A method to evaluate functional roots of $e^x$

I've an idea to find exact function $f(x)$ such that $f(f(x))=e^x$. But it involves solving complicated systems of non-linear equations, the skills for which I don't have.

Here's how I intend to do it:

The Taylor series of $e^x$ around $x=0$ is: $$1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+......$$ If we assume our required function $f(x)$ to be: $$f(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+......a_2x^2+a_1x+a_0$$ i.e. a polynomial of degree $n$. From that we can get $f(f(x))$.

I think it can be proved that the $f(f(x))$ that we get will have terms containing all the powers of $x$ ranging from $x^{(n^2)}$ to $x^0$ (i.e. a constant term). Now, when $x$ is small the higher power terms don't matter, so we ignore the terms containing $x^{n+1}$, $x^{n+2}$.........$x^{(n^2)}$. After ignoring these terms, the remaining terms contain these powers of $x$: $x^0$, $x$, $x^2$,........$x^n$. Now, we compare the coefficients of these terms with the coefficients of the like powers of $x$ in the Taylor series of $e^x$. In this way, we get $n+1$ equations in $n+1$ variables.

Now, we solve these equations to get the coefficients $a_0,a_1,a_2......a_n$. These will be functions of $n$ (I think). Now we apply $\lim_{n\rightarrow \infty}$ to these functions to get the exact value of coefficients.

It would be a stressful task to solve those non-linear equations to get the coefficients in terms of $n$. But would this method work?

EDIT: Even if $x$ is not small, the higher power terms can be ignored because the higher power terms will have very large factorials in their denominators. Also, we're applying $\lim_{n\rightarrow \infty}$ in the end and $\lim_{n\rightarrow \infty}\frac{x^n}{n!}=0$ for any finite $x$.

EDIT:By using a first degree polynomial, I got this function, which when applied twice to $c$ gives you exactly $a^c$(not approximately because I didn't have to ignore any terms in case of a first degree polynomial) : $$x\sqrt{a^c\log_ea}+\frac{a^c(1-c\log_ea)}{\sqrt{a^c\log_ea}+1}$$ Problem is it itself depends upon $c$, so I guess there's a family of these square-root functions. Similarly, the $n^{th}$ root of $a^x$ at $x=c$ is: $$x(a^c\log_ea)^{\frac{1}{n}}+\frac{a^c(1-c\log_ea)}{\sum_{k=0}^{n-1}(a^c\log_ea)^{\frac{k}{n}}}$$ $$=x(a^c\log_ea)^{\frac{1}{n}}+\frac{a^c(1-c\log_ea)((a^c\log_ea)^{\frac{1}{n}}-1)}{a^c\log_ea-1}$$

At $x=c$, my square-root function is: $$c\sqrt{a^c\log_ea}+\frac{a^c(1-c\log_ea)}{\sqrt{a^c\log_ea}+1}$$. Now substituting this in place of $x$ is my function gives, $$\left(c\sqrt{a^c\log_ea}+\frac{a^c(1-c\log_ea)}{\sqrt{a^c\log_ea}+1}\right)\sqrt{a^c\log_ea}+\frac{a^c(1-c\log_ea)}{\sqrt{a^c\log_ea}+1}$$

$$=ca^c\log_ea+a^c(1-c\log_ea)$$

$$=a^c$$

• The approximating polynomial of $f(f(x))$ is of degree $n^2$, not $n^n$. This said, the method will probably lead to intractable computation. – Yves Daoust Mar 9 '17 at 8:36
• @YvesDaoust Oh, Thanks for that. And I don't think either that any person alive can do these calculations. – Dove Mar 9 '17 at 8:38
• Maybe a person can't, but a computer might for finite (small enough) $n$ – vrugtehagel Mar 9 '17 at 8:57
• It is not straightforward for a function to have square roots (see e.g. this post) – Harry49 Mar 9 '17 at 9:58
• The taxing part about this, is that you'll get a formal power series, but it's essentially impossible to prove it has a nonzero radius of convergence. Look at this post here mathoverflow.net/questions/45608/… It derives a power series of $f(f(x)) = \sin$ in a similar manner that you're trying to derive one for $e^x$. However, this formal power series, DOES NOT CONVERGE. So it's essentially, not very useful. At least, in complex analysis, or as a Taylor series. – user335907 Mar 9 '17 at 18:34