What is $(\frac{d}{dx})^\frac{d}{dx}$?

Question

How I got interested:-

Yesterday I watched this new video from 3blue1brown.

And I realized that we can use the formula -

$\textstyle\displaystyle{e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}}$

to almost anything.

We can define matrix powers, matrix to the power of a matrix, exponential derivative etc.

So I started wondering about what $(\frac{d}{dx})^\frac{d}{dx}$ would mean.

My approach:-

Obviously in order to use the previously mentioned formula, we have to write $(\frac{d}{dx})^\frac{d}{dx}$ as a power of $e$. And we can do that very easily.

$\textstyle\displaystyle{\left(\frac{d}{dx}\right)^\frac{d}{dx}=e^{\frac{d}{dx}\ln(\frac{d}{dx})}}$

Now if we use the formula then we get

$\textstyle\displaystyle{\left(\frac{d}{dx}\right)^\frac{d}{dx}}$

$\textstyle\displaystyle{=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{d}{dx}\ln\left(\frac{d}{dx}\right)\right)^n}$

Well obviously $\frac{d^n}{dx^n}$ makes sense. But what about $\ln^n(\frac{d}{dx})$?

To make sense of it, I used the tailor series of $\ln(x)$.

$\textstyle\displaystyle{\ln(x)=\sum_{k=0}^{\infty}\frac{(-1)^{k-1}(x-1)^k}{k}}$

$\textstyle\displaystyle{\implies\ln^n\left(\frac{d}{dx}\right)=\left(\sum_{k=0}^{\infty}\frac{(-1)^{k-1}(\frac{d}{dx}-1)^k}{k}\right)^n}$

So,

$\textstyle\displaystyle{\left(\frac{d}{dx}\right)^\frac{d}{dx}}$ $\textstyle\displaystyle{=\sum_{n=0}^{\infty}\frac{1}{n!}\frac{d^n}{dx^n}\left(\sum_{k=0}^{\infty}\frac{(-1)^{k-1}\left(\frac{d}{dx}-1\right)^k}{k}\right)^n}$

My confusion:-

When I used the series expansion for $e^x$, it was very easy to understand because the radius of convergence in $\infty$.

But while using the series expansion for $ln(x)$ it doesn't make much of a sense to me, because the taylor series converges if $0\lt x\leq 2$.

But what does $0\lt\frac{d}{dx}\leq 2$ mean?

I mean differential operator is not a number which we can compare with other numbers.

So I am stuck here.

Questions:-

(1) Is my approach correct?

(2) How to make sense of $\ln(\frac{d}{dx})$? Or can we just ignore the convergence condition of the taylor series of $\ln(x)$?

(3) Are there other easier approaches for defining $(\frac{d}{dx})^\frac{d}{dx}$? And what they are?

Answer 1

Here's a interpretation of the symbol $(d/dx)^{d/dx}$ that is independent of summation. The steps I'll take are entirely formal. From the Euler representation of the gamma function, $$ a^s = \frac{1}{\Gamma{(-s)}} \int_0^\infty e^{-a u}u^{-s}\frac{du}{u} $$ let $a \to d/dx$ and $s \to d/dx.$ Let this operator act on a function $f(x).$ (We say nothing about the properties of $f,$ or even if such an $f$ can exist.) Then

$$(d/dx)^{d/dx} f(x) = \frac{1}{\Gamma{(-d/dx)} } \int_0^\infty f(x- (u + \log{u}) \ ) \frac{du}{u} $$ where we have used the Taylor series in operator form,

$$ (1) \quad \exp{(a \ d/dx)} f(x) = f(x + a).$$

The reciprocal gamma function containing the $d/dx$ is closely related to the inverse Laplace transform. Let's use the Hankle contour representation of the reciprocal gamma,

$$ (2) \quad \frac{1}{\Gamma(z)} = \frac{1}{2\pi i} \int_c e^s s^{-z} ds .$$ Then we can say

$$(d/dx)^{d/dx} f(x) = \int_c \frac{ds}{2 \pi i} \ e^s s^{d/dx} \int_0^\infty f(x- (u + \log{u}) \ ) \frac{du}{u} $$ $$= \int_c \frac{ds}{2 \pi i} \ e^s \int_0^\infty f(x- (u + \log{u}) + \log s \ ) \frac{du}{u}$$ where (1) has been used again.

For fun, try $f(x) = e^{a \ x} - e^{ b \ x}.$ Then the above prescription gives us

$$(d/dx)^{d/dx} f(x) = \int_c \frac{ds}{2 \pi i} \ e^s \Big( a^a \ e^{a \ x} \Gamma(-a) s^a - b^b \ e^{b \ x} \Gamma(-b) s^b ) \Big) =$$ $$ = a^a \ e^{a \ x} - b^b \ e^{b \ x} $$ where the innermost integral was done by Mathematica (first line of previous expression), and outer one by using (2) (final expression).

All this is forma manipulation. For more rigor, you'd have to define a class of functions for validity, define the Hankle contour, etc. Even the first step is problematic: should the reciprocal gamma come before the integral, or after it?

Answer 2

Before answering your question, I want to point out an important note, which is rarely mentioned by the YouTubers revealing the secrets of $i^i$.

When $x$ isn't a real number, a function $\exp(x)$, which is often denoted as $e^x$ is not a number $e$ times itself $x$ times which is a literal definition of power. It makes sense to write it as $e^x$, because it often has one or both of the properties: $\exp(x+y) = \exp(x)\exp(y)$ and $\frac{d}{dx}\exp x = \exp x$. You can define $x^y=\exp(y\log x)$. But this definition is problematic in a broad sense (discontinuity, multi-branching and so on).

So in reality, when $x$ is not a real number, you naturally meet either $x^p$, where $p$ is either integer or real, or $\exp(x)$. Writing $x^y$ is unjustified abuse of notation.

Having said that, if $p,q$ are element of some algebra, you can define: $$ \exp(p) =1+p+\frac{p^2}{2!}+\frac{p^3}{3!}+\ldots,\\ \log(1+p) =1-p+\frac{p^2}2-\frac{p^3}3+\ldots,\\ p^q = \exp(q\log(p)). $$ Or use any other definition of log you are offered in the comments. At it would be possible to express $p^p$ as a polynomial in $p$. Will it converge for $p=d/dx$ when applied to some function $f(x)$? Who knows? even $e^{d/dx}$ doesn't have this property (consider $f(x)=e^{e^{x^2}}$).

In general, when one gets a question “how can you define $\nabla^\nabla$” ($0/0$, $0^0$, $1+2+3+4\dots$, you name it), it's always good to ask “in what kind of problem one would need this concept or notation?”. Because the answer usually will depend on it. And without this question, any definition is as good.

Answer 3

To answer question 3: You could try using the power series of $x^x$ at $x=1$ \begin{align} x^x &= 1 + (x-1) + (x-1)^2 + \frac{1}{2}(x-1)^3 + \frac{1}{3}(x-1)^4 + \mathrm{O}((x-1)^5) \\ &= \sum_{n=0}^\infty \frac{a_n}{n!} (x-1)^n \end{align} (where $a_n$ is A005727) and plugging in the derivative operator $x = \mathrm{D}$, noting that \begin{align} \left(\mathrm{D}-1\right)^n = \sum_{k=0}^n \binom{n}{k} (-1)^{n-k} \mathrm{D}^k \end{align}