In hyperreal field, can ln(ε) and ln(ω) be expressed as infinite sums?

Question

In the hyperreal field, we can use Taylor series to express e^(ε) and e^(ω) as:

e^(ε) = 1 + ε + (ε^2)/2! + ...

e^(ω) = 1 + ω + (ω^2)/2! + ...

Is it similarly possible to express ln(ε) and ln(ω) as infinite sums in terms of ε and ω?

Since 0 = ln(1) = ln(ε*ω) = ln(ε) + ln(ω), we know ln(ε) = -ln(ω), so once we find one expression we can find the other.

For instance:

e^(ω) = 1 + ω + (ω^2)/2! + ...

ln(e^(ω)) = ln(1 + ω + (ω^2)/2! + ...)

ω = ln(1 + ω + (ω^2)/2! + ...)

from which we can derive a cheap answer:

ln(ω) = ln(ln(1 + ω + (ω^2)/2! + ...))

But I'm curious if ln(ω) can be expressed in terms of ω without the logarithm?

We see ln(ω) < ln(1 + ω + (ω^2)/2! + ...) = ω

which implies ln(ω) is possibly an infinite sum of roots of ω, such as:

ln(ω) = ω^(1/2) + ω^(1/3) + ...

or some such series.

Answer 1

To help avoid any misunderstanding that may arise for readers of this question, let me say that when understood in the usual sense, there are no nontrivial convergent sequences or series at all in the hyperreal numbers (and neither in the surreal numbers).

The reason is that the hyperreal numbers are usually understood to be countably saturated. This has a model-theoretic definition, meaning that every finitely realized type with countably many parameters is realized. But in the case of ordered fields, it can be expressed as the following property: whenever $$x_0\leq x_1\leq x_2\leq\cdots \leq y_2\leq y_1\leq y_0$$ with $x_n<y_n$ for all $n\in\mathbb{N}$, then there is a number $z$ in between $$x_0\leq x_1\leq x_2\leq\cdots <z< \cdots \leq y_2\leq y_1\leq y_0.$$

This saturation property prevents any nontrivial convergent sequence $x_n\to x$, unless it is eventually constant, since there will be various numbers $z$ in between, on each side.

Indeed, every countable set of hyperreal numbers is discrete in the hyperreal order. For this reason, one cannot apply any of the usual treatment of sequences and series from the real numbers in the hyperreal field.

Meanwhile, there is nevertheless a robust theory of sequences and series in the hyperreals, but using sequences and series indexed instead by $\mathbb{N}^*$, the nonstandard natural numbers, instead of merely $\mathbb{N}$. With this change, one can form hyperreal analogues of all the familiar sequences and series in the reals. The reason is that by the transfer principle, every assertion made in the real numbers about any such sequence or series is also true exactly the same in the nonstandard realm.

Thus, the radius of convergent of the nonstandard analogue of a power series remains the same as what it was.

For example, the series $$\sum_{n\in\mathbb{N}}\frac 1{2^n}=\frac12+\frac14+\cdots+\frac1{2^n}+\cdots$$ does not converge in the hyperreals, since one is using the natural numbers only. But the much longer (uncountably so) series $$\sum_{n\in\mathbb{N}^*}\frac 1{2^n}=\frac12+\frac14+\cdots+\frac1{2^n}+\cdots+\cdots=1$$ does converge to $1$.

In my essay on the surreal numbers, an elementary introduction, I mentioned this situation as a possible red flag for surreal calculus. To do calculus with the surreal numbers (or the hyperreals), one simply cannot expect to use sequences and series and convergence ideas in the same manner that one does with the real numbers. Rather, one must in a sense remount calculus entirely bottom-up on the new setting, using infinitesimals, saturation, and transfer.

Some accounts of nonstandard analysis seem to provide a way to avoid the issue by simply treating $\mathbb{N}^*$ as the genuine natural numbers, and then talking some the standard numbers as a special kind of number. In this way of speaking, the subject of nonstandard analysis is concerned with sequences and series, even though the index set is uncountable by the usual way of talking. In my experience, however, this practice often leads to confusion when mathematicians from other areas join in. And so I find it best to be clear about what one means.

(Lastly, let me add that unless one adopts additional axioms, it is not usually correct to refer to "the" hyperreals, since we have no categorical such structure. But see my recent paper How CH could have been a fundamental axiom for more about how it could have been different.)

Answer 2

For positive $\epsilon$, the expression $\ln \epsilon$ will be equal to its power series at $x=1$ (in the $\delta, N$ sense).

To help avoid any misunderstanding that may arise for readers of this question, let me say that when understood in the usual $\delta, N$ sense, there are indeed nontrivial convergent sequences in the hyperreal numbers (though perhaps not in the surreal numbers).

Indeed, since $\epsilon>0$ is in the open disk of convergence at the point $1$, the usual condition will hold namely that for every $\delta>0$ there exists and $N>0$ such that partial sums of length at least $N$ will be within $\delta$ of the value $\ln \epsilon$. Of course, for nonstandard points one may have to choose a nonstandard $N$, as in our case.

This follows by applying the transfer principle. As far as $\ln \omega$ is concerned, it certainly cannot be given by a power series because $\omega$ falls outside the radius of convergence, seeing that $\omega>2$.

Answer 3

I will propose an answer my question. Please let me know if I have spoken accurately.

Within the hyperreal field (that is, the most traditional hyperreal field as described by some of the original texts on the subject by Robinson, Goldblatt, Keisler, and Henle), can we express ln(ϵ) and ln(ω) as a power series in terms of ϵ and/or ω? (That is, where ϵ is infinitesimal, ω is infinite, and ϵ = 1/ω.)

Yes:

ln(ω) = (1-ϵ) + ((1-ϵ)^2)/2 + ((1-ϵ)^3)/3 + ((1-ϵ)^4)/4 + ....

ln(ϵ) = -(1-ϵ) + -((1-ϵ)^2)/2 + -((1-ϵ)^3)/3 + -((1-ϵ)^4)/4 + ....

These results are indeed related by the log identity ln(ω) = -ln(ϵ), but can both be derived independently from the Taylor series for natural logarithm

ln(1 + x) = x - (x^2)/2 + (x^3)/4 - (x^4)/4 + ...

whose radius of convergence is |x| < 1 and x = 1.

PROOF:

First, find ln(ϵ).

Since the infinitesimal ϵ is in the radius of convergence, we can directly let x = -1 + ϵ:

ln(ϵ) = ln(1 + (-1+ϵ)) = (-1+ϵ) - ((-1+ϵ)^2)/2 + ((-1+ϵ)^3)/3 - ((-1+ϵ)^4)/4 + ...

We can rewrite this expression more neatly as:

ln(ϵ) = -(1-ϵ) - ((1-ϵ)^2)/2 - ((1-ϵ)^3)/3 - ((1-ϵ)^4)/4 - ...

After expanding the right side and gathering all the standard terms, we see an expression that clearly diverges to negative infinity as expected:

ln(ϵ) = -1 - 1/2 - 1/3 - 1/4 - .... + (all of the non-standard terms)

Second, find ln(ω).

To do this, I will derive a transformed Taylor series for ln(m) whose radius of convergence is any real number m > 1.
Assuming m > 1, we have 1 > 1/m > 0, which can be rewritten as 0 > -1 + 1/m > -1.
Therefore, (-1 + 1/m) is within the radius of convergence of the Taylor series used above.
Thus, we can rewrite ln(m) as:

ln(m) = ln((1/m)^-1) = -ln(1/m) = -ln(1 + (-1 + 1/m))

and can now use the given Taylor series using x = -1 + 1/m:

ln(m) = -ln( 1 + (-1 + 1/m)) = (1-1/m) + ((1-1/m)^2)/2 + ((1-1/m)^3)/3 + ((1-1/m)^4)/4 + ....

This is a convenient result in and of itself because radius of convergence of this Taylor series for ln(m) is any real m > 1.

By transfer principle, since ω > 1, we can use ln(m) to find:

ln(ω) = (1-1/ω) + ((1-1/ω)^2)/2 + ((1-1/ω)^3)/3 + ((1-1/ω)^4)/4 + ....

and since 1/ω = ϵ, we have:

ln(ω) = (1-ϵ) + ((1-ϵ)^2)/2 + ((1-ϵ)^3)/3 + ((1-ϵ)^4)/4 + ....

which is indeed the opposite sign of the result for ln(ϵ):

ln(ϵ) = -(1-ϵ) + -((1-ϵ)^2)/2 + -((1-ϵ)^3)/3 + -((1-ϵ)^4)/4 + ....

In both cases, after expanding the terms and gathering all the standard terms, we see expressions that clearly diverge to infinity as expected:

ln(ϵ) = -1 - 1/2 - 1/3 - 1/4 - .... + (all of the non-standard terms in terms of ϵ)

ln(ω) = 1 + 1/2 + 1/3 + 1/4 + .... + (all of the non-standard terms in terms of ϵ)