Results 1 to 2 of 2

Math Help - Fixed Points

  1. #1
    Senior Member
    Joined
    Apr 2008
    From
    Vermont
    Posts
    318

    Fixed Points

    Suppose f is a real valued function on (-infinity; infinity): Call x a fixed point of f if f(x) = x:
    (a) If f is differentiable and f'(t) is not 1 for every t, prove that f has at most one fixed
    point.
    (b) Show that the function f defined by
    f(t) = t + (1 + e^t)^-1
    has no fixed point, although 0 < f'(t) < 1 for all t.
    (c) However, if there is a constant A < 1 such that |f'(t)| < A for all t, prove that a
    fixed point x of f exists, and that x = lim xn, where x1 is an arbitrary number and
    xn+1 = f(xn); n = 1; 2; 3; ::::

    b) is the only one where I've really gotten anywhere with
    f(t)=t
    Then t=t+(1 + e^t)^-1
    0=(1 + e^t)^-1
    0((1 + e^t)=1
    0=1
    impossible
    no fixed point

    a)f'(t)=[(f(x)-f(t))/(x-t)] is not 1
    Then f(x)-f(t) is not x-t


    c)|f'(t)| < A means there is a max value
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by kathrynmath View Post
    Suppose f is a real valued function on (-infinity; infinity): Call x a fixed point of f if f(x) = x:
    (a) If f is differentiable and f'(t) is not 1 for every t, prove that f has at most one fixed
    point.


    Suppose f(x_i)=x_i\,,\,\,i=1,2\,,\,\,x_1\neq x_2 , and use the mean value theorem:

    \displaystyle{\exists c\in (x_1,x_2)\,\,s.t.\,\,\frac{f(x_1)-f(x_2)}{x_1-x_2<br />
}=f'(c) ...



    (b) Show that the function f defined by
    f(t) = t + (1 + e^t)^-1
    has no fixed point, although 0 < f'(t) < 1 for all t.
    (c) However, if there is a constant A < 1 such that |f'(t)| < A for all t, prove that a
    fixed point x of f exists, and that x = lim xn, where x1 is an arbitrary number and
    xn+1 = f(xn); n = 1; 2; 3; ::::


    So you have a sequence \{x_n\} , where x_{n+1}=f(x_n) . Now, this seq. is Cauchy

    because [tex]|x_n-x_m|=|f(x_{n-1})-f(x_{m-1})|=|f'(d)||x_{n-1}-x_{m-1}| , using once again

    the MVT, and now use that the derivative is bounded.

    Thus, there exists \lim\limits_{n\to\infty}x_n=:x . Prove now this x fulfills the claim.

    Tonio


    b) is the only one where I've really gotten anywhere with
    f(t)=t
    Then t=t+(1 + e^t)^-1
    0=(1 + e^t)^-1
    0((1 + e^t)=1
    0=1
    impossible
    no fixed point

    a)f'(t)=[(f(x)-f(t))/(x-t)] is not 1
    Then f(x)-f(t) is not x-t


    c)|f'(t)| < A means there is a max value
    .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Fixed points of the Mandelbrot set
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: March 22nd 2010, 01:26 PM
  2. Fixed points
    Posted in the Calculus Forum
    Replies: 0
    Last Post: March 22nd 2010, 12:14 PM
  3. Fixed Points
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: February 21st 2010, 08:55 AM
  4. fixed points
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 7th 2009, 02:25 PM
  5. fixed points
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 25th 2009, 09:59 AM

Search Tags


/mathhelpforum @mathhelpforum