Functional Equations with Taylor Series

I am stuck on the functional equation

$$f(f(x))+f(x)=e^x$$
Somebody has suggested to me in the past that I can use Taylor Series to solve some of the harder functional equations, but I'm not sure how to use that to solve this. Help?

If Taylor Series is not the way to go for this functional equation, can somebody tell me what method I should use?

In general, can anybody recommend a few different methods that would be good to keep in mind when dealing with functional equations like this in which functional iteration is present?

Thanks!

Answer

I don't know whether there is a simple solution for the function, but I can show how one would use a Taylor expression here.

First, let us suppose there is a value $q$ which is a fixed point of the function f(x)
$$f(q)=q$$
If we fill this in the relation we find:
$$f(f(q)) + f(q) = f(q) + q = 2 q = e^q$$
The numerical value would be $q\approx0.693147$.

Since we are interested in a smooth function $f(x)$ we know that we could in principle make a Taylor expansion at any point $x$. If we use the fixed point we would have something like:

$$f(q+\delta) = q + \sum_{i=1}^\infty a_i \delta^i$$
with some still unknown coefficients $a_i$. Having assumed a smooth function $f(x)$ there is at least a small range of values for $\delta$ where the series converges.

We can now insert this expansion into the relation we have and work all terms out up to some order ${\cal O}(\delta^n)$ and group terms with the same power of $\delta$. The first few terms would be :

$$2 q + \left(a_1 + a_1^2\right) \delta + \left(a_2 + a_1 a_2 + a_1^2 a_2\right) \delta^2 + \dots = e^{q+\delta} = e^q + e^q \delta + e^q \frac{\delta^2}{2} + \dots$$
and since for any small $\delta$ this should be the same on both sides we get a lot of equations in the unknown values $a_i$. The lowest order is
$$2 q = e^q,$$
which already assumed. The next one is
$$a_1 + a_1^2 = e^q = 2q$$
from which we find
$$a_1 = \frac{1 \pm \sqrt{1 + 8 q}}{2}$$
The sign is not too important and only determines whether we choose $x$ to go to the left or the right. The next equation we get is
$$a_2 + a_1 a_2 + a_1^2 a_2 = \frac{1}{2} e^q = 2 q$$
We can now insert the solution obtained for $a_1$ (or combine the equations) and determine that

$$a_2 = \frac{q}{1 + 2 q}$$
This procedure can be continued arbitrarily far and in this particular problem(!) the first equation you encounter for $a_k$ is always linear and easy to solve.

Unfortunately going from a "determined" function in terms of a local Taylor expansion to a closed expression, is in general not so easy.

A typical strategy to solve this type of problems would be to start looking for fixed points or symmetries that can be exploited.