Two years ago, I wrote two articles titled "Operators and Linear Ordinary Differential Equations", briefly introducing the operatorization of linear ordinary differential equations and the method of finding general solutions by inverting the operators. In this article, I intend to introduce similar content regarding operators: difference operators, differential operators, and the related Bernoulli numbers.
We denote D = \frac{d}{dx}, so Df = \frac{df}{dx}. At the same time, we define \Delta_t f(x) = f(x+t) - f(x), and denote \Delta \equiv \Delta_1 = f(x+1) - f(x). Here, the functions f(x) we study are all 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} If we ignore the fact that D is an operator and treat it as a number, then the "function" in the parentheses is precisely the expansion of the exponential function! Thus: f(x+t) = \exp(tD)f(x) \tag{1} In this way, we have expressed the Taylor expansion in a concise formula. As we will see later, this formula is deeply significant.
Readers might ask how to calculate \exp(tD)? Let us reflect on how the ordinary real number e^x is calculated. Of course, if x=2, we just need to multiply two e’s. But how is e itself calculated? In fact, e^x can be considered defined by its power series (this definition is perhaps the most convenient). Similarly, for the exponential function of an operator \exp(tD), it is calculated using its series expansion (formal power series). That is to say, \exp(tD) is merely a notation for 1 + tD + \frac{1}{2!}t^2 D^2 + \dots. However, the operations of operators have many similarities with the operations of numbers. The uniqueness of operators lies in their non-commutativity, but many properties remain unchanged. Moreover, for D here, t is a constant, so D and t are commutative. Therefore, the properties of the symbol \exp(tD) are almost identical to those of the real function e^x.
Note that according to (1), we have: \Delta_t f(x) = f(x+t) - f(x) = [\exp(tD) - 1]f(x) Looking only at the operator part, we have: \Delta_t = \exp(tD) - 1 \tag{2} This is the connection between difference and differential (the relationship between discrete and continuous)—how concise and profound! In particular, we have: \Delta = \exp(D) - 1 \tag{3}
Equation (2) has not only formal significance but also practical value. We know that the inverse operation of differentiation is integration. While calculating integrals is not always easy, we have accumulated a vast amount of knowledge about them. However, for a general g(x), if we want to find f(x) such that \Delta f = g, there is no general method. For example, if g(x) = (x+1)^m, then solving \Delta f = g is equivalent to solving f(n) = 1 + 2^m + \dots + n^m, which is an application of series summation.
According to \Delta f = g, we have: f = \Delta^{-1} g The question is, what is \Delta^{-1}? This is where equation (3) comes into play. According to (3), we have: \Delta^{-1} = \frac{1}{\exp(D) - 1} What is the expression on the right-hand side? We can simply use the Taylor series expansion to calculate it! In fact, we have: \begin{aligned} \Delta^{-1} &= \frac{1}{\exp(D) - 1} = \frac{1}{D} \left( \frac{D}{\exp(D) - 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 transformed into a problem of infinite series. Here, B_n is the n-th Bernoulli number, and \frac{x}{e^x - 1} is precisely the generating function of Bernoulli numbers.
For g(x) = (x+1)^m, the derivatives will eventually become zero after a certain term, so the series is finite. For example, if m=2, we have: \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. 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, we find C=0.
For more information, please refer to Wikipedia:
https://en.wikipedia.org/wiki/Bernoulli_number
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