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Finding fixed points non-algebraically

  1. Jan 28, 2012 #1
    1. The problem statement, all variables and given/known data

    let f = [itex]\mu[/itex]ex
    let 0 < [itex]\mu[/itex] < 1/e

    Show that f has two fixed points q and p with q < p

    2. Relevant equations

    a fixed point p is a point such that f(p) = p

    3. The attempt at a solution

    solving f(x) = x:
    f(x) - x = 0
    [itex]\mu[/itex]ex - x = 0

    Now I want to take logarithms but ln(0) is undefined.

    This is the 'normal', algebraic way of solving for fixed points. Is there another way to solve for fixed points? I tried looking at the derivatives based on a suggestion but f'(x) = f(x) which doesnt tell me very much, and I also tried iterating the function, but it just seems to get messy (ie f(f(x)) = [itex]\mu[/itex]e[itex]\mu[/itex]ex )

    Any suggestions for finding fixed points would be helpful. I would prefer not to use the newton-raphson method to find a root of g(x) = f(x) - x, so if there are any other strategies please let me know
     
  2. jcsd
  3. Jan 28, 2012 #2

    morphism

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    Let g(x)=f(x)-x. A fixed point of f is a zero of g. Now try using the intermediate value theorem.
     
  4. Jan 28, 2012 #3

    LCKurtz

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    If you solve ##\mu e^x = x## for ##\mu## you get ##\mu = xe^{-x}##. What are the min and max values of ##xe^{-x}## for ##x\ge 0##? And if you can conclude there is one fixed point and the slopes aren't equal there, the graphs of ##x## and ##xe^{-x}## must cross. Then maybe you can get the other crossing by a concavity argument. Your post doesn't say you must calculate the crossings, just show they exist.
     
  5. Jan 29, 2012 #4
    morphism and LCKurtz thanks for the replies.

    Gonna try IVT in a few minutes then ill update, but for LCKurtz suggestion:

    Im not really sure what you mean. I found the max of x*e^-x to be 1/e at x = 1 and the min to be 0 at x = 0 (which is what we would expect, given the statement 0<mu<1/e). Im not sure how this ties in with:

    the slopes of x and x*e^-x? So if I can find one fixed point of mu*e^x then I can use the fact that the slopes of mu = x*e^-x and x are not equal to conclude that the graphs will cross again and therefore there will be another fixed point? Im having trouble understanding because mu is the parameter and I thought we wanted to look at the original function mu*e^x for varying mu, not solving for mu explicitly, or does it not make a difference?

    Thanks for the help
     
  6. Jan 29, 2012 #5
    Okay, I determined a solution using IVT, thanks for the tip morphism. There is another part of the question that asks which point is attracting and which is repelling, and to do this part I want to use a theorem that says if the value of the derivative at a fixed point is <1 then the point is attracting; if it is greater than 1 it is repelling. To do this I need to find the points explicitly.

    How would I go about doing that as the IVT doesnt give me anything more than the existence of points, and any other method I can think of to do this would involve newton method. Should I just use Newton method in this case?

    edit: Okay im stupid, just realized newton method will approximate the answers for me, but I still need to find them explicitly if I want to use the theorem i quoted. Any ideas on how to find these fixed points? or better yet any different method for determining which point is attracting and repelling without explicitly finding the points? I can use a graphical argument but would prefer to have a mathematical one if possible
     
    Last edited: Jan 29, 2012
  7. Jan 29, 2012 #6

    Dick

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    Nah. I would look at it this way. The fixed points are where [itex]g(x)=\mu e^x - x=0[/itex]. g(x) has a negative minimum at [itex]x_0 = log(\frac{1}{\mu})[/itex] and approaches [itex]\mu[/itex] as x->0 and infinity as x->infinity. Sketch a graph. So there are two zeros. One is at an x value less than [itex]x_0[/itex] and one is greater. What's [itex]f'(x_0)[/itex]? What can you say about the derivatives at the two zeros?
     
  8. Jan 29, 2012 #7
    Thanks for the reply Dick.

    So if Im understanding you we have a minimum between the two zeroes, at ln(1/mu), and the value of the derivative of mu*e^x at this point is 1; Since we have two values on either side of this point which are higher than it, we can say the following about the derivative:

    For the fixed point to the left of the minimum the derivative will have negative slope (since the graph of f is higher/larger at this point). Similarly for the zero to the right it will have positive slope. And there is the answer without explicitly finding the points.

    I think thats it, right? Thanks a ton!
     
  9. Jan 29, 2012 #8

    Dick

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    No. That's not quite it. The slope is [itex]f'(x) = \mu e^x[/itex]. That's an increasing function and always positive since [itex]\mu > 0[/itex]. Please rethink that and try again.
     
  10. Jan 29, 2012 #9
    Oops, guess I was thinking about g(x) = mu*e^x - x.

    I guess I want to say that for f(x) = mu*e^x there is some fixed point a < x_0 (x_0 is the minimum of g(x)) and some fixed point b > x_0

    Since the function is positive and increasing, and at the point x_0 its derivative is equal to 1, then for all input values less than x_0 (including a) its derivative will be less than one, so a will be an attracting fixed point.

    Similarly, its increasing, so for values greater than x_0 (including b) it will have slope greater than one -> repelling
     
    Last edited: Jan 29, 2012
  11. Jan 29, 2012 #10

    Dick

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    That's more like it.
     
  12. Jan 29, 2012 #11
    Awesome, thanks!
     
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