# Composition of a function with itself

• Andre' Quanta
From your earlier post:If f : \mathbb{R} \to \mathbb{R} and k is real, then taking the modulus of both sides yields |f(x)| = |c|. If f is to be continuous then either f(x) = |c| or f(x) = -|c|, ie. f is constant.If instead f: \mathbb{C} \to \mathbb{C} then you might be able to find a non-constant f which satisfies that equation.In summary, the conversation discusses the possibility of studying the composition of a function f with itself as the number of compositions goes to infinity. The focus is on finding functions that can be written as a complex exponential
Andre' Quanta
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?

Andre' Quanta said:
Is possible to study the composition of a function f with itself when the number of compositions goes to infinity?

This is studied in dynamical systems and chaos theory.

I am interessed in the functions that can be written as a complex exponential of the function itself.

Like functions such that ##f(x) = e^{if(x)}## for each ##x##?

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?

Andre' Quanta said:
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 the function is differentiable with continuous derivative, then it must be constant, as you can see by differentiating both sides.

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

Andre' Quanta said:
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

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

If instead $f: \mathbb{C} \to \mathbb{C}$ then you might be able to find a non-constant $f$ which satisfies that equation.

Andre' Quanta
pasmith said:
If $f : \mathbb{R} \to \mathbb{R}$ and $k$ is real, then taking the modulus of both sides yields $|f(x)| = |c|$. If $f$ is to be continuous then either $f(x) = |c|$ or $f(x) = -|c|$, ie. $f$ is constant.

If instead $f: \mathbb{C} \to \mathbb{C}$ then you might be able to find a non-constant $f$ which satisfies that equation.[/QUOTE
Thanks, you are right :)
Now i am really interessed in find those complex functions that satisfies that condition

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.

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

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?

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

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.

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.

wabbit said:
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.

I can't believe I made that mistake twice.

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.

wabbit
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 ##

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

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.

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.

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

Andre' Quanta said:
To sum up: my function is only a costant everywhere?

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.

Andre' Quanta

## 1. What is the definition of "Composition of a function with itself"?

The composition of a function with itself is a mathematical operation where the output of one function is used as the input for another function, resulting in a new function. This is also known as a function composition or a composite function.

## 2. Why is "Composition of a function with itself" important in mathematics?

Composition of a function with itself is important in mathematics because it allows us to combine multiple functions into one, which can help simplify complex equations and solve problems more efficiently. It also helps us understand the relationship between different functions and how they can affect each other.

## 3. How do you represent "Composition of a function with itself"?

The composition of a function with itself is represented using the notation f o f, where f represents the function and the o stands for "composed with". So, f o f means "f composed with f".

## 4. Can you provide an example of "Composition of a function with itself"?

Yes, for example, let's say we have the functions f(x) = 2x + 3 and g(x) = x^2. The composition of f with itself would be f o f, which can be written as f(f(x)). This means we first apply f to x, and then apply f again to the result. So, f(f(x)) = f(2x + 3) = 2(2x + 3) + 3 = 4x + 9.

## 5. What is the difference between "Composition of a function with itself" and "Inverse of a function"?

The composition of a function with itself involves using the output of one function as the input for another function. On the other hand, the inverse of a function is a function that "undoes" the original function, meaning it takes the output of the original function and returns the input. Inverse functions can also be composed with themselves, but the result will always be the original function.

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