divided difference table

divided difference table

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

f

in a table like so:

x

0

f

⁢

(

x

0

)

Δ

1

⁢

f

⁢

[

x

0

,

x

1

]

x

1

f

⁢

(

x

1

)

Δ

2

⁢

f

⁢

[

x

0

,

x

1

,

x

2

]

Δ

1

⁢

f

⁢

[

x

1

,

x

2

]

Δ

3

[

x

0

,

x

1

,

x

2

.

x

3

]

x

2

f

⁢

(

x

2

)

Δ

2

⁢

f

⁢

[

x

1

,

x

2

,

x

3

]

⋮

⋱

Δ

1

⁢

f

⁢

[

x

2

,

x

3

]

⋮

x

3

f

⁢

(

x

3

)

⋮

⋮

⋮

The arrangement of this table makes it easy to compute
the divided differences. Also, once such a table has
been computed, one can read off the coefficients in the

divided difference interpolation formula

as the top
entries in the various 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:

*

(

0

,

0

)

*

(

0

,

1

)

*

(

1

,

1

)

(

0

,

2

)

*

(

1

,

2

)

(

0

,

3

)

*

(

2

,

2

)

(

1

,

3

)

⋮

⋱

*

(

2

,

3

)

⋮

*

(

3

,

3

)

⋮

⋮

⋮

For convenience, introduce the notation

Δ

i

⁢

j

to
denote the entry of the

difference table

at location

(

i

,

j

)

.
Then, because of the

recursion

Δ

n

+

1

⁢

f

⁢

[

x

0

,

x

1

,

…

,

x

n

+

1

]

=

Δ

n

⁢

f

⁢

[

x

1

,

x

2

,

…

,

x

n

+

1

]

-

Δ

n

⁢

f

⁢

[

x

0

,

x

1

,

…

,

x

n

]

x

n

+

1

-

x

0

,

we have

Δ

j

⁢

j

=

f

⁢

(

x

j

)

Δ

i

⁢

j

=

Δ

i

-

1

⁢

j

-

Δ

i

⁢

j

-

1

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

f

and

x

1

:

f

⁢

(

x

)

=

x

2

-

4

⁢

x

+

1

x

0

=

2

x

1

=

3

x

2

=

5

We may write down our first two columns:

2

-

3

3

-

2

5

6

Now, we start filling in the next column, starting with

Δ

0 1

.
We take the

difference

of

-

2

and

-

3

and divide it by

x

1

-

x

0

. Since

(

-

2

)

-

(

-

3

)

3

-

2

=

1

1

=

1

,

we have

2

-

3

1

3

-

2

5

6

.

Next we fill in the entry

Δ

1 2

. We take the difference of

6

and

-

2

and divide it by

x

2

-

x

1

. Since

6

-

(

-

2

)

5

-

3

=

8

2

=

4

,

we have

2

-

3

1

3

-

2

4

5

6

.

Finally, we fill in the entry

Δ

0 2

. We take the difference of

4

and

1

and divide it by

x

2

-

x

0

. Since

4

-

1

5

-

2

=

3

3

=

1

,

we have

2

-

3

1

3

-

2

1

4

5

6

.

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

-

3

,

1

,

1

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

f

⁢

(

x

)

=

-

3

+

(

x

-

2

)

+

(

x

-

2

)

⁢

(

x

-

3

)

=

1

-

4

⁢

x

+

x

2

.

Since

f

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.

Title

divided difference table

Canonical

name

DividedDifferenceTable

Date of creation

2013-03-22 16:48:33

Last modified on

2013-03-22 16:48:33

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

15

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 39A70

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