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In Penrose's book (The Road to Reality, chapter 21) he gives an example of Oliver Heaviside's observation that you can treat differential operators like numbers:

The differential equation $(1+D^2)y = x^5$ can be solved by dividing by $(1+D^2)$ then taking the power series expansion: $$y = (1-D^2+D^4-D^6+\cdots)x^5$$ which evaluates to $y = x^5 - 20x^3 + 120x$.

Apparently this can be made perfectly rigorous!

How is this done? and do you know where I can read more details about this idea?

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You may want to read about the Fourier transform. – José Figueroa-O'Farrill Jun 26 '10 at 12:26
See, and for an application that motivates the generalized Euler-Maclaurin formula see… – Steve Huntsman Jun 26 '10 at 12:31
1… and you can count on Terry to have buzzed about it. – Willie Wong Jun 26 '10 at 13:19
Heaviside is often credited with this discovery but this [1] paper argues it was a standard method at Heaviside's time. Boole wrote a treatise in 1880 on solving difference equations where he uses the same methods [2]. For example on page 257 he manipulates $e^{-{d\over dx}}$ [1] [2] – Dan Piponi Jun 26 '10 at 18:52
up vote 11 down vote accepted

This is just a fact from linear algebra: if $T$ is a nilpotent transformation of a vector space $V$, then $(1-T)^{-1} = 1 + T + T^2 + \dots$. More generally, the same is true in any commutative Banach algebra (such as the endomorphism ring of a normed complex vector space) if $T$ is of norm less than 1.

In your case, the differential operator $D$ is a nilpotent operator on the vector space of polynomials, and consequently this formula applies.

As people have pointed out in the comments, you may want to look at the Fourier transform, which realizes differentiation as multiplication, which allows you to treat differentiation kind of like a number. (More precisely, there is an isomorphism $F:L^2 \to L^2$ such that $F D F^{-1}$ is equal to multiplication by $x$ on sufficiently nice (e.g. Schwarz) functions.)

In higher dimensions, for instance, this means that if $\Delta$ is the Laplacian, then multiplication by $(1 + |x|^2)^k$ corresponds to applying $(I-\Delta)^k$, and it is thus possible to define an operator $(I - \Delta)^r$ for any real $r$.

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Just a bit of a nit-pick: you probably mean $(1 - \triangle)^r$, as the Laplacian (as usually defined to be $\sum \partial_i^2$) is a negative operator. – Willie Wong Jun 26 '10 at 14:10
Corrected, thanks. – Akhil Mathew Jun 26 '10 at 19:01
You can be sure Heaviside didn't restrict himself to spaces of functions where the derivative operator was nilpotent! – Dan Piponi Jun 26 '10 at 19:08
@sigfpe: Yes, that's why I included the remark about its generalization to commutative Banach algebras. – Akhil Mathew Jun 26 '10 at 19:39

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