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Composition of a function with itself

  1. May 25, 2015 #1
    Is possible to study the composition of a function f with itself when the number of compositions goes to infinity? I am interessed in the functions that can be written as a complex exponential of the function itself. Where i can study this kind of things?
     
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  3. May 25, 2015 #2

    micromass

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    This is studied in dynamical systems and chaos theory.

    Like functions such that ##f(x) = e^{if(x)}## for each ##x##?
     
  4. May 26, 2015 #3

    Ssnow

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    The function ##f(x)## is a real function, is holomorphic or ... other assumptions?
     
  5. May 26, 2015 #4
    Thank you for the answer.
    I am interesed in a function that can be written in this way f(x)=c*exp{i*k*x*f(x)} where c and k are costants and f is a function fom R to R and x is the variabile.
    Does exist a similar function? If the answer is positive, does it have special propriety?
     
  6. May 30, 2015 #5

    disregardthat

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    If the function is differentiable with continuous derivative, then it must be constant, as you can see by differentiating both sides.
     
  7. Jun 2, 2015 #6
    Why it is a costant? If i take the derivative of that expression i get a differential equation for f(x), easy to resolve only respect to x: so it doesn' t seems to me that the solution is a costant
     
  8. Jun 2, 2015 #7

    pasmith

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    From your earlier post:

    If [itex]f : \mathbb{R} \to \mathbb{R}[/itex] and [itex]k[/itex] is real, then taking the modulus of both sides yields [itex]|f(x)| = |c|[/itex]. If [itex]f[/itex] is to be continuous then either [itex]f(x) = |c|[/itex] or [itex]f(x) = -|c|[/itex], ie. [itex]f[/itex] is constant.

    If instead [itex]f: \mathbb{C} \to \mathbb{C}[/itex] then you might be able to find a non-constant [itex]f[/itex] which satisfies that equation.
     
  9. Jun 2, 2015 #8
     
  10. Jun 2, 2015 #9

    disregardthat

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    If f(x) is differentiable, even as a complex function, yields
    ##f'(x) = ikf'(x)e^{ikf(x)} \Rightarrow f'(x)(1-ike^{ikf(x)}) = 0##
    So we either have ##f'(x) = 0##, or ##1 = ike^{ikf(x)} \Rightarrow log(ik)+ ikf(x) = 0##
    If it has a continuous derivative, then this implies that f is constant.

    If you're dealing with analytical functions for example, then your function is infinitely differentiable hence has a continuous derivative, leaving only constant functions.
     
  11. Jun 2, 2015 #10
    That is not my function: mine is f(x)=exp{ik x f(x)} you have forgotten the x
     
  12. Jun 3, 2015 #11

    wabbit

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    Your equation can be written ## x=\frac{1}{ikf(x)}\ln\frac{f(x)}{c}## so you are trying to find an inverse to the function ##g(y)=\frac{1}{iky}\ln\frac{y}{c}##, which will not exist globally but is possible for well chosen domains.

    I wonder if you might be able to relate f to Lambert's W function, the equation is somewhat similar?
     
    Last edited: Jun 3, 2015
  13. Jun 5, 2015 #12

    disregardthat

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    Oops, I was looking at the post above yours. Wabbit is right, however. The Lambert's W function is defined as the inverse of ##y = xe^x##, so that ##W(x)e^{W(x)} = x##. Since ##y = xe^x## is neither injective nor surjective, W is really "doubled valued", or we have to choose a branch of W. Now, the formula

    ##f(x) = ce^{ikf(x)}## really asks for solutions to ##y = ce^{iky}##. Rearranging, we get ##-ikye^{-iky} = -ikc##. Substituting ##z = -iky##, we get ##ze^{z} = -ikc##. Thus ##z = W(-ikc)##. I.e. ##f(x) = y = \frac{iz}{k} = \frac{iW(-ikc)}{k}##. While the Lambert W function is double valued, any continuous solution for f(x) must be constant. The constant value of f(x) may be chosen to be any value of ##\frac{iW(-ikc)}{k}## (if it exists). I am unsure if such a complex value always exists.
     
  14. Jun 5, 2015 #13

    wabbit

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    OP's equation however is ## f(x)=c e^{ikxf(x)} ##, not ## f(x)=c e^{ikf(x)} ##, so the possible link with Lambert's W function isn't as direct as that, and the solution is not constant if complex valued functions are allowed.
     
  15. Jun 5, 2015 #14

    disregardthat

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    I can't believe I made that mistake twice.
     
  16. Jun 5, 2015 #15

    disregardthat

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    Returning to the correct type of function: ##f(x) = ce^{ikxf(x)}##. Like above, we are asking for a solution to ##y = ce^{ikxy}##. Rearranging, we get
    ##ye^{-ikxy} = c##. So ##-ikxye^{-ikxy} = -ikcx##. Substituting ##z = -ikxy## we get ##ze^z = -ikcx##. Thus ##z = W(-ikcx)##. Plugging back yields ##-ikxy = W(-ikcx)##, so ##y = \frac{W(-ikcx)}{-ikx}##.

    Hence ##f(x) = \frac{W(-ikcx)}{-ikx}##. Thus any branching of ##W## will yield this unique solution for a continuous ##f##. Here of course we assume that both x and k are non-zero. It is likely possible to extend the domain of this function to include x = 0, by putting f(0) = 0, possibly depending on the branch chosen for W. In any case, it will depend on the behavior of f(x) as x approaches 0.
     
  17. Jun 5, 2015 #16

    wabbit

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    Indeed ! I missed that, the link with W is much closer than I was suspecting.

    ## f(0)=c ## for any solution however, from the equation itself.

    And to wrap it up, if ## k ## is real then the only real-valued continuous solution is the constant ## f(x)=c ## since ##\forall x, |f(x)|=|c| ## and ## f(0)=c ##
     
  18. Jun 5, 2015 #17

    disregardthat

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    Right, so depending on the behavior of W around x = 0, we may possibly continuously extend the domain to x = 0 by putting f(0) = c.
     
  19. Jun 10, 2015 #18

    disregardthat

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    Just to add to this, We have from the equation ##W(z)e^{W(z)} = z## that ##W'(z)e^{W(z)}+W'(z)W(z)e^{W(z)} = 1##, so that ##W'(z) = \frac{1}{z+e^{W(z)}}##. Since ##W(0) = 0##, we may apply l'hopital to the expression

    ##\lim_{x \to 0} \frac{W(-ikcx)}{-ikx} = \lim_{z \to 0} \frac{W(cz)}{z} = \lim_{z \to 0} \frac{cW'(cz)}{1} = \lim_{z \to 0} \frac{c}{z+e^{W(z)}} = c.##

    So we may extend the definition of f(x) to x = 0 by putting f(0) = c.
     
  20. Jun 13, 2015 #19
    To sum up: my function is only a costant everywhere?
     
  21. Jun 13, 2015 #20

    disregardthat

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    Not constant, it will have the form ##f(x) = \frac{W(-ikcx)}{-ikx}## for any choice of branching of the lambert W-function. You may also extend it to x = 0 by putting f(0) = c. It is only constant if c = 0, or k = 0.
     
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