TITLE: Half iteration of exponential function
QUESTION [8 upvotes]: I'm working on the half iteration of the exponential function. No one has any idea what fractional iterations could mean but I think intuitively it should be a function $f(x)$ such that $f(f(x))=e^x$.
Here's how I'm finding $f(x)$ when $x\approx 0$:
If $x\approx 0$, then, we have,
$$e^x\approx 1+x+\frac{x^2}{2}$$.  ...(1)
Now, if we assume the required function $f(x)$ to be of the form $ax^2+bx+c$, then $$f(f(x))= a^3x^4+2a^2bx^3+(2a^2c+ab^2+ab)x^2+(2abc+b^2)x+ac^2+bc+c$$
But, since $x\approx 0$ therefore,
$$f(f(x))=e^x\approx ac^2+bc+c+(2abc+b^2)x+(2a^2c+ab^2+ab)x^2$$.        ....(2)
Comparing coefficients of like powers of $x$ in equation (1) and (2), we get,
$$ac^2+bc+c=1 \tag {3.1}$$
$$2abc+b^2=1 \tag {3.2}$$
$$2a^2c+ab^2+ab=\frac{1}{2} \tag {3.3}$$
The problem is solving these equations. I've tried substitution but they get reduced to a polynomial of very high degree which I don't know how to solve. Is there some way to solve these to get $a$, $b$, and $c$ and hence get the required half iteration function of $e^x$ as $ax^2+bx+c$? Please tell me how to solve these three equations.

REPLY [2 votes]: Looking at the equations $$ac^2+bc+c=1 \tag 1$$ $$2abc+b^2=1 \tag 2$$ $$2a^2c+ab^2+ab=\frac{1}{2} \tag 3$$ we can eliminate $b$ from $(1)$ $$b=\frac{1-c-a c^2}{c}$$ Replacing in $(2)$ and solving for $a$ leads to $$a=\frac{\sqrt{1-2 c}}{c^2}$$ Replacing in $(3)$ leads to $$-c \left(c^3+12 c+6 \sqrt{1-2 c}-14\right)+4 \sqrt{1-2 c}=4$$ After squaring steps, this reduces to $$c^7+24c^5-28c^4+152c^3-264c^2+160c-32=0$$ which has only one real root close to $c=\frac 12$. 
Using Newton method for finding the zero of the equation in $c^7$ leads to $$a=0.261795456735753$$ $$b=0.878112905194437$$ $$c=0.497894079064888$$ as already given in Gottfried Helms's answer. These numbers can be rationalized as $$a=\frac{37409}{142894}\qquad b=\frac{77821}{88623}\qquad c=\frac{18323}{36801}$$
Edit
Back to the problem eighteen months later, we could get expressions for the solution using $[1,n]$ Padé approximants for the septic equation (built around $c=\frac 12$). For different values of $n$, this would give
$$a_1=\frac{146336 \sqrt{352121}}{331786225}\qquad b_1=\frac{18369-4 \sqrt{352121}}{18215}\qquad c_1=\frac{18215}{36584}$$
$$a_2=\frac{3257213 \sqrt{44685705147}}{2630063332009}\qquad b_2=\frac{1635466-\sqrt{44685705147}}{1621747}\qquad c_2=\frac{1621747}{3257213}$$