Consider infinite matrices of the form

$$\left( \begin{array}{ccccc} a_0 & a_1 & a_2 & a_3 & . \\ 0 & a_0 & a_1 & a_2 & . \\ 0 & 0 & a_0 & a_1 & . \\ 0 & 0 & 0 & a_0 & . \\ . & . & . & . & . \\ \end{array} \right)$$

The elements on each diagonal coincide.

My questions are:

Do they form a commutative ring?

Can they be extended to form a field?

Now, let define an operation $\operatorname{reg} A=\sum_{k=0}^\infty B_k a_k,$

where $B_k$ are Bernoulli numbers.

What are the properties of this operation?

Let's define another operation $\det' A=\exp(\Re \operatorname{reg} \log A)$.

What are the properties of this operation?

**Motivation part.**

This is meant to be a matrix representation of divergent integrals and series. For instance,

$\sum_{k=1}^\infty 1= \left( \begin{array}{ccccc} 0 & 1 & 0 & 0 & . \\ 0 & 0 & 1 & 0 & . \\ 0 & 0 & 0 & 1 & . \\ 0 & 0 & 0 & 0 & . \\ . & . & . & . & . \\ \end{array} \right)$

$\sum_{k=0}^\infty 1= \left( \begin{array}{ccccc} 1 & 1 & 0 & 0 & . \\ 0 & 1 & 1 & 0 & . \\ 0 & 0 & 1 & 1 & . \\ 0 & 0 & 0 & 1 & . \\ . & . & . & . & . \\ \end{array} \right)$

$\sum_{k=0}^\infty k= \left( \begin{array}{ccccc} 1/12 & 1/2 & 1/2 & 0 & . \\ 0 & 1/12 & 1/2 & 1/2 & . \\ 0 & 0 & 1/12 & 1/2 & . \\ 0 & 0 & 0 & 1/12 & . \\ . & . & . & . & . \\ \end{array} \right)$

$\int_0^\infty x dx=\int_0^\infty \frac 2{x^3}=\left( \begin{array}{ccccc} 1/6 & 1/2 & 1/2 & 0 & . \\ 0 & 1/6 & 1/2 & 1/2 & . \\ 0 & 0 & 1/6 & 1/2 & . \\ 0 & 0 & 0 & 1/6 & . \\ . & . & . & . & . \\ \end{array} \right)$

There are also some expressions that include divergent integrals that can be represented this way:

$(-1)^{\int_0^\infty dx}=\left( \begin{array}{ccccccc} i & -\pi & -\frac{i \pi ^2}{2} & \frac{\pi ^3}{6} & \frac{i \pi ^4}{24} & -\frac{\pi ^5}{120} & . \\ 0 & i & -\pi & -\frac{i \pi ^2}{2} & \frac{\pi ^3}{6} & \frac{i \pi ^4}{24} & . \\ 0 & 0 & i & -\pi & -\frac{i \pi ^2}{2} & \frac{\pi ^3}{6} & . \\ 0 & 0 & 0 & i & -\pi & -\frac{i \pi ^2}{2} & . \\ 0 & 0 & 0 & 0 & i & -\pi & . \\ 0 & 0 & 0 & 0 & 0 & i & . \\ . & . & . & . & . & . & . \\ \end{array} \right)$

The $\operatorname{reg}$ operation gives the regularized value of the integral or series.

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