Backward Difference Operator

DEFINITION 11.2.10 (First Backward Difference Operator)   The FIRST BACKWARD DIFFERENCE OPERATOR, denoted by $\nabla,$ is defined as

$$\nabla f(x) =f(x)-f(x-h).$$

Given the step size $h,$ note that this formula uses the values at $x$ and $x- h,$ the point at the previous step. As it moves in the backward direction, it is called the backward difference operator.

DEFINITION 11.2.11 ( $r^{\mbox{th}}$ Backward Difference Operator)   The $r^{\mbox{th}}$ backward difference operator, $\nabla^r,$ is defined as

$$\nabla^{r} f(x) = \nabla^{r-1} f(x)- \nabla^{r-1}f(x-h), \qquad\qquad r=1,2, \ldots,$$

$${\mbox{with }} \nabla^0 f(x)=f(x).$$

In particular, for $x=x_k,$ we get

$$\nabla y_k =y_k - y_{k-1}\; {\mbox{ and }} \; \nabla^2 y_k =y_k -2y_{k-1}+y_{k-2}.$$

Note that $\nabla^2 y_k = \Delta^2 y_{k-2}.$

EXAMPLE 11.2.12   Using the tabulated values in Example 11.2.5, find $\nabla y_4$ and $\nabla ^3 y_3.$

Solution: We have $\nabla y_4=y_4-y_3 = 0.49-0.35=0.14,$ and

$$\nabla ^3 y_3 = \nabla ^2 y_3 -\nabla^2 y_2=(y_3-2y_2+y_1)-(y_2-2y_1+y_0)$$

$$= y_3-3y_2+3y_1-y_0$$

$$= 0.35-3\times 0.26+3\times 0.11-0.05 =-0.15.$$

EXAMPLE 11.2.13   If $f(x)=x^2+ax+b,$ where $a$ and $b$ are real constants, calculate $\nabla^r f(x).$

Solution: We first calculate $\nabla f(x)$ as follows:

$$\nabla f(x) = f(x)-f(x-h)=\left[x^2 + ax +b \right]- \left[ (x-h)^2+a(x-h)+b \right]$$

$$= 2xh-h^2 + ah.$$

Now,

$$\nabla^2f(x) = \nabla f(x)-\Delta f(x-h) = [2xh-h^2 + ah]-[2(x-h)h- h^2+ ah] = 2 h^2,$$

$${\mbox{and }} \hspace{0.15in} \nabla ^3f(x) = \nabla^2 f(x)-\nabla^2 f(x) = 2 h^2- 2 h^2=0.$$

Thus, $\nabla ^r f(x)=0$   for all $\,\, r \geq 3.$

Remark 11.2.14   For a set of tabular values, backward difference table in the horizontal form is written as:

$x_0$	$y_0$
$x_{1}$	$y_{1}$	$\nabla y_{1}= y_1 - y_{0}$
$x_2$	$y_2$	$\nabla y_2= y_2 - y_1$	$\nabla^2 y_2 = \nabla y_2 - \nabla y_1$
&vellip#vdots;
$x_{n-2}$	$y_{n-2}$	$\cdots$	$\cdots$
$x_{n-1}$	$y_{n-1}$	$\nabla y_{n-1} = y_{n-1} - y_{n-2}$	$\cdots$	$\cdots$
$x_n$	$y_n$	$\nabla y_n = y_n - y_{n-1}$	$\nabla^2 y_n = \nabla y_n - \nabla y_{n-1}$	$\cdots$	$\nabla^n y_n = \nabla^{n-1} y_n - \nabla^{n-1} y_{n-1}$

EXAMPLE 11.2.15   For the following set of tabular values $(x_i,y_i),$ write the forward and backward difference tables.

$x_i$	9	10	11	12	13	14
$y_i$	5.0	5.4	6.0	6.8	7.5	8.7

Solution: The forward difference table is written as

$x$	$y$	$\Delta y$	$\Delta^2 y$	$\Delta^3 y$	$\Delta^4 y$	$\Delta^5 y$
9	5	0.4 = 5.4 - 5	0.2 = 0.6 - 0.4	0= 0.2-0.2	-.3 = -0.3 - 0.0	0.6 = 0.3 - (-0.3)
10	5.4	0.6	0.2	-0.3	0.3
11	6.0	0.8	-0.1	0.0
12	6.8	0.7	-0.1
13	7.5	0.6
14	8.1

In the similar manner, the backward difference table is written as follows:

$x$	$y$	$\nabla y$	$\nabla^2 y$	$\nabla^3 y$	$\nabla^4 y$	$\nabla^5 y$
9	5
10	5.4	0.4
11	6	0.6	0.2
12	6.8	0.8	0.2	0.0
13	7.5	0.7	-0.1	- 0.3	-0.3
14	8.1	0.6	-0.1	0.0	0.3	0.6

Observe from the above two tables that $\;\Delta^3 y_1 =\nabla^3 y_{4}, \;\Delta^2 y_3 =\nabla^2 y_{5}$ , $\;\Delta^4 y_1 =\nabla^4 y_{5}$ etc.

EXERCISE 11.2.16  

1. Show that $\Delta^3 y_4 =\nabla^3 y_7.$
2. Prove that $\Delta(\nabla y_k) = \Delta^2 y_{k+1} = \nabla^2 y_{k-1}.$
3. Obtain $\nabla^k y_k$ in terms of $y_0, y_1,y_2,\ldots,y_k.$ Hence show that $\;\nabla^k y_k = \Delta^k y_0.$

Remark 11.2.17   In general it can be shown that $\;\Delta^k f(x) =\nabla^k f(x+kh)$ or $\;\Delta^k y_m =\nabla^k y_{k+m}$

Remark 11.2.18   In view of the remarks (11.2.8) and (11.2.17) it is obvious that, if $y=f(x)$ is a polynomial function of degree $n,$ then $\nabla^n f(x)$ is constant and $\nabla^{n+r}f(x)= 0$ for $r> 0.$