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What's the super logarithmic inverse of tetration for $\bf{^{2}{x}}$?

Is it $slog^{x}_{2}$?

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I presume you've already looked into previous work by Ioannis Galidakis? –  J. M. Apr 18 '12 at 8:31
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1 Answer

up vote 1 down vote accepted

The inverse function of ${^x{2}}$ is the base-2 super-log of x, which is what you wrote. The inverse function of ${^2{x}}$ is the 2nd super-root of x.

While the super-log seems to have a more commonly accepted notation (slog), there is no such notation for super-root (or, if you wish, dozens of notations). There are also some places (like Wikipedia's Tetration page) that call the 2nd super-root the "super-sqrt" function. But that's OK, because lots of things in mathematics don't have unique notations. Take $e$ for example. It is used for basis vectors, physical constants, dimensional units, other variables, and the "base of the natural logarithm".

So if you're looking for terminology, the inverse of ${^2{x}}$ is called "super-root", but if you're looking for notation, just use $f(x)$, along with the explanation: "where $f$ is the second super-root".

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Thanks Andrew, I think that helped solve my problem. One question though: you mentioned that the inverse of $\bf{^2}{x}$ to be the 2$^{nd}$ super-root of $x$. Would my $\bf{slog^x_2}$ still satisfy? –  A T Apr 18 '12 at 17:57
No, super-roots and super-logs are different functions, they do not satisfy each other's equations. –  Andrew Robbins Apr 18 '12 at 18:00
Apologies, it's 4AM here. Only after rereading your answer did I realise you were talking about super roots on the same paragraph that you discussed super logarithms. –  A T Apr 18 '12 at 18:01
+1 and accepted. Thanks for your help –  A T Apr 18 '12 at 18:06
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