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Infinite tetration and superroot of infinitesimal
#9
Excuse me for continuing along the same lines- redefining complex numbers in terms of hyperreals:

In non-standard analysis, there are new "numbers" added to real line, which is subset of hyperreal set.

Let us denote any infinitesimal in hyperreal set such that 0< infinitesimal<any real (let us leave out negatives for time being) as I. There is no contradiction, as by definition any hyperreal has real part=0, as does imaginary unit.

Than infinity of any order ( same as infinitesimal defined by I- but there are MANY orders) would be 1/I. That completes the other end of extension of Real set.

Now, we know that hyperreals contain INFINITE integers > than any INTEGER ( that follows from defintion that for any real there exists integer bigger than any real, and in non-standard analysis, all statements about reals are applied to hypperreals) .

Let us define first of these integers as 1=i^4, second as 1=i^8 etc.

Now we have :

1) hyperreal infinitesimals = I ( but the I includes too much information, however, giving correctly via complex analysis the general properties of infinitesimals related to reals)

2) hyperreal infinities 1/I ( the same comment applies here, they are all different, but 1/I captures their properties as long as You stay within the same order of infinitesimals)

3) Hyperreal infinite integers = I^4, I^8, I^12 ...........etc.

4) Then statement like x+I*Y is true and finite as long as I is infinitesimal, Y= Y*1, where 1- hyperreal infinite integer, Y- just a number , coefficient.

So far I do not see any problems with proceeding with this construct.
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Messages In This Thread
RE: Infinite tetration and superroot of infinitesimal - by Ivars - 12/20/2007, 08:23 AM

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