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.

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    $\begingroup$ It's basically the same as writing $\frac{6}{2}=3$ when working in $\mathbb{Z}$, even though $2$ is not invertible in $\mathbb{Z}$. $\endgroup$ Commented Jul 3, 2020 at 17:57
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    $\begingroup$ 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: $C\div A=B$. $\endgroup$
    – Lubin
    Commented Jul 3, 2020 at 18:10

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Commented Jul 3, 2020 at 18:14
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    $\begingroup$ @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 $$\sum_{n\ge 0}a_nx^n=\sum_{n=0}^{k-1}a_nx^n+x^k\sum_{n\ge k}a_nx^{n+k}\;.$$ $\endgroup$ Commented Jul 3, 2020 at 18:28

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