PlanetMath: divided difference table

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divided difference table

(Definition)

In numerical work involving divided differences, when computing the divided differences of a tabulated function, it is convenient to arrange the divided differences of a function in a table like so:

$\displaystyle \begin{matrix} x_0 & f(x_0) &&& \ && \Delta^1 f[x_0, x_1] && \\... ...3] & \vdots & \ x_3 & f(x_3) & \vdots && \ \vdots & \vdots &&& \end{matrix}$

The arrrangement of this table makes it easy to compute the divided differrences. Also, once such a table has been computed, one can read off the coefficients in the divived difference interpolation formula as the top entries in the vatious columns.

To explain the computation, as well as to program it on a computer, it is convenient to label the locations in our table with pairs of integers like so:

$\displaystyle \begin{matrix} * & (0, 0) &&&& \ * & & (0, 1) &&& \ * & (1, 1... ...) & \vdots && \ * & (3, 3) & \vdots &&& \ \vdots & \vdots &&&& \end{matrix}$

For convenence, introduce the notation $\Delta_{ij}$ to denote the entry of the difference table at location

. Then, because of the recursion

$\displaystyle \Delta^{n+1} f [x_0, x_1,\ldots, x_{n+1}] = {\Delta^n f [x_1, x_2, \ldots ,x_{n+1}] - \Delta^n f [x_0, x_1 ,\ldots, x_{n}] \over x_{n+1} - x_0},$

we have

$\displaystyle \Delta_{jj}$	$\displaystyle = f(x_j)$
$\displaystyle \Delta_{ij}$	$\displaystyle = {\Delta_{i-1\,j} - \Delta_{i\,j-1} \over x_j - x_i}.$

Using these formulae, we may systematically compute the divided difference table as follows: The first and second column are just the tabulation of our function, so we may write them down immediately. Then we fill out the table one column at a time by using the formula.

Let us illustrate with a simple example. Consider the following choices for and :

$\displaystyle f(x)$	$\displaystyle = x^2 - 4 x + 1$
$\displaystyle x_0$	$\displaystyle = 2$
$\displaystyle x_1$	$\displaystyle = 3$
$\displaystyle x_2$	$\displaystyle = 5$

We may write down our first two columns:

$\displaystyle \begin{matrix} 2 & -3 \\ \ 3 & -2 \\ \ 5 & 6 \end{matrix}$

Now, we start filling in the next column, starting with $\Delta_{0\,1}$ . We take the difference of

and

and divide it by

. Since

$\displaystyle {(-2) - (-3) \over 3 - 2} = {1 \over 1} = 1,$

we have

$\displaystyle \begin{matrix} 2 & -3 &\ & & 1\ 3 & -2 & \ & & \ 5 & 6 & \end{matrix} \qquad.$

Next we fill in the entry $\Delta_{1\,2}$ . We take the difference of

and

and divide it by

. Since

$\displaystyle {6 - (-2) \over 5 - 3} = {8 \over 2} = 4,$

we have

$\displaystyle \begin{matrix} 2 & -3 &\ & & 1 \ 3 & -2 & \ & & 4 \ 5 & 6 & \end{matrix} \qquad.$

Finally, we fill in the entry $\Delta_{0\,2}$ . We take the difference of

and

and divide it by

. Since

$\displaystyle {4 - 1 \over 5 - 2} = {3 \over 3} = 1,$

we have

$\displaystyle \begin{matrix} 2 & -3 & & \ & & 1 & \ 3 & -2 & & 1 \ & & 4 &\ 5 & 6 & & \end{matrix} \qquad.$

Thus, we have constructed our difference table. The top entries in the columns are

so, as per our earlier remark, the divided difference interpolation formula reads

$\displaystyle f(x)$	$\displaystyle = -3 + (x - 2) + (x - 2) (x - 3)$
	$\displaystyle = 1 - 4 x + x^2.$

Since

is a second order polynomial, this interpolation to second order is exact. There is no remainder and, upon simplifying the expression, we recover our original polynomial.

"divided difference table" is owned by rspuzio.

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Cross-references: expression, remainder, polynomial, second order, divided difference interpolation formula, divide, simple, integers, label, columns, interpolation, difference, coefficients, function, divided differences
There are 3 references to this entry.

This is version 11 of divided difference table, born on 2007-03-07, modified 2007-03-07.
Object id is 9043, canonical name is DividedDifferenceTable.
Accessed 1482 times total.

Classification:

AMS MSC:

39A70 (Difference and functional equations :: Difference equations :: Difference operators)

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