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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 $$ f_1(x)=(x^4+4)^{4x-6x^2+3x^3-2x^4},\;f_2(x)=\cot(4x-5x^2+3x^3+6x^4). $$

Alex Bogatskiy
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  • What does $\left(\frac{d}{dx}\right)^n$ mean? What about $e^{\frac{d}{dx}}$, $\sin\left(\frac{d}{dx}\right)$, $\ln\left(\frac{d}{dx}\right)$, etc. ? – Michael Albanese Jan 08 '14 at 09:09
  • @MichaelAlbanese that is also unclear. The n-th power is clearly the n-th derivative. I would imagine that the exponential or sine are defined by their Taylor series as usual, but with logarithms it's really ambiguous. Maybe someone familiar with QTF might help. – Alex Bogatskiy Jan 08 '14 at 12:43
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    For example, $\exp(a\frac{d}{dx})$ is well-defined by the Taylor series and coincides with the shift operator $Tf(x)=f(x+a)$. However, $\exp(\frac{d^2}{dx^2})$ gives a series that is not similar to anything like the shift operator. – Alex Bogatskiy Jan 08 '14 at 12:59
  • Is any finite closed form known for $\exp(\frac{d^2}{dx^2})$? – Sidharth Ghoshal Nov 16 '23 at 05:43
  • Also those functions seem pretty intractable considering this is itself a tricky problem. Did your friend find closed forms for those?? – Sidharth Ghoshal Nov 16 '23 at 05:47

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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.

Paul Siegel
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    There is a typo in the formula above. The correct version is: $$f\left(\frac{d}{dx}\right)g(x) = \int_{\mathbb{R}} f(ik) \hat{g}(k) e^{ikx}, dx$$ – Mirar Sep 28 '22 at 06:16
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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.

Pgram
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