08/11/2007, 09:47 AM
Actually, I have an unpublished solution to base-\( e^{1/e} \) tetration as well, and I was saving it for publication, but I suppose I wouldn't mind explaining it here. Let \( DE_b(x) = b^x - 1 \), (I call them decremented exponentials). Then look at the iterates of DE with the -1 first, instead of last as it is usually written:
And let \( \exp_b^{[n]}(x) = b \uparrow (\cdots \uparrow (b \uparrow x)) \) represent iterated exponentials. We can then represent the relationship between these two functions as a simple linear equation:
First, assume that \( DE_b^{[n]}(-x) = -1 + b^{-1} \exp_{(b^{1/b})}^{[n-1]}(x) \). It then follows that:
Now that that's proven, anything thats true about one function should be true about the other function, since the relationship is linear. However, Trappmann said that \( e^x - 1 \) has a continuous iterate that fails to converge for non-integers. This could be disastrous for tetration (another reason I didn't want to post this). But it also means that since the function \( y = b^{1/b} \) is unique for b=e, it means that base-\( e^{1/e} \) tetration is uniquely defined, even if its series doesn't converge.
Andrew Robbins
- \( DE_b^{[2]}(-x) = -1 + b^{-1 + b^{-x}} = -1 + b^{-1}\left(b^{1/b}\right)^x \)
- \( DE_b^{[3]}(-x) = -1 + b^{-1 + b^{-1 + b^{x}}} = -1 + b^{-1}\left(b^{1/b}\right)^{\left(b^{1/b}\right)^x} \)
- \( DE_b^{[4]}(-x) = -1 + b^{-1 + b^{-1 + b^{-1 + b^{-x}}}} = -1 + b^{-1}\left(b^{1/b}\right)^{\left(b^{1/b}\right)^{\left(b^{1/b}\right)^x}} \)
And let \( \exp_b^{[n]}(x) = b \uparrow (\cdots \uparrow (b \uparrow x)) \) represent iterated exponentials. We can then represent the relationship between these two functions as a simple linear equation:
Theorem
\(
\exp_{(b^{1/b})}^{[n]}(x) = b(1 + DE_b^{[n+1]}(-x))
\)
\exp_{(b^{1/b})}^{[n]}(x) = b(1 + DE_b^{[n+1]}(-x))
\)
Proof
We already have several base cases above; the rest is proved by induction. First, assume that \( DE_b^{[n]}(-x) = -1 + b^{-1} \exp_{(b^{1/b})}^{[n-1]}(x) \). It then follows that:
\(
\begin{array}{rl}
DE_b^{[n+1]}(-x)
& = DE_b[-1 + b^{-1} \exp_{(b^{1/b})}^{[n-1]}(x)] \\
& = -1 + b^{[-1 + b^{-1} \exp_{(b^{1/b})}^{[n-1]}(x)]} \\
& = -1 + b^{-1} b^{[b^{-1} \exp_{(b^{1/b})}^{[n-1]}(x)]} \\
& = -1 + b^{-1} (b^{1/b})^{[\exp_{(b^{1/b})}^{[n-1]}(x)]} \\
& = -1 + b^{-1} {\exp_{(b^{1/b})}^{[n]}(x)}
\end{array}
\)
which proves our assumption, rearranging gives the above.\begin{array}{rl}
DE_b^{[n+1]}(-x)
& = DE_b[-1 + b^{-1} \exp_{(b^{1/b})}^{[n-1]}(x)] \\
& = -1 + b^{[-1 + b^{-1} \exp_{(b^{1/b})}^{[n-1]}(x)]} \\
& = -1 + b^{-1} b^{[b^{-1} \exp_{(b^{1/b})}^{[n-1]}(x)]} \\
& = -1 + b^{-1} (b^{1/b})^{[\exp_{(b^{1/b})}^{[n-1]}(x)]} \\
& = -1 + b^{-1} {\exp_{(b^{1/b})}^{[n]}(x)}
\end{array}
\)
Conclusion
Now that that's proven, anything thats true about one function should be true about the other function, since the relationship is linear. However, Trappmann said that \( e^x - 1 \) has a continuous iterate that fails to converge for non-integers. This could be disastrous for tetration (another reason I didn't want to post this). But it also means that since the function \( y = b^{1/b} \) is unique for b=e, it means that base-\( e^{1/e} \) tetration is uniquely defined, even if its series doesn't converge.
Andrew Robbins
