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For each $n$, define $f_n:\mathbb R^+\rightarrow \mathbb R^+$ by $f_n(x) = \underbrace{x^{x^{x^{...^{x^x}}}}}_n$

I want to find a function $f:\mathbb R^+\rightarrow \mathbb R^+$ such that for any given $n$, $f$ is eventually greater than $f_n$.

Here $\mathbb R^+$ means the non-negative reals.

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    see also http://en.wikipedia.org/wiki/Tetration and note that andres answer is like tetration of a number with itself, kind of like how squaring is multiplication of a number with itself2012-09-05

1 Answers 1

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To make notation smoother, write $f(n,x)$ for $f_n(x)$. Let $f(x)=f(\lceil x\rceil, x).$ Here $\lceil x\rceil$ is the "ceiling" function that gives the smallest integer $\ge x$.

Remark: This is a typical "diagonalization" argument. Basically the same idea seems to have been first used by du Bois-Reymond to deal with orders of growth of functions. He did it a few years before Cantor used diagonalization in Set Theory.

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    oh... and I fogot to mention... hopefully we can prove that our function / power series has d^2y/dx^2 < 0 for all$x$> 0. Also, hopefully our power series can be written concisely2012-09-06