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Mathematics LibreTexts

3.4: Finite Difference Calculus

  • Page ID
    13607
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    In this section, we will explore further to the method that we explained at 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.

    circle.png

    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 \(\PageIndex{1}\).

    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 Mathematics2006(3), 3-8.