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Show the iterates of tan(x)/2 converge to 0

  1. Jan 11, 2012 #1
    1. The problem statement, all variables and given/known data

    let f(x) = tan(x)/2. show the iterates of x converge to 0 whenever x is in (-pi/3, pi/3)
    (note this is for a course in chaos/fractals)


    2. Relevant equations

    A similar problem in the book used mean value theorem which says that for a function continuous on [a,b] and differentiable on (a,b), there exists c such that: f'(c) = (f(b)-f(a))/b-a


    3. The attempt at a solution

    First of all, tan(-x) = -tan(x) for the given interval so I restrict the problem to [0,pi/3)
    also tan(0) = 0 so restrict the interval further to (0,pi/3)

    now we want to show that the iterates of the function (ie the sequence {f(x), f(f(x)), f(f(f(x))), ...} ) converges to zero for the given interval

    I used the mean value theorem as was done in the textbook in hopes of getting down to tan(x)/2 < x so that I can say f[n+1](x) = f(f[n](x)) = tan(f[n](x))/2 < f[n](x) so that I can then conclude that every iterate of the function is less than the previous one, and since the function is bounded below by 0 (on our interval) I can finally conclude that the iterates converge to 0.

    First question: Would you fine people be convinced that:
    since tan(x) is strictly increasing on the given interval, and since tan(pi/3)/2 < pi/3 then tan(x)/2 < x for the given interval? If so, I think I can proceed as Ive outlined above.

    If not, here is how ive proceeded with MVT, on interval (0,x):
    f'(c)(b-a) = f(b)-f(a)
    -> ((sec^2(c))/2) * (x-0) = tan(x)/2 - tan(0)/2 = tan(x)/2
    then on the interval (0,pi/3) I get:
    1/2 < sec^2(x)/2 < 2
    -> x/2 < x*sec^2(x)/2 = tan(x)/2 < 2x

    And now im stuck because I cant get tan(x)/2 < x
    I can, however, get x/4 < tan(x)/4 < x
    which, using the method I wanted to above, would give me:
    f[n+1](x) = f(f[n](x)) = 1/2 *tan(f[n](x))/2 < f[n](x)

    but Im not sure that 1/2 *tan(f[n](x))/2 < f[n](x) is a strong enough statement to conclude that the iterates of tan will always be decreasing.

    Any hints, help, insight, comments, corrections, etc... are welcomed.

    thanks for you help.

    EDIT: Is this (showing f(x) < x) a good method to use when attempting to show that the iterates of an arbitrary function converge to a specific value in a specific interval? Is there a standard way to attempt such a problem?
     
    Last edited: Jan 11, 2012
  2. jcsd
  3. Jan 11, 2012 #2

    micromass

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    That inequality is certainly true! But your proof that it is true (that you stated later) is not correct. It is not enough to show tan(x)/2 is increasing. You should try finding another proof of this.

    No. Now you proved that the sequence [itex](f^{(n)}(x))_n[/itex] is decreasing and bounded below by 0. This does NOT mean that the sequence actually converges to 0!!

    For example: [itex]1+\frac{1}{n}[/itex] is bounded below by 0 and is decreasing. But it converges to 1.

    However, it is a good first step. Indeed, you've proven it to converge. Let y be its limit. Try to prove that y is a fixed point of f. I.e. that f(y)=y.
     
  4. Jan 14, 2012 #3
    Thanks for the reply micromass.

    Okay, well, if I evaluate f at 0 I get zero, which is the definition of a fixed point. Are you asking me to show that tan(x)/2 is 0 at 0, or are you asking me to try and show that f[n](0) = 0?
     
  5. Jan 14, 2012 #4
    Okay, so i tried a few times and i was unable to show algebraically that tan(x)/2 < x for (0,pi/3).

    To solve the problem I ended up graphing the two with maple and observing that tan(x)/2 < x is true. Then i used a corollary in my textbook which says that if a sequence is bounded and monotone then it converges to a fixed point; then i see that tan(0)=0 so 0 is the fixed point which the sequence converges to.

    Not sure how legit this technique of graphing is (if I were writing an exam I wouldnt be able to do it) so at this point I am looking for any hints:

    I want to show that for f(x) = tan(x)/2, the sequence {f[n](x)} from n = 0 to infinity (that is, {f(x), f(f(x), f(f(f(x))), ...}) is decreasing, then I can take care of the rest of the question using the corollary from the text
     
  6. Jan 16, 2012 #5

    lanedance

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    so we want to find the interval where f(x)<x, as a start we know
    [tex] f(x)=\frac{1}{2}tan(x) [/tex]
    [tex] f'(x)=\frac{1}{2}sec^2(x)\frac{1}{2cos^2(x)}[/tex]

    both f & f' are monotonically increasing functions on [0,pi/2) with
    [tex] f(0)=0 [/tex]
    [tex] f'(0)=\frac{1}{2}[/tex]

    [tex] f(\pi/3)= \frac{1}{2}\frac{sin(\pi/3)}{cos(\pi/3)}= \frac{1}{2}\frac{\sqrt{3}}{2}\frac{2}{1}= \frac{sqrt{3}}{2}[/tex]

    This, with f(0)=0 and being monotinically increasing, is enough to show f(x)<x on [0,pi/3] and using tan(-x)=-tan(x) extends the iteration to [-pi/3,pi/3]

    In fact tan(x)/2<x on the interval 0 to where tan(x)=2x which does not have an analytical solution, but is ~1.16>pi/3~1.047
     
  7. Jan 16, 2012 #6

    lanedance

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    Call f applied n times f_n. Are you sure you have to prove f_n(x) converges to 0?

    If so here's a real rough sketch idea

    Assume for 0<x<b, that 0<f(x)<x and f(0)=0.
    The sequence given by f_n(x) is monotonically decreasing and bounded below by zero
    Now, assume lim n-> inf, f_n(x)=c>0,
    But by def'n f(c)<c which is a contradiction
     
    Last edited: Jan 16, 2012
  8. Jan 16, 2012 #7
    Yeah, the problem asks for that specifically, but its restricted to the interval (-pi/3, pi/3).

    I wonder if there is a much easier way to solve this. We have a theorem in the text that says "suppose f is differentiable at a fixed point p. then if |f'(p)| < 1 then p is an attracting fixed point..."

    We know that 0 is a fixed point, and f'(0) = 1/2 so 0 is an attracting fixed point, so can I conclude that f_n will tend to 0 as n tends to infinity? The only reason Im not sure is because there could be other fixed points in the region, but since you showed that tan(x)/2 < x for (0,pi/3) then there cant be any other fixed points in the region. This seems right to me, what do you think?

    Note that an attracting fixed point of a function f is a point p such that f(p) = p and f_n(p) -> p as n -> infinity, which is really close to what Im trying to show
     
  9. Jan 17, 2012 #8

    lanedance

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    The sketch shows that any starting point for a sequence tends to zero, so I think we've done enough. Though you will need to flesh out the monotonicity. You could add in that any bounded

    That theorem you quote applies to the behaviour in the nighbourhood of the fixed point. So it shows your point is attracting, but says nothing about the global interval that will be attracted.
     
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