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Homework Help: Wether a point is stable unstable or other

  1. Aug 5, 2008 #1
    got a question Xn+1=X2n-4Xn+6

    so from there we play arround with Xs and get roots of Xs =2 or 3 then we sub that into F'(x) giving answers of 2therefore unstable and 0therefore stable i understand all of the question apart from the stable and unstable part could someone please tell me why these points are stable and or unstable and where else they are stable or unstable and any other possible results if there are any. thanks
  2. jcsd
  3. Aug 5, 2008 #2


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    The fact that x= x2- 4x+ 6 when x= 2 or x= 3 is what told you that 2 and 3 are "fixed points" for this iteration. That is, if x0= 2 or 3, It remains that: if x0= 2, then x1= 22- 4(2)+ 6= 2, x2= 2, etc. Similarly, if x0= 3, then x1= 32- 4(3)+ 6= 3, etc.

    Now, what happens is x0 is close to 2 but not equal to 2? That is, suppose x= 2.1. Then x1= 2.12- 4(2.1)+ 6= 2.01, x2= 2.012- 4(2.01)+ 6= 2.001, etc. Similarly, if x0= 1.9, x1= 1.92- 4(1.9)+ 6= 2.01 and again that goes back toward 2. x= 2 is a "stable" point because if you start with a value slightly off 2, it doesn't matter- it goes quickly back to 2.

    If x0 is close to 3 but not equal to 3, what happens? If x0= 3.1, then x1[/sub]= 3.21, x2= 3.4641, etc. going away from 3. x= 3 is an "unstable" point.

    Those are the only two possibilities for this case and only fixed points (more generally called "equilibrium points") can be "stable" or "unstable".

    A ball sitting at the bottom of a hole is in a 'stable' position because if it is disturbed slightly it will roll back to the bottom. A ball sitting at the top of a hill is in an 'unstable' position because if it is disturbed slightly it will roll AWAY from the hill.

    Of course, looking at a few values doesn't really guarantee what happens "as n goes to infinity" but notice that, taking f(x)= x2- 4x+ 6, f'(2)< 1 while f'(3)> 1. That is what tells you one is "stable" and the other "unstable".
  4. Aug 7, 2008 #3
    for the question n+1=X2n-2Xn+2

    one of the roots is 1, now if i take 1.1 and put it in i get a negitive and if i take 0.9 i get a positive does this make is a saddle point? also F'(x)=0 what do the F'(x) values indicate? something about being smaller or larger than 1
  5. Aug 7, 2008 #4


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    Okay, fixed points are given by x= x2- 2x+ 2 or x2- 3x+ 2= (x- 1)(x- 2)= 0 and so are x= 1 and x= 2.

    However, if you set x0= 1.1 you get x1= 1.12- 2(1.1)+ 2= 1.01, not a negative number.

    For your second question, yes, as I said before, if |f'(x)|> 1 at a fixed point, then the point is unstable, if |f'(x)|< 1, it is stable. In this case, f'(x)= 2x- 2 so |f'(1)= 0< 1 and f'(2)= 2> 1. x= 1 is stable, x= 2 is unstable.
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