto it!
Mark

### #6YIC

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Posted 18 May 2006 - 02:49 PM

I haven't been taught it I don't think :|

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