This question is motivated by model theory, but it's really an analysis question (which means it may have an easy analysis answer that I just don't have the background for). Here's the main question, followed by some definitions and motivation:



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> Is there a partial function $f:\mathbb R\to \mathbb R$, defined on an unbounded interval $(a,\infty)$, which is both superexponential and Pfaffian?
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I suspect if there is a solution to the above, there is also a totally defined function, but that's the form of the question as it's useful to me.



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> A function $f$ is *superexponential* if, for all $n$, there is an $a$ where for all $x>a$, $f(x)>\exp(\exp(\cdots \exp(x)\cdots))$, where $\exp(x)$ is $e^x$, and there are $n$ iterates of $exp$ in the preceding expression.
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Equivalently (for model theorists), a function is superexponential if and only if it is not eventually bounded by any function definable in $(\mathbb R, +,\cdot,0,1,\exp)$. It is possible to construct (real)-analytic superexponential functions without too much trouble; see this [answer by fedja](https://math.stackexchange.com/questions/493078/a-non-exponentially-bounded-analytic-function) to a Math.StackExchange question to this effect.



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> A Pfaffian chain is a sequence of unary functions $(f\_1,\ldots,f\_n)$ such that for each $k$, there is a real polynomial $p\_k(x,y\_1,\ldots,y\_k)$ where $f\_k'=p\_k(x,f\_1,\ldots,f\_k)$ on a tail of $\mathbb R$. A function is *Pfaffian* if it is part of a Pfaffian chain.
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So for example, all polynomials are Pfaffian, as is the exponential function. However, $\sin$ and $\cos$ are not Pfaffians, since each one would need to "reference the other." 


For motivation see [another question](https://mathoverflow.net/questions/121010/is-there-any-o-minimal-expansion-of-the-real-field-with-functions-of-growth-high), where this question was raised in a comment by Joel David Hamkins; if such a function exists, it would generate a superexponential o-minimal theory, resolving an interesting open question. However, a negative answer, while interesting, would not resolve that question. I fiddled with this formulation for a few months but eventually gave up.



Are there superexponential Pfaffian functions?
No, [1] proved more generally that the Pfaffian closure of any o-minimal polynomially bounded structure is exponentially bounded.


[1] Jean-Marie Lion, Chris Miller, and Patrick Speissegger, [*Differential equations over polynomially bounded o-minimal structures*](http://dx.doi.org/10.1090/S0002-9939-02-06509-7), Proc. Amer. Math. Soc. 131 (2003), 175–183.


