Infinite tetration and superroot of infinitesimal
I have stumbled upon Time calculus while searching for substructures of infinitesimals and how to include time in mathematics.

I thought it might give a handle on infinitesimals, since, how I understand, it, structureless linear infinitesimal h->0 in derivative definition is replaced in general case by graininess \( \mu(t) ->0 \) to obtain normal continuous derivative.

That means that infinite tetration etc of infinitesimal can be written(?):

lim \( \mu(t)->0 \): \( \mu(t) [4]\infty \)

And superroot :lim \( \mu(t)->0 \): \( q(t)^{q(t)}= \mu(t) \)

And self root: lim \( \mu(t)->0 \): \( \mu(t)^{1/\mu(t)} \)

The limit may depend on how \( \mu \) depends on t since it is not obvious what will happen faster- tetration or changes in \( \mu(t) \).

Gut feeling is, these speeds may be comparable.

The next question is, what will happen with this tetration and time calculus if \( \mu(t) \) itself is hyperoperation?

for example, \( \mu(t) = b[4]t \) ? or \( \mu(t) = t[4]\infty \)? Or \( \mu(t) = t[x]y \)?

And generally, hyperoperations in time scales might be interesting.

Please point out obvious mistakes. Thank You in advance,

Ivars
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Messages In This Thread
RE: Infinite tetration and superroot of infinitesimal - by Ivars - 05/10/2008, 08:33 AM

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