Understanding the functionness of a generating function

Question

A function, at its bare bones, takes an input and produces an output. For example, $f(x) = x^2$ can take a value $x = 4$ and produce an output of $16$. Contrarily, the $x$ in a generating function such as $1 + x + x^2 \; + \; ...$ generally serves the purpose of a placeholder, differentiating itself from an input-output process. In light of the above, I have seen advice encouraging me to ignore the standard notion of a function when dealing with generating functions.

To now illustrate my question, I will consider the proof that $1 + x + x^2 + x^3 \; + \; ... = \frac{1}{1-x}$.

From the finite case of a geometric series, it follows that $$\sum_{k=0}^\infty x^k = \lim_{n\to\infty} \frac{1-x^n}{1-x} \color{red}{= \frac{1}{1-x}}$$

But why does it follow? If the idea of $x$ having a value is disconnected from that of a generating function, shouldn't the $\color{red}{\text{red}}$ conclusion be unsubstantiated? If there is no notion of $x \in (-1,1)$, why is one allowed to reasonably assume that $\lim_{n \to \infty} x^{n} = 0$?

On a broader note, my question is directed toward why we see a specific representation of a generating function fit if the idea of it being a function is itself disregarded. For example, if a hypothetical generating function has varying representations for different domains of $x$, how would one reasonably select the universal representation, that is, the correct domain?

In my mind, if we are transforming clay from one shape to another, the transformation should always hold to conclude that the shapes are equivalent.

I suspect that my confusion is particularly related to the way in which a generating function acts as a function. I welcome all insights.

Answer 1

When talking about generating functions we have counting functions $f:\mathbb{N}_0\to\mathbb{C}$ in mind, a simple being \begin{align*} &f:\mathbb{N}_0\to\mathbb{C}\tag{1}\\ &f(n)=1 \end{align*} We see in (1) the domain of counting functions are natural numbers and we have just a sequence \begin{align*} \left(f(n)\right)_{n\geq 0}=(1,1,1,\ldots)\tag{2.1} \end{align*} We associate to $f$ the formal series \begin{align*} F(z)=\sum_{n=0}^\infty f(n)z^n=\sum_{n=0}^\infty z^n=1+z+z^2+\cdots\tag{2.2} \end{align*} and call this $F$ the generating function of $f$.

The notion formal means we consider the series $F$ as algebraic object. The argument $z$ is an indeterminate not considered to be evaluated at a specific value. The term $z^n$ is a formal placeholder indicating that the coefficient $f_n$ is at the $n$-th position (when starting from zero) of the sequence $(f(n))_{n\geq 0}$.

• In fact there is besides representational issues no difference between (2.1) and (2.2). They address the same object.

• Since $f$ is a function with discrete domain $\mathbb{N}$ and not with domain $\mathbb{R}$ or $\mathbb{C}$ the common usage of limit as we use it in analysis is based on a continuum and is not applicable here. Moreover also when using the formal power series notation, the variable $z$ is an indeterminate which can't take any values approaching a limiting value.

Note, generating functions do not disregard any concepts from analytical functions we know from analysis. They are rather a different species of functions. Nevertheless we borrow notational conventions from analysis and mimic calculations we know from power series in analysis, as far as these algebraic properties can be consistently used in the realm of generating functions.

Sum and product of generating functions:

We define sum and product of generating functions purely algebraically via \begin{align*} A(z)+B(z)=\sum_{n\geq 0}\left(a_n+b_n\right)z^n\qquad\qquad A(z)B(z)=\sum_{n\geq 0}\left(\sum_{k=0}^na_kb_{n-k}\right)z^n\tag{3} \end{align*} and can easily check that usual properties such as associativity and distributivity are satisfied, so that generating functions are forming a commutative ring $\mathbb{C}[[z]]$ over $\mathbb{C}$. In fact we have an integral domain, since $A(z)B(z)=0$ implies either $A(z)=0$ or $B(z)=0$.

The multiplicative inverse of a generating function:

Now we want to go one step further and ask for a multiplicative inverse $B(z)$ with \begin{align*} A(z)B(z)=1 \end{align*} The answer to this question is \begin{align*} A(z)=\sum_{n\geq 0}a_nz^n\qquad\text{has an inverse if and only if }a_0\neq 0 \end{align*} Since $A(z)B(z)=1$ this implies $a_0b_0=1$ and the condition $a_0\neq 0$ is necessary. We can now determine the coefficients $b_n$ inductively. Assuming $b_0,b_1,\ldots,b_{n-1}$ are already determined, we obtain \begin{align*} 0=\sum_{k=0}^na_kb_{n-k}=a_0b_n+\sum_{k=1}^na_kb_{n-k} \end{align*} and \begin{align*} b_n=-a_0^{-1}\sum_{k=1}^na_kb_{n-k}\tag{4} \end{align*} is uniquely determined.

The geometric series: We consider now the geometric series \begin{align*} F(z)=\sum_{n=0}^\infty z^n \end{align*} as generating function and find the multiplicative inverse from (3) and (4) as \begin{align*} F(z)(1-z)=\left(1+z+z^2+\cdots\right)(1-z)=1 \end{align*} As usual we write the multiplicative inverse $B(z)$ of $A(z)$ with $A(z)B(z)=1$ as \begin{align*} B(z)=\frac{1}{A(z)}\qquad\text{or}\qquad B(z)=\left(A(z)\right)^{-1} \end{align*} and obtain finally in the case of the geometric series also when considering it as generating function \begin{align*} \color{blue}{\sum_{n=0}^\infty z^n=\frac{1}{1-z}}\tag{5} \end{align*} Note there was no limit used to derive the identity (5).

Some final notes:

• There are some crucial caveats and differences when working with formal power series. One of them is we have to take care that the indeterminate $z$ is only allowed to be involved with finitely many operations. A consequence of this is that composition of generating functions $A(B(z))$ is only admissible when the constant term $b_0$ of $B(z)$ is equal to zero.

• Even when working with generating functions we have a notion of nearness and can define some kind of limits. But this is fundamentally different to the limits we use in analysis. Here they lead to ultrametrics. See for instance Appendix A.5 in Analytic Combinatorics by P. Flajolet and R. Sedgewick.

• Sometimes it is convenient to change the view and consider a generating function as analytic object and study it in the context of complex analysis. This is especially useful when analyzing the asymptotic behaviour of the coefficients of generating functions. (See this answer for more details.)

• Another important aspect is the so-called transfer principle which says that if two power series are equal as analytic functions, then they are also equal as formal power series. See for instance section 2.4 The Transfer Principle in The Concrete Tetrahedron by M. Kauers and P. Paule. Thanks to this transfer principle the identity (5) follows immediately.

• This answer followed mainly section 2.1 Algebra of Formal Series in A Course in Enumeration by M. Aigner.

Answer 2

The first thing to understand is that it is not the 'functionness' of the generating function we care about mostly but rather their algebra (*). In sense that, every practical thing we do from Generating functions stems out of their algebra rather than by them being a function itself.

Now, what do I mean by algebra? Well, it is how the coefficients of the expanded form of the product of two power series relate to the coefficients of the multiplicand power series.

Now, usually counting problems have some notion of consecutive selection of some $n$ terms, we choose some power series such that the multiplication give out a power series whose degree $n$ term has coefficient as the number of ways to consecutively select $n$ terms.

Now how do we find a power series for a counting problem? Well there are sometimes systematic ways to construct it (see Generating functionlogy chapter-3) but one can do it with a bit of guess work as well (see here for some hints) or maybe this blog I wrote can be helpful

On the point about Clay, I guess this is like the time humans realize "Oh wait, mud is not just for coating things, we can use it to make clay and then build clay pots or something" , The point I want to emphasize is that the claypots are of totally different than coating things.

Answer 3

Formal power series are not functions. They have their own topology, which is different from the topology of convergent or analytic series; the usual choice is that a sequence of power series converges if and only if the coefficient of $x^k$ is eventually constant for all $k$. We do have $\lim_{n \to \infty} x^n = 0$ in this topology, but for totally different reasons as in the analytic case.

Answer 4

To shed a different light on the other answers, with a flavor of the way we use formal power series in combinatorics (this short read is very enlightening : http://algo.inria.fr/flajolet/Publications/book.pdf )

If $S = \sum_{k\geq 0} x^k$ then $S$ obeys the recursion $$ S = 1 + xS.$$ Hence $S(1-x)=1$, hence your result.

You don't need to take limits to write these equalities. (Actually even in the space of real power series, factoring out an $x$ is perfectly allowed all of the time). But in the space of formal power series you are moreover dispensed of the job of proving $S$ is convergent. $S$ exists by mere virtue of you writing it.