Given this operator what is inverse operator?

Question

Given operator

$$\Delta_{sym}[f(x)]=\frac{f(x+\varepsilon)-f(x-\varepsilon)}{2\varepsilon}$$

what is inverse operator in terms of summations?

For instance, given operator

$$\Delta_{full}[f(x)]=\frac{f(x+\varepsilon)-f(x)}{\varepsilon}$$

the inverse operator is

$$\Delta_{full}^{-1}=\varepsilon \lim_{t\to x/\varepsilon} \sum_{s=0}^{t-1}f(\varepsilon s)$$

What would be a similar expression for $\Delta_{sym}^{-1}$?

Answer 1

Well, I can answer myself now. With simple substitution we have

$$\Delta_{sym}^{-1}[f(x)]=2\varepsilon \lim_{t\to x/(2\varepsilon)} \sum_{s=1/2}^{t-1/2}f(2\varepsilon s)$$

Answer 2

Let $g(x)=[Δ_{sym}f](x)$. Then obviously \begin{align} 2ε·g(x+ε)&=f(x+2ε)-f(x)\\ 2ε·g(x+3ε)&=f(x+4ε)-f(x+2ε)\\ 2ε·g(x+5ε)&=f(x+6ε)-f(x+4ε)\\ 2ε·g(x+7ε)&=f(x+8ε)-f(x+6ε)\\ &etc. \end{align} so that $$ 2ε·\sum_{k=0}^{n-1}g\bigl(x+(2k+1)·ε\bigr)=f(x+2n·ε)-f(x) $$ So, assuming that $\lim_{x\to\infty}f(x)=0$, then $$ f(x)=-2ε·\sum_{k=0}^{\infty}g\bigl(x+(2k+1)·ε\bigr) $$ Switching the sign of $ε$ and assuming that $\lim_{x\to-\infty}f(x)=0$, one also gets $$ f(x)=2ε·\sum_{k=0}^{\infty}g\bigl(x-(2k+1)·ε\bigr) $$