# What is the meaning of division of a formal power series by $x$?

Background: a formal power series is defined as an expression of the form $$\sum_{n\geq 0} a_n x^n$$. If $$f =\sum_{n=0}^\infty a_n x^n$$ then, we write $$\{a_n\}_{n\geq 0} \leftrightarrow f$$. Two formal power series are equal if each of the components match. The sum and difference of two formal series is defined component-wise. Also the product of two formal power series $$\sum_n a_n, \ \sum_n b_n$$ is defined as the formal power series $$\sum_n c_n$$ where $$c_n = \sum_k a_k b_{n-k}$$. Two formal power series $$\sum_n a_n,\ \sum_n b_n$$ are called reciprocal if $$\sum_n a_n \sum_n b_n = \sum_n b_n \sum_n a_n = 1$$.

Now, in generatingfunctionology (2.2), Wilf mentions that for $$k \geq 0$$, we have, if $$\{a_n\}_{n \geq 0} \leftrightarrow f$$, then $$\{a_{n+k}\}_{n \geq 0} \leftrightarrow \frac{f - a_0 - \dots - a_{k-1}x^{k-1}}{x^k}$$. In particular, $$\{a_{n+1}\}_{n \geq 0} \leftrightarrow \frac{f-a_0}{x}$$. In fact, in (2.1) it is discussed that a formal power series has a reciprocal if and only if the constant term is non-zero.

What is confusing to me is what it means to divide a formal power series by $$x$$. In fact, I am not sure what exactly $$\frac{1}{x}$$ is the context of formal power series. Although we can interpret $$\frac{1}{x}$$ as a formal power series that results in $$1$$ when multiplied by $$x$$, there is no expression of $$\frac{1}{x}$$ in the form $$\sum_n a_n$$.

So my question is: what does $$\{a_n\}_{n \geq 0} \leftrightarrow f$$ $$\implies$$ $$\{a_{n+k}\}_{n \geq 0} \leftrightarrow \frac{f - a_0 - \dots - a_{k-1}x^{k-1}}{x^k}$$ actually mean in the context of formal power series? The explantion should not depend on any analytical property of $$f$$, as we are treating $$f$$ as only an algebraic object without any analytical property.

• It's basically the same as writing $\frac{6}{2}=3$ when working in $\mathbb{Z}$, even though $2$ is not invertible in $\mathbb{Z}$. Commented Jul 3, 2020 at 17:57
• In this situation where multiplication is commutative, and there are no zero-divisors, I like to think of division as “dismultiplication”, in the sense that if $C=AB$, then we may always understand $C\div A$ simply as what you get by erasing the $A$ from the product: $C\div A=B$. Commented Jul 3, 2020 at 18:10

First you need to correct your definition of $$f$$: $$f\leftrightarrow\langle a_n:n\ge 0\rangle$$ means that $$f(x)=\sum_{n\ge 0}a_n\color{red}{x^n}\;.$$ Then

\begin{align*} f(x)-a_0-a_1x-\ldots-a_{k-1}x^{k-1}&=\sum_{n\ge k}a_nx^n\\ &=x^k\sum_{n\ge k}a_nx^{n-k}\\ &=x^k\sum_{n\ge 0}a_{n+k}x^n\; \end{align*}

dividing by $$x^k$$ now has a clear formal meaning and results in the series

$$\sum_{n\ge 0}a_{n+k}x^n=a_k+a_{k+1}x+a_{k+2}x^2+\ldots\;.$$

We can read off the coefficients and see that by definition

$$\sum_{n\ge 0}a_{n+k}x^n\leftrightarrow\langle a_k,a_{k+1},a_{k+2},\ldots\rangle=\langle a_{n+k}:n\ge 0\rangle\;.$$

In short, it’s a straightforward formal algebraic manipulation.

Added: Don’t think of it as division: think of

$$\frac{\sum_{n\ge 0}a_nx^n-a_0-a_1x-\ldots-a_{k-1}x^{k-1}}{x^k}=\sum_{n\ge 0}a_{n+k}x^n$$

as an alternative way to write

$$\sum_{n\ge 0}a_nx^n=\sum_{n=0}^{k-1}a_nx^n+x^k\sum_{n\ge k}a_nx^{n+k}\;,$$

one that emphasizes the nature of the transformation from $$\sum_{n\ge 0}a_nx^n$$ to $$\sum_{n\ge 0}a_{n+k}x^n$$, the fact that it corresponds to a left shift of the associated sequence.

• I am afraid your explanation does not answer my question. You said "... dividing by $x^k$... . How can you divide by $x^k$ when it does not have a reciprocal in the ring of formal power series? That is exactly my question. Commented Jul 3, 2020 at 18:14
• @MutasimMim: Because, as I thought I’d made reasonably clear, it’s not an operation in that ring: it’s a purely formal manipulation that self-evidently works. Don’t think of it as division: think of it as an alternative way of writing $$\sum_{n\ge 0}a_nx^n=\sum_{n=0}^{k-1}a_nx^n+x^k\sum_{n\ge k}a_nx^{n+k}\;.$$ Commented Jul 3, 2020 at 18:28