# Find a function f(z) such that f(f(z)) = exp(z)

1. Oct 26, 2007

### Count Iblis

Find a function $$f\left(z\right)$$ such that $$f\left(f\left(z\right)\right) = \exp\left(z\right)$$

2. Oct 28, 2007

### coomast

This is an interesting question, but it seems to be a difficult one. It's been on my mind for two days and I tried to solve it for a number of hours now. No result though.

I used the chain rule after converting it to a more general form as:

$$f\left(g(z)\right)=w(z)$$

I assumed that after some work I could then set $$g(z)=f(z)$$ of some kind, but it gets me nowhere.

Then I tried to convert it to a system. The problem seems to be that there is no way of solving this system. From:

$$\frac{df}{dg} \cdot \frac{dg}{dz} = \frac{dw}{dz}$$

and assuming the function to be looked for is $$k(z)$$

I get:

$$\frac{df}{dg} =k(g)$$
$$\frac{dg}{dz} =k(z)$$

I don't see a solution for this.

I assume that there are better ways of handling this problem and maybe even a simple solution exists, but I don't see it.
I'm very sorry that I can't help you with this. Hopefully someone will post the solution. Anyway I'm looking forward to it.

3. Oct 28, 2007

### Chris Hillman

Indeed, functional equations can be difficult. Two quick thoughts:

1. Did you think about whether some conditions on an initial value might enable you to construct a series solution? (I have in mind a condition which should make it possible to at least attempt such an analysis, which interated exponentials otherwise preclude.)

2. Did you notice that if the rhs is replaced by a function which is a diffeomorphism $R \rightarrow R$ (exp doesn't qualify!), we would be trying to find a "square root" in some group of transformations on the real line? That's a big group, but by imposing enough conditions on the transformation to be rooted, and by restricting the group to a proper subgroup, clearly you can eventually obtain a solvable problem.

Last edited: Oct 28, 2007
4. Oct 28, 2007

### Manchot

Was this given to you as a problem? That is to say, are you sure that a solution even exists?

Last edited: Oct 28, 2007
5. Oct 29, 2007

### Galileo

$f(z)=z$ if $z=\exp(z)$, and $f(z)=\exp(z)$ otherwise.

...

...

...

... what?

6. Oct 29, 2007

### coomast

Galileo,

Does that mean that the solution (for the first part) is only valid for a fixed number(s), fullfilling $$e^{z}=z$$?

For the second part it seems strange. If $$f(z)=e^{z}$$ and $$z\neq e^{z}$$, isn't the following true $$f(f(z))=e^{e^{z}}$$ ? Then it is not correct.

In case the first part is indeed valid, I would think it is the only solution and would not split it up in cases.

Can you give some explanation on how you got these results?

7. Oct 29, 2007

### Galileo

Yeah, so it doesn't work out right. It was meant as a joke anyway, hope that was obvious.

8. Oct 29, 2007

### coomast

No, it wasn't obvious. I'm upset and will not come back to this tread. You know how much time I spent on it? What were you thinking? Let's pull the stupid belgian's leg? How do think Count Iblis is feeling when he finds out his question is not taken serious?

9. Oct 29, 2007

### CRGreathouse

Cool off... you know, this is a very difficult question, and I'm not sure a solution is known. It seems related to the problem of a continuous extension of the hyperpowers to the real numbers, and that makes me suspect it's difficult.

10. Oct 29, 2007

### Zurtex

Edit: All the below is wrong read on if you want to watch me fail horrible, read ahead if you want to just read why it's wrong and can't be salvaged.

Riiight, I think I might of come up with a solution, I've only tacked it for real numbers, I'm not too fussed about extending it to complex number unless someone wants too

I tackled the problem like this, guess f(x) looks something like this:

$$f(x) = e^{a + \ln x}$$

$$f(f(x)) = e^{a + \ln e^{a + \ln x}} = e^{\left(\ln x\right) + 2 a} = e^{2a}$$

Which holds when a = x/2, but a is a constant, so you simply have to define the function at each point to be:

$$\lim_{x \rightarrow 2a} f(x) = e^{a + \ln x}$$

We could define as the limit of a sequence of functions:

$$\left{\{} a_i \right{\}}_{i=1}^{\infty} = \left{\{} 0, \frac{2}{n}, -\frac{2}{n}, \frac{4}{n}, -\frac{4}{n}, \ldots, \left(-1\right)^{i} \frac{2 \left[ \frac{i}{2} \right]}{n}, \ldots \right{\}}$$

Where [] is the floor function. We then define fn:

$$f_n(x) = e^{a_i + \ln x} \quad \text{for} \quad 2a_i - \frac{1}{n} < x \leq 2a_i + \frac{1}{n} \quad \forall a_i$$

We then define f(x) to be:

$$\lim_{n \rightarrow \infty} f_n (x) = f(x)$$

I've probabily done something wrong, but that's my best go at it for the moment.

Last edited: Oct 29, 2007
11. Oct 29, 2007

### d_leet

eln(x)=x, so eln(x)+2a=xe2a

12. Oct 29, 2007

### Zurtex

Oh my god, thanks, I don't know why my brain broke like that! But the princaple still holds I think. I'll retype it all again tomorrow morning when I'm not being an idiot.

Bah, I realised my whole method was fundamentally boring anyway, in more simplified way this is what I was trying to do:

f(x) = ea

f(f(x)) = ea

For each point let x = a

13. Oct 29, 2007

### d_leet

I don't think this makes sense, or at least it doesn't make sense to me as an answer for the question the OP posed.

14. Oct 30, 2007

### christianjb

Can it be done using Taylor series?

2 terms

f(x)=a+bx
f(f(x))=a+b(a+bx)=1+x -> a=1/2 b=1

3 terms

f(x)=a+bx+cxx
ff(x)=a+b(a+bx+cxx)+c(a+bx+cxx)^2
=(a+ba+caa)+(bb+2abc)x+(bc+2acc+bbc)xx+(2bcc)xxx+(ccc)xxxx

a+ba+caa=1.
bb+2abc=1
bc+2acc+bbc=1/2!
2bcc=1/3!
ccc=1/4!

Overdefined?- 3 unknowns/4 eqns.

15. Oct 30, 2007

### Zurtex

No, I don't think it makes sense either.

I tried a tayloer series before but did it with a full infinite series, I like this method a lot, keep trying it for larger and larger terms, will try it later when I have time to mess about with it. Perhaps only bother trying to approximate it for the first number of terms you have, for example in your last one there, ignore everything x^4 and above.

16. Oct 30, 2007

### Count Iblis

Thanks everyone for their input. I thought of this problem a long time ago, when I was discussing conformal invariance with friends and the discussion strayed off topic (as often happens when I'm involved in discussons ) . Anyway, I never seriously thought about how actually to attack this problem until I saw some responses here.

In particular, CrGreathouse's suggesting looks interesting. One can think of the functions $$g_{n}\left(z\right)$$ defined as $$g_{n+1}\left(z\right)=\exp\left[g_{n}\left(z\right)\right]$$ and $$g_{0}\left(z\right)=z$$. For fixed z, one can consider the sequence $$g_{n}\left(z\right)$$ and try to analytically continue it somehow to and consider the case $$n=\frac{1}{2}$$.

This may work better for the inverse of $$f\left(z\right)$$. Then we need to analytically continue the sequence $$\log\left(z\right)$$, $$\log\left(\log\left(z\right)\right)$$, $$\log\left(\log\left(\log\left(z\right)\right)\right)$$, ...

This sequence usually converges to some solution of the equation $$\log\left(z\right)=z$$. Suppose that a sequence starting at $$z_{0}$$ converges to $$p$$, then presumably there exists a region around $$z_{0}$$ that will also converge to $$p$$. At least, if I iterate the logarithm with various starting values on my calculator I always end up at $$p\approx 0.318 + 1.337 i$$...

So, it seems that there are limit points that have some region that are attracted to it. It seems that the (unique) analytical continuation to complex $$n$$ could be a nicely behaved function as a function of $$z_{0}$$. If we denote the k-th term in the sequence with starting value $$z_{0}=z$$ by $$\log\left(k,z\right)$$, and use this notation for the analytical continuation to complex k also, then we need to verify that:

$$\begin{equation*} \log\left[a,\log\left(b,z\right)\right] = \log\left(a+b,z\right)$$

17. Nov 1, 2007

### Zurtex

As mentioned earlier, I thought it might be a good idea to try the taylor series approach. However unfortunately this quickly breaks down, I tried to approach it my doing this:

fn(x) = a0 + a1x + ... + anxn

Then we have:

fn(fn) = g0(a0, a1, ... , an) + g1(a0, a1, ... , an)x + ... + gn-1(a0, a1, ... , an)xn-1 + O(xn) = 1 + x + x/2! + ... + xn-1/(n-1)! + O(xn)

This gives n-equations in n-variables, I was hoping this would produce a unique real solution. n = 0 and n = 1 are pretty easy to work out, and n = 2 gives:

a0 is the unique real root of:

x7 + 24x5 - 28x4 + 152x3 - 264x2 + 160x - 32 = 0

a1 = 1/32 (-1264 + 4536 a0 - 3888 a02 + 316 a03- 660 a04 - 18 a05 - 27 a06)

a2 = -1/a02 (a0a1 + a0 - 1)

Unfortunately, when you increase to n = 3, you get 4 real roots in a0. This seems to imply to me that if f(x) exists, it may not be analytical on the complex plane, but it's been a while since I did any complex analysis, so I'm likely completely wrong. Another thought occurs to me, that it may be possible to slowly increase the number of equations and see if its possible to keep the number of real roots down to 1. Will have to play about with it later.

Edit: Just tested it, trying 5 equations for n = 3, gives you no solutions at all. Oh well, was worth a try.

Last edited: Nov 1, 2007
18. Nov 1, 2007

### christianjb

19. Nov 1, 2007

### Count Iblis

Thanks Christian! That is very interesting!

An expansion about a solution of exp(z) = z is much easier to handle for numerical work. I just did this to fourth order with Mathematica and it does seem to give sensible answers (e.g. it is real within the expected accuracy for real z, even though we are expanding around a complex point). Anyway, it is a piece of cake to write a simple Mathematica program to compute the first hundreds of terms...

A related problem is to consider f(f(f......N terms f(z))) = exp(z)

and study what happens for large N. You would expect that:

f(z) = z + g(z)/N + higher order terms in 1/N

So, g(z) can be called a generator of the exponential function. Is there a simple way to see what function g(z) should be?

20. Nov 1, 2007

### Count Iblis

Note: I forgot to tell that if you expand around a solution of Exp(z) = z then solving for the expansion coefficients is trivial. If you have found the first n coefficients, then solving for the n+1-st involves solving a linear equation.