Find a function $f(x)$ such that $$f(f(x))=e^x$$
Whenever I encounter these type of questions, for instance $f(f(x))=x$, I take the help of graphs. But in this question, when I draw a graph the problem becomes exceedingly complicated, because the graph of $e^x$ is not a straight line. So, I thought of a different approach.
Consider the power series expansion of $f(x)$
$$f(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+\cdots$$
Substituting this expression into itself for $x,$ and collecting terms by powers of $x,$ we can equate the resulting coefficients to the corresponding coefficients of $e^x.$ This gives a set of conditions on the derivatives of $f$ at $0$ and $a_0.$ Letting $f_n(x)$ denote the $nth$ derivative of $f$, we have the conditions,
$$f_0(a_0) = 1$$
$$ f_1(0) f_1(a_0) = 1$$
$$ f_2(0) f_1(a_0) + f_1(0)^2 f_2(a_0) = 1$$
and so on.
After this, I'm stuck. I'm not able to think how to continue.
Any help would be greatly appreciated.
