Find inverse operator

Question

Let $D=\dfrac{d}{dx}.$ Consider the operator $$ D_{h,x}=\frac{e^{hD}-1}{h}. $$

Question. What is explicit form of the operator $D^{-1}_{h,x}?$

I think that $$ D^{-1}_{h,x}=\frac{h}{e^{hD}-1}, $$ is wrong answer.

Answer 1

If $\mathcal{I} = D^{-1}$ then wouldn't $$ D^{-1}_{h,x}[f] = \mathcal{I} \left[ \frac{\ln(fh+1)}{h} \right], $$

because then $$ D^{-1}_{h,x}\left[D_{h,x}[f]\right] = f? $$

Answer 2

$D^{-1}_{h,x}(f) = e^{-h\frac{d}{dx}}(hf+1)$

Answer 3

I think it is a correct answer: $$e^{hD}f = \sum \frac{h^n}{n!}D^n f = f(x+h)$$ so $$D_{h,x}f = \frac{f(x+h)-f(x)}{h}$$

Given $$g(x) = \frac{f(x+h)-f(x)}{h}$$ we may reconstruct $f(x)$ as $$f(x) = h\sum_{n=1} g(x - nh)$$

Now $g(x - nh) = e^{-nhD}g$, that is $$f = h\sum e^{-nhD}g = h\sum (e^{-hD})^n g= h(\frac{1}{1 - e^{-hD}} - 1)g = \frac{h}{e^{hD} - 1}g$$

All convergence considerations aside, of course.

PS: I don't know what I was thinking before. Thank @Leox for pointing out.

Answer 4

The op $D^{-1}D_{h,x}$ and its inverse $DD_{h,x}^{-1}$ are classic operators associated with the Bernoulli numbers (and polynomials) and the Euler-Maclaurin formula.