Past Paper help
Posted 18 April 2005 - 11:25 AM
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?.
Posted 18 April 2005 - 07:02 PM
The one thing we have done related to recurrence relations is iterative processes.
Posted 18 April 2005 - 08:36 PM
Basically, at fixed points the function has no effect. ie f(x) = x
Now, in the recurrence relation notation, xn+1 = xn
So for your problem:
Just let me know if you didn't follow that
Posted 19 April 2005 - 07:00 AM
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.
Posted 10 May 2005 - 07:27 PM
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