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My physics professor likes to write the following:

Let $D = \frac{\mathrm{d}}{\mathrm{d}x}$ therefore, $$ x - Dx = (1-D)x $$ This sort of makes my skin crawl; however, beyond the inelegance of the math, is there anything actually wrong about it?

Kalcifer
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This is not inelegant or wrong. It is simply a definition of the symbol $(1-D)$ as a differential operator. The set of all differential operators of the form $$ T=a_nD^n+a_{n-1}D^{n-1}+\cdots+a_1D+a_0 $$ defined by $$T(f) = a_nf^{(n)}+\cdots+a_1f'+a_0f$$ where $f^{(n)}$ denotes the $n$th derivative forms a ring, called the ring of differential operators. It is useful and often studied in analysis.

YiFan Tey
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  • What would be the use of factoring the differential operator/"differentiand" out of a derivative? To be able to evaluate the derivative, you would still have to factor it back in wouldn't you? – Kalcifer Oct 06 '21 at 03:54
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    @Kalcifer this is an abstract viewpoint that is useful when thinking of these differential operators as objects in and of themselves, instead of being bounded to any "differentiand". For instance, when we want to solve an ordinary differential equation $y''+3y'+2y=0$, we can write this as $Ty=0$ where $T=D^2+3D+2$. This allows one to use the tools of linear algebra to solve the problem, because $T$ is a linear transformation from the (vector) space of functions to itself. Of course this is not necessary in this case, but it is useful for more general complicated situations. – YiFan Tey Oct 06 '21 at 03:56
  • Dr. Peyam has a lovely video of this technique: https://youtu.be/NutSdXr1it0 – Sammy Black Oct 06 '21 at 04:10