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I’m working through some of the exercises in *generatingfunctionology*. One of the questions is to find the generating function where the $n$th term $a_n=P(n)$ for $P$ a polynomial. The answer is $P(xD)(1/(1-x))$. I thought it meant something like if $P(n)=3n+n^2$, then $P(xD)=3xD+x^2D^2$, and then you would apply that to $1/(1-x)$. However, I know the generating function for $a_n=n^2$ is $$\frac{x(x+1)}{(1-x)^3}$$

which is not the same as $x^2D^2(1/(1-x))$. So what does it mean? Thanks.

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The problem is that you’ve mistakenly assumed commutativity. Here $\rm\:x\:$ and $\rm\:D\:$ do not commute. Indeed $\rm\ D\ x = x\ D + 1\ $ since $\rm\ (D\ x)\ f\ =\ D\ (x\ f\:)\ =\ x\ f\:’+f\ =\ (x\ D + 1)\ f\:.\ $ So it’s not true that $\rm\ (x\ D)^2\: =\ x^2\ D^2\:.\:$ Instead $\rm\ x\ D\ x\ D\ =\ x\ (x\ D + 1)\ D\ =\ x^2\ D^2 + x\ D\:.\ $ Generally

$$\rm (x\ D)^n\ =\ \sum_{k\ =\ 0}^n\ S(n,k)\ x^k\ D^k $$

where $\rm\ S(n,k)\ $ are the Stirling numbers of the second kind.

Beware that because this operator algebra is noncommutative, many familiar polynomial identities do not hold true. For example, consider the proof of the difference of squares identity

$$\rm\ (x-y)\ (x+y)\ =\ x^2 + x\ y – y\ x – y^2\ =\ x^2 – y^2$$

Notice how commutativity is assumed to cancel the middle terms. Thus the identity needn’t remain true if we substitute elements from some noncommutative ring, e.g. we cannot substitute $\rm\:D\:$ for $\rm\:y\:.\:$ Said slightly more technically, the polynomial evaluation map is a ring homomorphism iff coefficients and indeterminates commute. See my post here for further remarks on this topic.

Without checking out the book, this would be my guess:

Let for instance $P(x) = x^2 + 2x$, then $P(xD) = (xD)^2 + 2xD$, which is an operator.

Let this operator work on your function $f(x) = \frac{1}{1-x}$. It’s obvious that $(2xD)(f) = 2xf'$. (Here $f'$ is a bit an abuse of notation for $f'(x)$.) The other term is probably the confusing one. But if you write it as $xDxDf$ and work everything out from the right to the left you’ll get

$$ xD(xf') = x(1f'+xf'') = xf' + x^2 f'' .$$

And similarly for higher order terms, so with my example polynomial $P$ we have that $P(xD)(f) = 2xf' + xf' + x^2 f''$.

So in summary your problem might be that $(xD)^2 = xDxD = xD(xD) = xD + x^2D^2 \neq x^2 D^2$, in other words $x$ and $D$ don’t commute.

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