Comparing sizes of sets of natural numbers

Question

It seems natural to consider $\lim_{q \rightarrow 1^-} \sum_{n \in S} q^n - \sum_{n \in T} q^n$, when it exists, as a way of comparing the sizes of two sets $S,T \subseteq {\bf N}$ that have the same density; for instance, $\{0,2,4,...\}$ and $\{1,3,5,...\}$ both have density 1/2, but the first set might be said to contain "half an element more" than the second, based on the fact that $1-q+q^2-q^3+...$ converges to 1/2 as $q$ goes to 1. Has this (non-Archimedean, non-translation-invariant) refinement of the concept of density, or anything similar to it, been developed anywhere?

Put $|S|_q = \sum_{n \in S} q^n$. Just as $|\{0,2,4,6,...\}|_q-|\{1,3,5,7,...\}|_q \rightarrow 1/2$ as $q \rightarrow 1^-$, it seems empirically that $$|\{0,1,4,9,16,...\}|_q-|\{0,2,6,12,20,...\}|_q \rightarrow 1/2,$$ $$|\{0,1,5,12,22,...\}|_q-|\{0,2,7,15,26,...\}|_q \rightarrow 1/3,$$ and $$|\{0,1,3,6,10,..,\}|_q - \sqrt{2} |\{0,1,4,9,16,...\}|_q \rightarrow \sqrt{2}/2.$$ Is there a place in the literature where results like this can be found, as part of a general setup for measuring sets that "sees" things at a finer level than mere density? To give one last example, racing "evil" versus "odious" numbers, it is easy to prove that $$|\{0,3,5,6,9,10,12,15,...\}|_q-|\{1,2,4,7,8,11,13,14,...\}|_q \rightarrow 0.$$

[ADDED ON MAY 17, 2017: There is now (at least) one article that uses analytic tools to generalize the notion of cardinality of sets of natural numbers: my preprint "One-Dimensional Packing: Maximality Implies Rationality", available at https://arxiv.org/abs/1704.08785 ]

Answer 1

Different notions of summability have indeed been considered before (see in particular Abel summation). For example, in the study of primes a famous observation of Chebyshev is that there are more primes of the form $3\pmod 4$ than $1\pmod 4$. Usually this is interpreted in the sense of the number of primes up to $x$ that are $3\pmod 4$ is more than the number of primes that are $1\pmod 4$; this is true most of the time, but not always. Chebyshev's original formulation was along the lines of your question: he wanted to know if $$ \lim_{q \to 1^-} \Big( \sum_{p \equiv 3\pmod 4} q^{p} - \sum_{p\equiv 1\pmod 4} q^p\Big) = \infty. $$ Hardy and Littlewood and Landau already noted a hundred years back that this is equivalent to the Riemann hypothesis for $L(s,\chi_{-4}) =1/1^s-1/3^s+1/5^s-\ldots$. The same question for natural density is more subtle, involving not just RH but also relations among the zeros. In general the smooth sums in the question would be easier to handle than corresponding problems for the counting functions of the sets. For some references on the particular example given here, see Ford and Konyagin, Rubinstein and Sarnak, or Granville and Martin.

Answer 2

Your post expresses the same idea about comparison of infinite sets as in my previous post with the difference being that you use Abel summation while I did use Ramanujan's summation and Zeta regularization. To be more precise, I used Faulhaber's formula for summation because it gives the same results as Ramanujan's summation (I do not know a proper term for this summation method):

$$\sum _{x\ge0}^\Re f(x)=-\sum _{n=1}^\infty \frac {f^{(n-1)}(0)}{n!} B_n(0)$$

Now, addressing your concerns, expressed in the comments to the other answer, there is a set of similar summation methods that are mostly compatible with each other. One can see that some of them depend only on the values of the series at the integer points while others involve integrals and derivatives. So, it clearly means that there can be constructed examples where these methods give different results.

My strong conviction, even if I do not have a proof, is that all these methods should give the same results for "well-behaving" functions. By "well-behaving" I mean functions that are equal to their Newton series:

$$f(x) = \sum_{k=0}^\infty \binom{x-a}k \Delta^k f\left (a\right)$$

I call such functions "discrete-analytic", "Newton-analytic" or "Newtonian".

The criterion can also be written in the following form:

$$f(x)=\lim_{n\to\infty}\frac{\sum_{k=0}^{n} \frac{(-1)^k f(k)}{(x-k)k!(n-k)!}}{\sum_{k=0}^{n} \frac{(-1)^k }{(x-k) k!(n-k)!}}$$

Some functions would have this series divergent, but their two-directional expansion would converge:

$$f(x)=\lim_{n\to\infty}\frac{\sum _{k=-n}^n \frac{(-1)^k f(k)}{(x-k) (k+n)! (n-k)!}}{\sum _{k=-n}^n \frac{(-1)^k}{(x-k) (k+n)! (n-k)!}}$$

I would include such functions into "Newtonian" as well.

Answer 3

Well, lots of time have passed and now I have an explicit formula for numerosity. It gives the same (up to an infinitesimal) differences between numerosities of sets as your formula, but can express the numerosities in a precise form.

There is a clear way to express numerosities of sequences via $\omega$, which is the germ of the identity function at infinity or half the numerosity of $\mathbb{Z}$. It also can be considered a surreal number, given the canonical embedding of Hardy fields into surreals. This measure is more accurate than asymptotic density, because it gives the exact value.

Suppose you have a strictly increasing sequence $a_k\ge0$, where $k\in\mathbb{Z}, k\ge0$.

To find the numerosity, you have to apply to your sequence the operator $N(a_k)=\left(D\Delta^{-1}a_k\right)^{[-1]}(\omega)$, where $f^{[-1]}$ is the inverse function.

The following Wolfram Language code does the thing:

a[k] := k^2
SolveValues[D[Sum[a[k], k], k] == \[Omega], k] /. C[1] -> 0 //
   FullSimplify // Expand

Inverse code to find a sequence with desired numerosity:

S = Log[\[Omega]];
DifferenceDelta[Integrate[Normal[SolveValues[S == k, \[Omega]]], k], 
     k] /. C[1] -> 0 // Last // FullSimplify // Expand

Basically, we represent a surreal number as a germ at infinity, the germ as a divergent integral, then divide the integral into segments of area $1$, and the centers of mass of these segments are the set of the desired numerosity.

Let us apply the formula to the examples from your question and see what comes.

• $N(\{0,2,4,6,...\})-N(\{1,3,5,7,...\})=N(2k)-N(2k+1)=\frac{\omega }{2}+\frac{1}{2}-\frac{\omega }{2}=1/2$

This coincides with your result.

• $N(\{0,1,4,9,16,...\})-N(\{0,2,6,12,20,...\})=N(k^2)-N(k(k+1))=\sqrt{\omega +\frac1{12}}+\frac{1}{2}-\sqrt{\omega +\frac{1}{3}}\simeq \frac12$

This is infinitesimally close to your result.

• $N(\{0,1,5,12,22,...\})-N(\{0,2,7,15,26,...\})=N({\frac {3k^{2}-k}{2}})-N(\frac {3 k^2 + k } 2)=\frac{1}{3} \sqrt{6 \omega +1}+\frac{2}{3}-\left(\frac{1}{3} \sqrt{6 \omega +1}+\frac{1}{3}\right)=\frac13$

This coincides with your result.

• $N(\{0,1,3,6,10,..,\}) - \sqrt{2} N(\{0,1,4,9,16,...\})=N(\frac12k(k+1))-\sqrt{2}N(k^2)=\sqrt{2 \omega +\frac{1}{3}}-\left(\sqrt{2 \omega +\frac{1}{6}}+\frac{1}{\sqrt{2}}\right)\simeq -\frac1{\sqrt{2}}$

This is infinitesimally close to your result but has opposite sign. I think, you made a mistake.

Perhaps before calling a surreal number a numerosity we should exclude the ifinitesimal part from it, which in surreal numbers can be done via decomposing the number into Conway normal form. If we do so, our results will completely coincide.

Some other examples:

• $N(1/3+k+k^2)=\sqrt{\omega}$

• $N(k^4)=\frac{1}{30} \sqrt{30 \sqrt{900 \omega +30}+225}+\frac{1}{2}$

• $N(7^k)=\log_7 \left(\frac{6 \omega }{\ln (7)}\right)$

Update. I made a preprint, which includes these formulas for numerosity: https://arxiv.org/abs/2411.00296