A nonlinear recurrence relation

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Wuberdall
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Hi Physics Forums,

I am stuck on the following nonlinear recurrence relation
$$a_{n+1}a_n^2 = a_0,$$
for ##n\geq0##.

Any ideas on how to defeat this innocent looking monster?

I have re-edited the recurrence relation
 
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Can you provide some context here? Is this homework? In what book / course did you come across this? Is this for a particular fractal graph?

It seems that it would oscillate from very small to very large until you're dividing by zero or by infinity.

##a_{n+1} = 1 / { a_n^2 } ## where ##a_n \neq 0##
 
Wuberdall said:
Hi Physics Forums,

I am stuck on the following nonlinear recurrence relation
$$a_{n+1}a_n^2 = 1,$$
for ##n\geq0##.

Any ideas on how to defeat this innocent looking monster?
What do you want to know about it? Have you tried isolating ##a_{n+1}## on one side of the equation?
 
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Wuberdall said:
I am stuck
In what way ?
This homework ? Please use the template and provide a full problem description and an attempt at solution tohat shows where you are stuck ...
 
Hi, this is not homework or course related. I am trying to determine if a fixed point for a certain dynamical system is unique. In doing so I come across the above recurrence relation.

So what I am really looking for, is a solution and whether or not this solution is unique
 
Wuberdall said:
Hi, it is not homework or course related. I am trying to determine if a fixed point for a certain dynamical system is unique. In doing so I come across the above recurrence relation.

So what I am really looking for, is a solution and whether or nor this solution is unique
It has exactly one fixed point at ##a_n=1##, though it is not a stable fixed point.
 
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tnich said:
It has exactly one fixed point at ##a_n=1##, though it is not a stable fixed point.
Thanks, this is exactly what I was looking for and also what my intuition told me.

How do you conclude that their is exactly one fixed point ?
 
Wuberdall said:
Thanks, this is exactly what I was looking for and also what my intuition told me.

How do you conclude that their is exactly one fixed point ?
If ##a_n## is not 1, then the sequence does not converge, so there can be no other fixed point.
 
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tnich said:
If ##a_n## is not 1, then the sequence does not converge, so there can be no other fixed point.
A fixed point must satisfy ##a_{n+1}=a_n##. In this case that results in ##a_n a_n^2=1## which has three solutions (two of which are complex), but only ##a_n=1## results in a fixed point.
 
tnich said:
If ##a_n## is not 1, then the sequence does not converge, so there can be no other fixed point.

Thanks, for your time.

I have figured it out now. It turned out that I was a bit rusty. So I found my old and dusty book by Strogatz on my bookshelf. All your comments make complete sense now and I see why they are true.

I wish you a happy and sunny weekend.
 
tnich said:
A fixed point must satisfy ##a_{n+1}=a_n##. In this case that results in ##a_n a_n^2=1## which has three solutions (two of which are complex), but only ##a_n=1## results in a fixed point.
Oops, no I think the two complex cube roots of 1 also are fixed points.
 
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