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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.

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  • $\begingroup$ "By inspection my attempts are always wrong" Such as? $\endgroup$ – Did Aug 6 '18 at 13:35
  • $\begingroup$ Is this called avoiding a question by asking another question or what? For the record, I do not call the lines you just added, "an attempt". $\endgroup$ – Did Aug 6 '18 at 13:41
  • $\begingroup$ Where did you find this problem? $\endgroup$ – TheSimpliFire Aug 6 '18 at 13:47
  • $\begingroup$ @TheSimpliFire: I just created it for fun. $\endgroup$ – Field Medalist Aug 6 '18 at 13:48
  • $\begingroup$ I might be wrong but if you take the derivative of both sides then you get $f'(f(x)) \cdot f'(x) = x^x(\ln(x) + 1)$. Maybe you can solve this DE or prove that i has no solutions? $\endgroup$ – Shrey Joshi Aug 6 '18 at 13:59
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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). \ $

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    $\begingroup$ You might consider adding the reasoning behind the assertion that $f(1)=1$. Obviously, $f(f(1))=1$, but how does that imply that $f(1)=1$? $\endgroup$ – Mark Viola Aug 6 '18 at 14:54
  • $\begingroup$ @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) $\endgroup$ – Simply Beautiful Art Aug 6 '18 at 15:04
  • $\begingroup$ 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 math.stackexchange.com/questions/208996/half-iterate-of-x2c/… and mathoverflow.net/questions/45608/… $\endgroup$ – Will Jagy Aug 6 '18 at 16:36
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    $\begingroup$ Your series for $g$ has radius of convergence $0 \; .$ $\endgroup$ – Will Jagy Aug 6 '18 at 16:39
  • $\begingroup$ @SimplyBeautifulArt All that shows is that $f(1)=f(1)^{f(1)}$. $\endgroup$ – Mark Viola Aug 6 '18 at 17:19

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