March 2022 > Math

# Wilf's Generatingfunctionology: Section 1.2

Section 1.1
This is discussion on section 1.2. of Herbert S. Wilf's book Generatingfunctionology.

This section is a work-through of the following recurrence relation: $a_{n+1} = 2a_n + n \qquad \qquad a_0=1, n\geq 0\,.$ As with the previous section, we'll do it the "generating-function" way and then explore what the sequence space version looks like.

The principal difference between this calculation and the last one is a bit more technical difficulty. Again, we take combine the set of recurrence relations $$\{a_{n+1}=2a_n+n:n\geq 0\}$$ into a summation: \begin{align*} \sum_{n=0}^\infty a_{n+1}x^n &= \sum_{n=0}^\infty (2a_n+n)x^n \\ &=2\sum_{n=0}^\infty a_nx^n + \sum_{n=0}^\infty nx^n\,. \end{align*} Here we're using convergence of the sum $$\sum_{n=0}^\infty a_n x^n$$, for some radius of convergence, which can be established easily enough by proving (inductively) the bound $$a_n \leq 3^n$$. For $$f(x)=\sum_{n=0}^\infty a_nx^n$$, the left-hand side is $$(f(x)-a_0)/x = (f(x)-1)/x$$, and the right-hand side $$f(x)+\sum_{n=0}^\infty nx^n$$. The last sum is fairly common. Wilf uses the usual derivative trick, but to avoid the excess machinery (I would feel compelled to bring up the relevant theorem regarding differentiation of infinite sums), I will do it with more elementary tools: \begin{align*} \sum_{n=0}^N nx^n = \sum_{n=1}^N nx^n &= \sum_{n=1}^N \sum_{m=1}^n x^n \\ &= \sum_{m=1}^N \sum_{n=m}^N x^n \\ &= \sum_{m=1}^N \frac{x^{N+1}-x^m}{1-x} = \frac{Nx^{N+1}}{1-x}-\frac{1}{1-x}\sum_{m=1}^N x^m\,. \end{align*} Thus $\sum_{n=0}^\infty nx^n=\lim_{N\to\infty} \left(\frac{Nx^{N+1}}{1-x}-\frac{1}{1-x}\sum_{m=1}^N x^m\right)=\frac{x}{(1-x)^2}\,.$

With this, our original recurrence sum turns into $$(f(x)-1)/x = 2f(x)+x/(1-x)^2$$. Solving for $$f$$ yields the generating function for $$a_n$$: $f(x)=\frac{1-2x+2x^2}{(1-x)^2(1-2x)} = \frac{-1}{(1-x)^2} + \frac{2}{1-2x}\,.$ In case you need a quick refresher, here's the justification for the "quick way" that Wilf mentions for partial fraction decomposition:

$$f(x)$$ is meromorphic on $$\mathbb{C}$$ with poles at $$x=1$$ and $$x=1/2$$, of degree 1 and 2, respectively. If we can remove poles of $$f$$ by subtracting a function $$g$$ such that $$\lim_{z\to\infty} g(z)=0$$, then $$f-g$$ is a bounded entire function by Liouville. Since $$\lim_{z\to\infty} f(z)-g(z)=0$$, we would have $$f=g$$. It is easy to do this when we express $$f$$ as a Laurent series. Indeed, the usual residue calculations yield \begin{align*} f(x)&=\frac{-1}{(x-1)^2}+\frac{0}{x-1}+\ldots \\ f(x)&=\frac{-1}{x-1/2}+\ldots \end{align*} so that $$f(x)=g(x)=\frac{-1}{(x-1)^2}-\frac{1}{x-1/2}$$.

After this, we write \begin{align*} f(x)&= \frac{-1}{(1-x)^2} + \frac{2}{1-2x} \\ &= -\sum_{n=0}^\infty (n+1)x^n + 2\sum_{n=0}^\infty (2x)^n &\text{First sum from above} \\ &= \sum_{n=0}^\infty x^n(2^{n+1}-n-1)\,. \end{align*} We have then recovered our series expression, $$a_n=2^{n+1}-n-1$$.

Next, we'll consider the "sequence space" method. See section 1.1 for the full exposition on this method. Let $$A=(\ldots,0,0,[a_0],a_1,a_2,\ldots)$$ be the sequence in question (recall from part 1.1 that we use brackets [] to distinguish the zero'th element of the bidirectionally-infinite series $$A$$). Then the relation $$a_{n+1}=2a_n+n$$ becomes $\mathcal{L}(A) = 2A+(\ldots,0,1,[0],1,2,3,\ldots)\,.$ Here the 1 to the left of [0] is to deal with the 0th element of $$A$$, which instead of being cut-off by the left-shift operator (as we might wish) is shifted to the -1 position instead. Rearranging, we have $\mathcal{L}(A)-2A=(\ldots,0,1,[0],1,2,3,\ldots) \iff A\star (\ldots, 1, [-2], 0,\ldots)=(\ldots,0,1,[0],1,2,3,\ldots)\,.$ In our last post, we computed the convolution-inverse $$(\ldots, 1,[-2],0,\ldots)^{-1}=(\ldots,[0],1,2,4,\ldots)$$ so that $A=(\ldots,0,1,[0],1,2,3,\ldots)\star (\ldots,[0],1,2,4,\ldots)$ and \begin{align*} a_k&=2^{k}+\sum_{n=0}^{k-1} n 2^{k-n-1} \\ &=2^k + 2^{k-1}\sum_{n=0}^{k-1}n (1/2)^n \\ &=2^k+2^k-k-1 = 2^{k+1}-k-1\,. \end{align*} Here we used our expression for $$\sum_{n=0}^N nx^n$$ found above.

Curiously, the sequence version only goes beyond the problem in section 1.1 in the form of $$\sum_{n=0}^N nx^n$$. The generating-function method requires this and an additional partial fraction decomposition.