1
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Could you add some motivation for the parts of the question on Bernoulli polynomials? $\endgroup$ Sep 21, 2021 at 23:10
  • $\begingroup$ @MarkWildon yes, just a moment. $\endgroup$
    – Anixx
    Sep 21, 2021 at 23:11
  • $\begingroup$ @MarkWildon done. By the way, a typo: it's Bernoulli numbers. $\endgroup$
    – Anixx
    Sep 21, 2021 at 23:32
  • $\begingroup$ Thanks for the examples, but I still don't understand the motivation. How does one use Bernoulli numbers to express the definite integral as a matrix / formal power series? Could you give an explicit example of the summation? Is it using Euler's summation formula? $\endgroup$ Sep 21, 2021 at 23:49
  • $\begingroup$ @MarkWildon Bernoulli numbers are used for regularization. The integrals can be transformed using these formulas: mathoverflow.net/questions/403704/… If you have a MathML-enabled browser (Firefox-based), here is my wiki: exnumbers.miraheze.org/wiki/… Also, look here: mathoverflow.net/questions/115743/an-algebra-of-integrals/… $\endgroup$
    – Anixx
    Sep 21, 2021 at 23:54

1 Answer 1

5
$\begingroup$

If the matrices have entries from a (unital) ring $R$ then the set of such matrices is isomorphic to $R[[x]]$, the ring of formal power series over $R$. To see this, observe that the map sending the infinite matrix with $a_0 = 0$, $a_1 = 1$ and $a_k = 0$ for $k \ge 2$ to $x$ is a ring isomorphism.

This also answers the second question: if $R$ is an integral domain then set of matrices embeds canonically in the field of fractions of $R[[x]]$ and this is the smallest field containing $R[[x]]$. In particular, if $R$ is a field then this field is $\{ \sum_{k=-m}^\infty a_k x^k : a_k \in R, m \in \mathbb{N}_0 \}$.

I'm uncertain how $\mathrm{reg}$ is (well)-defined, but certainly one can take $R$ to be the polynomial ring $\mathbb{C}[z]$ and then something like $\sum_{k=0}^\infty B_k(z) x^k$ is a well-defined element of $R[[x]] = \mathbb{C}[z][[x]]$. If, as in the correction then one wants Bernoulli numbers rather than the polynomials, just specialize to $\mathbb{C}[[x]]$ by evaluating at $z=0$.

$\endgroup$
1
  • 1
    $\begingroup$ Yes, I came to this from formal power series. The entries are complex numbers. The matrices represent divergent integrals. $\endgroup$
    – Anixx
    Sep 21, 2021 at 23:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.