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    Consider the iterated map xn+1=rxn(1-xn2).
    Show that for 0<r<3√3/2 if 0 ≤xn≤1 then 0≤xn+1≤1. Show that if
    r<1 then the only fixed point in (0,1) is zero, and that this is stable. When
    r>1 there is another fixed point in (0,1). Find the value of this fixed point
    (as a function of r). For which values of r is it stable, and for which values is it unstable? What would you expect to happen when r>2?
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    Hi there from Warwick, my name is Jesse a handsome man, I'm sure you've seen me in your lectures , I'm stuck on the exact same question would you believe it.
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    (Original post by Dilzo999)
    Hi there from Warwick, my name is Jesse a handsome man, I'm sure you've seen me in your lectures , I'm stuck on the exact same question would you believe it.
    Everyone that I've asked is stuck on it and I was hoping I would be that one person with the brilliant answer .. But nope. I hate this module so much :-(
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    (Original post by ellemay96)
    Consider the iterated map xn+1=rxn(1-xn2).
    Show that for 0<r<3√3/2 if 0 ≤xn≤1 then 0≤xn+1≤1. Show that if
    r<1 then the only fixed point in (0,1) is zero, and that this is stable. When
    r>1 there is another fixed point in (0,1). Find the value of this fixed point
    (as a function of r). For which values of r is it stable, and for which values is it unstable? What would you expect to happen when r>2?
    I have not done difference equations since I was at university and sometimes the vocabulary is different, but my suggestion is

    in a difference equation you can find limits, say L, by setting un = un+1 =L

    say here

    L=rL(1-L2)

    either L = 0

    or 1=r(1-L2)
    L=+/- ROOT(1 - 1/r)

    but if r is between 0 and 1 there will be no more solutions except L =0

    etc


    (PS: I stand corrected if I am suggesting non sensible things)
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    (Original post by ellemay96)
    Consider the iterated map xn+1=rxn(1-xn2).
    Show that for 0<r<3√3/2 if 0 ≤xn≤1 then 0≤xn+1≤1.
    For the first part, for r in the range specified, in order for x_{n+1} to be in [0,1], what is the restriction on the value of x_n(1-x_n^2), i.e. what must it's range be restricted to?

    Now show that  x_n\in[0,1] implies x_n(1-x_n^2) lies in that range, and hence x_{n+1}\in[0,1]
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    stability implies that \left | f ' (x_{0}) \right |&lt;1

    (UNstability implies that \left | f ' (x_{0}) \right |&gt;1) for fixed point x_{0}

    here, stability condition gives: \left | r(1-3x^{2}) \right |&lt;1

    if r&lt;1 then x MUST =0 for stability to hold (since we KNOW r is +ve and then we have: .  r \left | (1-3x^{2}) \right |&lt;1 , which must mean that x=0)


    if you solve the equation: x=rx(1-x^{2}) you get the fixed points - one of which is zero, the other is the +ve root TeeEm has mentioned, (both in the range we wish) of x in terms of r.

    The +ve root value of x (fixed point) you have found as a function of r goes into the:

    1). stability equation (simplifies (verify this) to: \left |3-2r \right | &lt;1 =&gt; 1&lt;r&lt;2 (3root(3)/2 approx = 2.6)

    2). UNstability equation (simplifies (verify this) to: \left |3-2r \right | &gt;1 =&gt; r&gt;2 OR r&lt;1 for THIS point. We reject the latter value, as anything

    satisfying it gives us, if we substitute back into x=sqrt(1-1/r).., a complex number.

    I will admit, it`s a while since i have done this, and the last question kind of confuses me - it`s apparently leading to chaotic iterations! (although i would have said i expect it to diverge to infinity)

    This plot is just 30 iterations for r=3
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