User:Vel!/Tetration

This, gentlemen, is tetration:

\begin{eqnarray*} ^0x &=& 1 \\ ^{n + 1}x &=& x^{^nx} \end{eqnarray*}

Note that \(x\) is any real number (or even a complex number) and \(n\) is any nonnegative integer.

If \(x = 0\), there will be some problems with \(0^0\). But I won't worry about them for now._

Making \(n\) negative
What if \(n = -1\)?

\begin{eqnarray*} ^{-1 + 1}x &=& x^{^{-1}x} \\ 0^x &=& x^{^{-1}x} \\ ^{-1}x &=& 0 \end{eqnarray*}

Or \(n = -2\)?

\begin{eqnarray*} ^{-2 + 1}x &=& x^{^{-2}x} \\ (-1)^x &=& x^{^{-2}x} \\ 0 &=& x^{^{-2}x} \end{eqnarray*}

There's no way to resolve this, so \(-2\) is our limit. We've extended \(n\) to all integers \(> -2\).

Making \(n\) rational
What if \(n = 1/2\)? It would make sense to have \(^{1/2}x = k\) be a solution to the equation \(^2k = x\). Similar things hold for \(^{1/3}x\), \(^{1/4}x\), etc.

\[^{1/n}x = k \implies ^nk = x\]