Divided Differences

DEFINITION 12.2.1 (First Divided Difference)   The ratio

$$\frac{f(x_i) - f(x_j)}{x_i - x_j}$$

for any two points $x_i$ and $x_j$ is called the FIRST DIVIDED DIFFERENCE of $f(x)$ relative to $x_i$ and $x_j.$ It is denoted by $\delta[x_i, x_j].$

Let us assume that the function $y=f(x)$ is linear. Then $\delta[x_i,x_j]$ is constant for any two tabular points $x_i$ and $x_j,$ i.e., it is independent of $x_i$ and $x_j.$ Hence,

$$\delta[x_i, x_j] = \frac{f(x_i) - f(x_j)}{x_i - x_j}= \delta[x_j,x_i].$$

Thus, for a linear function $f(x),$ if we take the points $x, x_0$ and $x_1$ then, $\delta[x_0,x] = \delta[x_0,x_1],$ i.e.,

$$\frac{f(x) - f(x_0)}{x - x_0}= \delta[x_0,x_1].$$

Thus, $f(x) = f(x_0) + (x - x_0) \delta[x_0, x_1].$

So, if $f(x)$ is approximated with a linear polynomial, then the value of the function at any point $x$ can be calculated by using $f(x) \approx P_1(x) = f(x_0) + (x - x_0) \delta[x_0, x_1],$ where $\delta[x_0,x_1]$ is the first divided difference of $f$ relative to $x_0$ and $x_1.$

DEFINITION 12.2.2 (Second Divided Difference)   The ratio

$$\delta[x_i, x_j,x_k] = \frac{\delta[x_j,x_k] - \delta[x_i,x_j]}{x_k - x_i}$$

is defined as SECOND DIVIDED DIFFERENCE of $f(x)$ relative to $x_i, x_j$ and $x_k.$

If $f(x)$ is a second degree polynomial then $\delta[x_0, x]$ is a linear function of $x.$ Hence,

$$\delta[x_i, x_j,x_k] = \frac{\delta[x_j,x_k] - \delta[x_i,x_j]}{x_k - x_i} \;\;{\mbox{ is constant}}.$$

In view of the above, for a polynomial function of degree 2, we have $\delta[x,x_0,x_1] = \delta[x_0,x_1,x_2].$ Thus,

$$\frac{\delta[x,x_0] - \delta[x_0,x_1]}{x - x_1} = \delta[x_0,x_1,x_2].$$

This gives,

$$\delta[x,x_0] = \delta[x_0,x_1] + (x - x_1) \delta[x_0,x_1,x_2].$$

From this we obtain,

$$f(x) = f(x_0) + (x - x_0) \delta[x_0,x_1] + (x - x_0)(x -x_1)\delta[x_0,x_1,x_2].$$

So, whenever $f(x)$ is approximated with a second degree polynomial, the value of $f(x)$ at any point $x$ can be computed using the above polynomial, which uses the values at three points $x_0, x_1$ and $x_2.$

EXAMPLE 12.2.3   Using the following tabular values for a function $y = f(x),$ obtain its second degree polynomial approximation.

$i$	0	1	2
$x_i$	0.1	0.16	0.2
$f(x_i)$	1.12	1.24	1.40

Also, find the approximate value of the function at $x = 0.13.$
Solution: We shall first calculate the desired divided differences.

$$\delta[x_0,x_1] = (1.24 - 1.12)/(0.16 - 0.1) = 2,$$

$$\delta[x_1,x_2] = (1.40-1.24)/(0.2-0.16) = 4, \;\; {\mbox{ and }}$$

$$\delta[x_0,x_1,x_2] = \frac{\delta[x_1,x_2] - \delta[x_0,x_1]}{x_2 - x_0} = (4-2)/(0.2-0.1) = 20.$$

Thus,

$$f(x) \approx P_2(x) = 1.12 + 2(x - 0.1) + 20 (x-0.1)(x-0.16).$$

Therefore

$$f(0.13) \approx 1.12 + 2(0.13 - 0.1) + 20 (0.13-0.1)(0.13-0.16)=1.162.$$

EXERCISE 12.2.4  

1. Using the following table, which gives values of $\log(x)$ corresponding to certain values of $x,$ find approximate value of $\;\;\log(323.5)$ with the help of a second degree polynomial.

   $x$	322.8	324.2	325
   $\log(x)$	2.50893	2.51081	2.5118

2. Show that
   $$\delta[x_0,x_1,x_2] = \frac{f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{f(x_1)}{(x_1-x_0)(x_1-x_2)}+\frac{f(x_2)}{(x_2-x_0)(x_2-x_1)}.$$
   So, $\delta[x_0,x_1,x_2] = \delta[x_0,x_2,x_1] = \delta[x_1,x_0,x_2] = \delta[x_1,x_2,x_0] = \delta[x_2,x_0,x_1] = \delta[x_2,x_1,x_0].$ That is, the second divided difference remains unchanged regardless of how its arguments are interchanged.
3. Show that for equidistant points $x_{0},$ $x_1$ and $x_2,$ $\;\delta[x_0,x_1,x_2] = \displaystyle\frac{\Delta^2 y_0}{2 h^2}=\displaystyle\frac{\nabla^2 y_2}{2 h^2},$ where $y_k = f(x_k),$ and $h = x_1 - x_0 = x_2 - x_1.$
4. Show that for a linear function, the second divided difference with respect to any three points, $x_{i},x_{j}$ and $x_{k},$ is always zero.

Now, we define the $k^{\mbox{th}}$ divided difference.

DEFINITION 12.2.5 ( $k^{\mbox{th}}$ Divided Difference)   The #MATH5307# MATHEND000# DIVIDED DIFFERENCE of $f(x)$ relative to the tabular points $x_0, x_1, \ldots, x_k,$ is defined recursively as

$$\delta[x_0,x_1,\ldots,x_k] = \frac{\delta[x_1,x_2,\ldots,x_k] - \delta[x_0,x_1, \ldots,x_{k-1}]}{x_k - x_0}.$$

It can be shown by mathematical induction that for equidistant points,

$$\delta[x_0,x_1, \ldots, x_k] = \frac{\Delta^k y_0}{k! h^k}=\frac{\nabla^k y_k}{k! h^k}$$ (12.2.1)

where, $y_0 = f(x_0),$ and $h = x_1 - x_0 = x_2 - x_1 = \cdots = x_k - x_{k-1}.$

In general,

$$\delta[x_i,x_{i+1}, \ldots, x_{i+n}] = \frac{\Delta^n y_i}{n! h^n},$$

where $y_i = f(x_i)$ and $h$ is the length of the interval for $i=0, 1, 2, \ldots.$

Remark 12.2.6   In view of the remark (11.2.18) and (12.2.1), it is easily seen that for a polynomial function of degree $n,$ the $n^{\mbox{th}}$ divided difference is constant and the $(n+1)^{\mbox{th}}$ divided difference is zero.

EXAMPLE 12.2.7   Show that $f(x)$ can be written as

$$f(x)=f(x_{0})+\delta[x_0,x_1](x-x_{0})+\delta[x,x_0,x_1](x-x_0)(x-x_1).$$

Solution:By definition, we have

$$\delta[x,x_0,x_1]=\frac{\delta[x,x_0]-\delta[x_0,x_1]}{(x-x_1)},$$

so, $\delta[x,x_0]=\delta[x_0,x_1]+(x-x_0)\delta[x,x_0,x_1].$ Now since,

$$\delta[x,x_0]=\frac{f(x)-f(x_0)}{(x-x_0)},$$

we get the desired result.

EXERCISE 12.2.8   Show that $f(x)$ can be written in the following form:

$$f(x) = P_{2}(x)+ R_{3}(x),$$

where, $P_{2}(x)=f(x_{0})+\delta[x_0,x_1](x-x_{0})+\delta[x_0,x_1,x_2](x-x_0)(x-x_1)$
and $R_{3}(x)=\delta[x,x_0,x_1,x_2](x-x_0)(x-x_1)(x-x_2).$

Further show that $P_{2}(x_{i})=f(x_i)$ for $i=0 , 1.$

Remark 12.2.9   In general it can be shown that $f(x)= P_{n}(x)+R_{n+1}(x),$ where,

$$P_{n}(x) = f(x_{0})+\delta[x_0,x_1](x-x_{0})+\delta[x_0,x_1,x_2] (x-x_0)(x-x_1) +\cdots$$

$$+ \delta[x_0,x_1,x_2,\ldots,x_{n}](x-x_0)(x-x_1)(x-x_2)\cdots(x-x_{n-1}),$$

and $R_{n+1}(x)=(x-x_0)(x-x_1)(x-x_2)\cdots(x-x_n)\delta[x,x_0,x_1,x_2,\ldots,x_{n}].$

Here, $R_{n+1}(x)$ is called the remainder term.

It may be observed here that the expression $P_{n}(x)$ is a polynomial of degree $'n'$ and $P_{n}(x_{i})=f(x_i)$ for $i=0 , 1, \cdots, (n-1).$

Further, if $f(x)$ is a polynomial of degree $n,$ then in view of the Remark 12.2.6, the remainder term, $R_{n+1}(x)=0,$ as it is a multiple of the $(n+1)^{\mbox{th}}$ divided difference, which is 0 .