Today I read the answer to this post, in which the poster integrates $x^5e^x$ by making these manipulations with the differential operator $D$: $$\frac1Dx^5e^x=e^x\frac{1}{1+D}x^5=e^x(1-D+D^2+...)x^5$$

which I was amazed by but yet suspicious of. After reading a bit on differential operators, I know a few properties. For example, $D+a$ (where $a$ is constant) is a polynomial differential operator which comes from the differential equation $y'+ay = q(x)$. Also, for polynomials $f$ and $g$, $f(x)\cdot g(x)=h(x)$ implies $h(D)u=f(D) \circ [g(D)u]$ with $u$ being a function and $f(D), g(D), h(D)$ being polynomial differential operators. Since $(1+x) \cdot \frac1{1+x} = 1$, I know that if we applied the operator $(1+D)$ to some function, $\frac{1}{1+D}$ would invert it back to the original function.

However, this still doesn't help me make sense of the meaning of $\frac{1}{1+D}$. For example:

  1. If $1+D$ comes from the differential equation $y'+ay+q$, where does $\frac{1}{1+D}$ come from? In other words, if $(1+D)y = 1\cdot y + Dy = y' + y$, how do we compute $\frac{1}{1+D}y$? (Is it even possible, and how does it relate to integration?)
  2. How do we justify the power series of the operator? How do we know there are no convergence issues?
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    $\begingroup$ I hope you don't mind that added the abstract-algebra tag. It's accurate, and I'm really close to getting the gold badge, so, you know. $\endgroup$ Jun 12, 2016 at 22:07
  • $\begingroup$ I'm not experienced in algebra but I'm willing to wade through if necessary to understand what's going on with the manipulations and such. Is there something I can read up on for it? i.e. what Henning Makholm mentioned. $\endgroup$
    – 986
    Jun 12, 2016 at 23:40
  • $\begingroup$ The algebra required to understand this is standard for a math PhD student in any (pure) field.. I've only heard about this application (I saw it in a number theory book when I was a kid), and learning the background for it will not be nearly as romantic. What is your current background/level? $\endgroup$ Jun 12, 2016 at 23:48
  • $\begingroup$ I am still in high school, so anything at the graduate level is much beyond my reach. I was just curious about how (and if) the operator could be manipulated like that, since if I could generalize it, that'd seem like a neat trick at my disposal in calculus. It seems I'm not ready yet for the algebra behind it. Though, thanks anyway :) $\endgroup$
    – 986
    Jun 13, 2016 at 2:31
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    $\begingroup$ Some undergraduates are already at this level by the time they get to grad school (I was). It also can't hurt to read advanced things even if you don't understand them fully. Try the Wikipedia article on formal power series (I haven't looked, but at least the topic is relevant). $\endgroup$ Jun 13, 2016 at 12:24

2 Answers 2


$\frac1{1+D}$ is the multiplicative inverse of $1+D$. In the ring of formal power series in an indeterminate $A$ over an integral domain, the multiplicative inverse of $1+A$ is $$\sum_{n=0}^\infty (-1)^nA^n$$ You can convince yourself that this is true by simply multiplying it out. When we consider formal power series, there's no need to worry about convergence.

When we actually apply the differential operator, convergence becomes an issue, but whenever the sum converges it always gives the right answer; this follows from continuity of the addition operation.

Notice that the operator was applied to a polynomial. Eventually each successive derivative will become zero, so there are really only finitely many terms in the sum. Thus the manipulation is valid in this case. Certainly there are cases where it is not; for example, applying this operator to $\sin(nx)$ for $n>1$ doesn't make much sense.


It's probably best to view it as "just suggestive notation that has no real right to work" and then justify it by the fact that the end result can be verified to be correct without doing such dodgy things.

If one really wanted to give the manipulation some formal existence, one would probably do something like taking the subring of the operator algebra generated by $D$, noting it is commutative (in contrast to the entire operator algebra), extend it into a ring of formal power series in $D$, taking the field of fractions over that, bla bla bla ...

Making sure that all of the i's are dotted and t's are crossed along the way is going to be a lot more work than simply shrugging and consider the whole thing a heuristic, yielding a possible result that needs to be checked properly afterwards.

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    $\begingroup$ I'd agree that it's a lot more work, but to me it'd be worth it. This is probably because algebra is my favorite thing in the world, and if there's an algebraic way to do it I'll probably do it that way. $\endgroup$ Jun 12, 2016 at 21:55

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