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?

Answer

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:
Ls(Ltf(x))=Lsf(x+t)=f(x+s+t)=Ls+tf(x).
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
Af(x):=lim
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.

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

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