Find $f(x)$ satisfying $f(f(x))=x^x$ By inspection my attempts are always wrong. I really have no idea and given up. 
How to find $f(x)$ satisfying $f(f(x))=x^x$?
My attempts:


*

*$f(x)=x^x$

*$f(x)=x^{1/x}$

*$f(x)=\frac{1}{x^x}$


My profession is not a mathematician so I am not well  trained in mathematics beyond high school mathematics. If you know the solution, please give me a hint.
 A: The key is to identify a fixed point of the function $x^x$. An obvious fixed point is $x=1.$ To simplify the work, we assume that $f$ has the same fixed point. We define $\ g(x) := f(1+x)-1 \ $ where $\ g(0) = 0 \ $ because $\ f(1) = 1. \ $ Now using $\ f(f(x)) = x^x \ $ we have
 $$ g(g(x)) = f(1\!+\!g(x)) \!-\! 1 = f(f(1\!+\!x)) \!-\! 1 = (1\!+\!x)^{1+x} - 1 = x + x^2 + \frac{x^3}2 + \frac{x^4}3 + O(x^5).  $$
Assuming a power series expansion for $\ g(x), \ $ we can solve for its coefficients and get
 $$ g(x) = x + \frac{x^2}2 + 0x^3 + \frac{5}{48}x^4 - \frac{11}{96}x^5 + \frac{257}{1920}x^6 - \frac{851}{5760}x^7 + \frac{15751}{107520}x^8 + O(x^9). $$
Now $\ f(x) = 1 + g(x-1). $ I am not sure about the radius of convergence of the series. It may be zero. All the coefficients up to $x^{18}$ are less than $1$ in absolute value, but then they grow very rapidly. Still, for  $\ .7 < x < 1.2 \ $ the $\ f(f(x)) \ $ is a close approximation to $\ x^x $ but adding more terms in the series makes it worse. It reminds me of the asymptotic expansion of $\ \log \Gamma (x). \ $
