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If $u(x)=x^x$ then we can form $$ u^2(x) = \left(x^x \right)^{x^x} = x^{x^{x+1}} $$ some simplification occurs, but the further iterates are a typographical challenge to mathjax.

Writing $E_k$ for $u^k(x)$ we have $$ E_0 =x \\ E_{n+1} = u(E_n) = E_n^{E_n}. $$ However, each term in the sequence is a power of $E_0 = x$ so we may also write $$ E_n = x^{P_n} $$ where $$\begin{align} P_0 &= 1\\ P_1 &= x \\ P_2 &= x^{x+1} \end{align} $$ the iteration gives $$ E_{n+1} = E_n^{E_n} = \left(x^{P_n}\right)^{x^{P_n}} = x^{P_nx^{P_n}} $$ so that $$ P_{n+1}=P_n x^{P_n}. $$ We note that $P_k$ is also a power of $x$, say $P_n=x^{Q_n}$ which gives $$ P_{n+1} = x^{P_n+Q_n} $$ it seems that each iteration takes us one step further up the ladder of tetration, but i do not know the notation to express this in a precise symbolic form. can anyone help?

Davide Giraudo
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David Holden
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  • apology - due to imminent travel, i may not be able to respond to comments until tomorrow – David Holden Aug 15 '14 at 13:52
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    I am not precisely sure what you are after, but you might want to look up "Knuth's up-arrow" notation. $a\uparrow b$ means $a^b$ while $a\uparrow\uparrow b$ means taking $a$ to its own power $b$-times. For example, $a\uparrow\uparrow2=a^{a^a}$. – user1729 Aug 15 '14 at 15:42
  • @user1729 yes, thx, that is the sort of thing i'm looking for, the $\uparrow\uparrow$ symbol seems a little clumsy visually, perhaps only because i am not used to writing $a^b$ as $a \uparrow b$. however one specific notational problem in the present case is that we have $x \uparrow (x \uparrow (x ... \uparrow f(x))...))$ where all "exponents" are $x$ except the final (innermost) one, which is a function of $x$ (or, rather, that is what i would like to show. but i can't even write this claim down at the moment, apart from describing it informally in ordinary language – David Holden Aug 15 '14 at 22:58
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    If I recall correctly then Heryk Trappmann had made his PhD thesis (in german) about symmetric algebraic operations, and $x \to x^x$ was one of its subjects. Try to find it via "Tetration Forum" (Sorry, I'm lazy to search it for you) – Gottfried Helms Aug 16 '14 at 21:36
  • thx Gottfried, interesting link. – David Holden Aug 17 '14 at 08:17
  • If you look at $g(y)=(1+y)^{(1+y)}-1$ instead you can expand it into a power series and study the emerging pattern of coefficients when iterating (however I don't know whether this helps really). After iterating you use $f(x,h)=g(x-1,h)+1$ where $h$ is the iteration-height to evaluate the original iteration – Gottfried Helms Sep 12 '14 at 03:55

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Extending my earlier comment. First we write $f(x)=x^x$ and $f(x,h)=f(f(x,h-1))$ , $f(x,0)=x$ and $f(x,1)=f(x)$ where $h$ indicates the "iteration-height". Then we consider the function $g(x,h)=f(x+1,h)-1$ and get a nicely iterable power series for $g(x)$ . The coefficients of $g(x,h)$ depend on $h$ and are in fact polynomials in $h$ of increasing order. The first few coefficients are $$ \begin{array} {rrrlrrrr} g(x,h) &= & x \cdot & (1 &&&&) \\ & +& x^2 \cdot & (&1 \cdot h&&&) \\ & +& x^3 \cdot & (&-1/2 \cdot h & + 1 \cdot h^2&&) \\ & +& x^4 \cdot & (& 7/12 \cdot h & -5/4 \cdot h^2 & + 1 \cdot h^3&) \\ & \vdots & \vdots \end{array}$$ After that, the actual evaluation to any arbitrary iteration-height (I think: even fractional height) can be done by evaluating $g(x)$ at $x-1$ because $$f(x,h)=g(x-1,h)+1$$

However, the next step must be to determine the range of convergence by analyzing the pattern of the coefficients in the polynomial to determine the "general term" for the polynomials in $g(x,h)$.
The convergence-radius might easily be zero, especially for the fractional iterates, but if this is the case then possibly a summation-method for divergent series can still help.

Gottfried Helms
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