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Need Help with a mathematical modeling question

  1. Mar 27, 2012 #1
    This question deals with a 1-parameter equation:

    fc(x)= (4/x)+(x/2)-c for 0≤c≤ 2.3

    a.) Use algebra to find the positive fixed point p(c) of fc(x) (in terms of c) and identify its exact interval of existence

    a.) Ok so for the fixed point I got p(c)= -c+√(c2+8) I'm not sure if this is correct though. I used the quadratic formula to find this after setting the initial equation equal to x and solving in terms of c. Now for the interval of existence I need to take the first derivative of the initial equation which I have as -4x-2+(1/2). Is this right? And I think I'm supposed to plug in the fixed point i have into this first derivative to get the interval of stability for c by setting an inequality up like: -1<c<1. This would be the interval of existence for the parameter c right? Please correct me if i'm making any mistakes. I can't do this last step because i end up getting ridiculous numbers.
     
  2. jcsd
  3. Mar 28, 2012 #2
    bump, anyone?
     
  4. Jul 6, 2012 #3
    Hi,
    I do not know if you still need the help, but i found it interesting to see the answer :)

    so from
    f(x) = x
    or
    4/x+x/2-c = x
    we have
    x^2+2cx-8 = 0
    or
    x = p(c) = -c+-sqrt{c^2+8}
    and since we are only looking for the positive solution we choose
    p(c) = -c+sqrt{c^2+8}


    for the interval we choose the c such that the solution of x=f(x) exists
    or in other words that
    -1 < f'(x) < 1 , where x = p(c)

    Since
    f'(x) = -4/(x^2)+1/2
    we have

    -1 < -4/(2 c^2+8-2c sqrt{c^2+8})+1/2 < 1

    from where we obtain
    0 < c < 2 sqrt{2/3}

    I think this is correct, but if you stil need it I could check and make a better calculation.
     
    Last edited: Jul 6, 2012
  5. Jul 6, 2012 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    I don't understand the question. You have found a solution of x = f(x) in terms of c, and it exists and is positive for all real values of c.

    If you don't trust algebra and logic, try instead to use plotting: you can plot f(x) for any given value of c and see right away that the line y = x cuts the graph y = f(x) in just one point x > 0.

    RGV
     
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