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## Basic ideaEdit

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

- The quantity of complex integers ordered lexicographically is $\Omega^2=4\tau^2$

Some sets and their quantities:

## Definition using series Edit

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

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

The quantity of a subset of natural numbers is equivalent to the series summing up the membership function of that subset. For instance, given a set $S \subseteq \mathbb{N}$ and a function $p(n)$ such that $p(n)=1$ if $n\in S$ and $p(n)=0$ otherwise,

$q(S)=\sum_{k=1}^\infty p(n)$

## Exponentiation of non-standard numbers Edit

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

$\operatorname{reg}\omega_-^n=B^{-}_n$
$\operatorname{reg}\omega_+^{n}=B^{+}_n$

Where $B_n$ are the first Bernoulli numbers and $B^{'}_n$ are the second Bernoulli numbers.

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

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

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

This allows to represent zeta function in exponential form:

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

or, more generally,

$\operatorname{reg}(\omega_-+z)^n= B_n(z)$
$\operatorname{reg}(\tau+y)^x=-x\zeta(1-x,1/2+y)$

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

There are the following relations:

$\operatorname{reg}\tau^x = \operatorname{reg}\omega_+^x (2^{1 - x} - 1)$
$\operatorname{reg}(2\tau)^x = \operatorname{reg}\omega_+^x (2 - 2^x)$
$\operatorname{reg}\frac{(2\tau)^s}{\omega_+}=-2(s-1)\eta(2-s)$

where $\eta(s)$ is Dirichlet Eta function.

$\operatorname{reg}\frac{\tau^s}{\omega_+^{2s}}=-s\lambda(s+1)$

where $\lambda(s)$ is Dirichlet Lambda function.

For $x>1$,

$\operatorname{reg}\omega_-^x = \operatorname{reg}\omega_+^x$

From the Riemann functional equation it follows:

$\operatorname{reg}\omega_+^{-x}=\operatorname{reg}\frac{\omega_+^{x+1} 2^x\pi^{x+1}}{\sin(\pi x/2)\Gamma(x)(x+1)}$

## Expression for derivative using regular partEdit

If $f(x)$ is analytic, the following holds:

$f'(x)=\operatorname{reg}(f(\omega_++x)-f(\omega_-+x))=\operatorname{reg} \Delta f(\omega_-+x)$
$f'(x)=\operatorname{reg}(f(\omega_++x)-f(-\omega_++x))$
$f'(x)=\operatorname{reg}(f(-\omega_-+x)-f(\omega_-+x))$

If $f(x)$ is odd,

$\operatorname{reg}f(\omega_+)=-\operatorname{reg}f(\omega_-)= \frac12 f'(0)$

Another consequence of Faulhaber's formula connects integral with the sum:

$\operatorname{reg}\sum_{k=0}^\infty f(k)=-\operatorname{reg}\int_0^{\omega_-} f(x) dx$

## Expression for integralEdit

If $\lim_{x\to -\infty} f(x)=\lim_{x\to +\infty}f(x)$,

$2\tau\int_{-\infty}^\infty f(x)dx=\int_{-\infty}^\infty \sum_{k=-\infty}^{+\infty} f(x+k)dx$

## Equivalence of some improper integrals and seriesEdit

Integrals can be transformed into sets of weighted dots using the following principle:

an integral over an interval $(a_i,b_i)$ can be replaced with a weighted dot at the center of mass of the figure under the graphic of the integrated function, with weight of the figure's area, if the figure is symmetric against its center of mass.

### ConsequencesEdit

If $f(x)$ is periodic and integral over period is zero, then

$\int_{-\infty}^{+\infty} f(x) dx=0$

If such function is also even, then

$\int_0^{+\infty} f(x) dx=0$

## Standard part of improper integralsEdit

Using the definition of integral as a limit of the sum, we can derive the formula for standard part of an improper integral.

$\operatorname{reg}\int_0^\infty f(x)\,dx=\lim_{s\to0} \operatorname{reg} s\sum_{k=1}^\infty f(sk)$

This can be implemented in Mathematica using the following code:

Sum[f[s x],{x,1,Infinity},Regularization->"Borel"]//FullSimplify 

Limit[s %,s→0]

Particularly,

$\operatorname{reg}\int_0^\infty e^x dx=-1$

## Norm-like measureEdit

One can define the norm-like measure of non-standard numbers, by analogy with complex numbers:

$\|w\|=\exp(\operatorname{reg}\ln w)$

If so, the following holds:

$\|\omega_+\|=e^{-\gamma}$

## Distribution formEdit

Integral form of Dirac Delta function is

$\delta\left(z\right)=\frac1{2\pi}\int_{-\infty}^{+\infty}e^{-ixz}dx$

It becomes $\tau/\pi$ at $x=0$.

Thus we can interpret extended numbers as derivatives of step-functions at the point of discontinuity, with step size determining the non-standard part and limits of derivatives from right and left determine the standard part.

Particularly,

$\delta(0)=\tau/\pi$

$\operatorname{sign}'(0)=2\tau/\pi$

## Standard parts of some expressions Edit

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

$\operatorname{reg} \cos (z\omega_-)=\operatorname{reg} \cos (z\omega_+)=\frac z2 \cot \left(\frac z2\right)$
$\operatorname{reg} \cosh (z\omega_-)=\operatorname{reg} \cosh (z\omega_+)=\frac z2 \coth \left(\frac z2\right)$
$\operatorname{reg} \cos (z\tau)=\frac z2 \csc \left(\frac z2\right)$
$\operatorname{reg} \cosh (z\tau)=\frac{z}{2} \operatorname{csch}\left(\frac{z}{2}\right)$
$\operatorname{reg} e^{z\omega_-}=\frac{z}{e^{z}-1}$
$\operatorname{reg} \left(\frac{1}{\pi^2 \tau+\pi x}+\frac{1}{\pi^2 \tau-\pi x}\right)=(\sec x)^2$
$\operatorname{reg}\ln (\omega_-+z)=\psi(z)$

Particularly,

$\operatorname{reg}\ln \omega_+=-\gamma$
$\operatorname{reg}\frac1{\pi }\ln \left(\frac{\omega _-+\frac{z}{\pi }}{\omega _+-\frac{z}{\pi }}\right)=-\cot z$
$\operatorname{reg} \frac1\pi\ln \left(\frac{\tau +\frac{z}{\pi }}{\tau -\frac{z}{\pi }}\right)=\tan z$
$\operatorname{reg}\sin (\omega_-+x) = \frac{1}{2} \cot \left(\frac{1}{2}\right) \sin x -\frac{1}{2} \cos x$

Particularly,

$\operatorname{reg}\sin \omega_-=-1/2$,
$\operatorname{reg}\sin \omega_+=1/2$,
$\operatorname{reg}\sin \tau=0$
$\operatorname{reg}\cos (\omega_-+x) = \frac{1}{2} \csc \left(\frac{1}{2}\right) \cos \left(\frac{1}{2}- x \right)$

$\operatorname{reg}\sin (a\omega_-+x) = \frac{a}{2} \cot \left(\frac{a}{2}\right) \sin x -\frac{a}{2} \cos x$
$\operatorname{reg}\cos (a\omega_-+x) = \frac{a}{2} \csc \left(\frac{a}{2}\right) \cos \left(\frac{a}{2}- x\right)$
$\operatorname{reg}\cos (\pi\tau+x)=-\frac\pi{2}\cos x$
$\operatorname{reg}\sin (\pi\tau+x)=-\frac\pi{2}\sin x$

## Relations between standard parts of trigonometric and inverse trigonometric functionsEdit

$\operatorname{reg}\left(\cosh \left(2 x \omega _\pm\right)-1\right)=\operatorname{reg}\frac{x}{\pi} \operatorname{artanh}\left(\frac{x}{\pi\omega _\pm}\right)=\operatorname{reg}\frac{x}{\pi} \operatorname{arcoth}\left(\frac{\pi \omega _\pm}{x}\right)=x \coth (x)-1$
$\frac{z}{2\pi }\ln \left(\frac{\omega _+-\frac{z}{2\pi }}{\omega _-+\frac{z}{2\pi }}\right)=\operatorname{reg} \cos (z\omega_-)=\operatorname{reg} \cos (z\omega_+)=\frac z2 \cot \left(\frac z2\right)$

For regular z the following holds:

$\tan z=\frac2\pi\operatorname{artanh} \frac{z}{\tau\pi}$

## Improper integrals of monomialsEdit

Following from Faulhaber's formula (for $n\ge0$),

$\int_0^\infty x^n dx=\frac{\left(\tau +\frac{1}{2}\right)^{n+2}-\left(\tau -\frac{1}{2}\right)^{n+2}}{(n+1)(n+2)}=\frac{\omega _+^{n+2}-\omega _-^{n+2}}{(n+1)(n+2)}$

Interpreting Fourier transform formally, for even $n$ we also have

$\int_0^\infty x^n dx=i^n\pi\delta^{(n)}(0)$

For $n>1$

$\int_0^\infty \frac1{x^n} dx=\frac1{(n-1)!}\int_0^\infty x^{n-2} dx=\frac{\omega _+^{n}-\omega _-^{n}}{(n-1)n!}$

Particularly,

$\int_0^\infty 1 dx =\tau$
$\int_0^\infty x dx=\frac{\tau^2}2+\frac1{24}$
$\int_0^\infty x^2 dx=\frac{\tau^3}3+\frac{\tau}{12}$

Using generalized limits (see below) one can write down the general formula for powers of $\omega_\pm$, for $n>0$:

$\omega_-^n=\operatorname{gen}\lim_{x\to\infty}B_n(x)$
$\omega_+^n=\operatorname{gen}\lim_{x\to\infty}B_n(x+1)$

Or in integral form:

$\omega_-^n=B_n+n\int_0^\infty B_{n-1}(x)dx$
$\omega_+^n=B^*_n+n\int_0^\infty B_{n-1}(x+1)dx$

They give us the general formulas:

$(\omega_-+a)^n=\operatorname{gen}\lim_{x\to\infty}B_n(x+a)$
$(\omega_-+a)^n=B_n(a)+n\int_0^\infty B_{n-1}(x+a)dx$

This in turn gives the formula for the exponential function:

$e^{a+\omega _-}=\int_{-\infty}^{\infty } \frac{e^{a+t}}{e-1} \, dt+\frac{e^a}{e-1}$

## Generalized limitsEdit

We introduce generalized limits in the following way:

$\operatorname{gen}\lim_{x\to u^+}f(x)=f(a)-\int_u^a f'(x)dx$

where $a>u$

and

$\operatorname{gen}\lim_{x\to u^-}f(x)=f(a)+\int_a^u f'(x)dx$

where $a<u$

This value can serve as a measure of the growth rate of a function.

Particularly, for $n\ge0$

$\operatorname{gen}\lim_{x\to\infty}x^n=\frac{\omega _+^{n+1}-\omega _-^{n+1}}{n+1}$

For odd $n$,

$\operatorname{gen}\lim_{x\to\infty}x^n=i^{n-1}\pi n\delta^{(n-1)}(0)$

And

$\operatorname{gen}\lim_{x\to 0^+} \frac1{x^n}=\frac{\omega _+^{n+1}-\omega _-^{n+1}}{(n+1)!}$

Following Urs Graf, p.36, for odd $n$,

$\operatorname{gen}\lim_{x\to 0^+} \frac1{x^n}=\frac{i^{n-1}\pi\delta^{(n-1)}(0)}{(n-1)!}$

The Cauchy principal value of an analytic function at a pole corresponds to the regular part of its generalized limits at the pole. Thus,

$\operatorname{gen}\lim_{x\to0^\pm}\Gamma(x)=-\gamma\pm\tau$

$\operatorname{gen}\lim_{x\to{-1}^\pm}\Gamma(x)=\gamma-1\mp\tau$

$\operatorname{gen}\lim_{x\to{-2}^\pm}\Gamma(x)=\frac{3}{4}-\frac{\gamma }{2}\pm\frac\tau 2$

$\operatorname{gen}\lim_{x\to{-3}^\pm}\Gamma(x)=\frac{\gamma }{6}-\frac{11}{36}\mp\frac\tau 6$

## Other identitiesEdit

$\int_{-\infty}^{+\infty}\cos(1/x)dx=2\tau-\pi$
$\gamma(x)=\operatorname{reg}\left( \omega _- \ln \left(\frac{x+\omega _-}{x+\omega _+}\right)\right)$

(generalized Euler's constant function)

## Some extended numbersEdit

$\begin{array}{cccccc} \text{Delta form} & \text{In terms of } \tau, \omega_+,\omega_- & \text{Regular part} & \text{Integral or series form} & \text{Generalized limit form} & \text{Numerocity form} \\ \pi \delta (0) & \tau & 0 & \int_0^{\infty } \, dx;\int_0^{\infty } \frac{1}{x^2} \, dx & \operatorname{gen}\lim_{x\to\infty}x;\operatorname{gen}\lim_{x\to0^+}\frac1x & q(2 \mathbb{Z}) \\ \pi \delta (0)-\frac{1}{2} & \omega _-;\tau-\frac{1}{2} & -\frac{1}{2} & \sum _{k=1}^{\infty } 1 & \operatorname{gen}\lim_{x\to\infty} (x-1/2) & q(\mathbb{N}) \\ \pi \delta (0)+\frac{1}{2} & \omega _+;\tau+\frac{1}{2} & \frac{1}{2} & \sum _{k=0}^{\infty } 1 & \operatorname{gen}\lim_{x\to\infty} (x+1/2) & q\left(\mathbb{Z}^*\right) \\ 2 \pi \delta (i) & e^{\omega_+}-e^{\omega_-}-1 & 0 & \int_{-\infty }^{\infty } e^x \, dx & \operatorname{gen}\lim_{x\to\infty} e^x & \\ & \frac{\tau ^2}{2}+\frac{1}{24};\frac{\omega_+^3-\omega_-^3}6 & 0 & \int_0^{\infty} x \, dx;\int_0^\infty \frac2{x^3}dx & \operatorname{gen}\lim_{x\to\infty}\frac{x^2}2;\operatorname{gen}\lim_{x\to{0^+}} \frac1{x^2} & \\ & \frac{\tau ^2}{2}-\frac{1}{24} & -\frac1{12} & \sum _{k=0}^{\infty } k & \operatorname{gen}\lim_{x\to\infty} \left(\frac{x^2}2-\frac1{12}\right) & \\ -\pi \delta''(0) &\frac {\tau^3}3 +\frac\tau{12};\frac{\omega_+^4-\omega_-^4}{12}& 0 & \int_0^\infty x^2dx;\int_0^\infty\frac6{x^4}dx&\operatorname{gen}\lim_{x\to\infty}\frac{x^3}3;\operatorname{gen}\lim_{x\to{0^+}} \frac2{x^3}& &\\ \pi^2\delta(0)^2-\pi\delta(0)+1/4&\omega_-^2&\frac16&2 \int_0^{\infty } \left(x-\frac{1}{2}\right) \, dx+\frac{1}{6}&\operatorname{gen}\lim_{x\to\infty}B_2(x)\\ \pi^2\delta(0)^2+\pi\delta(0)+1/4&\omega_+^2&\frac16&2 \int_0^{\infty } \left(x+\frac{1}{2}\right) \, dx+\frac{1}{6}&\operatorname{gen}\lim_{x\to\infty}B_2(x+1)\\ \pi^2\delta(0)^2&\tau^2&-\frac1{12}&2 \int_0^{\infty } x \, dx-\frac{1}{12}&\operatorname{gen}\lim_{x\to\infty}B_2(x+1/2)\\ ?&?&0&\int_1^\infty \frac{dx}x;\sum_{k=1}^\infty \frac1x -\gamma&\operatorname{gen}\lim_{x\to\infty}\ln x\\ -\pi\delta''(0)-\frac14 \pi\delta(0);\pi^3\delta(0)^3&\tau^3&0&\int_0^\infty \left(3x^2-\frac1{4}\right)dx&\operatorname{gen}\lim_{x\to\infty}B_3(x+1/2)\\ \frac{2\pi\delta(i)+1}{e-1}&e^{\omega_-}&\frac1{e-1}&\frac1{e-1}+\frac1{e-1}\int_{-\infty}^\infty e^x dx&&\\ \frac{2\pi\delta(i)+1}{1-e^{-1}}&e^{\omega_+}&\frac1{1-e^{-1}}&\frac1{1-e^{-1}}+\frac1{1-e^{-1}}\int_{-\infty}^\infty e^x dx&&\\ \end{array}$

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