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Given a hyperreal infinitesimal number $\epsilon$ , is it meaningful to take its square root, $\sqrt{\epsilon}$ or any other root? What about using it as an exponent, as in $2^{\epsilon}$ ? And what about something like $\epsilon^{\epsilon}$ .

I've looked in a few sources like Henle's Infinitesimal Calculus but couldn't find anything on how or whether exponentiation works with hyperreal infinitesimals.

bowsersenior
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1 Answers1

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As mentioned in the comments, any real function extends canonically to a hyperreal function, sometimes called the natural extension. You can then apply Łoś's theorem/the transfer principle to apply any identities you know for the real functions to their hyperreal extension.

From the book you mentioned, Henle's Infinitesimal Calculus, Chapter 3:

Suppose that $f$ is a function on the reals in our language $L.$ How do we define $f$ on the hyperreals? Suppose that $j$ is a hyperreal, What is $f(j)$?

Well, $j$ is the sequence $j(1), j(2), j(3), j(4), \dots,$ so we define $f(j)$ to be the new sequence

$$f(j(1)), f(j(2)), f(j(3)), f(j(4)), \dots$$

Here is a possible problem: suppose that $j$ and $k$ represent the same hyperreal. Then $f(j)$ and $f(k)$ should represent the same hyperreal. Do they? Certainly, for let

$$A=\{n\mid j(n)=k(n)\}$$

and

$$B=\{n\mid f(j(n))=f(k(n))\}.$$

We know $A$ is quasi-big and we can easily check that

$$A\subseteq B.$$

Thus $B$ must be quasi-big so so $f(j)$ and $f(k)$ represent the same number. In this case $f$ was a 1-place function, but 2-place, 3-place, and other functions work in the same way.

Dap
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  • Thanks for the answer. Can you say anything about the values of exponential functions using infinitesimals? It seems to me that $2^{\epsilon}$ would be an infinite number while $\sqrt{\epsilon}$ would be an infinitesimal greater than $\epsilon$ . I have no guess about $\epsilon^{\epsilon}$ . Would that be an infinitesimal as well? – bowsersenior Jan 13 '18 at 20:49
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    @bowsersenior: the functions $2^x$ and $x^x$ defined for real $x\geq 0$ have a limit of $1$ at $x=0$ (from above). This implies $1<2^\epsilon<1+\delta$ for every standard $\delta>0$ and similarly for $\epsilon^\epsilon,$ which means the standard part of $2^\epsilon$ and of $\epsilon^\epsilon$ are both $1.$ – Dap Jan 16 '18 at 13:32