# Derivatives and exponential derivatives quotient operators on two variables

I consider for example the following function of two variables given by $$f(x,y)=\sum_{n=0}^{+\infty}\frac{\delta_{x}^{-m} \exp(\delta_{x})}{\delta_{y}^{-m}\exp(\delta_{y})}\left(\frac{x}{y}\right)^{n}$$ Where $$\delta_{x}=\frac{d}{dx}$$ and $$\delta_{y}=\frac{d}{dy}$$ and $$m$$ is a positive integer. My question is: when $$x < y$$ can we write f as $$f(x,y)=\frac{\delta_{x}^{-m} \exp(\delta_{x})}{\delta_{y}^{-m}\exp(\delta_{y})}\sum_{n=0}^{+\infty} \left(\frac{x}{y}\right)^{n}=\frac{\delta_{x}^{-m} \exp(\delta_{x})}{\delta_{y}^{-m}\exp(\delta_{y})}g(x,y)?$$ such that $$g(x,y)=\frac{1}{1-\frac{x}{y}}=\frac{y}{y-x}$$. If the formula above is correct can we prove it more explicitly. Precisely, constructing a corollary for which $$\frac{\delta_{x}^{-m} \exp(\delta_{x})}{\delta_{y}^{-m}\exp(\delta_{y})}$$ is a well-defined operator denoted by $$A_{x,y}^{m}$$. I need a response, if someone have an idea.

Best regards and thank you.

• $\delta_{x}^{-1}$ is an integral operator. Those usually need a boundary condition? The exponentials are just shift operators. Sep 19, 2021 at 21:24
• for example $\delta_{x}^{-2}T(x)$ is equal to $$\int_{0}^{+ \infty} \int_{0}^{+ \infty}T(x) dx?$$ for $x \in R^{+}$. Sep 22, 2021 at 16:10
• Ok, so your boundary condition is that you start integrating at 0, that's fine. The upper limit of course must be variable - your result must again depend on $x$. So, $\delta^{-2}_{x} T(x) = \int_0^x dx' \int_0^{x'} dx'' T(x'')$. Sep 22, 2021 at 16:37
• Thank you. It is simple to have $\delta_{x}^{m} x^{m}= m!$ But if we have $\delta_{x}^{-m} x^{-m}$ what can we say in this case? Sep 22, 2021 at 16:42
• If you want to treat something like $x^{-m}$, you should probably use a different boundary condition. How would you integrate $x^{-m}$ starting at $x=0$? That's not well defined. Sep 22, 2021 at 16:46

To make sense of this, you might Fourier transform $$A_m=\frac{\delta_{x}^{-m} \exp(\delta_{x})}{\delta_{y}^{-m}\exp(\delta_{y})}=\delta_{x}^{-m} \delta_{y}^{m}e^{\delta_x-\delta_{y}},$$ which would give the operator $$\hat{A}_m=(k_y/k_x)^m e^{i(k_x-k_y)}.$$ Then $$A_m$$ acting on a function $$f(x,y)$$ could be obtained by inverse Fourier transformation, $$A_m f(x,y)=\int_{-\infty}^\infty \frac{dk_x}{2\pi}\int_{-\infty}^\infty \frac{dk_y}{2\pi}(k_y/k_x)^m \hat{f}(k_x,k_y)e^{i(x+1)k_x+i(y-1)k_y}.$$ The question whether $$A_m$$ is a well-defined operator would then boil down to whether this integration is well-defined. The function $$\hat{f}(k_x,k_y)$$ would need to vanish when $$k_x\rightarrow 0$$ to avoid the pole.
• Firstly, $\frac{\delta_{x}^{-m} exp(\delta_{x})}{\delta_{y}^{-m} exp(\delta_{y})} \neq \delta_{x}^{-m} \delta_{y}^{m} exp(\delta_{x}-\delta_{y})$. Secondly, where those calculations came from? I don't understand your explications. Sep 20, 2021 at 16:51
• $\delta_y$ and $\delta_x$ commute, so you have two commuting operators $A=\delta_x^{-m}e^{\delta_x}$ and $B=\delta_y^{-m}e^{\delta_y}$; the ratio $\frac{A}{B}=AB^{-1}=\left(\delta_x^{-m}e^{\delta_x}\right)\left(e^{-\delta_y}\delta_y^{m}\right)=\delta_x^{-m}\delta_y^{m}e^{\delta_x-\delta_y}$, where in the last step I again used the commutation of $\delta_x$ and $\delta_y$. Sep 20, 2021 at 17:07
• But the operator $e^{- \delta_{y}}= \sum_{k=0}^{+ \infty} \frac{(-1)^{k} \frac{d^{k}}{dy^{k}}}{k!}$ and $\frac{1}{e^{ \delta_{y}}}= \frac{1}{\sum_{k=0}^{+ \infty} \frac{\frac{d^{k}}{dy^{k}}}{k!}}$ they are not the same. Can we deduce a commutator here. Is there any propertie that is explaining this formula in a reference? Sep 21, 2021 at 10:53
• $e^{-\delta_y}=\frac{1}{e^{\delta_y}}$, for the reason that for any operator $A$ such that the sum converges one has $\left(\sum_{n=0}^\infty (-A)^n/n!\right)\left(\sum_{m=0}^\infty A^m/m!\right)=1$. Sep 21, 2021 at 10:57