FOM: answer to Percival

[Joe Shipman]

David Ross wrote:

> I don't
> understand your answer (to Percival's question);
> of course infinite products can be done using NSA, just as they can be
> done standardly.
> The NS machinery even helps simplify a little; see the appendix of my
> paper http://www.math.hawaii.edu/~ross/papers/infprod.pdf
>
> - David R.

Thanks for giving me the opportunity to clarify things here.

Are we talking about using NSA to define infinite products on standard
reals, or about defining an omega-ary operation on the entire structure of
(standard and nonstandard) reals?

In traditional analysis, an infinite product can be defined on the
structure of real numbers (actually, partially defined, because some
products won't converge).  In NSA, you can't do the same thing for the
entire structure of hyperreals, because convergence of sequences doesn't
work the same way (it is only the first-order properties that are the same
in the hyperreals, the least-upper-bound property doesn't hold anymore).
The product of a countably infinite sequence of STANDARD reals can be
defined as a standard real, using NSA-style definitions of limit and
convergence, but that's not the same thing as the product existing as a
partial omega-ary function on the set of all hyperreals.

Percival asked whether, "in NSA", the product (0.5 * 0.5 * 0.5 * 0.5 *
...), or "one-half to the omega power", is an infinitesimal (I assume
intending an infinitesimal greater than 0), and whether (0.6 * 0.6 * 0.6 *
0.6 * ...) is a LARGER infinitesimal.  So his question is asking for an
infinite product operation defined on the entire structure of hyperreals,
not just a way to use hyperreals to define an infinite product on the
reals.

My reply pointed out that although you can't do this in NSA, you CAN have
logarithmic and exponential operations, which allow you to define a binary
power function x^y for x>0, which allows you to define an omega-power
operation, which is as good as an infinite product for Percival's examples
where all the multiplicands were identical.  Because the binary power
function satisfies the usual first-order properties, it will be the case
that 0 < (0.5^omega) < (0.6^omega), which answers Percival's question.  On
the other hand, you can't say WHICH infinitesimal is the value of
0.5^omega, because there is an arbitrary choice of an infinite hyperreal
integer to play the role of omega, not to mention all the arbitrary
choices which went into picking an ultrafilter to construct the hyperreals
in the first place!

If you want something well-defined without arbitrary choices, you can use
Conway's "surreal numbers", which do not share all the first-order
properties of the real numbers but which naturally include the ordinals
and which support well-defined integrals, powers, logarithms, and even
countably infinite products.  (Usually, though, (x*x*x*x*...) and
(x^omega) will not be equal.)

-- Joe Shipman