TITLE: Is the derivative the natural logarithm of the left-shift?
QUESTION [107 upvotes]: (Disclaimer: I'm a high school student, and my knowledge of mathematics extends only to some elementary high school calculus. I don't know if what I'm about to do is valid mathematics.)
I noticed something really neat the other day.
Suppose we define $L$ as a "left-shift operator" that takes a function $f(x)$ and returns $f(x+1)$. Clearly, $(LLL\ldots LLLf)(x)=f(x+(\text{number of $L$s}))$, so it would seem a natural extension to denote $(L^hf)(x)=f(x+h)$.
Now, by the definition of the Taylor series, $f(x+h)=\sum\limits_{k=0}^\infty \frac{1}{k!}\frac{d^kf}{dx^k}\bigg|_{x}h^k$. Let's rewrite this as $\sum\limits_{k=0}^\infty \left(\frac{\left(h\frac{d}{dx}\right)^k}{k!}f\right)(x)$. Now, we can make an interesting observation: $\sum\limits_{k=0}^\infty \frac{\left(h\frac{d}{dx}\right)^k}{k!}$ is simply the Taylor series for $e^u$ with $u=h\frac{d}{dx}$. Let's rewrite the previous sum as $\left(e^{h\frac{d}{dx}}f\right)(x)$. This would seem to imply that $(L^hf)(x)=\left(e^{h\frac{d}{dx}}f\right)(x)$, or equivalently, $L=e^\frac{d}{dx}$. We might even say that $\frac{d}{dx}=\ln L$.
My question is, does what I just did have any mathematical meaning? Is it valid? I mean, I've done a bit of creative number-shuffling, but how does one make sense of exponentiating or taking the logarithm of an operator? What, if any, significance does a statement like $\frac{d}{dx}=\ln L$ have?

REPLY [9 votes]: how does one make sense of exponentiating or taking the logarithm of an operator?

The operator is linear, and therefore so are its positive integer powers, hence any power series in that operator has a chance of making sense.  At least the series is a limit of linear operators, and the series makes perfect sense without any limiting process when applied to polynomials, because almost all terms are zero.  
In the 17-18th century, functions were believed to have power series representations, so that formal manipulations that were provably correct for polynomials ought to somehow justify the same calculation in general.  This is not quite right but it works spectacularly as a heuristic.    

What, if any, significance does a statement like d/dx=ln L have?

It adds precision to the analogy between calculus and the "calculus of finite differences".  As you might guess, what is true for differentiation extends to integration. The corresponding relationship between finite sums and integrals is the Euler-Maclaurin summation formula and it is derived by taking the reciprocal of a power series equivalent to the one you wrote down.