Forward Difference Operator

DEFINITION 11.2.1 (First Forward Difference Operator)  

We define the FORWARD DIFFERENCE OPERATOR, denoted by $\Delta,$ as

$$\Delta f(x)=f(x+h)-f(x).$$

The expression $f(x+h) - f(x)$ gives the FIRST FORWARD DIFFERENCE of $f(x)$ and the operator $\Delta$ is called the FIRST FORWARD DIFFERENCE OPERATOR. Given the step size $h,$ this formula uses the values at $x$ and $x+h,$ the point at the next step. As it is moving in the forward direction, it is called the forward difference operator.

DEFINITION 11.2.2 (Second Forward Difference Operator)   The second forward difference operator, $\Delta^2,$ is defined as

$$\Delta^2 f(x) = \Delta \bigl( \Delta f(x) \bigr) = \Delta f(x+h) - \Delta f(x).$$

We note that

$$\Delta^2 f(x) = \Delta f(x+h)- \Delta f(x)$$

$$= \bigl( f(x+2h)- f(x+h) \bigr) - \bigl( f(x+h) - f(x) \bigr)$$

$$= f(x+ 2 h) - 2 f(x+h) + f(x).$$

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

$$\Delta y_k=y_{k+1}-y_k$$

and

$$\Delta^2 y_k=\Delta y_{k+1}-\Delta y_k = y_{k+2}-2y_{k+1} + y_{k}.$$

DEFINITION 11.2.3 ( $r^{\mbox{th}}$ Forward Difference Operator)   The $r^{\mbox{th}}$ forward difference operator, $\Delta^r,$ is defined as

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

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

EXERCISE 11.2.4   Show that $\Delta^3 y_k = \Delta^2 (\Delta y_k) = \Delta (\Delta^2 y_k).$ In general, show that for any positive integers $r$ and $m$ with $r > m,$

$$\Delta^r y_k = \Delta^{r-m} (\Delta^m y_k) = \Delta^m (\Delta^{r-m} y_k).$$

EXAMPLE 11.2.5   For the tabulated values of $y=f(x)$ find $\Delta y_3$ and $\Delta^3 y_2$

$$\begin{tabular}{c\vert c\vert c\vert c\vert c\vert c\vert c} $i$ ... ... & 0.5 \\ $y_i$ & 0.05 & 0.11 & 0.26 & 0.35 & 0.49 & 0.67 \\ \end{tabular}.$$

Solution: Here,

$$\Delta y_3=y_4-y_3=0.49-0.35=0.14, \;\; {\mbox{ and }}$$

$$\Delta^3 y_2 = \Delta (\Delta^2 y_2) = \Delta ( y_4 - 2 y_3 + y_2)$$

$$= (y_5 - y_4) - 2 ( y_4 - y_3) + (y_3 - y_2)$$

$$= y_5 - 3y_4 + 3y_3 - y_2$$

$$= 0.67 - 3 \times 0.49 + 3 \times 0.35-0.26= - 0.01.$$

Remark 11.2.6   Using mathematical induction, it can be shown that

$$\Delta ^r y_k =\sum\limits_{j=0}^r (-1)^{r-j}{r \choose j } y_{k+j}.$$

Thus the $r^{\mbox{th}}$ forward difference at $y_k$ uses the values at $y_k, y_{k+1},\ldots, y_{k+r}.$

EXAMPLE 11.2.7   If $f(x)=x^2+ax+b,$ where $a$ and $b$ are real constants, calculate $\Delta^r f(x).$

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

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

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

Now,

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

$${\mbox{ and}} \hspace{0.5in} \Delta ^3f(x) = \Delta^2 f(x)-\Delta^2 f(x) = 2 h^2- 2 h^2=0.$$

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

Remark 11.2.8   In general, if $f(x)=x^n+a_1 x^{n-1}+a_2 x^{n-2}+ \cdots+ a_{n-1} x + a_n$ is a polynomial of degree $n,$ then it can be shown that

$$\Delta^n f(x)= n! \, h^n \;\; {\mbox{ and }} \;\; \Delta ^{n+r}f(x)=0 \qquad {\mbox{ for }}\,\, r=1,2,\ldots.$$

The reader is advised to prove the above statement.

Remark 11.2.9  

1. For a set of tabular values, the horizontal forward difference table is written as:

   $x_0$	$y_0$	$\Delta y_0 = y_1 - y_0$	$\Delta^2 y_0 = \Delta y_1 - \Delta y_0$	$\cdots$	$\Delta^n y_0 = \Delta^{n-1} y_1 - \Delta^{n-1} y_0$
   $x_1$	$y_1$		$\Delta^2 y_1 = \Delta y_2 - \Delta y_1$	$\cdots$
   $x_2$	$y_2$	$\Delta y_2= y_3 - y_2$	$\Delta^2 y_2 = \Delta y_3 - \Delta y_2$
   &vellip#vdots;
   $x_{n-1}$	$y_{n-1}$	$\Delta y_{n-1}= y_n - y_{n-1}$
   $x_n$	$y_n$

2. In many books, a diagonal form of the difference table is also used. This is written as:

   $x_0$	$y_0$
   		$\Delta y_0$
   $x_1$	$y_1$		$\Delta^2 y_0$
   		$\Delta y_1$		$\Delta^3 y_0$
   $x_2$	$y_2$		$\Delta^2 y_1$
   &vellip#vdots;								$\Delta y_{n-1}$
   $x_{n-2}$	$y_{n-2}$		$\Delta^2 y_{n-3}$
   		$\Delta y_{n-2}$		$\Delta^3 y_{n-3}$
   $x_{n-1}$	$y_{n-1}$		$\Delta^2 y_{n-2}$
   		$\Delta y_{n-1}$
   $x_n$	$y_n$

However, in the following, we shall mostly adhere to horizontal form only.