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In mcs.pdf 16.1.1, it says:

$G(x)-xG(x)=1$

Solving for $G(x)$ gives

$$ \frac{1}{1-x}=G(x)::=\sum_{n=0}^{\infty}x^n\tag{16.3} $$

...

But in the context of generat- ing functions, we regard infinite series as formal algebraic objects. Equations such as (16.3) and (16.5) define symbolic identities that hold for purely algebraic rea- sons.

It is done by canceling terms. Since that is one infinite series, we can leave only 1 after canceling forever.

But why do we not need to care about convergence and say that the equation holds? Is there one better explanation for the reasons behind?

The book says later in 16.1.1 (Sorry for not continuing reading and asking one question instead here):

We’ll explain this further in Section 16.5 at the end of this chapter, but for now, take it on faith that you don’t need to worry about convergence.

After reading chapter 16.5, I thought I understood this question.

The key part is

It simply means that $G(x)$ really refers to its infinite sequence of coefficients $(g_0, g_1, \ldots)$ in the ring of formal power series.

In other words, the powers of the variable x just serve as a place holders—and as reminders of the definition of convolution.

The above means same as what PrincessEev says "we associate with sequences in a more intuitive way", the 1st paragraph of Theo Bendit's answer and wikipedia (See this lecture point 1 of "Conventions and Notation" for the similarity with "power series")

power series can be viewed as a generalization of polynomials, where the number of terms is allowed to be infinite, with no requirements of convergence. Thus, the series may no longer represent a function of its variable, merely a formal sequence of coefficients

The section 16.5 almost says same as Theo Bendit's answer.

Although after reading these, why the definition is helpful seems to be not very obvious. Maybe it is better to rethink after learning abstract algebra and knowing the whole knowledge realm of that topic. Thanks for all.

(This question is edited after the comments and the answer)

An5Drama
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    At least to me, I like to think of formal power series (in the GF context) as more of a visual tool for handling operations we associate with sequences in a more intuitive way. We could care about convergence if we wanted to: but ultimately it's a detail that is rarely if ever of any concern, especially if we're more focused on the associated sequence, so why waste time on it? – PrincessEev May 18 '24 at 03:22
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    Are you familiar with rings and related concepts? If so, then Wikipedia may help you. – Jyrki Lahtonen May 18 '24 at 03:41
  • There are a few pleasant surprises and a few pitfalls, if your exposure to power series is only from calculus. Take a look at for example this and this. – Jyrki Lahtonen May 18 '24 at 03:48
  • @JyrkiLahtonen Thanks for your comment. I intended to learn abstract algebra, but currently I haven't learned that. My current understanding about ring is merely the ring of integer modulo operation which is referred to in mcs.pdf. I will return to this question after learning abstract algebra. – An5Drama May 18 '24 at 03:48
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    There is a related concept of convergence here, with respect to the so called $x$-adic topology. A sequence $(p_n(x)){n\in\Bbb{N}}$ of polynomials converges $x$-adically (to a formal power series), if and only if to each exponent $M\in\Bbb{N}$ we can find a threshold $n(M)$ with the property that $p_i(x)-p_j(x)$ only contains terms of degree $>M$ whenever $i,j>n(M)$. In other words, the sequence of coefficients $c{m,i}$ of $x^m$ in $p_i(x)$ is eventually constant as $i\to\infty$. – Jyrki Lahtonen May 18 '24 at 03:53
  • (cont'd) Looked at that way, you can use some usual limit concepts. The above can be seen like a Cauchy convergence test, but the metric is unlike anything you have seen in calculus. – Jyrki Lahtonen May 18 '24 at 03:55
  • @PrincessEev Thanks for your comment. But unfortunately I haven't learnt formal power series and GF (maybe you mean Galois field). So I can't understand why we can define as what it is. – An5Drama May 18 '24 at 03:56
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    In other courses (like combinatorics) students learn to manipulate such formal power series without much background in abstract algebra. Even engineers are exposed to $z$-transforms! Pure mathematicians do feel uneasy before they learn a few definitions (adding rigor). It's a good sign that you feel uncomfortable about them without a due amount of convergence and theory! – Jyrki Lahtonen May 18 '24 at 04:00
  • @An5Drama By "GF" I meant "generating function" (A common abbreviation) – PrincessEev May 18 '24 at 04:06
  • @JyrkiLahtonen Thanks for your additional information about $x$-adic topology although I can't understand since as you says currently my understanding about series is only from the calculus course. – An5Drama May 18 '24 at 05:17
  • Basically it says that the sequence of polynomials $1$, $1+x$, $1+x+x^2$, $1+x+x^2+x^3$, $\ldots$, is a Cauchy sequence, because the coefficients of various powers of $x$ stabilize. Hence it converges to a formal power series $F(x)$. OTOH, when we multiply the polynomials in the sequence by $1-x$, we end up with $1-x$, $1-x^2$, $1-x^3$, $1-x^4$, that similarly converges - this time to the constant $1$. Therefore the series $F(x)$ has all the arithmetic properties of $1/(1-x)$. – Jyrki Lahtonen May 18 '24 at 05:32
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    (cont'd). Similarly, the product of two Cauchy sequences is Cauchy. If we square the polynomials in the previous example we get $1^2=1$, $(1+x)^2=1+2x+x^2$, $(1+x+x^2)^2=1+2x+3x^2+2x^3+x^4$, $(1+x+x^2+x^3)^2=1+2x+3x^2+4x^3+3x^4+2x^5+x^6$ etc. Here the coefficients of fixed powers again stabilize, and the resulting power series is $$F(x)^2=1+2x+3x^2+4x^3+\cdots=\sum_{n=0}^\infty(n+1)x^n.$$ You probably recognizer this as the Taylor series of $1/(1-x)^2$, which fits perfectly. – Jyrki Lahtonen May 18 '24 at 05:37
  • @JyrkiLahtonen Sorry, I missed noticing mcs.pdf has chapter 16.5 about Formal Power Series. I will rethink about this question after reading that. – An5Drama May 18 '24 at 07:11
  • @PrincessEev Please see the last comment since I can't mention multiple people in one comment https://meta.stackexchange.com/q/357415/1355014. – An5Drama May 18 '24 at 07:12

1 Answers1

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Formal power series, like formal polynomials, aren't really functions. You can think of them as sequences $a_0, a_1, a_2, \ldots$ written in a more suggestive format: $$a_0 + a_1 x + a_2 x^2 + \ldots$$ The $x$ here is not a variable, it's really just part of the notation. They look a lot like the power series we deal with in analysis, where $x$ would be a variable in which you can substitute values into (so long as they lie in the interval/disc of convergence), but this is not the foundation here. They are simply sequences, written differently.

Why do we notate them in this way? It suggests a particular multiplication operation:

$$\left(\sum a_n x^n\right)\left(\sum b_n x^n\right) := \sum \left(\sum_{i=0}^n a_i b_{n-i}\right) x^n.$$

This mimics what you'd get if you multiplied two not-so-formal power series (that are functions defined on intervals/discs), but again, it's just a fancy way of multiplying two sequences: $$(a_n)_n \cdot (b_n)_n = \left(\sum_{i=0}^n a_i b_{n-i}\right)_n.$$ Writing it as sequences, you'd wonder why anyone would choose to multiply two sequences this way. Writing as formal power series, you'd wonder how else anyone would define multiplication.

Now, this multiplication between formal power series has an identity. An identity is a series $I(x)$ such that $F(x)I(x) = I(x)F(x) = F(x)$ for all formal series $F(x)$. (Note that forcing both $F(x)I(x)$ and $I(x)F(x)$ to be equal to $F(x)$ is redundant, because multiplying power series is commutative, i.e. $F(x)G(x) = G(x)F(x)$ for all power series $F$ and $G$. But, the definition of an identity applies to non-commutative operations too.) In particular, the identity is: $$I(x) = 1 + 0x + 0x^2 + \ldots = 1,$$ or as a sequence, $(1, 0, 0, \ldots)$. That is, $1 \cdot F(x) = F(x) \cdot 1 = F(x)$... surely not a revelation.

Since there is an identity $I(x) = 1$, we can also investigate the concept of inverses. The inverse to a formal power series $F(x)$ is a formal power series $G(x)$ such that $F(x) \cdot G(x) = G(x) \cdot F(x) = I(x) = 1$. Again, in this context, $F(x) \cdot G(x) = 1$ would suffice. Usually, in abstract algebra, we would denote the inverse of $F(x)$ by $(F(x))^{-1}$. In this case, we are choosing to write it as $\frac{1}{F(X)}$.

This is what we mean by $\frac{1}{1 - x}$. It is the multiplicative inverse of the formal power series $1 - x + 0x^2 + 0x^3 + \ldots$. It is the (unique) formal power series $G(x)$ such that $(1 - x)G(x) = 1$. So, the established fact that $G(x) - xG(x) = 1$ implies that $G(x)$ is the multiplicative inverse of $1 - x$, and so we denote it $\frac{1}{1 - x}$.

Essentially, $\frac{1}{1 - x}$ is defined by the property that $(1 - x) \cdot \frac{1}{1 - x} = 1$.

Theo Bendit
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  • Thanks for your explanation without assuming much abstract algebra background. IMHO the key part here the "multiplicative inverse" which is said in mcs.pdf 16.2.1. But you added the preface about what $x$ is and the info about identity which are very helpful. 2. Sorry, I missed noticing mcs.pdf has 16.5 about Formal Power Series. I will rethink about your answer after reading that.
    • – An5Drama May 18 '24 at 05:34