Introduction to Umbral Calculus

Definition: Discrete Derivative (or Forward Difference Operator)

Given $f:\mathbb{Z}\to \mathbb{R}$, define

$$\Delta f(n):=f(n+1)-f(n)$$

which is “as infinitesimal” as the domain can get.

Examples

$$\begin{array}{rl}\Delta (n)=(n+1)-n& =1\\ \Delta (n^2)=(n+1{)}^2-n^2& =2n+1\\ \Delta (n^3)=(n+1{)}^3-n^3& =3n^2+3n+1\\ ⋮& \\ \Delta (n^k)=(n+1{)}^k-n^k& =k{n}^{k-1}+\mathrm{etc}\end{array}$$

which follow like a Pascal’s triangle. Let $c_k^m$ ($k\le m$) be the coefficient multiplying the $k$-th power of the discrete derivative of $n^m$. Then:

$$\begin{array}{rl}\Delta (n)=1n^0& =c_0^0\\ \Delta (n^2)=2n^1& +1n^0=c_1^1{n}^1+c_0^1{n}^0\\ \Delta (n^3)=3n^2& +3n^1+1n^0=c_2^2{n}^2+c_1^2{n}^1+c_0^2{n}^0\\ \Delta (n^4)=4n^3& +6n^2+4n^1+1n^0=c_3^3{n}^3+c_2^3{n}^2+c_1^3{n}^1+c_0^3{n}^0\\ ⋮& \end{array}$$

Then (for $m>2$)

$$\Delta (n^m)=m{n}^{m-1}+\sum _{l=1}^{m-2}(c_l^{m-1}+c_{l-1}^{m-1})n^l+1$$

Note that it’s the same relation as in the Pascal triangle¹, for which

$$\begin{gathered} \text{pmatrix}m \\ l=\text{pmatrix}m-1 \\ l+\text{pmatrix}m-1 \\ l-1 \end{gathered}$$

Definition: Falling Factorial

For $n,k\in \mathbb{N}$, define

$$n^{\underline{k}}:=\prod _{l=0}^{k-1}(n-l)=n(n-1)(n-2)\dots (n-k+1)$$

Note that

$$n^{\underline{k}}=k!\left(\begin{array}{c}n\\ k\end{array}\right)$$

Examples

$$\begin{array}{rl}{n}^{\underline{0}}& =1\\ n^{\underline{1}}& =n\\ n^{\underline{2}}& =n^2-n\\ ⋮& \\ n^{\underline{k}}& =n^k+\mathrm{etc}\end{array}$$

Theorem 1: Discrete derivative for falling factorials = Calculus for natural powers

For $n,k\in \mathbb{N}$, we have

$$\Delta n^{\underline{k}}=k{n}^{\underline{k-1}}$$

Proof:

$$\begin{array}{rl}\Delta n^{\underline{k}}& =(n+1{)}^{\underline{k}}-n^{\underline{k}}\\ & =(n+1)(n+1-1)(n+1-2)\dots (n+1-k+2)(n+1-k+1)-n(n-1)(n-2)\dots (n-k+1)\\ & =(n+1)\underbrace{n(n-1)(n-2)\dots (n+1-(k-1))}-\underbrace{n(n-1)\dots (n-k+2)}(n-(k-1))\\ & =\underbrace{n(n-1)\dots (n-k+2)}\left[n+1-(n-k+1)\right]\\ & =k{n}^{\underline{k-1}}\end{array}$$

In a more elucidative form:

$$\begin{array}{rl}\Delta n^{\underline{k}}& =(n+1{)}^{\underline{k}}-n^{\underline{k}}\\ & =k!\left(\left(\begin{array}{c}n+1\\ k\end{array}\right)-\left(\begin{array}{c}n\\ k\end{array}\right)\right)\\ & =k!\left(\begin{array}{c}n\\ k-1\end{array}\right)\\ & =k(k-1)!\left(\begin{array}{c}n\\ k-1\end{array}\right)\\ & =k{n}^{\underline{k-1}}\end{array}$$

(Third line is Pascal’s Triangle’s relationship)

Derivative of Exponential Functions

Given $0\ne a\in \mathbb{R}$, we have (let $f:\mathbb{N}\to \mathbb{R}\mid f(n)=a^n$)

$$\Delta a^n=a^{n+1}-a^n=(a-1)a^n$$

Note that it follows that

$$\Delta 2^n=2^n$$

Definition: Discrete Antiderivative

Given $g,G:\mathbb{Z}\to \mathbb{R}$, we say that $G$ is the antiderivative of $g$ if

$$\Delta G(n)=g(n)$$

Denote the antiderivative operator as

$$\sum g(n)=G(n)$$

Corollaries: Antiderivative of falling factorials and real numbers

$$\sum n^{\underline{k}}=\frac{1}{k+1}{n}^{\underline{k+1}}$$ $$\sum a^n=\frac{1}{a-1}{a}^n$$

Example: Fibonacci numbers!

$F_0=0,F_1=1,F_{n+1}=F_n+F_{n-1}$.

Observe that

$$\Delta F_n=F_{n+1}-F_n=F_{n-1}$$

and so it acts as a “step-down” operator.

Likewise, the discrete antiderivative acts as a “step-up” operator, since it yields the function that, when derived, results in $F_n$ (i.e. $F_{n+1}$):

$$\sum F_n=F_{n+1}$$

Theorem 2: Discrete Fundamental Theorem of Calculus

Given $g,G:\mathbb{Z}\to \mathbb{R}$, where $G$ is the antiderivative of $g$. Then

$$\sum _{n=a}^{b}g(n)=G(b+1)-G(a)$$

Proof: It’s essentially a telescopic sum. Note that the same proof sketch appears in the Fundamental Theorem of Calculus, and even in Stokes’ Theorem.

$$\begin{array}{rl}\sum _{n=a}^{b}g(n)& =\sum _{n=a}^b\Delta G(n)\\ & =\sum _{n=a}^b(G(n+1)-G(n))\\ & =(G(b+1)-G(b))+(G(b)-G(b-1))+\cdots +(G(a+1)-G(a))\\ & =G(b+1)-G(a)\end{array}$$

Note that it’s essentially the same as the fundamental theorem of calculus, except for the $+1$, which is analogous to an infinitesimal $dx$: all natural numbers are at least 1 unit apart from each other, which doesn’t really happen in the real numbers (nor in the rationals), since it’s a dense set.

Product Rule for Discrete Derivative

$$\Delta ((f\cdot g)(n))=\Delta f(n)g(n)+f(n)\Delta g(n)+\Delta f(n)\Delta g(n)$$

Proof:

$$\begin{array}{rl}\Delta ((f\cdot g)(n))& =f(n+1)g(n+1)-f(n)g(n)\pm f(n+1)g(n)\\ & =f(n+1)\cdot (g(n+1)-g(n))+(f(n+1)-f(n))\cdot g(n)\\ & =f(n+1)\Delta g(n)+\Delta f(n)g(n)\\ & =(\Delta f(n)+f(n))\Delta g(n)+\Delta f(n)g(n)\\ & =\Delta f(n)g(n)+f(n)\Delta g(n)+\Delta f(n)\Delta g(n)\end{array}$$

Note that the term $\Delta f(n)\Delta g(n)$ appears here, and not in regular Calculus, since there is a $\underset{dx\to 0}{\lim }f(x+dx)\frac{g(x+dx)}{dx}$ in place. In Discrete Calculus, we can only go as close as +1 from any integer, so these terms that would vanish in Standard Calculus don’t vanish here.

Summation by Parts

$$\sum _a^{b}f(n)\Delta g(n)=f(b+1)g(b+1)-f(a)g(a)-\sum _a^b\Delta f(n)g(n+1)$$

Proof:

$$\begin{array}{rl}\sum _a^{b}f(n)\Delta g(n)& =\sum _a^b\left(\Delta (f\cdot g)(n)-\Delta f(n)g(n)-\Delta f(n)\Delta g(n)\right)\\ & =\sum _a^b\Delta (f\cdot g)(n)-\sum _a^b\Delta f(n)\left(g(n)+\Delta g(n)\right)\\ & =f(b+1)g(b+1)-f(a)g(a)-\sum _a^b\Delta f(n)g(n+1)\\ & =[f(n){]}_a^b-\sum _a^b\Delta f(n)g(n+1)\end{array}$$

Note that it’s very similar to its analogue in Differential Calculus, in which

$${\int }_a^{b}f(x)\partial g(x)dx=f(b)g(b)-f(a)g(a)-{\int }_a^b\partial f(x)g(x)dx$$

Applications: Evaluating (some) series!

$$\begin{array}{rl}\sum _1^{N}n& =[n\cdot n{]}_1^{N+1}-\sum _1^{N}1\cdot (n+1)\\ & =(N+1{)}^2-1-\sum _1^N(n+1)\\ & =N^2+N-\sum _1^{N}n\\ \therefore \sum _1^{N}n& =\frac{N(N+1)}{2}\end{array}$$

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Referências

• https://www.youtube.com/watch?v=FNh-Xkfnu1M

Footnotes

1. Check Wikipedia’s article on Pascal’s Triangle. ↩