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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.

Simply Beautiful Art
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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). \ $

Somos
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    You might consider adding the reasoning behind the assertion that $f(1)=1$. Obviously, $f(f(1))=1$, but how does that imply that $f(1)=1$? – Mark Viola Aug 06 '18 at 14:54
  • @MarkViola Note that $f(1)^{f(1)}=f(f(f(1)))=f(1^1)$, so we don't have to assume this is the case (as per the last edit) – Simply Beautiful Art Aug 06 '18 at 15:04
  • The procedure of Ecalle gives $f \in C^\infty$ for $x > \frac{1}{e}$ and, except for the point $x=1$ itself, actually $C^\omega$ See http://math.stackexchange.com/questions/208996/half-iterate-of-x2c/209653#209653 and https://mathoverflow.net/questions/45608/does-the-formal-power-series-solution-to-ffx-sin-x-converge/46765#46765 – Will Jagy Aug 06 '18 at 16:36
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    Your series for $g$ has radius of convergence $0 ; .$ – Will Jagy Aug 06 '18 at 16:39
  • @SimplyBeautifulArt All that shows is that $f(1)=f(1)^{f(1)}$. – Mark Viola Aug 06 '18 at 17:19
  • @MarkViola Yes, hence $f(1)=1$, since $1$ is the only fixed-point of $x^x$... – Simply Beautiful Art Aug 06 '18 at 17:20
  • @SimplyBeautifulArt And how does one show that? The OP states "I am not well trained in mathematics beyond high school mathematics." – Mark Viola Aug 06 '18 at 17:22
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    @MarkViola I would've thought it obvious that if one restricts to positive real numbers, then we would have $$x=x^x\implies1=x^{x-1}$$hence either the base is 1, or the exponent 0. I don't think this requires anything beyond high school? – Simply Beautiful Art Aug 06 '18 at 17:24
  • @SimplyBeautifulArt Yes, I know. But the OP might not be able to form that argument. – Mark Viola Aug 06 '18 at 17:27
  • @MarkViola That's an argument based on what we don't know. But based on what we do, I think this is reasonable. – Simply Beautiful Art Aug 06 '18 at 17:30