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Help! Determining an interval of attraction for a fixed point

  • 15-05-2009 03:24PM
    #1
    Monkey61
    Registered Users, Registered Users 2 Posts: 1,137 ✭✭✭


    Hi folks, I'm in need of a bit of help again,

    I have the function f(x) which is a quadratic equation and I have found the fixed points, one attracting and one repelling.

    I have to use the gradient criterion to determine an interval of attraction for one of the fixed points. Using this and the attracting point I have gotten a set of values of x for which the gradient is less than 1. This is the open interval (1/3, 2/3) and from this I am meant to use the value closest to my fixed point and my fixed point as a midpoint, to find the Interval of attraction.

    But my attracting fixed point is 2/3. So I can't use it as a midpoint if it actually lies on the point 2/3...Or is the resulting interval of attraction for that fixed point (2/3, 2/3)? Is that a possible answer? Or have I made a mistake somewhere along the way?

    Thanks in advance!


Comments

  • #2
    LeixlipRed
    Closed Accounts Posts: 6,081 ✭✭✭LeixlipRed


    I've no idea what the interval of attraction is. Can you explain some of the terminology? I'm sure the maths behind it is quite simple


  • #3
    MathsManiac
    Registered Users, Registered Users 2 Posts: 1,595 ✭✭✭MathsManiac


    Sounds like you're talking about iterated functions and attractors, and about finding the maximal interval in the basin of attraction. Is that correct?

    If so, then can you give a bit more detail about how exactly you were told to use the fixed points and the interval for which the gradient is less than 1 in order to find "the interval of attraction"? I'm not familiar enough with the topic to have encountered such a procedure. e.g., can you show us a lecturer's or tutor's example, and then give us the details of the one you're stuck on? Is it f(x) = 3x(1-x), by any chance?


  • #4
    spring21
    Registered Users, Registered Users 2 Posts: 95 ✭✭spring21


    Monkey61 wrote: »
    Hi folks, I'm in need of a bit of help again,

    I have the function f(x) which is a quadratic equation and I have found the fixed points, one attracting and one repelling.

    I have to use the gradient criterion to determine an interval of attraction for one of the fixed points. Using this and the attracting point I have gotten a set of values of x for which the gradient is less than 1. This is the open interval (1/3, 2/3) and from this I am meant to use the value closest to my fixed point and my fixed point as a midpoint, to find the Interval of attraction.

    But my attracting fixed point is 2/3. So I can't use it as a midpoint if it actually lies on the point 2/3...Or is the resulting interval of attraction for that fixed point (2/3, 2/3)? Is that a possible answer? Or have I made a mistake somewhere along the way?

    Thanks in advance!
    Hi , i have the same problem , I just cant get this would u mind showimg me the way u worked out your tma2? please. thanks


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