Function of a differential operator.

Question

Friends of mine who study Quantum Field Theory asked me about the following problem. The task is to simplify the expression $$ f_1(\frac{d}{dx})f_2(x) $$ so that it doesn't contain derivatives, but only a finite number of integrals or finite sums. The functions are, for example: $$ f_1(x)=e^{x-2x^2+x^3-3x^4},\; f_2(x)=\sin(2x-x^2+4x^3+2x^4), $$ or $$ f_1(x)=\ln(6x-2x^2+3x^3-5x^4),\;f_2(x)=\cos(3x-x^2-6x^3+5x^4), $$ or even singular $f_2$ $$ f_1(x)=\sin(2x-4x^2+2x^3-5x^4),\;f_2(x)=\cot(x-4x^2+8x^3+3x^4). $$ How should one interpret the expression above? Does it have anything to do with pseudo-differential operators (although $f_1(x)$ doesn't satisfy the definition of a symbol)?

Edit: one more "bad" example $$ f_1(x)=(x^4+4)^{4x-6x^2+3x^3-2x^4},\;f_2(x)=\cot(4x-5x^2+3x^3+6x^4). $$

Answer 1

There might not be a general solution to it. Assume f1 is a linear operator i.e.

f1() = Sum of first order derivatives in different independent variables.

In that case, it's just the derivative (f2())

Now if f1() has higher order derivatives and linear, even then it's pretty straight forward.

Answer 2

I can tell you what this sort of thing would probably mean to an analyst. I will assume that $\frac{d}{dx}$ is acting on smooth functions on $\mathbb{R}$. The functions $\{e^{ikx}\}$, $k \in \mathbb{R}$, form a generalized eigenbasis for this operator in the sense that we can write $$g(x) = \int_{\mathbb{R}} \hat{g}(k) e^{ikx}\, dx$$ via the Fourier transform and then obtain $$\frac{d}{dx}g(x) = \int_{\mathbb{R}} ik \hat{g}(k) e^{ikx}\, dx$$ for suitable $g$. For suitable functions $f$ (say, in $C_0(\mathbb{R})$), one defines $$f\left(\frac{d}{dx}\right)g(x) = \int_{\mathbb{R}} i f(k) \hat{g}(k) e^{ikx}\, dx$$ In general, given an operator $D$ one defines $f(D)$ by first diagonalizing $D$ and then applying the function $f$ to the spectrum of $D$. These operators are often pseudodifferential, though you don't typically need much pseudodifferential operator theory to analyze them.