I am interested in using series of the form $\sum_{n=-\infty}^{\infty} a_nt^n$ (where $a_n\in\mathbb C$) as generating functions. In general, multiplication of such series goes against the "formal power series" philosophy since the coefficients of the product are themselves infinite sums. However, the results are sometimes intelligible. Here is an example:

Let $G(t) = \frac 1{2-t} = \frac 12 + \frac 14t + \frac 18t^2 + \cdots$ be the usual probability-generating function for a random variable with Geometric($\frac 12$) distribution. If we have two independent Geometric($\frac 12$) random variables, $X$ and $Y$, then we can correctly read the distribution of $X-Y$ from the expansion of $G(t)\cdot G(t^{-1})$: $$G(t)\cdot G(t^{-1}) = \cdots + \frac 1{12}t^{-2} + \frac 16t^{-1} + \frac 13 + \frac 16t + \frac 1{12}t^2 + \cdots$$ Although each coefficient is an infinite sum, those sums are absolutely convergent and so there is no doubt about how to evaluate them.

We can go further with this example. Suppose we write \begin{align*} G(t)\cdot G(t^{-1}) &= \frac 1{2-t}\cdot\frac 1{2-\frac 1t} \\ &= \frac 1{5-2t-\frac 2t} \\ &= \frac 15\cdot\frac 1{1-\frac 25(t+\frac 1t)} \\ &= \frac 15\sum_{n=0}^\infty \left[\frac 25\left(t+\frac 1t\right)\right]^n \end{align*} and attempt to extract the coefficients. Then the results appear to be accurate; for example, the constant term is $\frac 15\sum_{n=0}^\infty \binom{2n}n\left(\frac 25\right)^{2n}$, which does equal $\frac 13$. But this is a delicate game: if we multiply by $\frac tt$ at some point in the computation above, and try to expand $\frac t{5t-2t^2-2}$ as an ordinary power series, the results are (obviously) different.

Here's my question: Is there a theory that formalizes useful computations with series in $\mathbb C[[t,t^{-1}]]$ while excluding contradictory computations (much like the usual theory of formal power series does for computations in $\mathbb C[[t]]$)? References gratefully accepted.

  • $\begingroup$ In your particular example nothing really interesting happens other than the observation that the Laurent series of a meromorphic function does depend on the circle (or annulus) on which the decomposition is made (once you pass through a pole, the corresponding part flips). So, you just want the expansion near the unit circle rather than near the origin or near infinity. I'm not sure if you had something more interesting in mind, so I'll stop here. $\endgroup$
    – fedja
    Sep 17, 2016 at 17:43
  • $\begingroup$ fedja, I didn't mean to put the emphasis on why multiplying by $t/t$ in the example is problematic---though your comment does make the reason very clear. But in the treatments of formal Laurent series I've seen, the left tail is required to be finite so that multiplication is well-defined. What I want to know is whether there is a formal theory that allows the multiplication I did in my example ($G(t)\cdot G(t^{-1})$) without consideration of whether the series converge for particular values of $t$ (though we obviously do need to consider whether the terms in the product can be collected). $\endgroup$
    – tuna
    Sep 17, 2016 at 18:56
  • $\begingroup$ Sadly I am not competent to make it an answer but such formalism is very essentially used in the mathematical treatment of operator product expansions, vertex algebras, etc. I've first seen it in Kac's book, then in several other papers and books. Hopefully somebody can explain it in an answer. $\endgroup$ Sep 18, 2016 at 6:14
  • $\begingroup$ ...more specifically it is (e. g.) in V. G. Kac, "Vertex algebras for beginners" Chapter 2 (Calculus of formal distributions) and in J. Lepowsky & H. Li "Introduction to Vertex Operator Algebras and Their Representations" Chapter 2 (Formal calculus) $\endgroup$ Sep 18, 2016 at 12:22
  • $\begingroup$ Let me also add that in mathoverflow.net/q/220938/41291 I encountered something very similar $\endgroup$ Sep 18, 2016 at 19:08

1 Answer 1


There is a natural space in which such operations can be carried out; however, necessarily, its definition is functional-analytic rather than algebraic.

Let $H=L^2(S^1)$. Any $f\in H$ has a unique Fourier expansions of the form $$ f(\theta)=\sum_{n\in\Bbb{Z}}a_n t^n, \quad \theta\in\Bbb{R}/2\pi\Bbb{Z}, \quad t=e^{i\theta}\tag1 $$ whose coefficients satisfy $$ \sum_{n\in\Bbb{Z}}|a_n|^2<\infty.\tag2 $$ Conversely, every formal expansion of the form (1) whose coefficients satisfy (2) defines $f\in H$. In fact, $H$ is a commutative Banach algebra and it is a standard fact that addition and multiplication can be carried out at the level of formal expansions (1). Thus for
$$ g(\theta)=\sum_{n\in\Bbb{Z}}b_n t^n,\tag3 $$ the expansion of the product $fg$ is given by $$ f(\theta)g(\theta)=\sum_{n\in\Bbb{Z}}\left(\sum_{k\in\Bbb{Z}}a_k b_{n-k}\right)t^n,\tag4 $$ where each coefficient in the expansion has a well-defined complex value due to the absolute convergence of the corresponding series (the latter property follows from the Cauchy-Schwartz inequality). Clearly, the multiplication is associative, due to the corresponding property in $H=L^2(S^1)$ and uniqueness of the Fourier expansion.

On the other hand, let us start with a general expansion of the form $(1)$ and impose a fairly mild requirement that the constant term of the product $$ \left(\sum_{n\in\Bbb{Z}}a_n t^n\right)\left(\sum_{n\in\Bbb{Z}}a_n t^{-n}\right),\tag5 $$ where the second factor is obtained by substituting $t^{-1}$ for $t$ in $(1)$, be given by an absolutely convergent series. Using $(4)$ with $b_n=a_{-n}$, one easily sees that this requirement translates precisely to condition $(2)$ on the coefficients. So if we are interested in the products of type $(5)$, as in the example in the original posting, we might as well restrict attention to the elements of $H$ (i.e. impose condition $(2)$ on the coefficients) from the get-go. This indicates that $H$ is indeed a natural realm for usual aritmetic operations with such doubly-infinite expansions.

To summarize, we found a natural space $H$ where addition and multiplication of formal expansions of type $(1)$ and $(3)$ are well-defined. The definition of $H$ involves analytic condition $(2)$. The resulting object is a commutative Banach algebra.


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