Bridging fractional iteration and fractional calculus

[JmsNxn]

I thought I'd add here a relationship between "iterated matrices" and "iterated derivatives". This is a secondary thought to most of my work on fractional calculus; but it's masterful as a bridge between these operations. I am going to refer to this as Ramanujan; and "little circle method" mathematicians would look at it.

Let's let the Matrix \(A\) be a non-singular matrix; so that \(A^{-1}\) exists. To be simple; let's assume that \(A : \mathbb{C}^n \to \mathbb{C}^n\). And let's write that:

\[
e^{Ax} = \sum_{n=0}^\infty A^n \frac{x^n}{n!}\\
\]

We can safely assume that \(e^{Ax} : \mathbb{C}^n \to \mathbb{C}^n\). Where \(x\) produces a semi-group structure. From here, we can write:

\[
\frac{d^s}{dx^s} e^{Ax} = A^s e^{Ax}\\
\]

Where this is a linear operator applying \(\mathbb{C}^n \to \mathbb{C}^n\). We can set \(x=0\); and I'm just rewriting Ramanujan's master theorem as he wrote it:

\[
\Gamma(s) A^{-s} = \int_0^\infty e^{-Ax}x^{s-1}\,dx\\
\]

And we've fractionally iterated the matrix \(A: \mathbb{C}^n \to \mathbb{C}^n\). I avoided a lot of "singular moments" here. But if we can map the matrix \(A\) well enough; this discussion is entirely rigorous. It relates Daniel's work; Sheldon's work; Bo's work; Tommy's work; and all that matrix shit in a quantum physics./hilbert space shit.

Fractional calculus is just:

\[
\frac{d^s}{dA^s} : \mathbb{C}_{\Re(s) > 0} \times \mathcal{H} \to \mathcal{H}\\
\]

Where

\[
\mathcal{H} = \{ A : \mathbb{C}^n \to \mathbb{C}^n\,|\, A^{-1} :\mathbb{C}^n \to \mathbb{C}^n\}\\
\]

Then:

\[
\frac{d^s}{dA^s} e^{Ax} = x^s e^{Ax}
\]

And we can differentiate across \(A\) or \(x\); and still have the same rules.

This is the language I think in; and it's just a translation of much of the standard "tetration forum" language; and current literature.

In theory yeah.

But in practice ...

Differentiating a noninteger amount of times with respect to a nontrivial infinite square matrix that might not be diagonalizable ?!
That gives a non-unique infinite tensor with divergent norm ?! 

Or did you mean differentiating with respect to a vector ?

And that is just the last line of your answer.


regards

tommy1729

Oh yes! Tommy! I apologize; by "A" I meant non-singular; which implies diagonalizable. I can write the math for you if you'd like. But every FINITE diagonalizable matrix can be differentiated as:

\[
A^s e^{Ax} = \frac{d^s}{dx^s} e^{Ax}\\
\]

But we must choose a path of the differintegral so that it converges. For example; let:

\[
A x_j = \lambda_j x_j\\
\]

Where \(x_j\) is an eigenvector, and \(\lambda_j\) is an eigenvalue. And find a path \(\gamma\) such that \(\gamma(0) = 0\) and \(\gamma(\infty) = \infty\). Then assuming that:

\[
\int_\gamma \left|e^{-\lambda_j x}\right|\,dx < \infty\,\,\text{for all}\,\, 1 \le j \le n\\
\]

Then the differintegral always converges. Finding \(\gamma\) can be tricky. But if for the sake of the argument we assume that \(-\pi/2 < -\kappa< \arg(\lambda_j) < \kappa < \pi/2\); then choosing \(\gamma = [0,\infty]\) works fine. It gets much much trickier in a more general sense; but the idea still holds--at least in a general sense.

The differentiation by \(A\); is a formal operation. But it is entirely rigorous, though not common notation. If I take a function:

\[
f(A) = \sum_{k=0}^\infty f_k A^k\\
\]

Then:

\[
\frac{d}{dA} f(A) = \sum_{k=1}^\infty kf_k A^{k-1}\\
\]

Since \(A\) is an n'th order matrix; by the Cayley-Hamilton theorem; this is still a finite polynomial; and we're just differentiating a polynomial in \(A\).

The idea is that \(\frac{d^s}{dA^s}\) exists in the dual space; where as \(\frac{d^s}{dx^s}\) exists in the normal space. I could never be bothered to work out too many of the details; but much of it holds weight in numerical calculations.

You'll probably see this more often in functional analysis but the operation:

\[
x^{-s} e^{Ax} = \frac{1}{\Gamma(s)} \int_0^\infty e^{-Ax}A^{s-1}\,dA\\
\]

Is a perfectly valid operation.

Also, I apologize; I was mostly just spitballing a lot. So I may have screwed up some details. But the idea looks a lot like this--but still, roughly looks like this.