# Composition of a function with itself

1. May 25, 2015

### 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?

2. May 25, 2015

### micromass

Staff Emeritus
This is studied in dynamical systems and chaos theory.

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

3. May 26, 2015

### Ssnow

The function $f(x)$ is a real function, is holomorphic or ... other assumptions?

4. May 26, 2015

### Andre' Quanta

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?

5. May 30, 2015

### disregardthat

If the function is differentiable with continuous derivative, then it must be constant, as you can see by differentiating both sides.

6. Jun 2, 2015

### Andre' Quanta

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

7. Jun 2, 2015

### pasmith

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.

8. Jun 2, 2015

9. Jun 2, 2015

### disregardthat

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.

10. Jun 2, 2015

### Andre' Quanta

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

11. Jun 3, 2015

### wabbit

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
12. Jun 5, 2015

### disregardthat

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.

13. Jun 5, 2015

### wabbit

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.

14. Jun 5, 2015

### disregardthat

I can't believe I made that mistake twice.

15. Jun 5, 2015

### disregardthat

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.

16. Jun 5, 2015

### 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 $f(0)=c$

17. Jun 5, 2015

### disregardthat

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.

18. Jun 10, 2015

### disregardthat

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.

19. Jun 13, 2015

### Andre' Quanta

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

20. Jun 13, 2015

### disregardthat

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.