# Non-standard numbers and exponential form of Zeta function [closed]

Basic idea

For a long time I was looking for a numerical system that would allow to compare infinite sets. In contrast to Cantor's approach that empathizes the possibility of on-to-one correspondence between sets, I want that adding an element to even an infinite set should increase its "quantity" and removing an element should decrease. For instance, 1,2,3,4,5... should have greater quantity than 1,2,4,5... Similarly I want more dense sets having greater quantity, that is, 1,2,3,4,5... having greater quantity than 1,3,5,7... and slmaller than 1,3/2,2,5/2,3,...

This way I came to the following considerations.

First of all, we extend the real numbers with non-standard numbers. Each non-standard number consists of a standard part and non-standard part. The numbers whose standard part is zero we call pure non-standard. For instance, if p is a pure non-standard number, then p+1 has standard part 1.

Now we introduce the notion of quantity of a subset of real numbers q(S).

• If the set of reals is finite, then the quantity of that set is equal to the number of its members.

• The quantity of all integers we designate as $\Omega=2\tau$. It is a pure non-standard number.

• If two sets differ by only the presence or absence of finite number of elements then the non-standard parts of their quantities are equal.

• If two sets differ by only the position of finite number of elements, their quantities are equal.

• For non-intersecting sets $S_1$ and $S_2$, $q(S_1\cup S_2)=q(S_1)+q(S_2)$

• Quantities of sets symmetric against zero are equal.

• Quantities of uniformly distributed sets are proportional to their densities

Given these properties, lets find the quantity of the natural numbers $q(\mathbb{N})$.

We know that $\mathbb{Z}=\{-1,-2,-3,...\}\cup\{0\}\cup\{1,2,3,...\}=\mathbb{N^-}\cup\{0\}\cup\mathbb{N}$. Now $q(\{0\})=1$ (it is a finite set) and $q(\mathbb{N^-})=q(\mathbb{N})$.

So, $\Omega=2q(\mathbb{N})+1$. We designate $q(\mathbb{N})$ as $\omega_-$, so $\omega_-=\frac12 \Omega-\frac12=\tau-1/2$. It is not a pure non-standard number, its standard part is $-1/2$. The quantity of all non-negative integers is greater by one, so we designate it $\omega_+=\omega_-+1=\tau+1/2$

Here are some other examples:

• The quantity of even numbers is equal to the quantity of odd numbers, is equal to $\Omega/2=\tau$

• The quantity of the numbers of the form $\frac{2n-1}2$ with natural $n$ $(1/2, 3/2, 5/2,...)$ is $\frac{\Omega}2=\tau$

• The quantity of positive even numbers is $\tau/2$, the quantity of positive odd numbers is $\tau/2-1/2$, the quantity of non-negative odd numbers is $\tau/2+1/2$.

• The quantity of complex numbers ordered lexicographically is $\Omega^2=4\tau^2$

Definition using series

Another problem that for a long time interested me was the meaning of generalized sums of diverging series. I was looking for a non-Archimedian number system that would give the meaning to those sums. And now it seems that Ramanujan's summation of diverging series finally got its place.

Now we define that to any divergent series there corresponds a non-standard number. The standard part of that number is given by the Ramanujan's sunmmation of the series.

That way we see that $$\operatorname{st} q(\mathbb{N})= \operatorname{st} \omega_-=\sum_{n\ge1}^{\Re}1=-1/2$$

Insight on exponentiation of non-standard numbers

Examining the Faulhaber's formula for Ramanujan's summation one can come to the following striking insight on the exponentiation of non-standard numbers.

$$\operatorname{st}\omega_-^n=B_n$$ $$\operatorname{st}\omega_+^n=B^*_n$$

Where $B_n$ are the first Bernoulli numbers and $B^*_n$ are the second Bernoulli numbers.

Indeed, we can see that $\operatorname{st}\omega_-=-1/2$, $\operatorname{st}\omega_+=1/2$, $\operatorname{st}\omega_-^2=1/6$, $\operatorname{st}\omega_-^3=0$ etc.

Given that Bernoulli numbers can be expressed through Hurwitz Zeta function, we can generalize:

$$\operatorname{st}\omega_-^x=-x\zeta(1-x,0)$$

$$\operatorname{st}\omega_+^x=-x\zeta(1-x,1)=-x\zeta(1-x)$$

This allows to represent zeta function in exponential form:

$$(x-1)\zeta(x)= \operatorname{st}\omega_-^{1-x}$$

or, more generally,

$$\operatorname{st}(\omega_-+z)^n= B_n(z)$$

$$\operatorname{st}(\tau+y)^x=-x\zeta(1-x,1/2+y)$$

Moreover, now any series containing Bernoulli numbers can be represented as power series over non-standard numbers.

Connection between trigonometric functions on hypernumbers

Given the above definitions, we have a lot of relations between trigonometric functions, for instance,

$$\operatorname{st} \cos (z\omega_-)=\frac z2 \cot \left(\frac z2\right)$$

$$\operatorname{st} \cosh (z\omega_-)=\frac z2 \coth \left(\frac z2\right)$$

$$\operatorname{st} e^{z\omega_-}=\frac{z}{e^{z}-1}$$

$$\operatorname{st} e^{z\tau}=\frac{z}{2} \operatorname{csch}\left(\frac{z}{2}\right)$$

$$\operatorname{st} \left(\frac{1}{\pi^2 \tau+\pi x}+\frac{1}{\pi^2 \tau-\pi x}\right)=(\sec x)^2$$

$$\operatorname{st}\ln (\omega_-+z)=\psi(z)$$

Some sets and their quantities:

Now I have the following questions

• How the suggested system of non-standard numbers related to the surreals and other non-standard numerical system? Can it be incorporated into surreals?

• Is there a way to represent the non-standard part of these numbers in a more convinient way, similar to the surreals?

• Why are the downvotes? – Anixx Aug 27 '15 at 8:38
• I'm curious, why is $\Omega$ assumed to have the form $2\tau$ for some "purely nonstandard" $\tau$? – Noah Schweber Aug 30 '15 at 2:02
• Seems like according to your nomenclature one of my tasks is to compute$$\operatorname{st}\left(e^{\frac{\omega_-^2}z}\left(1+\operatorname{Erf}\left(\frac{\omega_-}{{\sqrt z}}\right)\right)\right)$$:D – მამუკა ჯიბლაძე Aug 30 '15 at 7:20
• Actually your representation of $\zeta$ is somehow reminiscent of $\zeta_{\mathscr O}(s)=\operatorname{tr}({\mathscr O}^{-s})$ – მამუკა ჯიბლაძე Aug 30 '15 at 8:07
• I don't know much of operator theory but seems like it has some ingredients to deal with something like your version of working with infinities and standard part extraction via relevant "operator infinities" (unbounded - Schatten - trace class - compact...). Such approach to infinities is, I believe, also extensively used by Connes in his noncommutative measure theory/topology/differential geometry... – მამუკა ჯიბლაძე Aug 30 '15 at 8:20

3. Yes. You can use functions of the cutoff and big-O notation. For example, $\sum_{n=1}^\infty 1$ can be smoothed and cut off at some large $N$ to yield the asymptotic $N - 1/2 + O(1/N)$. Then your non-standard part is the increasing part of the function of $N$, and the standard part is the constant part.