February 2022 > Math

# Wilf's Generatingfunctionology: Intro and Section 1.1

Section 1.2

### Introduction

This series will be a read-through of Herbert S. Wilf's short (text?)book Generatingfunctionology, third edition. The book is indeed all about generating functions, and since I did not receive a formal introduction to the topic at the University of Minnesota, I thought it a decent target to clean out the math-brain cobwebs since graduation.

The book appears to fall cleanly in the "combinatorics" category. Leafing through the pages, it is clear that it is accessible (at least in bulk) to anyone with a few university math classes under their belt.

I plan to make this series most readable to those with at least an undergraduate degree (in math).

### Introduction

Generating functions are introduced in the context of a sequence $$(a_n)_{n=1}^\infty$$. Presented as a kind of stand-in for an explicit formula $$a_n=\ldots$$, Wilf lists a bunch of ways that generating functions can be useful. The basic idea is "generating functions are useful for proving things," which is all the "real-world applicability" that I need to hear. But let's cut to the chase. What is a generating function?

(Generating Function) A generating function $$f$$ for a sequence $$(a_n)_{n=0}^\infty$$ is a complex-valued function on $$\mathbb{C}$$ so that $$f(x)=\sum_{n=0}^\infty a_n x^n$$.

Here Wilf pulls the age-old trick of not specifying the domain of either $$f$$ or $$(a_n)_n$$. I believe the intent is to allow readers to specify the most general domain that they are comfortable with. I'll attempt to keep intentional ambiguity to a minimum. (But hey, we'll give Wilf a break; it's just the first chapter). EDIT: So it appears that it is standard to view $$f$$ as a formal power series. Probably should have guessed this, but it won't matter for this introductory post.

The idea of a generating function is immediately attractive. After all, the much-revered "explicit form" $$a_n = f(n)$$ for some hopefully well-behaved function $$f$$ is a similar embedding of $$(a_n)_n$$ into a function space. It is often the case that $$f$$ has an unusually short description, at least in the context of a first-year university student's homework. A generating function is a non-trivial embedding of $$(a_n)_n$$ into a function space, which, if you know anything about power series, should at least feel more natural than the usual explicit form. Indeed, since we know a power series is smooth within its radius of convergence, even relatively fast-growing sequences have smooth generating functions. Moreover, power series in $$\mathbb{C}$$ are holomorphic functions, and that is one heck of a place to land for your back-of-the-napkin scribblings! Right away we can see that this embedding is at the very least a group-homomorphism under addition.

Anyway, before getting into it (which, going by Wilf, is to jump into some examples), it's worth at least mentioning the origin of the power series. To be precise, let's say $$\varphi_0((a_n)_n)$$ is the "embedding" operator that yields the generating function $$f_0$$ for $$(a_n)_n$$ defined above. Let $$\varphi_c((a_n)_n)$$ yield the function $$f_c$$ that satisfies $f_c(x)=\sum_{n=0}^\infty a_n (x-c)^n\,.$ The question is whether $$f_c$$ is any different than $$f_{c'}$$, or whether $$c=0$$ is canonical in some way. A moment's thought (or maybe a little longer, if you're as slow as me) shows that this question does not contain a lot of content. Any holomorphic function $$f$$ is the embedding $$\varphi_0((a_n)_n)$$ for some suitable sequence, and since holomorphic functions form quite a broad class, we can't expect $$f_0$$ and $$f_{c'}$$ to be related other than being simple translates of each other. Moreover, the relationship between $$f_0$$ and $$f_c$$ is so un-mysterious that it is clear that, convenience aside, the choice of origin doesn't matter in the slightest.

But while we're on the topic of complex analysis, we know that a generating function $$f=\varphi_0((a_n)_n)$$ extends to a holomophic function on some domain $$D\subset\mathbb{C}$$. Say $$(a_n)_n\sim (b_n)_n$$ if there exists some $$z\in D$$ so that $$f'=\varphi_z((b_n)_n)$$ coincides with $$f$$ within its disk of convergence. By the identity principle, we know that $$f'=f$$ so that $$\sim$$ is an equivalence relation, and indeed complex sequences that grow sufficiently slowly have classes corresponding to the unique holomophic function which they describe(!). Before I get lost in this rabbit hole (perhaps recalling some tidbit from my complex analysis days which makes this observation embarrassingly trivial), let's get a move-on to our first example.

### Section 1.1

Wilf asks us to find an explicit form for the recurrence relation $\begin{cases} a_{n+1} = 2a_n + 1 \\ n\geq 0,\,\, a_0 = 0 \end{cases}\,.$ We are to suppose that $$f(x)=\sum_{n=0}^\infty a_n x^n$$ converges when $$|x|<r$$ for sufficiently small, positive $$r$$. Wilf chooses to ignore convergence issues altogether (probably because it won't be an issue with the introduction of formal power series), but it isn't particularly difficult to show that $$f$$ converges (uniformly) on such a disk. It is clear that $$a_n$$ is non-negative for all $$n$$, and in that case $$a_{n+1}\geq 3a_n$$ whenever $$n\geq 1$$. In particular, the sequence $$\{ b_0=1, \,\, b_{n+1}=3b_n \}$$ is element-wise strictly greater than $$a_n$$. Since we know the explicit form of $$(b_n)_n$$ to be $$b_n=3^n$$, we can see for $$|x|<1/6$$ $\left|f(x)\right|= \left| \sum_{n=0}^\infty a_nx^n \right| \leq \sum_{n=0}^\infty \left|a_n\right|x^n \leq \sum_{n=0}^\infty (1/2)^n=2 \,.$ The way to proceed here is slightly tricky, but once you know it you won't forget: \begin{align*} a_{n+1}=2a_n+1&\implies \sum_{n=0}^\infty a_{n+1}x^n=\sum_{n=0}^\infty (2a_n+1)x^n &\text{for all }|x|\text{ small} \\ &\implies (1/x)\sum_{n=1}^\infty a_nx^n=2\sum_{n=0}^\infty a_nx^n + \sum_{n=0}^\infty x^n \\ &\implies f(x)/x=2f(x)+1/(1-x) \\ &\implies f(x)=\frac{x}{(1-x)(1-2x)}\,. \end{align*} You can perhaps tell right away that all the convergence issues are not potent since we have an exponential bound for $$a_n$$. Here, we just showed that if there is some sequence $$(a_n)_n$$ satisfying the aforementioned recurrence relation, AND if $$f$$ is a well-defined and exists, then $$f(x)=\frac{x}{(1-x)(1-2x)}$$. Hence, we have found our generating function.

NB: Wilf, again deploying another masterful move, completely ignores the "issues" of whether $$f$$ is well-defined or $$(a_n)_n$$ exists. This is a genius move because 1) they are both true, and 2) Wilf doesn't have to simultaneously sound like an idiot and go over the heads of his readers by proving these two facts.

More manipulation reminiscent of your calc I (or intro complex analysis?) days provides a path to an explicit form: \begin{align*} \sum_{n=0}^\infty a_n x^n=f(x)=\frac{x}{(1-x)(1-2x)}&=x\left(\frac{-1}{1-x}+\frac{2}{1-2x}\right) \\ &= -x\sum_{n=0}^\infty x^n + 2x\sum_{n=0}^\infty (2x)^n \\ &= \sum_{n=1}^\infty (2^n-1)x^n\,. \end{align*} Thus we have $$a_n = 2^n -1$$.

### Discrete Convolution and Shift Operators

Now this calculation was not especially painful. But before putting this to bed, I find it extremely instructive to "pull-back" operations done in some sort of dual space back to the origin space. This serves a few purposes. First it shows when a map to a dual space is not very deep. This is the case when the pull-back operation is something you should have seen but didn't. Secondly, it has the possibility of expanding your knowledge of the original space, by porting features of the dual space. Here, we will rediscover the discrete convolution and relate it to the "shift" operator.

Recall our first step above was a summation: $$a_{n+1}=2a_n+1\implies \sum_n a_{n+1}x^n=2\sum_n a_n x^n+\sum_n x^n$$. Broadly speaking, this doesn't really accomplish anything; it takes the implicit "for all $$n$$" contained in $$a_{n+1}=2a_n+1$$ and makes it explicit in a sum. Instead, we can write our recurrence relation element-wise: $\mathcal{L}(A)=2A+\mathbf{1}\,.$ Here $$\mathcal{L}$$ is the left-shift operator taking a sequence $$A=(a_0, a_1, a_2,\ldots)$$ to $$(a_1,a_2,a_3,\ldots)$$ and $$\mathbf{1}$$ is the constant sequence $$(1,1,1,\ldots)$$. We take multiplication and addition to be element-wise. (For a more formal methodology for this "pull-back", apply the map $$\sum_n a_n x^n\mapsto (a_0,a_1,\ldots)$$ to both sides of each step.) Keeping in mind what addition and scalar multiplication mean in $$\mathbb{C}^\infty$$, we can rearrange and "factor" using a unique kind of convolution that we denote $$\star$$: \begin{align*} \mathcal{L}(A)=2A+\mathbf{1} &\implies \mathcal{L}(A)-2A=\mathbf{1} \\ &\implies A\star(-2,1,0,0,\ldots)=\mathbf{1}\,. \end{align*} Our star operator ($$\star$$) satisfies $$A\star B=\sum_{n=0}^\infty b_n \mathcal{L}^{(n)}(A)$$. Unfortunately, $$\star$$ is not associative, meaning we can't apply $$(-2,1,0,0,\ldots)^{-1}$$ to both sides above (if such an inverse even existed). To get a useful (i.e. associative) operator, we fall back to the wonderful (regular-old) discrete convolution. Avoiding strenuous bookkeeping requires extending $$A$$ and $$B$$ to two-sided sequences: $A=(\ldots, a_{-2}, a_{-1}, [a_0], a_1, a_2,\ldots) \\ B=(\ldots, b_{-2}, a_{-1}, [b_0], b_1, b_2,\ldots)\,.$ We surround the zero'th entry with brackets to distinguish its location. Then the discrete convolution $$A\star B$$ is defined $(A\star B)_k = \sum_{n=-\infty}^\infty a_n b_{k-n}\,.$ This operation is commutative and associative -- not to mention distributive over addition -- and we may rewrite $$\mathcal{L}(A)=2A+\mathbf{1}$$ as \begin{align*} \mathcal{L}(A)-2A=\mathbf{1} &\iff (\ldots,0,[a_0],a_1,a_2,\ldots)\star (\ldots, 0, 1,[-2],0,\ldots)=(\ldots,0,[1],1,\ldots) \\ &\iff A\star (\ldots, 0, 1,[-2],0,\ldots)=\mathbf{1}\,. \end{align*} Since $$\star$$ is distributive over addition, we can relate the left shift operator to convolution with $$(\ldots,1,[0],0,\ldots)$$ and scalar multiplication with $$(\ldots,0,[\alpha],0,\ldots)$$. Then it's not too hard to construct at least this particular inverse: $(\ldots, 0, 1,[-2],0,\ldots)\star (\ldots,[0],1,2,4,8,\ldots)=(\ldots,0,0,[1],0,0,\ldots)$ so that \begin{align*} a_k&=((\ldots,[1],1,\ldots)\star(\ldots,0,[0],1,2,4,\ldots))_k \\ &= \sum_{n=0}^{k-1} 2^{k-n-1} = \sum_{n=0}^{k-1} 2^n \\ &= 2^k - 1\,. \end{align*} Interestingly, we encountered no convergence issues at all with this method. In addition, the partial fraction decomposition completely vanished, though this is certainly due to the explicit computation of the convolution-wise inverse. If instead we had expanded $$1/(1-2x)$$ and $$1/(1-x)$$ as power series we would have recovered a similar sum for $$a_k$$ without the decomposition.

Much could be said about the above calculation; here are three starting points:

1. The need to expand $$A$$ and $$B$$ to two-sided infinite sequences is immediately reminiscent of the expansion of a regular (holomorphic) power series into a (meromorphic) Laurent series. Indeed, the expansion is necessary if one wishes to have a multiplicative inverse to $$x^k$$.
2. The duality of convolution on the discrete side with multiplication on the function side is suspiciously similar to some convolution theorems in Fourier analysis.
3. What does partial fraction decomposition look like in terms of sequence convolution?

### Conclusion

For this simple example, the "generating function" method is friendly because the necessary manipulations are familiar. The equivalent computation acting on the sequence itself is not more difficult. Moreover, I find that the sequence version propels the mind more immediately than the generating function method. Nonetheless, as Wilf acknowledges, this is a particularly simple example amenable to many methods. We'll have to wait and see about the more subtle uses of generating functions.