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This is studied in dynamical systems and chaos theory.Is possible to study the composition of a function f with itself when the number of compositions goes to infinity?

Like functions such that ##f(x) = e^{if(x)}## for each ##x##?I am interessed in the functions that can be written as a complex exponential of the function itself.

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Ssnow

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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?

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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.

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?

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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.

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?

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.

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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.[/QUOTE

Thanks, you are right :)

Now i am really interessed in find those complex functions that satisfies that condition

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disregardthat

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##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.

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That is not my function: mine is f(x)=exp{ik x f(x)} you have forgotten the x

##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.

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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?

I wonder if you might be able to relate f to Lambert's W function, the equation is somewhat similar?

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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 formulaThat is not my function: mine is f(x)=exp{ik x f(x)} you have forgotten the x

##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.

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wabbit

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disregardthat

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I can't believe I made that mistake twice.

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disregardthat

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##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.

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wabbit

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## 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 ##

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

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disregardthat

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##\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.

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To sum up: my function is only a costant everywhere?

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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.To sum up: my function is only a costant everywhere?

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