# Exponential derivative operator and continuous functions

I would like to know how to write down the following expression $$f(y)=\frac{1}{y^{n} e^{\frac{d}{dy}}g(y)}$$ in the form of $$e^{-\frac{d}{dy}}y^{-n}(\frac{1}{g(y)})$$ where $$n$$ is an integer and $$f,g: R \longrightarrow R$$ are two continuous functions, that are not operators. My question is: is it possible to do that? If it is false what are the other suggestions for the explicit formula for $$f(y)$$? Please i need a response. Thank you and best regards.

• As stated it's hard to make mathematical sense of your expression, since the derivative operator is not defined on continuous functions. Its exponential should of course be precomposition with translation by $1$, so your formula seems to want to link the values at $g(y\pm 1)$, which is not possible using only derivative operations. Sep 29, 2021 at 10:32
• Ok. Now suppose that g it is not continuous and if i considerfor example $$h(y)= \frac{1}{y^{n} e^{-(d/dy)y (d/dy)}g(y)}$$ what is the expression of $h(y)$. Can we say $$h(y)=e^{(d/dy)y (d/dy)}y^{-n}(\frac{1}{g(y)}) \; ?$$ Sep 29, 2021 at 11:16
• We’ve discussed this before, but to understand what it is you are after could you answer this question: which of these two expressions do you want: $(d/dy)g(y)=g’(y)$ or $(d/dy)g(y)=g’(y)+g(y)(d/dy)$. Sep 29, 2021 at 12:25
• I am not talking about the derivative it is obvious that i will choose $(d/dy)g(y)=g′(y)$ . I am talking about the exponential derivative in quotients this is a very hard example and it is very confusing that is why i am asking a question because i need help from experts like you. Sep 29, 2021 at 12:33

If $$(d/dy)g(y)=g’(y)$$, so no operator commutator, then $$e^{d/dy}g(y)=g(y+1),$$ so $$f(y)=y^{-n}/g(y+1).$$ If instead you choose the operator identity $$(d/dy)g(y)=g’(y)+g(y)d/dy$$, then $$f(y)=(1/g(y))e^{-d/dy}y^{-n}.$$
• You have answered to my question correctly, and i am thankful for that. But i forgot to ask about this example $h(y)=\frac{1}{y^{n}e^{-(d/dy)y(d/dy)) }g(y)}$ where g is not an operator. What can we say in this case? Sep 29, 2021 at 14:00
• there is no closed form expression for $h(y)$ in the case of general $g(y)$, but if you define the function $G(y)=e^{-(d/dy)y(d/dy)}g(y)$, say via the Taylor series of the exponential, then $h(y)=y^{-n}/G(y)$; in the special case that $g(y)=e^{ay}$ one has $G(y)=-ae^{-ay}$, so then $h(y)=-a^{-1}y^{-n}e^{ay}$. [search for "Laguerre derivative"] Sep 29, 2021 at 21:53