3.4: Finite Difference Calculus

In this section, we will explore further to the method that we explained in the introduction of Quadratic sequences.

Example \(\PageIndex{1}\)

Create a sequence of numbers by finding a relationship between the number of points on the circumference of a circle and the number of regions created by joining the points.

Table \(\PageIndex{1}\): Dividing circle

Number of points on the circle	0	1	2	3	4	5	6	7	8
Number of regions	1	1	2	4	8	16	31	57	99

Difference operator

Notation:

Let \(a_n, n=0,1,2,\cdots\) be a sequence of numbers. Then the first difference is defined by \( \Delta a_n = a_{n+1}-a_n, n=0,1,2,\cdots\).

The second difference is defined by \(\Delta (\Delta a_n)= \Delta^2 a_n =\Delta a_{n+1}- \Delta a_n, n=0,1,2,\cdots\).

Further, \(k^{th}\) difference is denoted by \(\Delta^k a_n, n=k,\cdots\). We shall denote that \(\Delta^0 a_n=a_n\).

Example \(\PageIndex{2}\)

Let \(a_n=n^2\), for all \(n \in \mathbb{N}\). Show that \(\Delta^2 a_n=2\).

Solution

\(\Delta a_n =(n+1)^2-n^2= 2n+1,\)

\(\Delta^2 a_n=(2(n+1)+1)-(2n+1)= 2.\)

Rule

Let \(k \in \mathbb{Z_+}\). Let \(a_n=n^k\), for all \(n \in \mathbb{N}\). Then \(\Delta^k a_n=k! \).

Example \(\PageIndex{3}\)

Let \(a_n=c^n\), for all \(n \in \mathbb{N}\). Show that \(\Delta a_n=(c-1) a_n\).

Solution

\(\Delta a_n= c^{n+1}-c^{n}= c^n(c-1) = (c-1) a_n\).

Below are some properties of the difference operator.

Theorem \(\PageIndex{1}\): Difference operator is a Linear Operator

Let \(a_n\) and \(b_n\) be sequences, and let \(c\) be any number. Then

1. \( \Delta (a_n+b_n)= \Delta a_n+ \Delta b_n\), and
2. \( \Delta ( c a_n) =c \Delta a_n \).

Proof

Very simple calculation.

Definition: Falling Powers

The falling power is denoted by \(x^{\underline{m}} \) and it is defined by \( x (x-1)(x-2)(x-3) \cdots (x-m+1)\), with \(x^{\underline{0}}=1\).

Falling powers are useful in the difference calculus because of the following property:

Theorem \(\PageIndex{2}\)

\( \Delta n^{\underline{m}} =m \,n^{\underline{m-1}} \).

Proof

\( \Delta n^{\underline{m}}= (n+1)^{\underline{m}}-n^{\underline{m}} \)

\(=(n+1) n (n-1)(n-2)(n-3) \cdots (n-m+2)- n (n-1)(n-2)(n-3) \cdots (n-m+1)\)

\(=n (n-1)(n-2)(n-3) \cdots (n-m+2) ( (n+1)-(n-m+1))\)

\(= n (n-1)(n-2)(n-3) \cdots (n-m+2) \,m\)

\(=m\, n^{\underline{m-1}} \).

Let us now consider the sequence of numbers in the example \(\PageIndex{1}\).

Table \(\PageIndex{2}\): Falling Powers

Number of points on the circle	0	1	2	3	4	5	6	7	8
Number of regions	1	1	2	4	8	16	31	57	99
\(\Delta a_n\)		0	1	2	4	8	15	26	42
\(\Delta^2 a_n\)			1	1	2	4	7	11	16
\(\Delta^3 a_n\)				0	1	2	3	4	5
\(\Delta^4 a_n\)					1	1	1	1	1

Since the fourth difference is constant, \(a_n\) should be polynomial of degree \(4\). Let's explore how to find this polynomial.

Definition

\[ {n \choose k } =\displaystyle \frac{ n!}{k! (n-k)!} , k \leq n, n \in \mathbb{N} \cup \{0\} .\]

Theorem \(\PageIndex{2}\)

\[\Delta \displaystyle {n \choose k } = \displaystyle {n \choose k-1}, k \leq n, n \in \mathbb{N} \cup \{0\} .\]

Proof

Note that \[ \displaystyle {n \choose k }= \frac{1}{k!} n^{\underline{k}}.\]

Consider \( \Delta {n \choose k } = \Delta ( \frac{1}{k!} n^{\underline{k}}) \)

\(= \frac{1}{k!} \Delta( n^{\underline{k}})\)

\( = \frac{k}{k!} n^{\underline{(k-1)}}\)

\(= \frac{1}{(k-1)!} n^{\underline{(k-1)}}\)

\(= {n \choose k-1}\).

We can also see this by considering Pascal’s triangle.

Theorem \(\PageIndex{3}\) Newton's formula

Let \(a_0, a_1, a_2, \cdots \) be sequence of numbers such that \(\Delta^{k+1} a_0=0. \) Then the \(n\) ^th term of the original sequence is given by

\[ a_n= \frac{1}{k!}(\Delta^k a_0)n^{\underline{k}} + \cdots+ a_0= a_0+ {n \choose 0 } \Delta a_1 +{n \choose 2 } \Delta^2 a_2+ \cdots + {n \choose k } \Delta^k a_k. \]

Proof

Exercise.

If \(n\) ^th term of the original sequence is linear, then the first difference will be a constant. If \(n\) ^th term of the original sequence is quadratic, then the second difference will be a constant. A cubic sequence has the third difference constant.

Source

• Thanks to Olivia Nannan for the diagram.
• Reference: Kunin, George B. "The finite difference calculus and applications to the interpolation of sequences." MIT Undergraduate Journal of Mathematics 232.2001 (2001): 101-9.
• Reference: Samson, D. (2006). Number patterns, cautionary tales and finite differences. Learning and Teaching Mathematics, 2006(3), 3-8.