Is the derivative the natural logarithm of the left-shift?

(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, (LLLLLLf)(x)=f(x+(number of Ls)), so it would seem a natural extension to denote (Lhf)(x)=f(x+h).

Now, by the definition of the Taylor series, f(x+h)=k=01k!dkfdxk|xhk. Let’s rewrite this as k=0((hddx)kk!f)(x). Now, we can make an interesting observation: k=0(hddx)kk! is simply the Taylor series for eu with u=hddx. Let’s rewrite the previous sum as (ehddxf)(x). This would seem to imply that (Lhf)(x)=(ehddxf)(x), or equivalently, L=eddx. We might even say that ddx=lnL.

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 ddx=lnL have?


Well, it seems that you have just discovered a beautiful theory of (semi)group generators by yourself. To give some basics of it, let us consider a collection of “nice” functions on real values – e.g. bounded and having continuous derivatives. The action of operators Lh on this space has a semigroup structure:
Also, you have that L0f(x)=f(x), so L0 is the identity operator -it does not change its argument. You can see, that although there are a lot of operators in the collection (Lh)h, they have to satisfy the semigroup property Ls+t=LsLt, and so there is no much freedom in choosing them. Even more, one can define the generator of the semigroup (also sometimes called the derivative of it) by
which in your case exactly coincides with the derivative of the function. However, if you would consider the semigroup K^hf(x) = f(x + v\cdot h) for some constant v, you’ll see that the generator will be a bit different. Anyways, under certain condition – if you don’t know the semigroup L, but you’re just given the generator \mathscr A, it is possible to reconstruct L from \mathscr A by the so-called exponential map, that is

L^h:=\mathrm e^{h\mathscr A}

where the definition of the exponent of the operator indeed is given by the Taylor series where e.g. \mathscr A^2 f(x) = \mathscr A(\mathscr A f(x)). As a result, you can indeed consider \mathscr A to be a certain logarithm of L^h and it comes as no surprise that it has a similar Taylor expansion. However, although the name “exponential map” from \mathscr A to L^h is commonly used, I haven’t heard of the inverse being called the “logarithm”. One rather uses the “generator” or “derivative.” If you are further interested, I would suggest you reading the linked wikipedia article.

Source : Link , Question Author : David Zhang , Answer Author : Ilya

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