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If we set $f(x):=x^x$, then the iterations of $f(x)$ follow a steadily increasing in the complexity of the function as shown by https://www.desmos.com/calculator/wpgpilljhg.

The iterations also show a very nice order to them, with their steadily increasing.

This begs me to ask the question, what is the general function $g(n,x)=f^n(x)$? Not sure what tags to use, if you know, add them and delete this part please.

Mark Viola
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Jacob Claassen
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    Iterated exponentiation is called tetration. See: https://en.wikipedia.org/wiki/Tetration – David May 11 '17 at 05:34
  • That's not exactly what I had in mind. If you look at the Desmos page you'll see it isn't just $x^{x^x}...$ – Jacob Claassen May 11 '17 at 05:43
  • $g(2,x)=(x^x)^{x^x}\neq x^{x^{x^x}}$ – helloworld112358 May 11 '17 at 05:58
  • @helloworld112358 yes, I misunderstood the question slightly. The proof that this function is not elementary would follow in the exact same manner as that of tetration though. So at the end of the day, there isn't going to be a 'nice' form for $g$. – David May 11 '17 at 06:04
  • Could there be a way to calculate g for specific n's? – Jacob Claassen May 11 '17 at 06:06
  • "what is the general function": it cannot be expressed in terms of elementary functions and you must do with the recursive definition $g(n,x)=g(n-1,x)^{g(n-1,x)}$. –  May 11 '17 at 08:30

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I believe that your reference is wrong from $f_3(x)$ on. The correct expression is $$ f_3(x) = \left(x^{x^{1+x}}\right)^{x^{x^{1+x}}}=x^{x^{1+x}\cdot x^{x^{1+x}}}=x^{x^{1+x+x^{1+x}}} $$ I think that the best you can have is a recursive formula: the general term is $$ f_n=x^{x^{p_n}} $$ with $p_1=1$; then $$ p_{n+1}=p_n+x^{p_n} $$ so that the first terms are \begin{align} p_1=&\ 1\\ p_2=&\ 1+x\\ p_3=&\ 1+x+x^{1+x}\\ p_4=&\ 1+x+x^{1+x}+x^{1+x+x^{1+x}}\\ \vdots&\\ \end{align}

lesath82
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