The Art of Operators: Difference, Differentiation, and Bernoulli Numbers

算符的艺术：差分、微分与伯努利数 苏剑林 · 2014-10-27 · Mathematics operator operator difference

Two years ago I wrote two posts called Operators and Linear Ordinary Differential Equations, giving a brief introduction to turning linear ordinary differential equations into operator form, and then solving them by inverting the operator. In this post I want to introduce a related topic in the same spirit: the difference operator, the differentiation operator, and the Bernoulli numbers connected to them.

Let us write $D=\frac{d}{dx}$, so that $Df=\frac{df}{dx}$, and also define $\Delta_t f(x)=f(x+t)-f(x)$, denoting $\Delta \equiv \Delta_1 =f(x+1)-f(x)$; here all the functions $f(x)$ we study are assumed to be well-behaved. We know that the Taylor expansion of $f(x+t)$ around $t=0$ is

$$\begin{aligned}f(x+t)&=f(x) + \frac{df(x)}{dx}t + \frac{1}{2!}\frac{d^2 f(x)}{dx^2}t^2 + \frac{1}{3!}\frac{d^3 f(x)}{dx^3}t^3 + \dots\\ &=\left(1+t\frac{d}{dx}+\frac{1}{2!}t^2\frac{d^2}{dx^2}+\dots\right)f(x)\\ &=\left(1+tD+\frac{1}{2!}t^2 D^2+\dots\right)f(x)\end{aligned}$$more

If we ignore for a moment that $D$ is an operator and just treat it as an ordinary number, then the "function" in the parentheses is exactly the expansion of the exponential function! So

$$f(x+t)=\exp\left(tD\right)f(x)\tag{1}$$

In this way we've packed the Taylor expansion into a single compact expression—and as we'll see below, this expression carries much more meaning than it first appears. You might ask: how do we actually compute $\exp\left(tD\right)$? Let's first think back to how we compute $e^x$ for an ordinary real number $x$. Of course, if $x=2$, we just multiply two copies of $e$ together. But how do we compute $e$ in the first place? In fact, $e^x$ can be regarded as being defined by its series expansion (this is arguably the most convenient definition). By the same token, the exponential of an operator, $\exp\left(tD\right)$, is computed via its (formal) series expansion — in other words, $\exp\left(tD\right)$ is simply a notation for $1+tD+\frac{1}{2!}t^2 D^2+\dots$. However, the arithmetic of operators shares many similarities with the arithmetic of numbers; the special feature of operators is their non-commutativity, but many other properties still carry over unchanged. Moreover, since here $D$ is a constant with respect to $t$, $D$ and $t$ do commute, so the symbol $\exp(tD)$ behaves almost exactly like the real-valued function $e^x$.

Note that, from $(1)$, we have

$$\Delta_t f(x)=f(x+t)-f(x)=\left[\exp\left(tD\right)-1\right]f(x)$$

Looking only at the operator part, we get

$$\Delta_t=\exp\left(tD\right)-1\tag{2}$$

This is the connection between difference and differentiation (the relation between the discrete and the continuous), and how elegant and profound it is! In particular, we have

$$\Delta=\exp\left(D\right)-1\tag{3}$$

Equation $(2)$ is not merely a formal curiosity—it has real practical value. We know that the inverse operation of differentiation is integration; integrals aren't always easy to compute, but we've already accumulated a great deal of knowledge about them. For a general $g(x)$, however, if we want to find $f(x)$ such that $\Delta f=g$, there is no general method to do so. For example, if $g(x)=(x+1)^m$, then solving for $\Delta f=g$ is equivalent to solving for $f(n)=1+2^m+\dots+n^m$, which is precisely the problem of summing a sequence.

From $\Delta f=g$, we have

$$f=\Delta^{-1} g$$

The question is: what is $\Delta^{-1}$? This is exactly where equation $(3)$ comes into play. Using equation $(3)$, we get

$$\Delta^{-1}=\frac{1}{\exp\left(D\right)-1}$$

We still don't know what the expression on the right-hand side is—but we can work it out via a Taylor series expansion! Indeed, we have

$$\begin{aligned}\Delta^{-1}&=\frac{1}{\exp\left(D\right)-1}=\frac{1}{D}\left(\frac{D}{\exp\left(D\right)-1}\right)\\ &=\frac{1}{D}\sum_{n=0}^{\infty}\frac{B_n}{n!}D^n\\ &=D^{-1}-\frac{1}{2}+\frac{1}{12}D-\frac{1}{720}D^3+\frac{1}{30240}D^5-\dots\end{aligned}$$

In this way, the problem of solving a difference equation is turned into a problem about an infinite series, where $B_n$ is precisely the $n$-th Bernoulli number, and $\frac{x}{e^x-1}$ is exactly the generating function of the Bernoulli numbers.

For $g(x)=(x+1)^m$, taking derivatives eventually reduces everything to zero after some term, so the series terminates and is finite. For example, for $m=2$, we find

$$\left\{\begin{aligned}&D^{-1} g(x)=\frac{1}{3}(x+1)^3+C\\ &g(x)=(x+1)^2\\ &D g(x)=2(x+1)\\ &D^3 g(x)=0 \end{aligned}\right.$$

and therefore

$$\begin{aligned}f(x)&=\frac{1}{3}(x+1)^3+C-\frac{1}{2}(x+1)^2+\frac{1}{6}(x+1)\\ &=C+\frac{x^3}{3}+\frac{x^2}{2}+\frac{x}{6}\end{aligned}$$

Substituting $x=0$ then gives us $C=0$.

For more details, see the Wikipedia article:

http://zh.wikipedia.org/zh-cn/Bernoulli numbers